Instructor's Solutions Manual For Modern Control Systems, 12th Ed Systems Edition

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MODERN CONTROL SYSTEMS
SOLUTION MANUAL
Richard C. Dorf Robert H. Bishop
University of California, Davis Marquette University
A companion to
MODERN CONTROL SYSTEMS
TWELFTH EDITION
Richard C. Dorf
Robert H. Bishop
Prentice Hall
Upper Saddle River Boston Columbus San Francisco New York
Indianapolis London Toronto Sydney Singapore Tokyo Montreal Dubai
Madrid Hong Kong Mexico City Munich Paris Amsterdam Cape Town
© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained
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Instructor's Solutions Manual
Richard C. Dorf, University of California, Davis
Robert H. Bishop, University of Texas at Austin
ISBN-10: 013602498X
ISBN-13: 9780136024989
Publisher: Prentice Hall
Copyright: 2011
Format: On-line Supplement
Published: 08/16/2010
Educator Home | eLearning & Assessment | Support/Contact Us | Find your rep | Exam copy bookbag
© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained
from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying,
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for
Modern
Control Systems, 12/E
P R E F A C E
In each chapter, there are five problem types:
Exercises
Problems
Advanced Problems
Design Problems/Continuous Design Problem
Computer Problems
In total, there are over 1000 problems. The abundance of problems of in-
creasing complexity gives students confidence in their problem-solving
ability as they work their way from the exercises to the design and
computer-based problems.
It is assumed that instructors (and students) have access to MATLAB
and the Control System Toolbox or to LabVIEW and the MathScript RT
Module. All of the computer solutions in this Solution Manual were devel-
oped and tested on an Apple MacBook Pro platform using MATLAB 7.6
Release 2008a and the Control System Toolbox Version 8.1 and LabVIEW
2009. It is not possible to verify each solution on all the available computer
platforms that are compatible with MATLAB and LabVIEW MathScript
RT Module. Please forward any incompatibilities you encounter with the
scripts to Prof. Bishop at the email address given below.
The authors and the staff at Prentice Hall would like to establish an
open line of communication with the instructors using Modern Control
Systems. We encourage you to contact Prentice Hall with comments and
suggestions for this and future editions.
Robert H. Bishop rhbishop@marquette.edu
iii
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T A B L E - O F - C O N T E N T S
1. Introduction to Control Systems ..................................1
2. Mathematical Models of Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
3. State Variable Models ...........................................85
4. Feedback Control System Characteristics . . . . . . . . . . . . . . . . . . . . . . . 133
5. The Performance of Feedback Control Systems . . . . . . . . . . . . . . . . . 177
6. The Stability of Linear Feedback Systems . . . . . . . . . . . . . . . . . . . . . . 234
7. The Root Locus Method .......................................277
8. Frequency Response Methods .................................. 382
9. Stability in the Frequency Domain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 445
10. The Design of Feedback Control Systems . . . . . . . . . . . . . . . . . . . . . . .519
11. The Design of State Variable Feedback Systems . . . . . . . . . . . . . . . . 600
12. Robust Control Systems .......................................659
13. Digital Control Systems ........................................714
iv
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C H A P T E R 1
Introduction to Control Systems
There are, in general, no unique solutions to the following exercises and
problems. Other equally valid block diagrams may be submitted by the
student.
Exercises
E1.1 A microprocessor controlled laser system:
Controller
Error Current i(t) Power
out
Desired
power
output
Measured
power
-Laser
Process
processor
Micro-
Power
Sensor
Measurement
E1.2 A driver controlled cruise control system:
Desired
speed
Foot pedal Actual
auto
speed
Visual indication
of speed
Controller
-
Process
Measurement
Driver Car and
Engine
Speedometer
E1.3 Although the principle of conservation of momentum explains much of
the process of fly-casting, there does not exist a comprehensive scientific
explanation of how a fly-fisher uses the small backward and forward mo-
tion of the y rod to cast an almost weightless fly lure long distances (the
1
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2CHAPTER 1 Introduction to Control Systems
current world-record is 236 ft). The fly lure is attached to a short invisible
leader about 15-ft long, which is in turn attached to a longer and thicker
Dacron line. The objective is cast the fly lure to a distant spot with dead-
eye accuracy so that the thicker part of the line touches the water first
and then the fly gently settles on the water just as an insect might.
Desired
position of
the y
Actual
position
of the y
Visual indication
of the position of
the y
Fly-sher
Wind
disturbance
Controller
-
Process
Measurement
Mind and
body of the
y-sher
Rod, line,
and cast
Vision of
the y-sher
E1.4 An autofocus camera control system:
One-way trip time for the beam
Distance to subject
Lens focusing
motor
K 1
Lens
Conversion factor
(speed of light or
sound)
Emitter/
Receiver
Beam
Beam return Subject
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Exercises 3
E1.5 Tacking a sailboat as the wind shifts:
Desired
sailboat
direction
Actual
sailboat
direction
Measured sailboat direction
Wind
Error
-
Process
Measurement
Actuators
Controller
Sailboat
Gyro compass
Rudder and
sail adjustment
Sailor
E1.6 An automated highway control system merging two lanes of traffic:
Desired
gap Actual
gap
Measured gap
Error
-
Process
Measurement
Actuators
Controller
Active
vehicle
Brakes, gas or
steering
Embedded
computer
Radar
E1.7 Using the speedometer, the driver calculates the difference between the
measured speed and the desired speed. The driver throotle knob or the
brakes as necessary to adjust the speed. If the current speed is not too
much over the desired speed, the driver may let friction and gravity slow
the motorcycle down.
Desired
speed
Visual indication of speed
Actual
motorcycle
speed
Error
-
Process
Measurement
Actuators
Controller
Throttle or
brakes
Driver Motorcycle
Speedometer
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4CHAPTER 1 Introduction to Control Systems
E1.8 Human biofeedback control system:
Measurement
Desired
body
temp
Actual
body
temp
Visual indication of
body temperature
Message to
blood vessels
-
Process
Controller
Body sensor
Hypothalumus Human body
TV display
E1.9 E-enabled aircraft with ground-based flight path control:
Corrections to the
ight path
Controller
Gc(s)
Aircraft
G(s)
-
Desired
Flight
Path
Flight
Path
Corrections to the
ight path
Controller
Gc(s)
Aircraft
G(s)
-
Desired
Flight
Path
Flight
Path
Ground-Based Computer Network
Health
Parameters
Health
Parameters
Meteorological
data
Meteorological
data
Optimal
ight path
Optimal
ight path
Location
and speed
Location
and speed
E1.10 Unmanned aerial vehicle used for crop monitoring in an autonomous
mode:
Gc(s)G(s)
-
Camera
Ground
photo
Controller UAV
Specified
Flight
Trajectory
Location with
respect to the ground
Flight
Trajectory
Map
Correlation
Algorithm
Trajectory
error
Sensor
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Exercises 5
E1.11 An inverted pendulum control system using an optical encoder to measure
the angle of the pendulum and a motor producing a control torque:
Error Angle
Desired
angle
Measured
angle
-Pendulum
Process
Optical
encoder
Measurement
Motor
Actuator
Torque
Voltage
Controller
E1.12 In the video game, the player can serve as both the controller and the sen-
sor. The objective of the game might be to drive a car along a prescribed
path. The player controls the car trajectory using the joystick using the
visual queues from the game displayed on the computer monitor.
Error Game
objective
Desired
game
objective -Video game
Process
Player
(eyesight, tactile, etc.)
Measurement
Joystick
Actuator
Player
Controller
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6CHAPTER 1 Introduction to Control Systems
Problems
P1.1 An automobile interior cabin temperature control system block diagram:
Desired
temperature
set by the
driver
Automobile
cabin temperature
Measured temperature
Error
-
Process
Measurement
Controller
Automobile
cabin
Temperature
sensor
Thermostat and
air conditioning
unit
P1.2 A human operator controlled valve system:
Desired
uid
output *
Error * Fluid
output
* = operator functions
Visual indication
of uid output *
-
Process
Measurement
Controller
Valve
Meter
Tank
P1.3 A chemical composition control block diagram:
Desired
chemical
composition
Error
Chemical
composition
Measured chemical
composition
-
Process
Measurement
Controller
Valve Mixer tube
Infrared analyzer
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Problems 7
P1.4 A nuclear reactor control block diagram:
Desired
power level
Output
power level
Error
Measured chemical
composition
-
Process
Measurement
Controller
Ionization chamber
Reactor
and rods
Motor and
amplier
P1.5 A light seeking control system to track the sun:
Ligh
intensity
Desired
carriage
position
Light
source Photocell
carriage
position
Motor
inputs
Error
-
ProcessController
Motor,
carriage,
and gears
K
Controller
Trajectory
Planner
Dual
Photocells
Measurement
P1.6 If you assume that increasing worker’s wages results in increased prices,
then by delaying or falsifying cost-of-living data you could reduce or elim-
inate the pressure to increase worker’s wages, thus stabilizing prices. This
would work only if there were no other factors forcing the cost-of-living
up. Government price and wage economic guidelines would take the place
of additional “controllers” in the block diagram, as shown in the block
diagram.
Initial
wages
Prices
Wage increases
Market-based prices
Cost-of-living
-
Controller
Industry Government
price
guidelines
K1
Government
wage
guidelines
Controller
Process
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8CHAPTER 1 Introduction to Control Systems
P1.7 Assume that the cannon fires initially at exactly 5:00 p.m.. We have a
positive feedback system. Denote by ∆tthe time lost per day, and the
net time error by ET. Then the follwoing relationships hold:
t= 4/3 min.+ 3 min.= 13/3 min.
and
ET= 12 days ×13/3 min./day .
Therefore, the net time error after 15 days is
ET= 52 min.
P1.8 The student-teacher learning process:
Desired
knowledge
Error Lectures
Knowledge
Measured knowledge
-
Controller Process
Teacher Student
Measurement
Exams
P1.9 A human arm control system:
Visual indication of
arm location
z
y
u e
d
s
-
Controller Process
Measurement
Desired
arm
location
Arm
location
Nerve signals
Eyes and
pressure
receptors
Brain Arm &
muscles
Pressure
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Problems 9
P1.10 An aircraft flight path control system using GPS:
Desired
ight path
from air trac
controllers
Flight
path
Measured ight path
Error
-
Process
Measurement
Actuators
Controller
Aircraft
Global Positioning
System
Computer
Auto-pilot Ailerons, elevators,
rudder, and
engine power
P1.11 The accuracy of the clock is dependent upon a constant flow from the
orifice; the flow is dependent upon the height of the water in the float
tank. The height of the water is controlled by the oat. The control system
controls only the height of the water. Any errors due to enlargement of
the orifice or evaporation of the water in the lower tank is not accounted
for. The control system can be seen as:
Desired
height of
the water
in oat tank
Actual
height
-
Process
Controller
Flow from
upper tank
to oat tank
Float level
P1.12 Assume that the turret and fantail are at 90, if θw6=θF-90. The fantail
operates on the error signal θw-θT, and as the fantail turns, it drives the
turret to turn.
x
y
Wind
*
*
qW
qT
qF
Fantail
Turret
= Wind angle
= Fantail angle
= Turret angle
qW
qT
qF
Torque
qT
qW
Error
-
Process
Controller
Gears & turret
Fantail
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10 CHAPTER 1 Introduction to Control Systems
P1.13 This scheme assumes the person adjusts the hot water for temperature
control, and then adjusts the cold water for flow rate control.
Desired water
temperature
Actual
water temperature
and ow rate
Cold
water
Desired water
ow rate
Measured water ow
Measured water temperature
Error
-
ProcessController
-
Measurement
Human: visual
and touch
Valve adjust
Valve adjust Hot water
system
Cold water
system
Hot
water
P1.14 If the rewards in a specific trade is greater than the average reward, there
is a positive influx of workers, since
q(t) = f1(c(t)r(t)).
If an influx of workers occurs, then reward in specific trade decreases,
since
c(t) = f2(q(t)).
-
Error
-
Process
Controller
f1(c(t)-r(t)) f2(q(t))
q(t)Total of
rewards
c(t)
Average
rewards
r(t)
P1.15 A computer controlled fuel injection system:
Desired
Fuel
Pressure
Fuel
Pressure
Measured fuel pressure
-
Process
Measurement
Controller
Fuel Pressure
Sensor
Electronic
Control Unit
High Pressure Fuel
Supply Pump and
Electronic Fuel
Injectors
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Problems 11
P1.16 With the onset of a fever, the body thermostat is turned up. The body
adjusts by shivering and less blood flows to the skin surface. Aspirin acts
to lowers the thermal set-point in the brain.
Body
temperature
Desired temperature
or set-point from body
thermostat in the brain
Measured body temperature
-
Process
Measurement
Controller
Internal sensor
Body
Adjustments
within the
body
P1.17 Hitting a baseball is arguably one of the most difficult feats in all of sports.
Given that pitchers may throw the ball at speeds of 90 mph (or higher!),
batters have only about 0.1 second to make the decision to swing—with
bat speeds aproaching 90 mph. The key to hitting a baseball a long dis-
tance is to make contact with the ball with a high bat velocity. This is
more important than the bat’s weight, which is usually around 33 ounces
(compared to Ty Cobb’s bat which was 41 ounces!). Since the pitcher can
throw a variety of pitches (fast ball, curve ball, slider, etc.), a batter must
decide if the ball is going to enter the strike zone and if possible, decide
the type of pitch. The batter uses his/her vision as the sensor in the feed-
back loop. A high degree of eye-hand coordination is key to success—that
is, an accurate feedback control system.
P1.18 Define the following variables: p= output pressure, fs= spring force
=Kx,fd= diaphragm force = Ap, and fv= valve force = fs-fd.
The motion of the valve is described by ¨y=fv/m where mis the valve
mass. The output pressure is proportional to the valve displacement, thus
p=cy , where cis the constant of proportionality.
Screw
displacement
x(t)
y
Valve position
Output
pressure
p(t)
fs
-
Diaphragm area
c
Valve
Constant of
proportionality
A
K
Spring
fv
fd
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12 CHAPTER 1 Introduction to Control Systems
P1.19 A control system to keep a car at a given relative position offset from a
lead car:
Throttle Position of
follower
uReference
photo
Relative
position
Desired relative position
Position
of lead
-
Controller Video camera
& processing
algorithms
Follower
car
Actuator
Fuel
throttle
(fuel)
Lead car
-
P1.20 A control system for a high-performance car with an adjustable wing:
Desired
road
adhesion
Road
adhesion
Measured road adhesion
Road
conditions
-
Process
Measurement
Controller
Tire internal
strain gauges
Race Car
K
Actuator
Adjustable
wing
Computer
P1.21 A control system for a twin-lift helicopter system:
Desired altitude Altitude
Measured altitude
Separation distance
Desired separation
distance
Measured separation
distance
-
-
Measurement
Measurement
Radar
Altimeter
Controller
Pilot
Process
Helicopter
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from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying,
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Problems 13
P1.22 The desired building deflection would not necessarily be zero. Rather it
would be prescribed so that the building is allowed moderate movement
up to a point, and then active control is applied if the movement is larger
than some predetermined amount.
Desired
deection
Deection
Measured deection
-
Process
Measurement
Controller
K
Building
Hydraulic
stieners
Strain gauges
on truss structure
P1.23 The human-like face of the robot might have micro-actuators placed at
strategic points on the interior of the malleable facial structure. Coopera-
tive control of the micro-actuators would then enable the robot to achieve
various facial expressions.
Desired
actuator
position
Voltage Actuator
position
Measured position
Error
-
Process
Measurement
Controller
Amplier
Position
sensor
Electro-
mechanical
actuator
P1.24 We might envision a sensor embedded in a “gutter” at the base of the
windshield which measures water levels—higher water levels corresponds
to higher intensity rain. This information would be used to modulate the
wiper blade speed.
Desired
wiper speed
Wiper
blade
speed
Measured water level
-
Process
Measurement
Controller
KWater depth
sensor
Wiper blade
and motor
Electronic
Control Unit
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14 CHAPTER 1 Introduction to Control Systems
P1.25 A feedback control system for the space traffic control:
Desired
orbit position
Actual
orbit position
Measured orbit position
Jet
commands
Applied
forces
Error
-
Process
Measurement
Actuator
Controller
Satellite
Reaction
control jets
Control
law
Radar or GPS
P1.26 Earth-based control of a microrover to point the camera:
Microrover
Camera position
command
Controller
Gc(s)
Camera position command
Camera
Position
Receiver/
Transmitter Rover
position
Camera
Measured camera position
G(s)
Measured camera
position Sensor
P1.27 Control of a methanol fuel cell:
Methanol water
solution
Controller
Gc(s)
Recharging
System
GR(s)
Fuel Cell
G(s)
-Charge
Level
Desired
Charge
Level
Measured charge level
Sensor
H(s)
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from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying,
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Advanced Problems 15
Advanced Problems
AP1.1 Control of a robotic microsurgical device:
Controller
Gc(s)
Microsurgical
robotic manipulator
G(s)
-End-effector
Position
Desired
End-effector
Position
Sensor
H(s)
AP1.2 An advanced wind energy system viewed as a mechatronic system:
WIND ENERGY
SYSTEM
Physical System Modeling
Signals and Systems
Sensors and Actuators
Computers and
Logic Systems
Software and
Data Acquisition
COMPUTER EQUIPMENT FOR CONTROLLING THE SYSTEM
SAFETY MONITORING SYSTEMS
CONTROLLER ALGORITHMS
DATA ACQUISTION: WIND SPEED AND DIRECTION
ROTOR ANGULAR SPEED
PROPELLOR PITCH ANGLE
CONTROL SYSTEM DESIGN AND ANALYSIS
ELECTRICAL SYSTEM DESIGN AND ANALYSIS
POWER GENERATION AND STORAGE
SENSORS
Rotor rotational sensor
Wind speed and direction sensor
ACTUATORS
Motors for manipulatiing the propeller pitch
AERODYNAMIC DESIGN
STRUCTURAL DESIGN OF THE TOWER
ELECTRICAL AND POWER SYSTEMS
AP1.3 The automatic parallel parking system might use multiple ultrasound
sensors to measure distances to the parked automobiles and the curb.
The sensor measurements would be processed by an on-board computer
to determine the steering wheel, accelerator, and brake inputs to avoid
collision and to properly align the vehicle in the desired space.
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16 CHAPTER 1 Introduction to Control Systems
Even though the sensors may accurately measure the distance between
the two parked vehicles, there will be a problem if the available space is
not big enough to accommodate the parking car.
Error Actual
automobile
position
Desired
automobile
position -Automobile
Process
Ultrasound
Measurement
Steering wheel,
accelerator, and
brake
Actuators
On-board
computer
Controller
Position of automobile
relative to parked cars
and curb
AP1.4 There are various control methods that can be considered, including plac-
ing the controller in the feedforward loop (as in Figure 1.3). The adaptive
optics block diagram below shows the controller in the feedback loop, as
an alternative control system architecture.
Compensated
image
Uncompensated
image Astronomical
telescope
mirror
Process
Wavefront
sensor
Measurement
Wavefront
corrector
Actuator & controller
Wavefront
reconstructor
Astronomical
object
AP1.5 The control system might have an inner loop for controlling the acceler-
ation and an outer loop to reach the desired floor level precisely.
Elevator Floor
Desired
acceleration
Desired
floor
Elevator
motor,
cables, etc.
Controller #2 Controller #1
Error
-
Error
-
Acceleration
Measurement
Measured acceleration
Outer
Loop
Inner
Loop
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Advanced Problems 17
AP1.6 An obstacle avoidance control system would keep the robotic vacuum
cleaner from colliding with furniture but it would not necessarily put the
vacuum cleaner on an optimal path to reach the entire floor. This would
require another sensor to measure position in the room, a digital map of
the room layout, and a control system in the outer loop.
Desired
distance
from
obstacles
Distance
from
obstacles
Error
-
Infrared
sensors
Measured distance from obstacle
Controller
Process
Robotic
vacuum
cleaner
Motors,
wheels, etc.
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18 CHAPTER 1 Introduction to Control Systems
Design Problems
The machine tool with the movable table in a feedback control configu-CDP1.1
ration:
Desired
position
x
Measured position
Actual
position
x
Error
-
Process
Measurement
Actuator
Controller
Position sensor
Machine
tool with
table
Amplier Positioning
motor
DP1.1 Use the stereo system and amplifiers to cancel out the noise by emitting
signals 180out of phase with the noise.
Desired
noise = 0
Noise
signal
Noise in
cabin
-
Process
Measurement
Controller
Machine
tool with
table
Positioning
motor
Microphone
Shift phase
by 180 deg
DP1.2 An automobile cruise control system:
Desired
speed
of auto
set by
driver
Desired
shaft
speed Actual
speed
of auto
Drive shaf t speed
Measured shaft speed
-
Process
Measurement
Controller
Automobile
and engine
Valve
Electric
motor
Shaft speed
sensor
K
1/K
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Design Problems 19
DP1.3 An automoted cow milking system:
Location
of cup Milk
Desired cup
location
Measured cup location
Cow location
-
Measurement
Vision system
Measurement
Vision system
Controller Process
Motor and
gears Robot arm and
cup gripper
Actuator
Cow and
milker
DP1.4 A feedback control system for a robot welder:
Desired
position
Voltage
Weld
top
position
Measured position
Error
-
Process
Measurement
Controller
Motor and
arm
Computer and
amplier
Vision camera
DP1.5 A control system for one wheel of a traction control system:
Brake torque
Wheel
speed
Actual slip Measured
slip
Vehicle speed
Rw= Radius of wheel
-
-
Sensor
Vehicle
dynamics
Sensor
-
Antiskid
controller
-
Wheel
dynamics
Engine torque Antislip
controller
1/Rw
++
+
+
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20 CHAPTER 1 Introduction to Control Systems
DP1.6 A vibration damping system for the Hubble Space Telescope:
Signal to
cancel the jitter Jitter of
vibration
Measurement of 0.05 Hz jitter
Desired
jitter = 0
Error
-
Process
Measurement
Actuators
Controller
Rate gyro
sensor
Computer Gyro and
reaction wheels
Spacecraft
dynamics
DP1.7 A control system for a nanorobot:
Error Actual
nanorobot
position
Desired
nanorobot
position -Nanorobot
Process
External beacons
Measurement
Plane surfaces
and propellers
Actuators
Bio-
computer
Controller
Many concepts from underwater robotics can be applied to nanorobotics
within the bloodstream. For example, plane surfaces and propellers can
provide the required actuation with screw drives providing the propul-
sion. The nanorobots can use signals from beacons located outside the
skin as sensors to determine their position. The nanorobots use energy
from the chemical reaction of oxygen and glucose available in the human
body. The control system requires a bio-computer–an innovation that is
not yet available.
For further reading, see A. Cavalcanti, L. Rosen, L. C. Kretly, M. Rosen-
feld, and S. Einav, “Nanorobotic Challenges n Biomedical Application,
Design, and Control,” IEEE ICECS Intl Conf. on Electronics, Circuits
and Systems, Tel-Aviv, Israel, December 2004.
DP1.8 The feedback control system might use gyros and/or accelerometers to
measure angle change and assuming the HTV was originally in the vertical
position, the feedback would retain the vertical position using commands
to motors and other actuators that produced torques and could move the
HTV forward and backward.
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Design Problems 21
Desired angle
from vertical (0o)
Angle from
vertical
Error
-
Gyros &
accelerometers
Measured angle from vertical
Controller
Process
HTV
Motors,
wheels, etc.
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C H A P T E R 2
Mathematical Models of Systems
Exercises
E2.1 We have for the open-loop
y=r2
and for the closed-loop
e=ryand y=e2.
So, e=re2and e2+er= 0 .
0 0.5 1 1.5 2 2.5 3 3.5 4
0
2
4
6
8
10
12
14
16
r
y
open-loop
closed-loop
FIGURE E2.1
Plot of open-loop versus closed-loop.
For example, if r= 1, then e2+e1 = 0 implies that e= 0.618. Thus,
y= 0.382. A plot yversus ris shown in Figure E2.1.
22
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Exercises 23
E2.2 Define
f(T) = R=R0e0.1T
and
R=f(T)f(T0),T=TT0.
Then,
R=f(T)f(T0) = f
T T=T0=20
T+···
where
f
T T=T0=20
=0.1R0e0.1T0=135,
when R0= 10,000Ω. Thus, the linear approximation is computed by
considering only the first-order terms in the Taylor series expansion, and
is given by
R=135∆T .
E2.3 The spring constant for the equilibrium point is found graphically by
estimating the slope of a line tangent to the force versus displacement
curve at the point y= 0.5cm, see Figure E2.3. The slope of the line is
K1.
-3
-2.5
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
-2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 3
y=Displacement (cm)
Force (n)
Spring compresses
Spring breaks
FIGURE E2.3
Spring force as a function of displacement.
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24 CHAPTER 2 Mathematical Models of Systems
E2.4 Since
R(s) = 1
s
we have
Y(s) = 4(s+ 50)
s(s+ 20)(s+ 10) .
The partial fraction expansion of Y(s) is given by
Y(s) = A1
s+A2
s+ 20 +A3
s+ 10
where
A1= 1 , A2= 0.6 and A3=1.6.
Using the Laplace transform table, we find that
y(t) = 1 + 0.6e20t1.6e10t.
The final value is computed using the final value theorem:
lim
t→∞ y(t) = lim
s0s4(s+ 50)
s(s2+ 30s+ 200) = 1 .
E2.5 The circuit diagram is shown in Figure E2.5.
vin v0
+
--
++
-
R2
R1
v-
A
FIGURE E2.5
Noninverting op-amp circuit.
With an ideal op-amp, we have
vo=A(vin v),
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Exercises 25
where Ais very large. We have the relationship
v=R1
R1+R2
vo.
Therefore,
vo=A(vin R1
R1+R2
vo),
and solving for voyields
vo=A
1 + AR1
R1+R2
vin.
Since A1, it follows that 1 + AR1
R1+R2AR1
R1+R2. Then the expression for
vosimplifies to
vo=R1+R2
R1
vin.
E2.6 Given
y=f(x) = ex
and the operating point xo= 1, we have the linear approximation
y=f(x) = f(xo) + f
x x=xo
(xxo) + ···
where
f(xo) = e, df
dxx=xo=1
=e, and xxo=x1.
Therefore, we obtain the linear approximation y=ex.
E2.7 The block diagram is shown in Figure E2.7.
+I(s)
R(s)
-
H(s)
G2(s)
G1(s)
Ea(s)
FIGURE E2.7
Block diagram model.
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26 CHAPTER 2 Mathematical Models of Systems
Starting at the output we obtain
I(s) = G1(s)G2(s)E(s).
But E(s) = R(s)H(s)I(s), so
I(s) = G1(s)G2(s) [R(s)H(s)I(s)] .
Solving for I(s) yields the closed-loop transfer function
I(s)
R(s)=G1(s)G2(s)
1 + G1(s)G2(s)H(s).
E2.8 The block diagram is shown in Figure E2.8.
Y(s)
G2(s)
G1(s)
R(s)
-
H3(s)
--
H1(s)
K1
s
-
H2(s)
A(s)
W(s)
Z(s)
E(s)
FIGURE E2.8
Block diagram model.
Starting at the output we obtain
Y(s) = 1
sZ(s) = 1
sG2(s)A(s).
But A(s) = G1(s) [H2(s)Z(s)H3(s)A(s) + W(s)] and Z(s) = sY (s),
so
Y(s) = G1(s)G2(s)H2(s)Y(s)G1(s)H3(s)Y(s) + 1
sG1(s)G2(s)W(s).
Substituting W(s) = KE(s)H1(s)Z(s) into the above equation yields
Y(s) = G1(s)G2(s)H2(s)Y(s)G1(s)H3(s)Y(s)
+1
sG1(s)G2(s) [KE(s)H1(s)Z(s)]
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Exercises 27
and with E(s) = R(s)Y(s) and Z(s) = sY (s) this reduces to
Y(s) = [G1(s)G2(s) (H2(s) + H1(s)) G1(s)H3(s)
1
sG1(s)G2(s)K]Y(s) + 1
sG1(s)G2(s)KR(s).
Solving for Y(s) yields the transfer function
Y(s) = T(s)R(s),
where
T(s) = KG1(s)G2(s)/s
1 + G1(s)G2(s) [(H2(s) + H1(s)] + G1(s)H3(s) + KG1(s)G2(s)/s.
E2.9 From Figure E2.9, we observe that
Ff(s) = G2(s)U(s)
and
FR(s) = G3(s)U(s).
Then, solving for U(s) yields
U(s) = 1
G2(s)Ff(s)
and it follows that
FR(s) = G3(s)
G2(s)U(s).
Again, considering the block diagram in Figure E2.9 we determine
Ff(s) = G1(s)G2(s)[R(s)H2(s)Ff(s)H2(s)FR(s)] .
But, from the previous result, we substitute for FR(s) resulting in
Ff(s) = G1(s)G2(s)R(s)G1(s)G2(s)H2(s)Ff(s)G1(s)H2(s)G3(s)Ff(s).
Solving for Ff(s) yields
Ff(s) = G1(s)G2(s)
1 + G1(s)G2(s)H2(s) + G1(s)G3(s)H2(s)R(s).
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28 CHAPTER 2 Mathematical Models of Systems
R(s)G1(s)
H2(s)
-
+
G2(s)
G3(s)
H2(s)
-
Ff (s)
FR(s)
U(s)
U(s)
FIGURE E2.9
Block diagram model.
E2.10 The shock absorber block diagram is shown in Figure E2.10. The closed-
loop transfer function model is
T(s) = Gc(s)Gp(s)G(s)
1 + H(s)Gc(s)Gp(s)G(s).
+
-
R(s)
Desired piston
travel
Y(s)
Piston
travel
Controller
Gc(s)
Plunger and
Piston System
G(s)
Sensor
H(s)
Gear Motor
Gp(s)
Piston travel
measurement
FIGURE E2.10
Shock absorber block diagram.
E2.11 Let fdenote the spring force (n) and xdenote the deflection (m). Then
K=f
x.
Computing the slope from the graph yields:
(a) xo=0.14m K= ∆f /x= 10 n / 0.04 m = 250 n/m
(b) xo= 0m K= ∆f /x= 10 n / 0.05 m = 200 n/m
(c) xo= 0.35m K= ∆f /x= 3n / 0.05 m = 60 n/m
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Exercises 29
E2.12 The signal flow graph is shown in Fig. E2.12. Find Y(s) when R(s) = 0.
Y (s)
-1
K2G(s)
-K 1
1
Td(s)
FIGURE E2.12
Signal flow graph.
The transfer function from Td(s) to Y(s) is
Y(s) = G(s)Td(s)K1K2G(s)Td(s)
1(K2G(s)) =G(s)(1 K1K2)Td(s)
1 + K2G(s).
If we set
K1K2= 1 ,
then Y(s) = 0 for any Td(s).
E2.13 The transfer function from R(s), Td(s), and N(s) to Y(s) is
Y(s) = K
s2+ 10s+KR(s)+1
s2+ 10s+KTd(s)K
s2+ 10s+KN(s)
Therefore, we nd that
Y(s)/Td(s) = 1
s2+ 10s+Kand Y(s)/N(s) = K
s2+ 10s+K
E2.14 Since we want to compute the transfer function from R2(s) to Y1(s), we
can assume that R1= 0 (application of the principle of superposition).
Then, starting at the output Y1(s) we obtain
Y1(s) = G3(s) [H1(s)Y1(s) + G2(s)G8(s)W(s) + G9(s)W(s)] ,
or
[1 + G3(s)H1(s)] Y1(s) = [G3(s)G2(s)G8(s)W(s) + G3(s)G9(s)] W(s).
Considering the signal W(s) (see Figure E2.14), we determine that
W(s) = G5(s) [G4(s)R2(s)H2(s)W(s)] ,
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30 CHAPTER 2 Mathematical Models of Systems
G2(s)
G1(s)
-
H1(s)
G3(s)
G5(s)
G4(s)
-
H2(s)
G6(s)
R1(s)
R2(s)
Y1(s)
Y2(s)
+
+
G7(s) G8(s) G9(s)
+
+
+
+
W(s)
FIGURE E2.14
Block diagram model.
or
[1 + G5(s)H2(s)] W(s) = G5(s)G4(s)R2(s).
Substituting the expression for W(s) into the above equation for Y1(s)
yields
Y1(s)
R2(s)=G2(s)G3(s)G4(s)G5(s)G8(s) + G3(s)G4(s)G5(s)G9(s)
1 + G3(s)H1(s) + G5(s)H2(s) + G3(s)G5(s)H1(s)H2(s).
E2.15 For loop 1, we have
R1i1+L1
di1
dt +1
C1Z(i1i2)dt +R2(i1i2) = v(t).
And for loop 2, we have
1
C2Zi2dt +L2
di2
dt +R2(i2i1) + 1
C1Z(i2i1)dt = 0 .
E2.16 The transfer function from R(s) to P(s) is
P(s)
R(s)=4.2
s3+ 2s2+ 4s+ 4.2.
The block diagram is shown in Figure E2.16a. The corresponding signal
flow graph is shown in Figure E2.16b for
P(s)/R(s) = 4.2
s3+ 2s2+ 4s+ 4.2.
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Exercises 31
v1(s)
-
R(s)P(s)
7
v2(s)0.6
s
q(s)1
s2+2s+4
(a)
V2
0.6
s
R(s) P (s)
-1
1 7
V1
1
s2+ 2 s + 4
(b)
FIGURE E2.16
(a) Block diagram, (b) Signal flow graph.
E2.17 A linear approximation for fis given by
f=f
x x=xo
x= 2kxox=kx
where xo= 1/2, ∆f=f(x)f(xo), and ∆x=xxo.
E2.18 The linear approximation is given by
y=mx
where
m=y
x x=xo
.
(a) When xo= 1, we nd that yo= 2.4, and yo= 13.2 when xo= 2.
(b) The slope mis computed as follows:
m=y
x x=xo
= 1 + 4.2x2
o.
Therefore, m= 5.2 at xo= 1, and m= 18.8 at xo= 2.
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32 CHAPTER 2 Mathematical Models of Systems
E2.19 The output (with a step input) is
Y(s) = 15(s+ 1)
s(s+ 7)(s+ 2) .
The partial fraction expansion is
Y(s) = 15
14s18
7
1
s+ 7 +3
2
1
s+ 2 .
Taking the inverse Laplace transform yields
y(t) = 15
14 18
7e7t+3
2e2t.
E2.20 The input-output relationship is
Vo
V=A(K1)
1 + AK
where
K=Z1
Z1+Z2
.
Assume A1. Then,
Vo
V=K1
K=Z2
Z1
where
Z1=R1
R1C1s+ 1 and Z2=R2
R2C2s+ 1 .
Therefore,
Vo(s)
V(s)=R2(R1C1s+ 1)
R1(R2C2s+ 1) =2(s+ 1)
s+ 2 .
E2.21 The equation of motion of the mass mcis
mc¨xp+ (bd+bs) ˙xp+kdxp=bd˙xin +kdxin .
Taking the Laplace transform with zero initial conditions yields
[mcs2+ (bd+bs)s+kd]Xp(s) = [bds+kd]Xin(s).
So, the transfer function is
Xp(s)
Xin(s)=bds+kd
mcs2+ (bd+bs)s+kd
=0.7s+ 2
s2+ 2.8s+ 2 .
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Exercises 33
E2.22 The rotational velocity is
ω(s) = 2(s+ 4)
(s+ 5)(s+ 1)2
1
s.
Expanding in a partial fraction expansion yields
ω(s) = 8
5
1
s+1
40
1
s+ 5 3
2
1
(s+ 1)213
8
1
s+ 1 .
Taking the inverse Laplace transform yields
ω(t) = 8
5+1
40e5t3
2tet13
8et.
E2.23 The closed-loop transfer function is
Y(s)
R(s)=T(s) = K1K2
s2+ (K1+K2K3+K1K2)s+K1K2K3
.
E2.24 The closed-loop tranfser function is
Y(s)
R(s)=T(s) = 10
s2+ 21s+ 10 .
E2.25 Let x= 0.6 and y= 0.8. Then, with y=ax3, we have
0.8 = a(0.6)3.
Solving for ayields a= 3.704. A linear approximation is
yyo= 3ax2
o(xxo)
or y= 4x1.6, where yo= 0.8 and xo= 0.6.
E2.26 The equations of motion are
m1¨x1+k(x1x2) = F
m2¨x2+k(x2x1) = 0 .
Taking the Laplace transform (with zero initial conditions) and solving
for X2(s) yields
X2(s) = k
(m2s2+k)(m1s2+k)k2F(s).
Then, with m1=m2=k= 1, we have
X2(s)/F (s) = 1
s2(s2+ 2) .
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34 CHAPTER 2 Mathematical Models of Systems
E2.27 The transfer function from Td(s) to Y(s) is
Y(s)/Td(s) = G2(s)
1 + G1G2H(s).
E2.28 The transfer function is
Vo(s)
V(s)=R2R4C
R3
s+R2R4
R1R3
= 24s+ 144 .
E2.29 (a) If
G(s) = 1
s2+ 15s+ 50 and H(s) = 2s+ 15 ,
then the closed-loop transfer function of Figure E2.28(a) and (b) (in
Dorf & Bishop) are equivalent.
(b) The closed-loop transfer function is
T(s) = 1
s2+ 17s+ 65 .
E2.30 (a) The closed-loop transfer function is
T(s) = G(s)
1 + G(s)
1
s=10
s(s2+ 2s+ 20) where G(s) = 10
s2+ 2s+ 10 .
0 1 2 3 4 5 6
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
Time sec
Amplitude
FIGURE E2.30
Step response.
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Exercises 35
(b) The output Y(s) (when R(s) = 1/s) is
Y(s) = 0.5
s0.25 + 0.0573j
s+ 1 4.3589j+0.25 0.0573j
s+ 1 + 4.3589j.
(c) The plot of y(t) is shown in Figure E2.30. The output is given by
y(t) = 1
21etcos 19t1
19 sin 19t
E2.31 The partial fraction expansion is
V(s) = a
s+p1
+b
s+p2
where p1= 4 19.6jand p2= 4 + 19.6j. Then, the residues are
a=10.2j b = 10.2j .
The inverse Laplace transform is
v(t) = 10.2je(4+19.6j)t+ 10.2je(419.6j)t= 20.4e4tsin 19.6t .
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36 CHAPTER 2 Mathematical Models of Systems
Problems
P2.1 The integrodifferential equations, obtained by Kirchoff’s voltage law to
each loop, are as follows:
R1i1+1
C1Zi1dt +L1
d(i1i2)
dt +R2(i1i2) = v(t) (loop 1)
and
R3i2+1
C2Zi2dt +R2(i2i1) + L1
d(i2i1)
dt = 0 (loop 2) .
P2.2 The differential equations describing the system can be obtained by using
a free-body diagram analysis of each mass. For mass 1 and 2 we have
M1¨y1+k12(y1y2) + b˙y1+k1y1=F(t)
M2¨y2+k12(y2y1) = 0 .
Using a force-current analogy, the analagous electric circuit is shown in
Figure P2.2, where CiMi, L11/k1, L12 1/k12 ,and R1/b .
FIGURE P2.2
Analagous electric circuit.
P2.3 The differential equations describing the system can be obtained by using
a free-body diagram analysis of each mass. For mass 1 and 2 we have
M¨x1+kx1+k(x1x2) = F(t)
M¨x2+k(x2x1) + b˙x2= 0 .
Using a force-current analogy, the analagous electric circuit is shown in
Figure P2.3, where
CM L 1/k R 1/b .
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Problems 37
FIGURE P2.3
Analagous electric circuit.
P2.4 (a) The linear approximation around vin = 0 is vo= 0vin, see Fig-
ure P2.4(a).
(b) The linear approximation around vin = 1 is vo= 2vin 1, see Fig-
ure P2.4(b).
-1 -0.5 0 0.5 1
-0.4
-0.3
-0.2
-0.1
0
0.1
0.2
0.3
0.4
(a)
vin
vo
linear approximation
-1 0 1 2
-1
-0.5
0
0.5
1
1.5
2
2.5
3
3.5
4
(b)
vin
vo
linear approximation
FIGURE P2.4
Nonlinear functions and approximations.
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38 CHAPTER 2 Mathematical Models of Systems
P2.5 Given
Q=K(P1P2)1/2.
Let δP =P1P2and δPo= operating point. Using a Taylor series
expansion of Q, we have
Q=Qo+Q
δP δP =δPo
(δP δPo) + ···
where
Qo=KδP 1/2
oand Q
δP δP =δPo
=K
2δP 1/2
o.
Define ∆Q=QQoand ∆P=δP δPo. Then, dropping higher-order
terms in the Taylor series expansion yields
Q=mP
where
m=K
2δP 1/2
o
.
P2.6 From P2.1 we have
R1i1+1
C1Zi1dt +L1
d(i1i2)
dt +R2(i1i2) = v(t)
and
R3i2+1
C2Zi2dt +R2(i2i1) + L1
d(i2i1)
dt = 0 .
Taking the Laplace transform and using the fact that the initial voltage
across C2is 10v yields
[R1+1
C1s+L1s+R2]I1(s) + [R2L1s]I2(s) = 0
and
[R2L1s]I1(s) + [L1s+R3+1
C2s+R2]I2(s) = 10
s.
Rewriting in matrix form we have
R1+1
C1s+L1s+R2R2L1s
R2L1s L1s+R3+1
C2s+R2
I1(s)
I2(s)
=
0
10/s
.
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Problems 39
Solving for I2yields
I1(s)
I2(s)
=1
L1s+R3+1
C2s+R2R2+L1s
R2+L1s R1+1
C1s+L1s+R2
0
10/s
.
or
I2(s) = 10(R1+ 1/C1s+L1s+R2)
s
where
∆ = (R1+1
C1s+L1s+R2)(L1s+R3+1
C2s+R2)(R2+L1s)2.
P2.7 Consider the differentiating op-amp circuit in Figure P2.7. For an ideal
op-amp, the voltage gain (as a function of frequency) is
V2(s) = Z2(s)
Z1(s)V1(s),
where
Z1=R1
1 + R1Cs
and Z2=R2are the respective circuit impedances. Therefore, we obtain
V2(s) = R2(1 + R1Cs)
R1V1(s).
V1(s)V2(s)
+
--
+
+
-
C
R1
R2
Z1Z2
FIGURE P2.7
Differentiating op-amp circuit.
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40 CHAPTER 2 Mathematical Models of Systems
P2.8 Let
∆ =
G2+Cs Cs G2
Cs G1+ 2Cs Cs
G2Cs Cs +G2
.
Then,
Vj=ij
I1or or V3
V1
=13I1/
11I1/.
Therefore, the transfer function is
T(s) = V3
V1
=13
11
=
Cs 2Cs +G1
G2Cs
2Cs +G1Cs
Cs Cs +G2
-3
-2
-1
0
1
2
3
-8 -7 -6 -5 -4 -3 -2 -1 0
x x
o
o
Real Axis
Imag Axis
Pole-zero map (x:poles and o:zeros)
FIGURE P2.8
Pole-zero map.
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Problems 41
=C2R1R2s2+ 2CR1s+ 1
C2R1R2s2+ (2R1+R2)Cs + 1 .
Using R1= 0.5, R2= 1, and C= 0.5, we have
T(s) = s2+ 4s+ 8
s2+ 8s+ 8 =(s+ 2 + 2j)(s+ 2 2j)
(s+ 4 + 8)(s+ 4 8) .
The pole-zero map is shown in Figure P2.8.
P2.9 From P2.3 we have
M¨x1+kx1+k(x1x2) = F(t)
M¨x2+k(x2x1) + b˙x2= 0 .
Taking the Laplace transform of both equations and writing the result in
matrix form, it follows that
Ms2+ 2kk
k Ms2+bs +k
X1(s)
X2(s)
=
F(s)
0
,
-0.03 -0.025 -0.02 -0.015 -0.01 -0.005 0
-0.4
-0.3
-0.2
- 0.1
0
0.1
0.2
0.3
0.4
Real Axis
Imag Axis
Pole zero map
FIGURE P2.9
Pole-zero map.
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42 CHAPTER 2 Mathematical Models of Systems
or
X1(s)
X2(s)
=1
Ms2+bs +k k
k Ms2+ 2k
F(s)
0
where ∆ = (Ms2+bs +k)(Ms2+ 2k)k2.So,
G(s) = X1(s)
F(s)=Ms2+bs +k
.
When b/k = 1, M= 1 , b2/M k = 0.04, we have
G(s) = s2+ 0.04s+ 0.04
s4+ 0.04s3+ 0.12s2+ 0.0032s+ 0.0016 .
The pole-zero map is shown in Figure P2.9.
P2.10 From P2.2 we have
M1¨y1+k12(y1y2) + b˙y1+k1y1=F(t)
M2¨y2+k12(y2y1) = 0 .
Taking the Laplace transform of both equations and writing the result in
matrix form, it follows that
M1s2+bs +k1+k12 k12
k12 M2s2+k12
Y1(s)
Y2(s)
=
F(s)
0
or
Y1(s)
Y2(s)
=1
M2s2+k12 k12
k12 M1s2+bs +k1+k12
F(s)
0
where
∆ = (M2s2+k12)(M1s2+bs +k1+k12)k2
12 .
So, when f(t) = asin ωot, we have that Y1(s) is given by
Y1(s) = aM2ωo(s2+k12/M2)
(s2+ω2
o)∆(s).
For motionless response (in the steady-state), set the zero of the transfer
function so that
(s2+k12
M2
) = s2+ω2
oorω2
o=k12
M2
.
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Problems 43
P2.11 The transfer functions from Vc(s) to Vd(s) and from Vd(s) to θ(s) are:
Vd(s)/Vc(s) = K1K2
(Lqs+Rq)(Lcs+Rc),and
θ(s)/Vd(s) = Km
(Js2+fs)((Ld+La)s+Rd+Ra) + K3Kms.
The block diagram for θ(s)/Vc(s) is shown in Figure P2.11, where
θ(s)/Vc(s) = θ(s)
Vd(s)
Vd(s)
Vc(s)=K1K2Km
∆(s),
where
∆(s) = s(Lcs+Rc)(Lqs+Rq)((Js +b)((Ld+La)s+Rd+Ra) + KmK3).
-
+1
(L d
+L a)s+R d+R a
1
Js+f 1
s
Km
K3
1
Lcs+R c
1
Lq
s+R q
K1K2
Vc
IcVqVdIdTm
Vb
Iqw
q
FIGURE P2.11
Block diagram.
P2.12 The open-loop transfer function is
Y(s)
R(s)=K
s+ 20 .
With R(s) = 1/s, we have
Y(s) = K
s(s+ 20) .
The partial fraction expansion is
Y(s) = K
20 1
s1
s+ 20,
and the inverse Laplace transform is
y(t) = K
20 1e20t,
As t→ ∞, it follows that y(t)K/20. So we choose K= 20 so that y(t)
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44 CHAPTER 2 Mathematical Models of Systems
approaches 1. Alternatively we can use the final value theorem to obtain
y(t)t→∞ = lim
s0sY (s) = K
20 = 1 .
It follows that choosing K= 20 leads to y(t)1 as t→ ∞.
P2.13 The motor torque is given by
Tm(s) = (Jms2+bms)θm(s) + (JLs2+bLs)L(s)
=n((Jms2+bms)/n2+JLs2+bLs)θL(s)
where
n=θL(s)m(s) = gear ratio .
But
Tm(s) = KmIg(s)
and
Ig(s) = 1
(Lg+Lf)s+Rg+Rf
Vg(s),
and
Vg(s) = KgIf(s) = Kg
Rf+LfsVf(s).
Combining the above expressions yields
θL(s)
Vf(s)=KgKm
n1(s)∆2(s).
where
1(s) = JLs2+bLs+Jms2+bms
n2
and
2(s) = (Lgs+Lfs+Rg+Rf)(Rf+Lfs).
P2.14 For a field-controlled dc electric motor we have
ω(s)/Vf(s) = Km/Rf
Js +b.
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Problems 45
With a step input of Vf(s) = 80/s, the final value of ω(t) is
ω(t)t→∞ = lim
s0(s) = 80Km
Rfb= 2.4 or Km
Rfb= 0.03 .
Solving for ω(t) yields
ω(t) = 80Km
RfJL11
s(s+b/J)=80Km
Rfb(1e(b/J)t) = 2.4(1e(b/J)t).
At t= 1/2, ω(t) = 1, so
ω(1/2) = 2.4(1 e(b/J)t) = 1 implies b/J = 1.08 sec .
Therefore,
ω(s)/Vf(s) = 0.0324
s+ 1.08 .
P2.15 Summing the forces in the vertical direction and using Newton’s Second
Law we obtain
¨x+k
mx= 0 .
The system has no damping and no external inputs. Taking the Laplace
transform yields
X(s) = x0s
s2+k/m ,
where we used the fact that x(0) = x0and ˙x(0) = 0. Then taking the
inverse Laplace transform yields
x(t) = x0cos sk
mt .
P2.16 Using Cramer’s rule, we have
1 1.5
2 4
x1
x2
=
6
11
or
x1
x2
=1
41.5
2 1
6
11
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46 CHAPTER 2 Mathematical Models of Systems
where = 4(1) 2(1.5) = 1 .Therefore,
x1=4(6) 1.5(11)
1= 7.5 and x2=2(6) + 1(11)
1=1.
The signal flow graph is shown in Figure P2.16.
1/4
1
-1.5
X1
11
6-1/2 X2
FIGURE P2.16
Signal flow graph.
So,
x1=6(1) 1.5(11
4)
13
4
= 7.5 and x2=11(1
4) + 1
2(6)
13
4
=1.
P2.17 (a) For mass 1 and 2, we have
M1¨x1+K1(x1x2) + b1( ˙x3˙x1) = 0
M2¨x2+K2(x2x3) + b2( ˙x3˙x2) + K1(x2x1) = 0 .
(b) Taking the Laplace transform yields
(M1s2+b1s+K1)X1(s)K1X2(s) = b1sX3(s)
K1X1(s) + (M2s2+b2s+K1+K2)X2(s) = (b2s+K2)X3(s).
(c) Let
G1(s) = K2+b2s
G2(s) = 1/p(s)
G3(s) = 1/q(s)
G4(s) = sb1,
where
p(s) = s2M2+sf2+K1+K2
and
q(s) = s2M1+sf1+K1.
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Problems 47
The signal flow graph is shown in Figure P2.17.
X3X1
K1
G3
K1
G2
G1
G4
FIGURE P2.17
Signal flow graph.
(d) The transfer function from X3(s) to X1(s) is
X1(s)
X3(s)=K1G1(s)G2(s)G3(s) + G4(s)G3(s)
1K2
1G2(s)G3(s).
P2.18 The signal flow graph is shown in Figure P2.18.
V1V2
Z4
Y3
Z2
Y1
-Y 3
-Y 1
Ia
Va
I1
-Z 2
FIGURE P2.18
Signal flow graph.
The transfer function is
V2(s)
V1(s)=Y1Z2Y3Z4
1 + Y1Z2+Y3Z2+Y3Z4+Y1Z2Z4Y3
.
P2.19 For a noninerting op-amp circuit, depicted in Figure P2.19a, the voltage
gain (as a function of frequency) is
Vo(s) = Z1(s) + Z2(s)
Z1(s)Vin(s),
where Z1(s) and Z2(s) are the impedances of the respective circuits. In
the case of the voltage follower circuit, shown in Figure P2.19b, we have
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48 CHAPTER 2 Mathematical Models of Systems
vin
v0
+
-
vin
v0
+
-
Z2
Z1
(a)(a) (b)
FIGURE P2.19
(a) Noninverting op-amp circuit. (b) Voltage follower circuit.
Z1=(open circuit) and Z2= 0. Therefore, the transfer function is
Vo(s)
Vin(s)=Z1
Z1
= 1.
P2.20 (a) Assume RgRsand RsR1. Then Rs=R1+R2R2, and
vgs =vin vo,
where we neglect iin, since RgRs. At node S, we have
vo
Rs
=gmvgs =gm(vin vo) or vo
vin
=gmRs
1 + gmRs
.
(b) With gmRs= 20, we have
vo
vin
=20
21 = 0.95 .
(c) The block diagram is shown in Figure P2.20.
gmRs
-
vin(s)vo(s)
FIGURE P2.20
Block diagram model.
P2.21 From the geometry we find that
z=kl1l2
l1
(xy)l2
l1
y .
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Problems 49
The flow rate balance yields
Ady
dt =pzwhich implies Y(s) = pZ(s)
As .
By combining the above results it follows that
Y(s) = p
As kl1l2
l1(X(s)Y(s)) l2
l1
Y(s).
Therefore, the signal flow graph is shown in Figure P2.21. Using Mason’s
XY
p/As
k
1
DZ
-l / l
2 1
(l - l
1 2)/l 1
-1
FIGURE P2.21
Signal flow graph.
gain formula we find that the transfer function is given by
Y(s)
X(s)=
k(l1l2)p
l1As
1 + l2p
l1As +k(l1l2)p
l1As
=K1
s+K2+K1
,
where
K1=k(l1l2)p
l1Apand K2=l2p
l1A.
P2.22 (a) The equations of motion for the two masses are
ML2¨
θ1+Mg1+kL
22
(θ1θ2) = L
2f(t)
ML2¨
θ2+Mg2+kL
22
(θ2θ1) = 0 .
With ˙
θ1=ω1and ˙
θ2=ω2, we have
˙ω1=g
L+k
4Mθ1+k
4Mθ2+f(t)
2ML
˙ω2=k
4Mθ1g
L+k
4Mθ2.
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50 CHAPTER 2 Mathematical Models of Systems
1/2ML
1/s 1/s
a
b
1/s 1/s
a
F (t) w1q1
q2
w2
(a)
-
+ j g
L+k
2M
+ j g
L
+ j g
L+k
4M
X
X
Re(s)
Imag(s)
O
(b)
FIGURE P2.22
(a) Block diagram. (b) Pole-zero map.
(b) Define a=g/L +k/4Mand b=k/4M. Then
θ1(s)
F(s)=1
2ML
s2+a
(s2+a)2b2.
(c) The block diagram and pole-zero map are shown in Figure P2.22.
P2.23 The input-output ratio, Vce/Vin, is found to be
Vce
Vin
=β(R1) + hieRf
βhre +hie(hoe +Rf).
P2.24 (a) The voltage gain is given by
vo
vin
=RLβ1β2(R1+R2)
(R1+R2)(Rg+hie1)+R1(R1+R2)(1 + β1) + R1RLβ1β2
.
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Problems 51
(b) The current gain is found to be
ic2
ib1
=β1β2.
(c) The input impedance is
vin
ib1
=(R1+R2)(Rg+hie1) + R1(R1+R2)(1 + β1) + R1RLβ1β2
R1+R2
,
and when β1β2is very large, we have the approximation
vin
ib1RLR1β1β2
R1+R2
.
P2.25 The transfer function from R(s) and Td(s) to Y(s) is given by
Y(s) = G(s)R(s)1
G(s)(G(s)R(s) + Td(s))+Td(s) + G(s)R(s)
=G(s)R(s).
Thus,
Y(s)/R(s) = G(s).
Also, we have that
Y(s) = 0 .
when R(s) = 0. Therefore, the effect of the disturbance, Td(s), is elimi-
nated.
P2.26 The equations of motion for the two mass model of the robot are
M¨x+b( ˙x˙y) + k(xy) = F(t)
m¨y+b( ˙y˙x) + k(yx) = 0 .
Taking the Laplace transform and writing the result in matrix form yields
Ms2+bs +k(bs +k)
(bs +k)ms2+bs +k
X(s)
Y(s)
=
F(s)
0
.
Solving for Y(s) we find that
Y(s)
F(s)=
1
mM (bs +k)
s2[s2+1 + m
Mb
ms+k
m].
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52 CHAPTER 2 Mathematical Models of Systems
P2.27 The describing equation of motion is
m¨z=mg ki2
z2.
Defining
f(z, i) = gki2
mz2
leads to
¨z=f(z, i).
The equilibrium condition for ioand zo, found by solving the equation of
motion when
˙z= ¨z= 0 ,
is
ki2
o
mg =z2
o.
We linearize the equation of motion using a Taylor series approximation.
With the definitions
z=zzoand ∆i=iio,
we have ˙
z= ˙zand ¨
z= ¨z. Therefore,
¨
z=f(z, i) = f(zo, io) + f
z z=zo
i=io
z+f
i z=zo
i=io
i+···
But f(zo, io) = 0, and neglecting higher-order terms in the expansion
yields
¨
z=2ki2
o
mz3
o
z2kio
mz2
o
i .
Using the equilibrium condition which relates zoto io, we determine that
¨
z=2g
zo
zg
io
i .
Taking the Laplace transform yields the transfer function (valid around
the equilibrium point)
Z(s)
I(s)=g/io
s22g/zo
.
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Problems 57
-
+
K1
s (s+1)
1 + K 2s
R(s ) Y (s)
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2
time(sec)
y(t)
<---- time to 90% = 0.39 sec
FIGURE P2.35
The equivalent block diagram and the system step response.
P2.36 (a) Given R(s) = 1/s2, the partial fraction expansion is
Y(s) = 24
s2(s+ 2)(s+ 3)(s+ 4) =3
s+ 2 8/3
s+ 3 +3/4
s+ 4 +1
s213/12
s.
Therefore, using the Laplace transform table, we determine that the
ramp response is
y(t) = 3e2t8
3e3t+3
4e4t+t13
12 , t 0.
(b) For the ramp input, y(t)0.21 at t= 1.second (see Figure P2.36a).
(c) Given R(s) = 1, the partial fraction expansion is
Y(s) = 24
(s+ 2)(s+ 3)(s+ 4) =12
s+ 2 24
s+ 3 +12
s+ 4 .
Therefore, using the Laplace transform table, we determine that the
impulse response is
y(t) = 12e2t24e3t+ 412e4t, t 0.
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Problems 53
P2.28 The signal flow graph is shown in Figure P2.28.
P D
M
C
+f
+g
+e
+a
GB
+b +c
S
-m -k
-d
+h
FIGURE P2.28
Signal flow graph.
(a) The PGBDP loop gain is equal to -abcd. This is a negative transmis-
sion since the population produces garbage which increases bacteria
and leads to diseases, thus reducing the population.
(b) The PMCP loop gain is equal to +efg. This is a positive transmis-
sion since the population leads to modernization which encourages
immigration, thus increasing the population.
(c) The PMSDP loop gain is equal to +ehkd. This is a positive trans-
mission since the population leads to modernization and an increase
in sanitation facilities which reduces diseases, thus reducing the rate
of decreasing population.
(d) The PMSBDP loop gain is equal to +ehmcd. This is a positive
transmission by similar argument as in (3).
P2.29 Assume the motor torque is proportional to the input current
Tm=ki .
Then, the equation of motion of the beam is
J¨
φ=ki ,
where Jis the moment of inertia of the beam and shaft (neglecting the
inertia of the ball). We assume that forces acting on the ball are due to
gravity and friction. Hence, the motion of the ball is described by
m¨x=mgφ b˙x
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54 CHAPTER 2 Mathematical Models of Systems
where mis the mass of the ball, bis the coefficient of friction, and we
have assumed small angles, so that sin φφ. Taking the Laplace transfor
of both equations of motion and solving for X(s) yields
X(s)/I(s) = gk/J
s2(s2+b/m).
P2.30 Given
H(s) = k
τs + 1
where τ= 4µs = 4 ×106seconds and 0.999 k < 1.001. The step
response is
Y(s) = k
τs + 1 ·1
s=k
sk
s+ 1.
Taking the inverse Laplace transform yields
y(t) = kket/τ =k(1 et).
The final value is k. The time it takes to reach 98% of the final value is
t= 15.6µs independent of k.
P2.31 From the block diagram we have
Y1(s) = G2(s)[G1(s)E1(s) + G3(s)E2(s)]
=G2(s)G1(s)[R1(s)H1(s)Y1(s)] + G2(s)G3(s)E2(s).
Therefore,
Y1(s) = G1(s)G2(s)
1 + G1(s)G2(s)H1(s)R1(s) + G2(s)G3(s)
1 + G1(s)G2(s)H1(s)E2(s).
And, computing E2(s) (with R2(s) = 0) we find
E2(s) = H2(s)Y2(s) = H2(s)G6(s)G4(s)
G2(s)Y1(s) + G5(s)E2(s)
or
E2(s) = G4(s)G6(s)H2(s)
G2(s)(1 G5(s)G6(s)H2(s))Y1(s).
Substituting E2(s) into equation for Y1(s) yields
Y1(s) = G1(s)G2(s)
1 + G1(s)G2(s)H1(s)R1(s)
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Problems 55
+G3(s)G4(s)G6(s)H2(s)
(1 + G1(s)G2(s)H1(s))(1 G5(s)G6(s)H2(s)) Y1(s).
Finally, solving for Y1(s) yields
Y1(s) = T1(s)R1(s)
where
T1(s) =
G1(s)G2(s)(1 G5(s)G6(s)H2(s))
(1 + G1(s)G2(s)H1(s))(1 G5(s)G6(s)H2(s)) G3(s)G4(s)G6(s)H2(s).
Similarly, for Y2(s) we obtain
Y2(s) = T2(s)R1(s).
where
T2(s) =
G1(s)G4(s)G6(s)
(1 + G1(s)G2(s)H1(s))(1 G5(s)G6(s)H2(s)) G3(s)G4(s)G6(s)H2(s).
P2.32 The signal flow graph shows three loops:
L1=G1G3G4H2
L2=G2G5G6H1
L3=H1G8G6G2G7G4H2G1.
The transfer function Y2/R1is found to be
Y2(s)
R1(s)=G1G8G61G2G5G62
1(L1+L2+L3) + (L1L2),
where for path 1
1= 1
and for path 2
2= 1 L1.
Since we want Y2to be independent of R1, we need Y2/R1= 0. Therefore,
we require
G1G8G6G2G5G6(1 + G1G3G4H2) = 0 .
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56 CHAPTER 2 Mathematical Models of Systems
P2.33 The closed-loop transfer function is
Y(s)
R(s)=G3(s)G1(s)(G2(s) + K5K6)
1G3(s)(H1(s) + K6) + G3(s)G1(s)(G2(s) + K5K6)(H2(s) + K4).
P2.34 The equations of motion are
m1¨y1+b( ˙y1˙y2) + k1(y1y2) = 0
m2¨y2+b( ˙y2˙y1) + k1(y2y1) + k2y2=k2x
Taking the Laplace transform yields
(m1s2+bs +k1)Y1(s)(bs +k1)Y2(s) = 0
(m2s2+bs +k1+k2)Y2(s)(bs +k1)Y1(s) = k2X(s)
Therefore, after solving for Y1(s)/X(s), we have
Y2(s)
X(s)=k2(bs +k1)
(m1s2+bs +k1)(m2s2+bs +k1+k2)(bs +k1)2.
P2.35 (a) We can redraw the block diagram as shown in Figure P2.35. Then,
T(s) = K1/s(s+ 1)
1 + K1(1 + K2s)/s(s+ 1) =K1
s2+ (1 + K2K1)s+K2
.
(b) The signal flow graph reveals two loops (both touching):
L1=K1
s(s+ 1) and L2=K1K2
s+ 1 .
Therefore,
T(s) = K1/s(s+ 1)
1 + K1/s(s+ 1) + K1K2/(s+ 1) =K1
s2+ (1 + K2K1)s+K1
.
(c) We want to choose K1and K2such that
s2+ (1 + K2K1)s+K1=s2+ 20s+ 100 = (s+ 10)2.
Therefore, K1= 100 and 1 + K2K1= 20 or K2= 0.19.
(d) The step response is shown in Figure P2.35.
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58 CHAPTER 2 Mathematical Models of Systems
(d) For the impulse input, y(t)0.65 at t= 1 seconds (see Figure P2.36b).
0123
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
Time (sec)
y(t)
(a) Ramp input
0123
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
Time (sec)
y(t)
(b) Impulse input
FIGURE P2.36
(a) Ramp input response. (b) Impulse input response.
P2.37 The equations of motion are
m1
d2x
dt2=(k1+k2)x+k2yand m2
d2y
dt2=k2(xy) + u .
When m1=m2= 1 and k1=k2= 1, we have
d2x
dt2=2x+yand d2y
dt2=xy+u .
P2.38 The equation of motion for the system is
Jd2θ
dt2+b
dt +kθ = 0 ,
where kis the rotational spring constant and bis the viscous friction
coefficient. The initial conditions are θ(0) = θoand ˙
θ(0) = 0. Taking the
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Problems 59
Laplace transform yields
J(s2θ(s)o) + b((s)θo) + kθ(s) = 0 .
Therefore,
θ(s) = (s+b
Jθo)
(s2+b
Js+K
J)=(s+ 2ζωn)θo
s2+ 2ζωns+ω2
n
.
Neglecting the mass of the rod, the moment of inertia is detemined to be
J= 2Mr2= 0.5 kg ·m2.
Also,
ωn=sk
J= 0.02 rad/s and ζ=b
2Jωn
= 0.01 .
Solving for θ(t), we find that
θ(t) = θo
p1ζ2eζωntsin(ωnq1ζ2t+φ),
where tan φ=p1ζ2). Therefore, the envelope decay is
θe=θo
p1ζ2eζωnt.
So, with ζωn= 2 ×104,θo= 4000oand θf= 10o, the elapsed time is
computed as
t=1
ζωn
ln θo
p1ζ2θf
= 8.32 hours .
P2.39 When t < 0, we have the steady-state conditions
i1(0) = 1A , va(0) = 2Vand vc(0) = 5V ,
where vc(0) is associated with the 1F capacitor. After t0, we have
2di1
dt + 2i1+ 4(i1i2) = 10e2t
and
Zi2dt + 10i2+ 4(i2i1)i1= 0 .
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60 CHAPTER 2 Mathematical Models of Systems
Taking the Laplace transform (using the initial conditions) yields
2(sI1i1(0))+ 2I1+ 4I14I2=10
s+ 2 or (s+ 3)I1(s)2I2(s) = s+ 7
s+ 2
and
[1
sI2vc(0)]+10I2+4(I2I1) = I1(s) or 5sI1(s)+(14s+1)I2(s) = 5s .
Solving for I2(s) yields
I2=5s(s2+ 6s+ 13)
14(s+ 2)∆(s),
where
∆(s) =
s+ 3 2
5s14s+ 1
= 14s2+ 33s+ 3 .
Then,
Vo(s) = 10I2(s).
P2.40 The equations of motion are
J1¨
θ1=K(θ2θ1)b(˙
θ1˙
θ2) + Tand J2¨
θ2=b(˙
θ1˙
θ2).
Taking the Laplace transform yields
(J1s2+bs +K)θ1(s)bsθ2(s) = Kθ2(s) + T(s)
and
(J2s2+bs)θ2(s)bsθ1(s) = 0 .
Solving for θ1(s) and θ2(s), we find that
θ1(s) = (Kθ2(s) + T(s))(J2s+b)
∆(s)and θ2(s) = b(Kθ2(s) + T(s))
∆(s),
where
∆(s) = J1J2s3+b(J1+J2)s2+J2Ks +bK .
P2.41 Assume that the only external torques acting on the rocket are control
torques, Tcand disturbance torques, Td, and assume small angles, θ(t).
Using the small angle approximation, we have
˙
h=V θ
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Problems 61
J¨
θ=Tc+Td,
where Jis the moment of inertia of the rocket and Vis the rocket velocity
(assumed constant). Now, suppose that the control torque is proportional
to the lateral displacement, as
Tc(s) = KH(s),
where the negative sign denotes a negative feedback system. The corre-
sponding block diagram is shown in Figure P2.41.
-
+
1
Js2V
s
K+
+
Tc
Td
Hdesired
=0 H( s)
FIGURE P2.41
Block diagram.
P2.42 (a) The equation of motion of the motor is
J
dt =Tmbω ,
where J= 0.1, b= 0.06, and Tmis the motor input torque.
(b) Given Tm(s) = 1/s, and ω(0) = 0.7, we take the Laplace transform
of the equation of motion yielding
(s)ω(0) + 0.6ω(s) = 10Tm
or
ω(s) = 0.7s+ 10
s(s+ 0.6) .
Then, computing the partial fraction expansion, we find that
ω(s) = A
s+B
s+ 0.6=16.67
s15.97
s+ 0.6.
The step response, determined by taking the inverse Laplace trans-
form, is
ω(t) = 16.67 15.97e0.6t, t 0.
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62 CHAPTER 2 Mathematical Models of Systems
P2.43 The work done by each gear is equal to that of the other, therefore
Tmθm=TLθL.
Also, the travel distance is the same for each gear, so
r1θm=r2θL.
The number of teeth on each gear is proportional to the radius, or
r1N2=r2N1.
So,
θm
θL
=r2
r1
=N2
N1
,
and
N1θm=N2θL
θL=N1
N2
θm=m,
where
n=N1/N2.
Finally,
Tm
TL
=θL
θm
=N1
N2
=n .
P2.44 The inertia of the load is
JL=πρLr4
2.
Also, from the dynamics we have
T2=JL˙ω2+bLω2
and
T1=nT2=n(JL˙ω2+bLω2).
So,
T1=n2(JL˙ω1+bLω1),
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Problems 63
since
ω2=1.
Therefore, the torque at the motor shaft is
T=T1+Tm=n2(JL˙ω1+bLω1) + Jm˙ω1+bmω1.
P2.45 Let U(s) denote the human input and F(s) the load input. The transfer
function is
P(s) = G(s) + KG1(s)
∆(s)U(s) + Gc(s) + KG1(s)
∆(s)F(s),
where
= 1 + GH(s) + G1KBH(s) + GcE(s) + G1KE(s).
P2.46 Consider the application of Newton’s law (PF=m¨x). From the mass
mvwe obtain
mv¨x1=Fk1(x1x2)b1( ˙x1˙x2).
Taking the Laplace transform, and solving for X1(s) yields
X1(s) = 1
1(s)F(s) + b1s+k1
1(s)X2(s),
where
1:= mvs2+b1s+k1.
From the mass mtwe obtain
mt¨x2=k2x2b2˙x2+k1(x1x2) + b1( ˙x1˙x2).
Taking the Laplace transform, and solving for X2(s) yields
X2(s) = b1s+k1
2(s)X1(s),
where
2:= mts2+ (b1+b2)s+k1+k2.
Substituting X2(s) above into the relationship fpr X1(s) yields the trans-
fer function
X1(s)
F(s)=2(s)
1(s)∆2(s)(b1s+k1)2.
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64 CHAPTER 2 Mathematical Models of Systems
P2.47 Using the following relationships
h(t) = Z(1.6θ(t)h(t))dt
ω(t) = ˙
θ(t)
J˙ω(t) = Kmia(t)
va(t) = 50vi(t) = 10ia(t) + vb(t)
˙
θ=Kvb
we find the differential equation is
d3h
dt3+1 + Km
10JK d2h
dt2+Km
10JK
dh
dt =8Km
Jvi.
P2.48 (a) The transfer function is
V2(s)
V1(s)=(1 + sR1C1)(1 + sR2C2)
R1C2s.
(b) When R1= 100 kΩ, R2= 200 kΩ, C1= 1 µF and C2= 0.1µF , we
have
V2(s)
V1(s)=0.2(s+ 10)(s+ 50)
s.
P2.49 (a) The closed-loop transfer function is
T(s) = G(s)
1 + G(s)=6205
s3+ 13s2+ 1281s+ 6205 .
(b) The poles of T(s) are s1=5 and s2,3=4±j35.
(c) The partial fraction expansion (with a step input) is
Y(s) = 1 1.0122
s+ 5 +0.0061 + 0.0716j
s+ 4 + j35 +0.0061 0.0716j
s+ 4 j35 .
(d) The step response is shown in Figure P2.49. The real and complex
roots are close together and by looking at the poles in the s-plane we
have difficulty deciding which is dominant. However, the residue at
the real pole is much larger and thus dominates the response.
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Problems 65
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
00.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2
Time (secs)
Amplitude
FIGURE P2.49
Step response.
P2.50 (a) The closed-loop transfer function is
T(s) = 14000
s3+ 45s2+ 3100s+ 14500 .
(b) The poles of T(s) are
s1=5 and s2,3=20 ±j50.
(c) The partial fraction expansion (with a step input) is
Y(s) = 0.9655
s1.0275
s+ 5 +0.0310 0.0390j
s+ 20 + j50 +0.0310 + 0.0390j
s+ 20 j50 .
(d) The step response is shown in Figure P2.50. The real root dominates
the response.
(e) The final value of y(t) is
yss = lim
s0sY (s) = 0.9655 .
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66 CHAPTER 2 Mathematical Models of Systems
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
00.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2
Time (secs)
Amplitude
FIGURE P2.50
Step response.
P2.51 Consider the free body diagram in Figure P2.51. Using Newton’s Law
and summing the forces on the two masses yields
M1¨x(t) + b1˙x(t) + k1x(t) = b1˙y(t)
M2¨y(t) + b1˙y(t) + k2y(t) = b1˙x(t) + u(t)
M1
M2
k1
b1
k2
u(t)
x
y
M1
M2
k1x
k2
u(t)
x
y
b1(x - y)
. .
b1(y - x)
. . y
FIGURE P2.51
Free body diagram.
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Advanced Problems 67
Advanced Problems
AP2.1 The transfer function from V(s) to ω(s) has the form
ω(s)
V(s)=Km
τms+ 1 .
In the steady-state,
ωss = lim
s0sKm
τms+ 15
s= 5Km.
So,
Km= 70/5 = 14 .
Also,
ω(t) = VmKm(1 et/τm)
where V(s) = Vm/s. Solving for τmyields
τm=t
ln(1 ω(t)ss).
When t= 2, we have
τm=2
ln(1 30/70) = 3.57 .
Therefore, the transfer function is
ω(s)
V(s)=14
3.57s+ 1 .
AP2.2 The closed-loop transfer function form R1(s) to Y2(s) is
Y2(s)
R1(s)=G1G4G5(s) + G1G2G3G4G6(s)
where
= [1 + G3G4H2(s)][1 + G1G2H3(s)] .
If we select
G5(s) = G2G3G6(s)
then the numerator is zero, and Y2(s)/R1(s) = 0. The system is now
decoupled.
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68 CHAPTER 2 Mathematical Models of Systems
AP2.3 (a) Computing the closed-loop transfer function:
Y(s) = G(s)Gc(s)
1 + Gc(s)G(s)H(s)R(s).
Then, with E(s) = R(s)Y(s) we obtain
E(s) = 1 + Gc(s)G(s)(H(s)1)
1 + Gc(s)G(s)H(s)R(s).
If we require that E(s)0 for any input, we need 1 + Gc(s)G(s)(H(s)
1) = 0 or
H(s) = Gc(s)G(s)1
Gc(s)G(s)=n(s)
d(s).
Since we require H(s) to be a causal system, the order of the numerator
polynomial, n(s), must be less than or equal to the order of the denom-
inator polynomial, d(s). This will be true, in general, only if both Gc(s)
and G(s) are proper rational functions (that is, the numerator and de-
nominator polynomials have the same order). Therefore, making E0
for any input R(s) is possible only in certain circumstances.
(b) The transfer function from Td(s) to Y(s) is
Y(s) = Gd(s)G(s)
1 + Gc(s)G(s)H(s)Td(s).
With H(s) as in part (a) we have
Y(s) = Gd(s)
Gc(s)Td(s).
(c) No. Since
Y(s) = Gd(s)G(s)
1 + Gc(s)G(s)H(s)Td(s) = T(s)Td(s),
the only way to have Y(s)0 for any Td(s) is for the transfer function
T(s)0 which is not possible in general (since G(s)6= 0).
AP2.4 (a) With q(s) = 1/s we obtain
τ(s) = 1/Ct
s+QS+1/R
Ct·1
s.
Define
α:= QS + 1/R
Ct
andβ:= 1/Ct.
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Advanced Problems 69
Then, it follows that
τ(s) = β
s+α·1
s=β
s+α+β
s.
Taking the inverse Laplace transform yields
τ(t) = β
αeαt +β
α=β
α[1 eαt].
(b) As t→ ∞,τ(t)β
α=1
Qs+1/R .
(c) To increase the speed of response, you want to choose Ct,Q,Sand
Rsuch that
α:= Qs + 1/R
Ct
is ”large.”
AP2.5 Considering the motion of each mass, we have
M3¨x3+b3˙x3+k3x3=u3+b3˙x2+k3x2
M2¨x2+ (b2+b3) ˙x2+ (k2+k3)x2=u2+b3˙x3+k3x3+b2˙x1+k2x1
M1¨x1+ (b1+b2) ˙x1+ (k1+k2)x1=u1+b2˙x2+k2x2
In matrix form the three equations can be written as
M10 0
0M20
0 0 M3
¨x1
¨x2
¨x3
+
b1+b2b20
b2b2+b3b3
0b3b3
˙x1
˙x2
˙x3
+
k1+k2k20
k2k2+k3k3
0k3k3
x1
x2
x3
=
u1
u2
u3
.
AP2.6 Considering the cart mass and using Newton’s Law we obtain
M¨x=ub˙xFsin ϕ
where Fis the reaction force between the cart and the pendulum. Con-
sidering the pendulum we obtain
md2(x+Lsin ϕ)
dt2=Fsinϕ
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70 CHAPTER 2 Mathematical Models of Systems
md2(Lcos ϕ)
dt2=Fcos ϕ+mg
Eliminating the reaction force Fyields the two equations
(m+M)¨x+b˙x+mL ¨ϕcos ϕmL ˙ϕ2sin ϕ=u
mL2¨ϕ+mgL sin ϕ+mL¨xcos ϕ= 0
If we assume that the angle ϕ0, then we have the linear model
(m+M)¨x+b˙x+mL ¨ϕ=u
mL2¨ϕ+mg=mL¨x
AP2.7 The transfer function from the disturbance input to the output is
Y(s) = 1
s+ 20 + KTd(s).
When Td(s) = 1, we obtain
y(t) = e(20+K)t.
Solving for twhen y(t)<0.1 yields
t > 2.3
20 + K.
When t= 0.05 and y(0.05) = 0.1, we find K= 26.05.
AP2.8 The closed-loop transfer function is
T(s) = 200K(0.25s+ 1)
(0.25s+ 1)(s+ 1)(s+ 8) + 200K
The final value due to a step input of R(s) = A/s is
v(t)A200K
200K+ 8.
We need to select Kso that v(t)50. However, to keep the percent
overshoot to less than 10%, we need to limit the magnitude of K. Fig-
ure AP2.8a shows the percent overshoot as a function of K. Let K= 0.06
and select the magnitude of the input to be A= 83.3. The inverse Laplace
transform of the closed-loop response with R(s) = 83.3/s is
v(t) = 50 + 9.85e9.15te1.93t(59.85 cos(2.24t) + 11.27 sin(2.24t))
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Advanced Problems 71
The result is P.O. = 9.74% and the steady-state value of the output is
approximately 50 m/s, as shown in Figure AP2.8b.
0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1
0
5
10
15
20
25
K
Percent Overshoot (%)
Step Response
Time (sec)
Amplitude
0 0.5 1 1.5 2 2.5
0
10
20
30
40
50
60
System: untitled1
Peak amplitude: 54.9
Overshoot (%): 9.74
At time (sec): 1.15
FIGURE AP2.8
(a) Percent overshoot versus the gain K. (b) Step response.
AP2.9 The transfer function is
Vo(s)
Vi(s)=Z2(s)
Z1(s),
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72 CHAPTER 2 Mathematical Models of Systems
where
Z1(s) = R1
R1C1s+ 1 and Z2(s) = R2C2s+ 1
C2s.
Then we can write
Vo(s)
Vi(s)=Kp+KI
s+KDs
where
KP=R1C1
R2C2
+ 1, KI=1
R1C2
, KD=R2C1.
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Design Problems 73
Design Problems
The model of the traction drive, capstan roller, and linear slide followsCDP2.1
closely the armature-controlled dc motor model depicted in Figure 2.18
in Dorf and Bishop. The transfer function is
T(s) = rKm
s[(Lms+Rm)(JTs+bm) + KbKm],
where
JT=Jm+r2(Ms+Mb).
-
Va(s) X(s)
Kb
Back EMF
Km
Lms+Rm
1
JTs+bm
1
s
q
wr
DP2.1 The closed-loop transfer function is
Y(s)
R(s)=G1(s)G2(s)
1 + G1(s)H1(s)G2(s)H2(s).
When G1H1=G2H2and G1G2= 1, then Y(s)/R(s) = 1. Therefore,
select
G1(s) = 1
G2(s)and H1(s) = G2(s)H2(s)
G1(s)=G2
2(s)H2(s).
DP2.2 At the lower node we have
v1
4+1
3+G+ 2i220 = 0 .
Also, we have v= 24 and i2=Gv . So
v1
4+1
3+G+ 2Gv20 = 0
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74 CHAPTER 2 Mathematical Models of Systems
and
G=20 v1
4+1
3
3v=1
12 S .
DP2.3 Taking the Laplace transform of
y(t) = et1
4e2t3
4+1
2t
yields
Y(s) = 1
s+ 1 1
4(s+ 2) 3
4s+1
2s2.
Similarly, taking the Laplace transform of the ramp input yields
R(s) = 1
s2.
Therefore
G(s) = Y(s)
R(s)=1
(s+ 1)(s+ 2) .
DP2.4 For an ideal op-amp, at node a we have
vin va
R1
+vova
R1
= 0 ,
and at node b
vin vb
R2
=C˙vb,
from it follows that
1
R2
+CsVb=1
R2
Vin .
Also, for an ideal op-amp, VbVa= 0. Then solving for Vbin the above
equation and substituting the result into the node a equation for Vayields
Vo
Vin
=2
1
R2+Cs "1
R2
1
R2+Cs
2#
or
Vo(s)
Vin(s)=R2Cs 1
R2Cs + 1 .
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Design Problems 75
For vin(t) = At, we have Vin(s) = A/s2, therefore
vo(t) = A2
βeβt +t2
β
where β= 1/R2C.
DP2.5 The equation of motion describing the motion of the inverted pendulum
(assuming small angles) is
¨ϕ+g
Lϕ= 0 .
Assuming a solution of the form ϕ=kcos ϕ, taking the appropriate
derivatives and substituting the result into the equation of motion yields
the relationship
˙ϕ=rg
L.
If the period is T= 2 seconds, we compute ˙ϕ= 2π/T . Then solving for L
yields L= 0.99 meters when g= 9.81 m/s2. So, to fit the pendulum into
the grandfather clock, the dimensions are generally about 1.5 meters or
more.
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76 CHAPTER 2 Mathematical Models of Systems
Computer Problems
CP2.1 The m-file script is shown in Figure CP2.1.
pq =
1 9 24 20
P =
-5
-2
Z =
-2
value =
4
p=[1 7 10]; q=[1 2];
% Part (a)
pq=conv(p,q)
% Part (b)
P=roots(p), Z=roots(q)
% Part (c)
value=polyval(p,-1)
FIGURE CP2.1
Script for various polynomial evaluations.
CP2.2 The m-file script and step response is shown in Figure CP2.2.
numc = [1]; denc = [1 1]; sysc = tf(numc,denc)
numg = [1 2]; deng = [1 3]; sysg = tf(numg,deng)
% part (a)
sys_s = series(sysc,sysg);
sys_cl = feedback(sys_s,[1])
% part (b)
step(sys_cl); grid on
Transfer function:
s + 2
-------------
s^2 + 5 s + 5
Time (sec.)
Amplitude
Step Response
0 0.5 1 1.5 2 2.5 3 3.5 4
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
From: U(1)
To: Y(1)
FIGURE CP2.2
Step response.
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Computer Problems 77
CP2.3 Given
¨y+ 4 ˙y+ 3y=u
with y(0) = ˙y= 0 and U(s) = 1/s, we obtain (via Laplace transform)
Y(s) = 1
s(s2+ 4s+ 3) =1
s(s+ 3)(s+ 1) .
Expanding in a partial fraction expansion yields
Y(s) = 1
3s1
6(s+ 3) 1
2(s+ 1) .
Taking the inverse Laplace transform we obtain the solution
y(t) = 0.3333 + 0.1667e3t0.5et.
The m-file script and step response is shown in Figure CP2.3.
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
Step Response
Time (sec)
Amplitude
n=[1]; d=[1 4 3]; sys = tf(n,d);
t=[0:0.1:5];
y = step(sys,t);
ya=0.3333+0.1667*exp(-3*t)-0.5*exp(-t);
plot(t,y,t,ya); grid;
title('Step Response');
xlabel('Time (sec)');
ylabel('Amplitude');
FIGURE CP2.3
Step response.
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78 CHAPTER 2 Mathematical Models of Systems
CP2.4 The mass-spring-damper system is represented by
m¨x+b˙x+kx =f .
Taking the Laplace transform (with zero initial conditions) yields the
transfer function
X(s)/F (s) = 1/m
s2+bs/m +k/m .
The m-file script and step response is shown in Figure CP2.4.
m=10; k=1; b=0.5;
num=[1/m]; den=[1 b/m k/m];
sys = tf(num,den);
t=[0:0.1:150];
step(sys,t)
Time (sec.)
Amplitude
Step Response
0 50 100 150
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
From: U(1)
To: Y(1)
FIGURE CP2.4
Step response.
CP2.5 The spacecraft simulations are shown in Figure CP2.5. We see that as J
is decreased, the time to settle down decreases. Also, the overhoot from
10odecreases as Jdecreases. Thus, the performance seems to get better
(in some sense) as Jdecreases.
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Computer Problems 79
0
2
4
6
8
10
12
14
16
18
0 10 20 30 40 50 60 70 80 90 100
Time (sec)
Spacecraft attitude (deg)
Nominal (solid); O-nominal 80% (dashed); O-nominal 50% (dotted)
%Part (a)
a=1; b=8; k=10.8e+08; J=10.8e+08;
num=k*[1 a];
den=J*[1 b 0 0]; sys=tf(num,den);
sys_cl=feedback(sys,[1]);
%
% Part (b) and (c)
t=[0:0.1:100];
%
% Nominal case
f=10*pi/180; sysf=sys_cl*f;
y=step(sysf,t);
%
% O-nominal case 80%
J=10.8e+08*0.8; den=J*[1 b 0 0];
sys=tf(num,den); sys_cl=feedback(sys,[1]);
sysf=sys_cl*f;
y1=step(sysf,t);
%
% O-nominal case 50%
J=10.8e+08*0.5; den=J*[1 b 0 0];
sys=tf(num,den); sys_cl=feedback(sys,[1]);
sysf=sys_cl*f;
y2=step(sysf,t);
%
plot(t,y*180/pi,t,y1*180/pi,'--',t,y2*180/pi,':'),grid
xlabel('Time (sec)')
ylabel('Spacecraft attitude (deg)')
title('Nominal (solid); O-nominal 80% (dashed); O-nominal 50% (dotted)')
FIGURE CP2.5
Step responses for the nominal and off-nominal spacecraft parameters.
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80 CHAPTER 2 Mathematical Models of Systems
CP2.6 The closed-loop transfer function is
T(s) = 4s6+ 8s5+ 4s4+ 56s3+ 112s2+ 56s
∆(s),
num1=[4]; den1=[1]; sys1 = tf(num1,den1);
num2=[1]; den2=[1 1]; sys2 = tf(num2,den2);
num3=[1 0]; den3=[1 0 2]; sys3 = tf(num3,den3);
num4=[1]; den4=[1 0 0]; sys4 = tf(num4,den4);
num5=[4 2]; den5=[1 2 1]; sys5 = tf(num5,den5);
num6=[50]; den6=[1]; sys6 = tf(num6,den6);
num7=[1 0 2]; den7=[1 0 0 14]; sys7 = tf(num7,den7);
sysa = feedback(sys4,sys6,+1);
sysb = series(sys2,sys3);
sysc = feedback(sysb,sys5);
sysd = series(sysc,sysa);
syse = feedback(sysd,sys7);
sys = series(sys1,syse)
%
pzmap(sys)
%
p=pole(sys)
z=zero(sys)
p =
7.0709
-7.0713
1.2051 + 2.0863i
1.2051 - 2.0863i
0.1219 + 1.8374i
0.1219 - 1.8374i
-2.3933
-2.3333
-0.4635 + 0.1997i
-0.4635 - 0.1997i
z =
0
1.2051 + 2.0872i
1.2051 - 2.0872i
-2.4101
-1.0000 + 0.0000i
-1.0000 - 0.0000i
poles
Real Axis
Imag Axis
Polezero map
-8 -6 -4 -2 0 2 4 6 8
-2.5
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
2.5
FIGURE CP2.6
Pole-zero map.
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Computer Problems 81
where
∆(s) = s10 + 3s945s8125s7200s61177s5
2344s43485s37668s25598s1400 .
CP2.7 The m-file script and plot of the pendulum angle is shown in Figure CP2.7.
With the initial conditions, the Laplace transform of the linear system is
θ(s) = θ0s
s2+g/L.
To use the step function with the m-file, we can multiply the transfer
function as follows:
θ(s) = s2
s2+g/L
θ0
s,
which is equivalent to the original transfer function except that we can
use the step function input with magnitude θ0. The nonlinear response
is shown as the solid line and the linear response is shown as the dashed
line. The difference between the two responses is not great since the initial
condition of θ0= 30is not that large.
0246810
-30
-20
-10
0
10
20
30
Time (s)
θ (deg)
L=0.5; m=1; g=9.8;
theta0=30;
% Linear simulation
sys=tf([1 0 0],[1 0 g/L]);
[y,t]=step(theta0*sys,[0:0.01:10]);
% Nonlinear simulation
[t,ynl]=ode45(@pend,t,[theta0*pi/180 0]);
plot(t,ynl(:,1)*180/pi,t,y,'--');
xlabel('Time (s)')
ylabel('\theta (deg)')
function [yd]=pend(t,y)
L=0.5; g=9.8;
yd(1)=y(2);
yd(2)=-(g/L)*sin(y(1));
yd=yd';
FIGURE CP2.7
Plot of θversus xt when θ0= 30.
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82 CHAPTER 2 Mathematical Models of Systems
CP2.8 The system step responses for z= 5,10, and 15 are shown in Fig-
ure CP2.8.
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
0
0.5
1
1.5 z=5 (solid), z=10 (dashed), z=15 dotted)
Time (sec)
x(t)
FIGURE CP2.8
The system response.
CP2.9 (a,b) Computing the closed-loop transfer function yields
T(s) = G(s)
1 + G(s)H(s)=s2+ 2s+ 1
s2+ 4s+ 3 .
The poles are s=3,1 and the zeros are s=1,1.
(c) Yes, there is one pole-zero cancellation. The transfer function (after
pole-zero cancellation) is
T(s) = s+ 1
s+ 3 .
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Computer Problems 83
?-3 ?-2.5 ?-2 ?-1.5 ?-1 ?-0.5 0
?-1
?-0.8
?-0.6
?-0.4
?-0.2
0
0.2
0.4
0.6
0.8
1
Pole?Zero Map
Real Axi s
Imaginary Axi s
poles
ng=[1 1]; dg=[1 2]; sysg = tf(ng,dg);
nh=[1]; dh=[1 1]; sysh = tf(nh,dh);
sys=feedback(sysg,sysh)
%
pzmap(sys)
%
pole(sys)
zero(sys)
>>
Transfer function:
s^2 + 2 s + 1
-------------
s^2 + 4 s + 3
p =
-3
-1
z =
-1
-1
zeros
FIGURE CP2.9
Pole-zero map.
CP2.10 Figure CP2.10 shows the steady-state response to a unit step input and a
unit step disturbance. We see that K= 1 leads to the same steady-state
response.
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84 CHAPTER 2 Mathematical Models of Systems
0 1 2 3 4 5 6 7 8 9 10
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
K
Steady−state response
Disturbance Response Steady-State
Input Response Steady-State
K=1
K=[0.1:0.1:10];
sysg=tf([1],[1 20 20]);
for i=1:length(K)
nc=K(i); dc=[1];sysc=tf(nc,dc);
syscl=feedback(sysc*sysg,1);
systd=feedback(sysg,sysc);
y1=step(syscl);
Tf1(i)=y1(end);
y2=step(systd);
Tf2(i)=y2(end);
end
plot(K,Tf1,K,Tf2,'--')
xlabel('K')
ylabel('Steady-state response')
FIGURE CP2.10
Gain Kversus steady-state value.
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C H A P T E R 3
State Variable Models
Exercises
E3.1 One possible set of state variables is
(a) the current iL2through L2,
(b) the voltage vC2across C2, and
(c) the current iL1through L1.
We can also choose vC1, the voltage across C1as the third state variable,
in place of the current through L1.
E3.2 We know that the velocity is the derivative of the position, therefore we
have
dy
dt =v ,
and from the problem statement
dv
dt =k1v(t)k2y(t) + k3i(t).
This can be written in matrix form as
d
dt
y
v
=
0 1
k2k1
y
v
+
0
k3
i .
Define u=i, and let k1=k2= 1. Then,
˙
x=Ax +Bu
where
A=
0 1
11
,B=
0
k3
,and x=
y
v
.
85
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86 CHAPTER 3 State Variable Models
E3.3 The charactersitic roots, denoted by λ, are the solutions of det(λIA) =
0. For this problem we have
det(λIA) = det
λ1
1λ+ 2
=λ(λ+ 2) + 1 = λ2+ 2λ+ 1 = 0 .
Therefore, the characteristic roots are
λ1=1 and λ2=1.
E3.4 The system in phase variable form is
˙
x=Ax +Bu
y=Cx
where
A=
0 1 0
0 0 1
864
,B=
0
0
20
,C=h1 0 0 i.
E3.5 From the block diagram we determine that the state equations are
˙x2=(fk +d)x2+ax1+fu
˙x1=kx2+u
and the output equation is
y=bx2.
Therefore,
˙
x=Ax +Bu
y=Cx +Du ,
where
A=
0k
a(fk +d)
,B=
1
f
,C=h0biand D= [0] .
E3.6 (a) The state transition matrix is
Φ(t) = eAt=I+At+1
2!A2t2+···
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Exercises 87
But A2= 0, thus A3=A4=···= 0. So,
Φ(t) = eAt=I+At=
1 0
0 1
+
0 1
0 0
t=
1t
0 1
.
(b) The state at any time t0 is given by
x(t) = Φ(t)x(0)
and since x1(0) = x2(0) = 1, we determine that
x1(t) = x1(0) + tx2(0) = 1 + t
x2(t) = x2(0) = 1 .
E3.7 The state equations are
˙x1=x2
˙x2=100x120x2+u
or, in matrix form
˙
x=
0 1
100 20
x+
0
1
u .
So, the characteristic equation is determined to be
det(λIA) = det
λ1
100 λ+ 20
=λ2+ 20λ+ 100 = (λ+ 10)2= 0 .
Thus, the roots of the characteristic equation are
λ1=λ2=10 .
E3.8 The characteristic equation is
det(λIA) = det
λ1 0
0λ1
0 6 λ+ 3
=λ(λ2+ 3λ+ 6) = 0 .
Thus, the roots of the characteristic equation are
λ1= 0 , λ2=1.5 + j1.9365 and λ3=1.5j1.9365 .
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88 CHAPTER 3 State Variable Models
E3.9 Analyzing the block diagram yields
˙x1=x1+1
2x2+r
˙x2=x13
2x2r
y=x13
2x2r.
In state-variable form we have
˙
x=
11
2
13
2
x+
1
1
r , y =13
2x+h1ir .
The characteristic equation is
s2+5
2s+ 1 = (s+ 2)(s+1
2) = 0 .
E3.10 (a) The characteristic equation is
det[λIA] = det
λ6
1 (λ+ 5)
=λ(λ+5)+6 = (λ+2)(λ+3) = 0 .
So, the roots are λ1=2 and λ2=3.
(b) We note that
Φ(s) = [sIA]1=
s6
1s+ 5
1
=1
(s+ 2)(s+ 3)
s+ 5 6
1s
.
Taking the inverse Laplace transform yields the transition matrix
Φ(t) =
3e2t2e3t6e2t6e3t
e2t+e3t2e2t+ 3e3t
.
E3.11 A state variable representation is
˙
x=Ax +Br
y=Cx
where
A=
0 1
12 8
,B=
0
1
,C=h12 4 i.
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Exercises 89
E3.12 The equation of motion is
Ldi
dt +Ri +vc=vin
where
vc=1
CZi dt .
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35
−0.2
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
Time(sec)
State response
x1: capacitor voltage
x2: inductor current
Unit step response
FIGURE E3.12
State variable time history for a unit step input.
Selecting the state variables x1=vcand x2=i, we have
˙x1=1
Cx2
˙x2=R
Lx21
Lx1+1
Lvin .
This can be written in matrix form as
˙
x=
0 1/C
1/L R/L
x+
0
1/L
vin .
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90 CHAPTER 3 State Variable Models
When C= 0.001F,R= 4Ω, and L= 0.1H, we have
˙
x=
0 1000
10 40
x+
0
10
vin .
The step response is shown in Figure E3.12.
E3.13 (a) Select the state variables as x1=yand x2=ω.
(b) The corresponding state equation is
˙x1=x1ax2+ 2u
˙x2=bx14u
or, in matrix form
˙
x=
1a
b0
x+
2
4
uand x=
x1
x2
.
(c) The characteristic equation is
det[λIA] = det
λ+ 1 a
b λ
=λ2+λ+ab = 0 .
So, the roots are
λ=1
2±1
214ab .
E3.14 Assume that the mass decay is proportional to the mass present, so that
˙
M=qM +Ku
where qis the constant of proportionality. Select the state variable, x, to
be the mass, M. Then, the state equation is
˙x=qx +Ku .
E3.15 The equations of motion are
m¨x+kx +k1(xq) + b˙x= 0
m¨q+kq +b˙q+k1(qx) = 0 .
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Exercises 91
In state variable form we have
˙
x=
0100
(k+k1)
mb
m
k1
m0
0 0 0 1
k1
m0(k+k1)
mb
m
x
where x1=x, x2= ˙x, x3=qand x4= ˙q.
E3.16 The governing equations of motion are
m1¨x+k1(xq) + b1( ˙x˙q) = u(t)
m2¨q+k2q+b2˙q+b1( ˙q˙x) + k1(qx) = 0 .
Let x1=x, x2= ˙x, x3=qand x4= ˙q. Then,
˙
x=
0 1 0 0
k1
m1b1
m1
k1
m1
b1
m1
0 0 0 1
k1
m2
b1
m2(k1+k2)
m2(b1+b2)
m2
x+
0
1
m1
0
0
u(t).
Since the output is y(t) = q(t), then
y=h0 0 1 0 ix.
E3.17 At node 1 we have
C1˙v1=vav1
R1
+v2v1
R2
and at node 2 we have
C2˙v2=vbv2
R3
+v1v2
R2
.
Let
x1=v1
and
x2=v2.
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92 CHAPTER 3 State Variable Models
Then, in matrix form we have
˙
x=
1
R1C1+1
R2C11
R2C1
1
R2C21
R3C2+1
R2C2
x+
1
R1C10
01
R3C2
va
vb
.
E3.18 The governing equations of motion are
Ri1+L1
di1
dt +v=va
L2
di2
dt +v=vb
iL=i1+i2=Cdv
dt .
Let x1=i1, x2=i2, x3=v, u1=vaand u2=vb. Then,
˙
x=
R
L101
L1
0 0 1
L2
1
C
1
C0
x+
1
L10
01
L2
0 0
u
y=h0 0 1 ix+ [0] u.
E3.19 First, compute the matrix
sI A=
s1
3s+ 4
.
Then, Φ(s) is
Φ(s) = (sIA)1=1
∆(s)
s+ 4 1
3s
where ∆(s) = s2+ 4s+ 3, and
G(s) = h10 0 i
s+4
∆(s)
1
∆(s)
3
∆(s)
s
∆(s)
0
1
=10
s2+ 4s+ 3 .
E3.20 The linearized equation can be derived from the observation that sin θθ
when θ0. In this case, the linearized equations are
¨
θ+g
Lθ+k
m˙
θ= 0 .
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Exercises 93
Let x1=θand x2=˙
θ. Then in state variable form we have
˙
x=Ax
y=Cx
where
A=
0 1
g/L k/m
,C=h1 0 i,and x(0) =
θ(0)
˙
θ(0)
.
E3.21 The transfer function is
G(s) = C[sIA]1B+D=1
s2+ 2s+ 1 .
The unit step response is
y(t) = 1 + et+tet.
E3.22 The transfer function is
G(s) = s6
s27s+ 6 .
The poles are at s1= 1 and s2= 6. The zero is at s= 6. So, we see
that there is a pole-zero cancellation. We can write the system in state
variable form as
˙x=x2u
y=2
2x
and the transfer function is
G(s) = 1
s1.
E3.23 The system in state variable form can be represented by
˙
x=Ax +Bu
y=Cx +Du
where
A=
0 1 0
0 0 1
133
,B=
0
0
1
,C=h0 1 1i,D=h1i.
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94 CHAPTER 3 State Variable Models
U(s)
X(s)
x1
x2
s
1
x3
s
1
s
1
3
3
+
--
-
+
+
-
FIGURE E3.23
Block diagram.
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Problems 95
Problems
P3.1 The loop equation, derived from Kirchoff’s voltage law, is
di
dt =1
LvR
Li1
Lvc
where
vc=1
CZi dt .
(a) Select the state variables as x1=iand x2=vc.
(b) The corresponding state equations are
˙x1=1
LvR
Lx11
Lx2
˙x2=1
Cx1.
(c) Let the input u=v. Then, in matrix form, we have
˙
x=
R/L 1/L
1/C 0
x+
1/L
0
u .
1/L
v
-1/L
1/s
-R/L
x1
1/C 1/s
x2
FIGURE P3.1
Signal flow graph.
P3.2 Let
a11 =2
(R1+R2)C, a22 =2R1R2
(R1+R2)L,
b11 =b12 =1
(R1+R2)C, b21 =b22 =R2
(R1+R2)L.
The corresponding block diagram is shown in Figure P3.2.
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from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying,
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96 CHAPTER 3 State Variable Models
1/(R1+R2)C
1/s
-
x1
x2
-
2/(R1+R2)C
v1
v2
R2
1/(R1+R2)C
1/s
2R1R2/(R1+R2)C
x2
1/s
1/s
x1
v2
v1
b22
b21
b12
b11
a22
a11
(a)
(b)
FIGURE P3.2
(a) Block diagram. (b) Signal flow graph.
P3.3 Using Kirchoff’s voltage law around the outer loop, we have
LdiL
dt vc+v2v1= 0 .
Then, using Kirchoff’s current law at the node, we determine that
Cdvc
dt =iL+iR,
where iRis the current through the resistor R. Considering the right loop
we have
iRRv2+vc= 0 or iR=vc
R+v2
R.
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Problems 97
Thus,
dvc
dt =vc
RC iL
C+v2
RC and diL
dt =vc
dt +v1
Lv2
L.
In matrix form, the state equations are
˙x1
˙x2
=
0 1/L
1/C 1/RC
x1
x2
+
1/L 1/L
0 1/RC
v1
v2
,
where x1=iLand x2=vc. The signal flow graph is shown in Figure P3.3.
1/L
1/RC
1/s
1/L
x1
-1/C 1/s
x2
v1
v2
-1/L
-1/RC
FIGURE P3.3
Signal flow graph.
P3.4 (a) The block diagram model for phase variable form is shown in Fig-
ure P3.4a. The phase variable form is given by
˙
x=
0 1 0
0 0 1
10 64
x+
0
0
1
r
y=h10 2 1 ix.
(b) The block diagram in input feedforward form is shown in Figure P3.4b.
The input feedforward form is given by
˙
x=
4 1 0
6 0 1
10 0 0
x+
1
2
10
r(t)
y=h1 0 0 ix.
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98 CHAPTER 3 State Variable Models
R(s)Y(s)
10
x1
x2
s
1
10
x3
s
1
s
1
4
6
2
1
-
--
++
+
+
(a)
R(s)Y(s)
10 s
1
10
s
1
s
1
6
4
2
1
x1
.
x2
.
x3
.
--
-
+++++
(b)
FIGURE P3.4
(a)Block diagram model for phase variable form. (b) Block diagram model for input feedforward form.
P3.5 (a) The closed-loop transfer function is
T(s) = s+ 1
s3+ 4s211s+ 1 .
(b) A matrix differential equation is
˙
x=Ax +Bu
y=Cx
where
A=
0 1 0
0 0 1
1 11 4
,B=
0
0
1
,C=h1 1 0 i.
The block diagram is shown in Figure P3.5.
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Problems 99
R(s)Y(s)
1
x1
x2
s
1
1
x3
s
1
s
1
4
-11
1
-
--
++
+
FIGURE P3.5
Block diagram model.
P3.6 The node equations are
0.00025dv1
dt +iLviv1
4000 = 0
0.0005dv2
dt iL+v2
1000 i3= 0
0.002diL
dt +v2v1= 0 .
Define the state variables
x1=v1x2=v2x3=iL.
Then,
˙
x=Ax +Bu
where
A=
1 0 4000
02 2000
500 500 0
,B=
1 0
0 2000
0 0
P3.7 Given K= 1, we have
KG(s)·1
s=(s+ 1)2
s(s2+ 1) .
We then compute the closed-loop transfer function as
T(s) = s2+ 2s+ 1
3s3+ 5s2+ 5s+ 1 =s1+ 2s2+s3
3 + 5s1+ 5s2+s3.
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100 CHAPTER 3 State Variable Models
A state variable model is
˙
x=
010
001
1/35/35/3
x+
0
0
1/3
r
y=h1 2 1 ix.
P3.8 The state-space equations are
˙x1=x2
˙x2=ku
x3g
˙x3=u .
This is a set of nonlinear equations.
P3.9 (a) The closed-loop transfer function is
T(s) = 10
Js3+ (b+ 10J)s2+ 10bs + 10K1
=10s3
1 + 10.1s1+s2+ 5s3,
where K1= 0.5, J= 1, and b= 0.1.
(b) A state-space model is
˙
x=
0 1 0
0 0 1
5110.1
x+
0
0
10
r
ω=h1 0 0 ix.
(c) The characteristic equation is
det[sIA] = det
s1 0
0s1
5 1 s+ 10.1
=s3+ 10.1s2+s+ 5 = 0 .
The roots of the characteristic equation are
s1=10.05 and s2,3=0.0250 ±0.7049j .
All roots lie in the left hand-plane, therefore, the system is stable.
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Problems 101
P3.10 (a) From the signal flow diagram, we determine that a state-space model
is given by
˙
x=
K1K2
K1K2
x+
K1K2
K1K2
r1
r2
y=
y1
y2
=
1 0
0 1
x.
(b) The characteristic equation is
det[sIA] = s2+ (K2+K1)s+ 2K1K2= 0 .
(c) When K1=K2= 1, then
A=
1 1
11
.
The state transition matrix associated with Ais
Φ=L1n[sIA]1o=et
cos tsin t
sin tcos t
.
P3.11 The state transition matrix is
Φ(t) =
(2t1)et2tet
2tet(2t+ 1)et
.
So, when x1(0) = x2(0) = 10, we have
x(t) = Φ(t)x(0)
or
x1(t) = 10et
x2(t) = 10et
P3.12 (a) A state variable representation is given by
˙
x=
0 1 0
0 0 1
48 44 12
x+
0
0
1
r
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102 CHAPTER 3 State Variable Models
y= [40 8 0]x.
(b) The state transition matrix is
Φ(t) = Φ1(t).
.
2(t).
.
3(t),
where
Φ1(t) =
e6t3e4t+ 3e2t
6e6t+ 12e4t6e2t
36e6t48e4t+ 12e2t
Φ2(t) =
3
4e6t2e4t+5
4e2t
9
2e6t+ 8e4t5
2e2t
27e6t32e4t+ 5e2t
Φ3(t) =
1
8e6t1
4e4t+1
8e2t
3
4e6t+e4t1
4e2t
9
2e6t4e4t+1
2e2t
.
P3.13 (a) The RLC circuit state variable representation is
˙
x=
10 4
6 0
x+
4
0
u .
The characteristic equation is
s2+ 10s+ 24 = 0 .
All roots of the characteristic equation (that is, s1=4 and s2=6)
are in the left half-plane; therefore the system is stable.
(b) The state transition matrix is
Φ(t) =
3e6t2e4t2e6t+ 2e4t
3e6t+ 3e4t2e6t+ 3e4t
.
(c) Given
x1(0) = 0.1, x2(0) = 0 and e(t) = 0 ,
we have
i(t) = x1(t) = 0.3e6t0.2e4t
vc(t) = x2(t) = 0.3e6t+ 0.3e4t.
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Problems 103
(d) When x(0) = 0 and u(t) = E, we have
x(t) = Zt
0
Φ(tτ)Bu(τ)dτ ,
where
Bu(t) =
4E
0
.
Integrating yields
x1(t) = (2e6t+ 2e4t)E
x2(t) = (1 + 2e6t3e4t)E .
P3.14 A state space representation is
˙
x=Ax+Br , y =Cx
where
A=
0 1 0 0
0 0 1 0
0 0 0 1
50 34 10 12
,B=
0
0
0
1
,C= [50 1 0 0] .
P3.15 A state variable representation is
˙
x=
0 1 0
0 0 1
16 31 10
x+
0
0
1
r
y= [56 14 0]x.
The block diagram is shown in Figure P3.15.
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104 CHAPTER 3 State Variable Models
R(s)Y(s)
56
x1
x2
s
1
16
x3
s
1
s
1
10
31
14
-
--
++
+
FIGURE P3.15
Block diagram model.
P3.16 (a) The characteristic equation is
0 20 40 60 80 100
-3.5
-3
-2.5
-2
-1.5
-1
-0.5
0
0.5
Time (s)
Step response)
x1 - solid; x2 - dotted; x3 - dashed
FIGURE P3.16
Step response of magnitude 0.285.
det(sIA) = det
s1 0
0.0071 s+ 0.111 0.12
00.07 s+ 0.3
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Problems 105
=s3+ 0.411s2+ 0.032s+ 0.00213 = 0 .
The roots are
s1=0.3343 and s2,3=0.0383 ±0.0700j .
All the poles lie in the left half-plane, therefore, the system is stable.
(b) The solution of the system to a step of magnitude 0.285is given by
x1(t) = 2.66 0.11e0.33t+e0.038t(2.77 cos 0.07t+ 0.99 sin 0.07t)
x2(t) = 0.037e0.33te0.038t(0.037 cos 0.07t+ 0.23 sin 0.07t)
x3(t) = 0.069 0.075e0.33t+e0.038t(0.006 cos 0.07t0.06 sin 0.07t)
P3.17 The transfer function is
G(s) = C(sIA)1B=4s+ 12
s314s2+ 37s+ 20 .
P3.18 Define the state variables as
x1=φ1φ2
x2=ω1
ωo
x3=ω2
ωo
.
Then, the state equations of the robot are
˙x1=ωox2ωox3
˙x2=J2ωo
J1+J2
x1b
J1
x2+b
J1
x3+Km
J1ωo
i
˙x3=J1ωo
J1+J2
x2+b
J2
x2b
J2
x3
or, in matrix form
˙
x=ωo
0 1 1
a1b1b1
a b2b2
x+
0
d
0
i
where
a=J1
(J1+J2), b1=b
J1ωo
, b2=b
J2ωo
and d=Km
J1ωo
.
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106 CHAPTER 3 State Variable Models
P3.19 The state equation is given by
˙
x=
0 1
23
x
where x1(0) = 1 and x2(0) = 1. The state transition matrix is
Φ(t) =
e2t+ 2ete2t+et
2e2t2et2e2tet
.
The system response is
x1(t) = e2t+ 2etx1(0) + e2t+etx2(0)
x2(t) = 2e2t2etx1(0) + 2e2tetx2(0) .
The state response is shown in Figure P3.19.
0123456
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
Time (s)
System response
x1
x2
FIGURE P3.19
Response with x1(0) = 1 and x2(0) = 1.
P3.20 The state equation is given by
˙
x=
0.693
6.70
10.693
9.2
xwhere x(0) =
0.3×1016
7×1016
.
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Problems 107
The state transition matrix is
Φ(t) =
e0.103433t0
35.5786(e0.103433te0.0753261t)e0.075326t
.
The system response is
x1(t) = e0.103433tx1(0)
x2(t) = 35.5786 he0.103433te0.0753261tix1(0) + e0.075326tx2(0) .
The state response is shown in Figure P3.20.
0 10 20 30 40 50
-1
0
1
2
3
4
5
6
7
Time (hours)
Nucleide densities in atoms per unit volume
X=Xenon 135
I=Iodine 135
FIGURE P3.20
Nuclear reactor state response to initial conditions.
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108 CHAPTER 3 State Variable Models
P3.21 Referring to Figure P3.21 we have
Y(s) = 1
sW(s) = 1
sh1U(s) + 1
sQ(s)
=h1
sU(s) + 1
s2[h0U(s)a0Y(s)a1sY (s) + a1h1U(s)] .
Gathering like terms and re-arranging yields
1 + a1
s+a0
s2Y(s) = h1
s+h0
s2+a1h1
s2U(s)
or
Y(s) = h1s+h0+a1h1
s2+a1s+a0U(s).
Computing the transfer function from the state variable representation
yields
G(s) = C(sIA)1B
=h1 0 i
s+a1
s2+a1s+a0
1
s2+a1s+a0
a0
s2+a1s+a0
s
s2+a1s+a0
h1
h0
=h1s+h0+a1h1
s2+a1s+a0
.
U(s)
W(s)
Q(s)
s
1Y(s)
-
+ +
s
1
-
+
a0
a1
h0
h1
FIGURE P3.21
Block diagram with labeled signals.
P3.22 The governing equations are
Ldi
dt =v2
C1
dv1
dt +1
R1
(v1v) + 1
R2
(v1v2) = 0
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Problems 109
C2
dv2
dt +1
R2
(v2v1) + i+v2
R3
= 0 .
Let u=v, x1=i, x2=v1and x3=v2. Then,
˙
x=
0 0 1
L
01
a1
R1+1
R21
C1R2
1
C2
1
R2C21
R2C2+1
R3C2
x+
0
1
R1C1
0
u
y= [0 0 1]x.
P3.23 A state variable representation is given by
˙
x=
0 1 0
0 0 1
30 31 10
x+
0
0
1
r
y= [1 0 0]x.
Other representations include the input feedforward representation
˙
x=
10 1 0
31 0 1
30 0 0
x+
0
0
1
r
y= [1 0 0]x,
the physical variable representation
˙
x=
310
02 1
0 0 5
x+
0
0
1
r
y= [1 0 0]x,
and the decoupled representation
˙
x=
300
02 0
0 0 5
x+
1
1
1
r
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110 CHAPTER 3 State Variable Models
y=1
6
1
31
2x.
P3.24 The matrix representation of the state equations is
˙
x=
3 0
0 2
x+
1 1
0 1
u1
u2
+
0
1
d .
When u1= 0 and u2=d= 1, we have
˙x1= 3x1+u2
˙x2= 2x2+ 2u2
So we see that we have two independent equations for x1and x2. With
U2(s) = 1/s and zero initial conditions, the solution for x1is found to be
x1(t) = L1{X1(s)}=L11
s(s3)
=L11
3s+1
3
1
s3=1
31e3t
and the solution for x2is
x2(t) = L1{X2(s)}=L12
s(s2)=L11
s+1
s2=1+e2t.
P3.25 Since Φ(s) = (sIA)1, we have
Φ(s) =
s+ 1 0
2s+ 3
1
=
s+ 3 0
2s+ 1
1
∆(s)
where ∆(s) = (s+ 1)(s+ 3). The state transition matrix is
Φ(t) = L1{Φ(s)}=
et0
ete3te3t
.
P3.26 The state variable differential equation is
˙
x=
0 1
25 6
x+
0
25
r
y= [1 0]x.
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Problems 111
and
Φ(s) = (sIA)1=
s+ 6 1
25 s
1
∆(s)
where ∆(s) = s2+ 6s+ 25.
P3.27 Equating the change in angular momentum to the sum of the external
torques yields
J¨
θHω cos θ=b˙
θkθ
where bis the damping coefficient, kis the spring constant, and Jis the
wheel moment of inertia. Defining the state variables x1=θand x2= ˙x
and the input u=ω, we can write the equations of motion as
˙x1=x2
˙x2=k
Jx1b
Jx2+H
Jucos x1
With a small angle assumption (that is, cos x11) we have
˙
x=
0 1
k/J b/J
x+
0
H/J
u
y=θ=h1 0 ix.
P3.28 The governing equations of motion are
m1¨y1+k(y1y2) + b˙y1=u
m2¨y2+k(y2y1) + b˙y2= 0
y=y2.
Let x1=y1, x2= ˙y1, x3=y2and x4= ˙y2. Then
˙
x=
0 1 0 0
k
m1b
m1
k
m10
0 0 0 1
k
m20k
m2b
m2
x+
0
1
m1
0
0
u
y=h0 0 1 0 ix.
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112 CHAPTER 3 State Variable Models
P3.29 The equations of motion are
I¨q1+M gL sin q1+k(q1q2) = 0
J¨q2k(q1q2) = u .
Let x1=q1,x2= ˙q1,x3=q2, and x4= ˙q2and linearize the equations
using small angle assumptions (i.e. sin q1q1). Then, we have
˙x1=x2
˙x2=M gL
Ix1k
I(x1x3)
˙x3=x4
˙x4=k
J(x1x3) + 1
Ju .
P3.30 Using Kirchoff’s current law, we find that
Cdvc
dt =i2+i3
where i3= current in R3. Let i1= current in R1. Using Kirchoff’s voltage
law, we have
LdiL
dt =v1R1i1
and
R1i1+R2i2+vc=v1.
But
i2=i1iL,
so
(R1+R2)i1=v1vc+R2iL.
Using Kirchoff’s voltage law once again, we calculate i3as
i3=v2vc
R3
.
Utilizing the above equations, we can solve for diL/dt and dvc/dt, as
follows:
diL
dt =R2
L(R1+R2)v1+R1
L(R1+R2)vcR1R2
L(R1+R2)iL
vc
dt =v1
C(R1+R2)vc
C(R1+R2)vc
CR3R1iL
C(R1+R2)+v2
CR3
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Problems 113
Define the state variables x1=vcand x2=iL. Then, in matrix form we
have
˙
x=
(R1+R2+R3)
CR3(R1+R2)R1
C(R1+R2)
R1
L(R1+R2)R1R2
L(R1+R2)
x+
1
C(R1+R2)
1
CR3
R2
L(R1+R2)0
v1
v2
y=i2=h1
(R1+R2)R1
(R1+R2)ix+h1
(R1+R2)0i
v1
v2
P3.31 A state variable representation is
˙
x=
0 1
34
x+
0
30
u .
The state transition matrix can be computed as follows:
Φ=L1n[sIA]1o=L1
1
∆(s)
s+ 4 1
3s
=
3
2et1
2e3t1
2et1
2e3t
3
2et+3
2e3t1
2et+3
2e3t
where
∆(s) = s2+ 4s+ 3 = (s+ 1)(s+ 3) .
P3.32 A state variable representation is
˙m1=k1m1+r
˙m2=k1m1k2m2
where k1and k2are constants of proportionality. In matrix form, we have
˙
x=Ax +Br=
k10
k1k2
x+
1
0
r
where x1=m1and x2=m2. Let k1=k2= 1 and assume that r(t) = 0
and x1= 1 and x2= 0. Then
x(t) = Φ(t)x(0) =
et0
tetet
x(0) =
et
tet
.
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114 CHAPTER 3 State Variable Models
The simulation is shown in Figure P3.32.
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
05 10
time (sec)
state history, x(t)
x1
x2
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0 0.5 1
x1
x2
t=0
FIGURE P3.32
Actual versus approximate state response.
P3.33 The system (including the feedback) is described by
˙
x=Ax =
0 1
1/21
x.
The charactersitic equation is
det[λIA] = det
λ1
1/2λ+ 1
=λ2+λ+1
2= 0 .
The roots of the characteristic equation are
λ1,2=1
2±j1
2.
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Problems 115
The system response is
x(t) = eAtx(0) =
et/2cos t
2+et/2sin t
22et/2sin t
2
et/2sin t
2et/2cos t
2et/2sin t
2
x(0)
=et/2
2 sin t
2
cos t
2sin t
2
where x1(0) = 0 and x2(0) = 1.
P3.34 (a) The state space representation is
˙
x=
0 1 0
0 0 1
611 6
x+
0
0
1
r
y= [6 0 0] x.
(b) The element φ11(t) of the state transition matrix is
φ11(t) = e3t3e2t+ 3et.
P3.35 The state equations are
˙
h= ˙x1=1
50[80θ50h] = x1+8
5x2
˙
θ= ˙x2=ω=x3
˙ω= ˙x3=Km
Jia=KmKb
JRa
ω+KmKa
JRa
vi=353
30 x3+25000
3vi.
In state variable form, we have
˙
x=
18
50
0 0 1
0 0 353
30
x+
0
0
25000
3
vi.
P3.36 Using Newton’s Law and summing the forces on the two masses yields
M1¨x(t) + b1˙x(t) + k1x(t) = b1˙y(t)
M2¨y(t) + b1˙y(t) + k2y(t) = b1˙x(t) + u(t)
Let
z1=x, z2= ˙x, z3=y, and z4= ˙y .
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116 CHAPTER 3 State Variable Models
Then we write the system in state variable form as
˙
z=
0100
k1
M1b1
M10b1
M1
0 0 0 1
0b1
M2k2
M2b1
M2
z+
0
0
0
1
M2
u
y=h1 0 0 0 iz.
P3.37 From the block diagram in Figure P3.37, we obtain
˙x1=x2
˙x2=x3
˙x3=10x14x23x3+u
y=x1+ 12x2+ 5x3
or
˙
x=
0 1 0
0 0 1
10 43
x+
0
0
1
u
y= [1 12 5] x.
The third-order differential equation model is
...
y+3¨y+ 4 ˙y+ 10y= 5¨u+ 12 ˙u+u .
10
3
4
12
5
∫ ∫
-
-
-
++
+
+
U(s)Y(s)
x3x2x1
FIGURE P3.37
Block diagram with states labeled.
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Advanced Problems 117
Advanced Problems
AP3.1 With the state variables are defined as
z=
x
˙x
i
,
the nonlinear equations of motion are
˙z1
˙z2
˙z3
=
z2
gK
m
(Io+z3)2
(Xo+z1)2
1
L(vRz3)
,
where the control is the voltage v. We assume that z1=xis measurable.
The linearized equations of motion are
˙
z=Az +Bv
y=Cz
where
A=
0 1 0
2K
m
I2
o
X3
o02K
m
Io
X2
o
0 0 R
L
,B=
0
0
1
L
,and C=h1 0 0 i.
The transfer function is
G(s) = C(sIA)1B.
With the constants
R= 23.2
L= 0.508
m= 1.75
K= 2.9×104
Io= 1.06
Xo= 4.36 ×103
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118 CHAPTER 3 State Variable Models
the transfer function is
G(s) = 36.38
s3+ 45.67s2+ 4493s+ 205195 .
AP3.2 The differential equation describing the motion of yis
m¨y+b˙y+ky =b˙u+ku .
Taking Laplace tranforms (with zero initial conditions) yields the transfer
function
Y(s)
U(s)=(b/m)s+ (k/m)
s2+ (b/m)s+ (k/m).
In state space form, we have
˙
x=
0 1
k/m b/m
x+
0
1
u
y=hk/m b/m ix.
AP3.3 The transfer function is
Y(s)
R(s)=2s2+ 6s+ 5
s3+ 4s2+ 5s+ 2 .
In (nearly) diagonal form, we have
A=
110
01 0
0 0 2
,B=
0
1
1
,and C=h1 1 1 i.
The matrix Ais not exactly diagonal due to the repeated roots in the
denominator of the transfer function.
AP3.4 The differential equations describing the motion of yand qare
m¨y+k2˙y+k1(yq) = f
b˙q+k1(yq) = f
where k1= 2 and k2= 1. Assume the mass m= 1. Then with the state
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Advanced Problems 119
variables defined as z=hy˙y q iT, we have the state variable model
˙
z=
0 1 0
3 0 2
2/b 02/b
z+
0
1
1/b
f
y=h1 0 0 iz
If we model a large bump at high speeds as an impulse and a small bump
at low speeds as a step, then b= 0.8 provides good performance. In both
cases, the ride settles out completely in about 10 seconds.
AP3.5 The differential equations describing the motion of xand θare
(M+m)¨x+ML cos θ¨
θML sin θ˙
θ2=kx
gsin θ+ cos θ¨x+L¨
θ= 0
Assuming θand ˙
θare small, it follows that
(M+m)¨x+ML¨
θ=kx
¨x+L¨
θ=gθ
Define the state variables as z=hx˙x θ ˙
θiT. Then, the state vari-
able model is
˙
z=
0 1 0 0
k/m 0gM/m 0
0 0 0 1
k/(Lm) 0 g(M+m)/(Lm) 0
z
AP3.6
AP3.7 Computing the closed-loop system yields
ABK =
1 1
K1K2
,B=
0
1
,and C=h2 1 i.
The characteristic polynomial is
|sI(ABK)|=s2+ (K2+ 1)s+K1+K2= 0.
The roots are in the left-half plane whenever K2+1 >0 and K1+K2>0.
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120 CHAPTER 3 State Variable Models
AP3.8 (a) A state variable representation is given by
˙x1=x2
˙x2=x3
˙x3=Kx112x26x3+Kr
y=x1
or, in matrix form
˙
x=
0 1 0
0 0 1
K12 6
x+
0
0
K
r
y=h1 0 0 ix
(b) The characteristic roots are found by solving
det [λIA] = 0
or
λ3+ 6λ2+ 12λ+K= 0
When K= 8, we have characteristic roots at λ1=2, λ2=2, and
λ3=2, as desired.
(c) The unit step response is given by
y(t) = 1 e2t2te2t2t2e2t.
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Design Problems 121
Design Problems
The transfer model of the traction drive, capstan roller, and linear slideCDP3.1
was given in CDP2.1 as
X(s)
Va(s)=rKm
s[(Lms+Rm)(JTs+bm) + KbKm],
where
JT=Jm+r2(Ms+Mb).
Define x1=x,x2= ˙x, and x3= ¨x. Then, a state variable representation
is
˙
x=Ax +Bva
y=Cx
where
A=
0 1 0
0 0 1
0Rmbm+KbKm
LmJTLmbm+RmJT
LmJT
,B=
0
0
rKm
LmJT
C=h1 0 0 i.
DP3.1 (a) The equation of motion of the spring-mass-damper is
m¨y+b˙y+ky =u
or
¨y=b
m˙yk
my+1
mu .
Select the state variables
x1=yand x2= ˙y .
Then, we have
˙
x=Ax +Bu
y=Cx
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122 CHAPTER 3 State Variable Models
where
A=
0 1
20 9
,B=
0
1
,C=h1 0 i.
Ais the system matrix. The characteristic equation is
det[λIA] = det
s1
20 s+ 9
=s2+ 9s+ 20 = 0 .
The roots of the characteristic equation are s1=4 and s2=5,
and the transistion matrix is
Φ(t) =
5e4t4e5te4te5t
20e4t+ 20e5t4e4t+ 5e5t
.
(b) Assume the initial conditions are x1(0) = 1 and x2(0) = 2. The zero-
input response is shown in Figure DP3.1.
(c) Suppose we redesign the system by choosing band kto quickly damp
out x2and x1. We can select band kto achieve critical damping.
0 0.5 1 1.5 2
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
Time(sec)
State response, x
x1
x2
b/m=9, k/m=20
0 0.5 1 1.5 2
-4
-3
-2
-1
0
1
2
Time(sec)
State response, x
x1
x2
Critical damping: b/m=20, k/m=100
FIGURE DP3.1
Zero input state response.
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Design Problems 123
If we desire the characteristic polynomial to be pd(s) = (s+ 10)2=
s2+ 20s+ 100, then we need b= 20 and k= 100.
DP3.2 The desired transfer function is
Y(s)
U(s)=6
s2+ 7s+ 10 .
The transfer function derived from the phase variable representation is
Y(s)
U(s)=d
s2+bs +a.
Therefore, we select d= 6, a= 10 and b= 7.
DP3.3 Assume the aircraft lands precisely on the centerline. The linearized equa-
tions of motion are
m3¨x3+KD˙x3+K2(x3x2) = 0
m2¨x2+K2(x2x3) + K1(x2x1) = 0
m1¨x1=2
2K2(x1x2)
where x1(0) = x2(0) = ˙x2(0) = ˙x3= 0 and ˙x1(0) = 60. The system
response is shown in Figure DP3.3 where KD= 215. The aircraft settles
out at 30 m, although initially it overshoots by about 10 m at 1 second.
0
5
10
15
20
25
30
35
40
45
0 1 2 3 4 5 6 7 8 9 10
Time (secs)
Amplitude
FIGURE DP3.3
Aircraft arresting gear response.
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124 CHAPTER 3 State Variable Models
DP3.4 We can model the bungi cord system as a mass-spring-damper. This is
actually an over-simplification because the bungi cord cannot “push” the
jumper down as a spring would—it can only exert a restoring force when
the cord is stretched (that is, when the jumper exceeds the length, L, of
the cord). The problem is nonlinear! When the distance of the jumper
from the platform is less than Lwe should model the cord spring constant
and damping as K= 0 and b= 0, respectively. Only gravity acts on the
jumper. Also, when ˙x(the jumper velocity) is negative (where we define
positive towards the ground), then we should model b= 0. A reasonable
set of equations of motion are
˙x1=x2
˙x2=K
mx1b
mx2+g
where x1is the distance measured from the top of the platform and x2
is the jumper velocity. For the initial conditions we have x1(0) = 10
and x2(0) = 0. A reasonable set of parameters for the bungi cord are
L= 40 m, K= 40 N/m and b= 20 kg/m. The system response is
shown in Figure DP3.4 for a person with m= 100 kg. The accelerations
experienced by the jumper never exceed 1.5 g.
global MASS GRAVITY LENGTH K b
MASS=100; HEIGHT=100; GRAVITY=9.806;
LENGTH=40; SPRINGCONSTANT=40; SPRINGDAMPING=20;
x0=[10;0];
t=0; dt=0.1;
n=round(120/dt);
for i=1:n;
if x0(1)<LENGTH
K=0; b=0;
elseif x0(2)<0
b=0;
else
K=SPRINGCONSTANT; b=SPRINGDAMPING;
end
tf=t+dt;
[T,x] = ode45('bungi',[t tf],x0);
xs(i,:)=x(length(x),:); t=tf;
x0=x(length(x),:); ts(i)=tf;
end
plot(ts,HEIGHT-xs(:,1)), grid
function [xdot] = bungi(t,x)
global MASS GRAVITY LENGTH K b
xdot(1)=x(2);
xdot(2)=-(K/MASS)*(x(1)-
LENGTH)-(b/MASS)*x(2)+GRAVITY;
xdot=xdot';
FIGURE DP3.4
(a) Bungi cord system response m-file script.
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Design Problems 125
0 20 40 60 80 100 120
0
10
20
30
40
50
60
70
80
90
Time (sec)
Distance (m)
FIGURE DP3.4
CONTINUED: Bungi cord system time history response.
DP3.5 Computing the closed-loop system yields
ABK =
0 1
2K13K2
,B=
0
1
,and C=h1 0 i.
The characteristic polynomial is
|sI(ABK)|=s2+ (K23)s+K1+ 2 = 0.
Suppose that the desired poles are in the left-half plane and are denoted
by p1and p2. Then the desired characteristic polynomial is
(s+p1)(s+p2) = s2+ (p1+p2)s+p1p2= 0.
Equating coefficients and solving for K= [K1K2] yields
K1=p1p22
K2=p1+p2+ 3.
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126 CHAPTER 3 State Variable Models
Computer Problems
CP3.1 The m-file script to compute the state-space models using the ss function
is shown in Figure CP3.1.
% Part(a)
num = [1]; den = [1 25];
sys = tf(num,den);
sys_ss = ss(sys)
% Part(b)
num = [3 10 3]; den = [1 8 5];
sys = tf(num,den);
sys_ss = ss(sys)
% Part(c)
num = [1 10]; den = [1 3 3 1];
sys = tf(num,den);
sys_ss = ss(sys)
a =
x1
x1 -25
b =
u1
x1 1
c =
x1
y1 1
d =
u1
y1 0
a =
x1 x2
x1 -8 -2.5
x2 2 0
b =
u1
x1 4
x2 0
c =
x1 x2
y1 -3.5 -1.5
d =
u1
y1 3 a =
x1 x2 x3
x1 -3 -1.5 -0.25
x2 2 0 0
x3 0 2 0
b =
u1
x1 2
x2 0
x3 0
c =
x1 x2 x3
y1 0 0.25 1.25
d =
u1
y1 0
FIGURE CP3.1
Script to compute state-space models from transfer functions.
For example, in part (c) the state-space model is
˙
x=Ax +Bu
y=Cx +Du ,
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Computer Problems 127
where D= [0] and
A=
31.50.25
2 0 0
0 2 0
,B=
2
0
0
,C=h0 0.25 1.25 i.
CP3.2 The m-file script to compute the transfer function models using the tf
function is shown in Figure CP3.2.
Transfer function:
1
-------------
s^2 - 4 s - 2
Transfer function:
6s - 48
-----------------------
s^3 - 11 s^2 + 4 s - 36
Transfer function:
s - 2
-------------
s^2 + 2 s + 1
% Part (a)
A=[0 1;2 4]; B=[0;1]; C=[1 0]; D=[0];
sys_ss=ss(A,B,C,D);
sys_tf = tf(sys_ss)
% Part (b)
A=[1 1 0;-2 0 4; 6 2 10]; B=[-1;0;1]; C=[0 1 0]; D=[0];
sys_ss=ss(A,B,C,D);
sys_tf = tf(sys_ss)
% Part (c)
A=[0 1;-1 -2]; B=[0;1]; C=[-2 1]; D=[0];
sys_ss=ss(A,B,C,D);
sys_tf = tf(sys_ss)
FIGURE CP3.2
Script to compute transfer function models from the state-space models.
CP3.3 For an ideal op-amp, the voltage gain (as a function of frequency) is
Vo(s) = Z2(s)
Z1(s)Vin(s),
where
Z1=R1+1
C1s
Z2=R2
1 + R2C2s
are the respective circuit impedances. Therefore, we obtain
Vo(s) = R2C1s
(1 + R1C1s)(1 + R2C2s)Vin(s).
The m-file script and step response is shown in Figure CP3.3.
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128 CHAPTER 3 State Variable Models
R1=1000; R2=10000; C1=0.0005; C2=0.0001;
numg=[R2*C1 0];
deng=conv([R1*C1 1],[R2*C2 1]);
sys_tf=tf(numg,deng)
% Part (a)
%
sys_ss=ss(sys_tf)
% Part (b)
%
step(sys_ss)
a =
x1 x2
x1 -3.00000 -1.00000
x2 2.00000 0
b =
u1
x1 4.00000
x2 0
c =
x1 x2
y1 2.50000 0
d =
u1
y1 0
Continuous-time system.
Time (sec.)
Amplitude
Step Response
0 1 2 3 4 5 6
0
0.5
1
1.5
2
2.5
FIGURE CP3.3
The m-file script using the step function to determine the step response.
CP3.4 The m-file script and state history is shown in Figure CP3.4. The transfer
function equivalent is
G(s) = 1
s3+ 5s2+ 2s+ 3 .
The computed state vector at t= 10 is the same using the simulation and
the state transition matrix.
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Computer Problems 129
0 1 2 3 4 5 6 7 8 9 10
- 1
- 0.8
- 0.6
- 0.4
- 0.2
0
0.2
0.4
0.6
0.8
1
time (sec)
x(t)
x1 solid; x2 dotted; x3 dashed
a=[0 1 0; 0 0 1; -3 -2 -5];
b=[0;0;1];
c=[1 0 0];
d=[0];
%
% Part (a)
%
sys_ss = ss(a,b,c,d)
sys_tf = tf(sys_ss)
%
% Part (b)
%
x0 = [0 -1 1];
t = [0:0.1:10];
u = 0*t;
[y,t,x] = lsim(sys_ss,u,t,x0);
plot(t,x(:,1),t,x(:,2),':',t,x(:,3),'--');
xlabel('time (sec)'), ylabel('x(t)'), grid
title('x1 - solid; x2 - dotted; x3 - dashed')
xf_sim = x(length(t),:)'
%
% Part (c)
%
dt = 10;
Phi = expm(a*dt);
xf_phi = Phi*x0'
Transfer function:
1
---------------------
s^3 + 5 s^2 + 2 s + 3
xf_sim =
-0.2545
0.0418
0.1500
xf_phi =
-0.2545
0.0418
0.1500
FIGURE CP3.4
The m-file script using the lsim function to determine the step response.
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130 CHAPTER 3 State Variable Models
CP3.5 The two state-space models represent the same transfer function, as shown
in Figure CP3.5. The transfer function in both cases is
G(s) = 4
s3+ 8s2+ 5s+ 4 .
We see that a state-space representation of a transfer function is not
unique.
a1=[0 1 0; 0 0 1; -4 -5 -8];
b1=[0;0;4];
c1=[1 0 0];
d1=[0];
%
% Part (a)
%
sys_ss = ss(a1,b1,c1,d1);
sys_tf = tf(sys_ss)
%
% Part (b)
%
a2=[ 0.5000 0.5000 0.7071;
-0.5000 -0.5000 0.7071;
-6.3640 -0.7071 -8.0000];
b2=[0;0;4];
c2=[0.7071 -0.7071 0];
d2=[0];
sys_ss = ss(a2,b2,c2,d2);
sys_tf = tf(sys_ss)
Transfer function:
4
---------------------
s^3 + 8 s^2 + 5 s + 4
Transfer function:
4
---------------------
s^3 + 8 s^2 + 5 s + 4
FIGURE CP3.5
Comparison of the transfer functions of two state-space models.
CP3.6 The m-file script and impulse response are shown in Figure CP3.6. The
controller state-space representation is
˙x=2x+u
y=x
and the plant state-space representation is
˙
x=
22
2 0
x+
0.5
0
u
y=h0 1 ix
The closed-loop system state variable representation is
˙
x=Ax +Bu
y=Cx +Du ,
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Computer Problems 131
where D= [0] and
A=
22 0.5
200
012
,B=
0
0
1
,C=h0 1 0 i.
numc=[1]; denc=[1 2]; sys_tfc = tf(numc,denc)
numg=[1]; deng=[1 2 4]; sys_tfg = tf(numg,deng)
%
% Part (a)
%
sys_ssc = ss(sys_tfc)
%
% Part (b)
%
sys_ssg = ss(sys_tfg)
%
% Part (c)
%
sys_s = series(sys_ssc,sys_ssg);
sys_cl = feedback(sys_s,[1]);
impulse(sys_cl)
Time (sec.)
Amplitude
Impulse Response
0123456
-0.02
0
0.02
0.04
0.06
0.08
0.1
From: U(1)
To: Y(1)
FIGURE CP3.6
Computing the state-space representations and the impulse response.
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132 CHAPTER 3 State Variable Models
CP3.7 The m-file script and system response is shown in Figure CP3.7.
a=[0 1;-2 -3]; b=[0;1]; c=[1 0]; d=[0];
sys = ss(a,b,c,d);
x0=[1;0];
t=[0:0.1:10]; u=0*t;
[y,t,x]=lsim(sys,u,t,x0);
plot(t,x(:,1),t,x(:,2),'--')
xlabel('Time (sec)')
ylabel('State Response')
legend('x1','x2',-1)
grid
0 1 2 3 4 5 6 7 8 9 10
-0.5
0
0.5
1
Time (sec)
State Response
x1
x2
FIGURE CP3.7
Using the lsim function to compute the zero input response.
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C H A P T E R 4
Feedback Control System
Characteristics
Exercises
E4.1 (a) The system sensitivity to τis given by
ST
τ=ST
GSG
τ.
In this case, we have
ST
G=1
1 + GH(s)=1
1 + 100
3s+1
=3s+ 1
3s+ 101
and
SG
τ=τs
τs + 1 =3s
3s+ 1 ,
where τ= 3. Therefore,
ST
τ=3s
3s+ 101 .
(b) The closed-loop transfer function is
T(s) = G(s)
1 + GH(s)=100
3s+ 101 =100/101
3
101 s+ 1 =0.99
τcs+ 1 ,
where the time-constant τc= 3/101 = 0.0297 second.
E4.2 (a) The system sensitivity to K2is
ST
K2=T
K2
K2
T=1
1 + K1K2
.
133
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134 CHAPTER 4 Feedback Control System Characteristics
(b) The transfer function from Td(s) to Vo(s) is
Vo(s) = K2
1 + K1K2
Td(s).
(c) We would select K11, so that the transfer function from Td(s) to
Vo(s) is small.
E4.3 (a) The tracking error, E(s) = R(s)Y(s), is given by
E(s) = R(s)
1 + G(s)=A/s
1 + K/(s+ 5)2.
The steady-state error (computed using the final value theorem) is
ess = lim
s0sE(s) = lim
s0
A
1 + K
(s+5)2
=A
1 + K/25 .
(b) A disturbance would be the wind shaking the robot arm.
E4.4 (a) The tracking error, E(s) = R(s)Y(s), is given by
E(s) = R(s)
1 + KG(s).
The steady-state position error is computed (using the final value
theorem) to be
ess = lim
s0sA/s
1 + KG(s)= lim
s0
A
1 + 10K
s(τs+1)
= 0 .
(b) The ramp input of 0.1 m/sec is given by
R(s) = 0.1
s2.
Then, using the final value theorem, we have
ess = lim
s0s
0.1/s2
1 + 10K
s(τs+1)
= lim
s0"0.1
s+10K
τs+1 #,
or
ess =0.1
10K=0.01
K.
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Exercises 135
We desire ess 0.0001 m, so
K0.01
0.0001 = 100 .
E4.5 (a) The sensitivity is
ST
p=T
p
p
T=p"s415s2+ 3s+ 10
(s2+ps + 10)2#T(s).
(b) The tracking error is
E(s) = [1 T(s)] R(s) = s3+ (2p1)s2+ (4 p)s7p
s3+ 2ps2+ 4s+ 3 pR(s)
Using the final value theorem with R(s) = 1/s we obtain the steady-
state tracking error as
ess = lim
s0sE(s) = 7p
3p.
E4.6 The closed-loop transfer function is
T(s) = 10K
s2+bs + 10K.
The tracking error is
E(s) = [1 T(s)] R(s) = s(s+b)
s2+bs + 10K
1
s2,
where we let R(s) = 1/s2. Using the final value theorem we obtain the
steady-state tracking error as
ess = lim
s0sE(s) = b
10K.
If we require that b < K then the steady-state error is less than 0.1 to
the ramp input.
E4.7 The light bounces off the surface of the slide and into a detector. If the
light fails to hit the detector dead center, the unbalanced electric signal
causes the motor to adjust the position of the light source, and simulta-
neously the lens.
E4.8 The closed-loop transfer function is
T(s) = 5(s+ 3)
s2+ 20s+ 15 .
The step response is shown in Figure E4.8.
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136 CHAPTER 4 Feedback Control System Characteristics
0 1 2 3 4 5 6 7 8 9 10
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Step Response
Time (sec)
y(t)/A
FIGURE E4.8
Step response.
E4.9 (a) The closed-loop transfer function is
T(s) = KK1
s+K1(K+K2).
(b) The sensitivities are
ST
K=T /T
K/K =s+K1K2
s+K1(K+K2)
and
ST
K1=s
s+K1(K+K2).
(c) The transfer function from Td(s) to Y(s) is
Y(s)
Td(s)=1
s+K1(K2+K).
Therefore, since E(s) = KY (s) (when R(s) = 0), we have
E(s) = K
s+K1(K2+K)Td(s)
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Exercises 137
and
ess = lim
s0sE(s) = K
K1(K+K2).
(d) With K=K2= 1, we have
T(s) = K1
s+ 2K1
.
Then,
Y(s) = K1
s+ 2K1
1
s
and
y(t) = 1
2h1e2K1tiu(t),
where u(t) is the unit step function. Therefore, select K1= 10 for the
fastest response.
E4.10 The closed-loop transfer function is
T(s) = 46.24K(s+ 50)(s+ 425)
(s+ 200)(s+ 425)(s2+ 16.7s+ 72.9) + 19652K(s+ 50) .
The steady-state error is determined to be
ess = lim
s0sE(s) = lim
s0s(1 T(s))1
s
= 1 lim
s0T(s) = 1 T(0)
=6.3
6.3 + K.
The plots of the steady-state error versus Kand the percent overshoot
P.O. versus Kare shown in Figure E4.10 for
40 K400.
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138 CHAPTER 4 Feedback Control System Characteristics
0 100 200 300 400
0
0.02
0.04
0.06
0.08
0.1
0.12
0.14
K
steady-state error
0 100 200 300 400
15
20
25
30
35
40
45
K
percent overshoot (%)
FIGURE E4.10
(a) Steady-state error. (b) Percent overshoot.
E4.11 (a) The closed-loop transfer function is
T(s) = G(s)
1 + G(s)H(s)=K(s2+ 5s+ 6)
s3+ 15s2+ 56s+ 60 + 14K
(b) With E(s) = R(s)Y(s) we obtain
E(s) = 1G(s)
1 + G(s)H(s)R(s) = 1G(s)(1 H(s))
1 + G(s)H(s)R(s)
=s3+ (15 K)s2+ (56 5K)s+ (60 + 8K)
s3+ 15s2+ 56s+ 60 + 14K·1
s.
Then, using the final value theorem we find
lim
s0sE(s) = (60 + 8K)
60 + 14K.
(c) The transfer function from the disturbance Td(s) to the output is
Y(s) = 1
1 + G(s)H(s)Td(s) = s3+ 15s2+ 56s+ 60
s3+ 15s2+ 56s+ 60 + 14KTd(s).
The steady-state error to a unit step disturbance is
lim
s0sY (s) = lim
s0ss3+ 15s2+ 56s+ 60
s3+ 15s2+ 56s+ 60 + 14K·1
s=60
60 + 14K.
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Exercises 139
(d) The sensitivity is
ST
K=T
K
K
T=T
G
G
K
K
T
=1
(1 + G(s)H(s))2K
s+ 101 + G(s)H(s)
G(s)=1
1 + G(s)H(s).
E4.12 (a) The closed-loop transfer function is
T(s) = Gc(s)G(s)
1 + Gc(s)G(s)H(s)=100K1(s+ 5)
s2+ 105s+ (500 + 100K1K2).
The steady-state tracking error is
E(s) = R(s)Y(s) = 1Gc(s)G(s)(1 H(s))
1 + Gc(s)G(s)H(s)R(s)
=s2+ (105 100K1)s+ 500 100K1(5 K2)
s2+ 105s+ 500 + 100K1K2·1
s
and
lim
s0sE(s) = 5K1(5 K2)
5 + K1K2
.
(b) The transfer function from the noise disturbance N(s) to the output
Y(s) is
Y(s) = Gc(s)G(s)H(s)
1 + Gc(s)G(s)H(s)N(s) = 100K1K2
s2+ 105s+ (500 + 100K1K2)N(s).
The steady-state error to a unit step N(s) = 1/s is
lim
s0sY (s) = lim
s0s100K1K2
s2+ 105s+ (500 + 100K1K2)·1
s=K1K2
5 + K1K2
.
(c) The design trade-off would be to make K1K2as large as possible
to improve tracking performance while keeping K1K2as small as
possible to reject the noise.
E4.13 The closed-loop transfer function is
T(s) = K
s2+ 20s+K.
The sensitivity is
ST
K=T /T
K/K =s2+ 20s
s2+ 20s+K.
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140 CHAPTER 4 Feedback Control System Characteristics
E4.14 (a) The closed-loop transfer function is
T(s) = Gc(s)G(s)
1 + Gc(s)G(s)H(s)=K
s2+K1s+K.
The sensitivity is
ST
K1=T /T
K1/K1
=sK1
s2+K1s+K.
(b) You would make Kas large as possible to reduce the sensitivity to
changes in K1. But the design trade-off would be to keep Kas small
as possible to reject measurement noise.
E4.15 (a) The closed-loop transfer function is
T(s) = Gc(s)G(s)
1 + Gc(s)G(s)H(s)=120
s2+ 10s+ 120 .
The steady-state tracking error is
E(s) = R(s)Y(s) = 1
1 + Gc(s)G(s)R(s)
=s2+ 10s
s2+ 10s+ 120 ·1
s
and
lim
s0sE(s) = 0 .
(b) The transfer function from the disturbance Td(s) to the output Y(s)
is
Y(s) = 1
s2+ 10s+ 120Td(s).
The steady-state error to a unit step Td(s) = 1/s is
lim
s0sY (s) = lim
s0s1
s2+ 10s+ 120·1
s=1
120 .
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Problems 141
Problems
P4.1 The tank level control block diagram is shown in Figure P4.1.
-
+G1(s)
K
DH
dH
DQ1
+
+
FIGURE P4.1
Tank level control block diagram.
(a) For the open-loop system the transfer function is
G1(s) = R
RCs + 1 .
Thus,
SG1
R=G1
R ·R
G1
=1
RCs + 1 .
For the closed-loop system, the transfer function is
T(s) = G1
1 + KG1
=R
RCs + 1 + KR .
Thus,
ST
R=T
R ·R
T=1
RCs + 1 + KR ,
and
ST
K=T
K ·K
T=KR
RCs + 1 + KR .
(b) For the open-loop system
H(s)
δH(s)= 1 .
All disturbances show up directly in the output, thus the open-loop
system has no capability to reject disturbances. On the other hand,
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142 CHAPTER 4 Feedback Control System Characteristics
for the closed-loop system we have
H(s)
δH(s)=1
1 + KG1(s)=RCs + 1
RCs + 1 + KR .
By selecting Klarge, we reduce the effects of any disturbances. For
example, consider a step disturbance. The steady-state error due to
the disturbance is
ess = lim
s0s(RCs + 1)
RCs + 1 + KR A
s=A
1 + KR .
As Kgets larger, the steady-state error magnitude gets smaller, as
desired.
(c) Consider the step input
Q1(s) = A
s.
Then, for the open-loop system we have
ess = lim
s0s(1 G1)A
s= (1 R)A .
The steady-state error is zero when R= 1, but is sensitive to changes
in R. For the closed-loop system we have
ess = lim
s0s1
1 + KG1A
s=A
1 + KR .
By selecting Klarge, the effect of the disturbance is reduced and is
relatively insensitive to changes in R.
P4.2 (a) The open-loop transfer function is
T(s) = KaG(s).
Therefore, ST
K1is undefined and
ST
Ka= 1 .
The closed-loop transfer function is
T(s) = KaG(s)
1 + KaK1G(s).
Therefore,
ST
K1=T
K1·K1
T=KaK1G(s)
1 + KaK1G(s)
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Problems 143
and
ST
Ka=1
1 + K1KaG(s).
(b) The tracking error, E(s) = θd(s)θ(s) = θ(s), since θd(s) = 0. The
transfer function from the wave disturbance to the output θ(s) is
θ(s) = G(s)
1 + K1KaG(s)Td(s).
Consider a step disturbance input for the open- and closed-loop sys-
tems. For the open-loop system, we have
ess =lim
s0sG(s)A
s=A .
Thus, we see that the open-loop system does not have the capability
to reduce the effect of disturbances. For the closed-loop system, we
have
ess = lim
s0sG(s)
1 + K1KaG(s)A
s=2
n
1 + K1Kaω2
n
.
We see that the larger we make K1Ka, that smaller the effect of the
wave disturbance on the output in steady-state.
P4.3 (a) The open-loop transfer function is
G(s) = K
τs + 1
where K=k1kaEb. Then, computing the sensitivity yields
SG
K= 1 .
The closed-loop system transfer function is
T(s) = K
τs +KKth + 1 .
Similarly, computing the sensitivity yields
ST
K=1
1 + KthG(s)=τs + 1
τs + 1 + KKth
.
(b) For the closed-loop system
T(s) = 1/(τs + 1)
1 + KKth/(τs + 1)Te(s)Te(s)
KKth
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144 CHAPTER 4 Feedback Control System Characteristics
when KKth 1. So, by choosing KKth large, we can reduce the
effect of the disturbance. This cannot be done with the open-loop
system.
(c) Consider the step input
Edes(s) = A
s.
The tracking error for the open-loop system is
E(s) = Edes(s)− T (s).
Thus,
ess = lim
s0s1K
τs + 1A
s= (1 K)A .
So, ess = 0 when K= 1, but is sensitive to changes in K. The tracking
error for the closed-loop system is
E(s) = τs + 1 + K(Kth 1)
τs + 1 + KKth
Edes(s)
and
ess = lim
s0sE(s) = lim
s0sτs + 1 + K(Kth 1)
τs + 1 + KKth A
s=A(1 + K(Kth 1))
1 + KKth
.
Selecting Kth = 1 and K1 reduces the steady-state error.
P4.4 (a) The overall transfer function is
T(s) = Y(s)
R(s)=MG(s) + UQG(s)
1 + QG(s).
(b) From Eq. (4.16) in Dorf & Bishop, we have
ST
G=SN
GSD
G.
In our case, we find that
SN
G= 1 ,
and
SD
G=QG(s)
1 + QG(s).
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Problems 145
Thus,
ST
G= 1 QG(s)
1 + QG(s)=1
1 + QG(s).
(c) The sensitivity does not depend upon U(s) or M(s).
P4.5 The closed-loop transfer function is
T(s) = G1G(s)
1 + G1G(s).
(a) The sensitivity of T(s) to changes in kais
ST
ka=1
1 + G1G(s).
(b) The transfer function from Td(s) to θ(s) is
θ(s) = G(s)
1 + G1G(s)Td(s).
Since we want θ(s) due to a disturbance, E(s) = θ(s) and
ess = lim
s0sE(s) = lim
s0sG(s)
1 + G1G(s)10
s=10
ka
.
Since our maximum desired error magnitude is
ess =0.10o·π
180 = 0.001745 rad ,
we select
ka5730 .
(c) The open-loop transfer function is
θ(s) = G(s)Td(s).
So,
ess =lim
s0sG(s)10
s→ ∞ .
P4.6 The closed-loop transfer function is
T(s) = G1G(s)
1 + G1G(s).
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146 CHAPTER 4 Feedback Control System Characteristics
(a) The sensitivity is
ST
Ke=1
1 + G1G(s)=(τ1s+ 1)(τes+ 1)
(τ1s+ 1)(τes+ 1) + K1Ke
.
(b) The speed is affected by the load torque through the transfer function
V(s) = KgG(s)
1 + GG1(s)Td(s).
(c) Let R(s) = 30/s , and KeK11 . When the car stalls, V(s) = 0.
Using the final value theorem, we find
lim
s0sKgG(s)
1 + GG1(s)d
s+ lim
s0sG1G(s)
1 + GG1(s)30
s
=d+KgKe
1 + KeK1+ 30 K1Ke
1 + K1Ke.
Since KeK11, we have
Vss =dKg
K1+ 30 .
When Vss = 0, we have
d=30K1
Kg
.
Thus, if
Kg
K1
= 2 ,
then ∆d= 15 percent grade ( i.e. ∆d= 15 ft rise per 100 ft horizon-
tally) will stall the car.
P4.7 (a) Let
G1(s) = k1, G2(s) = k2
s(τs + 1) ,and H(s) = k3+k4s .
Then the transfer function from TL(s) to Y(s) is
Y(s) = G2(s)
1 + G1G2H(s)TL(s) = k2
s(τs + 1) + k1k2(k3+k4s)TL(s).
(b) The sensitivity of the closed-loop system to k2is
ST
k2=1
1 + G1G2H(s),
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Problems 147
where T(s) is the closed loop transfer function
T(s) = G1G2(s)
1 + G1G2H(s)=k1k2
s(τs + 1) + k1k2(k3+k4s).
(c) The error is given by
E(s) = R(s)T(s)R(s).
With
R(s) = 1
s,
we have
ess = lim
s0s(1 T(s))1
s= 1 T(0) = 1 1
k3
.
P4.8 (a) The sensitivity is
ST
K=1
1 + GcG(s)=(0.1s+ 1)(s2+ 20s+ 180)
(0.1s+ 1)(s2+ 20s+ 180) + 180K.
(b) The transfer function from Td(s) to Y(s) is
Y(s) = G(s)
1 + GcG(s)Td(s) = 180(0.1s+ 1)
(0.1s+ 1)(s2+ 20s+ 180) + 180KTd(s).
P4.9 (a) Computing the derivative of Rwith respect to iyields
dR
di =0.201R
(i0.005)3/2.
When vout = 35 volts, we have
i=35
5000 = 7ma .
At the operating point i= 7 ma, we find from Figure P4.9(b) in Dorf
& Bishop that R20K(note: If we use the given formula, we find
that R8.2Kwhen i= 7 ma, thus we see that the formula is just
an approximation to the plot). Using R= 20K, we have
dR
di =0.402 ×104
0.896 ×104=45 kohms/ma .
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148 CHAPTER 4 Feedback Control System Characteristics
The transfer function (valid around the operating point) is
T(s) = Vout(s)
Vin(s)=K
(+ 1) + K45I
5
=K
+ 1 + 9KI .
The photosensor block diagram is shown in Figure P4.9.
-
+
K
ts+ 1
1
5
45 I
i (ma)
vin vout
FIGURE P4.9
Photosensor block diagram.
(b) The sensitivity of the system to changes in Kis
ST
K=τs + 1
τs + 1 + 9KI .
P4.10 (a) and (b) The paper tension control block diagram is shown in Fig-
ure P4.10.
-
+
T( s)
R(s) Km
ts + 1
21
s
2
k1
DT (s)
V1+DV1(s)
-
+
-
+
Y (s)
Eo( s) wo( s)
k3
k2
FIGURE P4.10
Paper tension control block diagram.
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Problems 149
(c) The closed-loop transfer function, Tc, is given by
Tc(s) = T(s)
R(s)=
2Km
s(τs+1)
1 + 4Kmk2
k1s(τs+1)
=2Km
τs2+s+4Kmk2
k1
.
The sensitivity of Tcto changes in Kmis
STc
Km=1
1 + 4Kmk2
k1s(τs+1)
=s(τs + 1)
τs2+s+4Kmk2
k1
.
(d) The transfer function from ∆V1(s) to T(s) is
T(s) = 1
s
1 + 4Kmk2
k1s(τs+1)
V1(s) = 1
s+4Kmk2
k1(τs+1)
V1(s).
When ∆V1(s) = A/s, we have
T(s) = (τs + 1)
τs2+s+4Kmk2
k1
A
s.
and
lim
t→∞ T(t) = lim
s0sT (s) = Ak1
4Kmk2
.
P4.11 (a) The closed-loop transfer function is
T(s) = Gc(s)G(s)
1 + Gc(s)G(s)=K
(8s+ 1)(3s+ 1) + K=K
24s2+ 11s+ 1 + K.
(b) The sensitivity ST
Kis
ST
K=T
K ·K
T=24s2+ 11s+ 1
24s2+ 11s+ 1 + K.
(c) Define E(s) = R(s)Y(s). Then
E(s) = R(s)
1 + Gc(s)G(s)="24s2+ 11s+ 1
24s2+ 11s+K+ 1#R(s).
With
R(s) = A
s,
we have
ess = lim
s0sE(s) = A
1 + K.
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150 CHAPTER 4 Feedback Control System Characteristics
(d) We want |e(t)| 0.02Aas t→ ∞ . So,
0.02AA
K+ 1
implies
K49 .
P4.12 (a) The two transfer functions are
T1(s) = K1K2
s2+ 3s4 + 6K1K2
and
T2(s) = K1K2
s2+ (3 2K1+ 2K2)s4 + 8K2+ 2K14K1K2
.
When K1=K2= 1, we find that
T1(s) = T2(s) = 1
s2+ 3s+ 2.
(b) The sensitivity ST1
K1is
ST1
K1=T1
K1·K1
T1
=s2+ 3s4
s2+ 3s4 + 6K1K2
=(s+ 4)(s1)
(s+ 2)(s+ 1) ,
when K1=K2= 1. The sensitivity ST2
K1is
ST2
K1=T2
K1·K1
T2
=s+ 4
s+ 4 2K1
=s+ 4
s+ 2 ,
when K1= 1. Thus,
ST1
K1=ST2
K1
s1
s+ 1 .
P4.13 (a) Let N(s) = G1(s) + kG2(s) and Td(s) = G3(s) + kG4(s). Then
ST
k=N
k ·k
ND
k ·k
D=G2k
G1+kG2G4k
G3+kG4
=k(G2G3G1G4)
(G1+kG2)(G3+kG4).
(b) The closed-loop transfer function is
T(s) = MG(s) + kUG(s)
1 + kGH(s)=G1(s) + kG2(s)
G3(s) + kG4(s).
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Problems 151
Then using result from (a), we have
ST
k=k(UG(s)MG2H(s))
(MG(s) + kUG(s))(1 + kGH(s)) .
P4.14 The closed-loop transfer function is
T(s) = G(s)
1 + G(s)=10(s+ 4)
s(s+a)(s+ 1) + 10(s+ 4) .
Then
ST
a=SN
aSD
a,
where Nis the numerator and Dis the denominator. We have
SN
a= 0 .
Let
G(s) = p(s)
q(s)(s+a),
where p(s) = 10(s+ 4) and q(s) = s(s+ 1). Then
T(s) = G(s)
1 + G(s)=p(s)
q(s)(s+a) + p(s),
and
ST
a=SD
a=dD
da ·a
D=aq(s)
q(s)(s+a) + p(s)=a
s+a·1
1 + G(s).
P4.15 (a) The closed-loop transfer function for the disturbance to the output is
Y(s)
Td(s)=G(s)
1 + KG(s),
with R= 0. The steady-state deviation is
yss = lim
s0sG(s)
1 + KG(s)1
s=G(0)
1 + KG(0) =1
1 + K.
So, with K= 10 we have yss = 1/11, and with K= 25 we have
yss = 1/26.
(b) Considering the rudder input, we have
Y(s) = G(s)Td(s) + KG(s)R(s)
1 + KG(s)=G(s)(Td(s) + KR(s))
1 + KG(s).
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152 CHAPTER 4 Feedback Control System Characteristics
Setting R(s) = Td(s)
Kyields Y(s) = 0.
P4.16 (a) Let
G1(s) = 1
(τ1s+ 1)(τ2s+ 1) and G2(s) = G1(s)
100 .
Then
T2(s) = G1(s)
1 + G2Gc(s)To(s) + G2Gc(s)
1 + G2Gc(s)T2d(s).
(b) We can equivalently consider the case of a step input, T2d=A/s,
To= 0, and zero initial conditions. Thus,
T2(s) = GcG2
1 + GcG2
T2d=5
500s2+ 60s+ 6
A
s,
where Gc(s) = 500. The transient response is shown in the Fig-
ure P4.16 for a unit step input (A= 1).
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0 10 20 30 40 50 60 70 80 90 100
time (sec)
T2
Tp=34.3 sec
p.o. = 12.8% Ts=66.7 sec
Unit step response, A=1
FIGURE P4.16
Two tank temperature control system response.
(c) With
E(s) = T2d(s)T2(s),
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Problems 153
we have
E(s) = T2d(s)M(s)T2d(s)
where
M(s) = G2Gc(s)
1 + G2Gc(s).
Then
ess = lim
s0s(1 M(s))A
s= (1 M(0))A= (1 5
6)A=A
6.
P4.17 (a) The closed-loop transfer function is
θ(s)
θd(s)=600
0.1s2+s+ 600 =6000
s2+ 10s+ 6000 .
The solution for a step input is
θ(t) = 1 1.0021e5.0349tsin(77.2962t+ 1.5058).
(b) The transfer function from the disturbance to the output is
θ(s)
Td(s)=1
0.1s2+s+ 600 .
Thus,
θss =lim
s0(s) = A
600 .
Therefore, the disturbance input magnitude reduced by 600 at the
output.
(c) Using the final value theorem we have (for θd(s) = 1/s2)
ess = lim
s0sE(s) = lim
s0s(1 T(s))θd(s)
= lim
s0s 0.1s2+s
0.1s2+s+ 600!1
s2=1
600 .
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154 CHAPTER 4 Feedback Control System Characteristics
Advanced Problems
AP4.1 The plant transfer function is
Gp(s) = R
RCs + 1 .
The closed-loop output is given by
H(s) = 1
1 + GGp(s)Q3(s) + GGp(s)
1 + GGp(s)Hd(s).
Therefore, with E(s) = Hd(s)H(s), we have
E(s) = 1
1 + GGp(s)Q3(s),
since Hd(s) = 0.
(a) When G(s) = K, we have
ess = lim
s0sE(s) = 1
1 + KR .
(b) When G(s) = K/s, we have
ess = lim
s0sE(s) = 0 .
AP4.2 Define
G(s) = KmGc(s)
s(Las+Ra)(Js +f) + KmKbGc(s).
Then,
θ(s)d(s) = nG(s)
1 + nG(s)
and
E(s) = 1
1 + nG(s)θd(s).
So,
ess = lim
s0sE(s) = lim
s0s1
1 + nG(s)
A
s=A
1 + nG(0) =AKb
Kb+n.
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Advanced Problems 155
When θd(s) = 0 and Td(s) = M/s, we have
θ(s)/Td(s) = n(Las+Ra)
s(Las+Ra)(Js +f) + KmKb+KmGcn.
If Gc(s) = K, then
ess =nMRa
Km(Kb+nK)
and if Gc(s) = K/s, we determine that ess = 0.
AP4.3 (a) The input R(s) is
R(s) = 1
s1
s2
and the disturbance is Td(s) = 0. So,
ess = lim
s0s1
1 + G(s)R(s) = lim
s0
11
s
1 + 10(2s+4)
7s(s+5)
=0.8750 .
(b) The error plot is shown in Figure AP4.3a.
0 2 4 6 8 10
?-1
?-0.5
0
0.5
1
1.5
2
Time (sec)
e(t)
FIGURE AP4.3
(a) Error plot with d(t)=0.
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156 CHAPTER 4 Feedback Control System Characteristics
0 2 4 6 8 10
-?2
?-1.5
?-1
?-0.5
0
Time (sec)
e(t)
FIGURE AP4.3
CONTINUED: (b) Error plot with r(t)=0.
(c) The transfer function from Td(s) to Y(s) (with R(s) = 0 ) is
Y(s)/Td(s) = 70
7s2+ 55s+ 40 .
The steady-state error due to a disturbance Td(s) = 1/s is
ess = lim
s0s70
7s2+ 55s+ 40
1
s=1.75 .
(d) The error e(t) is shown in Figure AP4.3b.
AP4.4 (a) The closed-loop transfer function is
ω(s)/V (s) = Km
RaJs2+KbKms+KmKKt
.
With v(t) = t, we have V(s) = 1/s2, and Td(s) = 0. Using the final
value theorem yields
ess = lim
s0sE(s) = lim
s0
1
s+KKm
RaJs+KmKb
=Kb
K=0.1
K.
We desire that
ess =0.1
K<0.1.
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Advanced Problems 157
Therefore, we should select K > 1. For example, we can take K= 8.
(b) The transfer function from Td(s) to ω(s) is given by
ω(s)
Td(s)=10s
s2+ 10s+ 100 .
The error plot is shown in Figure AP4.4, where e(s) = ω(s) (V(s) =
0.)
0
0.02
0.04
0.06
0.08
0.1
0.12
0.14
0.16
0.18
0.2
0 1 2 3 4 5 6 7 8 9 10
Time (sec)
e(t)
FIGURE AP4.4
Error plot with a ramp disturbance input.
AP4.5 (a) The transfer function from the disturbance Td(s) to the output Y(s)
is
Y(s)
Td(s)=s
s3+ 4s2+ 4s+K.
The steady-state error (when Td(s) = 1/s) is
ess = lim
s0ss
s3+ 4s2+4s+K
1
s= 0 .
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158 CHAPTER 4 Feedback Control System Characteristics
(b) The closed-loop transfer function is
Y(s)
R(s)=K
s3+ 4s2+ 4s+K.
The steady-state error (when R(s) = 1/s2) is
ess = lim
s0s(1 T(s)) 1
s2= lim
s0
s3+ 4s2+ 4s
s(s3+ 4s2+ 4s+K)=4
K.
(c) Let K= 8. Then,
Y(s)
Td(s)=s
s3+ 4s2+ 4s+ 8 .
The error plot is shown in Figure AP4.5, for r(t) = 0.
-0.1
-0.05
0
0.05
0.1
0.15
0 2 4 6 8 10 12 14 16 18 20
Time (sec)
e(t)
FIGURE AP4.5
Error plot with a step disturbance input and K=8.
AP4.6 (a) The transfer function is
Vo(s)
V(s)=1 + RCs
2 + RCs .
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Advanced Problems 159
(b) The system sensitivity is defined as
SG
C=G/G
C/C .
Therefore, the sensitivity is determined to be
SG
C=RCs
(2 + RCs)(1 + RCs)=1
1 + 2
RCs 1 + 1
RCs .
(c) Let V(s) = 1/s. Then
Vo(s) = 1 + RCs
2 + RCs
1
s=0.5
s+0.5RC
RCs + 2 .
Taking the inverse Laplace transform yields
vo(t) = 0.5(1 + e2t/RC )u(t)
where u(t) is the unit step function. A plot of vo(t) versus t/RC is
shown in Figure AP4.6.
0.5
0.55
0.6
0.65
0.7
0.75
0.8
0.85
0.9
0.95
1
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
t / RC
Vo
FIGURE AP4.6
Step response.
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160 CHAPTER 4 Feedback Control System Characteristics
AP4.7 (a) The transfer function from Td(s) to Y(s) is
Y(s)
Td(s)=s
s(s+ 1) + K.
(b) The transfer function from N(s) to Y(s) is
Y(s)
N(s)=K
s(s+ 1) + K.
(c) Let Td(s) = A/s and N(s) = B/s. Then,
ess =yss = lim
s0ss
s(s+ 1) + K
A
slim
s0sK
s(s+ 1) + K
B
s=B .
So, Khas no effect on the steady-state errors. However, choosing
K= 100 will minimize the effects of the disturbance Td(s) during the
transient period.
AP4.8 (a) The closed-loop transfer function is
T(s) = Kb
s+Kb + 2 .
(b) The sensitivity is determined to be
ST
b=T /T
b/b =s+ 2
s+Kb + 2 .
(c) The transfer function from Td(s) to Y(s) is
Y(s)
Td(s)=b
s+Kb + 2 .
So, choose Kas large as possible, to make Y(s)/Td(s) as “small” as
possible. Thus, select
K= 50 .
This also minimizes ST
bat low frequencies.
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Design Problems 161
Design Problems
The model of the traction drive, capstan roller, and linear slide was de-CDP4.1
veloped in CDP2.1:
θ(s)
Va(s)=Km
s[(Lms+Rm)(JTs+bm) + KbKm].
The step response for the closed-loop system (with the tachometer not in
the loop) and various values of the controller gain Kais shown below.
% System parameters
Ms=5.693; Mb=6.96; Jm=10.91e-03; r=31.75e-03;
bm=0.268; Km=0.8379; Kb=0.838; Rm=1.36; Lm=3.6e-03; Lm=0;
% Controller gain
Ka=100;
% Motor and slide model
Jt=Jm+r^2*(Ms+Mb);
num=[Km];
den=[Lm*Jt Rm*Jt+Lm*bm Kb*Km+Rm*bm 0];
sys=tf(num,den);
%Closed-loop tf and step response
sys_cl=feedback(Ka*sys,[1]);
step(sys_cl)
0 0.2 0.4 0.6 0.8 1
0
0.5
1
1.5
Time (sec)
Theta step response
Ka=2
Ka=5
Ka=10
Ka=100
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162 CHAPTER 4 Feedback Control System Characteristics
DP4.1 (a) The transfer function from the load disturbance to the output speed
is
ω(s)
Td(s)=G(s)
1 + GcG(s)=s
s2+ 4s+K.
Thus, the effect on ω(s) (of a unit step disturbance) at steady-state
is
lim
t→∞ ω(t) = lim
s0ss
s2+ 4s+K1
s= 0 .
We see that the load disturbance has no effect on the output at steady-
state.
(b) The system response for 10 K25 is shown in Figure DP4.1.
99.84
99.86
99.88
99.9
99.92
99.94
99.96
99.98
100
100.02
100.04
0 0.5 1 1.5 2 2.5 3
Time(sec)
w(t)
K=10
K=25
K=10,12,16,18,20,23,25
FIGURE DP4.1
Speed control system response.
For example , if we select K= 16, then ωn= 4, ζ =1
2, and the
response due to a unit step disturbance is
ω(s) = s
s2+ 4s+ 16 1
s=1
(s+ 2)2+ 12 .
Hence, if we are originally at ω(t) = 100 for t < τ, we have
ω(t) = 100 1
12e2tsin12t t τ .
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Design Problems 163
DP4.2 With θd= 0, we have
θ(s) = G(s)
1 + G(s)KK1
s
Td(s) = s
s3+ 4s2+ 9s+KK1
Td(s).
For Td=A/s, we have
θ(s) = A
s3+ 4s2+ 9s+KK1
.
The system response to a unit step disturbance for various values of KK1
are shown in Figure DP4.2. From the plot we see that when KK1is small
the response is slow but not oscillatory. On the other hand, when KK1
is large the response is fast but highly oscillatory. In fact, if KK1>35,
the system is unstable. Thus, we might select KK1= 10 as a reasonable
trade-off between fast performance and stability.
-0.06
-0.04
-0.02
0
0.02
0.04
0.06
0.08
0.1
0.12
01 2 3 4 5 6 7 8 9 10
time(sec)
q
Unit step response for KK1=1,5,10,15,20,25
KK1=1
KK1=5
KK1=25
FIGURE DP4.2
Aircraft roll angle control system response to a disturbance.
DP4.3 (a) The closed-loop transfer function is
T(s) = ω(s)
ωd(s)=K
s2+ 5s+KK1
.
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164 CHAPTER 4 Feedback Control System Characteristics
Then,
E(s) = (1 T(s))ωd(s) = s2+ 5s+K(K11)
s2+ 5s+KK1
1
s.
So, if
0.99 < K1<1.01 ,
then
|ess|<0.01 .
(b) The transfer function from Td(s) to ω(s) is
ω(s) = s
s2+ 5s+KK1
Td(s).
So, with E(s) = ω(s) and Td(s) = 2/s2, we have
lim
s0sE(s) = 2
KK1
.
Therefore, we select KK1>20 to obtain ess <0.1.
DP4.4 The steady-state error for a step input command is zero for any K1. The
transfer function from Td(s) to Y(s) is
Y(s)
Td(s)=G(s)
1 + KG(s)=2
s3+ 5s2+ 4s+ 2K.
Thus, the output at steady-state due to a step disturbance Td(s) = A/s
is
lim
s0sY (s) = A
K.
We want to maximize Kto reduce the effect of the disturbance. As we will
see in Chapter 6, we cannot select Ktoo high or the system will become
unstable. That is why the problem statement suggests a maximum gain
of K= 10. For the design we choose
K= 10 .
DP4.5 The transfer function from V(s) to Vo(s) is
Vo(s)/V (s) = ks
s+a
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Design Problems 165
where
k=R2+R3
R2
and a=1
R1C.
Computing the step response, we find that
vo(t) = keat = 5e100t.
Solving for R1, R2, R3and Cyields
R1C= 0.01 and R2
R3
= 4 .
DP4.6 (a) The closed-loop transfer function is
θ(s) = K/J
s2+K/J θd(s).
Since J > 0, the system is unstable when K < 0 and marginally
stable when K > 0.
(b) Since the system is marginally stable, the system response does not
have a steady-state value—it oscillates indefinitely.
(c) The closed-loop transfer function is
θ(s) = KDs+KP
Js2+KDs+KP
θd(s).
The system is stable for all KD>0 and KP>0, given that J > 0.
(d) The tracking error E(s) = θd(s)θ(s) is
E(s) = Js2
Js2+KDs+KP
.
Therefore, using the nal value theorem we obtain the steady-state
value
lim
s0sE(s) = lim
s0sJs2
Js2+KDs+KP·1
s= 0 .
DP4.7 (a) The closed-loop transfer function is
Y(s) = s
s2+Ks + 2KTd(s) = 1
s2+Ks + 2K
where the disturbance is a unit step Td(s) = 1/s. Considering the
poles of the closed-loop system, we find that when K > 8 the system
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166 CHAPTER 4 Feedback Control System Characteristics
has two real poles. In that case the disturbance step response is
y(t) = 1
K28Keαt eβt,
where
α=KK28K
2and β=K+K28K
2
Bounding the maximum y(t) yields the inequality
|y(t)|=1
K28Keαt eβt0.05.
We know that
eαt eβt1,
for any αand βcomputed as shown above where K > 8. So, if we
choose Ksuch that
1
K28K0.05.
we will guarantee that the maximum bound of 0.05 is not exceeded.
Solving for Kyields
K > 24.4.
For any K > 24.4 we know that the maximum value of the disturbance
step response will be less than 0.05. When K= 24.4 the maximum
unit step disturbance response is 0.035. Solving explicitly for Kso
that the maximum is 0.05 we find that K= 16.3 (this was found
numerically since it is very difficult to obtain analytically).
(b) Since the system is type 2, we know that the steady-state value of
the disturbance step response is zero for a unit step disturbance.
DP4.8 (a) The sensitivities are
ST
τ1=T/T
τ11
="s2(τ2s+ 1) τ1
K#T(s)
and
ST
τ2=T/T
τ22
="s2(τ1s+ 1) τ2
K#T(s)
where we assume that K6= 0.
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Design Problems 167
(b) Computing the closed-loop transfer function yields
Y(s) = s(τ1s+ 1)
s(τ1s+ 1) (τ2s+ 1) + KTd(s)
When Td(s) = 1/s, using the final value theorem we find that
lim
s0sY (s) = lim
s0
s(τ1s+ 1)
s(τ1s+ 1) (τ2s+ 1) + K= 0
as long as K6= 0. We assume here that final value theorem applies
(i.e., the system is stable, more on this in Chapter 6).
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168 CHAPTER 4 Feedback Control System Characteristics
Computer Problems
CP4.1 The step response and an m-file script which generates the step response
is shown in Figure CP4.1. The closed-loop transfer function is
T(s) = 12
s2+ 2s+ 22 .
The percent overshoot is P.O. = 50.2% and the steady-state error is
ess = 0.45.
num = [12]; den = [1 2 10];
sys = tf(num,den);
sys_cl = feedback(sys,[1])
step(sys_cl)
Step Response
Time (sec)
Amplitude
0123456
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
System: sys_cl
Peak amplitude: 0.82
Overshoot (%): 50.2
At time (sec): 0.67
FIGURE CP4.1
Step response.
CP4.2 The transfer function is
G(s) = 4
s2+ 2s+ 20 .
An m-file script which generates the step response is shown in Figure CP4.2.
The step response is also shown in Figure CP4.2. The step response is
generated using the step function. In the script, the transfer function
numerator is represented by num and the denominator is represented by
den. The steady-state value is yss = 0.2 and the desired value is 1.0.
Therefore, the steady-state error is
ess = 0.8.
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Computer Problems 169
num=[4]; den=[1 2 20];
sys = tf(num,den);
axis([0 6 0 1]);
t=[0:0.01:6];
step(sys,t)
y = step(sys,t);
yss = y(length(t))
yss =
0.20
0 1 2 3 4 5 6
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
Step Response
Time (sec)
Amplitude
FIGURE CP4.2
Step response.
CP4.3 The step responses and the m-file script which generates the step re-
sponses is shown in Figure CP4.3.
K=[10,200,500];
t=[0:0.01:7];
for i=1:3
num=5*K(i); den=[1 15 K(i)];
sys = tf(num,den)
y(:,i)= step(sys,t);
end
plot(t,y(:,1),t,y(:,2),'--',t,y(:,3),':')
legend('K=10','K=200','K=500',-1)
01234567
0
1
2
3
4
5
6
7
K=10
K=200
K=500
FIGURE CP4.3
Step responses for K= 10,100,500.
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170 CHAPTER 4 Feedback Control System Characteristics
CP4.4 (a,b) The m-file and plots are shown in Figure CP4.4.
ng=1;dg=[1 1.91 0];sysg=tf(ng,dg);
K=10;
syscl=feedback(K*sysg,1);
figure(1)
subplot(211)
step(syscl)
subplot(212)
syst=feedback(sysg,K)
step(syst)
0 1 2 3 4 5 6
0
0.5
1
1.5
Time (sec)
0 1 2 3 4 5 6
0
0.05
0.1
0.15
0.2
Time (sec)
Step response
Disturbance response
y(t)
y(t)
FIGURE CP4.4
Step response and disturbance response.
(c) The estimated steady-state tracking error due to a unit step input is
zero, and the estimated steady-state tracking error to a unit distur-
bance is 0.1.
(d) The estimated maximum tracking error due to a unit step input is
0.4, and the estimated maximum tracking error to a unit disturbance
is 0.14. The maximum occurs at approximately t= 1 s.
CP4.5 The step response and the m-file script which generates the step response
is shown in Figure CP4.5. The closed-loop transfer function is determined
to be
T(s) = 10
s2+ 3.7s+ 10 .
Using the m-file script, a trial-and-error search on kyields
k= 3.7.
The percent overshoot P.O. = 10% and the steady-state value is 1, as
expected.
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Computer Problems 171
k = 3.7; % Final value of k=3.7
numcg = [10]; dencg = [1 k 0];
sys_o = tf(numcg,dencg);
sys_cl = feedback(sys_o,[1])
t = [0:0.1:5];
[y,t] = step(sys_cl,t);
plot(t,y,[0 5],[1.1 1.1],'--'); grid
xlabel('Time (sec)'); ylabel('y(t)');
Transfer function:
10
----------------
s^2 + 3.7 s + 10
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
0
0.2
0.4
0.6
0.8
1
1.2
1.4
Time (sec)
y(t)
FIGURE CP4.5
Step response.
CP4.6 The closed-loop transfer function is
T(s) = K
sa+K
where K= 2. When a= 1 and R(s) = 1/s, the final value is
lim
s0sT (s)R(s) = lim
s0T(s) = K
Ka= 2 .
The output is within 2% of the final value at around t= 4.6 seconds. The
plot of the step responses for
a= 1,0.5,2,5
is shown in Figure CP4.6. The output is unstable for
a > 2.
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172 CHAPTER 4 Feedback Control System Characteristics
K=2; t=[0:0.1:5];
num=K*[1];
a=[1 0.5 2 5];
for i=1:4
den=[1 -a(i)]; sys = tf(num,den);
sys_cl = feedback(sys,[1]);
y(:,i)=step(sys_cl,t);
end
plot(t,y(:,1),t,y(:,2),':',t,y(:,3),'--',t,y(:,4),'-.')
axis([0 5 0 5]);
xlabel('Time (sec)'), ylabel('y(t)')
title('a=1 (solid); a=0.5 (dotted); a=2 (dashed); a=5 (dashdot)')
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
Time (sec)
y(t)
a=1 (solid); a=0.5 (dotted); a=2 (dashed); a=5 (dashdot)
FIGURE CP4.6
Step response for a=1, 0.5, 2, and 5.
CP4.7 The transfer function from the disturbance to the output is
T(s) = G(s)
1 + K0G(s)=1
Js2+bs +k+K0
.
The disturbance response is shown in Figure CP4.7. The compensated
system response is significantly reduced from the uncompensated system
response. The compensated system output is about 11 times less than the
uncompensated system output. So, closed-loop feedback has the advan-
tage of reducing the effect of unwanted disturbances on the output.
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Computer Problems 173
J=1; k=5; c=0.9;
num=[1/J]; den=[1 c/J k/J];
sys = tf(num,den);
t=[0:0.1:10];
%
yu=step(sys,t); % Part (a)
K0=50;
numk=[K0]; denk=[1]; sysk = tf(numk,denk);
sys_cl = feedback(sys,sysk);
yc=step(sys_cl,t); % Part (b)
plot(t,yu,t,yc,'--')
xlabel('Time (sec)'), ylabel('\theta')
title('Uncompensated response (solid) & Compensated response (dashed)')
0 1 2 3 4 5 6 7 8 9 10
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
Time (sec)
q
Uncompensated response (solid) & Compensated response (dashed)
FIGURE CP4.7
Disturbance responses for both the uncompensated and compensated systems.
CP4.8 The step responses for the proportional and PI controller are shown in
Figure CP4.8. The steady-state tracking error for the proportional con-
troller is
ess = 0.33 .
Increasing the complexity of the controller from a proportional controller
to a proportional plus integral (PI) controller allows the closed-loop sys-
tem to track the unit step response with zero steady-state error. The cost
is controller complexity, which translates into higher costs ($).
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174 CHAPTER 4 Feedback Control System Characteristics
numg=[10]; deng=[1 10]; sysg = tf(numg,deng);
t=[0:0.001:0.5];
% Part (a)
numc=[2]; denc=[1]; sysc = tf(numc,denc);
sys_o = series(sysc,sysg);
sys_cl = feedback(sys_o,[1]);
yk=step(sys_cl,t);
% Part (b)
numc=[2 20]; denc=[1 0]; sysc = tf(numc,denc);
sys_o = series(sysc,sysg);
sys_cl = feedback(sys_o,[1]);
yp=step(sys_cl,t);
%
plot(t,yk,t,yp,'--')
xlabel('Time (sec)'),ylabel('y(t)')
title('Proportional controller (solid) & PI controller (dashed)')
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Time (sec)
y(t)
Proportional controller (solid) & PI controller (dashed)
FIGURE CP4.8
Step response for proportional controller and PI controller.
CP4.9 (a) The closed-loop transfer function is
T(s) = G(s)
1 + G(s)H(s)R(s) = 10s2+ 500s
s2+ 200s+ 5000 R(s).
The step response is shown in Figure CP4.9a.
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Computer Problems 175
(b) The response of the system to the sinusoidal disturbance
N(s) = 100
s2+ 100
is shown in Figure CP4.9b.
(c) In the steady-state, the magnitude of the peak response is 0.095 and
the frequency is 10 rad/sec (see Figure CP4.9b).
0 1 2 3 4 5 6 7
−0.1
−0.08
−0.06
−0.04
−0.02
0
0.02
0.04
0.06
0.08
0.1
Step Response
Time (sec)
Amplitude
0 0.02 0.04 0.06 0.08 0.1 0.12
0
2
4
6
8
10
Step Response
Time
(
sec
)
Amplitude
% Part (a)
ng=10*[1 0]; dg=[1 100]; sysg=tf(ng,dg);
nh=[5]; dh=[1 50]; sysh=tf(nh,dh);
sys=feedback(sysg,sysh)
figure(1)
step(sys)
% Part (b)
sysn=-feedback(sysg*sysh,1)
syss=tf([100],[1 0 100]);
% This is the sinusoidal input
figure(2)
t=[0:0.001:7];
step(syss*sysn,t)
>>
Transfer function:
10 s^2 + 500 s
------------------
s^2 + 200 s + 5000
(a)
(b)
FIGURE CP4.9
(a) Unit step response. (b) Response to sinusoidal noise input at ω= 10 rad/sec.
CP4.10 (a) The closed-loop transfer function is
T(s) = Gc(s)G(s)
1 + G(s)Gc(s)R(s) = K(s+ 1)
(s+ 15)(s2+s+ 6.5) + K(s+ 1)R(s).
(b) The step responses are shown in Figure CP4.10a.
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176 CHAPTER 4 Feedback Control System Characteristics
(c) The unit disturbance response of the system is shown in Figure CP4.10b.
The steady-state value is 0.14.
Step Response
Time (sec)
Amplitude
0 1 2 3 4 5 6 7
0
0.02
0.04
0.06
0.08
0.1
0.12
0.14
0.16
0.18
0.2
System: syscl
Final Value: 0.14
(a) (b)
01234567
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
Time (s)
Step response
K=5
K=10
K=50
FIGURE CP4.10
(a) Unit step responses for K= [5,10,50]. (b) Disturbance unit step response.
CP4.11 The m-file is shown in Figure CP4.11a and the step responses in Fig-
ure CP4.11b.
0 5 10 15 20
−3
−2
−1
0
1
2
3
4
5
Time (s)
Step response
K=10
K=12
K=15
K=[10, 12, 15];
t=[0:0.1:20];
ng=[20]; dg=[1 4.5 64]; sysg=tf(ng,dg);
nh=[1]; dh=[1 1]; sysh=tf(nh,dh);
for i=1:length(K)
sys=K(i)*sysg;
syscl=feedback(sys,sysh)
y(:,i)= step(syscl,t);
end
plot(t,y(:,1),t,y(:,2),'--',t,y(:,3),':')
xlabel('Time (s)')
ylabel('Step response')
legend('K=10','K=12','K=15',-1)
(a) (b)
FIGURE CP4.11
(a) M-file script. (b) Unit step responses for K= [10,12,15].
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C H A P T E R 5
The Performance of Feedback
Control Systems
Exercises
E5.1 For a zero steady-state error, when the input is a step we need one inte-
gration, or a type 1 system. A type 2 system is required for ess = 0 for a
ramp input.
E5.2 (a) The closed-loop transfer function is
T(s) = Y(s)
R(s)=G(s)
1 + G(s)=240
(s+ 4)(s+ 6) + 240 =240
s2+ 2ζωns+ω2
n
.
The steady-state error is given by
ess =A
1 + Kp
,
where R(s) = A/s and
Kp= lim
s0G(s) = 240
24 = 10 .
Therefore,
ess =A
11 .
(b) The closed-loop system is a second-order system with natural fre-
quency
ωn=264 ,
177
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178 CHAPTER 5 The Performance of Feedback Control Systems
and damping ratio
ζ=10
2264 = 0.31 .
The percent overshoot is thus computed to be
P.O. = 100eπζ/1ζ2= 36% .
E5.3 The closed-loop transfer function is
Y(s)
I(s)=G(s)
1 + G(s)=K
s(s+ 14) + K=K
s2+ 14s+K.
Utilizing Table 5.6 in Dorf & Bishop, we find that the optimum coefficients
are given by
s2+ 1.4ωns+ω2
n.
We have
s2+ 14s+K ,
so equating coefficients yields ωn= 10 and K=ω2
n= 100 .We can also
compute the damping ratio as
ζ=14
2ωn
= 0.7.
Then, using Figure 5.8 in Dorf & Bishop, we find that P.O. 5%.
E5.4 (a) The closed-loop transfer function is
T(s) = G(s)
1 + G(s)=2(s+ 8)
s2+ 6s+ 16 .
(b) We can expand Y(s) in a partial fraction expansion as
Y(s) = 2(s+ 8)
(s2+ 6s+ 16)
A
s=A1
ss+ 4
s2+ 6s+ 16.
Taking the inverse Laplace transform (using the Laplace transform
tables), we find
y(t) = A[1 1.07e3tsin(7t+ 1.21)] .
(c) Using the closed-loop transfer function, we compute ζ= 0.75 and
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Exercises 179
ωn= 4. Thus,
a
ζωn
=8
3= 2.67 ,
where a= 8. From Figure 5.13(a) in Dorf & Bishop, we find (approx-
imately) that P.O. = 4% .
(d) This is a type 1 system, thus the steady-state error is zero and y(t)
Aas t→ ∞.
E5.5 The closed-loop transfer function is
Y(s)
R(s)=K
s2+ 4s+K.
Utilizing Table 5.6 in Dorf & Bishop, we find that the optimum coefficients
are given by
s2+ 1.4ωns+ω2
n.
We have
s2+ 4s+K ,
so equating coefficients yields ωn= 2.86 and K=ω2
n= 8.16 .We can
also compute the damping ratio as
ζ=4
2ωn
= 0.7.
Then, using Figure 5.8 in Dorf & Bishop, we find that P.O. 5%.
E5.6 (a) The closed-loop transfer function is
T(s) = Y(s)
R(s)=G(s)
1 + GH(s)=100
s2+ 100Ks + 100 ,
where H(s) = 1 + Ks and G(s) = 100/s2. The steady-state error is
computed as follows:
ess = lim
s0s[R(s)Y(s)] = lim
s0s[1 T(s)] A
s2
= lim
s0"1
100
s2
1 + 100
s2(1 + Ks)#A
s=KA .
(b) From the closed-loop transfer function, T(s), we determine that ωn=
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180 CHAPTER 5 The Performance of Feedback Control Systems
10 and
ζ=100K
2(10) = 5K .
We want to choose Kso that the system is critically damped, or
ζ= 1.0. Thus,
K=1
5= 0.20 .
The closed-loop system has no zeros and the poles are at
s1,2=50K±10p25K21.
The percent overshoot to a step input is
P.O. = 100e
5πK
125K2for 0 < K < 0.2
and P.O. = 0 for K > 0.2.
E5.7 The closed-loop transfer function is
T(s) = Y(s)
R(s)=KG(s)
1 + KG(s)=K(s+ 2)
s(s+ 1) + K(s+ 2) =K(s+ 2)
s2+s(K+ 1) + 2K.
Therefore, ωn=2Kand ζ=K+1
22K. So,
a
ζωn
=4
K+ 1 .
From Figure 5.13a in Dorf & Bishop, we determine that
a
ζωn1.5
when ζ= 0.707. Thus, solving for Kyields
4
K+ 1 = 1.5
or K= 1.67.
E5.8 The pole-zero map is shown in Figure E5.8. Since the dominant poles are
at s=2±2.45iwe have a damping ratio ζ= 0.63. We estimate the
percent overshoot to be
P.O. = 100eπζ/1ζ2= 7.69%
The step response is shown in Figure E5.8b. The actual overshoot is 8%.
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Exercises 181
Pole−Zero Map
Real Axis
Imaginary Axis
−25 −20 −15 −10 −5 0
−2.5
−2
−1.5
−1
−0.5
0
0.5
1
1.5
2
2.5
0 0.5 1 1.5 2 2.5 3
0
0.2
0.4
0.6
0.8
1
1.2
1.4
System: sys
Time (sec): 1.28
Amplitude: 1.08
Step Response
Time (sec)
Amplitude
FIGURE E5.8
(a) Pole-zero map. (b) Unit step response.
E5.9 (a) The closed-loop transfer function is
T(s) = K
s2+2Ks +K.
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182 CHAPTER 5 The Performance of Feedback Control Systems
The damping ratio is ζ=2/2 and the natural frequency is ωn=
K. Therefore, we compute the percent overshoot to be
P.O. = 100eπζ/1ζ2= 4.3%
for ζ= 0.707. The settling time is estimated via
Ts=4
ζωn
=8
2K.
(b) The settling time is less than 1 second whenever K > 32.
E5.10 The second-order closed-loop transfer function is given by
T(s) = ω2
n
s2+ 2ζωns+ω2
n
.
From the percent overshoot specification, we determine that P.O. 5%
implies ζ0.69. From the settling time specification, we find that Ts<4
implies ωnζ > 1. And finally, from the peak time specification we have
Tp<1 implies ωnp1ζ2> π. The constraints imposed on ζand ωnby
the performance specifications define the permissible area for the poles of
T(s), as shown in Figure E5.10.
Re(s)
Im(s)
desired
region
for poles
z wn = -1
wn1-z2= - P
wn1-z2=P
46o
z = 0.69
FIGURE E5.10
Permissible area for poles of T(s).
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Exercises 183
E5.11 The system is a type 1. The error constants are
Kp=and Kv= 1.0.
Therefore, the steady-state error to a step input is 0; the steady-state
error to a ramp input is 1.0A0, where A0is the magnitude (slope) of the
ramp input.
E5.12 (a) The tracking error is given by
E(s) = R(s)
1 + GcG(s)=(s+ 9)(s+ 2)(s+ 4)
(s+ 9)(s+ 2)(s+ 4) + K(s+ 6)R(s).
The steady-state tracking error (with R(s) = 1/s) is
lim
s0sE(s) = 72
72 + 6K.
We require ess <0.05, so solving for Kyields K > 228.
(b) The tracking error due to the disturbance is
E(s) = G(s)
1 + GcG(s)Td(s) = (s+ 9)(s+ 6)
(s+ 9)(s+ 2)(s+ 4) + K(s+ 6)Td(s).
The tracking error is shown in Figure E5.12.
-0.08
-0.07
-0.06
-0.05
-0.04
-0.03
-0.02
-0.01
0
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
Time (secs)
Amplitude
FIGURE E5.12
Tracking error due a step disturbance.
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184 CHAPTER 5 The Performance of Feedback Control Systems
E5.13 The system is a type 0. The error constants are Kp= 0.4 and Kv= 0.
The steady-state error to a ramp input is . The steady-state error to a
step input is
ess =1
1 + Kp
= 0.71.
E5.14 (a) The tracking error is given by
E(s) = [1 T(s)] R(s).
The steady-state tracking error (with R(s) = 1/s) is
ess = lim
s0s[1 T(s)] R(s) = lim
s0[1 T(s)] = 1 T(0) .
The closed-loop transfer function is
T(s) = K(s+ 0.1)
s(s+ 0.1)(s+ 2) + K(s+ 3) ,
and T(0) = 0.033. Therefore, ess = 1 T(0) = 0.967.
(b) Use Gp(s) = 30. Then,
ess = lim
s0s[1 T(s)Gp(s)] R(s) = 1lim
s0T(s)Gp(s) = 130 T(0) = 0 .
E5.15 The plot of y(t) is shown in Figure E5.15.
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2
0
0.2
0.4
0.6
0.8
1
1.2
1.4
Time (sec)
y(t)
Response using only dominate poles
Actual response
FIGURE E5.15
Plot of y(t)with T(s)(solid line) and approximate Ta(s)(dashed line).
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Exercises 185
Using the dominant poles, the approximate closed-loop transfer function
is
Ta(s) = 50
s2+ 10s+ 50 .
The actual transfer function is
T(s) = 500
(s+ 10)(s2+ 10s+ 50) .
E5.16 The partial fraction expansion is
y(t) = 10(z1)
7zet+10(z8)
56ze8t+ 1.25 .
The plot of y(t) for z= 2,4,6 is shown in Figure E5.16.
0 1 2 3 4 5 6
0
0.2
0.4
0.6
0.8
1
1.2
1.4
Time (sec)
y(t)
z=2 (solid) & z=4 (dashed) & z=6 (dotted)
FIGURE E5.16
Plot of y(t)for z=2, 4, 6.
E5.17 The desired pole locations for the 5 different cases are shown in Fig-
ure E5.17.
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186 CHAPTER 5 The Performance of Feedback Control Systems
Re(s)
Im(s)
desired
region
for poles
37 o
53
o
wn = 10
(a) 0.6 < z < 0.8 and wn <10
Re(s)
Im(s)
desired
region
for poles
45 o
60
o
wn = 10
(b) 0.5 < z < 0.707 and wn >10
(c) 0.5 < z and 5 < wn <10
Re(s)
Im(s)
desired
region
for poles
60
o
wn = 10
wn = 5
FIGURE E5.17
Desired pole locations.
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Exercises 187
(d) 0.707 > z and 5 < wn <10
Re(s)
Im(s)
desired
region
for poles
45
o
wn = 5
wn = 10
(e) 0.6 < z and wn < 6
Re(s)
Im(s)
desired
region
for poles
53
own = 6
FIGURE E5.17
CONTINUED: Desired pole locations.
E5.18 The output is given by
Y(s) = T(s)R(s) = KG(s)
1 + G(s)R(s).
When K= 1, the steady-state error is
ess = 0.2
which implies that
lim
s0sY (s) = 0.8.
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188 CHAPTER 5 The Performance of Feedback Control Systems
Since we want ess = 0, it follows that
lim
s0sY (s) = 1 ,
or
0.8K= 1 .
Therefore, K= 1.25.
E5.19 (a) The characteristic equation is
s2= 2ζωns+ω2
n=s2+ 3.17s+ 7 = 0 ,
from which it follows that
ωn=7 = 2.65, ζ =3.17
2ωn
= 0.6.
Therefore, we compute the percent overshoot and the estimated set-
tling time to be
P.O. = 100eπζ/1ζ2= 9.53% and Ts=4
ζωn
= 2.5 s .
(b) The unit step response is shown in Figure E5.19.
Step Response
Time (sec)
Amplitude
0 0.5 1 1.5 2 2.5 3
0
0.2
0.4
0.6
0.8
1
1.2
1.4
System: sys
Peak amplitude: 1.1
Overshoot (%): 9.53
At time (sec): 1.47 System: sys
Settling Time (sec): 2.25
FIGURE E5.19
Unit step response.
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Exercises 189
E5.20 (a) The closed-loop transfer function is
T(s) = K
s2+2Ks +K.
The damping ratio is
ζ=2
2
and the natural frequency is ωn=K. Therefore, we compute the
percent overshoot to be
P.O. = 100eπζ/1ζ2= 4.3%
for ζ= 0.707. The settling time is estimated via
Ts=4
ζωn
=8
2K.
(b) The settling time is less than 1 second whenever K > 32.
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190 CHAPTER 5 The Performance of Feedback Control Systems
Problems
P5.1 (a) The system error is
E(s) = 1
1 + KaKm
m+1
R(s)
where
R(s) = 25o/sec
s.
So,
lim
t0e(t) = lim
s0sE(s) = 25
1 + KaKm
.
(b) If we desire ess 1o/sec, then
25o/s
1 + KaKm1o/sec ,
and solving for KaKmyields
KaKm24 .
(c) The closed-loop transfer function is
T(s) = Vb(s)
Vc(s)=KaKm
m+ 1 + KaKm
.
The step response of the system (i.e. vc(t) = A) is
vb(t) = AKaKm
1 + KaKm1e
(KaKm+1)
τmt.
So, at settling time, we have
1e
(1+KaKm)
τmt0.98 ,
where τm= 0.4. Setting t= 0.03 and solving for KaKmyields
KaKm52 .
P5.2 (a) The settling time specification
Ts=4
ζωn
<0.6
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Problems 191
is used to determine that ζωn>6.67. The P.O. < 20% requirement
is used to determine
ζ < 0.45 which implies θ < 63o
and the P.O. > 10% requirement is used to determine
ζ > 0.60 which implies θ > 53o,
since cos θ=ζ. The desired region for the poles is shown in Fig-
ure P5.2.
Re(s)
Im(s)
s = -6.67
63o
desired
region
for poles
53o
FIGURE P5.2
Desired region for pole placement.
(b) The third root should be at least 10 times farther in the left half-
plane, so
|r3| ≥ 10|ζωn|= 66.7.
(c) We select the third pole such that r3=66.7. Then, with ζ= 0.45
and ζωn= 6.67, we determine that ωn= 14.8. So, the closed-loop
transfer function is
T(s) = 66.7(219.7)
(s+ 66.7)(s2+ 13.3s+ 219.7) ,
where the gain K= (66.7)(219.7) is chosen so that the steady-state
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192 CHAPTER 5 The Performance of Feedback Control Systems
tracking error due to a step input is zero. Then,
T(s) = G(s)
1 + G(s),
or
G(s) = T(s)
1T(s).
P5.3 Given the input
R(s) = 1
s3,
we compute the steady-state error as
ess = lim
s0s1
1 + G(s)1
s3= lim
s01
s2G(s)= lim
s0
1
s2K
s2
=1
K.
Since we require that ess 0.5 cm, we determine
K2.
P5.4 (a) The closed-loop transfer function is
T(s) = G(s)
1 + G(s)=K
s2+ 2s+K=ω2
n
s2+ 2ζωns+ω2
n
.
Thus,
ωn=Kand ζ= 1n= 1/K .
Our percent overshoot requirement of 5% implies that ζ= 1/2 ,
which in turn implies that ωn=2. However, the corresponding
time to peak would be
Tp=4.4
2= 3.15 .
Our desired Tp= 1.1—we cannot meet both specification simultane-
ously.
(b) Let Tp= 1.1∆ and P.O. = 0.05∆, where ∆ is the relaxation factor
to be determined. We have that K=ω2
nand ζωn= 1, so
ζ=1
K.
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Problems 193
Thus,
P.O. =eπζ/1ζ2=eπ/K1.
Also,
Tp=π
K1= 1.1∆ .
Therefore, from the proceeding two equations we determine that
P.O. = 0.05∆ = e1.1∆ .
Solving for ∆ yields
f(∆) = ln ∆ + ln(0.05) + 1.1∆ = 0 .
The plot of f(∆) versus ∆ is shown in Figure P5.4. From the plot we
-8
-7
-6
-5
-4
-3
-2
-1
0
1
2
0 0.5 1 1.5 2 2.5 3
*
D
f(D)
D
=2.07
FIGURE P5.4
Solving for the zeros of f.
see that ∆ = 2.07 results in f(∆) = 0. Thus,
P.O. = 0.05∆ = 10%
Tp= 1.1∆ = 2.3 sec.
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194 CHAPTER 5 The Performance of Feedback Control Systems
So, we can meet the specifications if they are relaxed by a factor of
about 2 (i.e. ∆ = 2.07).
P5.5 (a) The closed-loop transfer function is
T(s) = K1K2(s+ 1)
s2+K1K2s+K1K2
.
A percent overshoot less than 5% implies ζ0.69. So, choose ζ=
0.69. Then set 2ζωn=K1K2and ω2
n=K1K2. Then
2(0.69)ωn=ω2
n;
and solving for ωnyields
ωn= 1.38 .
Therefore K1K2=ω2
n= 1.9. When K1K21.9 it follows that
ζ0.69.
(b) We have a type 2 system, so the steady-state tracking error to both
a step and ramp input is zero.
(c) For a step input, the optimum ITAE characteristic equation is
s2+ 1.4ωns+ω2
n= 0 .
For a ramp input, the optimum ITAE characteristic equation is
s2+ 3.2ωns+ω2
n= 0 .
Thus, K1K2=ω2
n= 3.2ωn. So, ωn= 3.2 and K1K2= 10.24.
P5.6 We have a ramp input, r(t) = t. So
Kv= lim
s0sG(s) = lim
s0s75(s+ 1)
s(s+ 5)(s+ 25) =75
125 = 0.6,
and
ess =|R|
Kv
=1
0.6= 1.67 .
P5.7 (a) The closed-loop transfer function is
T(s) = K1K2
Is2+K1K2K3s+K1K2
.
The steady-state tracking error for a ramp input is
ess = lim
s0sE(s) = lim
s0s(1 T(s))R(s) = lim
s0s(1 T(s)) 1
s2
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Problems 195
= lim
s0
Is +K1K2K3
Is2+K1K2K3s+K1K2
=K3.
But we desire ess = 0.01 m, so K3= 0.01.
(b) For P.O. = 10%, we have ζ= 0.6. Also,
2ζωn=0.01K1K2
25
and
ω2
n=K1K2
25 .
Thus, solving for K1K2yields K1K2= 36 ×104.
P5.8 (a) The closed-loop transfer function is
T(s) = P(s)
R(s)=G(s)/s
1 + G(s)H(s)/s =20
s(s+ 40) .
Therefore, the closed-loop system time constant is τ= 1/40 sec.
(b) The transfer function from Td(s) to the output P(s) is
P(s)
Td(s)=G(s)
1 + G(s)H(s)/s =20
s+ 40 .
The response to a unit step disturbance is
p(t) = 1
2(1 e40t).
At settling time, p(t) = 0.98pss =0.49. Thus, solving for t(= Ts)
we determine that Ts= 0.098 sec.
P5.9 We need to track at the rate
ω=v
r=16000
2500 = 1.78 ×103radians/sec .
The desired steady-state tracking error is
ess 1
10 degree = 0.1754 ×102rad .
Therefore, with
ess =|ω|
Kv
,
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196 CHAPTER 5 The Performance of Feedback Control Systems
we compute Kvas
Kv=1.78 ×103
0.1754 ×102= 1.02 .
This assumes that the system is type 1.
P5.10 (a) The armature controlled DC motorblock diagram is shown in Fig-
ure P5.10.
-
+K
Kb
Km
Ra+ L as
1
J s + b
w(s)
R(s )
back emf
amplier
FIGURE P5.10
Armature controlled DC motor block diagram.
(b) The closed-loop transfer function is
T(s) = ω(s)
R(s)=KG(s)
1 + KKbG(s),
where
G(s) = Km
(Ra+Las)(Js +b).
Thus,
T(s) = K
s2+ 2s+ 1 + K,
where Ra=La=J=b=Kb=Km= 1. The steady-state tracking
error is
ess = lim
s0s(R(s)Y(s)) = lim
s0sA
s(1 T(s))
=A(1 T(0)) = 1K
1 + K=A
1 + K.
(c) For a percent overshoot of 15%, we determine that ζ= 0.5. From
our characteristic polynomial we have 2ζωn= 2 and ωn=1 + K.
Solving for ωnyields ωn= 2, thus K= 3.
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Problems 197
P5.11 (a) The closed-loop transfer function is
T(s) = Y(s)
R(s)=K
s+K.
To include the initial condition, we obtain the differential equation:
˙y(t) + Ky(t) = Kr(t).
Taking the Laplace transform yields:
sY (s)y(to) + KY (s) = KA
s,
where y(to) = Q. Computing the inverse Laplace transform, L1{Y(s)}
yields
y(t) = A(1 eKt) + QeKt .
Also, the tracking error is given by
e(t) = Ay(t) = eKt(AQ).
Thus, the performance index, Iis determined to be (for K > 0)
I=Z
0
(AQ)2e2Ktdt = (AQ)21
2Ke2Kt
0
=(AQ)2
2K.
(b) The minimum I is obtained when K=, which is not practical.
(c) Set K at the maximum value allowable such that the process does not
saturate. For example, if K= 50, then
I=(AQ)2
100 .
P5.12 The optimum ITAE transfer function for a ramp input is
T(s) = 3.25ω2
ns+ω3
n
s3+ 1.75ωns+ 3.25ω2
ns+ω3
n
.
The steady-state tracking error, ess = 0, for a ramp input. The step
response is shown in Figure P5.12 for ωn= 10. The percent overshoot is
P.O. = 39%, and the settling time is Ts= 0.72 s .
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198 CHAPTER 5 The Performance of Feedback Control Systems
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2
0
0.2
0.4
0.6
0.8
1
1.2
1.4
PO=39%
Ts=0.72s
time (sec)
y(t)
FIGURE P5.12
Step input system response.
P5.13 The step responses for the actual system and the approximate system
are shown in Figure P5.13. It can be seen that the responses are nearly
identical.
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Time (sec)
y(t)
FIGURE P5.13
Closed-loop system step response: Actual T(s) (solid line) and second-order approximation
(dashed line).
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Problems 199
P5.14 Consider
L(s) = 2(c1s+ 1)
(s+ 1)(s+ 2) .
After cancellation of like factors, we compute H(s)/L(s),
H(s)
L(s)=s3+ 7s2+ 24s+ 24
(s+ 3)(s+ 4)2(c1s+ 1) .
Therefore,
M(s) = s3+ 7s2+ 24s+ 24 ,and
∆(s) = 2[c1s3+ (7c1+ 1)s2+ (12c1+ 7)s+ 12] .
Then, following the procedure outlined in Section 5.10, we have
Mo(0) = 24, M1(0) = 24, M2(0) = 14, M3(0) = 6,and
0(0) = 24,1(0) = (12c1+ 7)2,2(0) = 2(2·(7c1+ 1)),3(0) = 12c1.
For q= 1:
M2= 240,and
2= 4[144c2
1+ 25] .
Then, equating ∆2and M2, we find c1,
c1= 0.493 .
So,
L(s) = 2(0.493s+ 1)
(s+ 1)(s+ 2) =0.986s+ 2
s2+ 3s+ 2 =0.986(s+ 2.028)
(s+ 1)(s+ 2) .
P5.15 The closed-loop transfer function is
T(s) = K(s+ 1)
(s+ 4)(s2+s+ 10) + K(s+ 1) .
The percent overshoot as function of the gain, K, is shown in Figure P5.15.
It can be seen that the percent overshoot decreases as the gain increases
approaching a minimum around 85%. The larger the gain, the smaller the
percent overshoot. For a gain K250, we have essentially minimized the
percent overshoot.
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200 CHAPTER 5 The Performance of Feedback Control Systems
0 50 100 150 200 250 300 350
80
90
100
110
120
130
140
150
160
K
Percent Overshoot (%)
FIGURE P5.15
Percent overshoot versus the gain, K.
P5.16 The open-loop transfer function is
G(s) = 10
(s+ 1)(50Cs + 1) .
Define τ= 50C. Then, the closed-loop transfer function is
Vo(s)
Vin(s)=10
(s+ 1)(τs + 1) + 10 =10
s2+τ+1
τs+11
τ
.
With
ω2
n=11
τand ζ=1
2=τ+ 1
2τωn
,
we can solve for τ, yielding
τ220τ+ 1 = 0 .
Therefore, τ= 19.95 and 0.05. For each value of τwe determine Cas
follows: τ= 19.95 = 50C, implies C= 0.399F, and τ= 0.05 = 50C,
implies C= 1mF .
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Problems 201
P5.17 (a) The closed-loop transfer function is
T(s) = Y(s)
R(s)=12K
s2+ 12s+ 12K.
The percent overshoot specification P.O. 10% implies ζ0.59.
From the characteristic equation we find that
ω2
n= 12Kand ζωn= 6 .
Solving for Kyields
2(0.59)12K= 12 which implies that K= 8.6.
So, any gain in the interval 0 < K < 8.6 is valid. The settling time is
Ts= 4ωn= 4/6 seconds and satisfies the requirement. Notice that
Tsis not a function of K.
(b) The sensitivity is
ST
K(s) = 1
1 + G(s)=s(s+ 12)
s2+ 12s+ 120
when K= 10.
(c) The sensitivity at DC (s= 0) is
ST
K(0) = 0 .
(d) In this case, s=j2π·1 beat/sec = j2π. So, the sensitivity at s= 2πj
is
|ST
K(j2π)|=85.1084
110.31 = 0.77 .
P5.18 We select L(s) as
L(s) = 1
αs + 1 ,
then
H(s)
L(s)=6αs + 6
s3+ 6s2+ 11s+ 6 .
Therefore,
M(s) = 6αs + 6 ,
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202 CHAPTER 5 The Performance of Feedback Control Systems
and Mo(0) = 6, M1(0) = 6α,M2(0) = 0. Also,
∆(s) = s3+ 6s2+ 11s+ 6 ,
and ∆o(0) = 6 , ∆1(0) = 11 , 2(0) = 12. So, computing M2and ∆2
yields
M2= 36α2,and
2= 49 .
Finally, equating M2= ∆2yields 36α2= 49 ,or
α= 1.167 .
Thus,
L(s) = 1
1.167s+ 1 =0.857
s+ 0.857 .
P5.19 (a) The closed-loop transfer function is
T(s) = 8
s3+ 6s2+ 12s+ 8 .
(b) The second-order approximation is
L(s) = 1
d2s2+d1s+ 1 ,
where d1and d2are to be determined using the methods of Section
5.10 in Dorf & Bishop. Given
M(s) = 8d2s2+ 8d1s+ 8
∆(s) = s3+ 6s2+ 12s+ 8
we determine that
M2=128d2+ 64d2
1
M4= 64d2
2
2= 48
4= 18 .
Equating M2= ∆2and M4= ∆4, and solving for d1and d2yields
d1= 1.35 and d2= 0.53 .
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Problems 203
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Time (sec)
y(t)
FIGURE P5.19
Closed-loop system step response: Actual T(s)(solid line) and second-order approximation
(dashed line).
Thus, the second-order approximation is
L(s) = 1
0.53s2+ 1.35s+ 1 .
(c) The plot of the step response for the actual system and the approxi-
mate system is shown in Figure P5.19.
P5.20 The steady-state error is
ess = lim
s0
(s+ 5)(s+ 11) + K(1 K1)
(s+ 5)(s+ 11) + K=55 + K(1 K1)
55 + K.
To achieve a zero steady-state tracking error, select K1as follows:
K1= 1 + 55
K.
P5.21 The closed-loop transfer function is
T(s) = s+a
s2+ (2k+a)s+ 2ak + 1 .
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204 CHAPTER 5 The Performance of Feedback Control Systems
(a) If R(s) = 1/s, we have the tracking error
E(s) = R(s)Y(s) = [1 T(s)]R(s)
or
E(s) = s2+ (2k+a1)s+ 2ak + 1 a
s2+ (2k+a)s+ 2ak + 1 ·1
s.
From the final value theorem we obtain
ess = lim
s0sE(s) = 2ak + 1 a
2ak + 1 .
Selecting k= (a1)/(2a) leads to a zero steady-state error due to a
unit step input.
(b) To meet the percent overshoot specification we desire ζ0.69. From
T(s) we find ω2
n= 2ak + 1 and 2ζωn= 2k+a. Therefore, solving for
aand kyields
a= 1.5978 and k = 0.1871
when we select ζ= 0.78. We select ζ > 0.69 to account for the zero
in the closed-loop transfer function which will impact the percent
overshoot. With aand k, as chosen, we have
T(s) = s+ 1.598
s2+ 1.972s+ 1.598
and the step response yields P.O. 4%.
P5.22 The closed-loop transfer function is
T(s) = 2(2s+τ)
(s+ 0.2K)(2s+τ) + 4 .
(a) If R(s) = 1/s, we have the unit step response
Y(s) = 2(2s+τ)
(s+ 0.2K)(2s+τ) + 4
1
s.
From the final value theorem we obtain
yss = lim
s0sY (s) = 2τ
0.2Kτ + 4 .
Selecting K= 1020leads to yss = 1 and a zero steady-state error
due to a unit step input.
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Problems 205
(b) The characteristic equation is
(s+ 0.2K)(2s+τ) + 4 = 2s2+ (0.4K+τ)s+ 0.2Kτ + 4 = 0 .
So, with K= 10 20/τ , the natural frequency and damping ratio
are:
ωn=τand ζ=τ2+ 4τ8
4τ3/2.
The settling time and percent overshoot are found using the standard
design formulas
Tp=π
ωnp1ζ2and P.O. = 100eζπ1ζ2
with ωnand ζgiven above (as a function of τ). Since the closed-loop
system has a zero at s=τ /2, the formulas for Tpand P.O. will
only be approximate. Also, note that for the closed-loop system poles
to be in the left half-plane (that is, all the poles have negative real
parts), we require that τ > 2321.4642. As seen in the next
chapter, this is the condition for stability. Having τ > 232 insures
that the damping ratio ζis positive.
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206 CHAPTER 5 The Performance of Feedback Control Systems
Advanced Problems
AP5.1 (a) The steady-state error is
ess = lim
s0s(1 T(s))R(s) = 1 T(0) = 1 108(3)
9(36) = 0 .
(b) Assume the complex poles are dominant. Then, we compute
a
ζωn
= 0.75 ,
since a= 3, ζ= 0.67 and ωn= 6. Using Figure 5.13 in Dorf & Bishop,
we estimate the settling time and percent overshoot to be
P.O. = 45% and Ts=4
ζωn
= 1 second .
(c) The step response is shown in Figure AP5.1. The actual settling time
and percent overshoot are
P.O. = 34.4% and Ts= 1.18 second .
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2
0
0.2
0.4
0.6
0.8
1
1.2
1.4
Step Response
Time (sec)
Amplitude
FIGURE AP5.1
Closed-loop system step response.
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Advanced Problems 207
AP5.2 The closed-loop transfer function is
T(s) = 5440(τzs+ 1)
s3+ 28s2+ (432 + 5440τz)s+ 5440 .
The closed-loop step responses are shown in Figure AP5.2. The perfor-
mance results are summarized in Table AP5.2.
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6
0
0.2
0.4
0.6
0.8
1
1.2
1.4
Time (sec)
y(t)
tau=0 (solid) & tau=0.05 (dashed) & tau=0.1 (dotted) & tau=0.5 (dot-dash)
FIGURE AP5.2
Closed-loop system step responses.
τzTrTsP.O. closed-loop poles
0 0.16 0.89 32.7% p=20,4±16j
0.05 0.14 0.39 4.5% p=10.4,8.77 ±21.06j
0.1 0.10 0.49 0% p=6.5,10.74 ±26.84j
0.5 0.04 1.05 29.2% p=1.75,13.12 ±54.16j
TABLE AP5.2 Performance summary.
As τzincreases from 0 to 0.1, the P.O. decreases and the response is faster
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208 CHAPTER 5 The Performance of Feedback Control Systems
and more stable. However, as τzis increased beyond 0.1, the P.O. and Ts
increase, although Trcontinues to decrease.
AP5.3 The closed-loop transfer function is
T(s) = 1
τps3+ (1 + 2τp)s2+ 2s+ 1 .
The closed-loop step responses for τp= 0,0.5,2,5 are shown in Fig-
ure AP5.3. The performance results are summarized in Table AP5.3.
0 5 10 15 20 25 30 35 40 45 50
0
0.5
1
1.5
Time (sec)
y(t)
tau=5 (solid) & tau=2 (dotted) & tau=0.5 (dashed) & tau=0 (dot-dash)
FIGURE AP5.3
Closed-loop system step responses.
τpTrTsP.O. closed-loop poles
0 4 5.8 0% p=1,1
0.5 3.6 7.4 4.75% p=2.84,0.58 ±0.6j
2 4.6 22.4 27.7% p=2.14,0.18 ±0.45j
5 6 45.8 46% p=2.05,0.07 ±0.3j
TABLE AP5.3 Performance summary.
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Advanced Problems 209
As τpincreases, the P.O.,Trand Tsalso increase; adding the pole makes
the system less stable with more overshoot.
AP5.4 The system transfer function is
Y(s) = 15K
(s+ 5)(s+ 7) + 15KR(s) + 15
(s+ 5)(s+ 7) + 15KTd(s).
When considering the input response, we set Td(s) = 0, and similarly,
when considering the disturbance response, we set R(s) = 0. The closed-
loop step input and disturbance responses for K= 1,10,100 are shown in
Figure AP5.4. The performance results are summarized in Table AP5.4.
0 0.2 0.4 0.6 0.8 1
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
Time (sec)
y(t)
Unit step input response
0 0.2 0.4 0.6 0.8 1
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
Time (sec)
y(t)
Unit step distrubance response
FIGURE AP5.4
Closed-loop system input and disturbance responses (K=1: solid line, K=10: dotted line,
and K=100:dashed line).
K ess TsP.O. |y/d|max
1 0.7 0.45 0% 0.3
10 0.19 0.6 17.3% 0.1
100 0.023 0.59 60.0% 0.01
TABLE AP5.4 Performance summary.
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210 CHAPTER 5 The Performance of Feedback Control Systems
The best value of the gain is K= 10, which is compromise between (i)
percent overshoot, and (ii) disturbance rejection and tracking error.
AP5.5 The system transfer function is
Y(s) = 50(s+α)(s+ 2)
s(s+ 3)(s+ 4) + 50(s+α)(s+ 2)R(s)
+50s(s+ 2)
s(s+ 3)(s+ 4) + 50(s+α)(s+ 2)Td(s).
0 0.05 0.1 0.15 0.2 0.25
-1
0
1
2
3
4
5
6
7
8
9
10
Time (sec)
y(t)
Disturbance response: alpha=0 (solid) & alpha=10 (dashed) & alpha=100 (dotted)
FIGURE AP5.5
Closed-loop system disturbance response.
When considering the input response, we set Td(s) = 0, and similarly,
when considering the disturbance response, we set R(s) = 0. The steady-
state tracking error is
ess = lim
s0s(1 T(s))R(s) = lim
s0150(s+α)(s+ 2)
s(s+ 3)(s+ 4) + 50(s+α)(s+ 2) .
When α= 0, we have
ess = 1 100
100 + 12 = 0.11 ,
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Advanced Problems 211
and, for α6= 0
ess = 0 .
The closed-loop step input and disturbance responses for α= 0,10,100
are shown in Figure AP5.5. For disturbance rejection and steady-state
tracking error the best value of the parameter is
α= 100 .
However, when considering both the disturbance and input response we
would select the parameter
α= 10 ,
since it offers a good compromise between input response overshoot (about
5% for α= 10) and disturbance rejection/tracking error.
AP5.6 (a) The closed-loop transfer function is
T(s) = KKm
KKm+s(s+KmKb+ 0.01) .
The steady-state tracking error for a ramp input R(s) = 1/s2is
ess = lim
s0s(1 T(s))R(s)
= lim
s0
s+KmKb+ 0.01
KKm+s(s+KmKb+ 0.01)
=KmKb+ 0.01
KKm
.
(b) With
Km= 10
and
Kb= 0.05 ,
we have
KmKb+ 0.01
KKm
=10(0.05) + 0.01
10K= 1 .
Solving for Kyields
K= 0.051 .
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212 CHAPTER 5 The Performance of Feedback Control Systems
(c) The plot of the step and ramp responses are shown in Figure AP5.6.
The responses are acceptable.
0 2 4 6 8 10 12 14 16 18 20
0
0.2
0.4
0.6
0.8
1
1.2
1.4
Time (sec)
y(t)
Step input response
0 2 4 6 8 10 12 14 16 18 20
0
5
10
15
20
Time (sec)
y(t)
Ramp input response
FIGURE AP5.6
Closed-loop system step and ramp responses.
AP5.7 The performance is summarized in Table AP5.7 and shown in graphical
form in Fig. AP5.7.
KEstimated Percent Overshoot Actual Percent Overshoot
1000 8.8 % 8.5 %
2000 32.1 % 30.2 %
3000 50.0 % 46.6 %
4000 64.4 % 59.4 %
5000 76.4 % 69.9 %
TABLE AP5.7 Performance summary.
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Advanced Problems 213
1000 2000 3000 4000 5000
0
10
20
30
40
50
60
70
80
K
Percent Overshoot (% )
Actual P.O .
Estimated P.O.
FIGURE AP5.7
Percent overshoot versus K.
The closed-loop transfer function is
T(s) = 100K
s(s+ 50)(s+ 100) + 100K.
The impact of the third pole is more evident as Kgets larger as the
estimated and actual percent overshoot deviate in the range 0.3% at K=
1000 to 6.5% at K= 5000.
AP5.8 The closed-loop transfer function is
T(s) = K(s+ 2)
s2+ (2
3+K)s+1
3+ 2K.
Comparing T(s) to a second-order system we have
ωn=q1/3 + 2K
ζ=2/3 + K
2p1/3 + 2K
For the closed-loop transfer function to have complex roots, we require
K2(20/3)K(8/9) <0. This occurs when 0.13 K6.8. When
K= 1/3, we have the minimum ζ= 0.5, as shown in Figure AP5.8.
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214 CHAPTER 5 The Performance of Feedback Control Systems
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5
0.5
0.51
0.52
0.53
0.54
0.55
0.56
0.57
0.58
K
ζ
FIGURE AP5.8
Damping ratio, ζ, versus K.
AP5.9 The closed-loop characteristic equation is
s4+ 40s3+ 375s2+KPs+KI= 0.
The desired characteristic equation is
(s+a)(s+b)(s2+2ωns+ω2
n) = 0.
Expanding the desired characteristic equation and equating terms to the
actual characteristic equation yields
abω2
n=KI, ω2
n(a+b) + 2abωn=KP
2(a+b)ωn+ab = 375,2ωn+a+b= 40
This represents 4 equations with 5 unknowns (a,b,KP,KI, and ωn).
We can choose one variable as part of the controller design. Let KI=
0.1KP. Then, solving the 4 equations for the remaining 4 variables yields
a= 29.15, b= 0.1, KP= 1720, KI= 172, and ωn= 7.6. The resulting
Ts= 1.1s and P.O. = 6.4%, as shown in Figure AP5.9.
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Advanced Problems 215
Step Response
Time (sec)
Amplitude
0 1 2 3 4 5 6 7 8 9
0
0.2
0.4
0.6
0.8
1
1.2
1.4
System: sysa
Settling Time (sec): 1.09
System: sysa
Peak amplitude: 1.06
Overshoot (%): 6.39
At time (sec): 0.618
FIGURE AP5.9
Step response for KP= 1720 and KI= 172.
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216 CHAPTER 5 The Performance of Feedback Control Systems
Design Problems
The plant model with parameters given in Table CDP2.1 in Dorf andCDP5.1
Bishop is given by:
θ(s)
Va(s)=26.035
s(s+ 33.142) ,
where we neglect the motor inductance Lm. The closed-loop transfer func-
tion from the disturbance to the output is
θ(s)
Td(s)=26.035
s2+ 33.142s+ 26.035Ka
.
For a unit step disturbance input the steady-state response is
θss =1
Ka
.
Therefore, we want to use the maximum Kawhile keeping the percent
overshoot less than 5%. The step response for the closed-loop system (with
the tachometer not in the loop) and Ka= 22 is shown below. Values of
Kagreater than 22 lead to overshoots greater than 5%.
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
0
0.2
0.4
0.6
0.8
1
1.2
1.4
Time (sec)
q(t)/A
Step response
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
0
0.01
0.02
0.03
0.04
0.05
Time (sec)
q(t)
Unit disturbance response
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Design Problems 217
DP5.1 (a) The closed-loop transfer function is
φ(s)
φd(s)=12.2K
s(s+ 2.2)(s+ 7) + 12.2K=12.2K
s3+ 9.2s2+ 15.4s+ 12.2K.
(b) For K= 0.7, we have the characteristic equation
s3+ 9.2s2+ 15.4s+ 8.54 = 0 ,
with roots s1=7.23 and s2,3=0.98 ±0.46j. For K= 3, we have
the characteristic equation
s3+ 9.2s2+ 15.4s+ 36.6 = 0 ,
with roots s1=7.83 and s2,3=0.68 ±2.05j. And for K= 6, we
have the characteristic equation
s3+ 9.2s2+ 15.4s+ 73.2 = 0 ,
with roots s1=8.4 and s2,3=0.4±2.9j.
(c) Assuming the complex conjugate pair are the dominant roots, we
expect the following:
(i) for K= 0.7: P.O.=0.13% and Tp= 6.8 sec
(ii) for K= 3: P.O.=35.0% and Tp= 1.5 sec
(iii) for K= 6: P.O.=65.2% and Tp= 1.1 sec
(d),(e) We select
K= 1.71
to have a P.O. = 16% and Tp= 2.18sec. All four cases (K=
0.7,3,6,1.71) are shown in Figure DP5.1. In each case, the approxi-
mate transfer function is derived by neglecting the non-dominant real
pole and adjusting the gain for zero steady-state error. The approxi-
mate transfer functions are
TK=0.7(s) = 1.18
s2+ 1.965s+ 1.18 =0.7908
(s+ 0.98 + 0.46j)(s+ 0.98 0.46j)
TK=3(s) = 4.67
s2+ 1.37s+ 4.67 =3.299
(s+ 0.68 + 2.05j)(s+ 0.68 2.05j)
TK=6(s) = 8.71
s2+ 0.796s+ 8.71 =6.399
(s+ 0.4 + 2.9j)(s+ 0.42.9j)
TK=1.71(s) = 2.77
s2+ 1.679s+ 2.77 =1.458
(s+ 0.83 + 1.43j)(s+ 0.83 1.43j)
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218 CHAPTER 5 The Performance of Feedback Control Systems
0 5 10
0
0.5
1
1.5
time (sec)
phi
K=0.7
0 5 10
0
0.5
1
1.5
time (sec)
phi
K=3
0 5 10
0
0.5
1
1.5
2
time (sec)
phi
K=6
0 5 10
0
0.5
1
1.5
K=1.71
time (sec)
phi
FIGURE DP5.1
Step responses (actual response:solid lines; approximate response: dotted lines).
DP5.2 The closed-loop transfer function is
T(s) = Kω2
n
s3+ 2ζωns2+ω2
ns+Kω2
n
,
where ζ= 0.6. From the second-order system approximation, we have
Tp=π
ωnp1ζ2.
So, with ζ= 0.6 given, we should select ωn“large” to make Tp“small.”
Also, from the problem hint, let
0.2< K/ωn<0.4.
As a first attempt, we can select ωn= 10. See Figure DP5.8 for various
values of K/ωn. Our final selection is
K= 3.33 and ωn= 10 .
This results in P.O. = 3.6% and Tp= 0.66 second.
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Design Problems 219
0 0.5 1 1.5 2 2.5 3
0
0.2
0.4
0.6
0.8
1
1.2
1.4
K/ωn=0.2
K/ωn=0.4
K/ωn=0.33
Time (sec)
y(t)
FIGURE DP5.2
Closed-loop system response.
DP5.3 The closed-loop transfer function is
T(s) = K
s2+qs +K.
From the ITAE specification, we desire
T(s) = ω2
n
s2+ 1.4ωns+ω2
n
.
But
2ζωn= 1.4ωnwhich implies ζ= 0.7.
Since we want Ts0.5, we require ζωn8. So,
ωn8
0.7= 11.4.
We can select ωn= 12. Then,
T(s) = 144
s2+ 16.8s+ 144 .
Therefore, K= 144 and q= 16.8. The predicted percent overshoot is
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220 CHAPTER 5 The Performance of Feedback Control Systems
P.O. = 4.5%.
DP5.4 The loop transfer function is
Gc(s)G(s) = 10K
(s+ 70)(s+ 3)(s+ 7) =10K/70
(s/70 + 1)(s+ 3)(s+ 7) .
The second-order approximation is obtained by neglecting the fastest first-
order pole. Thus,
Gc(s)G(s)K/7
(s+ 3)(s+ 7) .
The closed-loop transfer function is
T(s) = K/7
s2+ 10s+ 21 + K/7.
When ζ0.52, we have less than 15% overshoot. So, we have
2ζωn= 10
and
ωn=q21 + K/7.
Eliminating ωnand solving for K(with P.O. 15%) yields
K500.19 .
Also,
Kp= lim
s0GGc(s) = K
7(21)
and
ess =1
1 + Kp
=1
1 + K
147
<0.12
implies
K1078 .
Therefore, we have an inconsistency. We require 1078 Kto meet the
steady-state requirement and K500.18 to meet the percent overshoot
requirement. It is not possible to meet both specifications.
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Design Problems 221
DP5.5 The closed-loop characteristic equation is
1 + K1G1(s) + K2G1G2(s) = 1 + K1
s(s+ 1) 2K2
s(s+ 1)(s+ 2) = 0
or
s3+ 3s2+ (2 + K1)s+ 2(K1K2) = 0 .
Assuming that K1>0 and K2>0, the range of the gains for stability is
0< K2< K1.
DP5.6 The closed-loop transfer function is
T(s) = K1
s2+ (K1K2+ 1)s+K1
.
The percent overshoot specification P.O. 2% is satisfied when ζ > 0.78.
The peak time specification Tp0.5 s is satisfied when ωn= 10 and
ζ= 0.78. So, given
K1=ω2
nand K1K2+ 1 = 2ζωn,
we determine that the specifications are satisfied when
K1= 100
and
K2= 0.15 .
DP5.7 The plant is
G(s) = 2
s(s+ 1)(s+ 4)
and the PD controller is
Gc(s) = KDs+KP.
The characteristic equation is
s3+ 6s2+ (8 + 2KD)s+ 2KP= 0.
The desired characteristic equation is
(s+a)(s2+ 2ζωns+ω2
n) = s3+ (2ζωn+a)s2+ (ω2
n+ 2ζωna)s+2
n= 0.
Equating the desired characteristic equation to the actual characteristic
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222 CHAPTER 5 The Performance of Feedback Control Systems
equation yields
2ζωn+a= 6, ω2
n+ 2ζωna= 8 + 2KD, aω2
n= 2KP,
where ζ= 0.69 and ωn= 3 to meet the design specifications. This repre-
sents 3 equations in 3 unknowns (a,KD, and KP). Solving yields a= 1.86,
KD= 4.35 and KP= 8.37. The step response is shown in Figure DP5.7.
Step Response
Time (sec)
Amplitude
0 0.5 1 1.5 2 2.5 3
0
0.2
0.4
0.6
0.8
1
1.2
1.4
System: sys_cl
Peak amplitude: 1.04
Overshoot (%): 4.28
At time (sec): 1.48
System: sys_cl
Settling Time (sec): 1.96
FIGURE DP5.7
Step response withKD= 4.35 and KP= 8.37.
DP5.8 The closed-loop transfer function is
T(s) = K
s2+ 6s+ 5 + K
The damping ratio and natural frequency is
ζ=3
K+ 5 and ωn=K+ 5
Using the design formulas for second-order systems we have
P O = 100eζπ/1ζ2and Ts=4
ζωn
.
We know that the formula for Tsis approximate and that the formulas
apply only to systems with ζ < 1. For K= 1 the closed-loop poles are
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Design Problems 223
both real, so there is no overshoot and the design formula for settling time
does not apply. Thus we obtain the results shown in Table DP5.8. We can
choose K= 10 as a good trade-off between percent overshoot, settling
time, and steady-state tracking error. The disturbance response is shown
in Figure DP5.8.
TABLE DP5.8 Step response for K=1, 10, and 20.
K % P.O. Ts, sec Ts, sec ess
Estimated Actual
1 0 - 3.24 0.83
10 2.13 1.33 1.38 0.33
20 9.48 1.33 1.19 0.20
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Time (sec)
y(t)
FIGURE DP5.8
Closed-loop system disturbance response for K= 10.
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224 CHAPTER 5 The Performance of Feedback Control Systems
Computer Problems
CP5.1 With the impulse input we have R(s) = 1. The transfer function is
Y(s) = 15
(s+ 3)(s+ 5)R(s) = 15
(s+ 3)(s+ 5) .
Therefore, taking the inverse Laplace transforms yields the output re-
sponse:
y(t) = 15
2e3t15
2e5t.
The impulse response and the analytic response is shown in Figure CP5.1.
n=15; d=[1 8 15];
t=[0:0.1:6];
ya=(15/2)*exp(-3.*t)-(15/2)*exp(-5.*t);
sys = tf(n,d)
y=impulse(sys,t);
plot(t,y,t,ya,'o')
xlabel('Time (sec)'), ylabel('y(t)'), legend('Computer','Analytic',-1)
0123456
0
0.2
0.4
0.6
0.8
1
1.2
1.4
Time (sec)
y(t)
Computer
Analytic
FIGURE CP5.1
Impulse responses.
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Computer Problems 225
CP5.2 The ramp response is shown in Figure CP5.2. The unity feedback system
is type 2, so that the steady-state tracking error is lim
t→∞ ess 0.
n=[1 10]; d=[1 15 0 0];
t=[0:0.1:50];
sys= tf(n,d);
sys_cl = feedback(sys,[1]);
u=t;
lsim(sys_cl,u,t);
0 5 10 15 20 25 30 35 40 45 50
0
10
20
30
40
50
60
Time (sec)
Amplitude
FIGURE CP5.2
Ramp responses.
CP5.3 The m-file script and the four plots are shown in Figure CP5.3. The plots
can be compared to Figure 5.17 in Dorf & Bishop.
-2
-1
0
1
2
0 5 10 15 20
wn=2, zeta=0
-2
-1
0
1
2
0 5 10 15 20
wn=2, zeta=0.1
-1
-0.5
0
0.5
1
0 5 10 15 20
wn=1, zeta=0
-0.5
0
0.5
1
0 5 10 15 20
wn=1, zeta=0.2
FIGURE CP5.3
Impulse responses.
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226 CHAPTER 5 The Performance of Feedback Control Systems
w1=2; z1=0;
w2=2; z2=0.1;
w3=1; z3=0;
w4=1; z4=0.2;
t=[0:0.1:20];
%
num1=[w1^2]; den1=[1 2*z1*w1 w1^2];
sys1 = tf(num1,den1);
[y1,x1]=impulse(sys1,t);
%
num2=[w2^2]; den2=[1 2*z2*w2 w2^2];
sys2 = tf(num2,den2);
[y2,x2]=impulse(sys2,t);
%
num3=[w3^2]; den3=[1 2*z3*w3 w3^2];
sys3 = tf(num3,den3);
[y3,x3]=impulse(sys3,t);
%
num4=[w4^2]; den4=[1 2*z4*w4 w4^2];
sys4 = tf(num4,den4);
[y4,x4]=impulse(sys4,t);
%
clf
subplot(221),plot(t,y1),title('wn=2, zeta=0')
subplot(222),plot(t,y2),title('wn=2, zeta=0.1')
subplot(223),plot(t,y3),title('wn=1, zeta=0')
subplot(224),plot(t,y4),title('wn=1, zeta=0.2')
FIGURE CP5.3
CONTINUED: Impulse response m-file script.
CP5.4 The closed-loop system is
T(s) = 21
s2+ 2s+ 21 .
Therefore, the natural frequency is
ωm=21 = 4.58
and the damping ratio is computed as
2ζωn= 2 ,
which implies
ζ= 0.218 .
The percent overshoot is estimated to be
P.O. = 100eζπ/1ζ2= 50% ,
since ζ= 0.218. The actual overshoot is shown in Figure CP5.4.
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Computer Problems 227
numc=[21]; denc=[1 0];
sysc = tf(numc,denc);
numg=[1]; deng=[1 2];
sysg = tf(numg,deng);
sys_o = series(sysc,sysg);
sys_cl = feedback(sys_o,[1])
step(sys_cl)
Time (sec.)
Amplitude
Step Response
0123456
0
0.5
1
1.5
From: U(1)
To: Y(1)
FIGURE CP5.4
Impulse responses.
CP5.5 For a step input, the optimum ITAE characteristic equation is
s3+ 1.75ωs2+ 2.15ω2s+ω3= 0 .
Examining Figure 5.30 for n=3 in Dorf & Bishop, we estimate that
ωTs= 8.
So, once we decide on the desired Tswe can estimate ω. For this design we
let Ts=8 so that ω= 1. Computing the desired characteristic equation and
the actual characteristic equation and comparing the coefficients leads to
the following relationships:
p= 1.75ω2ζωn
K= (2.15ω2ω2
n2ζωnp)2
n
z= (ω32
n)/(Kω2
n)
where ζ= 0.59 and ωn= 0.45. The controller and prefilter are
Gc(s) = 6.42s+ 0.58
s+ 1.22 and Gp(s) = 1
1.3s+ 0.75 .
The unit step response is shown in Figure CP5.5.
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228 CHAPTER 5 The Performance of Feedback Control Systems
wn=0.45; zeta=0.59
ng=wn^2; dg=[1 2*zeta*wn wn^2];
sysg=tf(ng,dg);
Ts=8; w=8/Ts;
p=1.75*w-2*zeta*wn;
K=(2.15*w^2-wn^2-2*zeta*wn*p)/wn^2;
z=(w^3-p*wn^2)/(K*wn^2);
nc=K*[1 z]; dc=[1 p]; sysc=tf(nc,dc);
sys=series(sysc,sysg);
syscl=feedback(sys,1);
[num,den]=tfdata(syscl,'v');
sysp=tf([den(end)],num);
step(syscl*sysp)
Step Response
Time (sec)
Amplitude
0 2 4 6 8 10 12
0
0.2
0.4
0.6
0.8
1
1.2
1.4
System: untitled1
Peak amplitude: 1.02
Overshoot (%): 1.98
At time (sec): 4.68
System: untitled1
Settling Time (sec): 7.54
FIGURE CP5.5
Closed-loop system step response m-file script.
CP5.6 The unit step response is shown in Figure CP5.6. The performance num-
bers are as follows: Mp= 1.16, Tp= 0.73, and Ts= 1.62.
numg=[25]; deng=[1 5 0];
sys = tf(numg,deng);
sys_cl = feedback(sys,[1]);
t=[0:0.01:2];
step(sys_cl,t);
Step Response
Time (sec)
Amplitude
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2
0
0.2
0.4
0.6
0.8
1
1.2
1.4 System: sys_cl
Peak amplitude: 1.16
Overshoot (%): 16.3
At time (sec): 0.73
System: sys_cl
Settling Time (sec): 1.62
FIGURE CP5.6
Closed-loop system step response m-file script.
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Computer Problems 229
CP5.7 The m-file script and the simulations are shown in Figure CP5.7.
% Part (a)
numc=[2]; denc=[1]; sys_c = tf(numc,denc);
nums=[-10]; dens=[1 10]; sys_s = tf(nums,dens);
numg=[-1 -5]; deng=[1 3.5 6 0]; sys_g = tf(numg,deng);
sysa = series(sys_c,sys_s);
sysb = series(sysa,sys_g);
sys = feedback(sysb,[1]);
f=0.5*pi/180; % Convert to rad/sec
t=[0:0.1:10]; u=f*t;
[y,x]=lsim(sys,u,t);(y(length(t),1)-u(1,length(t)))*180/pi
subplot(211)
plot(t,y*180/pi,t,u*180/pi,'--'), grid
xlabel('Time (sec)'),ylabel('theta')
title('Constant gain C(s) = 2: theta (solid) & input (dashed)')
% Part (b)
numc=[2 1]; denc=[1 0]; sys_c = tf(numc,denc);
[numa,dena]=series(numc,denc,nums,dens);
sysa = series(sys_c,sys_s);
sysb = series(sysa,sys_g);
sys = feedback(sysb,[1]);
[y,x]=lsim(sys,u,t);(y(length(t),1)-u(1,length(t)))*180/pi
subplot(212), plot(t,y*180/pi,t,u*180/pi,'--'), grid
xlabel('Time (sec)'),ylabel('theta')
title('PI controller C(s) = 2 + 1/s: theta (solid) & input (dashed)')
0 1 2 3 4 5 6 7 8 9 10
0
1
2
3
4
5
Time (sec)
theta
Constant gain C(s) = 2: theta (solid) & input (dashed)
0 1 2 3 4 5 6 7 8 9 10
0
1
2
3
4
5
6
Time (sec)
theta
PI controller C(s) = 2 + 1/s: theta (solid) & input (dashed)
FIGURE CP5.7
Closed-loop system response to a ramp input for two controllers.
For the constant gain controller, the attitude error after 10 seconds is
ess =0.3 deg. On the other hand, the PI controller has a zero steady-
state error ess = 0 deg. So, we can decrease the steady-state error by
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230 CHAPTER 5 The Performance of Feedback Control Systems
using a more sophisticated controller, in this case a PI controller versus a
constant gain controller.
CP5.8 The closed-loop characteristic equation is
s3+ 12s2+ 610s+ 500 = (s+ 0.8324)(s2+ 11.1676s+ 600.7027) = 0 .
The natural frequency and damping ratio of the complex roots are ωn=
24.5 and ζ= 0.23. From this we predict Mp= 1.48, Ts= 0.72, and
Tp= 0.13. The actual response is shown in Figure CP5.8. The differences
numg=[100 100]; deng=[1 2 100];
sysg = tf(numg,deng);
numc=[0.1 5]; denc=[1 0];
sysc = tf(numc,denc);
sys_o = series(sysg,sysc);
sys_cl = feedback(sys_o,[1])
t=[0:0.01:3];
step(sys_cl,t);
ylabel('theta dot')
Time (sec.)
theta dot
Step Response
0 0.5 1 1.5 2 2.5 3
0
0.2
0.4
0.6
0.8
1
1.2
1.4
From: U(1)
To: Y(1)
FIGURE CP5.8
Missile rate loop autopilot simulation.
can be explained by realizing that the system is not a second-order system.
The closed-loop system actually has two zeros, one real pole, and two
complex-conjugate poles:
T(s) = (s+ 50)(s+ 1)
(s+ 0.8324)(s2+ 11.1676s+ 600.7027) .
The effect of the pole at s=0.8324 is diminished by the zero at s=1.
The third pole and the zeros affect the overall response such that the
analytic formulas for second-order systems are not exact predictors of the
transient response.
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Computer Problems 231
CP5.9 Figure CP5.9 shows an m-file to compute the closed-loop transfer function
and to simulate and plot the step response.
numg=[10]; deng=[1 10]; sysg = tf(numg,deng);
numh=[0.5]; denh=[10 0.5]; sysh = tf(numh,denh);
sys = feedback(sysg,sysh)
step(sys);
Transfer function:
100 s + 5
---------------------
10 s^2 + 100.5 s + 10
Step Response
Time (sec)
Amplitude
0 10 20 30 40 50 60
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
System: sys
Settling Time (sec): 39.1
System: sys
Peak amplitude: 0.979
Overshoot (%): 95.7
At time (sec): 0.533
FIGURE CP5.9
M-file to compute the transfer function and to simulate the step response.
CP5.10 Figure CP5.10 shows an m-file to compute the closed-loop transfer func-
tion and to simulate and plot the ramp response. The steady-state error
numg=[10]; deng=[1 20 75 0];
sysg = tf(numg,deng);
sys = feedback(sysg,1)
t=[0:0.1:100];
u=t; % Unit ramp input
lsim(sys,u,t);
0 10 20 30 40 50 60 70 80 90 100
0
10
20
30
40
50
60
70
80
90
100
Linear Simulation Results
Time (sec)
Amplitude
FIGURE CP5.10
M-file to compute the transfer function and to simulate the ramp response.
is 7.5.
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232 CHAPTER 5 The Performance of Feedback Control Systems
CP5.11 Figure CP5.11 shows an m-file to compute the closed-loop transfer func-
tion and to simulate and plot the impulse, step, and ramp responses.
Notice that the closed-loop system is unstable.
numg=[1]; deng=[1 2 0]; sysg = tf(numg,deng);
numc=[0.5 2]; denc=[1 0]; sysc = tf(numc,denc);
syss=series(sysg,sysc);
sys = feedback(syss,1)
t=[0:0.1:20];
subplot(311)
impulse(sys,t);
subplot(312)
step(sys,t);
subplot(313)
u=t; % Unit ramp input
lsim(sys,u,t);
0 2 4 6 8 10 12 14 16 18 20
-10
0
10
Impulse Response
Time (sec)
Amplitude
0 2 4 6 8 10 12 14 16 18 20
-10
0
10
Step Response
Time (sec)
Amplitude
0 2 4 6 8 10 12 14 16 18 20
0
20
40
Linear Simulation Results
Time (sec)
Amplitude
FIGURE CP5.11
M-file to compute the transfer function and to simulate the ramp response.
CP5.12 Figure CP5.12 shows an m-file to simulate and plot the step response
for the original system and the 2nd-order approximation. For the orig-
inal system, we find Ts= 2.28 and P.O. = 80.6%. For the 2nd-order
approximation we nd Ts= 2.16 and P.O. = 101%
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Computer Problems 233
num=77*[1 2]; den=conv([1 7],[1 4 22]);
sys = tf(num,den)
na=(77/7)*[1 2]; da=[1 4 22]; sysa=tf(na,da);
t=[0:0.01:5];
y=step(sys,t);
ya=step(sysa,t);
plot(t,y,t,ya,'--')
xlabel('Time (s)'), ylabel('Step response')
012345
0
0.5
1
1.5
2
2.5
Time (s)
Step response
2nd order approximation
3rd order system response
FIGURE CP5.12
Step response.
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C H A P T E R 6
The Stability of Linear Feedback
Systems
Exercises
E6.1 The Routh array is
s31K+ 1
s2K6
s1b0
so6
where
b=K(K+ 1) 6
K.
For stability, we require K > 0 and b > 0. Therefore, using the condition
that b > 0, we obtain
K2+K6>0,
and solving for Kyields K > 2 and K < 3. We select K > 2, since we
also have the condition that K > 0.
E6.2 The Routh array is
s31 2
s210 30
s1-1 0
so30
The system is unstable since the first column shows two sign changes.
234
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Exercises 235
E6.3 The Routh array is
s41 32 20
s310 37
s228.3 20
s129.9
s020
By the Routh-Hurwitz criterion, the system is stable (i.e., all the numbers
in the first column are positive).
E6.4 The closed-loop transfer function is
T(s) = K(s2)
s3+ 5s2+ (4 K)s+ 2K.
Therefore, the characteristic equation is
s3+ 5s2+ (4 K)s+ 2K= 0 .
The corresponding Routh array is given by
s31 4 K
s25 2K
s1b0
so2K
where
b=5(4 K)2K
5=20 7K
5.
For stability we require K > 0 and b > 0 . Thus, the range of Kfor
stability is 0 < K < 20/7.
E6.5 The closed-loop transfer function is
T(s) = K
s3+ 10s2+ 27s+ 18 + K.
When K= 20, the roots of the characteristic polynomial are
s1,2=1.56 ±j1.76
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236 CHAPTER 6 The Stability of Linear Feedback Systems
and
s3=6.88 .
E6.6 When K= 252, the roots of the characteristic equation are on the imag-
inary axis. The roots are
s1,2=±j5.2 and s3=10 .
E6.7 (a) The closed-loop system characteristic equation is
1 + GH(s) = 1 + K(s+ 2)
s(s1) = 0 ,
or
s2+ (K1)s+ 2K= 0 .
We have the relationships ωn=2Kand 2ζωn=K1, where
ζ= 0.707. Thus,
21
22K=K1,
or
2
22
=K1
2K2
,
and
K26K+ 1 = 0 .
Solving for Kyields K= 5.83 and K= 0.17. However, for stability
we require K > 1 (from the Routh array), so we select K= 5.83.
(b) The two roots on the imaginary axis when K= 1 are s1,2=±j2.
E6.8 The closed-loop system characteristic equation is
3+ 20s2+ (100 + K)s+ 20K= 0 .
The corresponding Routh array is
s31 (100 + K)
s220 20K
s1b0
so20K
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Exercises 237
where
b=20(100 + K)20K
20 =20(100)
20 = 100 .
Therefore, the system is stable for all K > 0.
E6.9 The characteristic equation is
s3+ 2s2+ (K+ 1)s+ 8 = 0 ,
and the Routh array is given by
s31K+ 1
s22 8
s1b0
so8
where
b=2(K+ 1) 8
2=K3.
Setting b= 0, yields
K3 = 0 or K > 3.
E6.10 Stable with your eyes open and (generally) unstable with your eyes closed.
E6.11 The system is unstable. The poles are s1=5.66, s2=0.90 and s3,4=
0.28 ±j0.714.
E6.12 The characteristic equation is
s2+as +b= 0,
so, the Routh array is
s21b
s1a0
sob
The system is stable if and only if a > 0 and b > 0. For a second-order
system, a necessary and sufficient condition for stability is that all the
coefficients have the same sign.
E6.13 The characteristic equation is
s2+ (KD+ 2)s+ 4KP= 0.
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238 CHAPTER 6 The Stability of Linear Feedback Systems
The Routh array is
s21 4KP
s1KD+ 2 0
so4KP
The system is stable if and only if KP>0 and KD>2.
E6.14 The characteristic equation associated with the system matrix is
s3+ 3s2+ 5s+ 6 = 0 .
The roots of the characteristic equation are s1=2 and s2,3=5±j1.66.
The system is stable.
E6.15 The roots of q(s) are s1=4, s2=3, s3,4=1±j2 and s5,6=±j0.5.
The system is marginally stable. The Routh array is
s61 31.25 67.75 15
s59 61.25 14.75
s424.44 66.11 15
s331.909 9.2273 0
s260 15
s10 0
so
The auxillary equation is
60s2+ 15 = 0 .
Solving the auxillary equation yields two roots at s1,2=±j0.5. After
accounting for the row of zeros, the completed Routh array verifies that
the system has no poles in the right half-plane.
E6.16 The Routh array is
s41 45 50
s39 87
s235.33 50
s174.26 0
so50
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Exercises 239
The system is stable. The roots of q(s) are s1,2=3±j4, s3=2 and
s4=1.
E6.17 The characteristic equation is
s3+ 7s2+ 36s+ 24 = 0 .
The system is stable. The roots of the characteristic equation are s1=
0.77, s2,3=3.12 ±4.64i.
E6.18 The roots of q(s) are s1=20 and s2,3=±j2.24. The system is marginally
stable. The Routh array is
s31 5
s220 100
s10 0
so
The auxillary equation is
20s2+ 100 = 0 .
The roots are s=±j2.24. So, the system has roots at s=±j2.24.
Completing the Routh array (after accounting for the row of zeros) verifies
that no poles lie in the right half-plane.
E6.19 (a) Unstable.
(b) Stable.
(c) Stable.
E6.20 (a) The roots are s1,2=2 and s3=1.
(b) The roots are s1,2,3=3.
E6.21 The characteristic equation is
(sn2)3+ 10(sn2)2+ 29(sn2) + K= 0
or
s3
n+ 4s2
n+sn26 + K= 0 .
The Routh array is
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240 CHAPTER 6 The Stability of Linear Feedback Systems
s31 1
s24K26
s130K
40
soK26
If K= 30, then the auxillary equation is 4s2
n+ 4 = 0 or sn=±j.
Therefore, s=sn2 implies s=2±j.
E6.22 This system is not stable. The output response to a step input is a ramp
y(t) = kt.
E6.23 The characteristic polynomial is
s3+ 4s2+ks + 8 = 0 .
The Routh array is
s31k
s24 8
s14k8
4
so8
So, k > 2 for stability.
E6.24 The transfer function is
G(s) = C(sIA)1B+D
= [ 1 0 0 ]
s1 0
0s1
k k s +k
0
0
1
= [ 1 0 0 ]
s2+ks +k s +k1
k s2+ks s
ks ks k s2
0
0
1
1
∆(s)
where ∆(s) = s3+ks2+ks +k. Thus, the transfer function is
G(s) = 1
s3+ks2+ks +k.
The Routh array is
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Exercises 241
s31k
s2k k
s1k1
sok
For stability k > 1.
E6.25 The closed-loop transfer function is
T(s) = Ks + 1
s2(s+p) + Ks + 1 .
Therefore, the characteristic equation is
s3+ps2+Ks + 1 = 0 .
The Routh array is
s31 K
s2p 1
s1(pK 1)/p
so1
We see that the system is stable for any value of p > 0 and pK 1>0.
E6.26 The closed-loop transfer function is
T(s) = 10
2s2+ (K20)s+ 10 10K.
Therefore, the characteristic equation is
2s2+ (K20)s+ 10 10K= 0 .
The Routh array is
s22 10-10K
s1K20
so10 10K
We see that the system is stable for any value of K > 20 and any K < 1.
Therefore, the system is unstable for all K > 0 since the gain Kcannot
be simultaneously greater than 20 and less than 1.
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242 CHAPTER 6 The Stability of Linear Feedback Systems
Problems
P6.1 (a) Given
s2+ 5s+ 2 ,
we have the Routh array
s21 2
s15 0
so2
Each element in the first column is positive, thus the system is stable.
(b) Given
s3+ 4s2+ 8s+ 4 ,
we have the Routh array
s31 8
s24 4
s17 0
so4
Each element in the first column is positive, thus the system is stable.
(c) Given
s3+ 2s26s+ 20 ,
we determine by inspection that the system is unstable, since it is
necessary that all coefficients have the same sign. There are two roots
in the right half-plane.
(d) Given
s4+s3+ 2s2+ 12s+ 10 ,
we have the Routh array
s41 2 10
s31 12 0
s2-10 10 0
s113 0
so10
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Problems 243
There are two sign changes in the rst column, thus the system is
unstable with two roots in the right half-plane.
(e) Given
s4+s3+ 3s2+ 2s+K ,
we have the Routh array
s41 3 K
s31 2 0
s21K
s12K0
soK
Examining the first column, we determine that the system is stable
for 0 < K < 2.
(f) Given
s5+s4+ 2s3+s+ 6 ,
we know the system is unstable since the coefficient of the s2term is
missing. There are two roots in the right half-plane.
(g) Given
s5+s4+ 2s3+s2+s+K ,
we have the Routh array
s51 2 1
s41 1 K
s31 1 K
s2K K
s1K0
soK
Examining the first column, we determine that for stability we need
K > 0 and K < 0. Therefore the system is unstable for all K.
P6.2 (a) The closed-loop characteristic polynomial is
s4+ 27.88s3+ 366.4s2+ 1500s+ 1500ka= 0 .
The Routh array is
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244 CHAPTER 6 The Stability of Linear Feedback Systems
s41 366.4 1500ka
s327.88 1500
s2312.6 1500ka
s1b
so1500ka
where
b= 1500 133.78ka.
Examining the first column of the Routh array, we find that b > 0
and 1500ka>0 for stability. Thus,
0< ka<11.21 .
(b) With
Ts= 1.5 = 4
ζωn
,
we determine that
ζωn= 2.67 .
So, shift the axis by s=so2.67, and
(so2.67)4+ 27.88(so2.67)3+ 366.4(so2.67)2+ 1500(so2.67) +
1500ka=s4
o+ 17.2s3
o+ 185.85s2
o+ 63.55so1872.8 + 1500ka.
The Routh array is
s41 185.85 1500ka-1872.8
s317.2 63.55
s2182.16 1500ka-1872.8
s1b
so1500ka-1872.8
where
b= 240.38 141.63ka.
Examining the first column of the Routh array, we find that b > 0
and 1500ka1872.8>0. Thus, 1.25 < ka<1.69.
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Problems 245
P6.3 (a) Given
G(s) = K
(s+ 1)(s+ 2)(0.5s+ 1) ,
and
H(s) = 1
0.005s+ 1 ,
the closed-loop transfer function is
T(s) = K(0.005s+ 1)
0.0025s4+ 0.5125s3+ 2.52s2+ 4.01s+ 2 + K.
Therefore, the characteristic equation is
0.0025s4+ 0.5125s3+ 2.52s2+ 4.01s+ (2 + K) = 0 .
The Routh array is given by
s40.0025 2.52 2 + K
s30.5125 4.01 0
s22.50 2 + K
s13.60.205K0
so2 + K
Examining the first column, we determine that for stability we require
2< K < 17.6.
(b) Using K= 9, the roots of the characteristic equation are
s1=200 , s2,3=0.33 ±2.23j , and s4=4.35 .
Assuming the complex roots are dominant, we compute the damping
ratio ζ= 0.15. Therefore, we estimate the percent overshoot as
P.O. = 100eπζ/1ζ2= 62% .
The actual overshoot is 27%, so we see that assuming that the complex
poles are dominant does not lead to accurate predictions of the system
response.
P6.4 (a) The closed-loop characteristic equation is
1 + GH(s) = 1 + K(s+ 40)
s(s+ 10)(s+ 20) = 0 ,
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246 CHAPTER 6 The Stability of Linear Feedback Systems
or
s3+ 30s2+ 200s+Ks + 40K= 0 .
The Routh array is
s31 200 + K
s230 40K
s1200 K
30
so40K
Therefore, for stability we require 200 K/3>0 and 40K > 0. So,
the range of Kfor stability is
0< K < 600 .
(b) At K= 600, the auxilary equation is
30s2+ 40(600) = 0 or s2+ 800 = 0 .
The roots of the auxiliary equation are
s=±j28.3.
(c) Let K= 600/2 = 300. Then, to the shift the axis, first define so=
s+ 1. Substituting s=so1 into the characteristic equation yields
(so1)3+30(so1)2+500(so1)+12000 = s3
o+27s2
o+443so+11529 .
The Routh array is
s31 443
s227 11529
s116 0
so11529
All the elements of the first column are positive, therefore all the
roots lie to left of s=1. We repeat the procedure for s=so2
and obtain
s3
o+ 24s2
o+ 392so+ 10992 = 0 .
The Routh array is
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Problems 247
s31 392
s224 10992
s1-66 0
so10992
There are two sign changes in the first column indicating two roots
to right of s=2. Combining the results, we determine that there
are two roots located between s=1 and s=2. The roots of the
characteristic equation are
s1=27.6250 and s2,3=1.1875 ±20.8082j .
We see that indeed the two roots s2,3=1.1875±20.8082jlie between
-1 and -2.
P6.5 (a) Given the characteristic equation,
s3+ 3s2+ 4s+ 2 = 0 ,
we compute the roots s1=1, and s2,3=1±j.
(b) The roots of the characteristic equation
s4+ 9s3+ 30s2+ 42s+ 20 = 0
are s1=1, s2=2, and s3,4=3±j1.
(c) The roots of the characteristic equation
s3+ 19s2+ 110s+ 200 = 0
are s1=4, s2=5, and s3=10.
P6.6 (a) The characteristic equation is
1 + G(s) = 0 ,
or
s3+s2+ 10s+ 2 = 0 .
The roots are: s1=0.2033, and s2,3=0.3984 ±j3.1112.
(b) The characteristic equation is
s4+ 10s3+ 35s2+ 50s+ 24 = 0 .
The roots are s1=1, s2=2, s3=3, and s4=4.
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248 CHAPTER 6 The Stability of Linear Feedback Systems
(c) The characteristic equation is
s3+ 11s2+ 29s+ 6 = 0 .
The roots are s1=0.2258, s2=3.8206 and s3=6.9536.
P6.7 (a) The closed-loop characteristic equation is
s3+ 101s2+ (100 + 10KKa)s+ 100KKa= 0 .
The Routh array is
s31 100 + 10KKa
s2101 100KKa
s1b
so100KKa
where
b= 100 + 910
101 KKa>0.
Thus, examing the first column, we determine that KKa>0 stabi-
lizes the system.
(b) The tracking error is
e(s) = lim
s0s(1 T(s))100
s2=100
KKa
.
We require E(s)<1o= 0.01745. So,
KKa>100
0.01745 = 5729 .
When KKa= 5729, the roots of the characteristic polynomial are
s1=10.15 and s2,3=45.43 ±j233.25 .
P6.8 (a) The closed-loop characteristic equation is
1 + K
(0.5s+ 1)(s+ 1)(1
4s+ 1) = 0 ,
or
s3+ 7s2+ 14s+ 8(1 + K) = 0 .
The Routh array is
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Problems 249
s31 14
s27 8(1 + K)
s1b
so8(1 + K)
where
b=7(14) 8(1 + K)
7.
For stability, we require b > 0 and 8(1 + K)>0. Therefore, the range
of Kfor stability is
1< K < 11.25 .
(b) Let K= 11.25/3 = 3.75. Then, the closed-loop transfer function is
T(s) = 3.37
s3+ 7s2+ 14s+ 38 .
The settling time to a step input is Ts6 seconds.
(c) We want Ts= 4 sec, so
Ts= 4 = 4
ζωn
implies ζωn= 1 .
Our desired characteristic polynomial is
(s+b)(s2+ 2ζωns+ω2
n) = s3+ (2 + b)s2+ (ω2
n+ 2b)s+2
n
where we have used the fact that ζωn= 1 and ωnand bare to be
determined. Our actual characteristic polynomial is
s3+ 7s2+ 14s+ 8(1 + K) = 0 .
Comparing the coefficients of the actual and desired characteristic
polynomials, we find the following relationships:
2 + b= 7
ω2
n+ 2b= 14
2
n= 8(1 + K).
Solving these three equations yields
b= 5 , ωn= 2 and K= 1.5.
The actual settling time is Ts= 4.17 sec. This is not exactly our
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250 CHAPTER 6 The Stability of Linear Feedback Systems
desired Tssince we have the contribution of the additional pole at
s=5. The closed-loop poles are
s1=5 and s2,3=1±1.73j .
P6.9 (a) The closed-loop characteristic equation is
1 + GH(s) = 1 + 10K
(s+ 100)(s+ 20)2,
or
s3+ 140s2+ 4400s+ 40000 + 10K= 0 .
The Routh array is
s31 4400
s2140 40000 + 10K
s1b
so40000 + 10K
where
b=140(4400) (40000 + 10K)
140 .
Examining the first column and requiring all the terms to be positive,
we determine that the system is stable if
4000 < K < 57600 .
(b) The desired characteristic polynomial is
(s+b)(s2+1.38ωns+ω2
n) = s3+(1.38ωn+b)s2+(ω2
n+1.38ωnb)s+2
n
where we have used the fact that ζ= 0.69 to achieve a 5% over-
shoot, and ωnand bare to be determined. The actual characteristic
polynomial is
s3+ 140s2+ 4400s+ 40000 + 10K= 0 .
Equating the coefficients of the actual and desired characteristic poly-
nomials, and solving for K,b, and ωnyields
b= 104.2, ωn= 25.9 and K= 3003 .
So, a suitable gain is K= 3003.
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Problems 251
P6.10 (a) The closed-loop characteristic equation is
s4+ 7s3+ 20s2+ (24 + K)s+ 10K= 0 .
The Routh array is
s41 20 10K
s37 24 + K0
s2116K
710K
s1b
so10K
where
b=116K
7(24 + K)70K
116K
7.
Setting b > 0 yields
2784 398KK2>0,
which holds when
404.88 < K < 6.876 .
Examining the first column, we also find that K < 116 and K > 0
for stability. Combining all the stability regions, we determine that
for stability
0< K < 6.876 .
(b) When K= 6.876, the roots are
s1,2=3.5±1.63j , and s3,4=±2.1j .
P6.11 Given
s3+ (1 + K)s2+ 10s+ (5 + 15K) = 0 ,
the Routh array is
s31 10
s21 + K5 + 15K
s1b
so5 + 15K
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252 CHAPTER 6 The Stability of Linear Feedback Systems
where
b=(1 + K)10 (5 + 15K)
1 + K=55K
1 + K.
Given that K > 0, we determine that the system is stable when 55K > 0
or
0< K < 1.
When K= 1, the s2row yields the auxilary equation
2s2+ 20 = 0 .
The roots are s=±j10. So, the system frequency of oscillation is 10
rads/sec.
P6.12 The characteristic equation is
s3+as2+bs +c= 0,
so, the Routh array is
s31b
s2ac
s1abc
a
s0c
For the system to be stable, we require that a > 0, ab c > 0 and c > 0.
When a > 0 and c > 0, we know that b > 0. So, a necessary condition for
stability is that all coefficients a,b, and cbe positive. The necessary and
sufficient conditions for stability also require that b > c/a, in addition to
a > 0 and c > 0.
P6.13 The characteristic equation is
s3+ (p+ 2ζωn)s2+ (2ζωnp+Kω2
n)s+Kω2
nz= 0.
The conditions for stability (see P6.12) are p+ 2ζωn>0, 2ζωnp+Kω2
n>
(Kω2
nz)/(p+ 2ζωn), and Kω2
nz > 0. Since we know that K > 0, ζ > 0,
and ωn>0, it follows that for stability z > 0, p > 2ζωn, and
2ζωnp+Kω2
n>Kω2
nz
p+ 2ζωn
.
P6.14 The system has the roots
s1,2=±j , s3,4=±j , and s5,6=1±3j ,
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Problems 253
Therefore, the system is not stable since there are repeated roots on the
jω-axis.
P6.15 (a) Neglecting the zeros and poles, we have the characteristic equation
s4+ 30s3+ 325s2+ 2500s+K= 0 .
The Routh array is
s41 325 K
s330 2500 0
s2241.67 K
s1b
soK
where
b=604166.67 30K
241.67 .
Therefore, the system is stable for 0 < K < 20139.
(b) Without neglecting the zeros and poles, the closed-loop characteristic
equation is
s6+ 90s5+ 5525s4+ 12400s3+ (1255000 + K)s2
+ (8500000 + 30K)s+ 1125K= 0 .
This is stable for
0< K < 61818 .
We see that the additional poles and zero makes the system stable
for a much larger gain K.
P6.16 (a) The Routh array is
s31 5
s25 6
s13.8
so6
Examining the first column of the Routh array, we see no sign changes.
So, the system is stable.
(b) The roots of the system are s1=0.3246 and s2,3=2.3377 ±
3.6080j.
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254 CHAPTER 6 The Stability of Linear Feedback Systems
(c) The step response is shown in Figure P6.16.
0 5 10 15
0
0.02
0.04
0.06
0.08
0.1
0.12
0.14
0.16
0.18
Step Response
Time
(
sec
)
Amplitude
FIGURE P6.16
Unit step response.
P6.17 The closed-loop transfer function is
T(s) = K+ 1
s3+ 3s2+ 3s+K+ 1 .
The Routh array is
s31 3
s23K+ 1
s18K
3
soK+ 1
So, for stability we require 1< K < 8.
P6.18 The system characteristic equation is
s2+ (hk)s+ab kh = 0 .
For stability we require h > k and ab > kh. If k > h, the system is
unstable.
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Problems 255
P6.19 (a) The characteristic equation is
s3+ 9s2+ (K10)s+ 2K= 0 .
The Routh array is
s31K10
s29 2K
s17K90
9
so2K
For stability
K > 90/7.
(b) When K= 90/7, the system is marginally stable. The roots are
s1,2=±jq20/7,
at the jω-axis crossing.
P6.20 The closed-loop characteristic equation is
q(s) = s5+s4+ 9s3+Ks2+ 2Ks +K .
The range of stability for the vertical-liftoff vehicle is
5.177 < K < 7.823 .
Therefore, for K= 6, the system is stable. When K= 6 we have
q(s) = s5+s4+ 9s3+ 6s2+ 12s+ 6
The Routh array is
s51 9 12
s41 6 6
s33 6
s24 6
s11.5
so6
All entries in the first column are positive, so the system is stable.
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256 CHAPTER 6 The Stability of Linear Feedback Systems
P6.21 The state transition matrix is
Φ(t, 0) = 1
p2p1
(k2p1)ep1t(k2p2)ep2tep1tep2t
k1ep1t+k1ep2tp1ep1t+p2ep2t
where p1p2=k1and p1+p2=k2. We assume that p16=p2. In the case
when p1=p2, the state transition matrix will change, but the factors ep1t
and ep2twill remain. The eigenvalues of Aare given by the solution to
det |λIA|=λ2+k2λ+k1= 0 .
Therefore, the eigenvalues are λ1,2=k2/2±qk2
24k1. If k2>0 and
k1>0, then the eigenvalues are in the left half-plane, and the system is
stable. The transfer function is given by
G(s) = C(sIA)1B=s1
s2+k2s+k1
.
Therefore the characteristic equation is s2+k2s+k1= 0 and the poles
are s1,2=k2/2±qk2
24k1. If k2>0 and k1>0, then the poles are in
the left half-plane, and the system is stable. Notice that the values of λ1,2
and s1,2are the same. Also, the eigenvalues are the same as the values of
p1and p2. So, if the eigenvalues are negative, then the elements of the
state transition matrix will decay exponentially.
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Advanced Problems 257
Advanced Problems
AP6.1 The Routh array is
s41K1K2
s320 4
s220K14
20 K2
s1b0
soK2
where
b=20K14100K2
5K11.
For stability, we require K2>0, K1>0.2, and b > 0. Therefore, using
the condition that b > 0, we obtain
K2<0.2K10.04 .
The stability region is shown in Figure AP6.1.
0.2 0.4 0.6 0. 8 1 1.2 1.4 1.6 1.8 2
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
K1
K2
STABLE REGION
FIGURE AP6.1
Stability region.
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258 CHAPTER 6 The Stability of Linear Feedback Systems
AP6.2 The Routh array is
s41 30 K
s39K40
s2310K
9K
s1b0
soK
where
b=(310 K)(K40) 81K
310 K.
Therefore, using the condition that b > 0, we obtain the stability range
for K:
59.07 < K < 209.94 .
AP6.3 (a) The steady-state tracking error to a step input is
ess = lim
s0s(1 T(s))R(s) = 1 T(0) = 1 α .
We want
|1α|<0.05 .
This yields the bounds for α
0.95 < α < 1.05 .
(b) The Routh array is
s31α
s21 + α1
s1b0
so1
where
b=α2+α1
1 + α.
Therefore, using the condition that b > 0, we obtain the stability
range for α:
α > 0.618 .
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Advanced Problems 259
(c) Choosing α= 1 satisfies both the steady-state tracking requirement
and the stability requirement.
AP6.4 The closed-loop transfer function is
T(s) = K
s3+ (p+ 1)s2+ps +K.
The Routh array is
s31p
s21 + p K
s1b0
soK
where
b=p2+pK
1 + p.
Therefore, using the condition that b > 0, we obtain the the relationship
K < p2+p .
The plot of Kas a function of pis shown in Figure AP6.4.
0
20
40
60
80
100
120
0 1 2 3 4 5 6 7 8 9 10
p
K
STABLE REGION
FIGURE AP6.4
Stability region.
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260 CHAPTER 6 The Stability of Linear Feedback Systems
AP6.5 The closed-loop transfer function is
T(s) = 30K1K2
(s+ 1 + K1K3)(s10)(2s+K2K34) + 30K1K2K4
.
The Routh array is
s32a
s2b c
s1d0
soc
where a=9K2K3+ 16 + K1K2K2
324K1K3,b= 2K1K3+K2K322,
and c=10K2K3+ 40 10K1K2K2
3+ 40K1K3and d= (ab 2c)/b .
The conditions for stability are
2K1K3+K2K322 >0
10K2K3+ 40 10K1K2K2
3+ 40K1K3>0
2(10K2K3+ 40 10K1K2K2
3+ 40K1K3) + (9K2K3
+16 + K1K2K2
324K1K3)(2K1K3+K2K322) >0
Valid values for the various gains are: K1= 50, K2= 30, K3= 1, and
K4= 0.3. The step response is shown in Figure AP6.5.
0 5 10 15 20 25 30
0
50
100
150
200
250
300
350
Step Response
Time (sec)
Amplitude
FIGURE AP6.5
Stability region.
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Advanced Problems 261
AP6.6 The characteristic equation is
s3+ 7s2+ (KD+ 14)s+KP+ 8 = 0.
For stability we require that KP>8 and
KD>KP+ 8
714.
The relationship between KDand KPis shown in Figure AP6.6.
0 10 20 30 40 50 60 70 80 90 100
−14
−12
−10
−8
−6
−4
−2
0
2
KP
KD
UNSTABLE REGION
STABLE REGION
FIGURE AP6.6
Stability region.
AP6.7 The characteristic equation is
0.1s4+ 2.05s3+s2+ 8KPs+ 8KI= 0.
From the Routh array we find the conditions for stability are
0< KI<0.3125
1.2812 p1.6416 5.2531KI< KP<1.2812 + p1.6416 5.2531KI
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262 CHAPTER 6 The Stability of Linear Feedback Systems
Design Problems
The plant model with parameters given in Table CDP2.1 in Dorf andCDP6.1
Bishop is given by:
θ(s)
Va(s)=26.035
s(s+ 33.142) .
In the above transfer function we have neglected the motor inductance
Lm. The closed-loop transfer function from the input to the output is
θ(s)
R(s)=26.035Ka
s2+ 33.142s+ 26.035Ka
.
The Routh array is
s21 26.035Ka
s133.142 0
s026.035Ka
Stability is achieved for any 0 < Ka<.
DP6.1 The closed-loop characteristic polynomial is
s3+s2(5 + p+1
5K) + s(1
5Kp +1
5K+ 5p) + K= 0 .
(i) When p= 2, we have
s3+s2(7 + 1
5K) + s(10 + 3
5K) + K= 0 .
The Routh array is
s31 10 + 3
5K
s27 + K
5K
s1b
soK
where
b=(7 + K/5)(10 + 3K/5) K
7 + 1
5K.
When 32.98 < K < 17.69, we find that b > 0. Examining the
other terms in the first column of the array, we find that the system
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Design Problems 263
is stable for any K > 0.
(ii) When p= 0, we have
s3+s2(5 + 1
5K) + s(1
5K) + K= 0 .
The Routh array is
s311
5K
s25 + 1
5K K
s1b
soK
where
b=(5 + 1
5K)1
5KK
(5 + K/5) =K2/25
(5 + K/5) .
Again, examination of the first column reveals that any K > 0 results
in a stable system. So, we just need to select any K > 0; e.g. K= 10.
DP6.2 (a) The closed-loop characteristic equation is
1 + 20(Ks + 1)
s2(s+ 20) = 0 ,
or
s3+ 20s2+ 20Ks + 20 = 0 .
The Routh array is
s31 20K
s220 20
s1b
so1
where
b=20K1
1.
For stability, we require K > 0.05.
(b) The desired characteristic polynomial is
(s2+as +b)(s+ 5) = s3+s2(a+ 5) + s(5a+b) + 5b= 0 .
Equating coefficients with the actual characteristic equation we can
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264 CHAPTER 6 The Stability of Linear Feedback Systems
solve for a, b and K, yielding b= 4, a= 15, and
K=5a+b
20 =79
20 .
(c) The remaining two poles are s1=14.73 and s2=0.27.
(d) The step response is shown in Figure DP6.2.
0 2 4 6 8 10 12 14 16 18 20
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
time (sec)
y(t)
FIGURE DP6.2
Mars guided vehicle step response.
DP6.3 (a) The closed-loop characteristic equation is
2τs3+ (τ+ 2)s2+ (K+ 1)s+ 2K= 0 .
The Routh array is
s32τ K + 1
s2τ+ 2 2K
s1b
so2K
where
b=(τ+ 2)(K+ 1) 4Kτ
(τ+ 2) .
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Design Problems 265
Examining the first column of the Routh array, we determine that for
stability τ > 0, K > 0 and setting b > 0 yields the relationships:
(1) K < τ+ 2
3τ2when τ > 2
3(2) K > 0 when 0 < τ 2
3.
The plot of τversus Kis shown in Figure DP6.3a.
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
0 1 2 3 4 5 6 7
K
tau
STABLE REGION
FIGURE DP6.3
(a) The plot of τversus K.
(b) The steady-state error is
ess =A
Kv
,where Kv= 2K .
So,
ess
A=1
2K.
We require that ess 0.25A, therefore
K2.
One solution is to select τ= 0.5, then we remain within the stable
region.
(c) The step response is shown in Figure DP6.3b. The percent overshoot
is P.O. = 57%.
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266 CHAPTER 6 The Stability of Linear Feedback Systems
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
0 2 4 6 8 10 12 14 16 18 20
time (sec)
y(t)
P. O . = 56.77 %
FIGURE DP6.3
CONTINUED: (b) Closed-loop system step response.
DP6.4 (a) The closed-loop characteristic polynomial is
s3+Ks2+ [(2 + m)K1]s+ 2mK = 0 .
The Routh array is
s31 2K+mK 1
s2K2mK
s1b
so2mK
Examining the first column of the Routh array, we see that for sta-
bility we require m > 0, K > 0, and b > 0, where
b=(2K+mK 1)K2mK
K= (2 + m)K(1 + 2m)>0,
or
K > 1 + 2m
2 + m.
The plot of Kvs mis shown in Figure DP6.4a.
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Design Problems 267
0.4
0.6
0.8
1
1.2
1.4
1.6
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
m
K
STABLE REGION
FIGURE DP6.4
(a) The plot of Kversus m.
0 1 2 3 4 5 6 7 8 9 10
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
time (sec)
y(t)
P. O . = 64.3208 %
FIGURE DP6.4
CONTINUED: (b) Shuttle attitude control step response.
(b) The steady-state error is
ess
A=1
Kv
=1
2mK <0.10,
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268 CHAPTER 6 The Stability of Linear Feedback Systems
or mK > 5. For example, we can select m= 0.5 and K= 2.
(c) See Figure DP6.4b for the step response where P.O. = 64.3%.
DP6.5 The closed-loop transfer function is
T(s) = K
s3+ 10s2+ 20s+K.
The range of Kfor stability is 0 < K < 200. If we let K=Km/N where
Km= 200, then N= 6.25 results in a step response with P.O. = 15.7%
and Ts= 1.96 seconds.
DP6.6 The closed-loop system is given by
˙
x=
0 1
2K12K2
x+
0
1
r
The characteristic polynomial is s2+(2+K2)s+K12 = 0. So the system
is stable for K1>2 and K2>2. Selecting K=h10 2 iresults in
closed-loop eigenvalues at s=2±2j. The closed-loop step response has
a settling time of 2.11 s and a percent overshoot of 4.32%.
Re(s)
Im(s)
desired
region
for eigenvalues
ζωn = -1
sin-1 ζ = sin-1 0.69=43.63ο
FIGURE DP6.6
Performance region.
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Design Problems 269
DP6.7 (a) The inner loop closed-loop transfer function is
Y(s)
U(s)=20s
s3+ 10s2+ 20s+ 20K1
.
The Routh array is
s31 20
s2ω20K1
s120020K1
10
so20K1
For stability 0 < K1<10.
(b) The fastest response (that is, the quickest settling time) occurs when
K1= 2.2
(c) With K1= 2.2, the closed-loop transfer function is
Y(s)
R(s)=20K2s
s3+ 10s2+ (20 + 20K2)s+ 44 .
The Routh array is
s31 20(K2+ 1)
s210 44
s1200K2+156
10
so44
For stability, we require
200K2+ 156 >0.
Therefore, K2>0.78.
DP6.8 The closed-loop characteristic equation is
s2+ 4KDs+ 4(KP+ 1) = 0.
So, it is possible to find KPand KDto stabilize the system. For example,
any KP>0 and KD>0 leads to stability. Choosing KP9 results in a
steady-state tracking error less than 0.1 due to a unit step input. Then,
the damping ratio ζ=2/2 is achieved by selecting
KD=2KP+ 1
2.
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270 CHAPTER 6 The Stability of Linear Feedback Systems
Computer Problems
CP6.1 The m-file script is shown in Figure CP6.1.
ans =
-0.6063 + 2.7322i
-0.6063 - 2.7322i
-1.7874
ans =
-2.0004
-2.0000 + 0.0004i
-2.0000 - 0.0004i
-1.9996
ans =
0.2267 + 1.4677i
0.2267 - 1.4677i
-0.4534
pa=[1 3 10 14]; roots(pa)
pb=[1 8 24 32 16]; roots(pb)
pc=[1 0 2 1]; roots(pc)
FIGURE CP6.1
Computing the polynomial roots with the rootsfunction.
CP6.2 The m-file script is shown in Figure CP6.2.
ans =
-2.5000e-01 + 1.1990e+00i
-2.5000e-01 - 1.1990e+00i
ans =
0 + 1.2910e+00i
0 - 1.2910e+00i
ans =
2.5000e-01 + 1.3307e+00i
2.5000e-01 - 1.3307e+00i
K1=1;K2=2;K3=5; den=[1 2 1];
num1=K1*[1 -1 2];num2=K2*[1 -1 2];num3=K3*[1 -1 2];
sys1 = tf(num1,den); sys2 = tf(num2,den); sys3 = tf(num3,den);
sys1_cl=feedback(sys1,[1]);
sys2_cl=feedback(sys2,[1]);
sys3_cl=feedback(sys3,[1]);
p1 = pole(sys1_cl), p2 = pole(sys2_cl), p3 = pole(sys3_cl)
FIGURE CP6.2
K= 1 is stable;K= 2 is marginally stable; and K= 5 is unstable.
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Computer Problems 271
CP6.3 The closed-loop transfer function and the roots of the characteristic equa-
tion are shown in Figure CP6.3.
numg=[1 1]; deng=[1 4 6 10];
sysg = tf(numg,deng);
sys = feedback(sysg,[1])
r=pole(sys)
Transfer function:
s + 1
----------------------
s^3 + 4 s^2 + 7 s + 11
r =
-2.8946
-0.5527 + 1.8694i
-0.5527 - 1.8694i
FIGURE CP6.3
Closed-loop transfer function and roots.
CP6.4 There are no poles in the right half-plane, but the system is unstable
since there are multiple poles on the jω-axis at s=±jand s=±j(see
Figure CP6.4).
num=[1]; den=[1 2 2 4 1 2];
sys = tf(num,den);
pole(sys)
t = 0:0.1:100;
step(sys,t)
ans =
-2.0000
0.0000 + 1.0000i
0.0000 - 1.0000i
-0.0000 + 1.0000i
-0.0000 - 1.0000i
Time (sec.)
Amplitude
Step Response
0 10 20 30 40 50 60 70 80 90 100
-25
-20
-15
-10
-5
0
5
10
15
20
25
From: U(1)
To: Y(1)
FIGURE CP6.4
Unstable system step response.
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272 CHAPTER 6 The Stability of Linear Feedback Systems
CP6.5 The closed-loop system poles for the slow/fast pilots are shown in Fig-
ure CP6.5. The maximum allowable time delay is 0.2045seconds. At the
maximum allowable time delay, the system has roots on the jω-axis at
s=±2.6j. The slow pilot destabilizes the aircraft.
closed-loop
system poles
nume=[-10]; dene=[1 10]; syse = tf(nume,dene);
numg=[-1 -6]; deng=[1 3 6 0]; sysg = tf(numg,deng);
%
% Fast pilot
%
tau=0.1; tau1=2; K=1; tau2=0.5;
nump=-K*[tau1*tau tau-2*tau1 -2];
denp=[tau2*tau tau+2*tau2 2];
sysp = tf(nump,denp);
sysa = series(sysp,syse);
sysb = series(sysa, sysg);
sys = feedback(sysb,[1]);
fast_pilot=pole(sys)
%
% Slow pilot
%
tau=0.6; tau1=2; K=1; tau2=0.5;
nump=-K*[tau1*tau tau-2*tau1 -2];
denp=[tau2*tau tau+2*tau2 2];
sysp = tf(nump,denp);
sysa = series(sysp,syse);
sysb = series(sysa, sysg);
sys = feedback(sysb,[1]);
slow_pilot = pole(sys)
%
% Maximum pilot time delay, tau = 0.2045 sec
%
tau=0.2045; tau1=2; K=1; tau2=0.5;
nump=-K*[tau1*tau tau-2*tau1 -2];
denp=[tau2*tau tau+2*tau2 2];
sysp = tf(nump,denp);
sysa = series(sysp,syse);
sysb = series(sysa, sysg);
sys = feedback(sysb,[1]);
max_pilot_delay=pole(sys)
fast_pilot =
-19.6267
-10.7712
-3.8885
-0.1697 + 2.7880i
-0.1697 - 2.7880i
-0.3742
slow_pilot =
-9.4526
-4.5228 + 2.2595i
-4.5228 - 2.2595i
0.2793 + 2.0314i
0.2793 - 2.0314i
-0.3937
max_pilot_delay =
-10.0433 + 2.2684i
-10.0433 - 2.2684i
-4.3153
0.0001 + 2.6040i
0.0001 - 2.6040i
-0.3783
FIGURE CP6.5
Closed-loop system poles for an aircraft with a pilot in-the-loop.
CP6.6 The closed-loop transfer function is
T(s) = 1
s3+ 5s2+ (K3)s+K+ 1 .
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Computer Problems 273
Utilizing the Routh-Hurwitz approach, for stability we determine that
K > 4.
When K= 4, the roots of the characteristic equation are
s1=5 and s2,3=±j .
The m-file script which generates a plot of the roots of the characteristic
equation as a function of Kis shown in Figure CP6.6.
K=[0:0.1:5];
n=length(K);
for i=1:n
numg=[1]; deng=[1 5 K(i)-3 K(i)];
sys_o = tf(numg,deng);
sys_cl = feedback(sys_o,[1]);
p(:,i)=pole(sys_cl);
end
plot(real(p),imag(p),'x'), grid
text(-0.9,0.95,'K=4 -->'); text(-0.2,1.3,'K=5'); text(0,0.2,'K=0')
% From a Routh-Hurwitz analysis we nd that
% minimum K for stability is K=4
Kmax=4;
numg=[1]; deng=[1 5 Kmax-3 Kmax]; sysg = tf(numg,deng);
sys_cl = feedback(sysg,[1]); pole(sys_cl)
-6 -5 -4 -3 -2 -1 0 1
-1.5
-1
-0.5
0
0.5
1
1.5
K=4 -->
K=5
K=0
FIGURE CP6.6
Roots of the characteristic equation as a function of K, where 0< K < 5.
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274 CHAPTER 6 The Stability of Linear Feedback Systems
CP6.7 The characteristic equation is
p(s) = s3+ 10s2+ 14s+ 12 .
A=[0 1 0;0 0 1;-12 -14 -10]; b=[0;0;12]; c=[1 1 0]; d=[0];
sys = ss(A,b,c,d);
%
% Part (a)
%
p=poly(A)
%
% Part (b)
%
roots(p)
%
% Part (c)
%
step(sys)
p =
1.0000 10.0000 14.0000 12.0000
ans =
-8.5225
-0.7387 + 0.9286i
-0.7387 - 0.9286i
0 1 2 3 4 5 6 7 8 9
0
0.2
0.4
0.6
0.8
1
1.2
1.4
Step Response
Time (sec)
Amplitude
FIGURE CP6.7
Characteristic equation from the state-space representation using the poly function.
The roots of the characteristic equation are
s1=8.5225 and s2,3=0.7387 ±0.9286j .
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Computer Problems 275
The system is stable since all roots of the characteristic equation are in
the left half-plane. The unit step response and associated m-file script are
shown in Figure CP6.7.
CP6.8 The characteristic equation is
s3+ 10s2+ 10s+ 5K1= 0 .
(a) The Routh array is
s31 10
s210 5K1
s11005K1
10
so5K1
From the Routh-Hurwitz criterion, we obtain the limits 0 < K1<20
for stability.
(b) The plot of the pole locations is 0 < K1<30 is shown in Fig-
ure CP6.8. As seen in Figure CP6.8, when K1>20, the pole locations
move into the right half-plane.
?-12 ?-10 ?-8 ?-6 ?-4 ?-2 0 2
?-4
?-3
?-2
?-1
0
1
2
3
4
Root Locus
Real Axi s
Imaginary Axi s
k=20
FIGURE CP6.8
Pole locations for 0< K1<30.
CP6.9 (a) The characteristic equation is
s3+ 2s2+s+k4 = 0 .
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276 CHAPTER 6 The Stability of Linear Feedback Systems
The Routh array is
s31 1
s22k4
s16k
2
sok4
For stability, we obtain 4 < k < 6.
(b) The pole locations for 0 < k < 10 are shown in Figure CP6.9. We see
that for 0 < k < 4 the system is unstable. Similarly, for 6 < k < 10,
the system is unstable.
?-3 ?-2 ?-1 0 1
?-2
?-1.5
?-1
?-0.5
0
0.5
1
1.5
2
Root Locus
Real Axi s
Imaginary Axi s
pole locations when k=0
pole location when k=0
increasing k
increasing k
k=4
k=6
k=10
k=10
FIGURE CP6.9
Pole locations for 0< k < 10.
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C H A P T E R 7
The Root Locus Method
Exercises
E7.1 (a) For the characteristic equation
1 + Ks(s+ 4)
s2+ 2s+ 2 = 0 ,
the root locus is shown in Figure E7.1.
-4
-3
-2
-1
0
1
2
3
4
-4 -3 -2 -1 0 1 2 3 4
x
x
oo
Real Axis
Imag Axis
FIGURE E7.1
Root locus for 1 + Ks(s+4)
s2+2s+2 = 0.
277
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278 CHAPTER 7 The Root Locus Method
(b) The system characteristic equation can be written as
(1 + K)s2+ (2 + 4K)s+ 2 = 0 .
Solving for syields
s=(1 + 2K)
(1 + K)±p(2 + 4K)28(1 + K)
2(1 + K).
When
(2 + 4K)28(1 + K) = 0 ,
then we have two roots at s1,2=(1+2K)
1+K. Solving for Kyields K=
0.31.
(c) When K= 0.31, the roots are
s1,2=(1 + 0.62)
(1.31) =1.24 .
(d) When K= 0.31, the characterisitc equation is
s2+ 2.472s+ 1.528 = (s+ 1.24)2= 0 .
Thus, ωn= 1.24 and ζ= 1, the system is critically damped. The
settling time is Ts4 sec.
E7.2 (a) The root locus is shown in Figure E7.2. When K= 6.5, the roots of
the characteristic equation are
s1,2=2.65 ±j1.23 and s3,4=0.35 ±j0.8.
The real part of the dominant root is 8 times smaller than the other
two roots.
(b) The dominant roots are
(s+ 0.35 + j0.8)(s+ 0.35 j0.8) = s2+ 0.7s+ 0.7625 .
From this we determine that
ωn= 0.873 and ζ=0.7
2(0.873) = 0.40 .
Thus, the settling time is
Ts=4
ζωn
=4
0.35 = 11.43 sec .
The percent overshoot is P.O. =eπζ/1ζ2= 25.4%.
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Exercises 279
-4
-3
-2
-1
0
1
2
3
4
-4 -3 -2 -1 0 1 2 3 4
x
x
x
x
Real Axis
Imag Axis
*
*
*
*
* K=6.5
FIGURE E7.2
Root locus for 1 + K1
s(s+2)(s2+4s+5) = 0.
E7.3 The root locus is shown in Figure E7.3. The roots are s1=8.7, s2,3=
1.3±j2.2 when K= 7.35 and ζ= 0.5.
-4
-3
-2
-1
0
1
2
3
4
-10 -8 -6 -4 -2 0 2 4
xxx
o
o
Real Axis
Imag Axis
*
*
*
<----- K=7.35
zeta=0.5
FIGURE E7.3
Root locus for 1 + Ks2+4s+8
s2(s+4) = 0.
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280 CHAPTER 7 The Root Locus Method
E7.4 The root locus is shown in Figure E7.4.
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
-4 -3.5 -3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1
x
x
o
Real Axis
Imag Axis
FIGURE E7.4
Root locus for 1 + Ks+1
s2+4s+5 = 0.
The departure angles and entry points are
θd= 225o,225o
and
σb=2.4.
E7.5 (a) The root locus is in Figure E7.5. The breakaway points are
σb1=13.0, σb2=5.89 .
(b) The asymptote centroid is
σcent =18 ,
and
φasym =±90o.
(c) The gains are K1= 1.57 and K2= 2.14 at the breakaway points.
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Exercises 281
-15 -10 -5 0 5
-40
-30
-20
-10
0
10
20
30
40
Real Axis
Imag Axis
< asymptote< asymptote
FIGURE E7.5
Root locus for 1 + Ks2+2s+10
(s4+38s3+515s2+2950s+6000) = 0.
E7.6 The system is unstable for K > 75.
Real Axis
Imaginary Axis
−10 −8 −6 −4 −2 0 2
−10
−8
−6
−4
−2
0
2
4
6
8
10
System: sys
Gain: 75
Pole: −0.000981 + 8.66i
Damping: 0.000113
Overshoot (%): 100
Frequency (rad/sec): 8.66
FIGURE E7.6
Root locus for 1 + K15K
s(s2+15s+75) = 0.
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282 CHAPTER 7 The Root Locus Method
E7.7 The root locus is shown in Figure E7.7. The characteristic equation has
Real Axis
Imaginary Axis
−25 −20 −15 −10 −5 0 5 10
−20
−15
−10
−5
0
5
10
15
20
asymptote −−−>
System: sys
Gain: 27.3
Pole: −1.44 + 1.11i
Damping: 0.792
Overshoot (%): 1.7
Frequency (rad/sec): 1.81
FIGURE E7.7
Root locus for 1 + Ks+8
s(s+4)(s+6)(s+9) = 0.
4 poles and 1 zero. The asymptote angles are φ= +60o,60o,180o
centered at σcent =3.7. When K= 27.35 then ζ= 0.8 for the complex
roots.
E7.8 The characteristic equation is
1 + K(s+ 1)
s2(s+ 9) = 0 ,
or
s3+ 9s2+Ks +K= 0 .
For all the roots to be equal and real, we require
(s+r)3=s3+ 3rs2+ 3r2s+r3= 0 .
Equating terms and solving for Kyields K= 27. All three roots are equal
at s=3, when K= 27. The root locus is shown in Figure E7.8.
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Exercises 283
-8
-6
-4
-2
0
2
4
6
8
-15 -10 -5 0 5
o xx
3 roots at s=-3
Real Axis
Imag Axis
FIGURE E7.8
Root locus for 1 + Ks+1
s2(s+9) = 0.
E7.9 The characteristic equation is
1 + K1
s(s2+ 2s+ 5) = 0
or
s3+ 2s2+ 5s+K= 0 .
(a) The system has three poles at s= 0 and 1±j2. The number of
asymptotes is npnz= 3 centered at σcent =2/3, and the angles
are φasymp at ±60o,180o.
(b) The angle of departure, θd, is 90o+θd+116.6o= 180o, so θd=26.6o.
(c) The Routh array is
s31 5
s22K
s1b
soK
where b= 5 K/2. So, when K= 10 the roots lie on the imaginary
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284 CHAPTER 7 The Root Locus Method
axis. The auxilary equation is
2s2+ 10 = 0 which implies s1,2=±j5.
(d) The root locus is shown in Figure E7.9.
-4
-3
-2
-1
0
1
2
3
4
-4 -3 -2 -1 0 1 2 3 4
x
x
x
Real Axis
Imag Axis
asymptote --->
FIGURE E7.9
Root locus for 1 + K1
s(s2+2s+5) = 0.
E7.10 (a) The characteristic equation is
1 + K(s+ 2)
s(s+ 1) = 0 .
Therefore,
K=(s2+s)
(s+ 2) ,
and
dK
ds =s2+ 4s+ 2
(s+ 2)2= 0 .
Solving s2+4s+2 = 0 yields s=0.586 and 3.414. Thus, the system
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Exercises 285
breakaway and entry points are at s=0.586 and s=3.414.
(b) The desired characteristic polynomial is
(s+ 2 + aj)(s+ 2 aj) = s2+ 4s+ 4 + a2= 0 ,
where ais not specified. The actual characteristic polynomial is
s2+ (1 + K)s+ 2K= 0 .
Equating coefficients and solving for Kyields K= 3 and a=2.
Thus, when K= 3, the roots are s1,2=2±2j.
(c) The root locus is shown in Figure E7.10.
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
-4 -3.5 -3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1
xxo
Real Axis
Imag Axis
*
*
K=3, s=-2+1.414j
s=-0.58s=-3.41
FIGURE E7.10
Root locus for 1 + Ks+2
s(s+1) = 0.
E7.11 The root locus is shown in Figure E7.11 for the characteristic equation
1 + K(s+ 2.5)
(s2+ 2s+ 2)(s2+ 4s+ 5) = 0 .
From the root locus we see that we can only achieve ζ= 0.707 when
K= 0.
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286 CHAPTER 7 The Root Locus Method
-5
-4
-3
-2
-1
0
1
2
3
4
5
-5 -4 -3 -2 -1 0 1 2 3 4 5
x
x
x
x
o
Real Axis
Imag Axis
<---- zeta=0.707 & K=0
FIGURE E7.11
Root locus for 1 + K(s+2.5)
(s2+2s+2)(s2+4s+5) = 0.
E7.12 (a) The root locus is shown in Figure E7.12 for the characteristic equation
1 + K(s+ 1)
s(s2+ 6s+ 18) = 0 .
(b) The roots of the characteristic equation are
(i) K= 10: s1,2=2.8064 ±4.2368jand s3=0.3872
(ii) K= 20: s1,2=2.7134 ±5.2466jand s3=0.5732
(c) The step response performance of the system is summarized in Ta-
ble E7.12.
K10 20
Ts(sec) 9.0 5.5
P.O. 0 0
Tr(sec) 4.8 2.6
TABLE E7.12 System performance when K= 10 and K= 20.
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Exercises 287
−3.5 −3 −2.5 −2 −1.5 −1 −0.5 0 0.5
−15
−10
−5
0
5
10
15
Root Locus
Real Axis
Imaginary Axis
FIGURE E7.12
Root locus for 1 + K(s+1)
s(s2+6s+18) = 0.
E7.13 (a) The characteristic equation is
s(s+ 1)(s+ 3) + 4s+ 4z= 0 .
Rewriting with zas the parameter of interest yields
1 + z4
s(s+ 1)(s+ 3) + 4s= 0 .
The root locus is shown in Figure E7.13a.
(b) The root locations for
z= 0.6,2.0,and 4.0
are shown in Figure E7.13a. When z= 0.6, we have ζ= 0.76 and
ωn= 2.33. Therefore, the predicted step response is
P.O. = 2.4% and Ts= 2.3 sec (ζ= 0.6) .
When z= 2.0, we have ζ= 0.42 and ωn= 1.79. Therefore, the
predicted step response is
P.O. = 23% and Ts= 5.3 sec (ζ= 2.0) .
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288 CHAPTER 7 The Root Locus Method
-4
-3
-2
-1
0
1
2
3
4
-4 -3 -2 -1 0 1 2 3 4
x
x
x
Real Axis
Imag Axis
*
*
*
o
o
o
+
+
+
* z=0.6
o z=2.0
+ z=4.0
FIGURE E7.13
(a) Root locus for 1 + z4
s(s+1)(s+3)+4s= 0.
Finally, when z= 4.0, we have ζ= 0.15 and ωn= 2.19. Therefore,
the predicted step response is P.O. = 62% and Ts= 12 sec.
(c) The actual step responses are shown in Figure E7.13b.
0 2 4 6 8 10 12 14 16
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
time (sec)
y(t)
___ z=0.6
- - - z=2.0
..... z=4.0
FIGURE E7.13
CONTINUED: (b) Step Responses for z= 0.6,2.0,and 4.0.
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Exercises 289
E7.14 (a) The root locus is shown in Figure E7.14 for the characteristic equation
1 + K(s+ 10)
s(s+ 5) = 0 .
The breakaway point is sb=2.93; the entry point is se=17.1.
-10
-8
-6
-4
-2
0
2
4
6
8
10
-20 -15 -10 -5 0 5
xxo
Real Axis
Imag Axis
*
K=5, s=-5+5j
s=-2.93s=-17.1
FIGURE E7.14
Root locus for 1 + K(s+10)
s(s+5) = 0.
(b) We desire ζ= 1/2 = 0.707. So, the desired characteristic polyno-
mial is
s2+ 2 1
2ωns+ω2
n= 0 .
Comparing the desired characteristic polynomial to the actual we find
the relationships
ω2
n= 10Kand 2ωn= 5 + K .
Solving for Kand ωnyields K= 5 and ωn= 7.07. The roots are
s1,2=5±j5 when K= 5.
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290 CHAPTER 7 The Root Locus Method
E7.15 (a) The characteristic equation
1 + K(s+ 10)(s+ 2)
s3= 0
has the root locus in Figure E7.15.
?-25 ?-20 ?-15 ?-10 -5 0 5
?-15
?-10
?-5
0
5
10
15
K=1.67
Root Locus
Real Axi s
Imaginary Axi s
FIGURE E7.15
Root locus for 1 + K(s+10)(s+2)
s3= 0.
(b) The Routh array is
s31 12K
s2K20K
s1b
so20K
when b= 12K20. For stability, we require all elements in the first
column to be positive. Therefore,
K > 1.67 .
(c) When K > 3/4, we have
ess = lim
s0sE(s) = lim
s0s1
1 + GH(s)·1
s2= lim
s0
s2
s3+K(s+ 1)(s+ 3) = 0 .
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Exercises 291
E7.16 The expansion for eT s is
eT s = 1 T s +(T s)2
2! ...
If (T s)<< 1, then
eT s 1T s =a+bs
c+ds ,
where a, b, c and dare constants to be determined. Using long division,
-40
-30
-20
-10
0
10
20
30
40
-40 -30 -20 -10 0 10 20 30 40 50 60
x x o
Real Axis
Imag Axis
*
K=21
FIGURE E7.16
Root locus for 1 + K(20s)
(s+1)(20+s)= 0.
we expand (a+bs)/(c+ds) and match as many coefficients as possible. In
this case, we determine that a=c= (2/T ) and also that b=d=1.
In this case, with T= 0.1, we have
eT s =20 s
20 + s=(s20)
(s+ 20) .
So, the characteristic equation is
1 + K(s20)
(s+ 1)(s+ 20) ,
and the root locus is shown in Figure E7.16. Using a Routh-Hurwitz
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292 CHAPTER 7 The Root Locus Method
analysis with the characteristic polynomial
s2+ (21 K)s+ 20 + 20K= 0 ,
we determine that the system is stable for 1< K < 21.
E7.17 (a) The root locus is in Figure E7.17a.
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
-2 -1.5 -1 -0.5 0 0.5 1 1.5 2
x x
Real Axis
Imag Axis
FIGURE E7.17
(a) Root locus for 1 + K
s(s1) = 0.
The root locus is always in the right half-plane; the system is unstable
for K > 0.
(b) The characteristic equation is
1 + K(s+ 2)
s(s1)(s+ 20) = 0 ,
and the root locus is shown in Figure E7.17b. The system is stable
for K > 22.3 and when K= 22.3, the roots are
s1,2=±j1.53 and s3=19 .
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Exercises 293
-10
-8
-6
-4
-2
0
2
4
6
8
10
-30 -25 -20 -15 -10 -5 0 5 10
x xx o
Real Axis
Imag Axis
*K=22.3
FIGURE E7.17
CONTINUED: (b) Root locus for 1 + K(s+2)
s(s+20)(s1) = 0.
E7.18 The root locus is shown in Figure E7.18.
-6
-4
-2
0
2
4
6
-6 -4 -2 0 2 4 6
x
x
x
x
Real Axis
Imag Axis
K=8.15
+
+
+
+
FIGURE E7.18
Root locus for 1 + K
s(s+3)(s2+2s+2) = 0.
When K= 8.15, the roots are s1,2=±j1.095 and s3,4=2.5±j0.74.
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294 CHAPTER 7 The Root Locus Method
E7.19 The characteristic equation is
1 + K
s(s+ 3)(s2+ 6s+ 64) = 0 ,
and the root locus is shown in Figure E7.19. When K= 1292.5, the roots
are
s1,2=±j4.62 and s3,4=4.49 ±j6.36 .
-15
-10
-5
0
5
10
15
-15 -10 -5 0 5 10 15
x
x
x
x
Real Axis
Imag Axis
K=1292.5
+
+
+
+
FIGURE E7.19
Root locus for 1 + K
s(s+3)(s2+6s+64) = 0.
E7.20 The characteristic equation is
1 + K(s+ 1)
s(s2)(s+ 6) = 0 ,
and the root locus is shown in Figure E7.20. The system is stable for
K > 16 .
The maximum damping ratio of the stable complex roots is
ζ= 0.25 .
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Exercises 295
0 1 2 3
0
5
10
15
20
Real Axis
Imaginary Axis
ζmax = 0.25
FIGURE E7.20
Root locus for 1 + K(s+1)
s(s2)(s+6) = 0.
E7.21 The gain is K= 10.8 when the complex roots have ζ= 0.66.
-10
-5
0
5
10
-10 -5 0 5 10
x
x
x
o
Real Axis
Imag Axis
K=10.8
+
+
+
FIGURE E7.21
Root locus for 1 + Ks
s3+5s2+10 = 0.
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296 CHAPTER 7 The Root Locus Method
E7.22 The root locus is shown in Figure E7.22. The characteristic equation is
1 + K(s2+ 18)(s+ 2)
(s22)(s+ 12) = 0 .
−14 −12 −10 −8 −6 −4 −2 0 2
−5
−4
−3
−2
−1
0
1
2
3
4
5
Root Locus
Real Axis
Imaginary Axis
FIGURE E7.22
Root locus for 1 + K(s2+18)(s+2)
(s22)(s+12) = 0.
E7.23 The characteristic equation is
5s2+as + 4 = 0 ,
which can rewritten as
1 + as
5s2+ 4 = 0 .
The roots locus (with aas the parameter) is shown in Figure E7.23.
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Exercises 297
-1.5
-1
-0.5
0
0.5
1
1.5
-1.5 -1 -0.5 0 0.5 1 1.5
x
x
o
Real Axis
Imag Axis
FIGURE E7.23
Root locus for 1 + as
5s2+4 = 0.
E7.24 The transfer function is
G(s) = C(sIA)1B+D
= [ 1 0 ]
s1
4s+k
1
0
1
=1
s2+ks + 4 .
Therefore, the characteristic equation is
s2+ks + 4 = 0 ,
or
1 + ks
s2+ 4 = 0 .
The root locus for 0 < k < is shown in Figure E7.24. The closed-loop
system is stable for all 0 < k < .
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298 CHAPTER 7 The Root Locus Method
5 5 5 5 0 0.5
5
5
5
0
0.5
1
1.5
2
2.5
Real Axis
Imaginary Axis
FIGURE E7.24
Root locus for 1 + ks
s2+4 = 0.
E7.25 The characteristic equation is
1 + K10
s(s+ 25) = 0 .
The root locus shown in Figure E7.25 is stable for all 0 < K < .
0 5
0
5
10
15
Real Axis
Imaginary Axis
FIGURE E7.25
Root locus for 1 + K10
s(s+25) = 0.
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Exercises 299
E7.26 The characteristic polynomial is
det
s1
s+K3s+K+ 2
= 0
or
1 + Ks+ 1
s2+ 2s3= 0 .
The root locus shown in Figure E7.26 is stable for all 0 < K < 3.
−12 −10 −8 −6 −4 −2 0 2
−0.8
−0.6
−0.4
−0.2
0
0.2
0.4
0.6
0.8
Root Locus
Real Axis
Imaginary Axis
FIGURE E7.26
Root locus for 1 + Ks+1
s2+2s3= 0.
E7.27 The characteristic equation is
1 + ps
s2+ 4s+ 40 = 0 .
The root locus shown in Figure E7.27 is stable for all 0 < p < .
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300 CHAPTER 7 The Root Locus Method
0 2
0
2
4
6
8
Real Axis
Imaginary Axis
FIGURE E7.27
Root locus for 1 + ps
s2+4s+40 = 0.
E7.28 The characteristic equation is
1 + Ks1
s(s2+ 2s+ 2) = 0 .
The system is stable for 1.33 < K < 0.
0 2 4 6 8
5
5
0
0.5
1
1.5
Real Axis
Imaginary Axis
FIGURE E7.28
Root locus for 1 + Ks1
s(s2+2s+2) = 0.
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Problems 301
Problems
P7.1
−50 −40 −30 −20 −10 0 10 20
−30
−20
−10
0
10
20
30
Root Locus
Real Axis
Imaginary Axis
−7 −6 −5 −4 −3 −2 −1 0 1 2
−5
−4
−3
−2
−1
0
1
2
3
4
5
Root Locus
Real Axis
Imaginary Axis
FIGURE P7.1
(a) Root locus for 1 + K
s(s+10)(s+8) = 0, and (b) 1 + K
(s2+2s+2)(s+1) = 0.
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302 CHAPTER 7 The Root Locus Method
−12 −10 −8 −6 −4 −2 0 2
−40
−30
−20
−10
0
10
20
30
40
Root Locus
Real Axis
Imaginary Axis
−4 −3.5 −3 −2.5 −2 −1.5 −1 −0.5 0 0.5
−4
−3
−2
−1
0
1
2
3
4
Root Locus
Real Axis
Imaginary Axis
FIGURE P7.1
CONTINUED: (c) Root locus for 1 + K(s+5)
s(s+2)(s+7) = 0, and (d)1 + K(s2+4s+8)
s2(s+7) = 0.
P7.2 The root locus is shown in Figure P7.2 for the characteristic equation
1 + 10Kv(s+ 10)
s(s+ 1)(s+ 100) = 0 .
The damping ratio is ζ= 0.6 when Kv= 0.8, 135 and 648. The roots of
the characteristic equation are:
(a) Kv= 0.8 : s1=99.9, s2,3=0.54 ±j0.71
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Problems 303
(b) Kv= 135 : s1=85.9, s2,3=7.5±j10
(c) Kv= 648 : s1=11.7, s2,3=44.6±j59.5
-30
-20
-10
0
10
20
30
-100 -80 -60 -40 -20 0
xxx o
Real Axis
Imag Axis
FIGURE P7.2
Root locus for 1 + 10Kv(s+10)
s(s+1)(s+100) = 0.
P7.3 (a) The breakaway point is s=0.88 at K= 4.06.
(b) The characteristic equation can be written as
s(s+ 2)(s+ 5) + K= 0 .
The Routh array is
s31 10
s27K
s1b0
soK
where
b=70 K
7.
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304 CHAPTER 7 The Root Locus Method
When K= 70, the system has roots on jω-axis at s=±j10.
(c) When K= 6, the roots are s1,2=0.83 ±j0.66, s3=5.34.
(d) The characteristic equation
1 + K
s(s+ 2)(s+ 5) = 0
has the root locus shown in Figure P7.3.
-10 -8 -6 -4 -2 0 2 4 6 8 10
-10
-8
-6
-4
-2
0
2
4
6
8
10
Real Axis
Imag Axis
FIGURE P7.3
Root locus for 1 + K
s(s+2)(s+5) = 0.
P7.4 The characteristic equation for the large antenna is
1 + G1G(s) = 1 + 100ka
(0.1s+ 1)(s2+ 14.4s+ 100) = 0 ,
or
1 + 1000ka
(s+ 10)(s2+ 14.4s+ 100) = 0 .
The root locus is shown in Figure P7.4. Using Routh’s criteria, we find
that the system is stable for
1< ka<4.83 .
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Problems 305
-20
-15
-10
-5
0
5
10
15
20
-20 -15 -10 -5 0 5 10 15 20
x
x
x
Real Axis
Imag Axis
*
*
<-- K=4.827
FIGURE P7.4
Root locus for 1 + 1000ka
(s2+14.14s+100)(s+10) = 0.
When ka= 4.83, we have s1,2=±j15.53.
P7.5 (a) The characteristic equation for hands-off control is
1 + 25K2(s+ 0.03)(s+ 1)
(s+ 0.4)(s20.36s+ 0.16)(s+ 9) = 0 .
The root locus is shown in Figure P7.5a. The damping ratio is ζ=
0.707 when K2= 1.6 or K2= 0.74.
(b) The transfer function from Td(s) to Y(s) is
Y(s) = G2(s)Td(s)
1 + G2(s)Gf(s),
where
Gf(s) = K2(s+ 1)
s+ 9 .
Using the final value theorem, we determine that
yss = lim
s0sG2(s)
1 + G2(s)Gf(s)
1
s=11.7
1 + 11.7K2
9= 3.8,
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306 CHAPTER 7 The Root Locus Method
-5
-4
-3
-2
-1
0
1
2
3
4
5
-10 -8 -6 -4 -2 0 2 4
o ox x
x
x
Real Axis
Imag Axis
<-- K=1.6
<-- K=0.74
FIGURE P7.5
(a) Root locus for 1 + 25K2(s+0.03)(s+1)
(s+9)(s20.36s+0.16)(s+0.4) = 0.
-20
-15
-10
-5
0
5
10
15
20
-20 -15 -10 -5 0 5 10 15 20
x
x
x
x xxo o
Real Axis
Imag Axis
FIGURE P7.5
CONTINUED: (b) Root locus for 1 + 25K1(s+0.03)(s+9)
(s+0.045)(s2+12s+1)(s+1.33)(s2+7.66s+29.78) = 0.
where we have selected K2= 1.6. For K2= 0.74, we find that yss =
5.96.
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Problems 307
(c) The closed-loop characteristic equation with the pilot loop added is
1 + 25K1(s+ 0.03)(s+ 9)
(s+ 0.045)(s+ 1.33)(s2+ 7.66s+ 29.78)(s2+ 12s+ 1) = 0 .
The root locus is shown in Figure P7.5b.
(d) Using K1= 2, we determine that
ess = 0.44 .
P7.6 (a) The characteristic equation is
1 + K(s+ 0.20)(s2+ 4s+ 6.25)
(s+ 0.9)(s0.6)(s0.1)(s+ 4) = 0 .
The root locus is shown in Figure P7.6.
-4
-3
-2
-1
0
1
2
3
4
-6 -5 -4 -3 -2 -1 0 1 2
x x xx
o
o
o
Real Axis
Imag Axis
zeta=0.5
zeta*wn=-1/3
K=4 -->
FIGURE P7.6
Root locus for 1 + K(s+0.2)(s2+4s+6.25)
(s+0.9)(s0.6)(s0.1)(s+4) = 0.
(b) For Ts<12 sec, we require ζωn>1/3. Also, we want ζ > 0.5. So,
we seek roots for a stable system with ζωn>1/3 and ζ > 0.5. This
occurs when K > 4.
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308 CHAPTER 7 The Root Locus Method
P7.7 (a) The characteristic equation for the speed control system is
1 + K
(s+ 4)2(s+δ)= 0 ,
where
K=0.004
Rand δ=0.75
4000 = 0.0001875 .
The root locus is shown in Figure P7.7. At ζ= 0.6, we have K= 19.1,
-6
-4
-2
0
2
4
6
-6 -5 -4 -3 -2 -1 0 1 2 3 4
xx x
Real Axis
Imag Axis
<-- K=19.1
FIGURE P7.7
Root locus for 1 + K
(s+4)2(s+1.875e04) = 0.
therefore
R= 0.00021 .
When K= 19.1 the roots are
s1,2=1.1±j1.43 and s3=5.80 .
(b) The steady-state error is
lim
s0sω(s) = lim
s0s(0.25s+ 1)2
(0.25s+ 1)2(Js +b) + 1/RL(s)
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Problems 309
=1
b+ 1/RLLR ,
when R < 0.1.
P7.8 (a) The characteristic equation for the speed control system with the
hydroturbine is
1 + K(s+ 1)
(s+ 4)(s+ 2)(s+δ)= 0 ,
where
K=0.002
Rand δ=0.75
4000 = 0.0001875 .
The root locus is shown in Figure P7.8. At ζ= 0.6, we have K= 2.85,
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
-6 -5 -4 -3 -2 -1 0 1 2 3 4
x x x o
Real Axis
Imag Axis
K=2.85 -->
FIGURE P7.8
Root locus for 1 + K(s+1)
(s+4)(s+2)(s+δ)= 0.
therefore
R= 0.0007 .
When K= 2.85 the roots are 0.45 ±j0.60, and -5.1.
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310 CHAPTER 7 The Root Locus Method
(b) The steady-state error is
lim
s0sω(s) = lim
s0s(0.25s+ 1)(0.5s+ 1)
(0.25s+ 1)(0.5s+ 1)(Js +f) + (s+ 1)/R L(s)
=1
f+ 1/RLLR ,
when R < 0.1.
P7.9 The characteristic equation is
1 + K(s+ 0.5)(s+ 0.1)(s2+ 2s+ 289)
s(s+ 30)2(s0.4)(s+ 0.8)(s2+ 1.45s+ 361) = 0
where K=K1K2. The root locus is shown in Figure P7.9. When
K= 4000 ,
the roots are
s1,2=0.82 ±j19.4
-35 -30 -25 -20 -15 -10 -5 0 5
-50
-40
-30
-20
-10
0
10
20
30
40
50
Real Axis
Imag Axis
FIGURE P7.9
Root locus for 1 + K(s+0.5)(s+0.1)(s2+2s+289)
s(s+30)2(s0.4)(s+0.8)(s2+1.45s+361) = 0.
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Problems 311
s3=39.8
s4=14.9
s5=5.0
s6=0.38
s7=0.14 .
P7.10 (a) The characteristic equation is
1 + K1K2(s+ 2)2
(s+ 10)(s+ 100)(s2+ 1.5s+ 6.25) = 0 .
The root locus is shown in Figure P7.10.
-10
-8
-6
-4
-2
0
2
4
6
8
10
-120 -100 -80 -60 -40 -20 0
x x
x
x
oo
Real Axis
Imag Axis
FIGURE P7.10
Root locus for 1 + K1K2(s+2)2
(s+10)(s+100)(s2+1.5s+6.25) = 0.
(b) The gain
K1K2= 1620
when ζ= 0.707. Therefore,
K2= 81000 ,
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312 CHAPTER 7 The Root Locus Method
since K1= 0.02 at medium weight cruise condition.
(c) At lightweight cruise condition
K1= 0.2.
Using K2= 81000, we find the roots are
s1,2=54 ±j119
s3,4=2±j0.6.
The roots s3,4become negligible and the roots at s1,2become highly
oscillatory. Hence, in this case
ζ= 0.41 .
P7.11 (a) The closed-loop characteristic equation is
1 + 20Ka(s2+s+ 0.02)
s(s+ 1)2(s2+ 2s+ 0.8) = 0 ,
where
K2= 10 .
Then, the root locus is shown in Figure P7.11a.
-3
-2
-1
0
1
2
3
-3 -2 -1 0 1 2 3
xxx xx o o
Real Axis
Imag Axis
Ka=0.035 -->
FIGURE P7.11
(a) Root locus for 1 + Ka20s2+20s+0.4
s(s+1)2(s2+2s+0.8) = 0, where K2= 10.
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Problems 313
(b) When
Ka<0.035 ,
all the roots have a damping greater than or equal to 0.60.
(c) Select
Ka= 0.035 .
Then, the characteristic equation with K2as the parameter is
1 + K2
0.07(s2+s)
s5+ 4s4+ 5.8s3+ 3.6s2+ 0.8s+ 0.014 = 0 .
The root locus is shown in Figure P7.11b.
-3
-2
-1
0
1
2
3
-3 -2 -1 0 1 2 3
x
x
x
x xoo
Real Axis
Imag Axis
Ka=0.035
FIGURE P7.11
CONTINUED: (b) Root locus for 1+K20.07s(s+1)
s(s+1)2(s2+2s+0.8)+0.014 = 0, where Ka= 0.035.
P7.12 (a) The closed-loop transfer function is
T(s) = KaKm(s+ 25)(s+ 15)
1.8s2(s+ 2) + KaKm(s+ 25)(s+ 15) + 1.6Kms(s+ 2) .
So, with E(s) = R(s)Y(s), we have E(s) = (1 T(s))R(s) and
ess = lim
s0sE(s) = 1 T(0) = 0 .
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314 CHAPTER 7 The Root Locus Method
Therefore, when the system is stable, it has zero steady-state error.
(b) The characteristic equation is
s3+ (3.6 + Ka)s2+ (3.2 + 40Ka)s+ 375Ka.
The Routh array is
s31 3.2 + 40Ka
s23.6 + Ka375Ka
s1b
so375K
Solving for b > 0 leads to 0 < Ka<0.05 or Ka>5.64 for stability.
(c) The characteristic equation can be written as
1 + Ka(s+ 25)(s+ 15)
s(s+ 2)(s+ 1.6) = 0 .
The root locus is shown in Figure P7.12.
(d) When
K > 40 ,
-40
-30
-20
-10
0
10
20
30
40
-70 -60 -50 -40 -30 -20 -10 0 10
xxxo o
Real Axis
Imag Axis
FIGURE P7.12
Root locus for 1 + Ka(s+25)(s+15)
s(s+2)(s+1.6) = 0, where Km= 1.8.
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Problems 315
the roots are
s1=123 and s2,3=15.6±j31.2.
From the step response we find
P.O. = 5%
Tp= 0.67 sec
Ts= 0.25 sec .
P7.13 (a) The characteristic equation is
1 + K
s(s+ 3)(s2+ 4s+ 7.84) = 0 .
The root locus is shown in Figure P7.13. The breakaway point is
s=1.09 at K= 9.72.
(b) When K= 13.5, the roots are
s1,2=0.84 ±j0.84
s3,4=2.66 ±j1.55 .
-6
-4
-2
0
2
4
6
-6 -4 -2 0 2 4 6
x
x
x
x
Real Axis
Imag Axis
+
+
+
+
FIGURE P7.13
Root locus for 1 + K
s(s+3)(s2+4s+7.84) = 0.
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316 CHAPTER 7 The Root Locus Method
(c) The roots
s=0.84 ±j0.84
are dominant roots.
(d) For the dominant roots, we determine that ζ= 0.7 and ωn= 1.19.
Therefore, the settling time is
Ts=4
ζωn
= 4.8 sec .
P7.14 The characteristic equation is
1 + K(s+ 2.5)(s+ 3.2)
s2(s+ 1)(s+ 10)(s+ 30) = 0 .
The root locus is shown in Figure P7.14. When K= 559.3, the roots are
s1=30.75 s2=8.48 s3=1.78 s4,5=±j3.11 .
When K= 4321, the roots are
s1=34.45 s2=4.35 s3=2.21 s4,5=±j10.23 .
The crossover points are
s=±j3.11 and s=±j10.23 .
0 5 10
0
5
10
15
20
25
Real Axis
Imaginary Axis
FIGURE P7.14
Root locus for 1 + K(s+2.5)(s+3.5)
s2(s+1)(s+10)(s+30) = 0.
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Problems 317
Therefore, the system is stable for
559.3< K < 4321 .
P7.15 The characteristic equation is
1 + K(s2+ 30s+ 625)
s(s+ 20)(s2+ 20s+ 200)(s2+ 60s+ 3400) .
The root locus is shown in Figure P7.15. When K= 30000, the roots are
s1=18.5s2=1.69 s3,4=9.8±j8.9s5,6=30.1±j49.9.
The real root near the origin dominates, and the step response is over-
damped.
-100
-80
-60
-40
-20
0
20
40
60
80
100
-100 -80 -60 -40 -20 0 20 40 60 80 100
x
x
x
x
x
x
o
o
Real Axis
Imag Axis
FIGURE P7.15
Root locus for 1 + Ks2+30s+625
s(s+20)(s2+20s+200)(s2+60s+3400) = 0.
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318 CHAPTER 7 The Root Locus Method
P7.16 (a) Let τ= 0. Then, first reduce the motor and rolls to an equivalent
G(s) as follows:
G(s) =
0.25
s(s+1)
1 + 0.25
s(s+1)
=0.25
s(s+ 1) + 0.25 =0.25
(s+ 0.5)2.
The loop transfer function is then
L(s) = 2(s+ 0.5)Ka(0.25)
s(s+ 1)2(s+ 0.5)2=0.5Ka
s(s+ 1)2(s+ 0.5) .
The characteristic equation is
1 + Ka
0.5
s(s+ 1)2(s+ 0.5) = 0 .
The root locus is shown in Figure P7.16.
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
-2 -1.5 -1 -0.5 0 0.5 1 1.5 2
xxx x
Real Axis
Imag Axis
+
+
+
+
FIGURE P7.16
Root locus for 1 + 0.5Ka
s(s+1)2(s+0.5) = 0.
(b) When K= 0.123, the roots of the characteristic equation are
s1,2=1.1±j0.27
s3,4=0.15 ±j0.15 .
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Problems 319
The roots at s=0.15 ±j0.15 have a damping ratio of ζ= 0.707.
(c) When τbecomes nonnegligible, the root locus will have an additional
pole, and the root locus will change accordingly.
P7.17 The characteristic equation is
(M1s2+bs +k1+k12)(M2s2+k12)k2
12 = 0 .
If we let M1=k1=b= 1, and assume k12 <1 so that k2
12 is negligible
and k1+k12 k1, then the characteristic equation is
(s2+s+ 1)(M2s2+k12) = 0 or 1 + k
s2= 0 ,
where
k=k12
M2
.
The root locus is shown in Figure P7.17. All the roots lie on the jω axis.
If we select
sk12
M2
=ωo,
then we cancel the vibration.
-3
-2
-1
0
1
2
3
-3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1
xx
Real Axis
Imag Axis
root locus -->
FIGURE P7.17
Root locus for 1 + k
s2= 0.
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320 CHAPTER 7 The Root Locus Method
P7.18 The characteristic equation is
βs3+ (1 + 2β)s2+ (2 + 4α)s+ 4 = 0 .
When β= 0 we have
1 + 4αs
s2+ 2s+ 4 = 0 .
The root locus for β= 0 is shown in Figure P7.18.
-3
-2
-1
0
1
2
3
-3 -2 -1 0 1 2 3
x
x
o
Real Axis
Imag Axis
beta=0<-- zeta=0.6
*
*
FIGURE P7.18
Root locus for 1 + α4s
s2+2s+4 = 0, where β= 0.
For α= 0.3, the poles are
s=1.6±j1.2.
Then, we have
1 + β(s+ 2)s2
s2+ (2 + 4α)s+ 4 = 0 .
When β= 0.121
s1,2=1.51 ±j1.51
s3=7.24 .
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Problems 321
Thus,
ζ= 0.707 and ζωn= 1.5.
So, the performance specs are met. Also,
Gc(s) = 0.3s+ 1
0.121s+ 1 =2.48(s+ 3.33)
(s+ 8.26) .
P7.19 The characteristic equation is
1 + Ka(s2+ 4s+ 100)
s(s+ 2)(s+ 6) = 0 .
The root locus is shown in Figure P7.19.
−10 −8 −6 −4 −2 0 2
−10
−8
−6
−4
−2
0
2
4
6
8
10
Real Axis
Imaginary Axis
FIGURE P7.19
Root locus for 1 + Kas2+4s+100
s(s+2)(s+6) = 0.
When Ka= 0.094;, the roots are
s1,2=0.85 ±j0.85
s3=6.38 .
Thus, the complex roots have a damping ratio of ζ= 0.707.
P7.20 The characteristic equation is
s3+ (2 + 1
β)s2+ ( 2
β+K)s+4
β= 0 ,
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322 CHAPTER 7 The Root Locus Method
where
K=4α
βα= 0.3β= 0.121 .
The root sensitivity to changes in Kis found to be
Sr1
K
=r1
K/K = 1.186149.75o.
The root sensitivity to changes in the pole at s=2 is found to be
Sr1
=r1
/2= 1.656137o,where the pole is s+ 2 + ∆.
P7.21 (a) Let the pole be (s+ 4 + ) and neglect 2terms. Then, the charac-
teristic equation is
1 + ∆ 2s2+ (8 + 2δ)s+ 8δ
s3+ (8 + δ)s2+ (16 + 8δ)s+ 16δ+K= 0
where δ= 0.000788 and K= 19.1.
-3
-2
-1
0
1
2
3
-10 -8 -6 -4 -2 0 2
x
x
x
o o
Real Axis
Imag Axis
FIGURE P7.21
Root locus for 1 + 2s2+(8+2δ)s+8δ
s3+(8+δ)s2+(16+8δ)s+16δ+K= 0, (δ= 0.000788 and K= 19.1).
The root sensitivity is determined to be
Sr1
=r1
/4= 3.3146132o.
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Problems 323
(b) Let R=Ro+ ∆R, where R= 0.00021. Then,
Sr1
R
=r1
R/R = 1.31 6107o.
P7.22 The characteristic equation is
s3+ 2s2+s+K ,
where K= 0.24 for ζ= 0.707. The root sensitivity to changes in the pole
at s=1 is found to be
Sr1
=r1
= 0.956126o,
where the pole is s+ 1 + ∆.
P7.23 The characteristic equation is
s3+ 5s2+ (6 + K)s+K ,
where K= 6.3 for ζ= 0.707. The root sensitivity to changes in the pole
at s=2 is found to be
Sr1
=r1
/2= 1.256169.4o,
where the pole is s+ 2 + ∆. The root sensitivity to changes in the zero at
s=1 is found to be
Sr1
=r1
= 0.55634.4o,
where the zero is s+ 1 + ∆.
P7.24 The root locus for each of the four cases shown is shown in Figure P7.24.
The four open-loop transfer functions are
(a)KF (s) = s2+ 7s+ 8.25
s3+ 6s2+ 5s
(b)KF (s) = s+ 8
s5+ 30s4+ 296s3+ 1170s2+ 1575s
(c)KF (s) = 1
s6+ 2s5+s4(d)KF (s) = s2+ 6s+ 6.75
s3+ 5s2+ 4s
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324 CHAPTER 7 The Root Locus Method
-5
0
5
-10 -5 0
xxxo o
Real Axis
Imag Axis
(a)
-10
-5
0
5
10
-20 -10 0
xx x xxo
Real Axis
Imag Axis
(b)
-2
-1
0
1
2
-2 0 2
xxxxxx
Real Axis
Imag Axis
(c)
-5
0
5
-10 -5 0
xxxo o
Real Axis
Imag Axis
(d)
FIGURE P7.24
Root locus for the four cases.
P7.25 The characteristic equation is
1 + KGc(s)G(s) = 0 ,
therefore,
KGc(s)G(s) = 1.
Squaring both sides yields K2G2
c(s)G2(s) = 1 and
1K2G2
c(s)G2(s) = 0 .
The root locus with 0 < K2<is shown in Figure P7.25. The value of
K2for which the locus crosses the imaginary axis is
K2= 2/3,
therefore K=p2/3 = 0.8165 corresponds to the jω-axis crossing (at
s= 0). You can check that 1 + KGc(s)G(s) = 0 for K= 0.8165 and
s= 0.
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Problems 325
-3 -2 -1 0 1 2 3
-3
-2
-1
0
1
2
3
Real Axis
Imag Axis
FIGURE P7.25
Root locus for the equation 1K2G2
c(s)G2(s) = 0.
P7.26 (a) The characteristic equation is
1 + K(s+ 2)2
s(s2+ 1)(s+ 8) = 0 .
The root locus is shown in Figure P7.26.
(b) Using Routh’s criteria, we determine that
K > 14
for stability.
(c) From the Routh array, we determine that for K= 14, we have two
purely imaginary poles at
s=±j8.
(d) When K > 50, the real part of the complex roots is approximately
equal to the real part of the two real roots and therefore the complex
roots are not dominant roots.
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326 CHAPTER 7 The Root Locus Method
-15 -10 -5 0 5 10 15
-15
-10
-5
0
5
10
15
Real Axis
Imag Axis
FIGURE P7.26
Root locus for 1 + K(s+2)2
s(s2+1)(s+8) = 0.
P7.27 The characteristic equation is
1 + K(s2+ 0.1)
s(s2+ 2) = 0 .
The root locus is shown in Figure P7.27a. The locus enters the axis at
s=1.26
and leaves the axis at
s=0.36 .
Define
p(s) = K=(s3+ 2s)
s2+ 0.1.
Then, a plot of p(s) vs sis shown in Figure P7.27b, where it can be seen
that p(s) has two inflection points at
s=1.28 and s=0.36 .
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Problems 327
−3 −2.5 −2 −1.5 −1 −0.5 0 0.5
−2
−1.5
−1
−0.5
0
0.5
1
1.5
2
Root Locus
Real Axis
Imaginary Axis
−2 −1.8 −1.6 −1.4 −1.2 −1 −0.8 −0.6 −0.4 −0.2 0
0
0.5
1
1.5
2
2.5
3
3.5
s
p(s)
FIGURE P7.27
(a) Root locus for 1 + Ks2+0.1
s(s2+2) = 0. (b) Plot of p(s) = s3+2s
s2+0.1versus s.
P7.28 The characteristic equation is
1 + L(s) = 1 + K(s2+ 12s+ 20)
s3+ 10s2+ 25s= 0 .
The root locus is shown in Figure P7.28. The breakaway point is s=5.0
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328 CHAPTER 7 The Root Locus Method
-20 -15 -10 -5 0
-6
-4
-2
0
2
4
6
Real Axis
Imag Axis
FIGURE P7.28
Root locus for 1 + K(s2+12s+20)
s3+10s2+25s= 0.
and the entry point is s=15.6. When K= 2, the roots are
s1=1.07
s2,3=5.46 ±j2.75 .
When K= 2, the roots are
s1=1.07
s2,3=4.36 ±j1.68 .
The predicted step response when K= 2 is Ts= 9 sec and P O 0%.
P7.29 The characteristic equation is
1 + Ks2+ 10s+ 30
s2(s+ 10) = 0 .
The root locus is shown in Figure P7.29. When ζ= 0.707, the necessary
gain is K= 16. The corresponding roots are s1=18.87 and s2,3=
3.56 ±j3.56.
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Problems 329
Root Locus
Real Axis
Imaginary Axis
−20 −15 −10 −5 0 5
−4
−3
−2
−1
0
1
2
3
4
System: sys
Gain: 16
Pole: −3.56 + 3.57i
Damping: 0.707
Overshoot (%): 4.34
Frequency (rad/sec): 5.04
FIGURE P7.29
Root locus for 1 + Ks2+10s+30
s2(s+10) = 0.
P7.30 The transfer function is
Z(s) = LCRs2+Ls
LCs2+CRs + 1 =Rs2+s
s2+Rs + 1 .
So,
r1=R
2+ R2
41!1
2
.
Thus, the nominal r1o=1
2. Simultaneously,
r2=R
2 R2
41!1
2
.
Thus, the nominal r2o=2. We see that there is a difference by a factor
of 4. Also,
Sri
R=r1
R Ro·Ro=Ro
2+R2
o
4 R2
o
41!1
2
=5
6,
where Ro= 2.5. And
Sr2
R=r2
R Ro
Ro=Ro
2R2
o
4 R2
o
41!1
2
=10
3.
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330 CHAPTER 7 The Root Locus Method
So, the magnitude of |Sr2
R|= 4|Sr1
R|.
P7.31 The characteristic equation is
1 + Ks+ 4
s(s+ 0.16)(s2+ 14.6s+ 148.999) = 0 .
The root locus is shown in Figure P7.31. When K= 1350, the roots are
-20
-15
-10
-5
0
5
10
15
20
-20 -15 -10 -5 0 5 10 15 20
x
x
x
xo
Real Axis
Imag Axis
*
*
*
*
+
+
+
+
<-- K=1350 (*)
(+) K=326 -->
FIGURE P7.31
Root locus for 1 + Ks+4
s(s+0.16)(s2+14.6s+148.999) = 0.
s1,2=±j9.6
s3,4=7.4±j1.9.
When K= 326, the roots are
s1,2=6.5±j8.7s3,4=0.9±j3.2.
P7.32 The characteristic equation is
1 + K(s+ 1)(s+ 5)
s(s+ 1.5)(s+ 2) = 0 .
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Problems 331
-4
-3
-2
-1
0
1
2
3
4
-10 -8 -6 -4 -2 0 2 4
xxxo o
Real Axis
Imag Axis
*
*
*
*
*
*
FIGURE P7.32
Root locus for 1 + K(s+1)(s+5)
s(s+1.5)(s+2) = 0.
K ζ Ts(sec) P.O. (%)
1.57 0.707 0.98 1.4
3.48 0.707 1.1 5.8
2.35 0.69 1.3 4.0
TABLE P7.32 Step Response Results for K= 1.57,K= 3.48, and K= 2.35.
(a) The breakaway point is s=1.73; the entry point is s=8.62.
(b) The damping ratio ζ= 0.707 when K= 1.57 and again when K=
3.46.
(c) The minimum damping ratio ζ= 0.69 is achieved when K= 2.35.
(d) The results are summarized in Table P7.32.The best choice of gain is
K= 1.57.
P7.33 (a) The root locus for the V-22 is shown in Figure P7.33a. The system is
stable when 0 < K < 0.48 and K > 136.5.
(b) The unit step input response (for K= 280) is shown in Figure P7.33b.
The step response has a P.O. = 90% and Ts50 sec.
(c) The plot of y(t) for a unit step disturbance is shown in Figure P7.33b.
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332 CHAPTER 7 The Root Locus Method
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
-3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1
xxx xo o
Real Axis
Imag Axis
0 10 20 30 40 50 60 70 80
0
0.5
1
1.5
2
Time (sec)
y(t)
(i) Unit step input response
y(t) w/o prelter .... (dotted line)
y(t) with prelter ____ (solid line)
0 10 20 30 40 50 60 70 80
-1
0
1
2
3
4x 10 -3
Time (sec)
y(t)
(ii) Unit step disturbance response
FIGURE P7.33
(a) Root locus for 1 + Ks2+1.5s+0.5
s(20s+1)(10s+1)(0.5s+1) = 0. (b) (i) Unit step input response with
and without prefilter; (ii) Unit step disturbance response.
The response to the disturbance is oscillatory, but the maximum value
of oscillation is about 0.003; so it is negligible.
(d) The effect of adding a prefilter can be seen in Figure P7.33b. With
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Problems 333
the prefilter we find P O = 7% and Ts40 sec.
P7.34 The characteristic equation is
1 + K(s+ 2)
(s+ 1)(s+ 2.5)(s+ 4)(s+ 10) = 0 .
The root locus is shown in Figure P7.34a. The roots, predicted and ac-
tual percent overshoot for K= 400,500, and 600 are summarized in
Table P7.34. The actual unit step input responses are shown in Fig-
ure P7.34b.
Kroots ζpredicted P.O. (%) actual P.O. (%)
400 -13.5,-1.00 ±5.71j,-1.98 0.173 57.6 51.6
500 -14.0,-0.75 ±6.24j,-1.98 0.120 68.4 61.2
600 -14.4,-0.53 ±6.71j,-1.98 0.079 77.9 69.6
TABLE P7.34 Summary for K= 400,500,600.
−30 −25 −20 −15 −10 −5 0 5 10
−20
−15
−10
−5
0
5
10
15
20
Root Locus
Real Axis
Imaginary Axis
FIGURE P7.34
(a) Root locus for 1 + Ks+2
(s+1)(s+2.5)(s+4)(s+10) = 0.
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334 CHAPTER 7 The Root Locus Method
0 2 4 6 8 10 12 14 16 18 20
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
Time (sec)
y(t)
K=400 .... (dotted line)
K=500 −−− (dashed line)
K=600 ___ (solid line)
FIGURE P7.34
CONTINUED (b) Unit step input responses for K= 400,500,600.
P7.35 (a) The root locus is shown in Figure P7.35 for the characteristic equation
1 + K(s+ 1)2
s(s2+ 1) = 0 .
-3
-2
-1
0
1
2
3
-5 -4 -3 -2 -1 0 1
x
x
x
oo
Real Axis
Imag Axis
K=4.52 *
*
*
FIGURE P7.35
Root locus for 1 + K(s+1)2
s(s2+1) = 0.
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Problems 335
(b) When K= 4.52, the roots are
s1=0.58
s2,3=1.96 ±j1.96 .
The complex roots have ζ= 0.707.
(c) The entry point is s=3.38 when K= 7.41.
(d) The predicted P.O. = 4.5% (ζ= 0.707) and the actual P.O. = 17%.
P7.36 The characteristic equation is
1 + K(s+ 1)(s+ 2)(s+ 3)
s3(s1) = 0 .
(a) The root locus is shown in Figure P7.36.
-8
-6
-4
-2
0
2
4
6
8
-10 -8 -6 -4 -2 0 2
xxx xo o o
Real Axis
Imag Axis
FIGURE P7.36
Root locus for 1 + K(s+1)(s+2)(s+3)
s3(s1) = 0.
(b) When K= 2.96, the roots are
s1,2=±j4.08
s3,4=0.98 ±j0.33 .
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336 CHAPTER 7 The Root Locus Method
(c) When K= 20, the roots are
s1=1.46
s2=1.07
s3,4=8.23 ±j2.99 .
When K= 100, the roots are
s1=92.65
s2=3.51
s3=1.82
s4=1.01 .
(d) When K= 20, the damping ratio is ζ= 0.94. Therefore, the predicted
P.O. = 0.02%. The actual overshoot is P.O. = 23%.
P7.37 Since we know that ess = 0 for a step input, we know that a= 0 or b= 0.
Select a= 0. Also, ωn= 2π/T = 20 rad/sec. The desired characteristic
polynomial is
(s+r1)(s+j20)(sj20) = s3+r1s2+ 400s+ 400r1= 0 .
The actual characteristic polynomial is
1 + 2K
s(s+b)(s+ 40) = 0 ,or s3+ (40 + b)s2+ 40bs + 2K= 0 .
Comparing the coefficients in the desired and actual characteristic poly-
nomials, we determine that b= 10, r1= 50, and K= 10000.
P7.38 (a) The characteristic equation is
1 + K(s+ 1)
s(s3) = 0 .
The system is stable for K > 3. When K= 3, the roots are s=±j3.
(b) The root locus is shown in Figure P7.38a.
(c) When K= 10 , the roots are
s1=2
s2=5.
Since both roots are real and stable, we expect that there will be
zero overshoot. The actual response has a 40% overshoot, as seen in
Figure P7.38b.
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Problems 337
-6
-4
-2
0
2
4
6
-6 -4 -2 0 2 4 6
x xo
Real Axis
Imag Axis
0
0.2
0.4
0.6
0.8
1
1.2
1.4
00.5 1 1.5 2 2.5 3
Time (secs)
Amplitude
FIGURE P7.38
(a) Root locus for 1 + Ks+1
s(s3) = 0. (b) Unit step response.
P7.39 The loop transfer function is
Gc(s)G(s) = 22K
(s+ 1)(s2+ 8s+ 22) .
When K= 0.529, the closed-loop poles are s1,2=3.34 ±1.83jand
s3=2.32 and have the maximum damping ζ= 0.877. The root locus is
shown in Figure P7.39a. The step response is shown in Figure P7.39b.
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338 CHAPTER 7 The Root Locus Method
−14 −12 −10 −8 −6 −4 −2 0 2
−10
−8
−6
−4
−2
0
2
4
6
8
10
Root Locus
Real Axis
Imaginary Axis
Step Response
Time (sec)
Amplitude
0 0.5 1 1.5 2 2.5 3 3.5
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
FIGURE P7.39
(a) Root locus for 22K
(s+1)(s2+8s+22) = 0. (b) Unit step response.
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Advanced Problems 339
Advanced Problems
AP7.1 The characteristic equation is
1 + Ks+ 6
s(s+ 4)(s2+ 4s+ 8) = 0 .
The root locus is shown in Figure AP7.1. The gain at maximum ζis
-10
-5
0
5
10
-10 -5 0 5 10
x
x
x
xo
Real Axis
Imag Axis
+
+
+
+
FIGURE AP7.1
Root locus for 1 + Ks(s+4)
s2+2s+2 = 0.
K= 3.7.
The roots at K= 3.7 are
s1=3.6424 s2,3=1.3395 ±+1.3553j s4=1.6786 .
Using Figure 5.13 in Dorf & Bishop, the predicted percent overshoot and
settling time are
P.O. = 5% and Ts= 3 sec ,
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340 CHAPTER 7 The Root Locus Method
since ζ= 0.7 and
a
ωnζ=6
1.9(0.7) = 4.5.
The actual percent overshoot and settling time are P.O. = 1% and Ts=
2.8 sec.
AP7.2 The characteristic equation is
1 + K(s+ 1)(s+ 4)
s(s1)(s+ 5)(s+ 10) = 0 .
The root locus is shown in Figure AP7.2a. The selected gain is K= 43.7.
−12 −10 −8 −6 −4 −2 0 2
−15
−10
−5
0
5
10
15
Real Axis
Imaginary Axis
Time (sec)
Amplitude
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5
0
0.5
1
1.5
System: syscl
Peak amplitude: 1.48
Overshoot (%): 48.3
At time (sec): 0.857
System: syscl
Settling Time (sec): 2.31
FIGURE AP7.2
(a) Root locus for 1 + K(s+1)(s+4)
s(s1)(s+5)(s+10) = 0; (b) Step response for K= 43.7.
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Advanced Problems 341
The actual percent overshoot (see Figure AP7.2b) is P.O. = 48.3%.
AP7.3 The characteristic equation (with pas the parameter) is
1 + ps(s+ 1)
s3+s2+ 10 = 0 .
The root locus is shown in Figure AP7.3.
-5
-4
-3
-2
-1
0
1
2
3
4
5
-5 -4 -3 -2 -1 0 1 2 3 4 5
x
x
x
oo
Real Axis
Imag Axis
+
+
FIGURE AP7.3
Root locus for 1 + ps(s+1)
s3+s2+10 = 0.
When p= 21 the dominant roots have a damping ratio of ζ= 0.707.
AP7.4 The characteristic equation (with αas the parameter) is
1 + αs(s+ 1)
s3+s2+ 1 = 0 .
The root locus is shown in Figure AP7.4a. The steady-state error is
ess = lim
s0sE(s) = lim
s0
1
1 + G(s)= 1 α .
To meet the steady-state error specification, we require
0.9< α < 1.1.
The step responses for α= 0.9,1 and 1.1 are shown in Figure AP7.4b.
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342 CHAPTER 7 The Root Locus Method
-3
-2
-1
0
1
2
3
-3 -2 -1 0 1 2 3
x
x
x
oo
Real Axis
Imag Axis
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
0 5 10 15 20 25 30 35 40 45 50
Time (sec)
Amplitude
alpha=0.9 (solid); alpha=1.0 (dashed); alpha=1.1 (dotted)
FIGURE AP7.4
(a) Root locus for 1 + ps(s+1)
s3+s2+10 = 0. (b) Step responses for α= 0.9,1and 1.1.
AP7.5 The root locus is shown in Figure AP7.5. When K= 20.45, ζ= 0.707. The
root sensitivity is Sr1
K
=r1/(∆K/20.45) = 3.15687.76o. When K= 88,
the complex roots lie on the jω-axis—a 330% increase in the gain.
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Advanced Problems 343
-5
-4
-3
-2
-1
0
1
2
3
4
5
-15 -10 -5 0 5
x x x
Real Axis
Imag Axis
+
+
+
FIGURE AP7.5
Root locus for 1 + K1
s3+10s2+7s18 = 0.
AP7.6 A gain of K= 13 provides an acceptable response of Ts<1 and P.O. <
7.5%. The root locus is shown in Figure AP7.6.
−3 −2.5 −2 −1.5 −1 −0.5 0 0.5
−2.5
−2
−1.5
−1
−0.5
0
0.5
1
1.5
2
2.5
Root Locus
Real Axis
Imaginary Axis
FIGURE AP7.6
Root locus for 1 + Ks2+3s+6
s3+2s2+3s+1 = 0.
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344 CHAPTER 7 The Root Locus Method
AP7.7 The root locus for the positive feedback system is shown in Figure AP7.7.
-15
-10
-5
0
5
10
15
-15 -10 -5 0 5 10 15
x x
Real Axis
Imag Axis
FIGURE AP7.7
Root locus for 1 + K1
s2+12s+32 = 0.
AP7.8 The root locus is shown in Figure AP7.8a. When k= 0.448, all the roots
-30
-20
-10
0
10
20
30
-30 -20 -10 0 10 20 30
x
x
x
o
Real Axis
Imag Axis
FIGURE AP7.8
(a) Root locus for 1 + k120s
s3+19s2+34s+120 = 0.
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Advanced Problems 345
of the characteristic equation are real—the step response is shown in
Figure AP7.8b.
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0 0.5 1 1.5 2 2.5 3
Time (secs)
Amplitude
FIGURE AP7.8
CONTINUED (b) Step response with k= 0.448.
AP7.9 The root locus for each controller is shown in Figure AP7.9.
−15 −10 −5 0 5
−5
0
5
(a)
Real Axis
Imaginary Axis
−15 −10 −5 0 5
−5
0
5
(b)
Real Axis
Imaginary Axis
−15 −10 −5 0 5
−15
−10
−5
0
5
10
15
(c)
Real Axis
Imaginary Axis
−15 −10 −5 0 5
−5
0
5
(d)
Real Axis
Imaginary Axis
FIGURE AP7.9
Root locus for the various controllers.
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346 CHAPTER 7 The Root Locus Method
AP7.10 The characteristic equation (with Kas the parameter) is
1 + Ks2+ 7s+ 20
s(s2+ 7s+ 10) = 0 .
The root locus is shown in Figure AP7.10. The steady-state value of the
-10 -8 -6 -4 -2 0 2 4 6 8 10
-10
-8
-6
-4
-2
0
2
4
6
8
10
Real Axis
Imag Axis
FIGURE AP7.10
Root locus for 1 + Ks2+7s+20
s(s2+7s+10) = 0.
step response for any Kis 0.5. With K= 15 the closed-loop transfer
function is
T(s) = 10s+ 150
s3+ 22s2+ 115s+ 300 .
The step response has the following characteristics:
P.O. = 4.8% and Ts= 2 seconds .
AP7.11 The root locus is shown in Figure AP7.11a. A suitable gain is
K= 500.
The step response is shown in Figure AP7.11b.
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Advanced Problems 347
−100 −80 −60 −40 −20 0 20 40
−80
−60
−40
−20
0
20
40
60
80
Root Locus
Real Axis
Imaginary Axis
FIGURE AP7.11
(a) Root locus for 1 + K(s+2)2
s(s+10)(s+20)(s2+3s+3.5) = 0.
Step Response
Time (sec)
Amplitude
0 0.5 1 1.5 2 2.5 3 3.5 4
0
0.2
0.4
0.6
0.8
1
1.2
1.4
System: sys_cl
Peak amplitude: 1.09
Overshoot (%): 9.01
At time (sec): 0.945
System: sys_cl
Settling Time (sec): 2.39
FIGURE AP7.11
CONTINUED: (b) Step response with K= 500.
AP7.12 The root locus is shown in Figure AP7.12a. The PI controller can be
written as
Gc(s) = Kps+KI
s
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348 CHAPTER 7 The Root Locus Method
-7 -6 -5 -4 -3 -2 -1 0 1 2
-8
-6
-4
-2
0
2
4
6
8
Real Axis
Imag Axis
Time (sec.)
Amplitude
Step Response
0 5 10 15
0
0.2
0.4
0.6
0.8
1
1.2
1.4
From: U(1)
To: Y(1)
FIGURE AP7.12
(a) Root locus for 1 + Kp(s+0.2)
s(s2+7s+10) = 0. (b) Step response with Kp= 5.54.
and setting KI= 0.2Kp, the characteristic equation can be written as
1 + Kp
(s+ 0.2)
s(s2+ 7s+ 10) = 0
A suitable gain is Kp= 5.55. The step response is shown in Figure AP7.12b.
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Advanced Problems 349
AP7.13 The characteristic equation is
1 + K1K2
1
(s+ 5)(s1) = 0.
The root locus is shown in Figure AP7.12a. The fastest expected settling
−6 −5 −4 −3 −2 −1 0 1 2
−4
−3
−2
−1
0
1
2
3
4
Root Locus
Real Axis
Imaginary Axis
FIGURE AP7.13
Root locus for 1 + K1K21
(s+5)(s1) = 0.
time is Ts= 4nζ= 2 seconds since maximum |ωnζ|= 2.
AP7.14 The root locus of the uncompensated transfer function is shown in Fig-
ure AP7.14a. It can be seen that the system is unstable for Ku= 131.25
with a period of Tu= 0.72, as illustrated in FigureAP7.14b. Using the
Ziegler-Nichols design formulas yields KP= 0.6Ku= 78.75, KI= 1.2Ku/Tu=
218.75, and KD= 0.6KuTu= 7.0875 where
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350 CHAPTER 7 The Root Locus Method
−40 −30 −20 −10 0 10 20
−30
−20
−10
0
10
20
30
System: sysg
Gain: 131
Pole: 0.0153 + 8.66i
Damping: −0.00176
Overshoot (%): 101
Frequency (rad/sec): 8.66
Root Locus
Real Axis
Imaginary Axis
FIGURE AP7.14
(a) Root locus for 1 + Ku10
s(s+10)(s+7.5) = 0.
0 2 4 6 8 10 12
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
Step Response
Time (sec)
Amplitude
FIGURE AP7.14
CONTIUED: (b) Step response at the ultimate gain Ku= 131.
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Advanced Problems 351
Step Response
Time (sec)
Amplitude
0 0.5 1 1.5 2 2.5 3 3.5
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
System: sys_cl
Peak amplitude: 1.6
Overshoot (%): 59.5
At time (sec): 0.445
System: sys_cl
Settling Time (sec): 2.1
FIGURE AP7.14
CONTINUED: (c) Step response with the Ziegler-Nichols tuned PID controller.
0 0.5 1 1.5 2 2.5 3 3.5
−4
−2
0
2
4
6
8
10
12
14 x 10−3 Step Response
Time (sec)
Amplitude
FIGURE AP7.14
CONTINUED: (d) Disturbance response with the Ziegler-Nichols tuned PID controller.
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352 CHAPTER 7 The Root Locus Method
Design Problems
The closed-loop transfer function from the input to the output isCDP7.1
θ(s)
R(s)=26.035Ka
s2+ (33.1415 + 26.035KaK1)s+ 26.035Ka
,
where we consider for the first time the tachometer feedback (see Figure
CDP4.1 in Dorf and Bishop). The characteristic equation is
1 + K1
26.035Kas
s2+ 33.1415s+ 26.035Ka
= 0 .
The root locus is shown below. In accordance with the discussion in Chap-
-30 -20 -10 0 10 20 30
-30
-20
-10
0
10
20
30
Real Axis
Imag Axis
ter 5, we continue to use Ka= 22. This allows us to meet the overshoot
specification (P.O. < 5%) without the tachometer feedback and to pro-
vides good steady-state tracking errors to a step input. To meet the design
specifications of both P.O. and Tswe want the closed-loop poles to the
left of ζω =4/0.3 = 13.33 and ζ > 0.69. A reasonable selection is
K1= 0.012. This places the closed-loop poles at s=20 ±j13.
DP7.1 (a) The characteristic equation is
1 + 18K(s+ 0.015)(s+ 0.45)
(s2+ 1.2s+ 12)(s2+ 0.01s+ 0.0025) = 0 .
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Design Problems 353
Since we want a negative feedback system, we have Gc(s) = K.
When ωn>2 and ζ= 0.15, the gain K= 0.12. The root locus is
shown in Figure DP7.1a.
-6
-4
-2
0
2
4
6
-4 -3.5 -3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1
x
x
x
x
o o
Real Axis
Imag Axis
FIGURE DP7.1
(a) Root locus for 1 + K18(s+0.015)(s+0.45)
(s2+1.2s+12)(s2+0.01s+0.0025) = 0.
(b) The unit step response is shown in Figure DP7.1b. The percent over-
shoot is
P.O. = 100% .
(c) The characteristic equation with the anticipatory controller is
1 + 18K(s+ 2)(s+ 0.015)(s+ 0.45)
(s2+ 1.2s+ 12)(s2+ 0.01s+ 0.002s)= 0 .
The root locus is shown in Figure DP7.1c. If we select
K= 9.2/18 ,
then the complex roots have a damping ζ= 0.90. The roots are at
s1=0.253
s2=0.019
s3,4=5.07 ±j2.50 .
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354 CHAPTER 7 The Root Locus Method
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
020 40 60 80 100 120 140 160 180 200
Time (secs)
Amplitude
FIGURE DP7.1
CONTINUED: (b) Unit step response for gain controller.
-6
-4
-2
0
2
4
6
-6 -4 -2 0 2 4 6
x
x
x
x
o o o
Real Axis
Imag Axis
FIGURE DP7.1
CONTINUED: (c) Root locus for 1 + K18(s+2)(s+0.015)(s+0.45)
(s2+1.2s+12)(s2+0.01s+0.0025) = 0.
(d) The unit step response for the system with the anticipatory controller
is shown in Figure DP7.1d.
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Design Problems 355
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
020 40 60 80 100 120 140 160 180 200
Time (secs)
Amplitude
FIGURE DP7.1
CONTINUED: (d) Unit step response for anticipatory controller.
DP7.2 The characteristic equation is
1 + 10K(s+ 1)
s(s2+ 4.5s+ 9) = 0 .
(a) The root locus is shown in Figure DP7.2a. When K= 0.435, we have
ζ= 0.6 and the roots are
s1=0.368
s2,3=2.1±j2.75 .
(b) The response to a step input is shown in Figure DP7.2b. The per-
formance results are
P.O. = 0%
Tss = 10 sec
ess = 0 .
(c) We have ζ= 0.41 when K= 1.51. The step response is shown in
Figure DP7.2b. The performance results to the step input are
P.O. = 0%
Ts= 4 sec
ess = 0 .
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356 CHAPTER 7 The Root Locus Method
-5
-4
-3
-2
-1
0
1
2
3
4
5
-5 -4 -3 -2 -1 0 1 2 3 4 5
x
x
x
o
Real Axis
Imag Axis
FIGURE DP7.2
(a) Root locus for 1 + K10(s+1)
s(s2+4.5s+9) = 0.
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
02 4 6 8 10 12 14 16
Time (sec)
Amplitude
K=0.435 ____ (solid line)
K=1.510 ---- (dashed line)
FIGURE DP7.2
CONTINUED: (b) Unit step responses for K= 0.425,1.51.
DP7.3 The characteristic equation is
1 + K(s2+ 6.5s+ 12)
s(s+ 1)(s+2) = 0 .
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Design Problems 357
(a) The root locus is shown in Figure DP7.3.
-6
-4
-2
0
2
4
6
-6 -5 -4 -3 -2 -1 0 1
xxx
o
o
Real Axis
Imag Axis
FIGURE DP7.3
Root locus for 1 + Ks2+6.5s+12
s(s+1)(s+2) = 0.
When K= 41, the roots are s1=37.12 and s2,3=3.44 ±j1.19 .
(b) The percent overshoot is P.O. 1% when ζ= 0.82 at K= 0.062.
(c) Select K > 300.
DP7.4 The characteristic equation is
1 + K10(0.01s+ 1)
s(s2+ 10s+ 10K1)= 0 .
If we choose K1= 2.5, then the root locus will start at s= 0,5 and
-5. This is shown in Figure DP7.4. The root locus then has a nice shape
so that we can select Kto place the complex poles where desired and
the one real root will be farther in the left half-plane; thus the notion
of dominant poles will be valid. So, if we desire a P.O. < 5%, we want
ζ > 0.69. This occurs when K3. Thus, our design is
K1= 2.5 and K= 3 .
The unit step response is shown in Figure DP7.4. The settling time is less
than 3.5 sec and the P O < 4%. The response to a unit step disturbance
is also shown in Figure DP7.4. The steady-state error magnitude to the
disturbance is 0.33.
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358 CHAPTER 7 The Root Locus Method
-20 -15 -10 -5 0 5 10
-4
-3
-2
-1
0
1
2
3
4
Real Axis
Imag Axis
K=3 -->
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
0
0.2
0.4
0.6
0.8
1
1.2
1.4
Time (sec)
y(t)
Input step response
Disturbance step response
FIGURE DP7.4
(a) Root locus for 1 + K10(0.01s+1)
s(s2+10s+25) = 0. (b) System response to step input and distur-
bance.
DP7.5 The characteristic equation is
1 + Ks+ 1
s(s0.1)(s2+ 10s+ 41) = 0 .
The root locus is shown in Figure DP7.5a.
The system is stable for 5 < K < 300. The step response with K=
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Design Problems 359
−12 −10 −8 −6 −4 −2 0 2
−10
−8
−6
−4
−2
0
2
4
6
8
10
System: sysgc
Gain: 90.5
Pole: −1.42 + 2.24i
Damping: 0.536
Overshoot (%): 13.6
Frequency (rad/sec): 2.66
Root Locus
Real Axis
Imaginary Axis
Step Response
Time (sec)
Amplitude
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
System: sys_cl
Peak amplitude: 1.57
Overshoot (%): 57
At time (sec): 1.24
System: sys_cl
Time (sec): 3.39
Amplitude: 0.98
FIGURE DP7.5
(a) Root locus for 1 + Ks+1
s(s0.1)(s2+10s+41) = 0. (b) Step response with K= 875.
90.5 is shown in Figure DP7.5b. We choose K= 90.5 to minimize the
settling time. The damping of the dominant poles is ζ= 0.54, so that the
estimated percent overshoot is P.O. = 13%. The actual percent overshoot
and settling time are P.O. = 57% and Ts= 3.4 seconds. The match
between the actual and predicted percent overshoot can be improved by
selecting a much higher gain K, but then the step response becomes
overy oscillatory and the settling time increases too much for a typical
high-performance aircraft.
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360 CHAPTER 7 The Root Locus Method
DP7.6 The characteristic equation is
1 + Ks+ 2
s(s+ 10)(s1) = 0 .
The maximum damping is ζ= 0.46 at K= 55. The root locus is shown in
Figure DP7.6a; the step response is shown in Figure DP7.6b. The percent
overshoot and settling time are P.O. = 61.3% and Ts= 2 seconds.
-20
-15
-10
-5
0
5
10
15
20
-20 -15 -10 -5 0 5 10 15 20
x xx o
Real Axis
Imag Axis
+
+
+
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
0 0.5 1 1.5 2 2.5 3
Time (secs)
Amplitude
FIGURE DP7.6
(a) Root locus for 1 + Ks+2
s(s+10)(s1) = 0. (b) Step response with K= 55.
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Design Problems 361
DP7.7 The loop transfer function is
Gc(s)G(s) = KPs+KI
s(s+ 1)(0.5s+ 1).
One possible set of PI controller gains are KP= 0.82 and KI= 0.9. The
step response is shown in Figure DP7.7.
Step Response
Time (sec)
Amplitude
0 1 2 3 4 5 6 7 8
0
0.2
0.4
0.6
0.8
1
1.2
1.4
System: syscl
Peak amplitude: 1.05
Overshoot (%): 4.59
At time (sec): 3.57
System: syscl
Settling Time (sec): 4.94
FIGURE DP7.7
Step response for with PI controller Gc(s) = (0.82s+ 0.9)/s.
DP7.8 The closed-loop transfer function is
T(s) = Vo(s)
V(s)=G(s)
1 + KG(s).
The dc gain is
T(0) = G(0)
1 + KG(0) 1
K.
The root locus is shown in Figure DP7.8. The maximum value of Kfor
stability is
K= 0.062 .
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362 CHAPTER 7 The Root Locus Method
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
x107
-2 -1.5 -1 -0.5 0 0.5 1 1.5 2
x107
xx x
Real Axis
Imag Axis
+
+
+
FIGURE DP7.8
Root locus for 1 + K3.142K1×1017
(s+3142)(s+107)2= 0.
Therefore, the minimum dc gain is about 1/0.062=16. Selecting
K= 0.05 and R1= 10 K
yields
R2= 19R1= 190 K .
DP7.9 The closed-loop transfer function (with Gp(s) = 1 and K= 1) is
T(s) = 2s3+ 6s2+ 14s+ 10
s4+ 6s3+ 13s2+ 26s+ 6 .
So, if we select Gp(s) = 1/T (0) = 0.6, the step response (with K= 1)
will have a zero steady-state tracking error. The root locus is shown in
Figure DP7.9a. The step responses for K= 1,1.5 and 2.85 are shown in
Figure DP7.9b. For K= 1, we have P.O. = 0%, Tr= 7.8 and Ts= 13.9;
for K= 1.5, we have P.O. = 0%, Tr= 5.4 and Ts= 9.6; and for K= 2.85,
we have P.O. = 5.2%, Tr= 0.5 and Ts= 7.3. The best gain selection is
K= 2.85.
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Design Problems 363
-8
-6
-4
-2
0
2
4
6
8
-8 -6 -4 -2 0 2 4 6 8
x
x
x
x o
Real Axis
Imag Axis
0
0.2
0.4
0.6
0.8
1
1.2
0 2 4 6 8 10 12 14 16 18 20
Time (sec)
Amplitude
K=1 (solid); K=1.5 (dashed); K=2.85 (dotted)
FIGURE DP7.9
(a) Root locus for 1 + K6(s+1)
s(s+4)(s2+2s+5) = 0. (b) Step responses with K= 1,1.5,2.85.
DP7.10 A suitable selection of the various parameters is
ζ= 0.5 and q= 3/5.
With q= 3/5, the open-loop zeros are real and equal. Then, it follows
that
λ=2q
1q= 3 .
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364 CHAPTER 7 The Root Locus Method
The root locus is shown in Figure DP7.10. A reasonable choice of gain is
K= 30 .
The resulting step response is extremely fast with no overshoot. The
closed-loop transfer function is approximately given by
T(s)1923
s+ 1923 .
-6
-4
-2
0
2
4
6
-6 -4 -2 0 2 4 6
x
x
x
oo
Real Axis
Imag Axis
FIGURE DP7.10
Root locus for 1 + K4s2+4s+1
0.0625s3+0.25s2+s= 0.
DP7.11 The characteristic equation (with Kas the parameter) is
1 + K10(s2+ 10)
s3+ 20s= 0 .
The root locus is shown in Figure DP7.11a. To maximize the closed-loop
system damping we choose
K= 0.513.
The step response is shown in Figure DP7.11b.
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Design Problems 365
-2 -1.5 -1 -0.5 0 0.5 1
-5
-4
-3
-2
-1
0
1
2
3
4
5
Real Axis
Imag Axis
Time (sec.)
Amplitude
Step Response
0123456
0
0.2
0.4
0.6
0.8
1
1.2
1.4
From: U(1)
To: Y(1)
FIGURE DP7.11
(a) Root locus for 1 + K10(s2+10)
s3+20s= 0. (b) Step response with K= 0.513.
DP7.12 The characteristic equation is
1 + Ks+ 1.5
(s+ 1)(s+ 2)(s+ 4)(s+ 10) = 0 .
The root locus is shown in Figure DP7.12a.
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366 CHAPTER 7 The Root Locus Method
-10
-8
-6
-4
-2
0
2
4
6
8
10
-15 -10 -5 0 5
x x x xo
Real Axis
Imag Axis
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
012345678910
Time (sec)
Amplitude
K=100 (solid); K=300 (dashed); K=600 (dotted)
FIGURE DP7.12
(a) Root locus for 1 + Ks+1.5
(s+1)(s+2)(s+4)(s+10) = 0. (b) Step response with K=
100,300,600.
The closed-loop system roots are:
K= 100 : s1=11.38 s2,3=2.09 ±3.10j s4=1.45
K= 300 : s1=12.94 s2,3=1.29 ±5.10j s4=1.48
K= 600 : s1=14.44 s2,3=0.53 ±6.72j s4=1.49
The step responses are shown in Figure DP7.12b.
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Design Problems 367
DP7.13 The closed-loop transfer function is
T(s) = Ka
s3+s2+KaK2s+Ka
.
A suitable choice of gains is
Ka= 0.52 and K2= 3 .
The step response is shown in Figure DP7.13.
0
0.2
0.4
0.6
0.8
1
1.2
0 2 4 6 8 10 12 14 16 18 20
Time (secs)
Amplitude
FIGURE DP7.13
Step response with Ka= 0.52 and K2= 3.
DP7.14 The characteristic equation is
s2+ 10KDs+ 10(KP+ 1) = 0 .
In the Evans form we have
1 + KD
10(s+τ)
s2+ 10 = 0 .
The root locus is shown in Figure DP7.14 for τ= 6. As τ0, the
dominant closed-loop pole approaches s= 0 as KD→ ∞. As τ10,
the dominant closed-loop pole approaches s=10 as KD→ ∞. A
viable controller is KP= 72 and KD= 12 when τ= 6.
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368 CHAPTER 7 The Root Locus Method
−25 −20 −15 −10 −5 0 5
−8
−6
−4
−2
0
2
4
6
8
Root Locus
Real Axis
Imaginary Axis
FIGURE DP7.14
Root locus when τ= 6.
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Computer Problems 369
Computer Problems
CP7.1 The root locus for parts (a)-(d) are shown in Figures CP7.1a - CP7.1d.
−40 −30 −20 −10 0 10 20
−30
−20
−10
0
10
20
30
num=[30]; den=[1 14 43 30]; rlocus(sys)
Real Axis
Imaginary Axis
−70 −60 −50 −40 −30 −20 −10 0 10
−20
−15
−10
−5
0
5
10
15
20
num=[1 20]; den=[1 4 20]; rlocus(sys)
Real Axis
Imaginary Axis
FIGURE CP7.1
(a) Root locus for 1 + k30
s3+14s2+43s+30 = 0.(b) Root locus for 1 + ks+20
s2+4s+20 = 0.
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370 CHAPTER 7 The Root Locus Method
−6 −5 −4 −3 −2 −1 0 1
−1.5
−1
−0.5
0
0.5
1
1.5
num=[1 1 2]; den=[1 6 10 0]; rlocus(sys)
Real Axis
Imaginary Axis
−7 −6 −5 −4 −3 −2 −1 0 1
−1.5
−1
−0.5
0
0.5
1
1.5
num=[1 4 6 10 6 4]; den=[1 4 4 1 1 10 1]; rlocus(sys)
Real Axis
Imaginary Axis
FIGURE CP7.1
CONTINUED: (c) Root locus for 1 + ks2+s+2
s(s2+6s+10) = 0.(d) Root locus for 1 +
ks5+4s4+6s3+10s2+6s+4
s6+4s5+4s4+s3+s2+10s+1 = 0.
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Computer Problems 371
CP7.2 The maximum value of the gain for stability is k= 0.791. The m-file
script and root locus is shown in Figure CP7.2.
num=[1 -2 2]; den=[1 3 2 0]; sys = tf(num,den);
rlocus(sys)
rlocnd(sys)
Select a point in the graphics window
selected_point =
-0.0025 + 0.6550i
ans =
0.8008
−3 −2.5 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2
−1
−0.8
−0.6
−0.4
−0.2
0
0.2
0.4
0.6
0.8
1
Real Axis
Imag Axis
FIGURE CP7.2
Using the rlocfind function.
The value of k= 0.8008 selected by the rlocfind function is not exact
since you cannot select the jω-axis crossing precisely. The actual value is
determined using Routh-Hurwitz analysis.
CP7.3 The partial fraction expansion of Y(s) is
Y(s) = s+ 6
s(s2+ 5s+ 4) =0.1667
s+ 4 1.6667
s+ 1 +1.5
s.
The m-file script and output is shown in Figure CP7.3.
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372 CHAPTER 7 The Root Locus Method
num=[1 6]; den=[1 5 4 0];
[r,p,k]=residue(num,den)
r =
0.1667
-1.6667
1.5000
p =
-4
-1
0
k =
[]
FIGURE CP7.3
Using the residue function.
CP7.4 The characteristic equation is
1 + ps1
s2+ 5s+ 10 = 0 .
The root locus is shown in Figure CP7.4. The closed-loop system is stable
for
0< p < 10 .
−8 −7 −6 −5 −4 −3 −2 −1 0 1 2
−2.5
−2
−1.5
−1
−0.5
0
0.5
1
1.5
2
2.5
Real Axis
Imaginary Axis
FIGURE CP7.4
Root locus for 1 + ps1
s2+5s+10 = 0.
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from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying,
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Computer Problems 373
CP7.5 The characteristic equation is
1 + ks+ 1
s2= 0 .
The root locus is shown in Figure CP7.5. For k= 2 we obtain s1,2=
1±j, that is, we have ζ= 0.707.
num=[1 1]; den=[1 0 0]; sys = tf(num,den);
hold o, clf
rlocus(sys);
hold on
plot([0 -2],[0 2*tan(acos(0.707))],'--')
plot([0 -2],[0 -2*tan(acos(0.707))],'--')
plot([-1 -1],[1 -1],'*')
-3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1
-2.5
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
2.5
Real Axis
Imag Axis
FIGURE CP7.5
Root locus for 1 + ks+1
s2= 0.
CP7.6 We choose a controller with two real poles and two real zeros selected to
meet the steady-state specification. The characteristic equation is
1 + K(s+ 5.5)(s+ 0.01)
(s+ 6.5)(s+ 0.0001)
10
s3+ 15s2+ 50s= 0 .
The m-file and root locus is shown in Figure CP7.4a. From the root lo-
cus we can select the value of the gain Kthat results in an estimated
P.O. 5% and a ζωn≤ −2 to meet the settling time specification. We
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from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying,
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374 CHAPTER 7 The Root Locus Method
select K= 8.58. The step response is shown in Figure CP7.6b showing
the percent overshoot and settling time specifications are satisfied. The
velocity constant is kv= 145.2 which implies a steady-state error to a
ramp input of ess = 1/kv= 0.0069.
−40 −35 −30 −25 −20 −15 −10 −5 0 5 10
−25
−20
−15
−10
−5
0
5
10
15
20
25
System: untitled1
Gain: 8.58
Pole:
Damping: 0.807
Overshoot (%): 1.37
Frequency (rad/sec): 2.58
Real Axis
Imaginary Axis
ng=10; dg=conv([1 10 0],[1 5]); s
ysg=tf(ng,dg);
nh=conv([1 0.01],[1 5.5]);
dh=conv([1 6.5],[1 0.0001]);
sysh=tf(nh,dh);
gure(1)
rlocus(sysg*sysh)
K=8.58; sysh=tf(K*nh,dh);
sys=series(sysg,sysh);syscl=feedback(sys,1)
gure(2)
subplot(121)
step(syscl);
Kv=10*8.58*0.01*5.5/10/6.5/0.0001/5
systd=feedback(sysg,sysh);
subplot(122)
step(systd)
Step Response
Time (sec)
Amplitude
0 200 400 600
0
0.02
0.04
0.06
0.08
0.1
0.12
0.14
Disturbance Response
Time (sec)
y(t)/Q
012345
0
0.2
0.4
0.6
0.8
1
1.2
1.4
System: syscl
Peak amplitude: 1.02
Overshoot (%): 1.77
At time (sec): 2.14
System: syscl
Settling Time (sec): 1.51
FIGURE CP7.6
(a) Root locus. (b) Step response and disturbance response.
CP7.7 The m-file script to generate the root locus for each controller in parts
(a)-(c) is shown in Figure CP7.7. The performance region is indicated on
each root locus in Figures CP7.7b - CP7.7d. For part (a), the controller
gain is found to be Gc(s) = 11.3920. The integral controller in part (b)
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Computer Problems 375
ÈSelect a point in the graphics window
selected_point =
-2.5030 + 3.3380i
ans =
11.3920
Select a point in the graphics window
selected_point =
-0.6690 + 0.8210i
ans =
4.0930
Select a point in the graphics window
selected_point =
-2.0695+ 2.7387i
ans =
9.2516
on graph.Plot performance region boundaries
numg=[1]; deng=[1 5 6]; sysg = tf(numg,deng);
t=[0:0.1:15];
%
% Part (a)
%
sys1 = sysg;
rlocus(sys1), grid
hold on
plot([-0.4 -0.4],[-6 6],'--',...
[0 -6*tan(36.2*pi/180)],[0 6],'--',...
[0 -6*tan(36.2*pi/180)],[0 -6],'--')
hold o
[kp,poles] = rlocnd(sys1)
%
% Part (b)
%
numc=[1]; denc=[1 0]; sysc = tf(numc,denc);
sys2 = series(sysc,sysg);
gure
rlocus(sys2), grid
hold on
plot([-0.4 -0.4],[-6 6],'--',...
[0 -6*tan(36.2*pi/180)],[0 6],'--',...
[0 -6*tan(36.2*pi/180)],[0 -6],'--')
hold o
[ki,poles] = rlocnd(sys2)
%
% Part (c)
%
gure
numc=[1 1]; denc=[1 0]; sysc = tf(numc,denc);
sys3 = series(sysc,sysg);
rlocus(sys3), grid
hold on
plot([-0.4 -0.4],[-6 6],'--',...
[0 -6*tan(36.2*pi/180)],[0 6],'--',...
[0 -6*tan(36.2*pi/180)],[0 -6],'--')
hold o
[kpi,poles] = rlocnd(sys3)
%
% Part (d)
%
gure
sys1_o = kp*sys1; sys1_cl = feedback(sys1_o,[1]);
sys2_o = ki*sys2; sys2_cl = feedback(sys2_o,[1]);
sys3_o = kpi*sys3; sys3_cl = feedback(sys3_o,[1]);
[y1,t]=step(sys1_cl,t);
[y2,t]=step(sys2_cl,t);
[y3,t]=step(sys3_cl,t);
plot(t,y1,t,y2,'--',t,y3,':'),grid
xlabel('time [sec]'),ylabel('y(t)')
title('Gc(s): proportional (solid), integral (dashed) & PI (dotted)')
FIGURE CP7.7
(a) Script to generate the root locus for each controller.
is determined to be
Gc(s) = 4.093
s.
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from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying,
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376 CHAPTER 7 The Root Locus Method
-4 -3 -2 -1 0 1 2
-6
-4
-2
0
2
4
6
Real Axis
Imag Axis
FIGURE CP7.7
CONTINUED: (b) Root locus for proportional controller with selected K= 11.3920.
The proportional integral (PI) controller in part (c) is
-4 -3 -2 -1 0 1 2
-6
-4
-2
0
2
4
6
Real Axis
Imag Axis
FIGURE CP7.7
CONTINUED: (c) Root locus for integral controller with selected K= 4.0930.
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Computer Problems 377
Gc(s) = 9.2516(s+ 1)
s.
The proportional controller is stable for all K > 0 but has a significant
-4 -3 -2 -1 0 1 2
-6
-4
-2
0
2
4
6
Real Axis
Imag Axis
FIGURE CP7.7
CONTINUED: (d) Root locus for PI controller with selected K= 9.2516.
steady-state error. The integral controller has no steady-state error,
but is stable only for K < 30. The PI controller has zero steady-state
error and is stable for all K > 0. Additionally, the PI controller has a
fast transient response. The step responses for each controller is shown in
Figure CP7.7e.
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378 CHAPTER 7 The Root Locus Method
0 5 10 15
0
0.2
0.4
0.6
0.8
1
1.2
1.4
time [sec]
y(t)
Gc(s): proportional (solid), integral (dashed) & PI (dotted)
FIGURE CP7.7
CONTINUED: (e) Step responses for each controller.
CP7.8 The loop transfer function can be written as
Gc(s)G(s) = K1+K2s
Js2=¯
K2
s+ 5
s2
where
¯
K2=K2/J .
The parameter of interest for the root locus is ¯
K2. The root locus is shown
in Figure CP7.8. The selected value of
¯
K2= 7.1075 .
Therefore,
K2
J= 7.1075 and K1
J= 35.5375 .
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Computer Problems 379
-10
-8
-6
-4
-2
0
2
4
6
8
10
-10 -8 -6 -4 -2 0 2 4 6 8 10
xxo
Real Axis
Imag Axis
num=[1 5]; den=[1 0 0]; sys=tf(num,den); rlocus(sys); rlocnd(sys)
+
+
FIGURE CP7.8
Root locus to determine ¯
K2.
CP7.9 The value of Kthat results in a damping ratio of ζ= 0.707 is K= 5.2.
The root locus is shown in Figure CP7.9.
−10 −5 0 5
−5
−4
−3
−2
−1
0
1
2
3
4
5
Root Locus
Real Axis
Imaginary Axis
s = -0.68 + 0.68j
s = -6.63
s = -0.68 - 0.68j
FIGURE CP7.9
Root locus for 1 + K1
s3+8s2+10s+1 = 0.
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380 CHAPTER 7 The Root Locus Method
CP7.10 (a) The characteristic equation is
s3+ (2 + k)s2+ 5s+ 1 = 0 .
(b) The Routh array is
s31 5
s22 + k1
s15k+9
2+k
so1
?-2.5 ?-2 ?-1.5 ?-1 ?-0.5 0
?-2
?-1.5
?-1
?-0.5
0
0.5
1
1.5
2
Root Locus
Real Axi s
Imaginary Axi s
FIGURE CP7.10
Root locus for 1 + ks2
s3+2s2+5s+1 = 0.
For stability we require
2 + k > 0 or k >2
and
5k+ 9 >0 or k >9/5.
Therefore, the stability region is defined by
k > 1.8.
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Computer Problems 381
(c) Rearranging the characteristic equation yields
1 + ks2
s3+ 2s2+ 5s+ 1 = 0 .
The root locus is shown in Figure CP7.10. We see that the system is
stable for all k > 0.
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C H A P T E R 8
Frequency Response Methods
Exercises
E8.1 Given the loop transfer function
L(s) = 4
(s+ 2)2,
we determine that
|L(jω)|=4
4 + ω2and φ(ω) = 2 tan1ω/2.
The frequency response is shown in Figure E8.1.
−80
−60
−40
−20
0
Magnitude (dB)
10−2 10−1 100101102
−180
−135
−90
−45
0
Phase (deg)
Bode Diagram
Frequency (rad/sec)
FIGURE E8.1
Frequency response for L(s) = 4
(s+2)2.
382
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Exercises 383
The magnitude and phase angle for ω= 0,0.5,1,2,4,are summarized
in Table E8.1.
ω0 0.5 1 2 4
|L(jω)|1 0.94 0.80 0.50 0.20 0
φ(deg) 0 -28.07 -53.13 -90 –126.87 -180
TABLE E8.1 Magnitude and phase for L(s) = 4
(s+2)2.
E8.2 The transfer function is
G(s) = 5000
(s+ 70)(s+ 500) .
The frequency response plot is shown in Figure E8.2. The phase angle is
computed from
φ=tan1ω
70 tan1ω
500 .
The phase angles for ω= 10,100 and 700 are summarized in Table E8.2.
ω10 200 700
|G(jω)|-16.99 -27.17 -41.66
φ(deg) -9.28 -92.51 -138.75
TABLE E8.2 Magnitude and phase for G(s) = 5000
(s+70)(s+500) .
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384 CHAPTER 8 Frequency Response Methods
−100
−80
−60
−40
−20
0
Magnitude (dB)
100101102103104
−180
−135
−90
−45
0
Phase (deg)
Bode Diagram
Frequency (rad/sec)
FIGURE E8.2
Frequency response for G(s) = 5000
(s+70)(s+500) .
E8.3 The loop transfer function is
L(s) = 300(s+ 100)
s(s+ 10)(s+ 40) .
The phase angle is computed via
φ(ω) = 90otan1ω
10 tan1ω
40 + tan1ω
100 .
At ω= 28.3, we determine that
φ=90o70.5o35.3o+ 15.8o= 180o.
Computing the magnitude yields
|L(jω)|=300(100)(1 + ( ω
100 )2)1
2
ω10(1 + ( ω
10 )2)1
240(1 + ( ω
40 )2)1
2
= 0.75 ,
when ω= 28.3. We can also rewrite L(s) as
L(s) = 75( s
100 + 1)
s(s
10 + 1)( s
40 + 1).
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Exercises 385
Then, the magnitude in dB is
20 log10 |L|= 20 log10(75) + 10 log10(1 + ( ω
100 )2)10 log10(1 + ( ω
10)2)
10 log10(1 + ( ω
40)2)20 log10 ω=2.5 dB ,
at ω= 28.3.
E8.4 The transfer function is
G(s) = Ks
(s+a)(s+ 10)2.
Note that φ= 0oat ω= 3, and that
φ= +90otan1ω
a2 tan1ω
10 .
Substituting ω= 3 and solving for ayields
a= 2 .
Similarly, from the magnitude relationship we determine that
K= 400 .
E8.5 The lower portion for ω < 2 is
20 log K
ω= 0 dB ,
at ω= 8. Therefore,
20 log K
8= 0 dB
which occurs when
K= 8 .
We have a zero at ω= 2 and another zero at ω= 4. The zero at ω= 4
yields
a= 0.25 .
We also have a pole at ω= 8, and a second pole at ω= 24. The pole at
ω= 24 yields
b= 1/24 .
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386 CHAPTER 8 Frequency Response Methods
Therefore,
G(s) = 8(1 + s/2)(1 + s/4)
s(1 + s/8)(1 + s/24)(1 + s/36) .
E8.6 The loop transfer function is
L(s) = 10
s(s/5 + 1)(s/100 + 1) .
The Bode diagram is shown in Figure E8.6. When 20 log10 |L(jω)|= 0 dB,
we have
ω= 9.4 rad/sec .
−150
−100
−50
0
50
Magnitude (dB)
10−1 100101102103104
−270
−225
−180
−135
−90
Phase (deg)
Bode Diagram
Frequency (rad/sec)
FIGURE E8.6
Bode Diagram for L(s) = 10
s(s/5+1)(s/100+1) .
E8.7 The transfer function is
T(s) = 4
(s2+s+ 1)(s2+ 0.4s+ 4) .
(a) The frequency response magnitude is shown in Figure E8.7.
The frequency response has two resonant peaks at
ωr1= 0.8 rad/sec and ωr2= 1.9 rad/sec .
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Exercises 387
-10
-5
0
5
10
10-1 100101
Gain dB
Frequency (rad/sec)
0
0.5
1
1.5
02 4 6 8 10 12 14 16 18 20
Time (secs)
Amplitude
FIGURE E8.7
(a) Bode Diagram for T(s) = 4
(s2+s+1)(s2+0.4s+4) . (b) Unit step response.
(b) The percent overshoot is
P.O. = 35% ,
and the settling time is
Ts16 sec .
(c) The step response is shown in Figure E8.7.
E8.8 (a) The break frequencies are
ω1= 1 rad/sec, ω2= 5 rad/sec,and ω3= 20 rad/sec .
(b) The slope of the asymptotic plot at low frequencies is 0 dB/decade.
And at high frequencies the slope of the asymptotic plot is -20 dB/decade.
(c) The Bode plot is shown in Figure E8.8.
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388 CHAPTER 8 Frequency Response Methods
−30
−20
−10
0
10
20
Magnitude (dB)
10−2 10−1 100101102103
−90
−45
0
45
90
135
180
Phase (deg)
Bode Diagram
Frequency (rad/sec)
FIGURE E8.8
Bode Diagram for Gc(s)G(s) = 100(s1)
s2+25s+100 .
E8.9 The Bode diagram for G(s) is shown in Figure E8.9.
-40
-20
0
20
40
10-1 100101102103
Frequency (rad/sec)
Gain dB
-50
0
50
10-1 100101102103
Frequency (rad/sec])
Phase deg
FIGURE E8.9
Bode Diagram for G(s) = (s/5+1)(s/20+1)
(s+1)(s/80+1) .
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Exercises 389
E8.10 The frequency response has two peaks; the first peak at f1.8 and the
second peak at f3.1. One possible G(jω) is
G(jω) = 1
(jωτ + 1) 1 + 2ζ1
ωn1jω +jω
ωn121 + 2ζ2
ωn2jω +jω
ωn22,
where
τ=1
2π(0.2) ,
ζ1= 0.15; ωn1= 2π(1.8×103)ζ2= 0.15; ωn2= 2π(3.1×103).
The damping ratios are estimated using Figure 8.10 in Dorf & Bishop.
E8.11 The Bode plot is shown in Figure E8.11. The frequency when 20 log10 |GCG(ω)|=
0 is ω= 9.9 rad/sec.
−120
−100
−80
−60
−40
−20
0
20
Magnitude (dB)
10−1 100101102103
−270
−225
−180
−135
−90
−45
0
Phase (deg)
Bode Diagram
Frequency (rad/sec)
FIGURE E8.11
Bode Diagram for Gc(s)G(s) = 1000
(s2+10s+100)(s+2) .
E8.12 (a) The transfer function is
G(s) = C(sIA)1B+D=5(s1)
s2+ 3s+ 2 .
(b) The Bode plot is shown in Figure E8.12.
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390 CHAPTER 8 Frequency Response Methods
-20
-10
0
10
Magnitude (dB)
10 -2 10 -1 100101102
90
180
270
Phase (deg)
Bode Diagram
Frequency (rad/sec)
FIGURE E8.12
Bode Diagram for G(s) = 5(s‘1)
s2+3s+2 .
E8.13 The closed-loop transfer function is
T(s) = 100
s3+ 11s2+ 20s+ 110 .
The Bode plot of T(s) is shown in Figure E8.13, where ωB= 4.9 rad/sec.
-100
-50
0
50
Magnitude (dB)
10 -1 100101102103
-270
-225
-180
-135
-90
-45
Phase (deg)
Bode Diagram
ωb=4.9
-3 dB
Frequency (rad/sec)
FIGURE E8.13
Bode Diagram for T(s) = 100
s3+11s2+20s+110 .
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Exercises 391
E8.14 The loop transfer function is
L(s) = 20
(s2+ 1.4s+ 1)(s+ 10) .
The Bode plot of L(s) is shown in Figure E8.14. The frequency when
20 log10 |L(ω)|= 0 is ω= 1.32 rad/sec.
−150
−100
−50
0
50
Magnitude (dB)
10−2 10−1 100101102103
−270
−225
−180
−135
−90
−45
0
Phase (deg)
Bode Diagram
Frequency (rad/sec)
FIGURE E8.14
Bode Diagram for L(s) = 20
(s2+1.4s+1)(s+10) .
E8.15 The closed-loop transfer function is
T(s) = 3s+ 5
s2+s+K+ 6 .
The bandwidth as a function of Kis shown in Figure E8.15. The band-
width as a function of Kis:
(a) K= 1 and ωb= 7.0 rad/sec.
(b) K= 2 and ωb= 7.9 rad/sec.
(c) K= 10 and ωb= 14.7 rad/sec.
The bandwidth increases as Kincreases.
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392 CHAPTER 8 Frequency Response Methods
0 2 4 6 8 10 12 14 16 18 20
6
8
10
12
14
16
18
20
22
24
K
ωb (rad/s)
FIGURE E8.15
Bandwith of T(s) = 3s+5
s2+s+K+6 .
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Problems 393
Problems
P8.1 (a) The transfer function is
Gc(s)G(s) = 1
(1 + 0.25s)(1 + 3s),
and
Gc(jω)G(jω) = 1
(1 0.75ω2) + j3.25ω.
The polar plot is shown in Figure P8.1a. A summary of the magni-
tude and phase angles for ω= 0,0.5,1,2,5 and can be found in
Table P8.1a.
−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1
−0.8
−0.6
−0.4
−0.2
0
0.2
0.4
0.6
0.8
Nyquist Diagram
Real Axis
Imaginary Axis
FIGURE P8.1
(a) Polar plot for Gc(s)G(s) = 1
(1+0.25s)(1+3s).
ω0 0.5 1 2 5
|Gc(jω)G(jω)|(dB) 1.00 0.55 0.31 0.15 0.04 0
φ(deg) 0 -63.4 -85.6 -107.1 -137.51 -180
TABLE P8.1 (a) Magnitudes and phase angles for Gc(s)G(s) = 1
(1+0.25s)(1+3s).
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394 CHAPTER 8 Frequency Response Methods
(b) The transfer function is
Gc(s)G(s) = 5(s2+ 1.4s+ 1)
(s1)2
and
Gc(jω)G(jω) = 5(1 ω2) + 1.4jω
(1 ω2)2jω .
The polar plot is shown in Figure P8.1b. A summary of the magnitude
and phase angles for ω= 0,0.25,0.5,1,2,8,16 and can be found in
Table P8.1b.
−4 −3 −2 −1 0 1 2 3 4 5 6
−5
−4
−3
−2
−1
0
1
2
3
4
5
Nyquist Diagram
Real Axis
Imaginary Axis
FIGURE P8.1
CONTINUED: (b) Polar plot for Gc(s)G(s) = 5(s2+1.4s+1)
(s1)2.
ω0 0.25 0.5 1 2 8 16
|Gc(jω)G(jω)|(dB) 5.00 4.71 4.10 3.50 4.10 4.92 4.98 5.00
φ(deg) 0 48.5 96.1 180.0 -96.2 -24.3 -12.2 0
TABLE P8.1 CONTINUED: (b) Magnitudes and phase angles for Gc(s)G(s) = 5(s2+1.4s+1)
(s1)2.
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Problems 395
(c) The transfer function is
Gc(s)G(s) = (s8
(s2+ 6s+ 8) .
The polar plot is shown in Figure P8.1c. A summary of the magnitude
and phase angles for
ω= 0,1,2,3,4,5,6,
can be found in Table P8.1c.
−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4
−0.8
−0.6
−0.4
−0.2
0
0.2
0.4
0.6
0.8
Nyquist Diagram
Real Axis
Imaginary Axis
FIGURE P8.1
CONTINUED: (c) Polar plot for Gc(s)G(s) = s8
s2+6s+8 .
ω0 1 2 3 4 5 6
|Gc(jω)G(jω)|(dB) 1.00 0.87 0.65 0.47 0.35 0.27 0.22 0.00
φ(deg) 180.0 132.3 94.4 66.3 45.0 28.5 15.3 -90.0
TABLE P8.1 CONTINUED: (c) Magnitudes and phase angles for Gc(s)G(s) = s8
s2+6s+8 .
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396 CHAPTER 8 Frequency Response Methods
(d) The transfer function is
Gc(s)G(s) = 20(s+ 8)
s(s+ 2)(s+ 4) .
The polar plot is shown in Figure P8.1d. A summary of the magnitude
and phase angles for
ω= 1,0.1,0.8,1.6,3.2,12.8,
can be found in Table P8.1d.
−20 −15 −10 −5 0 5
−20
−15
−10
−5
0
5
10
15
20
Nyquist Diagram
Real Axis
Imaginary Axis
FIGURE P8.1
CONTINUED: (d) Polar plot for Gc(s)G(s) = 20(s+8)
s(s+2)(s+4) .
ω0 0.1 0.8 1.6 3.2 12.8
|Gc(jω)G(jω)|(dB) 199.70 22.87 9.24 2.79 0.14 0.00
φ(deg) 0 -93.6 -117.4 -139.1 -164.8 174.3 180.0
TABLE P8.1 CONTINUED: (d) Magnitudes and phase angles for Gc(s)G(s) = 20(s+8)
s(s+2)(s+4) .
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Problems 397
P8.2 (a) The Bode plot is shown in Figure P8.2a. A summary of the magnitude
and phase angles for
ω= 0.25,0.5,1,2,4,8,16
can be found in Table P8.2a.
−80
−60
−40
−20
0
Magnitude (dB)
10−2 10−1 100101102
−180
−135
−90
−45
0
Phase (deg)
Bode Diagram
Frequency (rad/sec)
FIGURE P8.2
(a) Bode plot for Gc(s)G(s) = 1
(1+0.25s)(1+3s).
ω0.25 0.5 1.0 2.0 4.0 8.0 16.0
|Gc(jω)G(jω)|(dB) -1.95 -5.19 -10.26 -16.65 -24.62 -34.60 -45.93
φ(deg) -40.5 -63.4 -85.6 -107.1 -130.2 -151.0 -164.8
TABLE P8.2 (a) Magnitudes and phase angles for Gc(s)G(s) = 1
(1+0.25s)(1+3s).
(b) The transfer function is
Gc(s)G(s) = 5(s2+ 1.4s+ 1)
(s1)2
The Bode plot is shown in Figure P8.2b. A summary of the magnitude
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398 CHAPTER 8 Frequency Response Methods
and phase angles for
ω= 0.25,0.5,1,2,4,8,16
can be found in Table P8.2b.
10
11
12
13
14
Magnitude (dB)
10−1 100101
−360
−270
−180
−90
0
Phase (deg)
Bode Diagram
Frequency (rad/sec)
FIGURE P8.2
CONTINUED: (b) Bode plot for Gc(s)G(s) = 5(s2+1.4s+1)
(s1)2.
ω0.25 0.5 1.0 2.0 4.0 8.0 16.0
|Gc(jω)G(jω)|(dB) 13.46 12.26 10.88 12.26 13.46 13.84 13.95
φ(deg) 48.5 96.2 180.0 -96.2 -48.5 -24.3 -12.2
TABLE P8.2 CONTINUED: (b) Magnitudes and phase angles for Gc(s)G(s) = 5(s2+1.4s+1)
(s1)2.
(c) The transfer function is
Gc(s)G(s) = (s8)
(s2+ 6s+ 8) .
The Bode plot is shown in Figure P8.2c. A summary of the magnitude
and phase angles for ω= 0.6,1,2,3,4,5,6,can be found in
Table P8.2c.
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Problems 399
−60
−50
−40
−30
−20
−10
0
Magnitude (dB)
10−2 10−1 100101102103
−90
−45
0
45
90
135
180
Phase (deg)
Bode Diagram
Frequency (rad/sec)
FIGURE P8.2
CONTINUED: (c) Bode plot for Gc(s)G(s) = s8
s2+6s+8 .
ω0.6 1 2 3 4 5 6
|Gc(jω)G(jω)|(dB) -0.45 -1.17 -3.72 -6.49 -9.03 -11.26 -13.18 -120.00
φ(deg) 150.5 132.3 94.4 66.3 45.0 28.5 15.3 -90.0
TABLE P8.2 CONTINUED: (c) Magnitudes and phase angles for Gc(s)G(s) = s8
s2+6s+8 .
(d) A summary of the magnitude and phase angles for
ω= 0.2,0.8,3.2,6.4,12.8,25.6,51.2
can be found in Table P8.2d. The Bode plot is shown in Figure P8.2d.
ω0.2 0.8 3.2 6.4 12.8 25.6 51.2
|Gc(jω)G(jω)|(dB) 39.95 27.19 8.90 -3.98 -17.35 -30.0355 -42.28
φ(deg) -97.1 -117.4 -164.8 178.0 174.2 176.0 177.8
TABLE P8.2 CONTINUED: (d) Magnitudes and phase angles for Gc(s)G(s) = 20(s+8)
s(s+2)(s+4) .
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400 CHAPTER 8 Frequency Response Methods
−60
−40
−20
0
20
40
60
Magnitude (dB)
10−1 100101102
−225
−180
−135
−90
Phase (deg)
Bode Diagram
Frequency (rad/sec)
FIGURE P8.2
CONTINUED: (d) Bode plot for Gc(s)G(s) = 20(s+8)
s(s+2)(s+4) .
P8.3 (a) The bridged-T network we found has zeros at
s=±jωn
and poles at
s=ωn
Q±ωnq1/Q21.
The frequency response is shown in Figure P8.3 for Q= 10.
(b) For the twin-T network, we evaluate the magnitude at
ω= 1.1ωn
or 10% from the center frequency (see Example 8.4 in Dorf & Bishop).
This yields
|G| ≈ 2.1×0.1
3.9×1.1 = 0.05 .
Similarly, for the bridged-T network
|G|= 2.1×0.1
2.1×0.14 = 0.707 .
The bridged-T network possesses a narrower band than the twin-T
network.
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Problems 401
-40
-30
-20
-10
0
10-1 100101
w/wn
Gain dB
-100
-50
0
50
100
10-1 100101
w/wn
Phase deg
FIGURE P8.3
Bode plot for G(s) = s2+ω2
n
s2+(2ωn/Q)s+ω2
n, where ζ= 1/Q = 0.1.
P8.4 The transfer function is
G(s) = GcG1H(s)1
s
=30000(2s+ 1)
s(s+ 10)(s+ 20)(s2+ 15s+ 150) .
A summary of the magnitude and phase angles can be found in Table P8.4.
The Bode plot is shown in Figure P8.4.
ω1 3 5 8 10 15 24
|G(jω|dB 6.95 5.78 5.08 3.38 1.59 -5.01 -17.56
φ(deg)40.89o52.39o77.28o118.41o145.99o203.52o258.57o
TABLE P8.4 Magnitudes and phase angles for GH(s) = 30000(2s+1)
s(s+10)(s+20)(s2+15s+150) .
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402 CHAPTER 8 Frequency Response Methods
-150
-100
-50
0
50
Magnitude (dB)
10-2 10-1 100101102103
-360
-270
-180
-90
0
Phase (deg)
Bode Diagram
Frequency (rad/sec)
FIGURE P8.4
Bode plot for GH(s) = 30000(2s+1)
s(s+10)(s+20)(s2+15s+150) .
P8.5 The Bode plot is shown in Figure P8.5.
−250
−200
−150
−100
−50
0
50
Magnitude (dB)
10−2 10−1 100101102103104
−360
−270
−180
−90
0
Phase (deg)
Bode Diagram
Frequency (rad/sec)
FIGURE P8.5
Bode plot for G(s) = 10
(s/4+1)(s+1)(s/20+1)(s/80+1) .
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Problems 403
P8.6 (a) The transfer function is
GH(s) = 3.98(1 + s/1)
s(1 + s/10)2.
We have a zero at ω= 1 and two poles at ω= 10.0. The low frequency
approximation is K/s and at ω= 1 we have
20 log K
ω= 12dB .
Therefore,
K= 3.98
at ω= 1 (an approximation). The phase plot is shown in Figure P8.6a.
10-2 10-1 100101102
-180
-160
-140
-120
-100
-80
-60
-40
Phase deg
(a)
10-1 100101102103
-100
-50
0
50
100
Frequency (rad/sec)
Phase deg
(b)
FIGURE P8.6
Phase plots for (a) G(s) = 3.98(s/1+1)
s(s/10+1)2. (b) G(s) = s
(s/10+1)(s/50+1) .
(b) The transfer function is
GH(s) = s
(1 + s/10)(1 + s/50) .
The poles are located by noting that the slope is ±20 dB/dec. The
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404 CHAPTER 8 Frequency Response Methods
low frequency approximation is Ks, so
20 log Kω = 0dB .
At ω= 1 we determine that
K= 1 .
The phase plot is shown in Figure P8.6b.
P8.7 The loop transfer function is
L(s) = Kv
s(s/π + 1)2.
(a) Set Kv= 2π. The Bode plot is shown in Figure P8.7a.
-20
0
20
40
10-1 100101
Frequency (rad/sec)
Gain dB
-180
-160
-140
-120
-100
-80
10-1 100101
Frequency (rad/sec)
Phase deg
FIGURE P8.7
(a) Bode plot for L(s) = Kv
s(s/π+1) , where Kv= 2π.
(b) The logarithmic magnitude versus the phase angle is shown in Fig-
ure P8.7b.
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Problems 405
-20
-10
0
10
20
30
40
-170 -160 -150 -140 -130 -120 -110 -100 -90
Phase deg
Gain dB
FIGURE P8.7
CONTINUED: (b) Log-magnitude-phase curve for L(jω).
P8.8 The transfer function is
T(s) = K
s2+ 10s+K.
(a) When P.O. = 15%, we determine that ζ= 0.517 by solving
15 = 100eπζ/1ζ2.
So, 2ζωn= 10 implies that ωn= 9.67, hence K=ω2
n= 93.53. Also,
M= (2ζq1ζ2)1= 1.13 .
(b) For second-order systems we have
ωr=ωnq12ζ2= 6.59
when ζ= 0.517 and ωn= 9.67.
(c) We estimate ωBto be
ωB(1.19ζ+ 1.85)ωn= 11.94 rad/s.
P8.9 The log-magnitude phase curves are shown in Figure P8.9.
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406 CHAPTER 8 Frequency Response Methods
-45
-40
-35
-30
-25
-20
-15
-10
-5
0
-200 -150 -100 -50 0
Phase deg
Gain dB
(a)
-30
-20
-10
0
10
20
30
40
-180 -160 -140 -120 -100
Phase deg
Gain dB
(b)
FIGURE P8.9
Log-magnitude-phase curve for (a) Gc(s)G(s) = 1
(1+0.5s)(1+2s)and (b) Gc(s)G(s) =
1+0.5s
s2.
P8.10 The governing equations of motion are
F(s) = KfIf(s) and If(s) = Vf(s)
Rf+Lfs.
Without loss of generality we can let Kf= 1.0. Also, we have
F(s) = (Ms2+bs +K)Y(s).
Therefore, the transfer function is
GH(s) = KKf
(Rf+Lfs)(Ms2+bs +K)=50K
(s+ 0.5)(s2+ 2s+ 4) .
This is a type 0 system, therefore Kp= 25K.
(a) If we allow a 1% error , we have ess =|R|/(1 + Kp) = 0.01|R|. Thus
Kp= 25K= 99. Select
K= 4 .
(b) The Bode plot is shown in Figure P8.10a.
(c) The log-magnitude phase curve is shown in Figure P8.10b.
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Problems 407
-20
0
20
40
10-2 10-1 100101
Frequency (rad/sec)
Gain dB
-300
-200
-100
0
10-2 10-1 100101
Frequency (rad/sec)
Phase deg
FIGURE P8.10
(a) Bode plot for GH(s) = 200
(s2+2s+4)(s+0.5) .
-20
-10
0
10
20
30
40
-300 -250 -200 -150 -100 -50 0
Phase deg
Gain dB
FIGURE P8.10
CONTINUED: (b) Log-magnitude-phase curve for GH(s) = 200
(s2+2s+4)(s+0.5) .
(d) The closed-loop transfer function Bode plot is shown in Figure P8.10c.
We determine from the plot that M= 1.6, ωr= 4.4 and ωB= 6.8.
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408 CHAPTER 8 Frequency Response Methods
-15
-10
-5
0
5
10-1 100101
Frequency (rad/sec)
Gain dB
-200
-100
0
100
10-1 100101
Frequency (rad/sec)
Phase deg
FIGURE P8.10
CONTINUED: (c) Bode plot for closed-loop T(s) = Y(s)/R(s).
P8.11 The Bode plot is shown in Figure P8.11.
-100
0
100
200
10-4 10-3 10-2 10-1 100101
Frequency (rad/sec)
Gain dB
-200
-100
0
100
10-4 10-3 10-2 10-1 100101
Frequency (rad/sec)
Phase deg
FIGURE P8.11
Bode plot for G(s) = 0.164(s+0.2)(s+0.32)
s2(s+0.25)(s0.009) .
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Problems 409
P8.12 The three transfer functions are
G1(s) = 10 G2(s) = 1
s(s/0.6 + 1) G3(s) = 3s .
(a) When G3(s) is out of the loop, the characteristic equation is
1 + G1G2(s) = 1 + 10
s(s/0.6 + 1) = 0
or s2+ 0.6s+ 6 = 0. Thus, ζ= 0.6/(26) = 0.12.
(b) With G3(s), the characteristic equation is
1 + G1G2(s) + G2G3(s) = 1 + 6
s(s+ 0.6) +1.85
s(s+ 0.6) = 0 ,
or
s2+ 2.4s+ 6 = 0 .
Thus, ζ= 2.4/(26) = 0.49.
P8.13 By inspection of the frequency response, we determine
L(s) = Gc(s)G(s)H(s) = K
s(s/100 + 1)(s/1000 + 1)2.
For small ω, we have 20 log K/ω = 40 dB at ω= 10. So, K= 1000.
P8.14 The data we have are R1=R2= 1000Ω, c1= 107farad and c2= 106
farad. The governing equations are
V2(s)
V1(s)=
1
C1s
R1+1
C1s
,
and
Vo(s)
V2(s)=KR2
R2+1
C2s
.
So
Vo(s)
V1(s)=KR2C2s
(R1C1s+ 1)(R2C2s+ 1) =109s
(s+ 107)(s+ 1000) .
(a) The Bode plot is shown in Figure P8.14.
(b) The mid-band gain is = 40 dB.
(c) The -3 dB points are (rad/sec): ωlow 7 and ωhigh 1.5×109.
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410 CHAPTER 8 Frequency Response Methods
-20
0
20
40
1001011021031041051061071081091010
Frequency (rad/sec)
Gain dB
-200
-100
0
100
1001011021031041051061071081091010
Frequency (rad/sec)
Phase deg
FIGURE P8.14
Bode plot for G(s) = 109s
(s+107)(s+103).
P8.15 The data are plotted in Figure P8.15, denoted by an asterisk (*).
-100
-50
0
50
10-1 100101102
*
********
*
*
-300
-250
-200
-150
-100
-50
10-1 100101102
*******
*
*
**
FIGURE P8.15
Bode plot for G(s) = 809.7
s(s2+6.35s+161.3) ; tabular data is indicated by an asterick (*).
The low frequency slope is -20 dB/dec and the initial low frequency φis
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Problems 411
90o, so we have an integrator of the form K/s. The initial phase is 90o
and the final phase 270o, so we have a minimum phase G(s). Now, |G|
is 0.97 at ω= 8 and ω= 10 indicating two complex poles. We postulate
a transfer function of the form
G(s) = K
ss2
ω2
n+2ζs
ωn+ 1.
The phase angle φ=180oat ω=ωn. Then, from Figure 8.10 in Dorf &
Bishop, we determine that ωn= 12.7. At ω= 8, ω
ωn= 0.63 and φ, due to
the complex poles is 30o(subtract 90odue to the integrator). Again,
from Figure 8.10 in Dorf & Bishop, we estimate ζ= 0.25. To determine
K, note that when ω
ωn0.1, the effect of the complex poles on magnitude
is negligible, so at ω= 1 we have
|G|= 5.02
=
K
j1
.
So K= 5.02. Therefore,
G(s) = 5.02
ss2
161.3+0.5s
12.7+ 1=809.7
s(s2+ 6.35s+ 161.3) .
P8.16 (a) The unit step input response is shown in Figure P8.16. The step
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
0
0.2
0.4
0.6
0.8
1
1.2
1.4
Step Response
Time (sec)
Amplitude
FIGURE P8.16
Unit step input response for T(s) = 60.2
s2+12.1s+60.2.
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412 CHAPTER 8 Frequency Response Methods
response is given by
y(t) = 1 e6.05t(cos 4.85t+ 1.25 sin 4.85t).
(b) The system bandwidth is ωB= 4.95 rad/sec.
P8.17 The transfer function is
Gc(s)G(s) = 4(0.5s+ 1)
s(2s+ 1)(s2/64 + s/20 + 1) .
(a) The Bode plot is shown in Figure P8.17.
-100
-50
0
50
10-1 100101102
Frequency (rad/sec)
Gain dB
-300
-250
-200
-150
-100
-50
10-1 100101102
Frequency (rad/sec)
Phase deg
FIGURE P8.17
Bode plot for Gs(s)G(s) = 4(0.5s+1)
s(2s+1)(s2/64+s/20+1) .
(b) When the magnitude is 0 dB, we have
ω1= 1.6 rad/sec
and when φ=180o, we have
ω2= 7.7 rad/sec .
P8.18 The transfer function is
Gc(s)G(s) = 12(s+ 0.5)
(s+ 3)(s+ 10) =0.2(2s+ 1)
(s/3 + 1)(s/10 + 1) .
The Bode plot is shown in Figure P8.18. Near 0 dB, the frequency is
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Problems 413
ω= 5.4 rad/sec.
-20
-15
-10
-5
0
10-1 100101102
Frequency (rad/sec)
Gain dB
-200
-150
-100
-50
0
50
10-1 100101102
Frequency (rad/sec)
Phase deg
FIGURE P8.18
Bode plot for Gc(s)G(s) = 12(s+0.5)
s2+13s+30 .
P8.19 Examining the frequency response, we postulate a second-order transfer
function
θ(s)
I(s)=ω2
n
s2+ 2ζωns+ω2
n
.
From the data we see that φ=90oat ω= 2. Using Figure 8.10 in Dorf
& Bishop, we determine that ωn=ω= 2. We also estimate ζ= 0.4 from
Figure 8.10. Thus,
θ(s)
I(s)=4
s2+ 1.6s+ 4 .
P8.20 The transfer function is
Gc(s)G(s) = 823(s+ 9.8)
s2+ 22s+ 471 .
The Bode plot is shown in Figure P8.20. The maximum value of
20 log10 |Gc(jω)G(jω)|= 32.3 dB
occurs at ω= 20.6 rad/sec and the corresponding phase is φ=19.6o.
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414 CHAPTER 8 Frequency Response Methods
15
20
25
30
35
Magnitude (dB)
10−1 100101102
−90
−45
0
45
Phase (deg)
Bode Diagram
Frequency (rad/sec)
FIGURE P8.20
Bode plot for Gc(s)G(s) = 832(s+9.8)
s2+22s+471 .
P8.21 The Bode plot is shown in Figure P8.21. The gain is 24 dB when φ=
180o
-40
-20
0
20
40
10-1 100101102
Frequency (rad/sec)
Gain dB
-300
-200
-100
0
10-1 100101102
Frequency (rad/sec)
Phase deg
FIGURE P8.21
Bode plot for Gc(s)G(s) = 200s2
s3+14s2+44s+40 .
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Problems 415
P8.22 The transfer function is
G(s) = 10000(s+ 1)(s+ 80)
s(s+ 300)(s+ 9000) .
P8.23 The transfer function is
G(s) = 100(s+ 20)(s+ 8000)
(s+ 1)(s+ 80)(s+ 500) .
The system is type 0 and the steady-state error to a unit step input is
ess =1
1 + Kp
= 0.0025
since
Kp= lim
s0G(s) = 400 .
P8.24 (a) From the Bode plot we see that
20 log10 M= 12
or M= 3.981. For a second-order system we know that
M= (2ζq1ζ2)1.
Solving for ζ(with M= 3.981) yields ζ= 0.12. Also, from the Bode
plot,
ωr= 0.9rad/sec .
So,
ωn=ωr
p12ζ2= 0.91 .
Therefore, the second-order approximate transfer function is
T(s) = ω2
n
s2+ 2ζωns+ω2
n
=0.83
s2+ 0.22s+ 0.83 .
(b) The predicted overshoot and settling time are P.O. = 68% and Ts=
37 sec.
P8.25 The transfer function is
G(s) = 100(s+ 10)
s2(s+ 100) .
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416 CHAPTER 8 Frequency Response Methods
P8.26 The transfer function is
T(s) = Vo(s)
V(s)=1 + R1/R2
1 + RCs .
Substituting R= 10kΩ, C= 1µF ,R1= 9kΩ, and R2= 1kΩ yields
T(s) = 10
1 + 0.01s.
The frequency response is shown in Figure P8.26.
Frequency (rad/sec)
Phase (deg); Magnitude (dB)
Bode Diagrams
0
5
10
15
20
101102103
-100
-80
-60
-40
-20
0
FIGURE P8.26
Bode plot for T(s) = 1+R1/R2
1+RCs
P8.27 The frequency response is shown in Figure P8.27.
K0.75 1 10
|L(jω)|jω=0 ,dB 3.52 12.04 26.02
ωb,rad/s 8.3 14.0 33.4
ωc,rad/s 3.5 8.7 22.9
TABLE P8.27 System performance as Kvaries.
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Problems 417
−60
−50
−40
−30
−20
−10
0
10
Magnitude (dB)
10−1 100101102103
−135
−90
−45
0
Phase (deg)
Bode Diagram
Frequency (rad/sec)
K increases
K decreases
Phase plot remains unchanged as K varies
FIGURE P8.27
Bode plot for K= 1
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418 CHAPTER 8 Frequency Response Methods
Advanced Problems
AP8.1 The spring-mass-damper system is described by
m¨x+b˙x+kx =p .
Taking the Laplace transform (with zero initial conditions) yields
X(s)
P(s)=1
ms2+bs +k.
From Figure AP8.1(b) in Dorf & Bishop, we determine that
20 log
X(j0)
P(j0)
= 20 log
1
k
=26dB .
Solving for kyields
k= 19.96 N/m.
Also, ω2
n=k/m implies m=k/ω2
n, where ωn= corner frequency = 3.2
rad/sec. So,
m= 1.949 kg .
Comparing Figure AP8.1(b) in Dorf & Bishop to the known standard
Bode plot of a second-order system, we estimate ζ0.32. Therefore,
b= 2ωn= 2(1.949)(0.32)(3.2) = 3.992 N s/m.
AP8.2 The closed-loop transfer function is
T(s) = Y(s)
R(s)=Kb
s+ 1 + 0.2Kb .
WIth K= 5, we have
T(s) = 5b
s+ 1 + b.
The sensitivity is
ST
b=T
b
b
T=s+ 1
s+ 1 + b.
With the nominal value of b= 4, we have
ST
b=s+ 1
s+ 5 .
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Advanced Problems 419
The sensitivity plot is shown in Figure AP8.2.
-14
-12
-10
-8
-6
-4
-2
0
10-1 100101102
Frequency (rad/sec)
20*log(mag) (dB)
FIGURE AP8.2
Bode plot for ST
b(s) = s+1
s+5 .
AP8.3 The equation of motion is
m¨x+b˙x+Kx =b˙r+Kr .
Taking Laplace transforms yields
X(s)
R(s)=bs +K
ms2+bs +K.
Then, given the various system parameters m= 1 kg, b= 4 Ns/m, K= 18
N/m, we obtain the transfer function:
X(s)
R(s)=4s+ 18
s2+ 4s+ 18 .
Also, ωn= corner frequency = pK/m =18 = 4.243 rad/s and
ζ= damping ratio = b/m
2ωn
=4
2(4.243) = 0.471 .
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420 CHAPTER 8 Frequency Response Methods
The Bode plot is shown in Figure AP8.3.
-30
-20
-10
0
10
10-1 100101102
Frequency (rad/sec)
Gain dB
-200
-150
-100
-50
0
10-1 100101102
Frequency (rad/sec)
Phase deg
FIGURE AP8.3
Bode plot for G(s) = 4s+18
s2+4s+18 .
AP8.4 The Bode plot is shown in Figure AP8.4.
−120
−100
−80
−60
−40
−20
Magnitude (dB)
10−1 100101102
−270
−225
−180
−135
−90
−45
0
Phase (deg)
Bode Diagram
Frequency (rad/sec)
FIGURE AP8.4
Bode plot for L(s) = 1
(0.4s+1)(s2+3.9s+15) .
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Advanced Problems 421
AP8.5 The closed-loop transfer function with unity feedback is given by
T(s) = Gc(s)G(s)
1 + Gc(s)G(s)=10(s+ 1)
s2+ 9s+ 10 .
(a) Solving for Gc(s)G(s) yields
Gc(s)G(s) = 10(s+ 1)
s(s1) .
(b) A summary of the plot data (see Figure AP8.5) is presented in Ta-
ble AP8.5.
(c) The open-loop system is unstable; the closed-loop system is stable.
100 120 140 160 180 200 220 240 260 280
−40
−30
−20
−10
0
10
20
30
40
Phase, degrees
20 log|GcG(j ω)|, dB
FIGURE AP8.5
Log-magnitude-phase curve for GcG(jω).
ω1 10 50 110 500
20 log |GcG|40 4.85 -13.33 -20.61 -33.94
phase (deg) 101.42 250.17 267.53 268.93 269.77
TABLE AP8.5 Summary of magnitude and phase for ω= 1,10,50,110,500.
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422 CHAPTER 8 Frequency Response Methods
AP8.6 The transfer function is given by
T(s) = 1/m
s2+ (b/m)s+ (k/m).
Selecting k= 1 and b= 2 results in the Bode plot magnitude always
less than 0 dB. Choosing b=2/2 leads to a peak response with a sinu-
soidal input at ω= 0.66 rad/s. Figure AP8.6a shows the Bode plot and
Figure AP8.6b shows the response to a sinusiodal input with frequency
ω= 1 rad/s is less than 1 in the steady-state, as desired.
Bode Diagram
Frequency (rad/sec)
10−2 10−1 100101
−50
−40
−30
−20
−10
0
10
System: sys
Peak gain (dB): 6.3
At frequency (rad/sec): 0.661
Magnitude (dB)
0 100 200 300 400 500 600 700 800
−1.5
−1
−0.5
0
0.5
1
Impulse Response
Time (sec)
Amplitude
FIGURE AP8.6
(a) Bode plot for b/m = 1 and k/m = 1. (b) Response to a sinusiodal input.
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Advanced Problems 423
AP8.7 The transfer function is
G(s) = Vo(s)
Vi(s)=1 + R2C2s
1 + R1C1s.
Substituting C1= 0.1µF ,C2= 1 mF ,R1= 10 k, and R2= 10 Ω yields
G(s) = 0.01s+ 1
0.001s+ 1 .
The frequency response is shown in Figure AP8.7.
0
5
10
15
20
Magnitude (dB)
100101102103104105
0
30
60
Phase (deg)
Bode Diagram
Frequency (rad/sec)
FIGURE AP8.7
Bode plot for G(s) = 0.01s+1
0.001s+1
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424 CHAPTER 8 Frequency Response Methods
Design Problems
With the PI controller in the loop, the closed-loop transfer function fromCDP8.1
the input to the output is
θ(s)
R(s)=26.035K(s+ 2)
s2+ (33.1415 + 26.035K)s+ 52.07K,
where we switch off the tachometer feedback (see Figure CDP4.1 in Dorf
and Bishop). The Bode plot is shown below for K= 40. From the step
response we determine that P.O. = 0 and Ts= 0.19. With K= 40, the
closed-loop poles are both real roots with values of s1=1072.6 and
s2=1.9.
10-1 100101102
0
20
40
60
Frequency (rad/sec)
Gain dB
10-1 100101102
-30
-60
-90
0
Frequency (rad/sec)
Phase deg
DP8.1 The loop transfer function is
L(s) = Gc(s)G(s) = K(s+ 2)
s2(s+ 12) .
(a,b) Let K= 1. The Bode plot of the loop transfer function and the
closed-loop transfer functions are shown in Figure DP8.1a and Fig-
ure DP8.1b, respectively.
(c) Let K= 50. The Bode plot of the loop transfer function and the
closed-loop transfer functions are shown in Figure DP8.1c and Fig-
ure DP8.1d, respectively.
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Design Problems 425
10-1 100101102
-100
-50
0
50
Frequency (rad/sec)
Gain dB
10-1 100101102
-180
-160
-140
-120
Frequency (rad/sec)
Phase deg
FIGURE DP8.1
(a) Bode plot for the loop transfer function Gc(s)G(s) = (s+2)
s2(s+12) .
10-2 10-1 100101102
-100
-50
0
50
Frequency (rad/sec)
Gain dB
10-2 10-1 100101102
-90
-180
0
Frequency (rad/sec)
Phase deg
FIGURE DP8.1
CONTINUED: (b) Bode plot for the closed-loop T(s) = (s+2)
s3+12s2+s+2 .
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426 CHAPTER 8 Frequency Response Methods
10-1 100101102
-50
0
50
100
Frequency (rad/sec)
Gain dB
10-1 100101102
-180
-160
-140
-120
Frequency (rad/sec)
Phase deg
FIGURE DP8.1
CONTINUED: (c) Bode plot for the loop transfer function Gc(s)G(s) = 50(s+2)
s2(s+12) .
10-1 100101102
-60
-40
-20
0
20
Frequency (rad/sec)
Gain dB
10-1 100101102
-90
-180
0
Frequency (rad/sec)
Phase deg
FIGURE DP8.1
CONTINUED: (d) Bode plot for the closed-loop T(s) = 50(s+2)
s3+12s2+50s+100 .
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Design Problems 427
(d) The peak value of Mp2 occurs for 14 K350. The maximum
bandwidth is achieved for the largest gain K. Thus, we select K= 350
and the corresponding bandwidth is ωB= 29 rad/sec.
(e) The system is type 2—the steady-state error is zero for a ramp input.
DP8.2 The open-loop transfer function is
Gc(s)G(s) = 20(s+ 1)
s(s+ 4)(s2+ 2s+ 8) .
(a) The phase angle is φ=180owhen ω= 3.54 rad/sec. The magnitude
is 0 dB when ω= 0.87 rad/sec.
(b) The closed-loop transfer function is
T(s) = 20(s+ 1)
s4+ 6s3+ 16s2+ 52s+ 20 .
The closed-loop Bode plot is shown in Figure DP8.2.
−100
−80
−60
−40
−20
0
20
Magnitude (dB)
10−1 100101102
−270
−225
−180
−135
−90
−45
Phase (deg)
Bode Diagram
Gm = 6.71 dB (at 3.54 rad/sec) , Pm = 105 deg (at 0.869 rad/sec)
Frequency (rad/sec)
FIGURE DP8.2
Bode plot for closed-loop T(s) = 20(s+1)
s4+6s3+16s2+52s+20 .
(c) When K= 22, we have
Mpω= 4.84dB , ωr= 3.11 ,and ωB= 3.78 rad/sec .
When K= 25, we have
Mpω= 7.18 dB , ωr= 3.18 rad/sec ,and ωB= 3.94 rad/sec .
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428 CHAPTER 8 Frequency Response Methods
(d) Select K= 22.
DP8.3 The closed-loop transfer function is
T(s) = K(s+ 5)
s3+ 7s2+ 12s+ 10 + 5K.
When K= 4.2, we have 10 log10 Mpω= 3 dB. The system bandwidth is
ωb= 3.7178 rad/sec. The steady-state tracking error to a unit step input
is
ess = lim
s0sE(s) = lim
s01T(s).
So,
ess = 1 5K
10 + 5K= 0.322 ,
when K= 4.2. Since the system is unstable when K > 14.8, the steady-
state error does not exist after K= 14.8. The Bode plot is shown in
Figure DP8.3.
-80
-60
-40
-20
0
20
10-1 100101102
Frequency (rad/sec)
Gain dB
-200
-150
-100
-50
0
10-1 100101102
Frequency (rad/sec)
Phase deg
FIGURE DP8.3
Bode plot for closed-loop T(s) = K(s+5)
s3+7s2+12s+10+5K, where K= 4.2.
DP8.4 We have a second-order loop transfer function
Gc(s)G(s) = K
(0.3s+ 1)(0.6s+ 1) .
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Design Problems 429
With Mpω= 1.5, we determine that
M= (2ζq1ζ2)1or ζ= 0.3568 .
Now the characteristic equation is
s2+ 5s+ 5.56(1 + K) = 0 .
So, solving 2ζωn= 5 yields ωn= 7. Therefore,
K= 0.18ω2
n1 = 7.82 .
The closed-loop transfer function is
T(s) = K
K+ 1
5.56(K+ 1)
s2+ 5s+ 5.56(K+ 1) .
So, the overall gain of the standard second-order system will be attenuated
by the factor K/(K+ 1). To compensate, we amplify the gain by a small
factor. Thus we choose K= 10. The bandwidth is ωb= 11.25 rad/sec and
the peak magnitude is Mpω=1.5.
DP8.5 From the Bode plot of G(s) we find that there exists two pnoles, at ap-
proximately ω= 1 rad/sec and ω= 10 rad/sec. Then, by examining the
Bode plot we estimate
G(s) = 10
(s+ 1)(s+ 10) .
We use a scale factor of 10 because at low frequency the Bode plot has
magnitude 0 dB (or a DC gain of 1). With G(s) as above, we can utilize
the controller
Gc(s) = 500
s+ 20
yielding a crossover
ωc= 12.9 rad/sec
and a magnitude of at least 25 dB for ω < 0.1 rad/sec. Figure DP8.5
shows the compensator Bode plot of Gc(s)G(s).
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430 CHAPTER 8 Frequency Response Methods
-150
-100
-50
0
50
Magnitude (dB)
10 -2 10 -1 100101102103
-225
-180
-135
-90
-45
Phase (deg)
Bode Diagram
ωc=12.9
25 dB
Frequency (rad/sec)
0
FIGURE DP8.5
Bode Diagram for G(s)Gc(s) = 5000
(s+1)(s+10)(s+20) .
DP8.6 Let K=1 to meet the steady-state tracking error requirement and
p= 2ζ, where ζ= 0.69 to obtain a 5% overshoot. The system is given by
˙
x=Ax +Bu
where
A=
0 1
11.38
,B=
1
0
,and C=0 1 .
The characteristic polynomial is
s2+ 1.38s+ 1 = 0 .
The associated damping ratio is ζ= 0.69 and the natural frequency is
ωn= 1 rad/s. Using the approximation
ωb= (1.19ζ+ 1.85)ωn
we obtain ωb1.028 rad/s. The Bode plot is shown in Figure DP8.6.
The bandwidth is ωb= 1.023 rad/s.
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Design Problems 431
−80
−60
−40
−20
0
20
Magnitude (dB)
10−2 10−1 100101102
−180
−135
−90
−45
0
Phase (deg)
Bode Diagram
Frequency (rad/sec)
FIGURE DP8.6
Bode diagram for K=1and p= 1.38.
DP8.7 A viable controler is
Gc(s) = KP+KI
s+KDs= 5.5 + 3.33
s+ 3.5s.
The loop transfer function is
Gc(s)G(s) = 10.5s2+ 16.5s+ 10
s2(s2+ 4s+ 5)
and computing Kayields
Ka= lim
s0s2Gc(s)G(s) = 10
5= 2,
as desired. The phase margin is P.O. = 44.35and the bandwidth is
ωb= 4.5 rad/sec. The step response is shown in Figure DP8.7.
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432 CHAPTER 8 Frequency Response Methods
Step Response
Time (sec)
Amplitude
01234567
0
0.2
0.4
0.6
0.8
1
1.2
1.4
System: sys_cl
Settling Time (sec): 3.93
System: sys_cl
Peak amplitude: 1.32
Overshoot (%): 32.1
At time (sec): 1.11
FIGURE DP8.7
Step response for KP= 5.5, KI= 3.33, and KD= 3.5.
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Computer Problems 433
Computer Problems
CP8.1 The m-file script and Bode plot are shown in Figure CP8.1. The script
automatically computes Mpωand ωr.
mp =
14.0228
wr =
4.9458
num=[25]; den=[1 1 25];
sys = tf(num,den);
w=logspace(0,1,400);
[mag,phase]=bode(sys,w);
[y,l]=max(mag);
mp=20*log10(y), wr=w(l)
bode(sys,w);
Frequency (rad/sec)
Phase (deg); Magnitude (dB)
Bode Diagrams
- 10
- 5
0
5
10
15
From: U(1)
100101
- 200
- 150
- 100
- 50
0
To: Y(1)
FIGURE CP8.1
Generating a Bode plot with the bode function.
CP8.2 The m-file script to generate the Bode plots is shown in Figure CP8.2a.
The Bode plots are presented in Figures CP8.2b-CP8.2e. The transfer
functions are
(a) : G(s) = 1000
(s+ 10)(s+ 100) ; (b) : G(s) = s+ 100
(s+ 2)(s+ 25) ;
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434 CHAPTER 8 Frequency Response Methods
(c) : G(s) = 100
s2+ 2s+ 50 ; (d) : G(s) = s6
(s+ 3)(s2+ 12s+ 50) .
% Part (a)
num=[1000]; den=conv([1 10],[1 100]); sys1=tf(num,den);
sys = tf(num,den);
figure(1), bode(sys1), grid
% Part (b)
num=[1 100]; den=conv([1 2],[1 25]); sys2=tf(num,den);
sys = tf(num,den);
figure(2), bode(sys2), grid
% Part (c)
num=[100]; den=[1 2 50]; sys3=tf(num,den);
sys = tf(num,den);
figure(3), bode(sys3), grid
% Part (d)
num=[1 -6]; den=conv([1 3],[1 12 50]); sys4=tf(num,den);
sys = tf(sys);
figure(4), bode(sys4), grid
FIGURE CP8.2
(a) Script to generate the four Bode plots.
−100
−80
−60
−40
−20
0
Magnitude (dB)
10−1 100101102103104
−180
−135
−90
−45
0
Phase (deg)
Bode Diagram
Frequency (rad/sec)
FIGURE CP8.2
CONTINUED: (b) Bode plot for G(s) = 1000
(s+10)(s+100) .
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Computer Problems 435
−80
−60
−40
−20
0
20
Magnitude (dB)
10−1 100101102103104
−135
−90
−45
0
Phase (deg)
Bode Diagram
Frequency (rad/sec)
FIGURE CP8.2
CONTINUED: (c) Bode plot for G(s) = s+100
(s+2)(s+25) .
−40
−30
−20
−10
0
10
20
Magnitude (dB)
100101102
−180
−135
−90
−45
0
Phase (deg)
Bode Diagram
Frequency (rad/sec)
FIGURE CP8.2
CONTINUED: (d) Bode plot for G(s) = 100
s2+2s+50 .
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from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying,
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436 CHAPTER 8 Frequency Response Methods
−80
−70
−60
−50
−40
−30
−20
Magnitude (dB)
10−1 100101102
−180
−90
0
90
180
Phase (deg)
Bode Diagram
Frequency (rad/sec)
FIGURE CP8.2
CONTINUED: (e) Bode plot for G(s) = s6
(s+3)(s2+12s+50) .
CP8.3 The Bode plots are shown in Figure CP8.3(a-d) with the transfer functions
listed in the caption. The crossover frequency for (a) is 17 rad/sec.
−100
−80
−60
−40
−20
0
20
Magnitude (dB)
10−1 100101102103104
−180
−135
−90
−45
0
Phase (deg)
Bode Diagram
Frequency (rad/sec)
FIGURE CP8.3
(a) Bode plot for G(s) = 2000
(s+10)(s+100) .
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Computer Problems 437
The crossover frequency for (b) is 0.99 rad/sec.
−80
−60
−40
−20
0
20
Magnitude (dB)
10−2 10−1 100101102
−270
−225
−180
−135
−90
−45
0
Phase (deg)
Bode Diagram
Frequency (rad/sec)
FIGURE CP8.3
CONTINUED: (b) Bode plot for G(s) = 100
(s+1)(s2+10s+2) .
The crossover frequency for (c) is 70.7 rad/sec.
−30
−20
−10
0
10
20
30
40
Magnitude (dB)
10−2 10−1 100101102103
−135
−90
−45
0
Phase (deg)
Bode Diagram
Frequency (rad/sec)
FIGURE CP8.3
CONTINUED: (c) Bode plot for G(s) = 50(s+100)
(s+1)(s+50) .
The crossover frequency for (d) is 3.1 rad/sec.
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438 CHAPTER 8 Frequency Response Methods
−40
−30
−20
−10
0
10
20
Magnitude (dB)
10−1 100101102103104
−90
−45
0
Phase (deg)
Bode Diagram
Frequency (rad/sec)
FIGURE CP8.3
CONTINUED: (d) Bode plot for G(s) = 100(s2+14s+50)
(s+1)(s+2)(s+500) .
CP8.4 The m-file script and Bode plot are shown in Figure CP8.4a and b. The
bandwidth is ωb= 10 rad/sec.
−50
−40
−30
−20
−10
0
10
Magnitude (dB)
10−1 100101102
−180
−135
−90
−45
0
Phase (deg)
Bandwidth=10.0394 rad/sec
Frequency (rad/sec)
FIGURE CP8.4
(a) Bode plot for T(s) = 54
s2+6s+54 .
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Computer Problems 439
numg=[54]; deng=[1 6 0];
sys_o = tf(numg,deng);
sys_cl = feedback(sys_o,[1])
wb=bandwidth(sys_cl)
bode(sys_cl), grid
titlename=strcat('Bandwidth= ', num2str(wb), ' rad/sec')
title(titlename)
FIGURE CP8.4
CONTINUED: (b) M-file script to obtain the closed-loop Bode plot.
CP8.5 The Bode plot of the closed-loop system is shown in Figure CP8.5. The
closed-loop transfer function is
T(s) = 100
s2+ 6s+ 100 .
(a) From the Bode plot we determine that
Mpω5 dB and ωr9 rad/sec .
Frequency (rad/sec)
Phase (deg); Magnitude (dB)
Bode Diagrams
- 80
- 60
- 40
- 20
0
20
From: U(1)
10- 1100101102103
- 200
- 150
- 100
- 50
0
To: Y(1)
FIGURE CP8.5
Closed-loop system Bode plot.
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440 CHAPTER 8 Frequency Response Methods
(b) From Equations (8.36) and (8.37) in Dorf & Bishop, we find that
ζ0.28 and ωrn0.92
which implies that
ωn=ωr/0.92 = 9.8 rad/sec .
(c) From T(s) we find that
ωn= 10 rad/sec and ζ= 0.3.
The actual values and the estimated values compare very well.
CP8.6 The open-loop and closed-loop Bode plots are shown in Figure CP8.6a
and b. The open-loop and closed-loop transfers functions are
Gc(s)G(s) = 25
s3+ 3s2+ 27s+ 25
and
T(s) = Gc(s)G(s)
1 + Gc(s)G(s)=25
s3+ 3s2+ 27s+ 50 .
−100
−80
−60
−40
−20
0
Magnitude (dB)
10−2 10−1 100101102
−270
−225
−180
−135
−90
−45
0
Phase (deg)
Loop transfer function; bode(syso)
Frequency (rad/sec)
FIGURE CP8.6
(a) Open-loop system Bode plot for Gc(s)G(s) = 25
s3+3s2+27s+25 .
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Computer Problems 441
−100
−80
−60
−40
−20
0
Magnitude (dB)
10−1 100101102
−270
−225
−180
−135
−90
−45
0
Phase (deg)
Closed−loop system; bode(syscl)
Frequency (rad/sec)
FIGURE CP8.6
CONTINUED: (b) Closed-loop system Bode plot T(s) = 25
s3+3s2+27s+50 .
CP8.7 The m-file script and plot of ωbversus pare shown in Figure CP8.7a and
b.
p=[0:0.001:1];
w=logspace(-1,1,1000);
n=length(p);
for i=1:n
num=[1]; den=[1 2*p(i) 0];
sys = tf(num,den);
sys_cl = feedback(sys,[1]);
[mag,phase,w]=bode(sys_cl,w);
a=nd(mag<0.707); wb(i)=w(a(1));
end
plot(p,wb)
xlabel('p'), ylabel('Bandwidth (rad/sec)')
FIGURE CP8.7
(a) M-file script to generate plot of ωbversus p.
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442 CHAPTER 8 Frequency Response Methods
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
0.6
0.7
0.8
0.9
1
1.1
1.2
1.3
1.4
1.5
1.6
p
Bandwidth (rad/sec)
FIGURE CP8.7
CONTINUED: (b) Plot of ωbversus p.
CP8.8 The transfer function from Td(s) to θ(s) is
θ(s)/Td(s) = 0.01(s+ 10)
s3+ 10s2+ (0.01K10.791)s107.91 + 0.05K.
Using the final value theorem and Td(s) = 1/s, we determine that
lim
s0(s) = 0.1
107.91 + 0.05K.
The design specifications require that
|ess|<0.1o.
So, solving for Kyields
K > 3300 .
We can select
K= 3300
as the initial value of Kfor the design. The m-file script is shown in
Figure CP8.8a. For the design shown, the final selection for the gain is
K= 6000. The disturbance response is shown in Figure CP8.8b.
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Computer Problems 443
MpDb =
4.0003
wr =
4.7226
ts =
2.23
po =
32.75
meets specs
Mb=100; Ms=10; L=1; g=9.81; a=5; b=10;
%
K=6000; % Final design value of K
%
numg=[-1/Mb/L]; deng=[1 0 -(Mb+Ms)*g/Mb/L];
sysg = tf(numg,deng);
numc=-K*[1 a]; denc=[1 b];
sysc = tf(numc,denc);
%
% Part (a)
%
sys = feedback(sysg,sysc);
w=logspace(0,1,400);
bode(sys,w)
[mag,phase]=bode(sys,w);
[M,l]=max(mag);
MpDb=20*log10(M)-20*log10(mag(1)) % Mpw in decibels
wr=w(l) % Mpw and peak frequency
%
% Part (b)
%
% From Eqs. (8.35) and (8.37)
Mpw=10^(MpDb/20);zeta=sqrt((1-sqrt(1-(1/Mpw^2)))/2);
wn=wr/sqrt(1-2*zeta^2);
ts=4/zeta/wn
po=100*exp(-zeta*pi/sqrt(1-zeta^2))
%
% Part (c)
%
t=[0:0.1:10];
[y,x]=step(sys,t);
plot(t,y*180/pi)
xlabel('time [sec]')
ylabel('theta [deg]')
grid
-0.04
-0.035
-0.03
-0.025
-0.02
-0.015
-0.01
-0.005
0
0 1 2 3 4 5 6 7 8 9 10
time [sec]
theta [deg]
FIGURE CP8.8
(a) Design script. (b) Disturbance response - meets all specs!
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444 CHAPTER 8 Frequency Response Methods
CP8.9 A viable filter is
G(s) = 0.7(s+ 1000)(s+ 1)
(s+ 100)(s+ 10) .
The Bode plot is shown in Figure CP8.9
0
5
10
15
20
Magnitude (dB)
10 -2 100102104
-90
-45
0
45
Phase (deg)
Bode Diagram
Frequency (rad/sec)
90
-5
FIGURE CP8.9
Bode plot for G(s) = 0.7(s+1000)(s+1)
(s+100)(s+10) .
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C H A P T E R 9
Stability in the Frequency Domain
Exercises
E9.1 The Bode plot for the transfer function Gc(s)G(s) is shown in Figure E9.1,
where
Gc(s)G(s) = 2(1 + s/10)
s(1 + 5s)(1 + s/9 + s2/81) .
The gain and phase margins are
G.M. = 26.2 dB and P.M. = 17.5o.
−150
−100
−50
0
50
Magnitude (dB)
10−2 10−1 100101102
−315
−270
−225
−180
−135
−90
Phase (deg)
Bode Diagram
Gm = 26.2 dB (at 2.99 rad/sec) , Pm = 17.5 deg (at 0.618 rad/sec)
Frequency (rad/sec)
FIGURE E9.1
Bode Diagram for Gc(s)G(s) = 2(1+s/10)
s(1+5s)(1+s/9+s2/81) .
445
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446 CHAPTER 9 Stability in the Frequency Domain
E9.2 The loop transfer function is
Gc(s)G(s) = 10.5(1 + s/5)
s(1 + s/2)(1 + s/10) .
The Bode plot is shown in Figure E9.2. The phase margin is
P.M. = 40.4o
at ωc= 4.96 rad/sec.
−100
−50
0
50
Magnitude (dB)
10−1 100101102103
−180
−135
−90
Phase (deg)
Bode Diagram
Gm = Inf dB (at Inf rad/sec) , Pm = 40.4 deg (at 4.96 rad/sec)
Frequency (rad/sec)
FIGURE E9.2
Bode Diagram for Gc(s)G(s) = 10.5(1+s/5)
s(1+s/2)(1+s/10) .
E9.3 The phase margin P.M. 75oat 200 kHz. We estimate the 180ophase
angle at 2 MHz, so the gain margin is G.M. 25 dB.
E9.4 The loop transfer function is
Gc(s)G(s) = 100
s(s+ 10) .
The Nichols diagram is shown in Figure E9.4. When the gain is raised by
4.6 dB, Mpω= 3 and the resonant frequency is ωR= 11 rad/sec.
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Exercises 447
-40
-30
-20
-10
0
10
20
30
40
-350 -300 -250 -200 -150 -100 -50 0
-40
-20
-12
-6
-3
-1
0
0.25
0.5
1
3
6
Phase (deg)
Gain dB
Phase (deg)
Gain dB
------ K=100K=171 ------
FIGURE E9.4
Nichols Diagram for Gc(s)G(s) = K)
s(s+10) , where K= 100 and K= 171.
E9.5 (a) The G.M. 5 dB and the P.M. 10o. (b) Lower the gain by 10 dB
to obtain P.M. 60o.
E9.6 The Bode plot of the closed-loop transfer function is shown in Figure E9.6.
The value of Mpω= 3 dB. The phase margin is P.M. = 40owhen K= 50.
-45
-40
-35
-30
-25
-20
-15
-10
-5
0
5
10-1 100101102
Frequency (rad/sec)
Gain dB
FIGURE E9.6
Closed-loop Bode Diagram for T(s) = 50(s+100)
s3+50s2+450s+5000 .
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448 CHAPTER 9 Stability in the Frequency Domain
E9.7 The Nyquist plot is shown in Figure E9.7 for K= 5; the plot is a circle
with diameter= K/5. For K > 5, we have P= 1 and N=1 (ccw as
−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4
−0.5
−0.4
−0.3
−0.2
−0.1
0
0.1
0.2
0.3
0.4
0.5
Nyquist Diagram
Real Axis
Imaginary Axis
FIGURE E9.7
Nyquist Diagram for Gc(s)G(s) = K
s5, where K= 5.
shown). So Z=N+P=1 + 1 = 0 and the system is stable for K > 5.
E9.8 (a) When K= 4, the G.M. = 3.5 dB. This is illustrated in Figure E9.8.
−150
−100
−50
0
50
Magnitude (dB)
10−1 100101102
−270
−225
−180
−135
−90
Phase (deg)
Bode Diagram
Gm = 3.52 dB (at 1.41 rad/sec) , Pm = 11.4 deg (at 1.14 rad/sec)
Frequency (rad/sec)
FIGURE E9.8
Bode Diagram for Gc(s)G(s) = K
s(s+1)(s+2) , where K= 4.
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Exercises 449
(b) The new gain should be K= 1 for a gain margin G.M. = 16 dB.
E9.9 For K= 5, the phase margin P.M. = 5oas shown in Figure E9.9.
-150
-100
-50
0
50
100
Magnitude (dB)
10 -2 10 -1 100101102
-270
-225
-180
-135
-90
Phase (deg)
Bode Diagram
Gm = 1.58 dB (at 1.41 rad/sec) , Pm = 5.02 deg (at 1.29 rad/sec)
Frequency (rad/sec)
FIGURE E9.9
Bode Diagram for Gc(s)G(s) = K
s(s+1)(s+2) , where K= 5.
E9.10 The Bode plot is shown in Figure E9.10a. The closed-loop frequency
-100
-50
0
50
100
10-2 10-1 100101102
Frequency (rad/sec)
Gain dB
-300
-200
-100
0
10-2 10-1 100101102
Frequency (rad/sec)
Phase deg
PM=23.14 deg
GM=12.35 dB
FIGURE E9.10
(a) Bode Diagram for Gc(s)G(s) = 326s+1304
s4+14.76s3+151.3s2+23.84s.
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450 CHAPTER 9 Stability in the Frequency Domain
-70
-60
-50
-40
-30
-20
-10
0
10
10-1 100101102
Frequency (rad/sec)
Gain dB
FIGURE E9.10
CONTINUED: (b) Closed-loop frequency response: ωB= 6 rad/sec.
response is shown in Figure E9.10b. The bandwidth is ωB= 6 rad/sec.
E9.11 The Bode plot is shown in Figure E9.11. The system is stable.
−100
−50
0
50
100
Magnitude (dB)
10−2 10−1 100101102
−270
−225
−180
−135
−90
Phase (deg)
Bode Diagram
Gm = 3.91 dB (at 3.74 rad/sec) , Pm = 14.4 deg (at 2.76 rad/sec)
Frequency (rad/sec)
FIGURE E9.11
Bode Diagram for Gc(s)G(s) = 10(1+0.4s)
s(1+2s)(1+0.24s+0.04s2).
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Exercises 451
E9.12 We select the gain K= 10 to meet the 10% steady-state tracking error
specification for a ramp input. The Bode plot and Nichols chart are shown
in Figures E9.12a and E9.12b, respectively.
-150
-100
-50
0
50
10-1 100101102103
Frequency (rad/sec)
Gain dB
-300
-200
-100
0
10-1 100101102103
Frequency (rad/sec)
Phase deg
PM=31.79 deg
GM=14.82 dB
-40
-30
-20
-10
0
10
20
30
40
-350 -300 -250 -200 -150 -100 -50 0
-40
-20
-12
-6
-3
-1
0
0.25
0.5
1
3
6
8
Phase (deg)
Gain dB
FIGURE E9.12
(a) Bode Diagram for Gc(s)G(s) = 10
s(0.02s+1)(0.2s+1) . (b) Nichols chart for Gc(s)G(s) =
10
s(0.02s+1)(0.2s+1) .
E9.13 (a) The Nichols diagram is shown in Figure E9.13a and Mpω= 7.97 dB.
(b) The closed-loop Bode plot is shown in Figure E9.13b. The bandwidth
ωB= 18.65 rad/sec and the resonant frequency is ωr= 11.69 rad/sec.
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452 CHAPTER 9 Stability in the Frequency Domain
-40
-30
-20
-10
0
10
20
30
40
-350 -300 -250 -200 -150 -100 -50 0
-40
-20
-12
-6
-3
-1
0
0.25
0.5
1
3
6
Phase (deg)
Gain dB
8
-40
-30
-20
-10
0
10
10-1 100101102
Frequency (rad/sec)
Gain dB
-200
-150
-100
-50
0
10-1 100101102
Frequency (rad/sec)
Phase deg
FIGURE E9.13
(a) Nichols Diagram for Gc(s)G(s) = 150
s(s+5) . (b) Closed-loop Bode Diagram for T(s) =
150
s2+5s+150 .
(c) From Mpω= 8 dB, we estimate ζ= 0.2, so the expected P.O. = 52%.
E9.14 (a) The peak resonance Mpω= 6 dB.
(b) The resonant frequency is ωr=ω2= 3 rad/sec.
(c) The bandwidth is ωB=ω4= 10 rad/sec.
(d) The phase margin is P.M. = 30o.
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Exercises 453
E9.15 The loop transfer function is
Gc(s)G(s) = 100
s(s+ 20) ,
and the closed-loop transfer function is
T(s) = 100
s2+ 20s+ 100 .
The magnitude plot for the closed-loop system is shown in Figure E9.15.
With bandwidth defined as frequency at which the magnitude is reduced
10−1 100101
−7
−6
−5
−4
−3
−2
−1
0
Magnitude (dB)
Bode Diagram
Frequency (rad/sec)
FIGURE E9.15
Magnitude plot for the closed-loop T(s) = 100
s2+20s+100 .
-3 dB from the dc value, we determine that ωB= 6.4 rad/sec.
E9.16 The transfer function of the approximation is
G(jω) = 1jω/10
1 + jω/10 ,
and the magnitude is
|G(jω)|=
1jω/10
1 + jω/10
= 1 ,
which is equivalent to the actual time delay magnitude. The phase ap-
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454 CHAPTER 9 Stability in the Frequency Domain
proximation is
φ=tan1ω/10 + tan1(ω/10) = 2 tan1ω/10
and the actual phase is
φ=0.2ω .
The phase plots are shown in Figure E9.16. The approximation is accurate
for ω < 3 rad/sec.
10−2 10−1 100101
−120
−100
−80
−60
−40
−20
0
Frequency (rad/sec)
Phase deg
Actual _______ & Approximation −−−−−−−
FIGURE E9.16
Phase plots for time delay actual vs approximation.
E9.17 (a,b) The phase angle for P.M. = 30 is
φ=90o+ tan1ω
2tan12ω
15 ω2=150o.
Solving for ωyields ω= 4.7. Then, at ω= 4.7, we have K= 10.82
when
|GcG(jω)|=K(ω2+ 4)1
2
ω((2ω2)2+ (15 ω2)2)1
2
= 1 .
The Bode plot is shown in Figure E9.17.
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Exercises 455
Frequency (rad/sec)
Phase (deg); Magnitude (dB)
Bode Diagrams
- 100
- 50
0
50
Gm=3.5545 dB (at 4.3301 rad/sec), Pm=40 deg. (at 3.5147 rad/sec)
10- 1100101102
- 250
- 200
- 150
- 100
- 50
FIGURE E9.17
Bode Diagram for Gc(s)G(s) = K(s+2)
s3+2s2+15s, where K= 10.82.
(c) The steady-state error for a ramp is
ess =A
Kv
=A
10K
15
= 0.60A ,
where R(s) = A/s2.
E9.18 (a) The gain crossover is at ωc= 486 Hz, and the phase margin P.O. =
36.2o. So, ζ0.36. Then, the expected percent overshoot to a step
input is
P.O. = 100eζπ/1ζ2= 30% ,where ζ= 0.36 .
(b) The estimated bandwidth is ωB2π(600).
(c) Approximate
ωnωr= 2π(480) .
Then,
Ts=4
ζωn
=4
(0.36)2π(480) 4 ms .
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456 CHAPTER 9 Stability in the Frequency Domain
E9.19 The Bode plot is shown in Figure E9.19 for K= 16.75. The phase and
gain margins are P M = 50.0oand GM = 2.72 dB.
Frequency (rad/sec)
Phase (deg); Magnitude (dB)
- 40
- 30
- 20
- 10
0
10 Gm=2.7233 dB (at 20.618 rad/sec), Pm=50 deg. (at 13.434 rad/sec)
100101102103
- 500
- 400
- 300
- 200
- 100
0
FIGURE E9.19
Bode Diagram for Gc(s)G(s) = Ke0.1s
s+10 , where K= 16.75.
E9.20 The system response for both drivers is shown in Figure E9.20.
-8
-7
-6
-5
-4
-3
-2
-1
0
1
012345678910
Time (sec)
Automobile velocity change
T=1 sec (solid line) & T=1.5 sec (dashed line)
FIGURE E9.20
Change in automobile velocity due to braking for two drivers.
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Exercises 457
E9.21 The Bode plot is shown in Figure E9.21.
-150
-100
-50
0
50
10-1 100101102103
Frequency (rad/sec)
Gain dB
-300
-200
-100
0
10-1 100101102103
Frequency (rad/sec)
Phase deg
PM=16.85 deg
GM=12.04 dB
FIGURE E9.21
Bode Diagram for Gc(s)G(s) = 1300
s(s+2)(s+50) .
E9.22 When K= 10, the P.M. = 36.9o; the system is stable. Decreasing the
gain to K= 4 results in a P.M. = 60o.
E9.23 The Nichols chart is shown in Figure E9.23.
-40
-30
-20
-10
0
10
20
30
40
-350 -300 -250 -200 -150 -100 -50 0
-40
-20
-12
-6
-3
-1
0
0.25
0.5
1
3
6
8
Phase (deg)
Gain dB
FIGURE E9.23
Nichols chart for Gc(s)G(s) = 438
s(s+2)(s+50) .
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458 CHAPTER 9 Stability in the Frequency Domain
The actual values are
Mpω= 1.6598 (4.4 dB) ωr= 2.4228 rad/sec ωB= 4.5834 rad/sec .
E9.24 Using the Nyquist criterion, we have
P= 1 and N= 0
which implies
Z=N+P= 1 .
Therefore, the system has one root in the right half-plane.
E9.25 The Bode plot is shown in Figure E9.25.
-100
-50
0
50
10-1 100101102
Frequency (rad/sec)
Gain dB
-300
-200
-100
0
10-1 100101102
Frequency (rad/sec)
Phase deg
PM=27.73 deg at wc=8.29 rad/sec
FIGURE E9.25
Bode plot for Gc(s)G(s) = 11.7
s(0.05s+1)(0.1s+1) .
E9.26 The Nichols chart for
Gc(s)G(s) = 11.7
s(0.05s+ 1)(0.1s+ 1)
is shown in Figure E9.26, where we find that
Mpω= 6.76 dB ωr= 8.96 rad/sec ωB= 13.73 rad/sec .
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Exercises 459
-40
-30
-20
-10
0
10
20
30
40
-350 -300 -250 -200 -150 -100 -50 0
-40
-20
-12
-6
-3
-1
0
0.25
0.5
1
3
6
8
Phase (deg)
Gain dB
FIGURE E9.26
Nichols chart for Gc(s)G(s) = 11.7
s(0.05s+1)(0.1s+1) .
E9.27 The Bode plot for G(s) with K= 122.62 is shown in Figure E9.27.
Frequency (rad/sec)
Phase (deg); Magnitude (dB)
K=122.63
- 100
- 50
0
50
Gm=10.938 dB (at 6 rad/sec), Pm=40 deg. (at 2.7978 rad/sec)
10- 1100101102
- 300
- 250
- 200
- 150
- 100
- 50
FIGURE E9.27
Bode plot for Gc(s)G(s) = K
s(s+6)2,with K= 122.62.
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460 CHAPTER 9 Stability in the Frequency Domain
The phase margin is
P.M. = 40.0o
and the gain margin is
G.M. = 10.94 dB .
E9.28 The phase margin is P.M. = 28o. The estimated damping is
ζ=P.M.
100 = 0.28 .
The estimated percent overshoot is
P.O. = 100eπζ/1ζ2= 40% .
The actual overshoot is P.O. = 44.43%.
E9.29 The F(s)-plane contour is shown in Figure E9.29, where
F(s) = 1 + Gc(s)G(s) = s+ 3
s+ 2 .
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2
*
*
*
*
*
*
*
*
Re
Im
F(s)-plane
FIGURE E9.29
F(s)-plane contour, where F(s) = 1 + Gc(s)G(s) = s+3
s+2 .
E9.30 The Bode plot is shown in Figure E9.30.
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Exercises 461
-50
0
Magnitude (dB)
10 -2 100102104
-180
-135
-90
-45
0
Phase (deg)
Bode Diagram
50
Frequency (rad/sec)
-100
FIGURE E9.30
Bode plot for G(s) = C[sIA]1B+D=1000
s2+100s+10 .
E9.31 The Bode plot is shown in Figure E9.31. The phase margin is P.M. = 50.6
deg.
-40
-20
0
20
40
60
80
Magnitude (dB)
10-3 10-2 10-1 100101
-150
-120
-90
Phase (deg)
Bode Diagram
Gm = Inf , Pm = 50.6 deg (at 0.341 rad/sec)
Frequency (rad/sec)
FIGURE E9.31
Bode plot for L(s) = G(s)H(s) = 2s+1
10s2+s.
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462 CHAPTER 9 Stability in the Frequency Domain
E9.32 The Bode plot is shown in Figure E9.32. The phase margin is P.M. = 29.
−80
−60
−40
−20
0
20
Magnitude (dB)
10−1 100101102
−180
−135
−90
−45
0
Phase (deg)
Bode Diagram
Gm = Inf dB (at Inf rad/sec) , Pm = 29 deg (at 3.1 rad/sec)
Frequency (rad/sec)
FIGURE E9.32
Bode plot for G(s) = C[sIA]1B+D=6.4
s2+s+4 .
E9.33 The Bode plot is shown in Figure E9.33. The phase margin is P.M. =
17.7and the gain margin is G.M. = 5.45 dB.
−150
−100
−50
0
50
Magnitude (dB)
10−1 100101102103
−270
−225
−180
−135
−90
−45
0
Phase (deg)
Bode Diagram
Gm = 5.45 dB (at 5.68 rad/sec) , Pm = 17.7 deg (at 4.24 rad/sec)
Frequency (rad/sec)
FIGURE E9.33
Bode plot for L(s) = 200
(s2+2.83s+4)(s+10) .
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Problems 463
Problems
P9.1 (a) The loop transfer function is
Gc(s)G(s) = 1
(1 + 0.5s)(1 + 2s).
P= 0, N = 0; therefore Z=N+P= 0. The system is stable. (Note:
See P8.1 for the polar plots.)
(b) The loop transfer function is
1 + 0.5s
s2.
P= 0, N = 0, therefore Z=N+P= 0. The system is stable.
(c) The loop transfer function is
s+ 4
s2+ 5s+ 25 .
P= 0, N = 0, Z =N+P= 0. Therefore, the system is stable.
(d) The loop transfer function is
30(s+ 8)
s(s+ 2)(s+ 4) .
P= 0, N = 2 therefore Z=P+N= 2. Therefore, the system has
two roots in the right half-plane, and is unstable.
P9.2 (a) The loop transfer function is
Gc(s)G(s) = K
s(s2+s+ 6) ,
and
Gc(jω)G(jω) = K
jω(ω2+jω + 6) == K[ω2jω(6 ω2)]
[(6 ω2)2ω2+ω4].
To determine the real axis crossing, we let
Im{Gc(jω)G(jω)}= 0 = Kω(6 ω2)
or
ω=6.
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464 CHAPTER 9 Stability in the Frequency Domain
Then,
Re{Gc(jω)G(jω)}ω=6=Kω2
ω4ω=6
=K
6.
So, K/6>1 for stability. Thus K < 6 for a stable system.
(b) The loop transfer function is
Gc(s)G(s) = K(s+ 1)
s2(s+ 6) .
The polar plot never encircles the -1 point, so the system is stable for
all gains K(See Figure 10 in Table 9.6 in Dorf & Bishop).
P9.3 (a,b) The suitable contours are shown in Figure P9.3.
s
jw
Gs
q=cosz
q
r approaches
innity
r
(a)
s
jw
Gs
r
(b)
-s1
r approaches
innity
FIGURE P9.3
Suitable contours Γsfor (a) and (b).
(c) Rewrite the characteristic equation as
1 + 96
s(s2+ 11s+ 56) = 0 .
In this case, σ1=1. Therefore, we have one pole inside the contour
at s= 0, so P= 1. The polar plot yields N=1, so Z=N+P= 0.
Therefore, all three roots have real parts less than -1. In fact, the
roots are s1=3, and s2,3=4±j4.
P9.4 (a) P= 0, N = 2, therefore Z= 2. The system has two roots in the right
hand s-plane.
(b) In this case, N= +1 1 = 0, so Z= 0. Therefore the system is
stable.
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Problems 465
P9.5 (a) The loop transfer function is
L(s) = Gc(s)G(s)H(s) = K
(s+ 1)(3s+ 1)(0.4s+ 1) .
The steady-state error is
ess =|R|
1 + K.
We require ess = 0.1|R|, so K > 9.
(b) Use K= 9. The Nyquist plot is shown in Figure P9.5. We determine
that P= 0 and N= 0. Therefore, Z= 0 and the system is stable.
-8
-6
-4
-2
0
2
4
6
8
-2 0 2 4 6 8 10
Real Axis
Imag Axis
FIGURE P9.5
Nyquist Diagram for L(s) = Gc(s)G(s)H(s) = 9
(s+1)(3s+1)(0.4s+1) .
(c) The phase and gain margins are P.M. = 18oand G.M. = 5 dB.
P9.6 The rotational velocity transfer function is
ω(s)
R(s)=G(s) = K
1 + s
3.7(2π)s
68(2π)+1 .
At low frequency, we have the magnitude near 35 dB, so 20 log K= 35 dB
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466 CHAPTER 9 Stability in the Frequency Domain
and K= 56. Since the frequency response plot is for rotational velocity
ω(s), and we are interested in position control, we add an integrator. The
characteristic equation is
1 + G(s)1
s= 1 + 56(23)(427)
s(s+ 23)(s+ 427) = 0 .
The roots are
s1=430 and s2,3=10 ±j35 .
Thus, ωn= 36 and ζ= 0.28. The time constant of the closed-loop system
is
τ=1
ζωn
= 99.6 msec .
P9.7 The loop transfer function is
L(s) = Gc(s)G(s)H(s) = 10K1s(s+ 7)
(s+ 3)(s2+ 0.36) .
(a) The Bode plot is shown in Figure P9.7 for K1= 2.
-50
0
50
100
10-1 100101102
Frequency (rad/sec)
Gain dB
-150
-100
-50
0
50
100
10-1 100101102
Frequency (rad/sec)
Phase deg
FIGURE P9.7
Bode Diagram for Gc(s)G(s)H(s) = 10K1s(s+7)
(s+3)(s2+0.36) , where K1= 2.
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Problems 467
(b) The phase margin P.M. = 80oand the gain margin G.M. =, since
φnever crosses = 180o.
(c) The transfer function from Td(s) to θ(s) is
θ(s) = G(s)
1 + Gc(s)G(s)H(s)Td(s).
Then, for a step disturbance θ() = lims0(s) = G(0) = 10/0.36 =
27.8, since H(0) = 0.
(d) The system is so highly damped, there is very little resonant peak.
(e) The estimated ζ=P.M./100 = 0.80. The actual ζ= 0.97.
P9.8 (a) The loop transfer function is
Gc(s)G(s)H(s) = s2
ω2
1+2ζ1s
ω1+ 1
(0.02s+ 1) s2
ω2
2+2ζ2s
ω2+ 1,
where ω1= 20π= 62.8, ω2= 14π= 43.9, ζ1= 0.05 and ζ2= 0.05.
The Bode plot is shown in Figure P9.8a. The phase margin is P.M. =
9o. Therefore, the system is unstable.
-40
-20
0
20
100101102103
Frequency (rad/sec)
Gain dB
-200
-150
-100
-50
0
100101102103
Frequency (rad/sec)
Phase deg
FIGURE P9.8
(a) Bode Diagram for Gc(s)G(s)H(s) = s22
1+(0.11)s+1
(0.02s+1)(s22
2+(0.12)s+1) , where ω1= 20π
and ω2= 14π.
(b) In this case ζ2= 0.25, with all other parameters the same as before.
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468 CHAPTER 9 Stability in the Frequency Domain
-40
-30
-20
-10
0
10
100101102103
Frequency (rad/sec)
Gain dB
-200
-150
-100
-50
0
100101102103
Frequency (rad/sec)
Phase deg
FIGURE P9.8
CONTINUED: (b) Bode Diagram for Gc(s)G(s)H(s) = s22
1+(0.11)s+1
(0.02s+1)(s22
2+(0.52)s+1) ,
where ω1= 20πand ω2= 14π.
The Bode plot is shown in Figure P9.8b. The phase margin is P.M. =
86o. Therefore, the system is now stable.
P9.9 (a) The Bode plot is shown in Figure P9.9a
The phase margin is P.M. = 83oand the gain margin is G.M. =.
(b) With the compensator, the loop transfer function is
Gc(s)G(s)H(s) = K1
0.30(s+ 0.05)(s2+ 1600)(s+ 0.5)
s(s2+ 0.05s+ 16)(s+ 70) ,
where
K2/K1= 0.5.
Let K1= 1. The Bode plot is shown in Figure P9.9b. The phase
margin is P.M. = 80oand the gain margin is G.M. =, essentially
the same as in (a). But the system in (b) is a type one, so that ess = 0
to a step input or disturbance. We cannot achieve a G.M. = 10 dB
by increasing or decreasing K1.
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Problems 469
-60
-40
-20
0
20
40
10-3 10-2 10-1 100101102103
Frequency (rad/sec)
Gain dB
-300
-200
-100
0
10-3 10-2 10-1 100101102103
Frequency (rad/sec)
Phase deg
FIGURE P9.9
(a) Bode Diagram for Gc(s)G(s)H(s) = 0.3(s+0.05)(s2+1600)
(s+70)(s2+0.05s+16) .
-60
-40
-20
0
20
40
10-3 10-2 10-1 100101102103
Frequency (rad/sec)
Gain dB
-300
-200
-100
0
10-3 10-2 10-1 100101102103
Frequency (rad/sec)
Phase deg
FIGURE P9.9
CONTINUED: (b) Bode Diagram for Gs(s)G(s)H(s) = 0.15K1(s+0.05)(s2+1600)(s+0.5)
(s+70)(s2+0.05s+16) ,
K1= 1.
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470 CHAPTER 9 Stability in the Frequency Domain
P9.10 The equations of motion are
F(s) = 3I(s) and I(s) = Eo(s)
R+Ls =Eo(s)
0.1 + 0.2s.
So,
F(s) = 30
(2s+ 1)Eo(s).
The actuator without the spring (see Table 2.7, Number 9 in Dorf &
Bishop) is modeled via
X(s)
Y(s)=1
Ms2+Bs =Ka
τas2+s.
With the spring, we have
X(s)
Y(s)=Ka
τas2+s+Ks
or GA(s) = 1
0.4s2+s+ 1.5.
Then, the loop transfer function is
L(s) = 30K1
(2s+ 1)(0.4s2+s+ 1.5) .
(a) The Bode plot for K1= 0.2 in Fig. P9.10 shows the P.M. = 30o.
-60
-40
-20
0
20
10-2 10-1 100101
Frequency (rad/sec)
Gain dB
-300
-200
-100
0
10-2 10-1 100101
Frequency (rad/sec)
Phase deg
FIGURE P9.10
Bode Diagram for L(s) = 30K1
(2s+1)(0.4s2+s+1.5) , where K1= 0.2.
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Problems 471
(b) For K1= 0.2, we determine that Mpω= 7.8 dB, ωr= 1.9 rad/sec,
and ωB= 2.8 rad/sec.
(c) The estimated percent overshoot is P.O. = 51% and the estimated
settling time is Ts= 10 sec. This is based on ζ= 0.21 and ωnωr=
1.9 rad/sec.
P9.11 The loop transfer function is
Gc(s)G(s) = 5(K1s+K2)e1.5s
s(5s+ 1) .
(a) Let K1=K2= 1. Then
Gc(s)G(s) = 5(s+ 1)
s(5s+ 1)e1.5s.
The Bode plot is shown in Figure P9.11a. The phase margin is P.M. =
48o. The system is unstable.
(b) Let K1= 0.1 and K2= 0.04. Then, the loop transfer function is
Gc(s)G(s) = 5(0.1s+ 0.04)e1.5s
s(5s+ 1) .
The Bode plot shown in Figure P9.11b shows P.M. = 45o. Thus, the
system is stable.
-20
0
20
40
60
10-2 10-1 100101
Frequency (rad/sec)
Gain dB
-1000
-800
-600
-400
-200
0
10-2 10-1 100101
Frequency (rad/sec)
Phase deg
FIGURE P9.11
(a) Bode Diagram for Gc(s)G(s) = 5(s+1)esT
s(5s+1) , where T= 1.5.
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472 CHAPTER 9 Stability in the Frequency Domain
-40
-20
0
20
40
10-2 10-1 100101
Frequency (rad/sec)
Gain dB
-1000
-800
-600
-400
-200
0
10-2 10-1 100101
Frequency (rad/sec)
Phase deg
FIGURE P9.11
CONTINUED: (b) Bode Diagram for Gc(s)G(s) = 5(0.1s+0.04)esT
s(5s+1) , where T= 1.5.
(c) When K2= 0.1394, the phase margin is P.M. = 0oand G.M. = 0
dB. So, for stability we require K20.1394 when K1= 0.
P9.12 (a) The Bode plot is shown in Figure P9.12.
−100
−80
−60
−40
−20
0
20
Magnitude (dB)
10−1 100101102
−270
−225
−180
−135
−90
−45
0
Phase (deg)
Bode Diagram
Gm = 12 dB (at 3.46 rad/sec) , Pm = 67.6 deg (at 1.53 rad/sec)
Frequency (rad/sec)
FIGURE P9.12
Bode Diagram for Gc(s)G(s) = 2
(0.5s+1)3.
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Problems 473
The loop transfer function (without the time delay) is
Gc(s)G(s) = 2
(0.5s+ 1)3.
The phase margin is P.M. = 67.6o.
(b) With the delay, the loop transfer function is
Gc(s)G(s)H(s) = 2e0.5s
(0.5s+ 1)3.
The phase margin is now P.M. = 23.7o. So the 0.5 sec time delay has
reduced the phase margin by 43.9.
P9.13 The loop transfer function is
Gc(s)G(s) = Ka(Ks + 1)
se1.2s.
(a) Let Ka=K= 1. Without the time delay, the system has infinite
phase and gain margin. However, with the time delay, the system has
a negative gain margin, hence it is unstable.
(b) A plot of phase margin versus Kais shown in Figure P9.13.
-60
-40
-20
0
20
40
60
80
100
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
Ka
Phase margin deg
FIGURE P9.13
Phase margin as a function of Kafor Gc(s)G(s) = Ka(s+1)e1.2s
s.
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474 CHAPTER 9 Stability in the Frequency Domain
Let K= 1, and find Kafor a stable system. Then,
Gc(s)G(s) = Ka(s+ 1)e1.2s
s.
If Ka= 0.8, then the phase margin is P.M. = 50o.
P9.14 The loop transfer function is
Gc(s)G(s) = Ke0.2s
s(0.1s+ 1) .
(a) The Nichols diagram is shown in Figure P9.14 for K= 2.5.
-40
-30
-20
-10
0
10
20
30
40
-350 -300 -250 -200 -150 -100 -50 0
-40
-20
-12
-6
-3
-1
0
0.25
0.5
1
2
3
6
Phase (deg)
Gain dB
FIGURE P9.14
Nichols diagram for Gc(s)G(s) = Ke0.2s
s(0.1s+1) , for K= 2.5.
It can be seen that
Mpω= 2.0 dB .
The phase and gain margins are P.M. = 48.5oand G.M. = 7.77 dB.
(b) We determine that ζ= 0.43 (based on Mpω= 2 dB) and ζ= 0.48
(based on the phase margin P.M. = 48.5o).
(c) The bandwidth is
ωB= 5.4 rad/sec .
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Problems 475
P9.15 (a) The ship transfer function is
G(s) = 0.164(s+ 0.2)(s0.32)
s2(s+ 0.25)(s0.009) .
The closed-loop system is unstable; the roots are
s1=0.5467
s2,3= 0.2503 ±0.1893j
s4=0.1949
Therefore the ship will not track the straight track.
(b) The system cannot be stabilized by lowering the gain; this is verified
in the root locus in Figure P9.15, where it is seen that the locus has
a branch in the right half-plane for all K > 0.
(c) Yes, the system can be stabilized.
(d) When the switch is closed, we have a derivative feedback, which adds
90ophase lead. This is not enough to stabilize the system. Additional
lead networks are necessary.
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
-0.6 -0.4 -0.2 0 0.2 0.4 0.6
xxxx oo
Real Axis
Imag Axis
FIGURE P9.15
Root locus for 1 + GH(s) = 1 + K0.164(s+0.2)(s0.32)
s2(s+0.25)(s0.009) = 0.
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476 CHAPTER 9 Stability in the Frequency Domain
P9.16 The loop transfer function is
Gc(s)G(s) = K
(s/10 + 1)(s2+s+ 2) .
When K= 3.2, the phase margin is P.M. 30o. The Bode plot is shown
in Figure P9.16.
10-1 100101102
-100
-50
0
50
Frequency (rad/sec)
Gain dB
Gm=10.88 dB, (w= 3.464) Pm=29.91 deg. (w=2.083)
10-1 100101102
0
-90
-180
-270
-360
Frequency (rad/sec)
Phase deg
FIGURE P9.16
Bode plot for Gc(s)G(s) = K
(s/10+1)(s2+s+2) , where K= 3.2.
P9.17 (a) We require ess 0.05A, and we have
ess =A
1 + Kp
<0.05A
or Kp>19. But
Kp= lim
s0G1(s)G2(s)G3(s)G4(s) = lim
s0
20K1
(0.5s+ 1) 0.1
1 + 4s2
= 0.2K1.
So, Kp= 0.2K1>19, or K1>95.
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Problems 477
(b) Given
G1(s) = K1(1 + 1
s) = K1s+ 1
s,
we require 1.05 < MPt<1.30, or 0.70 > ζ > 0.36, or 70o> P.M. >
36o. Then,
G1(s)G2(s)G3(s)G4(s) = 0.2K1(s+ 1)
s(0.5s+ 1)(4s+ 1)2.
When K1= 0.8, the P.M. = 40o. The Bode plot is shown in Fig-
ure P9.17a.
-100
-50
0
50
10-2 10-1 100101
Frequency (rad/sec)
Gain dB
-300
-200
-100
0
10-2 10-1 100101
Frequency (rad/sec)
Phase deg
FIGURE P9.17
(a) Bode plot for G1(s)G2(s)G3(s)G4(s) = 0.2K1(s+1)
s(0.5s+1)(4s+1)2, where K1= 0.8and
P.M. = 40o.
(c) For part (a), we had
G1(s)G2(s)G3(s)G4(s) = 2.375
(s+ 2)(s+ 0.25)2.
The characteristic equation is
s3+ 2.5s2+ 1.06s+ 2.50 = (s+ 2.48)(s2+ 0.02s+ 1.013) .
The dominant complex roots are lightly damped since ζ= 0.01 and
ζωn= 0.01.
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478 CHAPTER 9 Stability in the Frequency Domain
Thus,
Ts=4
ζωn
= 400 sec .
For part (b), we had
G1(s)G2(s)G3(s)G4(s) = (0.2)(0.8)(s+ 1)
s(0.5s+ 1)(4s+ 1)2.
The characteristic equation is
8s4+ 20s3+ 8.5s2+ 1.16s+ 0.16 = 0 .
The roots are s1=2, s2=0.4 and s3,4=0.05 ±j0.15. Thus
ζ= 0.16 and ζωn= 0.05. So,
Ts=4
ζωn
=4
0.05 = 75 sec .
(d) Let U(s) be a unit step disturbance and R(s) = 0. Then
Y(s)
U(s)=G3(s)G4(s)
1 + G1(s)G2(s)G3(s)G4(s)=0.1
1+4s2
1 + 20K1(s+1)
s(0.5s+1)(4s+1)2
.
The disturbance response is shown in Figure P9.17b.
-2
-1
0
1
2
3
4
5
6
x10-3
0 10 20 30 40 50 60 70 80 90 100
Time (secs)
Amplitude
FIGURE P9.17
CONTINUED: (b) System response to a unit disturbance U(s).
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Problems 479
P9.18 The transfer function is
Gc(s)G(s)H(s) = 5.3(s2+ 0.8s+ 0.32)eT s
s3.
The Bode plot is shown in Figure P9.18.
-20
0
20
40
60
80
10-1 100101
Frequency (rad/sec)
Gain dB
100
150
200
250
300
10-1 100101
Frequency (rad/sec)
Phase deg
T=0 solid ___ & T=0.1 dashed ---- & T=0.2674 dotted ....
FIGURE P9.18
Bode diagram for Gc(s)G(s)H(s) = K(s2+0.8s+0.32)esT
s3, where T= 0 (solid line),
T= 0.1(dashed line), and T= 0.2674 (dotted line).
The following results are verified in the figure.
(a) The phase margin is P.M. = 81oat ω= 5.3 when T= 0.
(b) For T= 0.1, the added phase is φ=T ω (in radians). The phase
margin is P.M. = 51oat ω= 5.3 when T= 0.1.
(c) The system is borderline stable when T= 0.2674 sec. The phase
margin is P.M. = 0oat ω= 5.3.
P9.19 The transfer function is
Gc(s)G(s) = 0.5
s(1 + 2s)(4 + s).
(a) The Nichols diagram is shown in Figure P9.19. The gain margin is
G.M. = 31.4 dB.
(b) The phase margin is P.M. = 75oand Mpω= 0 dB. The bandwidth is
0.17 rad/sec.
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480 CHAPTER 9 Stability in the Frequency Domain
Nichols Chart
Open
0 0
0
20
40
System: sys
Gain (dB): 4
Phase (deg):
Frequency (rad/sec): 1.44
System: sys
Gain (dB):
Phase (deg):
Frequency (rad/sec): 0.122
6 dB
3 dB
1 dB
0.5 dB
0.25 dB 0 dB
Gain margin = 31.4 dB
Phase margin = 180-105=75o
FIGURE P9.19
Nichols diagram for Gc(s)G(s) = 0.5
s(2s+1)(s+4) .
P9.20 (a) Let K= 100. The Bode plot is shown in Figure P9.20a. The loop
transfer function is
Gc(s)G(s) = K(s2+ 1.5s+ 0.5)
s(20s+ 1)(10s+ 1)(0.5s+ 1) .
-50
0
50
100
10-3 10-2 10-1 100101
Frequency (rad/sec)
Gain dB
-300
-200
-100
0
10-3 10-2 10-1 100101
Frequency (rad/sec)
Phase deg
FIGURE P9.20
(a) Bode diagram for Gc(s)G(s) = K(s2+1.5s+0.5)
s(20s+1)(10s+1)(0.5s+1) , where K= 100.
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Problems 481
(b) The phase margin is P.M. =3.5oand the gain margin is G.M. = 2.7
dB.
(c) You must decrease Kbelow 100 to achieve a P.M. = 40o. For K= 0.1,
the phase margin P.M. = 37.9o.
(d) The step response is shown in Figure P9.20b for K= 0.1.
0
0.2
0.4
0.6
0.8
1
1.2
1.4
0 50 100 150 200 250 300 350 400
Time (secs)
Amplitude
FIGURE P9.20
CONTINUED: (b) Unit step response K= 0.1.
P9.21 The loop transfer function is
Gc(s)G(s) = K
s(s+ 1)(s+ 4) .
(a) The Bode plot is shown in Figure P9.21 for K= 4.
(b) The gain margin is
G.M. = 14 dB .
(c) When K= 5, the gain margin is
G.M. = 12 dB .
(d) We require Kv>3, but Kv=K
4. So, we need K > 12. This gain can
be utilized since K < 20 is required for stability.
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482 CHAPTER 9 Stability in the Frequency Domain
-150
-100
-50
0
50
10-1 100101102
Frequency (rad/sec)
Gain dB
-300
-200
-100
0
10-1 100101102
Frequency (rad/sec)
Phase deg
FIGURE P9.21
Bode diagram for Gc(s)G(s) = K
s(s+1)(s+4) , where K= 4.
P9.22 (a) The resonant frequency ωr= 5.2 rad/sec is point 6 on the Nichol’s
chart.
(b) The bandwidth is between points 8 and 9. We estimate the bandwidth
to be ωB= 7.5 rad/sec.
(c) The phase margin P.M. = 30o.
(d) The gain margin G.M. = 8 dB.
(e) Since we have P.M. = 30o, then we estimate ζ= 0.3. We can also
approximate
ωnωr= 5.2.ap9.1
Thus,
Ts=4
ζωn
=4
1.56 = 2.5sec .
P9.23 The phase margin is P.M. = 60 deg when K= 266. The gain margin is
G.M. = 17.2 dB .
The Bode plot is shown in Figure P9.23.
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Problems 483
−150
−100
−50
0
50
Magnitude (dB)
10−1 100101102103
−270
−225
−180
−135
−90
Phase (deg)
Bode Diagram
Gm = 17.2 dB (at 9.8 rad/sec) , Pm = 60 deg (at 2.58 rad/sec)
Frequency (rad/sec)
FIGURE P9.23
Bode diagram for Gc(s)G(s) = K
s(s+8)(s+12) , where K= 266.
P9.24 When K= 14.1, then P.M. = 45 deg, G.M. =dB and ωB= 29.3
rad/sec.
Frequency (rad/sec)
Phase (deg); Magnitude (dB)
- 50
0
50
100
Gm=356.59 dB (at 0 rad/sec), Pm=60 deg. (at 17.321 rad/sec)
10- 110010 1102
- 180
- 160
- 140
- 120
- 100
- 80
FIGURE P9.24
Bode diagram for G(s) = K(s+20)
s2, where K= 14.1.
P9.25 The phase margin is P.M. = 60 deg when K= 2.61 and T= 0.2 second.
The Bode plot is shown in Figure P9.25.
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484 CHAPTER 9 Stability in the Frequency Domain
10-1 100101102
-40
-20
0
20
40
K=2.61; PM=60.09 at wc=2.61 rad/sec
Frequency (rad/sec)
Gain dB
10-1 100101102
-1500
-1000
-500
0
Frequency (rad/sec)
Phase deg
FIGURE P9.25
Bode diagram for Gc(s)G(s) = Ke0.2s
s, where K= 2.61.
P9.26 The loop transfer function is
Gc(s)G(s) = K
s(0.25s+ 1)(0.1s+ 1) .
The Bode plot is shown in Figure P9.26a for K= 10. The Nichols chart
is shown in Figure P9.26b. The phase and gain margins are
P.M. = 9oand G.M. = 3 dB .
The system bandwidth is ωB= 8 rad/sec. From the P.M. = 9o, we
estimate ζ= 0.09. Therefore, the predicted overshoot is
P.O. = 100eπζ/1ζ2= 75% ,where ζ= 0.09 .
The resonant peak occurs at ωr= 5.5 rad/sec. If we estimate ωnωr=
5.5 rad/sec, then the settling time is
Ts=4
ζωn
= 8 sec .
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Problems 485
-100
-50
0
50
10-1 100101102
Frequency (rad/sec)
Gain dB
-300
-200
-100
0
10-1 100101102
Frequency (rad/sec)
Phase deg
-40
-30
-20
-10
0
10
20
30
40
-350 -300 -250 -200 -150 -100 -50 0
-40
-20
-12
-6
-3
-1
0
0.25
0.5
1
2
3
16
Phase (deg)
Gain dB
FIGURE P9.26
(a) Bode diagram for Gc(s)G(s) = K
s(0.25s+1)(0.1s+1) , where K= 10. (b) Nichols chart
for Gc(s)G(s) = K
s(0.25s+1)(0.1s+1) , where K= 10.
P9.27 The loop transfer function is
L(s) = Gc(s)G(s)H(s) = 4K
(s2+ 2s+ 4)(s+ 1) .
The plot of the phase margin versus the gain Kis shown in Figure P9.27.
As the gain increases towards Kmax = 3.5, the phase margin decreases
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486 CHAPTER 9 Stability in the Frequency Domain
towards zero.
1 1.5 2 2.5 3 3.5
0
20
40
60
80
100
120
140
160
180
K
Phase margin (deg)
FIGURE P9.27
Phase margin versus the gain K.
P9.28 The loop transfer function is
Gc(s)G(s) = KP
s(s+ 1) .
When KP= 1.414, we have P.M. 45. Using the approximation that
ζP.M./100 we estimate that ζ= 0.45. Then using the design formula
P.O. = 100eπζ/1ζ2= 20.5% .
The actual overshoot is 23.4%. The step input response is shown in Fig-
ure P9.28. The actual damping ratio is ζ= 0.42. This shows that the
approximation ζP.M./100 is quite applicable and useful in predicting
the percent overshoot.
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Problems 487
Step Response
Time (sec)
Amplitude
0 5 10 15
0
0.2
0.4
0.6
0.8
1
1.2
1.4
System: syscl
Peak amplitude: 1.23
Overshoot (%): 23.3
At time (sec): 2.97
FIGURE P9.28
Step response showing a 23.3% overshoot.
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488 CHAPTER 9 Stability in the Frequency Domain
Advanced Problems
AP9.1 The loop transfer function is
L(s) = Gc(s)G(s)H(s) = 236607.5(s+ 10)(s+ 5)
s(s+ 2)(s2+ 100s+ω2
n)(s+ 1) .
(a) The Bode plot for
ω2
n= 15267
is shown in Figure AP9.1a.
-100
-50
0
50
100
150
10-3 10-2 10-1 100101102103
Frequency (rad/sec)
Gain dB
-300
-200
-100
0
10-3 10-2 10-1 100101102103
Frequency (rad/sec)
Phase deg
FIGURE AP9.1
(a) Bode Diagram for L(s) = 236607.5(s+10)(s+5)
s(s+2)(s2+100s+ω2
n)(s+1) , where ω2
n= 15267.
The phase and gain margins are
P.M. = 48.6oand G.M. = 15.5 dB .
(b) The Bode plot for ω2
n= 9500 is shown in Figure AP9.1b. The gain
and phase margins are
P.M. = 48.5oand G.M. = 10.9 dB .
Reducing the natural frequency by 38% has the effect of reducing the
gain margin by 30%.
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Advanced Problems 489
-100
-50
0
50
100
150
10-3 10-2 10-1 100101102103
Frequency (rad/sec)
Gain dB
-300
-200
-100
0
10-3 10-2 10-1 100101102103
Frequency (rad/sec)
Phase deg
FIGURE AP9.1
CONTINUED: (b) Bode Diagram for L(s) = 236607.5(s+10)(s+5)
s(s+2)(s2+100s+ω2
n)(s+1) , where ω2
n= 9500.
AP9.2 (a) The Bode plot with T= 0.05 sec is shown in Figure AP9.2a. The
phase margin is P.M. = 47.7oand the gain margin is G.M. = 11.2
dB.
-40
-20
0
20
40
100101102
Gain dB
Frequency (rad/s)
-400
-300
-200
-100
100101102
Phase deg
Frequency (rad/s)
FIGURE AP9.2
(a) Bode Diagram for Gc(s)G(s)H(s) = 8 (s+5)
s(s+2) esT, where T= 0.05s.
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490 CHAPTER 9 Stability in the Frequency Domain
(b) The Bode plot with T= 0.1 sec is shown in Figure AP9.2b. The
-40
-20
0
20
40
100101102
Gain dB
Frequency (rad/s)
-800
-600
-400
-200
0
100101102
Phase deg
Frequency (rad/s)
FIGURE AP9.2
CONTINUED: (b) Bode Diagram for Gc(s)G(s)H(s) = 8 (s+5)
s(s+2) esT , where T= 0.1s.
phase margin is P.M. = 22.1oand the gain margin is G.M. = 4.18
dB. A 100% increase in time delay Tleads to a 50% decrease in phase
and gain margins.
(c) The damping ratio ζP.M./100 and
P.O. 100eπζ/1ζ2.
So, for T= 0.05 sec, ζ0.47 and P.O. 18.7%. Also, for T= 0.1
sec, ζ0.22 and P.O. 49.2%.
AP9.3 The loop transfer function is
L(s) = Gc(s)G(s)H(s) = 66K(1 + 0.1s)
(1 + 0.01s)(1 + 0.01s)(1 + 1.5s)(1 + 0.2s).
(a) When K= 1, the gain and phase margins are
G.M. = 18.4 dB and P.M. = 55o.
(b) When K= 1.5, the gain and phase margins are
G.M. = 14.9 dB and P.M. = 47.8o.
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Advanced Problems 491
(c,d) The bandwidth and settling time with K= 1 are ωB= 233.6
rad/sec and Ts= 0.4 second. When K= 1.5, we determine that
ωB= 294.20 rad/sec and Ts= 0.33 second.
AP9.4 The loop transfer function is
L(s) = Gc(s)G(s) = K(s+ 40)
s(s+ 15)(s+ 10) .
The gain K= 28.8 satisfies the specifications. The actual gain and phase
margins are
G.M. = 18.8 dB and P.M. = 45o.
The system bandwidth is ωB= 10.3 rad/sec. The step response is shown
in Figure AP9.4.
Step Response
Time (sec)
Amplitude
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2
0
0.2
0.4
0.6
0.8
1
1.2
1.4
System: sys_cl
Peak amplitude: 1.23
Overshoot (%): 23.4
At time (sec): 0.476
System: sys_cl
Settling Time (sec): 1.1
FIGURE AP9.4
Closed-loop system step response.
AP9.5 The loop transfer function is
L(s) = Gc(s)G(s) = Ks+ 0.4
s4+ 9s3+ 18s2.
The Bode plot for K= 1 is shown in Figure AP9.5. From the phase
response, we determine that the maximum P.M. 41o. From the mag-
nitude response (for K= 1), we find that the gain needs to be raised to
K= 14 to achieve maximum phase margin at ω= 0.826 rad/sec. The
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492 CHAPTER 9 Stability in the Frequency Domain
gain and phase margin with K= 14 are
G.M. = 19.3 dB and P.M. = 40.9o.
Also, the overshoot is P.O. = 38.3%.
−150
−100
−50
0
50
Magnitude (dB)
10−2 10−1 100101102
−270
−225
−180
−135
System: sys
Frequency (rad/sec): 0.865
Phase (deg): −139
Phase (deg)
Bode Diagram
Gm = 42.3 dB (at 3.79 rad/sec) , Pm = 16.7 deg (at 0.154 rad/sec)
Frequency (rad/sec)
FIGURE AP9.5
Bode plot for L(s) = Ks+0.4
s4+9s3+18s2with K= 1.
AP9.6 With D > 2m, the gain can be increased up to K= 100, while still
retaining stability.
AP9.7 The loop transfer function is
L(s) = Gc(s)G(s) = K(s+ 4)
s2.
We select
K= 22
for P.M. = 45o. The system bandwidth is
ωB= 5.88 rad/sec .
The disturbance response is shown in Figure AP9.7. The maximum output
due to a disturbance is y(t) = 0.11.
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Advanced Problems 493
0
0.02
0.04
0.06
0.08
0.1
0.12
0 0.5 1 1.5 2 2.5 3 3.5 4
Time (secs)
Amplitude
FIGURE AP9.7
Closed-loop system disturbance response.
AP9.8 A reasonable choice for the gain is K= 2680. The phase margin is P.M. =
42.8and the percent overshoot is P.O. = 18.9%. The Nichols chart is
shown in Figure AP9.8.
Nichols Chart
Open−Loop Phase (deg)
Open−Loop Gain (dB)
−360 −315 −270 −225 −180 −135 −90 −45 0
−120
−100
−80
−60
−40
−20
0
20
40
60
6 dB
3 dB
1 dB
0.5 dB
0.25 dB 0 dB
−1 dB
−3 dB
−6 dB
−12 dB
−20 dB
−40 dB
−60 dB
−80 dB
−100 dB
−120 dB
FIGURE AP9.8
Nichols chart.
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494 CHAPTER 9 Stability in the Frequency Domain
AP9.9 The loop transfer function is
L(s) = Gc(s)G(s) = Kp(s+ 0.2)
s2(s2+ 7s+ 10) .
At the maximum phase margin, Kp= 4.9 for P.M. = 48.6o. The Bode
diagram is shown in Figure AP9.9.
Frequency (rad/sec)
Phase (deg); Magnitude (dB)
Bode Diagrams
-150
-100
-50
0
50
100
Gm=21.788 dB (at 2.9326 rad/sec), Pm=48.457 deg. (at 0.50782 rad/sec)
10-3 10-2 10-1 100101102
-300
-250
-200
-150
-100
FIGURE AP9.9
Phase and gain margin.
AP9.10 The closed-loop transfer function is
T(s) = K
s2+ 3s+ 1 .
We require K= 1 a zero steady-state tracking error to a unit step. The
step response is shown in Figure AP9.10. Computing T(jω) = 0.707 it
follows that
1
(jω)2+ 3jω + 1
= 0.707 or ω4+ 7ω21 = 0 .
Solving for ωyields ω= 0.37 rad/s. This is the bandwidth of the system.
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Advanced Problems 495
0 2 4 6 8 10 12 14 16
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Step Response
Time (sec)
Amplitude
FIGURE AP9.10
Unit step response.
AP9.11 The phase margin versus time delay is shown in Figure AP9.11a.
0.5 1 1.5 2 2.5 3 3.5
−10
0
10
20
30
40
50
60
70
80
Time Delay (s)
Phase Margin (deg)
Time Delay=1 PM=58.5285
Time Delay=3.0455 PM=0.001
FIGURE AP9.11
Phase margin versus time delay.
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496 CHAPTER 9 Stability in the Frequency Domain
The maximum time delay is T= 3.04 s for stability. The step response is
shown in Figure AP9.11b. The percent overshoot is P.O. = 7.6%.
0 1 2 3 4 5 6 7 8 9 10
0
0.2
0.4
0.6
0.8
1
1.2
1.4
Time (sec)
Amplitude
FIGURE AP9.11
Unit step response.
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Design Problems 497
Design Problems
The plant model with parameters given in Table CDP2.1 in Dorf andCDP9.1
Bishop is given by:
θ(s)
Va(s)=26.035
s(s+ 33.142) ,
where we neglect the motor inductance Lmand where we switch off the
tachometer feedback (see Figure CDP4.1 in Dorf and Bishop). The closed-
loop system characteristic equation is
1 + 26.035Ka
s(s+ 33.142) = 0 .
The phase margin is P.M. = 70.4when Ka= 16. The step response with
K= 16 is shown below.
0 0.05 0.1 0.15 0.2 0.25 0.3
0
0.2
0.4
0.6
0.8
1
1.2
Time (secs)
Amplitude
DP9.1 (a) The gain and phase margins are G.M. = 7 dB and P.M. = 60o.
(b) The resonant peak and frequency are Mpω= 2 dB and ωr= 5 rad/sec.
(c) We have ωB= 20 rad/sec. From Mpω= 2 dB we estimate ζ= 0.45
(Figure 8.11 in Dorf & Bishop). Also, ωrn= 0.8, so ωn= 6.25.
Thus, Ts= 1.4.
(d) We need P.O. = 30oor ζ= 0.3 or P.M. 30o. So, we need to raise
the gain by 10 dB or K= 3.2.
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498 CHAPTER 9 Stability in the Frequency Domain
DP9.2 The loop transfer function is
L(s) = Gc(s)G(s) = K(s+ 0.5)
s2(s2+ 7.5s+ 9) .
When K= 6.25, we have the maximum phase margin. The phase mar-
gin maximum is P.M. = 23o. The plot of P.M. versus Kis shown in
Figure DP9.2a.
0123456789
4
6
8
10
12
14
16
18
20
22
24
K
Phase Margin deg
FIGURE DP9.2
(a) Phase margin versus Kfor L(s) = K(s+0.5)
s2(s2+7.5s+9) .
The predicted damping is ζ= 0.23. It then follows that the predicted
percent overshoot is
P.O. = 100eπζ/1ζ2= 48% .
The actual overshoot is 65%. The step input response is shown in Fig-
ure DP9.2b. The resonant peak occurs at ωr= 0.75 rad/sec. Approxi-
mating ωnωr= 0.75 rad/sec, we can estimate the settling time as
Ts=4
ζωn
= 23 sec .
The actual settling time is 20 sec.
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Design Problems 499
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
0 5 10 15 20 25 30
Time (secs)
Amplitude
FIGURE DP9.2
CONTINUE: (b) Closed-loop unit step response.
DP9.3 We want to select the gain Kas large as possible to reduce the steady-state
error, but we want a minimum phase margin of P.M. = 45oto achieve
good dynamic response. A suitable gain is K= 4.2, see Figure DP9.3.
10-2 10-1 100101
-300
-200
-100
0
Frequency (rad/sec)
Phase deg
10-2 10-1 100101
-40
-20
0
20
K=4.2; PM=45.34 at wc=0.102 rad/sec
Frequency (rad/sec)
Gain dB
FIGURE DP9.3
Bode plot for G(s) = Ke10s
40s+1 .
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500 CHAPTER 9 Stability in the Frequency Domain
DP9.4 We are given the loop transfer function
L(s) = Gc(s)G(s) = K
s(s+ 1)(s+ 4)
which can be written as
Gc(s)G(s) = Kv
s(s+ 1)(0.25s+ 1) .
The performance results are summarized in Table DP9.4.
KvG.M. P.M. ωBP.O. Ts
(dB) (deg) (rad/sec) (%) (sec)
0.40 21.9 64.2 0.62 4.4 9.8
0.75 16.5 49.0 1.09 19.0 10.1
TABLE DP9.4 Summary for Kv= 0.40 and Kv= 0.75.
When Kv= 0.40, we have
ess
A=1
0.40 = 2.5,
or 2 1/2 times the magnitude of the ramp. This system would be accept-
able for step inputs, but unacceptable for ramp inputs.
DP9.5 (a) With a time delay of T= 0.8 second, we determine that the propor-
tional controller
Gc(s) = K= 7
provides a suitable response with
P.O. = 8.3 % ess = 12.5 % Ts= 4.38 sec .
(b) A suitable proportional, integral controller is
Gc(s) = K1+K2/s = 6 + 0.6/s .
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Design Problems 501
The response to a unit step is
P.O. = 5.14 % ess = 0 % Ts= 6.37 sec .
The Nichols chart is shown in Figure DP9.5.
-40
-30
-20
-10
0
10
20
30
40
-350 -300 -250 -200 -150 -100 -50 0
-40
-20
-12
-6
-3
-1
0
0.25
0.5
1
3
6
8
Phase (deg)
Gain dB
FIGURE DP9.5
Nichols chart for Gc(s)G(s) = (K1s+K2)e0.8s
s(10s+1) , where K1= 6 and K2= 0.6.
DP9.6 With K= 170, at the two extreme values of b, we have
b= 80 P.M. = 91.62oG.M. = 13.66 dB
b= 300 P.M. = 75.23oG.M. = 25.67 dB .
Since reducing the value of Konly increases the P.M. and G.M., a value
of
K= 170
is suitable to meet P.M. = 40oand G.M. = 8 dB for the range of b.
DP9.7 A suitable gain is
K= 0.22 .
This results in P.M. = 60.17oand G.M. = 13.39 dB. The step reponse is
shown in Figure DP9.7.
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502 CHAPTER 9 Stability in the Frequency Domain
0
0.2
0.4
0.6
0.8
1
1.2
0 2 4 6 8 10 12 14 16 18 20
Time (secs)
Amplitude
FIGURE DP9.7
Lunar vehicle step response.
DP9.8 A gain of K= 315000 will satisfy the P.O. specification, while giving the
fastest response. The step response is shown in Figure DP9.8.
0
0.2
0.4
0.6
0.8
1
1.2
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
Time (secs)
Amplitude
FIGURE DP9.8
Steel rolling mill step response.
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Design Problems 503
DP9.9 The closed-loop transfer function is
Ts(2) = G1(s)
1 + Gc(s)G2(s)To(s) + Gc(s)G2(s)
1 + Gc(s)G2(s)T2d(s).
where
G1(s) = 1
(10s+ 1)(50s+ 1)
and
G2(s) = 0.01
(10s+ 1)(50s+ 1) .
The steady-state error (with Gc(s) = 500) to a unit step 2A(and after
the system has settled out subsequent to a step of magnitude A) is
ess = 2(0.167) = 0.33 .
The step response is shown in Figure DP9.9.
0
0.5
1
1.5
2
2.5
0 200 400 600 800 1000 1200 1400 1600
Time (sec)
T2/A
Gc=500 (solid); Gc=1/s (dashed); Gc=600+6/s (dotted)
FIGURE DP9.9
Two tank temperature control step response.
A suitable integral controller is
Gc(s) = 1
s.
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504 CHAPTER 9 Stability in the Frequency Domain
In this case, the steady-state tracking error is zero , since the system is a
type 1. The system response is shown in Figure DP9.9. With the integral
controller, the settling time is about Ts= 438 seconds and the P.O. = 7%.
A suitable PI controller is
Gc(s) = 600 + 6
s.
With the PI controller, the settling time is about Ts= 150 seconds and
the P.O. = 10%.
DP9.10 The system is given by
˙
x=Ax +Br
y=Cx
where
A=
0 1
2K13K2
,B=
0
1
,and C=1 0 .
The associated transfer function is
T(s) = 1
s2+ (K23)s+K12.
The characteristic polynomial is
s2+ (K23)s+K12 = 0 .
If we select K1= 3, then we have a zero-steady error to a unit step
response R(s) = 1/s, since
lim
s0s[1 T(s)] R(s) = lim
s0
s2+ (K23)s
s2+ (K23)s+K12= 0 .
Let
K=3 4.3.
The step response is shown in Figure DP9.10a. The bandwidth is ωb=
1.08 rad/s, as seen in Figure DP9.10b.
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Design Problems 505
0 1 2 3 4 5 6 7 8 9
0
0.2
0.4
0.6
0.8
1
1.2
1.4
Step Response
Time (sec)
Amplitude
Bode Diagram
Frequency (rad/sec)
10−1 100101
−40
−35
−30
−25
−20
−15
−10
−5
0
5
System: sys
Frequency (rad/sec): 1.08
Magnitude (dB): −3
Magnitude (dB)
FIGURE DP9.10
Step response with K= [3 4.3] and closed-loop Bode plot.
DP9.11 A time domain step response specification P.O. > 10% requires the dom-
inant poles to have a damping ration of ζ= 0.6. This time domain spec-
ification can be transformed to a frequency response specification using
the approximation P.M. 100ζ= 60. To keep the problem tractable,
we consider the controller with the form
Gc(s) = KP+KI
s=KP+1
s,
where we let KI= 1. The plot of the P.M. as a function of KPis shown
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506 CHAPTER 9 Stability in the Frequency Domain
in Figure DP9.11a. If we select KP= 0.07 we expect a phase margin
of approximately 60, hence a percent overshoot P.O. 10%. The step
response is shown in Figure DP9.11b. The actual phase margin is P.M. =
60.2, the percent overshoot is P.O. = 5.9% and the settling time is
Ts= 3.4 sec.
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5
55
60
65
70
75
80
85
KP
Phase Margin (deg)
0123456
0
0.2
0.4
0.6
0.8
1
1.2
1.4
Time (sec)
Amplitude
FIGURE DP9.11
(a) Phase margin versus controller gain KPand KI= 1. (b) Step response with KP= 0.07
and KI= 1.
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Computer Problems 507
Computer Problems
CP9.1 The m-file script to generate the Bode plot (from which the gain and
phase margin can be determined) is shown in Figure CP9.1. The transfer
function is
G(s) = 141
s2+ 2s+ 12 .
The gain margin is
G.M. =
and the phase margin is
P.M. = 10o.
num=141; den=[1 2 12];
sys = tf(num,den);
margin(sys);
−40
−20
0
20
40
Magnitude (dB)
10−1 100101102
−180
−135
−90
−45
0
Phase (deg)
Bode Diagram
Gm = Inf dB (at Inf rad/sec) , Pm = 10 deg (at 12.3 rad/sec)
Frequency (rad/sec)
FIGURE CP9.1
Gain and phase margin with the margin function.
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508 CHAPTER 9 Stability in the Frequency Domain
CP9.2 The Nyquist plots are shown in Figures CP9.2a-c.
−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1
−0.5
−0.4
−0.3
−0.2
−0.1
0
0.1
0.2
0.3
0.4
0.5
num=[5]; den=[1 5]; sys=tf(num,den); nyquist(sys)
Real Axis
Imaginary Axis
FIGURE CP9.2
(a) Nyquist plot for G(s) = 5
s+5 .
−1 −0.5 0 0.5 1 1.5 2
−1.5
−1
−0.5
0
0.5
1
1.5
num=[50]; den=[1 10 25]; sys=tf(num,den); nyquist(sys)
Real Axis
Imaginary Axis
FIGURE CP9.2
CONTINUED: (b) Nyquist plot for G(s) = 50
s2+10s+25 .
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Computer Problems 509
−5 0 5 10 15
−15
−10
−5
0
5
10
15
num=[15]; den=[1 3 3 1]; sys=tf(num,den); nyquist(sys)
Real Axis
Imaginary Axis
FIGURE CP9.2
CONTINUED: (c) Nyquist plot for G(s) = 15
s3+3s2+3s+1 .
CP9.3 The m-file script to generate the Nichols chart for part (a) is shown in
Figure CP9.3a. The Nichols charts for (b) and (c) are similiarly generated;
all plots are in Figure CP9.3a-c.
num = [1]; den = [1 0.2];
sys = tf(num,den);
nichols(sys)
ngrid
−360 −315 −270 −225 −180 −135 −90 −45 0
−20
−10
0
10
20
30
40
6 dB
3 dB
1 dB
0.5 dB
0.25 dB
0 dB
−1 dB
−3 dB
−6 dB
−12 dB
−20 dB
Nichols Chart
Open−Loop Phase (deg)
Open−Loop Gain (dB)
FIGURE CP9.3
(a) M-file script and Nichols chart for G(s) = 1
s+0.1.
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510 CHAPTER 9 Stability in the Frequency Domain
The gain and phase margin for each transfer function are as follows:
(a) G.M. =and P.M. = 102o
(b) G.M. =and P.M. =
(c) G.M. = 20 dB and P.M. =
−360 −315 −270 −225 −180 −135 −90 −45 0
−60
−50
−40
−30
−20
−10
0
10
20
30
40
6 dB
3 dB
1 dB
0.5 dB
0.25 dB
0 dB
−1 dB
−3 dB
−6 dB
−12 dB
−20 dB
−40 dB
−60 dB
Nichols Chart
Open−Loop Phase (deg)
Open−Loop Gain (dB)
FIGURE CP9.3
CONTINUED: (b) Nichols chart for G(s) = 1
s2+2s+1 .
−360 −315 −270 −225 −180 −135 −90 −45 0
−100
−80
−60
−40
−20
0
20
40
6 dB
3 dB
1 dB
0.5 dB
0.25 dB 0 dB
−1 dB
−3 dB
−6 dB
−12 dB
−20 dB
−40 dB
−60 dB
−80 dB
−100 dB
Nichols Chart
Open−Loop Phase (deg)
Open−Loop Gain (dB)
FIGURE CP9.3
CONTINUED: (c) Nichols chart for G(s) = 24
s3+9s2+26s+24 .
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Computer Problems 511
CP9.4 To obtain a phase margin P.M. = 40we select
K= 15
when T= 0.2 second. The variation in the phase margin for 0 T0.3
is shown in Figure CP9.4.
T=[0:0.01:0.3]; K=15;
num=K;den=[1 12]; sys = tf(num,den);
% w=logspace(-2,1,400);
for i=1:length(T)
[mag,phase,w]=bode(sys);
ph(1:length(phase))=phase(1,1,:); ph=ph';
ph2=ph-w*T(i)*180/pi;
[Gm,Pm,Wcg,Wcp]=margin(mag,ph2,w);
clear ph ph2
PMo(i)=Pm;
end
plot(T,PMo), grid
xlabel('Time delay (sec)')
ylabel('Phase margin (deg)')
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35
−20
0
20
40
60
80
100
120
140
160
Time delay (sec)
Phase margin (deg)
K=15
FIGURE CP9.4
Variation in the phase margin for 0T0.3with K= 15.
CP9.5 The loop transfer function is
L(s) = Gc(s)G(s) = K(s+ 50)
s(s+ 20)(s+ 10) .
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512 CHAPTER 9 Stability in the Frequency Domain
The plot of system bandwidth versus the gain Kis shown in Figure CP9.7.
K=[0.1:1:50];
w=logspace(-2,3,2000);
den=[1 30 200 0];
for i=1:length(K)
num=K(i)*[1 50]; sys = tf(num,den);
sys_cl = feedback(sys,[1]);
[mag,phase,w]=bode(sys_cl,w);
L=nd(mag<0.707); wb(i)=w(L(1));
end
plot(K,wb), grid
xlabel('Gain K')
ylabel('Bandwidth (rad/sec)')
0 5 10 15 20 25 30 35 40 45 50
0
5
10
15
Gain K
Bandwidth (rad/sec)
FIGURE CP9.5
Variation in the system bandwidth for 0K50.
CP9.6 The m-file script and Bode plot are shown in Figure CP9.6. The gain and
phase margin and ωcare determined to be G.M. = 2.23, P.M. = 26oand
ωc= 12.6 rad/sec. So, the maximum value of bois found to be
bomax = 2.13bo= 1.11 .
In this problem, there is also a minimum value of bo. Using the Routh-
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Computer Problems 513
numg = -0.5*[1 0 -2500]; deng = [1 47 850 -3000];
sysg = tf(numg,deng);
numc = 10*[1 3]; denc = [1 0]; sysc = tf(numc,denc);
sys_o = series(sysc,sysg);
bode(sys_o)
[mag,phase,w] = bode(sys_o);
[gm,pm,wg,wc] = margin(mag,phase,w)
gm =
2.2238
pm =
26.3187
wg =
26.1155
wc =
12.6487
-50
0
50
10-1 100101102103
Frequency (rad/sec)
Gain dB
50
100
150
200
250
10-1 100101102103
Frequency (rad/sec)
Phase deg
FIGURE CP9.6
Using the margin function to compute stability margins.
Hurwitz method, we determine that (for stability) the range of bois
0.14 < bo<1.11 .
CP9.7 The m-file script is shown in Figure CP9.7a. Since we do not have a value
for J, we write the loop transfer function as
Gc(s)G(s) = ¯
K1+¯
K2s
s2
where ¯
K1=K1/J and ¯
K2=K2/J. We work with ¯
K1and ¯
K2, then we
can always compute K1and K2whenever Jis specified. A PD controller
which meets the specs is given by
Gc(s) = 0.04 + 0.3s .
The step response is shown in Figure CP9.7b. The Bode plot is shown in
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514 CHAPTER 9 Stability in the Frequency Domain
%
% Part (a)
%
numc = [0.3 0.04]; denc = [1]; sysc = tf(numc,denc);
numg = [1]; deng = [1 0 0]; sysg = tf(numg,deng);
sys_o = series(sysc,sysg);
sys_cl = feedback(sys_o,[1]);
step(sys_cl), pause
%
% Part (b)
%
w = logspace(-1,1,400);
[mag,phase] = bode(sys_o,w);
[gm,pm,w1,w2] = margin(mag,phase,w);
margin(mag,phase,w), pause
%
% Part (c)
%
T = [1:0.1:5];
for i = 1:length(T)
[numd,dend] = pade(T(i),2); sysd = tf(numd,dend);
sys_o1 = series(sysd,sys_o);
sys_cl1 = feedback(sys_o1,sysd);
p(:,i) = pole(sys_cl1);
end
plot(real(p),imag(p),'*');grid
xlabel('Real Axis');
ylabel('Imag Axis')
FIGURE CP9.7
Script to assist in all three parts of the problem.
Figure CP9.7c. The phase margin is P.M. = 67.7oat ω= 0.32 rad/sec.
The loop transfer function is
Gc(s)G(s)H(s) = ¯
K1+¯
K2s
s2e2T s
where Tis the one-way time delay. If the phase lag introduced by the
delay is greater than 67.7oat ω= 0.32 rad/sec, then the system will
become unstable. So, since the phase lag due to the time delay ˜
Tis
φ(ω) = ω˜
T
we have 67.7oπ/180 = 0.32(2T) where ˜
T= 2T. Solving for Tyields
T= 1.82 seconds. This is the maximum allowable one-way time delay.
Executing the third part of the m-file script in Figure CP9.7a generates
the plot illustrating the movement of the closed-loop system roots as the
time delay is varied. The plot is shown in Figure CP9.7d. Examining
the root locations, we find that when T= 1.9, the closed-loop roots
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Computer Problems 515
0
0.2
0.4
0.6
0.8
1
1.2
0 5 10 15 20 25 30 35 40
Time (secs)
Amplitude
FIGURE CP9.7
CONTINUED: (b) Step response without time delays meets specs.
-40
-20
0
20
10-1 100101
Frequency (rad/sec)
Gain dB
-300
-200
-100
0
10-1 100101
Frequency (rad/sec)
Phase deg
FIGURE CP9.7
CONTINUED: (c) System Bode plot shows P.M. = 67.7o.
are s1=4.56, s2,3=0.94 ±2.02j,s4=0.19, and s5,6=±0.32j.
Therefore, the system is marginally stable when T= 1.9, and is unstable
as the time delay increases.
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516 CHAPTER 9 Stability in the Frequency Domain
-4
-3
-2
-1
0
1
2
3
4
-8 -7 -6 -5 -4 -3 -2 -1 0 1
*
*
*
*
*
**
*
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*
**
*
*
*
*
*
Real Axis
Imag Axis
FIGURE CP9.7
CONTINUED: (d) Closed-loop root locations as the time delay varies.
CP9.8 The Nyquist plot and associated m-file code are shown in Figure CP9.8.
a=[0 1;-1 -15]; b=[0;30]; c=[8 0]; d=[0];
sys=ss(a,b,c,d);
nyquist(sys)
−50 0 50 100 150 200 250
−150
−100
−50
0
50
100
150
Nyquist Diagram
Real Axis
Imaginary Axis
FIGURE CP9.8
Using the Nyquist function to obtain a Nyquist plot.
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from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying,
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Computer Problems 517
CP9.9 The Nichols chart is shown in Figure CP9.9. The phase and gain margins
are 37.1 degrees and dB, respectively.
-360 -315 -270 -225 -180 -135 -90 -45 0
-60
-40
-20
0
20
40
60
?-40 dB
?-20 dB
?-60 dB
?-1 dB
?-6 dB
0.5 dB
0 dB
6 dB ?-3 dB
3 dB
?-12 dB
1 dB
0.25 dB
Nichols Chart
O
p
en-Loop Phase
(
de
g)
Open-Loop Gain (dB)
a=[0 1;-1 -10]; b=[0;22]; c=[10 0]; d=[0];
sys=ss(a,b,c,d);
nichols(sys)
ngrid
FIGURE CP9.9
The Nichols chart for the system in CP9.8.
CP9.10 (a) The Nyquist plot is shown in Figure CP9.10. The phase margin is
P.M. = 18o.
(b) When the time delay is T= 0.05 seconds, the phase margin is P.M. =
9o.
(c) When the time delay is T= 0.1 seconds, the system is marginally
stable. So, for T > 0.1 seconds, the system is unstable.
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518 CHAPTER 9 Stability in the Frequency Domain
-10 -8 -6 -4 -2 0
-200
-150
-100
-50
0
50
100
150
200
Nyquist Diagram
Real Axi s
Imaginary Axi s
-1 point
FIGURE CP9.10
Nyquist plot for G(s)H(s) = 10
s(s+1) .
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C H A P T E R 1 0
The Design of Feedback Control
Systems
Exercises
E10.1 From the design specifications, we determine that our desired ζ= 0.69
and ωn= 5.79. The characteristic equation is
1 + Gc(s)G(s) = 1 + K(s+a)
s(s+ 2) = 0 ,
or
s2+ (2 + K)s+Ka = 0 .
Our desired characteristic polynomial is
s2+ 2ζωns+ω2
n=s2+ 8s+ 33.6 = 0 .
Thus, K+ 2 = 8, or
K= 6
and Ka = 33.6, so a= 5.6. The actual percent overshoot and settling
time will be different from the predicted values due to the presence of the
closed-loop system zero at s=a. In fact, the actual percent overshoot
and settling time are P.O. = 12.6% and Ts= 0.87s, respectively.
E10.2 The characteristic equation is
1 + Gc(s)G(s) = 1 + 400
s(s+ 40) K1+1
s= 1 + 400(K1s+ 1)
s2(s+ 40) = 0 ,
or
1 + K1
400s
s3+ 40s2+ 400 = 0 .
519
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520 CHAPTER 10 The Design of Feedback Control Systems
We desire ζ= 0.45 for an overshoot of 20%. The root locus is shown
in Figure E10.2. We select a point slightly inside the performance region
(defined by ζ= 0.45 ) to account for the zero. Thus,
K1= 0.5
and the closed-loop poles are
s1=35
s2,3=2.7±j2.
The actual
P.O. = 20.7% .
-50
-40
-30
-20
-10
0
10
20
30
40
50
-50 -40 -30 -20 -10 0
x
x
x
o
Real Axis
Imag Axis
**
*
FIGURE E10.2
Root locus for 1 + K1400s
s3+40s2+400 = 0.
E10.3 The step response is shown in Figure E10.3 for τ= 1 and K= 0.5. It can
be seen that the
P.O. = 4% ,
so this is a valid solution.
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Exercises 521
-0.2
0
0.2
0.4
0.6
0.8
1
1.2
01 2 3 4 5 6 7 8
Time (secs)
Amplitude
FIGURE E10.3
Step response for K= 0.5and τ= 1.
E10.4 The Bode plot is shown in Figure E10.4. The phase and gain margins
are marked on the plot, where it can be seen that P.M. = 75.4oand
G.M. = 28.6 dB.
-150
-100
-50
0
50
100
150
Magnitude (dB)
10-4 10-2 100102
-270
-225
-180
-135
-90
-45
Phase (deg)
Bode Diagram
Gm = 28.6 dB (at 11.8 rad/sec) , Pm = 75.4 deg (at 0.247 rad/sec)
Frequency (rad/sec)
FIGURE E10.4
Bode plot for Gc(s)G(s) = 100(s+0.15)(s+0.7)
s(s+5)(s+10)(s+0.015)(s+7) .
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522 CHAPTER 10 The Design of Feedback Control Systems
E10.5 We require that Kv2.7, ζ= 0.5 and ωn= 3 for the dominant roots.
We want to place a zero to left of the pole at -2, so the complex roots will
dominate. Set the zero at s=2.2. Then for the desired roots find the
location of pole pin compensator
Gc(s) = K1(s+ 2.2)
(s+p)
to satisfy 180ophase at the desired roots. This yields p= 16.4. Using
root locus methods, we find that KK1= 165.7, so with K1= 7.53, we
determine that K= 22, and
Gc(s) = 7.46(s+ 2.2)
(s+ 16.4) .
Then
Kv= 2.78 .
E10.6 The closed-loop transfer function is
T(s) = Gc(s)G(s)
1 + Gc(s)G(s)=326(s+ 4)
s4+ 14.76s3+ 151.3s2+ 349.8s+ 1304 .
The roots are
s1,2=0.87 ±j3.2
s3,4=6.5±j8.7.
Assuming s1,2dominates, then we expect overshoot
P.O. = 43% and Ts= 4.6 sec .
The discrepencies with the actual P.O. and Tsare due to the poles s3,4
and the zero at s=4.
E10.7 The loop transfer function is
L(s) = Ke0.6s
s(s+ 20) .
A plot of P.M. as a function of Kis shown in Figure E10.7. It can be
seen that
P.M. = 40o
when K= 26.93.
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Exercises 523
0 5 10 15 20 25 30
0
10
20
30
40
50
60
70
80
90
K
Phase Margin deg
phase margin versus K (PM=40º, K=26.93)
FIGURE E10.7
Plot of phase margin versus K.
E10.8 The open-loop transfer function is
G(s) = 2257
s(0.0028s+ 1) =806071.4
s(s+ 357.14) ,
and the compensator is
Gc(s) = K1(s+z)
s,
where z=K2/K1. The characteristic equation is
s3+ 357.14s2+K1s+K2= 0 .
Using Routh-Hurwitz methods, the system is stable for
0< K2<357.14 K1
or K2/K1<357.14. Select the zero zat s=10, then using root locus
methods we determine that K1= 0.08 and K2= 0.8. The roots of the
characteristic equation are
s1=10.6 and s2,3=175 ±j175 ,
and ζ= 0.707, as desired. The step response is shown in Figure E10.8.
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524 CHAPTER 10 The Design of Feedback Control Systems
0
0.2
0.4
0.6
0.8
1
1.2
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5
Time (secs)
Amplitude
FIGURE E10.8
Step response with K1= 0.08 and K2= 0.8.
E10.9 The loop transfer function is
L(s) = Gc(s)G(s) = K1(s+K2/K1)
s(s+ 1) ,
and
Kv= lim
s0sGc(s)G(s) = K2.
Select K2= 5. The characteristic equation is
s2+ (K1+ 1) + K2= 0 ,
and we want
s2+ 2ζωns+ω2
n= 0 .
Equating coefficients yields ωn=K2=5. Also, since we want P.O. =
5%, we require ζ= 0.69. Thus,
2ζωn=K1+ 1 implies K1= 2.08 .
The step response with K1= 2.08 and K2=5 yields a P.O. > 5%. This
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Exercises 525
is due to the zero at
s=1.08 .
So, we raise the gain K1= 3 and then the P.O. = 5%. The step response
is shown in Figure E10.9.
0
0.2
0.4
0.6
0.8
1
1.2
00.5 1 1.5 2 2.5 3
Time (secs)
Amplitude
FIGURE E10.9
Step response with K1= 3 and K2= 5.
E10.10 The loop transfer function is
L(s) = Gc(s)G(s) = (KPs+KI)
s(s+ 1)(s+ 2) .
Let KI= 2. Then, the plot of the phase margin as a function of KPis
shown in Figure E10.10, where it can be seen that
P.M. = 71.6o
is the maximum achievable phase margin. When KP= 1.54 and KI= 2
we have P.M. = 60o, as desired, and P.O. = 9% and Tp= 3.4 sec.
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526 CHAPTER 10 The Design of Feedback Control Systems
12345678910
50
55
60
65
70
75
KP
Phase Margin (deg)
FIGURE E10.10
Phase margin versus KPwith KI= 2.
E10.11 The Nichols diagram and the closed-loop Bode plot are shown in Fig-
ures E10.11a and E10.11b, respectively.
-40
-30
-20
-10
0
10
20
30
40
-350 -300 -250 -200 -150 -100 -50 0
-40
-20
-12
-6
-3
-1
0
0.25
0.5
1
2.3
Phase (deg)
Gain dB
FIGURE E10.11
(a) Nichols diagram for Gc(s)G(s) = 1350(1+0.25s)
s(s+2)(s+30)(1+0.025s).
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Exercises 527
−100
−80
−60
−40
−20
0
20
Magnitude (dB)
100101102103
−270
−225
−180
−135
−90
−45
0
Phase (deg)
Bode Diagram
Frequency (rad/sec)
FIGURE E10.11
CONTINUED: (b) Closed-loop Bode plot.
E10.12 The loop transfer function is
L(s) = Gc(s)G(s) = KK1s+1
2
s2(s+ 5) .
When KK1= 5.12, the roots are
s1,2=0.58 ±j0.58
s3=3.84 .
The complex poles have ζ= 0.707 and the predicted settling time is
Ts= 4/0.58 = 6.89 sec .
The actual settling time is Ts= 6.22 s.
E10.13 For the cascade compensator, we have
T1(s) = Gc(s)G(s)
1 + Gc(s)G(s)=8.1(s+ 1)
(s+r1)(s+ ˆr1)(s+r2),
where r1=1 + j2 and r2=1.67. For the feedback compensator, we
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528 CHAPTER 10 The Design of Feedback Control Systems
have
T2(s) = G(s)
1 + Gc(s)G(s)=8.1(s+ 3.6)
(s+r1)(s+ ˆr1)(s+r2),
where
G(s) = 8.1
s2
and
Gc(s) = s+ 1
s+ 3.6.
The response of the two systems differ due to different value of the zero of
T1and T2, however, both systems have the same characteristic equation.
E10.14 The Bode plot (with the lag network) is shown in Figure E10.14; the
phase margin is P.M. = 46o.
−150
−100
−50
0
50
100
Magnitude (dB)
10−4 10−3 10−2 10−1 100101102
−270
−225
−180
−135
−90
Phase (deg)
Bode Diagram
Gm = 21.9 dB (at 1.84 rad/sec) , Pm = 46.4 deg (at 0.344 rad/sec)
Frequency (rad/sec)
FIGURE E10.14
Bode plot for Gc(s)G(s) = 5(7.5s+1)
s(s+1)(0.25s+1)(110s+1) = 0.
E10.15 At the desired crossover frequency ωc= 10 rad/sec, we have
20 log |Gc(j10)G(j10)|=8.1 dB and 6Gc(j10)G(j10) = 169o.
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Exercises 529
Therefore, the phase margin is P.M. = 11o. So,
φ= 30o11o= 19oand M= 8.1 dB .
Since φ > 0 and M > 0, a lead compensator is required.
E10.16 At the desired crossover frequency ωc= 2 rad/sec, we have
20 log |Gc(j2)G(j2)|= 17 dB and 6Gc(j2)G(j2) = 134o.
Therefore, the phase margin is P.M. = 46o. So,
φ= 30o46o=16o
M=17 dB .
Since φ < 0 and M < 0, a lag compensator is required.
E10.17 Using a prefilter
Gp(s) = KI
KPs+KI
the closed-loop transfer function is
T(s) = KI
s2+ (KP+ 1)s+KI
.
The required coefficients for a deadbeat system are α= 1.82 and Ts=
4.82. Therefore,
KI=ω2
n
KP=αωn1.
Since we desired a settling time less than 2 seconds, we determine that
ωn=Ts/2 = 4.82/2 = 2.41 .
Then, the gains are
KP= 3.39
KI= 5.81 .
The step response (with the prefilter) is shown in Figure E10.17. The
percent overshoot is P.O. = 0.098% and the settling time is Ts= 1.99
seconds.
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530 CHAPTER 10 The Design of Feedback Control Systems
0
0.2
0.4
0.6
0.8
1
1.2
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
Time (secs)
Amplitude
FIGURE E10.17
Step response for the deadbeat system.
E10.18 Consider the PI controller
Gc(s) = Kp+KI
s=Kps+KI
s=30s+ 300
s
and the prefilter
Gp(s) = 10 .
Then, the closed-loop system is
T(s) = 300s+ 3000
s2+ 280s+ 3000 .
The percent overshoot is P.O. = 9.2% and the settling time Ts= 0.16
seconds. The steady-state tracking error to a unit step is zero, as desired.
E10.19 Consider the PID controller
Gc(s) = 29s2+ 10s+ 100
s.
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Exercises 531
The closed-loop transfer function is
T(s) = 29(s2+ 10s+ 100)
s3+ 24s2+ 290s+ 2900 .
The settling time to a unit step is Ts= 0.94 seconds.
E10.20 Consider the PD controller
Gc(s) = KDs+Kp= 3s+ 1 .
The loop transfer function is
L(s) = Gc(s)G(s) = 3s+ 1
s(s2) .
The Bode plot is shown in Figure E10.20. The phase margin is P.M. =
40.4. This is a situation where decreasing the gain leads to instability. The
Bode plot shows a negative gain margin indicating that the system gain
can be decreased up to -3.5 dB before the closed-loop becomes unstable.
−40
−20
0
20
40
Magnitude (dB)
10−2 10−1 100101102
−270
−225
−180
−135
−90
Phase (deg)
Bode Diagram
Gm = −3.52 dB (at 0.816 rad/sec) , Pm = 40.4 deg (at 2.28 rad/sec)
Frequency (rad/sec)
FIGURE E10.20
Bode plot for the loop transfer function L(s) = 3s+1
s(s2) .
E10.21 The transfer function from Td(s) to Y(s) is
T(s) = 1
s2+ 4.4s+K.
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532 CHAPTER 10 The Design of Feedback Control Systems
The tracking error is E(s) = R(s)Y(s). When R(s) = 0, then E(s) =
Y(s). The final value of the output to a unit step disturbance is ess =
1/K. If we want the tracking error to be less than 0.1, then we require
K > 10. When K= 10, we have the disturbance response shown in
Figure E10.21.
0 0.5 1 1.5 2 2.5 3
0
0.02
0.04
0.06
0.08
0.1
0.12
Step Response
Time (sec)
Amplitude
FIGURE E10.21
Disturbance response for K= 10.
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Problems 533
Problems
P10.1 (a) The loop transfer function is
L(s) = Gc(s)G(s)H(s) = (1 + ατs)K1K2
α(1 + τs)(Js2).
We desire ζ= 0.6, Ts2.5 or ζωn1.6. The uncompensated closed-
loop system is
T(s) = K
s2+K,
where K=K1K2/J and K=ω2
n. We can select K= 20, and then
ζωn>1.6. First, plot the Bode diagram for
G(s)H(s) = 20
s2
where K1K2/αJ = 20. The phase margin of the uncompensated sys-
tem is 0o. We need to add phase at ωc. After several iterations, we
choose to add 40ophase at ωc, so
sin 40o=α1
α+ 1 = 0.64 .
Therefore, α= 4.6. Then,
10 log α= 10 log 4.6 = 6.63dB .
We determine the frequency where magnitude is -6.63 dB to be ωm=
6.6 rad/sec. Then,
p=ωnα= 14.1 and z=p/α = 3.07 .
The compensated loop transfer function (see Figure P10.1a) is
Gc(s)G(s)H(s) = 20 s
3.07 + 1
s2s
14.1+ 1.
(b) Since we desire ζωn1.6, we place the compensator zero at z= 1.6.
Then, we place the compensator pole far in the left half-plane; in this
case, we selected p= 20. Thus, the compensator is
Gc(s) = s+ 1.6
s+ 20 .
The root locus is shown in Figure P10.1b. To satisfy the ζ= 0.6
requirement, we find K= 250, and the compensated loop transfer
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534 CHAPTER 10 The Design of Feedback Control Systems
-50
0
50
100
10-1 100101102
Frequency (rad/sec)
Gain dB
-180
-170
-160
-150
-140
10-1 100101102
Frequency (rad/sec)
Phase deg
FIGURE P10.1
(a) Compensated Bode plot for Gc(s)G(s)H(s) = 20(s/3.07+1)
s2(s/14.1+1) .
function is
Gc(s)G(s)H(s) = 250(s+ 1.6)
s2(s+ 20) =20 s
1.6+ 1
s2s
20 + 1.
-20
-15
-10
-5
0
5
10
15
20
-25 -20 -15 -10 -5 0 5
xxx o
Real Axis
Imag Axis
*
*
*
FIGURE P10.1
CONTINUED: (b) Root locus for Gc(s)G(s)H(s) = 1 + Ks+1.6
s2(s+20) .
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Problems 535
P10.2 The transfer function of the system is
G(s) = 1.0e+ 14
s3+ 2000s2+ 1e+ 11s,
where we use the system parameters given in P7.11 with the following
modifications: τ1=τ1= 0 and K1= 1. Also we have scaled the transfer
function so that the time units are seconds. The parameters in P7.11 are
given for time in milliseconds. A suitable compensator is
Gc(s) = s+ 500
s+ 1 .
The closed-loop system response is shown in Figure P10.2. The percent
overshoot is P.O. 20% and the time to settle is Ts<0.01 second.
0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05
0
0.2
0.4
0.6
0.8
1
1.2
1.4
Time (secs)
Amplitude
FIGURE P10.2
Step response.
P10.3 The loop transfer function is
Gc(s)G(s) = 16(s+ 1)
s(s2+ 2s+ 16)
K(s+z)
(s+p).
We desire dominant roots with Ts<5 sec and P.O. < 5%, so use ζ= 0.69
and ζωn= 0.8. One solution is to select z= 1.1 (i.e. to the left of the
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536 CHAPTER 10 The Design of Feedback Control Systems
existing zero at s=1) and determine the pole pand gain Kfor dominant
roots with ζ= 0.69. After iteration, we can select p= 100, so that the
root locus has the form shown in Figure P10.3. Then, we select K= 320,
-200 -150 -100 -50 0 50 100 150 200
-200
-150
-100
-50
0
50
100
150
200
Real Axis
Imag Axis
FIGURE P10.3
Root locus for 1 + K16(s+1)(s+1.1)
s(s2+2s+16)(s+100) = 0.
so that ζ= 0.69. The final compensator is
Gc(s) = 320(s+ 1.1)
s+ 100 .
The design specifications are satisfied with this compensator.
P10.4 The uncompensated loop transfer function is
G(s) = 1
s2(1
40 s+ 1) =40
s2(s+ 40) .
We desire 10% < P.O. < 20%, so 0.58 < ζ < 0.65, and Ts<2 implies
ζωn<2. We will utilize a PD compensator Ka(s+a). We select a= 2,
to obtain the root locus shown in Figure P10.4. Then with Ka= 23.5, we
have the desired root location, and
Gc(s) = 23.5(s+ 2) .
The design specifications are satisfied with the PD compensator.
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Problems 537
-30
-20
-10
0
10
20
30
-50 -40 -30 -20 -10 0
xxx o
Real Axis
Imag Axis
+
+
+
*
*
*
FIGURE P10.4
Root locus for 1 + Ka40(s+2)
s2(s+40) = 0.
P10.5 We desire P.O. < 10% and Ts<1.5 sec. The compensator is a PI-type,
given by
Gc(s) = K2+K3
s=K2s+K3
s=K2(s+a)
s
where a=K3/K2. So, ess = 0 for a step input and
G(s) = 3.75Ka
(s+ 0.15)(0.15s+ 1) =25Ka
(s+ 0.15)(s+ 6.67) .
The loop transfer function is
Gc(s)G(s) = 25KaK2(s+a)
s(s+ 0.15)(s+ 6.67) .
Using root locus methods, we select a= 0.2 (after several iterations) and
determine KaK2to yield ζ= 0.65. This results in KaK2= 1. The root
locus is shown in Figure P10.5. The design specifications are met. The
actual percent overshoot and settling time are P.O. = 7.4% and Ts= 1.3
s. The controller is
Gc(s) = 1 + 0.2
s.
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538 CHAPTER 10 The Design of Feedback Control Systems
-20
-15
-10
-5
0
5
10
15
20
-20 -15 -10 -5 0 5 10 15 20
xxx o
Real Axis
Imag Axis
+
+
+
*
*
*
FIGURE P10.5
Root locus for 1 + KaK225(s+0.2)
s(s+0.15)(s+6.67) = 0.
P10.6 As in P10.5, using root locus we find that placing z= 15 and p= 30 yields
a root locus shape (see Figure P10.6) where the loop transfer function is
-60
-40
-20
0
20
40
60
-60 -40 -20 0 20 40 60
x x xo
Real Axis
Imag Axis
+
+
+
*
*
*
FIGURE P10.6
Root locus for 1 + Ka25(s+15)
(s+0.15)(s+6.67)(s+30) = 0.
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Problems 539
Gc(s)G(s) = 25Ka(s+z)
(s+p)(s+ 0.15)(s+ 6.67) .
and where z, p and Kaare the parameters to be determined. Properly
choosing the parameter values allows us to increase ζωnof the dominant
roots (compared to the PI compensator of P10.5). Then, with Ka= 3.7,
the dominant roots have ζ= 0.65. The design specifications are met with
the compensator.
P10.7 The plant transfer function is
G(s) = e50s
(40s+ 1)2.
The steady-state error is
ess =A
1 + Kp
<0.1A .
Therefore, Kp>9. Insert an amplifier with the compensator with a dc
gain = 9, as follows
Gc(s)G(s) = 9e50s(s+ 2)
(40s+ 1)2(s+p).
The system is unstable without compensation, and it is very difficult to
compensate such a time delay system with a lead compensator. Consider
a lag network
Gc(s) = s+z
s+p
where z > p. Let
z= 10p.
Then, a plot of the P.M. versus pis shown in Figure P10.8a. Suitable
system performance can be obtained with P.M. > 45o, so choose
p= 0.0001.
The Bode plot of the compensated and uncompensated systems is shown
in Figure P10.7c, where we have selected z= 0.001 and p= 0.0001. The
compensated system has
P.M. = 62oand Ts= 9 minutes .
The step response is shown in Figure P10.7b.
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540 CHAPTER 10 The Design of Feedback Control Systems
0
0.5
1
1.5
0100 200 300 400 500 600 700 800
Time (secs)
Amplitude
(b)
-50
0
50
100
150
00.5 1 1.5 2 2.5 3
x10-3
p
Phase Margin (deg)
(a)
FIGURE P10.7
(a) Phase margin versus p. (b) Step response with p= 0.0001 and z= 0.001.
-30
-20
-10
0
10
20
10-4 10-3 10-2 10-1
Frequency (rad/sec)
Gain dB
-500
-400
-300
-200
-100
0
10-4 10-3 10-2 10-1
Frequency (rad/sec)
Phase deg
FIGURE P10.7
CONTINUED: (c) Bode plot for the compensated system (solid lines) and the uncompen-
sated system (dashed line).
P10.8 The transfer function is
G(s) = 5000
s(s+ 10)2.
To meet the steady-state accuracy, we need Kv>40. The uncompensated
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Problems 541
Kv= 50, so the steady-state accuracy can be met.
(a) Using the Bode method, we need P.M. = 70% (to meet P.O. < 5%
specification). Let
Gc(s) = bs + 1
as + 1 .
The plot of P.M. versus bis shown in Figure P10.8a, where we set
a= 50b. Choosing b= 20 should satisfy the P.O. specification. The
Bode plot is shown in Figure P10.8c. Thus,
0
0.5
1
1.5
05 10 15 20 25 30 35 40 45 50
Time (secs)
Amplitude
(b)
30
40
50
60
70
80
0 5 10 15 20 25 30
b
Phase Margin (deg)
(a)
FIGURE P10.8
(a) Phase margin versus b; (b) Step response for lag compensator designed with Bode where
a= 1000 and b= 20.
Gc(s)G(s) = 5000(20s+ 1)
s(s+ 10)2(1000s+ 1) .
The step response is shown in Figure P10.8b.
(b) We require that ζ= 0.7 to meet the P.O. specifications. Let
Gc(s) = K(bs + 1)
(as + 1) .
Using root locus methods, we fix aand b, and then determine Kfor
ζ= 0.7. Let a= 50band select b= 10 (other values will work). The
root locus is shown in Figure P10.8d. We find K= 2.5 when ζ= 0.7.
Now, Kv= 125, so the steady-state accuracy requirement is satisfied
for the step response as shown in Figure P10.8e.
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542 CHAPTER 10 The Design of Feedback Control Systems
-100
-50
0
50
100
150
10-4 10-3 10-2 10-1 100101102
Frequency (rad/sec)
Gain dB
-300
-200
-100
0
10-4 10-3 10-2 10-1 100101102
Frequency (rad/sec)
Phase deg
FIGURE P10.8
CONTINUED: (c) Bode plot for the compensated system with Gc(s) = 20s+1
1000s+1 .
-20
-15
-10
-5
0
5
10
15
20
-20 -15 -10 -5 0 5 10 15 20
xxx xo
Real Axis
Imag Axis
*
*
*
*
FIGURE P10.8
CONTINUED: (d) Root locus for 1 + K5000(10s+1)
s(s+10)2(500s+1) .
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Problems 543
0
0.2
0.4
0.6
0.8
1
1.2
0 2 4 6 8 10 12 14 16 18 20
Time (secs)
Amplitude
FIGURE P10.8
CONTINUED: (e) Step response for lag compensator designed with root locus methods,
where K= 2.5.
P10.9 We desire a small response for a disturbance at 6 rad/sec. The Bode plot
of Gc(s)G(s) is shown in Figure P10.9a where we consider a compensator
-150
-100
-50
0
10-1 100101102103
Frequency (rad/sec)
Gain dB
-300
-200
-100
0
10-1 100101102103
Frequency (rad/sec)
Phase deg
FIGURE P10.9
(a) Bode plot for the compensated system with Gc(s) = 10(s2+4s+10)
s2+36 .
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544 CHAPTER 10 The Design of Feedback Control Systems
of the form
Gc(s) = K(s2+as +b)
s2+ 36 .
Notice that the magnitude is large at ω= 6, as desired. We select
a= 4, b = 10 and K= 10 .
The response to a sinusoidal disturbance at 6 rad/sec is shown in Fig-
ure P10.9b. Notice that the effect of the disturbance is virtually eliminated
in steady-state.
-0.02
-0.015
-0.01
-0.005
0
0.005
0.01
0.015
0.02
0 10 20 30 40 50 60 70 80 90 100
Time (secs)
Amplitude
FIGURE P10.9
CONTINUED: (b) Disturbance response for a sinusoidal disturbance at 6 rad/sec.
P10.10 The step response with Gc(s) = 1 is shown in Figure P10.10. A suitable
lag compensator is
Gc(s) = s+ 0.05
s+ 0.005 .
The step response of the compensated system is also shown in Figure P10.10.
The settling time of the compensated system is
Ts= 28 seconds .
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Problems 545
0
5
10
15
20
25
30
0 5 10 15 20 25 30
Time (sec)
Amplitude
Input (dotted line)
Compensated system (solid) & Uncompensated system (dashed)
FIGURE P10.10
Step response of uncompensated and compensated systems.
P10.11 The root locus is shown in Figure P10.11 where a suitable lead-lag com-
-300
-200
-100
0
100
200
300
-300 -200 -100 0 100 200 300
xxxx oo
Real Axis
Imag Axis
+
+
+
+
FIGURE P10.11
Root locus for 1 + K160(s+17)(s+10)
s2(s+170)(s+1) = 0.
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546 CHAPTER 10 The Design of Feedback Control Systems
pensator is
Gc(s) = Ks+ 10
s+ 1
s+ 17
s+ 170 .
The selected gain is K= 57, so that the damping of the complex roots
is about ζ= 0.7. For this particular design, the closed-loop system zeros
will affect the system response and the percent overshoot specification
may not be satisfied. Some design iteration may be necessary or aprefilter
can be utilized. A suitable prefilter is
Gp(s) = 17
s+ 17 .
The acceleration constant is Ka= 9120.
P10.12 We choose K= 10. This yields a velocity constant Kv= 20K= 200, as
desired. A suitable two-stage lead compensaator is
Gc(s) = (0.05s+ 1)(0.05s+ 1)
(0.0008s+ 1)(0.0008s+ 1) .
The Bode plot is shown in Figure P10.12. The phase margin is P.M. =
75.06o.
-100
-50
0
50
100
10-1 100101102103104
Frequency (rad/sec)
Gain dB
-300
-200
-100
0
10-1 100101102103104
Frequency (rad/sec)
Phase deg
Phase margin=75.06 deg
FIGURE P10.12
Bode plot for 200(0.05s+1)2
s(0.1s+1)(0.05s+1)(0.0008s+1)2.
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Problems 547
P10.13 (a) When
Gc(s) = K= 0.288 ,
the phase margin is P.M. = 49.3oand the bandwidth is ωB= 0.95
rad/sec.
(b) A suitable lag compensator is
Gc(s) = 25s+ 1
113.6s+ 1 .
The compensated system phase margin is P.M. = 52.21oand Kv= 2,
as desired.
P10.14 A suitable lead compensator is
Gc(s) = 1.155s+ 1
0.032s+ 1 .
The compensated system phase margin is P.M. = 50oand Kv= 2, as
desired. The settling time is Ts= 3.82 seconds.
P10.15 One possible solution is
Gc(s) = K(s+ 12)(s+ 15)
(s+ 120)(s+ 150) ,
where K= 900. The disturbance response is shown in Figure P10.15.
0 0.1 0.2 0.3 0.4 0.5 0.6
0
0.02
0.04
0.06
0.08
0.1
Step Response
Time
(
sec
)
Amplitude
FIGURE P10.15
Compensated system disturbance response.
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548 CHAPTER 10 The Design of Feedback Control Systems
P10.16 The PI controller is given by
Gc(s) = K(s+b)
s,
where Kand bare to be determined. To meet the design specifications,
we need
ζ= 0.6 and ωn= 6.67 rad/sec .
The closed-loop transfer function is
T(s) = K(s+b)
s2+Ks +bK .
Solving for the gains yields K= 2ζωn= 8 and b=ω2
n/K = 5.55. A
suitable prefilter is
Gp(s) = 5.55
s+ 5.55 .
The step response, with and without the prefilter, is shown in Figure P10.16.
0
0.2
0.4
0.6
0.8
1
1.2
1.4
0 0.5 1 1.5 2 2.5 3
Without prelter (solid) & with prelter (dashed)
Time (sec)
Amplitude
FIGURE P10.16
Compensated system response with and without a prefilter.
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Problems 549
P10.17 The plant transfer function is
G(s) = K
s(s+ 10)(s+ 50) .
We desire ζωn>10 to meet Ts<0.4 sec and ζ= 0.65 to meet P.O. <
7.5%. Try a pole at s=120. The root locus is shown in Figure P10.17.
The gain K= 6000 for ζ= 0.65. Thus,
Gc(s)G(s) = 6000(s/15 + 1)
s(s+ 10)(s+ 50)(s/120 + 1) and Kv=6000
500 = 12 .
-200
-150
-100
-50
0
50
100
150
200
-200 -150 -100 -50 0 50 100 150 200
xxx xo
Real Axis
Imag Axis
* * *
*
FIGURE P10.17
Root locus for 1 + Ks/15+1
s(s+10)(s+50)(s/120+1) .
P10.18 (a) The loop transfer function is
L(s) = K1e2T s
0.25s+ 1
where T= 1.28. The phase angle is
φ=2.56ωtan 0.25ω .
So, ω= 1.12 rad/sec when φ=180o. However, the break frequency
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550 CHAPTER 10 The Design of Feedback Control Systems
is 4 rad/sec. Therefore, you cannot achieve P.M. = 30oand have the
system be stable for K1<1. The steady-state error is
ess =A
1 + Kp
=A
1 + K1
since K1=Kp.
(b) Set K1= 20, then Kp= 20 and this yields a 5% steady-state error.
Without compensation, the system is now unstable. Let
Gc(s) = s/b + 1
s/a + 1
where b= 5 and a= 0.01. Then, the system is stable with
P.M. = 63o.
The system response is shown in Figure P10.18.
-0.2
0
0.2
0.4
0.6
0.8
1
1.2
02 4 6 8 10 12 14 16 18 20
Time (secs)
Amplitude
FIGURE P10.18
Unit step response with Gc(s) = 20(s/5+1)
s/0.01+1 .
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Problems 551
P10.19 (a) The open-loop transfer function is
G(s) = KesT
(s+ 1)(s+ 3) ,
where T= 0.5 sec. We desire P.O. < 30%, thus ζ > 0.36. We will
design for ζ= 0.4, which implies P.M. = 40o. Then
φ=tan1ωtan1ω
30.5ω(57.3o).
At ωc= 1.75, the phase margin is P.M. = 40o, and solving
|G(jω)|=K
[(3 ω2)2+ (4ω)2]1
2
= 1
at ω= 1.75 yields K= 7. Then ess = 0.3.
(b) We want ess <0.12, so use ess = 0.10 as the goal. Then
Gc(s)G(s) = Ke0.5s(s+ 2)
(s+ 1)(s+ 3)(s+b),
and
ess =1
1 + Kp
where Kp=2K
3b. If b= 0.1 then Kp= 6.7Kand
ess =1
1 + 6.7K.
So, we need 6.7K= 9, or K= 1.35. We need a lag compensator (i.e.
b < 2) to meet ess <12% and have stability.
P10.20 We desire Kv= 20, P.M. = 45oand ωB>4 rad/sec. Thus, we set K= 20,
and
G(s) = 20
ss
2+ 1s
6+ 1.
Then, the Bode plot yields P.M. =21ouncompensated at ωc= 5.2
rad/sec. The phase lead compensator must add 66 oplus phase lead to
account for the shift of the crossover to a higher frequency with the phase
lead compensator. Consider
Gc(s) = 1 + ατs
1 + τs 2
.
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552 CHAPTER 10 The Design of Feedback Control Systems
One solution is to use
α= 10
τ= 1/67 .
Then
Gc(s) = 100(s+ 6.7)2
(s+ 67)2.
The compensator has two zeros at ω= 6.7, two poles at ω= 67 yielding
P.M. = 47o,ωc= 7.3 and ωB= 12 rad/sec.
P10.21 We desire Kv= 20, P.M. = 45oand ωB2. The lag compensator is
Gc(jω) = 1 + jωτ
1 + jωατ
where α > 1. From the Bode plot, φ=135oat ω
=1.3. So, at ω= 1.3,
we need to lower the magnitude by 22 dB to cause ω= 1.3 to be ω
c, the
new crossover frequency. Thus, solving
22 = 20 log α
yields α= 14. We select the zero one decade below ω
cor 1
τ= 0.13.
Therefore,
1
ατ =0.13
14 = 0.0093 .
Then, the lag compensator is given by
Gc(s) = 1 + s
0.13
1 + s
0.0093
=s+ 0.13
14(s+ 0.0093) .
The new crossover is ω
c= 1.3, and ωB= 2.14 rad/sec.
P10.22 We desire P.M. = 45o,Kv= 20 and 2 ωB10. The lead-lag compen-
sator is
Gc(s) = 1 + s
b
1 + s
10b·1 + s
10a
1 + s
a
.
Since ωB
=1.5ωc, we design for a new crossover frequency ω
cso that
1.4< ω
c<7.
Try for ω
c= 4. The phase φ=190oat ω= 4, so we need to add phase
lead of 55oplus phase to account for lag part of network at ω
c. Use α= 10
and bracket ω= 4 with the lead network. Put the zero at ω= 0.8 = b
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Problems 553
and the pole at ω= 8. For the lag compensator, put the zero at a lower
frequency than ω
c/10. So try a zero at ω= 0.2 = 10aand a pole at
ω= 0.02 = a. Then, the lead-lag compensator is
Gc(s) = 1 + s
0.81 + s
0.2
1 + s
81 + s
0.02 .
The compensated Bode plot yields
ω
c= 3.5 rad/sec, P.M. = 50oand ωB= 6.2 rad/sec .
P10.23 The steady-state error is
ess =1
1 + Kp
=1
1 + K/25 = 0.05 .
So, we need K/25 19 or K475. One possible solution is
Gc(s) = 4s+ 1
12s+ 1 and K= 475 .
The compensated Bode plot is shown in Figure P10.23. The phase margin
is P.M. = 46o.
−40
−20
0
20
40
Magnitude (dB)
10−3 10−2 10−1 100101102
−180
−135
−90
−45
0
Phase (deg)
Bode Diagram
Gm = Inf dB (at Inf rad/sec) , Pm = 46 deg (at 11.5 rad/sec)
Frequency (rad/sec)
FIGURE P10.23
Bode plot for Gc(s)G(s) = 475(4s+1)
(s+5)2(12s+1) .
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554 CHAPTER 10 The Design of Feedback Control Systems
P10.24 The arm-rotating dynamics are represented by
G(s) = 80
ss2
4900 +s
70 + 1.
We desire Kv= 20, and P.O. < 10%. One possible solution is the lead-lag
compensator
Gc(s) = (s+ 50)(s+ 0.48)
4(s+ 400)(s+ 0.06) .
With this compensator, we have
P.O. = 9.5% and Kv= 20 .
P10.25 Neglect the pole of the airgap feedback loop at s= 200. The characteristic
equation is
1 + ¯
K(s+ 20)(s+c)
s3= 0 ,
where
¯
K=K
K1+K2
c=K2b
K1+K2
.
Choose
c= 10
to attain the root locus structure shown in Figure P10.25. The gain
¯
K= 38.87
insures the damping ratio of ζ= 0.5. Then, solving for K1and byields
K1=K
38.87 K2
and
b=0.1K
38.87K2
.
For given values of Kand K2(unspecified in the problem), we can com-
pute K1and b.
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Problems 555
-40
-30
-20
-10
0
10
20
30
40
-40 -30 -20 -10 0 10 20 30 40
xxxo o
Real Axis
Imag Axis
+
+
+
FIGURE P10.25
Root locus for 1 + ¯
K(s+20)(s+10)
s3= 0.
P10.26 The loop transfer function is
Gc(s)G(s) = 0.15K(10as + 1)
s(s+ 1)(5s+ 1)(as + 1) ,
where Kand aare to be selected to meet the design specifications. Suit-
able values are
K= 6.25 and a= 0.15 .
Then, the phase margin is P.M. = 30.79oand the bandwidth is ωB=
0.746 rad/sec. The lead compensator is
Gc(s) = 6.25 1.5s+ 1
0.15s+ 1 .
P10.27 (a) Let Gc(s) = K= 11. Then the phase margin is P.M. = 50oand the
performance summary is shown in Table P10.27.
(b) Let
Gc(s) = K(s+ 12)
(s+ 20) ,
where K= 32. Then, the phase margin is P.M. = 50oand the per-
formance summary is given in Table P10.27.
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556 CHAPTER 10 The Design of Feedback Control Systems
compensator P.M. P.O. TpTsMpωωB
Gc(s) = K= 11 50o18% 0.34 sec 0.78 sec 1.5 dB 13.9 rad/sec
Gc(s) = 32(s+12)
s+20 50o18% 0.20 sec 0.47 sec 1.5 dB 26.3 rad/sec
TABLE P10.27 Performance Summary.
P10.28 The loop transfer function is
Gc(s)G(s) = K(as + 1)
s(s+ 10)(s+ 14)(10as + 1) ,
where Kand aare to be selected to meet the design specifications, and
we have set α= 10. The root locus is shown in Figure P10.28a. To satisfy
-30
-20
-10
0
10
20
30
-30 -20 -10 0 10 20 30
xxx xo
Real Axis
Imag Axis
*
*
*
*
FIGURE P10.28
(a) Root locus for 1 + K1400(s+1)
s(s+10)(s+14)(10s+1) = 0.
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Problems 557
the steady-state tracking error we must select
K > 1400 .
Suitable values for the lag compensator are
K= 4060 and a= 1 .
Then, the percent overshoot is P.O. = 31% and the settling time is Ts=
2.34 sec. The lag compensator is
Gc(s) = s+ 1
10s+ 1 .
The step response is shown in Figure P10.28b.
0
0.2
0.4
0.6
0.8
1
1.2
1.4
00.5 1 1.5 2 2.5 3
Time (secs)
Amplitude
FIGURE P10.28
CONTINUED: (b) Step response.
P10.29 The plant transfer function is
G(s) = 10e0.05s
s2(s+ 10) .
The lead network
Gc(s) = 16(s+ 0.7)
(s+ 9)
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558 CHAPTER 10 The Design of Feedback Control Systems
provides Mpω= 3.4 dB and ωr= 1.39 rad/sec. The step response is shown
in Figure P10.29. The overshoot is P.O. = 37% and Ts= 3.5 sec.
0
0.2
0.4
0.6
0.8
1
1.2
1.4
01 2 3 4 5 6
Time (secs)
Amplitude
FIGURE P10.29
Unit step response with Gc(s) = 16(s+0.7)
s+9 .
P10.30 The vehicle is represented by
G(s) = K
s(0.04s+ 1)(0.001s+ 1) K
s(0.04s+ 1) .
For a ramp input, we want
ess
A= 0.01 = 1
Kv
.
So, let
G(s) = 100
s(0.04s+ 1) .
The uncompensated P.M. = 28oat ωc= 47 rad/sec. We need to add 17o.
Case (1) Phase lead compensation:
Gc(s) = 1 + 0.021s
1 + 0.01s.
The phase margin is P.M. = 45o.
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Problems 559
Case (2) Phase lead compensation:
Gc(s) = 1 + 0.04s
1 + 0.005s.
The phase margin is P.M. = 65o.
For Case 1, we have
P.O. = 25% , Ts= 0.13 sec and Tp= 0.05 sec .
For Case 2, we have
P.O. = 4% , Ts= 0.04 sec and Tp= 0.03 sec .
P10.31 As in P10.30, the plant is given by
G(s) = 100
s(0.04s+ 1) .
The uncompensated P.M. = 28o. We need P.M. = 50o. The phase lag
compensator
Gc(s) = 1 + 0.5s
1 + 2.5s
results in P.M. = 50o. The P.O. = 21%, Ts= 0.72 sec and Tp= 0.17 sec.
P10.32 (a) To obtain Kv= 100, we have
Gc(s)G(s) = 43.33(s+ 500)
s(s+ 0.0325)(s2+ 2.57s+ 6667) .
With K= 43.33, we have
P.M. = 1.2o, Mpω= 26 dB , ωr= 1.8 rad/sec and ωB= 3.7 rad/sec .
The Bode plot is shown in Figure P10.32.
(b) Let
Gc(s) = 0.35s+ 1
0.001s+ 1 ,
and K= 43.33 (as before). Then,
P.M. = 36o, Mpω= 5.4 dB , ωr= 1.7 rad/sec and ωB= 3.0 rad/sec .
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560 CHAPTER 10 The Design of Feedback Control Systems
-150
-100
-50
0
50
100
10-2 10-1 100101102103
Frequency (rad/sec)
Gain dB
-350
-300
-250
-200
-150
-100
10-2 10-1 100101102103
Frequency (rad/sec)
Phase deg
FIGURE P10.32
Bode plot with Gc(s) = K= 43.33.
P10.33 The step response is shown in Figure P10.33, where
Gc(s) = 10(s+ 0.71)(s+ 0.02)
(s+ 0.0017)(s+ 10) .
0
0.2
0.4
0.6
0.8
1
1.2
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
Time (secs)
Amplitude
FIGURE P10.33
Step response with the lead-lag compensator Gc(s) = 10(s+0.71)(s+0.02)
(s+0.0017)(s+10) .
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Problems 561
Then,
Kv= 80
and P.O. = 17%, Ts= 1.8 sec, and ζ= 0.54.
P10.34 The process model is
G(s) = 1
s2(s+ 10) ,
and we consider the lead compensator
Gc(s) = K1 + sατ
1 + ,
where α= 100, τ= 0.4 and K= 0.5. Then, P.M. = 46.4o. The step
response is shown in Figure P10.34. The system performance is
P.O. = 22.7%
Ts= 5.2 sec
Tp= 1.72 sec .
0 1 2 3 4 5 6 7 8 9 10
0
0.2
0.4
0.6
0.8
1
1.2
1.4
Time (secs)
Amplitude
FIGURE P10.34
Step response with the lead compensator Gc(s) = 0.540s+1
0.4s+1 .
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562 CHAPTER 10 The Design of Feedback Control Systems
P10.35 The phase margin is shown in Figure P10.35. As the time delay increases,
the phase margin decreases. The system is unstable when T > 2.1843 s.
0 0.5 1 1.5 2 2.5
−20
0
20
40
60
80
100
120
140
Time delay (s)
Phase margin (deg)
Stability boundary
T=2.1843 s
FIGURE P10.35
Step response with Gc(s)G(s) = 2s+0.54
s(s+1.76) eT s, where 0T2.5.
P10.36 One possible solution is the integral controller Gc(s) = 2/s. The step
response is shown in Figure P10.36. The steady-state tracking error to a
-0.2
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
012345678910
Time (secs)
Amplitude
FIGURE P10.36
Step response with the integral controller Gc(s) = 2/s.
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Problems 563
step input is zero since the system is type-1. The phase margin is P.M. =
32.8and the bandwidth is ωB= 4.3 rad/s .
P10.37 One possible solution is
Gc(s) = 1600(s+ 1)
25s+ 1 .
The overshoot to a unit step is P.O. = 4.75% and the steady-state error
to a step input is ess = 1%. The system bandwidth is ωB= 9.7 rad/sec.
P10.38 The lead compensator is
Gc(s) = 2.88(s+ 2.04)
s+ 5.88 .
The Bode plot is shown in Figure P10.38. The phase margin is P.M. =
30.4oat ωc= 9.95 rad/sec and the bandwidth is ωB= 17.43 rad/sec.
-40
-20
0
20
40
60
10-1 100101102
Frequency (rad/sec)
Gain dB
-300
-200
-100
0
10-1 100101102
Frequency (rad/sec)
Phase deg
FIGURE P10.38
Bode plot for Gc(s)G(s) = 115.29(s+2.04)
s(s+2)(s+5.88) .
P10.39 The lag compensator is
Gc(s) = 1 + 1.48s
1 + 11.08s.
The Bode plot is shown in Figure P10.39. The steady-state error specifi-
cation is satisfied since Kv= 20. The phase margin is P.M. = 28.85oat
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564 CHAPTER 10 The Design of Feedback Control Systems
-50
0
50
100
10-3 10-2 10-1 100101
Frequency (rad/sec)
Gain dB
-300
-200
-100
0
10-3 10-2 10-1 100101
Frequency (rad/sec)
Phase deg
FIGURE P10.39
Bode plot for Gc(s)G(s) = 40(1+1.48s)
s(s+2)(1+11.08s).
ωc= 2 rad/sec and the bandwidth is ωB= 3.57 rad/sec.
P10.40 The lag compensator is
Gc(s) = 2.5(1 + 1.64s)
1 + 30.5s.
The Bode plot is shown in Figure P10.40. The steady-state error specifi-
cation is satisfied since
Kv= 50 .
The phase margin is
P.M. = 28.93o
at ωc= 1.98 rad/sec and the bandwidth is
ωB= 3.59 rad/sec.
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Problems 565
-50
0
50
100
10-3 10-2 10-1 100101
Frequency (rad/sec)
Gain dB
-300
-200
-100
0
10-3 10-2 10-1 100101
Frequency (rad/sec)
Phase deg
FIGURE P10.40
Bode plot for Gc(s)G(s) = 100(1+1.64s)
s(s+2)(1+30.5s).
P10.41 We use Table 10.2 in Dorf & Bishop to determine the required coefficients
α= 1.9 and β= 2.2.
Also,
ωnTr= 4.32 implies ωn= 4.32,
since we require
Tr= 1 second.
The characteristic equation is
s3+ 8.21s2+ 41.06s+ 80.62 = s3+ (1 + p)s2+ (K+p)s+Kz = 0 .
Equating coefficients and solving yields
p= 7.21 K= 33.85 z= 2.38 .
P10.42 From Example 10.4 in Dorf & Bishop, we have the closed-loop transfer
function
T(s) = 96.5(s+ 4)
(s2+ 8s+ 80)(s+ 4.83) .
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566 CHAPTER 10 The Design of Feedback Control Systems
A suitable prefilter is
Gp(s) = 4
s+ 4 .
The step response (with and without the prefilter) is shown in Figure P10.42.
0
0.2
0.4
0.6
0.8
1
1.2
1.4
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
With prelter (solid) & without prelter (dashed)
Time (sec)
Amplitude
FIGURE P10.42
Step response with and without the prefilter.
P10.43 Let
K= 100.
The Bode plot is shown in Figure P10.43a and the response to a simusoidal
noise input with
ω= 100 rad/s
is shown in Figure P10.43b.
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Problems 567
Bode Diagram
Frequency (rad/sec)
10−1 100101102103
−80
−60
−40
−20
0
20
40
60
System: sysg
Frequency (rad/sec): 100
Magnitude (dB): −40.1
Magnitude (dB)
012345678
−0.02
−0.01
0
0.01
0.02
0.03
0.04
0.05
0.06
0.07
Time (sec)
Amplitude
FIGURE P10.43
(a) Bode magnitude plot. (b) Response to a noise input.
P10.44 For 0.129 < K 69.87, the system is unstable. The percent overshoot is
shown in Figure P10.44 .
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568 CHAPTER 10 The Design of Feedback Control Systems
0 0.02 0.04 0.06 0.08 0.1 0.12 0.14
−50
0
50
100
K
Percent Overshoot
65 70 75 80 85 90 95 100
100
120
140
160
K
Percent Overshoot
FIGURE P10.44
Percent overshoot.
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Advanced Problems 569
Advanced Problems
AP10.1 (a) With
Gc(s) = K ,
the closed-loop transfer function is
T(s) = K
s3+ 5s2+ 4s+K.
When K= 2.05, the characteristic equation is
s3+ 5s2+ 4s+ 2.05 = 0
with poles at s=4.1563 and s=0.4219 ±j0.5615. Therefore
ζ= 0.6, and the predicted overshoot is
P.O. = 100eπ0.6/10.62= 9.5% <13% .
The actual overshoot is P.O. = 9.3% and Ts= 8.7 seconds.
(b) When
Gc(s) = 82.3(s+ 1.114)
s+ 11.46
the closed-loop transfer function is
T(s) = 82.3(s+ 1.114)
s4+ 16.46s3+ 61.3s2+ 128.14s+ 91.6822
=82.3(s+ 1.114)
(s+ 1.196)(s+ 12.26)(s+ 1.5±j2) .
Therefore ζ= 0.6 and the predicted overshoot is P.O. = 9.5% <13%.
The actual overshoot is P.O. = 12% and Ts= 2.5 seconds.
AP10.2 The lag network is given by
Gc=K(s+a1)
s+a2
.
The closed-loop transfer function is
T(s) = Ks+a1
s4+ (5 + a2)s3+ (4 + 5a2)s2+ (4a2+K)s+Ka1
.
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570 CHAPTER 10 The Design of Feedback Control Systems
Computing the steady-state tracking error yields
ess = lim
s0
s4+ (5 + a2)s3+ (4 + 5a2)s2+ 4a2s
s5+ (5 + a2)s4+ (4 + 5a2)s3+ (4a2+K)s2+Ka1s
=4a2
a1K<0.125 .
If we select K= 2.05 (as in AP10.1), then
a1>15.61a2.
So, take a2=a1/16. The lag compensator can now be written as
Gc(s) = 2.05 s+a1
s+a1/16 .
Select a1= 0.018. Then, the closed-loop transfer function is
T(s) = 2.05(s+ 0.018)
s4+ 5.0011s3+ 4.0056s2+ 2.0545s+ 0.0369 .
The performance results are P.O. = 13% and Ts= 29.6 seconds for a step
input, and ess = 0.12 for a ramp input.
AP10.3 The plant transfer function is
G(s) = 1
s(s+ 1)(s+ 4)
and the PI controller is given by
Gc(s) = Kps+KI
s.
The closed-loop transfer function is
T(s) = Kps+KI
s4+ 5s3+ 4s2+Kps+KI
.
For a unit ramp, the steady-state tracking error is
ess = lim
s0
s4+ 5s3+ 4s2
s5+ 5s4+ 4s3+Kps2+KIs= 0 .
Any KI>0 and Kp>0 (such that the system is stable) is suitable and
will track a ramp with zero steady-state error. Since we want P.O. <
13%, the damping of the dominant roots should be ζ0.6. One suitable
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Advanced Problems 571
solution is to place the zero at s=0.01 and select the PI controller
Gc(s) = 2.05(s+ 0.01)
s.
Therefore, Kp= 2.05 and KI= 0.0205. The closed-loop transfer function
is
T(s) = 2.05(s+ 0.01)
s4+ 5s3+ 4s2+ 2.05s+ 0.0205 .
The performance results are P.O. = 11.5% and Ts= 9.8 seconds for a
step input, and ess = 0 for a unit ramp.
AP10.4 The closed-loop transfer function is
T(s) = 10K1
s2+ 10(1 + K1K2)s+ 10K1
.
From the performance specifications, we determine that the natural fre-
quency and damping of the dominant poles should be ωn= 5.79 and
ζ= 0.69. So,
s2+ 8(1 + K1K2)s+ 8K1=s2+ 2ζωns+ω2
n=s2+ 7.99s+ 33.52 .
Solving for the gains yields K1= 4.19 and K2= 0. The closed-loop
transfer function is
T(s) = 33.52
s2+ 8s+ 33.52 .
The performance results are P.O. = 5% and Ts= 1 second.
AP10.5 (a) From the overshoot specification P.O. = 10%. The plant transfer
function is
G(s) = 1
s(s+ 1)(s+ 10) .
Let Gp= 1. A suitable compensator is
Gc=Ks+ 0.5
s+ 10 .
Using root locus methods, we determine that K= 45 yields P.O.
10%. The closed-loop poles are s1,2=2.5±j5.1, s3=15.48, and
s4=0.45.
(b) The closed-loop transfer function is
T(s) = 450(s+ 0.5)
s4+ 21s3+ 120s2+ 550s+ 225 .
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572 CHAPTER 10 The Design of Feedback Control Systems
The step response is shown in Figure AP10.5. The overshoot and
settling time are P.O. = 9.5% and Ts= 3.4 seconds.
(c) A suitable prefilter is
Gp(s) = 0.5
s+ 0.5.
The closed-loop transfer function is
T(s) = 225
s4+ 21s3+ 120s2+ 550s+ 225 .
The step response is shown in Figure AP10.5. The overshoot and
settling time are P.O. = 0% and Ts= 6.85 seconds.
012345678
0
0.2
0.4
0.6
0.8
1
1.2
1.4
Time (sec)
Amplitude
FIGURE AP10.5
Step response with prefilter (dashed line) and without prefilter (solid line).
AP10.6 From Example 10.12 in Dorf & Bishop, we have the relationship
ωnTs= 4.04 .
Thereore, minimizing Tsimplies maximizing ωn. Using Table 10.2 in Dorf
& Bishop, we equate the desired and actual characteristic polynomials
q(s) = s3+ 1.9ωns2+ 2.2ω2
ns+ω3
n=s3+ (1 + p)s2+ (K+p)s+Kz .
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Advanced Problems 573
Comparing coefficients yields
(1 + p) = 1.9ωn, K +p= 2.21 + p
1.92
, Kz =ω3
n.
So, from the first relationship we see that maximizing ωnimplies maxi-
mizing p. Solving for pwhile maintaining K < 52
K=2.2
3.61(p2+ 2p+ 1) p < 52
we determine that
9.3643 < p < 9.005 .
The largest p= 9. Therefore, K= 51.94 and z= 2.81. The step response
is shown in Figure AP10.6. The settling time is Ts= 0.77 second.
0
0.2
0.4
0.6
0.8
1
1.2
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2
Time (secs)
Amplitude
FIGURE AP10.6
Step response with minimum settling time.
AP10.7 Let Gp= 1. The closed-loop transfer function is
T(s) = K(s+ 3)
s4+ 38s3+ 296s2+ (K+ 448)s+ 3K.
When K= 311, the characteristic equation
s4+ 38s3+ 296s2+ 759s+ 933 = 0
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574 CHAPTER 10 The Design of Feedback Control Systems
has poles at s=1.619±j1.617 (ζ= 1/2), s=6.25, and s=28.51.
(a) When Gp(s) = 1 and K= 311, the closed-loop transfer function is
T(s) = 311(s+ 3)
s4+ 38s3+ 296s2+ 759s+ 933 .
The step input performance is P.O. = 6.5%, Ts= 2.5 seconds, and
Tr= 1.6 seconds. With the prefilter
Gp(s) = 3
s+ 3
and K= 311, the closed-loop transfer function is
T(s) = 933
s4+ 38s3+ 296s2+ 759s+ 933 .
In this case, the step response is P.O. = 3.9%, Ts= 2.8 seconds, and
Tr= 1.3 seconds.
(b) Now, consider the prefilter
Gp(s) = 1.8
s+ 1.8
and K= 311. The closed-loop transfer function is
(s) = 559.8(s+ 3)
s5+ 39.8s4+ 364.4s3+ 1291.8s2+ 2299.2s+ 1679.4.
The step input response is P.O. = 0.7%, Ts= 2.14 seconds and
Tr= 1.3 seconds.
AP10.8 The plant transfer function is
G(s) = 250
s(s+ 2)(s+ 40)(s+ 45) .
The performance specifications are P.O. < 20%, Tr<0.5 second, Ts<1.2
seconds and Kv10. A suitable lead compensator is
Gc= 1483.7s+ 3.5
s+ 33.75 .
The closed-loop transfer function is
T(s) = 250(1483.7)(s+ 35)
s(s+ 2)(s+ 40)(s+ 45)(s+ 33.75) + 250(1483.7)(s+ 3.5)
The actual step input performance (see Figure AP10.8) is P.O. = 18%,
Ts= 0.88 second, Tr= 0.18 second, and Kv= 10.7.
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Advanced Problems 575
0
0.2
0.4
0.6
0.8
1
1.2
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
Time (secs)
Amplitude
FIGURE AP10.8
Step response with lead compensator.
AP10.9 The frequency response is shown in Figure AP10.9.
Frequency (rad/sec)
Phase (deg); Magnitude (dB)
Bode Diagrams
-150
-100
-50
0
50
100
150
Gm=12.4 dB (Wcg=20.9); Pm=42.0 deg. (Wcp=9.0)
10-3 10-2 10-1 100101102103
-300
-250
-200
-150
-100
-50
FIGURE AP10.9
Bode plot with Gc(s) = (s+2.5)(s+0.9871)
(s+36.54)(s+0.0675)
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576 CHAPTER 10 The Design of Feedback Control Systems
One lead-lag compensator that satisfies the specifications is
Gc(s) = (s+ 2.5)(s+ 0.9871)
(s+ 36.54)(s+ 0.0675) .
The gain and phase margins are Gm = 12.35 dB and P m = 41.8, re-
spectively. The velocity error constant is Kv= 100. Therefore, all speci-
fications are satisfied.
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Design Problems 577
Design Problems
The plant model with parameters given in Table CDP2.1 in Dorf andCDP10.1
Bishop is given by:
θ(s)
Va(s)=26.035
s(s+ 33.142) ,
where we neglect the motor inductance Lmand where we switch off the
tachometer feedback (see Figure CDP4.1 in Dorf and Bishop). With a PD
controller the closed-loop system characteristic equation is
s2+ (33.142 + 26.035KD)s+ 26.035Kp= 0 .
Using Table 10.2 in Dorf and Bishop we determine that for a second-order
system with a deadbeat response we have α= 1.82 and ωnTs= 4.82. Since
we desire Ts<0.25 seconds, we choose ωn= 19.28. Equating the actual
characteristic equation with the desired characteristic equation we obtain
s2+ωnαs +ω2
n=s2+ (33.142 + 26.035KD)s+ 26.035Kp.
Solving for Kpand KDyields the PD controller:
Gc(s) = 14.28 + 0.075s .
The step response is shown below. The settling time is Ts= 0.24 second.
0 0.05 0.1 0.15 0.2 0.25 0.3
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Time (secs)
Amplitude
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578 CHAPTER 10 The Design of Feedback Control Systems
DP10.1 The plant is given as
G(s) = 20
s(s+ 2) .
One possible lead compensator is
Gclead (s) = 50(s+ 1)
s+ 20 .
Similarly, a suitable lag compensator is
Gclag (s) = s+ 0.1
s+ 0.022 .
The loop transfer function with the lead-lag compensator is
Gc(s)G(s) = 1000(s+ 1)(s+ 0.1)
s(s+ 2) (s+ 0.022)(s+ 20) .
The step response and ramp response are shown in Figure DP10.1. The
velocity constant is Kv= 50, so the steady-state error specification is
satisfied.
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
0
0.5
1
1.5
Time (sec)
Step response
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
0
0.2
0.4
0.6
0.8
1
Time (sec)
Ramp response
FIGURE DP10.1
Step response and ramp response.
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Design Problems 579
DP10.2 (a) When Gc(s) = K, we require K > 20 to meet the steady-state track-
ing specification of less than 5%.
(b) The system is unstable for K > 20.
(c) A single stage lead compensator is
Gc1(s) = 1 + 0.49s
1 + 0.0035s.
With this compensator, the bandwidth is
ωB= 68.9 rad/sec
and the phase margin is P.M. = 28.57o.
(d) A two stage lead compensator is
Gc2(s) = (1 + 0.0185s)(1 + 0.49s)
(1 + 0.00263s)(1 + 0.0035s).
With the two stage compensator, the bandwidth is
ωB= 83.6 rad/sec
and the phase margin is P.M. = 56.79o. The step response for the
two compensators is shown in Figure DP10.2.
0
0.2
0.4
0.6
0.8
1
1.2
1.4
00.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
Single stage (solid) & two stage (dashed)
Time (sec)
Amplitude
FIGURE DP10.2
Step response for one- and two-stage lead compensators.
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580 CHAPTER 10 The Design of Feedback Control Systems
DP10.3 The mast flight system is modeled as
G(s) = 6
s(s+ 1.5)(s+ 3.9) .
Consider the proportional controller
Gc(s) = K= 0.85 .
The system step response is shown in Figure DP10.3. The percent over-
shoot is
P.O. = 15.9%,
the rise time is
Tr= 3.63 seconds,
and the phase margin is
P.M. = 52o.
Step Response
Time (sec)
Amplitude
0 2 4 6 8 10 12
0
0.2
0.4
0.6
0.8
1
1.2
1.4 System: syscl
Peak amplitude: 1.16
Overshoot (%): 15.9
At time (sec): 3.63
FIGURE DP10.3
Step response for the mast flight system.
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Design Problems 581
DP10.4 One possible compensator is
Gc(s) = 5682s+ 12.6
s+ 87.3.
The step response is shown in Figure DP10.4. The performance results
0
0.2
0.4
0.6
0.8
1
1.2
00.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
Time (sec)
Amplitude
FIGURE DP10.4
Step response for the high speed train system.
are
P.O. = 4.44% Ts= 0.36 sec Kv= 14.1.
DP10.5 The design specifications are Kv>200; Ts<12 ms and percent overshoot
P.O. < 10%. The step response is shown in Figure DP10.5. A suitable
compensator is
Gc(s) = Ks+ 403
s+ 2336 ,
where
K= 1.9476e+ 13.
Then,
P.O. = 9.5% Ts= 10 ms Kv= 560 .
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582 CHAPTER 10 The Design of Feedback Control Systems
0
0.2
0.4
0.6
0.8
1
1.2
0 2 4 6 8 10 12 14 16 18 20
Time (ms)
Amplitude
FIGURE DP10.5
Step response for the tape transport system.
DP10.6 A solution to the problem is the PI controller
Gc(s) = 4.21s+ 1.2
s.
The step response is shown in Figure DP10.6.
0
0.2
0.4
0.6
0.8
1
1.2
0 1 2 3 4 5 6
Time (sec)
Amplitude
FIGURE DP10.6
Step response for the engine control system.
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Design Problems 583
The performance results are
P.O. = 8.8% and Ts= 2.14 .
The system is a type-1, so the steady-state error for a step input is zero,
as desired.
DP10.7 The jet aircraft roll angle motion is represented by the transfer function
G(s) = 10
(s+ 10)(s2+ 2s+ 20) .
A good controls solution is obtained with a PID controller
Gc(s) = 10s2+ 20s+ 150
s.
The system is type-1, so the steady-state tracking error is zero for a step
input. The performance results are
P.O. = 9.13% and Ts= 1.56 .
0 0.5 1 1.5 2 2.5 3 3.5
0
0.2
0.4
0.6
0.8
1
1.2
1.4
Step Response
Time (sec)
Amplitude
FIGURE DP10.7
Step response for the jet aircraft roll control system.
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584 CHAPTER 10 The Design of Feedback Control Systems
DP10.8 One good solution is obtained with the following PI controller
Gc(s) = 27.35(s+ 2)
s.
The system is type-1, so the steady-state tracking error is zero for a step
input. The step response is shown in Figure DP10.8.
Time (sec.)
Amplitude
Step Response
0 0.2 0.4 0.6 0.8 1 1.2
0
0.2
0.4
0.6
0.8
1
1.2
1.4
From: U(1)
To: Y(1)
FIGURE DP10.8
Step response for the windmill radiometer.
DP10.9 Consider the PID controller
Gc(s) = Kp+KDs+KI
s=1.554s2+ 1.08s+ 1
s
and the lead-lag controller
Gc(s) = Ks+a
s+bs+c
s+d= 6.04(s+ 10)(s+ 2)
(s+ 1)(s+ 5) .
Both are stabilizing in the presence of a T= 0.1 second time delay. For the
PID controller the phase margin is P.M. = 40o. For the lead-lag controller
the phase margin is P.M. = 45o. We find (for these particular designs)
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Design Problems 585
that the lead-lag controller is more able to remain stable in the process
of increasing time delay. For a time-delay of T= 0.2 seconds, the lead-lag
compensator has a phase margin of P.M. = 22o, while the PID controller
is unstable.
DP10.10 One solution is
Gc(s) = 50(s+ 0.01)
s+ 2 .
The Bode magnitude is shown in Figure DP10.10. You want high gain at
Bode Diagram
Frequency (rad/sec)
10−4 10−3 10−2 10−1 100101102
−100
−80
−60
−40
−20
0
20
40
60
80
System: sys
Frequency (rad/sec): 10
Magnitude (dB): −26.9
System: sys
Frequency (rad/sec): 0.101
Magnitude (dB): 26.9
Magnitude (dB)
FIGURE DP10.10
Step response for the windmill radiometer.
low frequency to improve disturbance rejection and decrease sensitivity to
plant changes and low gain at high frequency to attenuate measurement
noise.
DP10.11 One solution is the PD controller
Gc(s) = 0008(s+ 10) .
The step response is shown in Figure DP10.11. The closed-loop transfer
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586 CHAPTER 10 The Design of Feedback Control Systems
function is
T(s) = 4
s2+ 3.4s+ 4,
where we use the prefilter
Gp(s) = 4
0.36s+ 3.6.
Step Response
Time (sec)
Amplitude
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
0
0.2
0.4
0.6
0.8
1
1.2
1.4
System: sys_cl
Peak amplitude: 1.01
Overshoot (%): 0.637
At time (sec): 2.97
FIGURE DP10.11
Step response for the polymerase chain reaction system.
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Computer Problems 587
Computer Problems
CP10.1 The m-file script and step response is shown in Figure CP10.1. The phase
margin and percent overshoot are
P.M. = 50o
P.O. 18% ,
respectively.
nnumc=[110]; denc=[1 0]; sysc = tf(numc,denc);
numg=[1]; deng=[1 10]; sysg = tf(numg,deng);
syss = series(sysc,sysg);
[Gm,Pm]=margin(syss);
Pm
%
sys_cl = feedback(syss,1);
[y,t]=step(sys_cl);
step(sys_cl); grid
S=stepinfo(y,t);
PO=S.Overshoot
Pm =
49.9158
PO =
17.5724
0 0.2 0.4 0.6 0.8 1 1.2
0
0.2
0.4
0.6
0.8
1
1.2
1.4
Step Response
Time (sec)
Amplitude
FIGURE CP10.1
Phase margin and step response for the closed-loop system.
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588 CHAPTER 10 The Design of Feedback Control Systems
CP10.2 Using a proportional controller the closed-loop characteristic equation is
1 + K24.2
s2+ 8s+ 24.2.
A simple m-file script which computes the P.M. as a function of the gain
Kyields the proportional controller gain K= 6. Checking the phase
margin of the system reveals that P.M. 40, as desired.
n=24.2; d=[1 8 24.2]; sys = tf(n,d);
K=6;
margin(K*sys), grid
−80
−60
−40
−20
0
20
Magnitude (dB)
10−1 100101102103
−180
−135
−90
−45
0
Phase (deg)
Bode Diagram
Gm = Inf dB (at Inf rad/sec) , Pm = 39.9 deg (at 11.6 rad/sec)
Frequency (rad/sec)
FIGURE CP10.2
Bode plot with a proportional controller K= 6 in the loop.
CP10.3 The uncompensated system is type-1. To realize a zero steady-state error
to a ramp input we need to increase the system type by one. One controller
that does this is the PI controller:
Gc(s) = KPs+KD
s.
The step response is shown in Figure CP10.3 where it can be seen in the
tracking error plot that the settling time is Ts<5 seconds. The actual
settling time is
Ts= 3.6 seconds .
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Computer Problems 589
KP=20; KD=10;
nc=[KP KD]; dc=[1 0]; sysc = tf(nc,dc);
n=1; d=[1 2 0]; sys = tf(n,d);
sys_o = series(sysc,sys);
sys_cl = feedback(sys_o,[1]);
t=[0:0.001:10];
sys1 = tf([1],[1 0]); sys_cl1 = series(sys_cl,sys1);
subplot(121)
y=step(sys_cl1,t);
plot(t,y,t,t,'--'), grid
xlabel('Time (sec)'), ylabel('Ramp response')
e=y-t'; L=nd(abs(e)>0.02);
Ts=t(L(length(L)))
subplot(122)
plot(t,e,[0 10],[0.02 0.02],':',[0 10],[-0.02 -0.02],':')
xlabel('Time (sec)'), ylabel('Tracking error')
grid
0 5 10
0
1
2
3
4
5
6
7
8
9
10
Time (sec)
Ramp response
0 5 10
-0.25
-0.2
-0.15
-0.1
-0.05
0
0.05
0.1
Time (sec)
Tracking error
FIGURE CP10.3
Ramp response with a PI controller Gc(s) = 20s+10
sin the loop.
CP10.4 From the percent overshoot spec we determine that P.O. < 10% implies
ζ > 0.6. So, we target a phase margin P.M. = 100ζ= 60o. The m-file
script which generates the uncompensated Bode plot is shown in Fig-
ure CP10.4a.
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590 CHAPTER 10 The Design of Feedback Control Systems
È
Phi =
56.2111
Pm =
3.7889
alpha =
10.8408
numg = 100*conv([1 1],[1 0.01]);
deng = conv([1 10],conv([1 2 2],[1 0.02 0.0101]));
sysg = tf(numg,deng)
w=logspace(-1,2,200);
[mag,phase,w]=bode(sysg,w);
[Gm,Pm,Wcg,Wcp]=margin(mag,phase,w);
%
Phi=60-Pm
Pm
Phi=(60-Pm)*pi/180;
alpha=(1+sin(Phi))/(1-sin(Phi))
M=-10*log10(alpha)*ones(length(w),1);
[mag,phase,w]=bode(sysg,w);
for i = 1:length(w),
magdB(i) = 20*log10(mag(1,1,i));
end
semilogx(w,magdB,w,M), grid
xlabel('Frequency (rad/sec)'), ylabel('mag [dB]')
title('Uncompensated Bode Plot')
hold on
semilogx([.56072 5.6072 56.072 560.72],[20 0 -20 -40],'--')
-80
-60
-40
-20
0
20
40
60
10-1 100101102
Frequency (rad/sec)
mag [dB]
Uncompensated Bode Plot
FIGURE CP10.4
(a) Uncompensated Bode plot.
We assume that K= 1 and raise the gain at a later step to meet settling
time requirement. The uncompensated phase margin is P.M. = 3.7o, so
that the lead compensator needs to add φ= 56.2o. The script also cal-
culates α= 10.84. Following the design procedure outlined in Dorf &
Bishop, we locate the compensator zero at ω= 2 rad/sec (see dashed line
in Figure CP10.4a). Then, p=αz implies p= 21.68. After several iter-
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Computer Problems 591
ations, we converge on K= 4 as a “good” value. The lead compensator
is
Gc(s) = 4 s+ 2
s+ 22 .
The step response is shown in Figure CP10.4b. The compensated Bode is
shown in Figure CP10.4c.
K=4;
numg = 100*conv([1 1],[1 0.01]);
deng = conv([1 10],conv([1 2 2],[1 0.02 0.0101]));
sysg = tf(numg,deng)
numc=K*[1 2]; denc=[1 22]; sysc = tf(numc,denc);
sys_o = series(sysc,sysg);
sys_cl = feedback(sys_o,[1]);
t=[0:0.01:5];
f=10*pi/180;
[y,t,x]=step(f*sys_cl,t);
plot(t,y*180/pi), grid
xlabel('Time (sec)')
ylabel('Attitude rate (deg/sec)'), pause
w=logspace(-1,2,200);
[mag,phase,w]=bode(sys_o,w);
[Gm,Pm,Wcg,Wcp]=margin(mag,phase,w);
bode(sys_o)
title(['Gain Margin = ',num2str(Gm),' Phase Margin = ',num2str(Pm)])
0
2
4
6
8
10
12
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
Time (sec)
Attitude rate (deg/sec)
FIGURE CP10.4
CONTINUED: (b) Step response.
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592 CHAPTER 10 The Design of Feedback Control Systems
-100
-50
0
50
10-3 10-2 10-1 100101102
Frequency (rad/sec)
Gain dB
-300
-200
-100
0
100
10-3 10-2 10-1 100101102
Frequency (rad/sec)
Phase deg
Gain Margin = 14.96 Phase Margin = 60.49
FIGURE CP10.4
CONTINUED: (c) Bode plot with lead compensator.
CP10.5 The closed-loop transfer function is
θ(s)d(s) = ¯
K1+¯
K2s
s2+¯
K2s+¯
K1
where ¯
K1=K1/J and ¯
K2=K2/J. A percent overshoot P.O. 20%
requires ζ > 0.45. Select as the initial damping
ζ= 0.7 (initial selection) .
For a second-order system with ζ= 0.7, we find that ωn0.9 when
|θ(s)d(s)|= 0.7. So, we select
ωn=ωB/0.9
as a starting choice. Therefore, since ωB= 10, we have
ωn= 11 .
The m-file script is shown in Figure CP10.5a. After several iterations, we
find a set of “good” values for
ζ= 0.8 and ωn= 4.5 (final selection) .
The step response and closed-loop Bode plot are shown in Figures CP10.5b
and CP10.5c.
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Computer Problems 593
% Part (a)
wn=4.5; zeta=0.8; K2=2*zeta*wn; K1=wn^2;
% Part (b)
num=[K2 K1]; den=[1 0 0]; sys = tf(num,den);
sys_cl = feedback(sys,[1]);
f=10*pi/180; % set-up for 10 deg step input
t=[0:.05:3];
[y,t,x]=step(f*sys_cl,t);
plot(t,y*180/pi), xlabel('time [sec]'), ylabel('theta [deg]'),
grid, pause
% Part (c)
w=logspace(-1,2,400);
[mag,phase,w]=bode(sys_cl,w);
for i = 1:length(w),
magdB(i) = 20*log10(mag(1,1,i));
end
semilogx(w,magdB,[w(1) w(length(w))],[-3 -3]), grid
xlabel('Frequency (rad/sec)')
ylabel('Gain dB')
FIGURE CP10.5
(a) Script to generate the step response and the closed-loop Bode plot.
0
2
4
6
8
10
12
0 0.5 1 1.5 2 2.5 3
time [sec]
theta [deg]
FIGURE CP10.5
CONTINUED: (b) Step response.
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594 CHAPTER 10 The Design of Feedback Control Systems
-25
-20
-15
-10
-5
0
5
10-1 100101102
Frequency (rad/sec)
Gain dB
FIGURE CP10.5
CONTINUED: (c) Closed-loop Bode plot.
CP10.6 The settling time and phase margin specifications require that the dom-
inant closed-loop poles have natural frequency and damping of ζ0.45
and ωn1.78. The uncompensated roots locus is shown in Figure CP10.6a.
numg=[1 10]; deng=[1 2 20];
sysg = tf(numg,deng);
axis([-15,1,-10,10]);
rlocus(sysg); hold on
%
zeta=0.45; wn=1.7778;
x=[-10:0.1:-zeta*wn];
y=-(sqrt(1-zeta^2)/zeta)*x;
xc=[-10:0.1:-zeta*wn];
c=sqrt(wn^2-xc.^2);
plot(x,y,':',x,-y,':',xc,c,':',xc,-c,':')
rlocnd(sysg),
hold o
-10
-8
-6
-4
-2
0
2
4
6
8
10
-14 -12 -10 -8 -6 -4 -2 0
x
x
o
Real Axis
Imag Axis
+
+
K=10
FIGURE CP10.6
(a) Uncompensated root locus.
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Computer Problems 595
From the final value theorem, we determine that
lim
s0=sE(s)0.1Aimplies A
1 + GGc(s)= 0.1A .
Therefore, the compensated Kpcomp 9. With the compensator
Gc(s) = Ks+z
s+p
we find that
Kpcomp =Kz
pKpuncomp .
But Kpuncomp = 0.5 and (from the uncompensated root locus) a gain of
K= 10 results in roots of the characteristic equation in the desired region.
Solving for
z
p=1
K
Kpcomp
Kpuncomp 2.
Select z= 0.5 to minimize changing the root locus. Then, p= 0.25, and
the compensator is
Gc(s) = 10 s+ 0.5
s+ 0.25 .
The compensated root locus is shown in Figure CP10.6b and the step re-
sponse is shown in Figure CP10.6c. The phase margin of the compensated
numg=[1 10]; deng=[1 2 20];
sysg = tf(numg,deng);
numc=[1 0.5]; denc=[1 0.25];
sysc = tf(numc,denc);
sys_o = series(sysc,sysg);
axis([-15,1,-10,10]);
rlocus(sys_o); hold on
%
zeta=0.45; wn=1.7778;
x=[-10:0.1:-zeta*wn];
y=-(sqrt(1-zeta^2)/zeta)*x;
xc=[-10:0.1:-zeta*wn];
c=sqrt(wn^2-xc.^2);
plot(x,y,':',x,-y,':',xc,c,':',xc,-c,':')
rlocnd(sys_o)
hold o
-10
-8
-6
-4
-2
0
2
4
6
8
10
-14 -12 -10 -8 -6 -4 -2 0
x
x
xo o
Real Axis
Imag Axis
+
+
+
FIGURE CP10.6
CONTINUED: (b) Compensated root locus.
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596 CHAPTER 10 The Design of Feedback Control Systems
system is P.M. = 62.3oand the settling time Ts<5 seconds.
>>
pm =
62.3201
numg=[1 10]; deng=[1 2 20];
sysg = tf(numg,deng);
numgc=10*[1 0.5]; dengc=[1 0.25];
sysc = tf(numgc,dengc);
sys_o = series(sysc,sysg);
sys_cl = feedback(sys_o,[1]);
t=[0:0.1:5]; step(sys_cl,t)
[mag,phase,w]=bode(sys_o);
[gm,pm,w1,w2]=margin(mag,phase,w); pm
0
0.2
0.4
0.6
0.8
1
1.2
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
Time (secs)
Amplitude
FIGURE CP10.6
CONTINUED: (c) Step response and phase margin verification.
CP10.7 Both design specifications can be satisfied with an integral controller
Gc(s) = K1+K2
s=10
s.
The simulation results and m-file script are shown in Figures CP10.7a
and b.
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Computer Problems 597
0
0.5
1
1.5
00.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
Time (sec)
Phi dot
Unit Step Response
-0.15
-0.1
-0.05
0
00.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
Time (sec)
Tracking error
Unit Ramp Response
FIGURE CP10.7
(a) Simulation results.
K1=0; K2=10;
numc=[K1 K2]; denc=[1 0]; sysc = tf(numc,denc);
numg=[23]; deng=[1 23]; sysg = tf(numg,deng);
sys_o = series(sysc,sysg);
sys_cl = feedback(sys_o,[1]);
t=[0:0.01:1];
ys=step(sys_cl,t);
subplot(211)
plot(t,ys), xlabel('Time (sec)'), ylabel('Phi dot')
title('Unit Step Response'), grid
u=t;
yr=lsim(sys_cl,u,t);
subplot(212)
plot(t,yr-u','--')
xlabel('Time (sec)'), ylabel('Tracking error')
title('Unit Ramp Response'), grid
FIGURE CP10.7
CONTINUED: (b) M-file design script.
CP10.8 From Example 10.3, we have that the loop transfer function is
Gc(s)G(s) = 8.1(s+z)
s2(s+ 3.6),
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598 CHAPTER 10 The Design of Feedback Control Systems
where z= 1. We want to determine a value of zso the the percent
overshoot is reduced from 46% to less than 32%. A valid design is
Gc(s)G(s) = 8.1(s+ 0.45)
s2(s+ 3.6) .
The m-file script and step response are shown in Figure CP10.8. The
percent overshoot is P.O.=27.7 %.
K1 = 8.1;
numg = [K1]; deng = [1 0 0];
sysg = tf(numg,deng);
numc = [1 0.45]; denc = [1 3.6];
sysc = tf(numc,denc);
sys_o = series(sysc,sysg);
sys_cl = feedback(sys_o,[1]);
step(sys_cl)
y=step(sys_cl);
po=100*(max(y)-1)
Time (sec.)
Amplitude
Step Response
0 1.6 3.2 4.8 6.4 8
0
0.2
0.4
0.6
0.8
1
1.2
1.4
From: U(1)
To: Y(1)
FIGURE CP10.8
Response of system with new lead compensator design.
CP10.9 From AP10.10, we have the transfer function is
T(s) = Vo(s)
Vi(s)
=1 + R2C2s
1 + R1C1s.
Substituting C1= 0.1µF ,C2= 1 mF ,R1= 10 k, and R2= 10 Ω yields
T(s) = 1 + 0.01s
1 + 0.001s.
The frequency response is shown in Figure CP10.9.
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Computer Problems 599
c1=0.0000001; c2=0.001;
r1=10000; r2=10;
n=[c2*r2 1]; d=[c1*r1 1];
sys=tf(n,d)
bode(sys)
Frequency (rad/sec)
Phase (deg); Magnitude (dB)
Bode Diagrams
0
5
10
15
20
101102103104
0
10
20
30
40
50
60
FIGURE CP10.9
Op-amp circuit frequency response.
CP10.10 The plot of Kversus phase margin is shown in Figure CP10.10. The value
of Kthat maximizes the phase margin is K= 4.15.
K=[0.1:0.01:10];
T=0.2;
[np,dp]=pade(T,6); sysp=tf(np,dp);
for i=1:length(K)
ng=K(i)*[1 0.2]; dg=[1 6 0 0]; sysg=tf(ng,dg);
[gm,pm]=margin(sysg*sysp);
PM(i)=pm;
end
plot(K,PM), grid
[P,n]=max(PM); K(n)
xlabel('K'), ylabel('P.M.')
012345678910
15
20
25
30
35
40
45
50
55
60
K
P.M.
FIGURE CP10.10
Plot of Kversus phase margin.
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C H A P T E R 1 1
The Design of State Variable
Feedback Systems
Exercises
E11.1 The system is given by
˙
x=Ax +Bu
u=Kx
where
A=
0 1
1 0
B=
1 0
0 1
and K=
k0
02k
.
Then, with u=Kx, we have
˙
x=
k1
12k
x.
The characteristic equation is
det[sIA] = det
s+k1
1s+ 2k
=s2+ 3ks + 2k2+ 1
=s2+ 2ζωns+ω2
n= 0 .
Solving for kwhere ω2
n= 2k2+ 1 and ζ= 1 (critical damping) yields
k= 2.
600
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Exercises 601
E11.2 Let
u=k1x1k2x2+r .
Then,
˙
x=
0 1
9k1k2
x+
0
1
r ,
and
det(sIA) = s2+k2s+k19 = 0 .
We want ζ= 1, so the desired characteristic equation is
pd(s) = (s+co)2,
where cois to be determined to meet Ts= 4 and where k2= 2coand
k1=c2
o+ 9. Solving for the state response of x1(t) to a unit step input
we find
x1(t) = 1 ecotcotecot.
When tTs= 4 sec we want x1(t)0.98. Solving for coat t=Tsyields
co= 1.459, k1= 11.13,and k2= 2.92 .
E11.3 The controllability matrix is
Pc=hB AB i=
0 1
13
,
and det Pc6= 0, therefore the system is controllable. The observability
matrix is
Po=
C
CA
=
0 2
06
,
and det Po= 0; therefore the system is unobservable.
E11.4 The controllability matrix is
Pc=hB AB i=
0 0
24
,
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602 CHAPTER 11 The Design of State Variable Feedback Systems
and the det Pc= 0; therefore the system is uncontrollable. The observ-
ability matrix is
Po=
C
CA
=
1 0
10 0
,
and det Po= 0; therefore the system is also unobservable.
E11.5 The controllability matrix is
Pc=hB AB i=
12
2 3
,
and det Pc=16= 0; therefore the system is controllable. The observ-
ability matrix is
Po=
C
CA
=
1 0
0 1
,
and det Po= 1 6= 0; therefore the system is observable.
E11.6 The controllability matrix is
Pc=hB AB i=
0 1
12
,
and det Pc6= 0; therefore the system is controllable. The observability
matrix is
Po=
C
CA
=
1 0
0 1
,
and det Po6= 0; therefore the system is observable.
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Exercises 603
E11.7 The block diagram is shown in Fig. E11.7.
U(s)Y(s)
s
1s
1
5
3
-
-
+
12
2
2+-
FIGURE E11.7
The block diagram for E11.7.
E11.8 The block diagram is shown in Fig. E11.8.
U(s)Y(s)
s
1
s
1
s
1
1
3
--
-
+
9
10
8
2++
+
4
-
+
FIGURE E11.8
The block diagram for E11.8.
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604 CHAPTER 11 The Design of State Variable Feedback Systems
E11.9 The controllability matrix is
Pc=hB AB i=
k1k1k2
k2k1+k2
,
and det Pc=k2
1+k2
2. So, the condition for complete controllability is
k2
16=k2
2.
E11.10 A matrix differential equation representation is
˙
x=
010
001
10 63
x+
0
0
1
u
y= [3 4 2]x+ [0]u .
E11.11 The system is given by
˙
x=Ax +Bu
y=Cx +Du
where
A=
0 1 0
0 0 1
2 0 7
,B=
0
0
1
,C=h1 2 0 i,and D= [1] .
The controllability matrix is
Pc=hB AB A2Bi=
0 0 1
0 1 7
17 49
,
and det Pc=16= 0; therefore the system is controllable. The observ-
ability matrix is
Po=
C
CA
CA2
=
1 2 0
0 1 2
4 0 13
,
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Exercises 605
and det Po=29 6= 0; therefore the system is observable.
E11.12 The transfer function is
G(s) = 6
s2+ 5s+ 6 .
The response of the system to a unit step is
y(t) = 1 3e2t+ 2e3t.
The step response is shown in Figure E11.12
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Time (s)
Step Response
FIGURE E11.12
Unit step response.
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606 CHAPTER 11 The Design of State Variable Feedback Systems
Problems
P11.1 Consider the system
˙x=x+u
u=kx .
So,
˙x=xkx = (1 k)x
and
x(t) = e(1k)tx(0) .
The system is stable if k > 1. Computing the value of J(assuming k > 1)
yields
J=Z
0
e2(1k)tx2(0)dt =1
k1.
Thus, Jis minimum when k→ ∞. This is not physically realizable. Select
k= 35. Then, the value of the performance index Jis
J=1
34 .
The system is not stable without feedback.
P11.2 (a) The performance index is given
J=Z
0
(x2+λu2)dt .
The system is
˙x=x+u
u=kx .
So,
J=Z
0
(x2+λk2x2)dt =Z
0
(1 + λk2)x2dt = (1 + λk2)Z
0
x2dt .
Carrying out the integration (assuming k > 1) yields
J= (1 + λk2)1
k1.
We want to determine kthat minimizes J. Taking the partial of J
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Problems 607
with respect to kand setting the result to zero yields
J
k =λk22λk 1
(k1)2= 0 ,
or
λk22λk 1 = 0 .
Solving for kyields
k= 1 + r1 + 1
λ,
where we reject the solution k= 1 q1 + 1
λ, since we require k > 1.
(b) For λ= 2, we determine that k= 2.2 and Jmin = 8.9.
P11.3 The system is given by
˙
x=
1 0
1 2
x+
1
1
u
u=k(x1+x2) = k[1 1]x.
Then, with feedback applied, the system is
˙
x=
(1 k)k
(1 + k) (2 k)
x.
Solving
HTP+PH =I
yields
2p11(1 k)2p12(k+ 1) = 1
p12(3 2k)p11kp22(k+ 1) = 0
2kp12 + 2p22(2 k) = 1.
Solving for p11, p12 and p22 yields
p11 =(2k26k+ 7)
4(4k28k+ 3)
p12 =2k22k1
4(4k28k+ 3)
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608 CHAPTER 11 The Design of State Variable Feedback Systems
p22 =(2k26k+ 3)
4(4k28k+ 3) .
The performance index is computed to be
J=xT(0)Px(0) = p11 + 2p12 +p22 =1
2k1,
when x(0) = [1 1]T. So as k→ ∞,J0. The system is unstable without
feedback.
P11.4 The performance index is
J=xT(0)Px(0) = p11 2p12 +p22 .
From Example 11.12 in Dorf and Bishop, we determine that
J=2k2+ 1
2k2.
So, when k→ ∞, the performance index J1. The plot of Jversus k
is shown in Figure P11.4.
0 1 2 3 4 5 6 7 8 9 10
0
10
20
30
40
50
60
K
J
FIGURE P11.4
The performance index Jversus k.
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Problems 609
P11.5 The system is given by
˙
x=
0 1
0 0
x+
0 0
1 1
u
u=kx.
The performance index is
J=Z
0
(xTx+uTu)dt =Z
0
(1 + k2)(xTx)dt .
First, we solve
HTP+PH =(1 + k2)I,
yielding,
p12 =(1 + k2)
2k
p22 =k3+k2+k+ 1
2k2
p11 =2k3+k2+ 2k+ 1
2k.
The performance index is then given by
J=p11 + 2p12 +p22 =2k4+ 4k3+ 3k2+ 4k+ 1
2k2.
Taking the partial of Jwith respect to k, setting the result to zero and
solving for kyields
J
k =2k4+ 2k32k1
k3= 0
or
2k4+ 2k32k1 = 0 .
Solving for kyields k= 0.90. The plot of Jversus kis shown in Fig-
ure P11.5. The value of the performance index is
J= 6.95
when k= 0.90.
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610 CHAPTER 11 The Design of State Variable Feedback Systems
6.8
7
7.2
7.4
7.6
7.8
8
8.2
8.4
8.6
8.8
0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4
k
J
FIGURE P11.5
The performance index Jversus k.
P11.6 (a) For P11.3, we have
J=1
2k1.
So, as k→ ∞, then J0. But k=is not a practical solution, so
select k= 10. Then, J= 1/19, and
˙
x=
910
11 8
x=Ax .
The closed-loop system roots are determined by solving
det[sIA] = s2+ 17s38 = 0 ,
which yields s=19 and s2= 2. The system is unstable. The original
system was unstable, and it remains unstable with the feedback. In
general,
˙
x=
(1 k)k
(1 + k) (2 k)
x=Ax
and det[sIA] = s2+s(2k3) + (2 4k) = 0. A Routh-Hurwitz
analysis reveals that the system is unstable for all k.
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Problems 611
(b) For P11.4, we have
˙
x=
0 1
kk
x=Ax ,
and
det[sIA] = s2+ks +k= 0 .
The performance index was found to be
J= 1 + 4k+ 1
2k2.
As k→ ∞, we have J0. But k=is not a practical choice for
k. Select k= 10. Then,
det[sIA] = s2+ 10s+ 10 = (s+ 1.13)(s+ 8.87) .
The closed-loop system is stable.
(c) In P11.5, we found that k= 0.90 for Jmin. We are given
˙
x=
0 1
kk
x
and
det[sIA] = s2+ks +k=s2+ 0.9s+ 0.9
= (s+ 0.45 + j0.835)(s+ 0.45 j0.835) .
P11.7 The closed-loop system is
˙
x=
0 1
k1k2
x=Hx ,
and
det[sIH] = s2+k2s+k1=s2+ 2ζωns+ω2
n= 0 .
We desire ωn= 2, so set k1= 4. With xT(0) = [1,0], we have J=p11,
and solving
HTP+PH =I
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612 CHAPTER 11 The Design of State Variable Feedback Systems
yields
04
1k2
p11 p12
p12 p22
+
p11 p12
p12 p22
0 1
4k2
=
1 0
01
,
and
p11 =k2
8+20
8k2
=k2
2+ 20
8k2
.
Select
k2=20
for Jmin, where Jmin =5
2. Then
det[sIH] = s2+20s+ 4 = 0 ,
and ωn= 2 and ζ= 1.12. The system is overdamped.
P11.8 From Example 11.11 in Dorf and Bishop, we have
P=
k2
2+2
2k2
1
2
1
2
1
k2
.
So,
J=xT(0)Px(0) = k2
2+ 2
2k2
when xT(0) = [1 0]. Taking the partial of Jwith respect to k2and setting
the result to zero yields
J
k2
= 1 k2
2+ 2
2k2
2
= 0 .
Solving for the optimum value of k2yields
k2=2.
P11.9 Let x1=φand x2=ω. We have that
ω=
dt .
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Problems 613
The state equations are
˙x1=x2
˙x2=Ku .
Select a feedback such that
u=x1K1x2+r
when r(t) is the reference input. Then,
˙
x=
0 1
KKK1
x+
0
K
r ,
and
det[sI A] = s2+K1Ks +K .
We desire ζ=1
2, so that the overshoot is 4%. Since Ts= 1 = 4
ζωn, we
require ζωn= 4 or ωn= 42. Then, s2+ 8s+ 32 = s2+K1Ks +K, or
K= 32 and K1=8
32 =1
4.
P11.10 The system with feedback is given by
˙
x=Ax =
10 25
1 0
x,
where x1(0) = 1, and x2(0) = 1. The characteristic equation is
det[sIA] = det
s+ 10 25
1s
=s(s+ 10) + 25 = s2+ 10s+ 25 = 0 .
The roots are s1,2=5. The solution is
x(t) =
φ11 φ12
φ21 φ22
x(0) =
φ11 φ12
φ21 φ22
since x1(0) = 1 and x2(0) = 1. We compute the elements of the state
transition matrix as follows:
φ22(t) = (1 + 5t)e5tand φ21(t) = te5t,
therefore
x2(t) = (1 + 4t)e5t.
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614 CHAPTER 11 The Design of State Variable Feedback Systems
Similarly,
φ11(t) = (1 5t)e5tand φ12 = 25e5t.
Therefore,
x1(t) = (1 + 20t)e5t.
P11.11 Let
u=k1x1k2x2+αr
where r(t) is the command input. A state variable representation of the
plant is
˙
x=
52
2 0
x+
0.5
0
u
y=h0 1 ix+h0iu .
The closed-loop transfer function is
T(s) = α
s2+ (k1/2 + 5)s+ 4 + k2
.
To meet the performance specifications we need ωn= 4.8 and ζ= 0.826.
Therefore, the desired characteristic polynomial is
q(s) = s2+ 2(0.826)4.8s+ 23 = s2+ 8s+ 23 .
Equating coefficients and solving for k1and k2yields k2= 19 and k1= 6.
Select α= 23 to obtain zero steady-state error to a step input.
P11.12 A state variable representation of the dc motor is
˙
x=
320.75 0 0
3 0 0 0 0
0 2 0 0 0
0 0 1 0 0
0 0 0 2 0
x+
1
0
0
0
0
u
y= [0 0 0 0 2.75]x.
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Problems 615
The controllability matrix is
Pc=
13 3 4.518
0 3 9 9 13.5
00618 18
0 0 0 6 18
0 0 0 0 12
and the det Pc6= 0, so the system is controllable. The observability matrix
is
Po=
0 0 0 0 2.75
0 0 0 5.5 0
0 0 5.5 0 0
0 11 0 0 0
33 0 0 0 0
,
and the det Po6= 0, so the system is observable.
P11.13 To meet the Kv= 35 specification, we need K= 2450. A state variable
representation is
˙
x=
0 1
070
x+
0
2450
u
y= [1 0]x.
Let
u=k1x1k2x2.
Then, the closed-loop characteristic equation is
q(s) = s2+ (2450k2+ 70)s+ 2450k1= 0 .
The desired characteristic polynomial is
s2+ 72.73s+ 2644.63 = 0
where we select ζ= 0.707 and ωn= 51.42 to meet the performance
specifications. Equating coefficients and solving for the gains yields k1=
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616 CHAPTER 11 The Design of State Variable Feedback Systems
1.08 and k2= 0.0011.
P11.14 Let
u=k1x1k2x2k3r
where r(t) is the command input. Then, the closed-loop system in state
variable form is
˙
x=
10 k1k2
1 0
x+
1
0
r
y= [0 1]x.
To meet the performance specifications, we want the closed-loop charac-
teristic polynomial to be
q(s) = s2+ 8s+ 45.96 = 0
where ζ= 0.59 and ωn= 6.78. The actual characteristic polynomial is
det(sIA) = s2+ (10 + k1)s+k2= 0 .
Equating coefficients and solving for the gains yields k2= 45.96 and
k1=2. Select k3=k2= 45.96 to obtain a zero steady-state error to a
step input. This results in a settling time of Ts= 0.87 s and a percent
overshoot of P.O. = 10%.
P11.15 The transfer function is
G(s) = C(sIA)1B=1
s+ 1 .
The system is not controllable and not observable.
P11.16 Let
u=Kx .
Then, Ackermann’s formula is
K= [0,0, ..., 1]P1
cq(A)
where q(s) is the desired characteristic polynomial, which in this case is
q(s) = s2+ 2s+ 10 .
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Problems 617
A state-space representation of the limb motion dynamics is
˙
x=
4 0
11
x+
1
0
u .
The controllability matrix is
Pc= [B AB] =
14
0 1
and
P1
c=
1 4
0 1
.
Also, we have
q(A) = A2+ 2A+ 10I=
18 0
3 9
.
Using Ackermann’s formula, we have
K= [3 9] .
P11.17 The system is either uncontrollable or unobservable if a= 5 or a= 8.
Both of these values correspond to system real poles. So, if atakes on
either value, a pole-zero cancellation occurs in the transfer function.
P11.18 A matrix differential equation representation is
˙
x=
0 1
12
x+
0
1
u
y= [1 0]x.
Let u(t) = k1x1k2x2. Then, the closed-loop characteristic equation is
q(s) = s2+ (2 + k2)s+ 1 + k1= 0 .
We desire the characteristic equation
s2+ 22s+ 2 = 0 .
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618 CHAPTER 11 The Design of State Variable Feedback Systems
Equating coefficients and solving for the gains yields k1= 1 and k2=
222 = 0.828.
P11.19 A state space representation is
˙
x=
0 1
32
x+
0
1
r
y= [3 1]x.
The controllability matrix is
Pc=
0 1
12
,
and det Pc6= 0, so the system is controllable. The observability matrix is
Po=
3 1
3 1
,
and the det Po= 0, so the system is not observable.
P11.20 The characteristic equation associated with Ais
s2(s2+ 0.2s+ 0.0015) = 0 .
There are two roots at the origin, so the system is unstable. The system
can be stabilized with
δ=k1x1k3x3= 20x110x3.
P11.21 (a) Let x1=i1,x2=i2and u=v. Then, the state equation is
˙
x=
(R1+R3)
L1
R3
L1
R3
L2(R3+R2)
L2
x+
1
L1
0
u.
Also,
y=vo,
but
y= [R3R3]x.
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Problems 619
(b) The observability matrix is
Po=
C
CA
=
R3R3
R1R3
L1R2
31
L1+1
L2R2R3
L2+R2
31
L1+1
L2
and
det Po=R2
L2R1
L1R2
3.
So, when
R1
L1
=R2
L2
,
det Po= 0 and the system is not observable.
(c) Let
a=R1+R3
L1
,
and
b=R3+R2
L2
.
Then
det[sIA] = det
(s+a)R3
L1
R3
L2(s+b)
="(s+a)(s+b) + R2
3
L1L2#= (s+r)2
=s2+ (a+b)s+ab +R2
3
L1L2
.
The system has two equal roots when
(a+b)24 ab +R2
3
L1L2!
or
R1+R3
L1
+R3+R2
L22
4(R1+R3)(R3+R2)+R2
3
L1L2
= 0 .
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620 CHAPTER 11 The Design of State Variable Feedback Systems
P11.22 (a) Without state feedback the state differential equation is given by
˙
x=
0.41
1 0
x+
1
0
u
y=h0 1 ix.
The step response is shown in Figure P11.22a.
0
0.5
1
1.5
2
02 4 6 8 10 12 14 16 18 20
Time (sec)
x2
(a) Without state feedback
0
0.5
1
1.5
00.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2
Time (sec)
x2
(b) With state feedback
FIGURE P11.22
Step response (a) without state feedback, and (b) with state feedback.
(b) Consider state feedback
u=K(ax2+bx1) + cr
where ris the reference input and K, a, b and care to be determined.
Then, the state differential equation is
˙
x=
0.4Kb 1Ka
1 0
x+
c
0
r
y=h0 1 ix,
and det(sIA) = s2+ (0.4 + Kb)s+ (1 + Ka) = 0. Our specifications
are P.O. = 5% and Ts= 1.35 sec. So, ζ= 0.69 and ωn=4
ζ1.35 = 4.3.
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Problems 621
Solving for K, a and byields
Ka =ω2
n1
and
Kb = 2ζωn0.4.
Select K= 1. Then, a= 17.49 and b= 5.53. Select c= 1 + Ka to
achieve a zero steady-state tracking error.
(c) The step response is shown in Figure P11.22b for the system with
state feedback.
P11.23 Using the internal model design method for step inputs, we have
˙e
˙
z
=
0 1 0
0 0 1
0 0 0
e
z
+
0
0
1
w ,
where we choose
w=K1eK2z.
To place the poles at s=10 and s=2±jwe use Ackermann’s formula
to compute
K1= 50
K2= [45 14] .
The compensator has the form shown in Figure 11.14 in Dorf and Bishop.
P11.24 Using the internal model design method for ramp inputs, we have
˙e
¨e
˙
z
=
0 1 0 0
0 0 1 0
0 0 0 1
0 0 0 0
e
˙e
z
+
0
0
0
1
w
where we choose
w=K1eK2˙eK3z.
To place the poles at s=20 and s=2±2jwe can use Ackermann’s
formula. We also need an additional pole (must be a stable pole); select
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622 CHAPTER 11 The Design of State Variable Feedback Systems
s=20 as the fourth pole. Then,
K1= 3200
K2= 1920
K3= [568 44] .
The compensator has the form shown in Figure 11.16 in Dorf and Bishop.
P11.25 The observability matrix is
Po=
C
CA
=
14
21 36
,
and det Po= 48 6= 0; therefore the system is completely observable. The
desired poles of the observer are s1,2=1. This implies that the desired
characteristic polynomial is
pd(s) = s2+ 2s+ 1 .
The actual characteristic polynomial is
det |λI(ALC)|= det
λ1 + L144L1
5 + L2λ10 4L2
=λ2+ (L14L211)λ+ 10L1+ 8L2+ 30 = 0 .
Solving for L1and L2yields
L=
L1
L2
=
0.25
3.3125
.
Checking we find that det(λI(ALC)) = s2+ 2s+ 1. The response of
the estimation error is shown in Figure P11.25, where e(0) = [ 1 1 ]T.
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Problems 623
0
0.5
1
1.5
2
2.5
To: Out(1)
0 1 2 3 4 5 6
-1.5
?-1
-0.5
0
0.5
1
To: Out(2)
Response to Initial Conditions
Time
(
sec
)
Amplitude
FIGURE P11.25
Estimation error response to an initial condition.
P11.26 The observability matrix is
Po=
C
CA
CA2
=
24 0
0 2 4
32 20 14
.
The det Po= 728 6= 0, hence the system is observable. The gain matrix
L=
0.14
0.93
0.79
results in the observer poles at s1,2=1±jand s3=5, as desired.
P11.27 The observability matrix is
Po=
C
CA
=
1 0
1 0
.
The det Po= 0, hence the system is not completely observable. So, we
cannot find an observer gain matrix that places the observer poles at the
desired locations.
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624 CHAPTER 11 The Design of State Variable Feedback Systems
P11.28 Selecting K= 16 yields a zero steady-state error to a unit step input.
The step response is shown in Figure P11.28.
0 0.5 1 1.5 2 2.5
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Step Response
Time (sec)
Amplitude
FIGURE P11.28
Estimation error response to an initial condition.
P11.29 The system transfer function is
Y(s) = 2
s+ 3U(s).
The associated state variable model is
˙x=3x+ 2u
y=x .
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Advanced Problems 625
Advanced Problems
AP11.1 The closed loop system in state-space form is given by
˙x1
˙x2
˙x3
=
0 1 0
01 2
2KK12KK242KK3
x1
x2
x3
+
0
0
2K
u
y=h1 0 0 i
x1
x2
x3
.
The closed-loop transfer function is
T(s) = 4K
s3+ (2KK3+ 5)s2+ (4KK2+ 2KK3+ 4)s+ 4KK1
.
Setting the steady-state error to zero, we determine that
ess = 1 T(0) = 1 1
K1
.
Solving for K1yields
K1= 0.5.
Choosing
K2= 0.5 and K3= 1.5
results in a percent overshoot of P.O. = 2.82%.
AP11.2 A state variable representation is given by
˙
x=Ax +Bu
where
A=
311
4 0 0
0 1 0
,B=
3
0
0
.
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626 CHAPTER 11 The Design of State Variable Feedback Systems
Let
u=Kx .
Then, with
K=h4.00 24.33 39.67 i,
the closed-loop system poles are s=4,5, and 6.
AP11.3 Given
A=
0 1
12
,and B=
b1
b2
,
we compute the determinant of the controllability matrix as
det Pc= det[B AB] = (b1+b2).
The system is controllable if and only if the determinant is non-zero. So,
for the system to be controllable, we require that b26=b1.
AP11.4 Consider the state variable feedback law
u=Kx .
Using Ackermann’s formula, we determine that
K= [14.2045 17.0455 94.0045 31.0455]
results in the closed-loop system characteristic roots at s=2±j,s=5
and s=5.
AP11.5 The closed-loop transfer function for the system is
T(s) = 2Kp
s3+ (9 + 2K3)s2+ (26 + 2K2+ 10K3)s+ (26 + 6K2+ 12K3).
Setting the steady-state error for a step input to zero yields
ess = 1 2Kp
26 + 6K2+ 12K3
= 0 .
Solving for Kpin terms of K2and K3yields
Kp= 13 + 3K2+ 12K3.
Now, choosing
K2= 5
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Advanced Problems 627
K3= 2
results in the closed-loop characteristic roots at
s1=4s2=4s3=5.
Also, the prefilter gain is
Kp= 52 .
AP11.6 (a) A state variable representation is given by
A=
0 1
12
,B=
0
1
,
C=h1 0 i.
Since the determinant of the controllability matrix det[B AB]6= 0,
the system is controllable.
(b) The state variable representation is
˙
x=Ax +Bu ,
or
˙x1
˙x2
=
0 1
12
x1
x2
+
1
1
u.
The determinant of the controllability matrix
det Pc= det[B AB] = 0 .
Therefore, the system is uncontrollable.
AP11.7 The closed-loop transfer function is
T(s) = 120
s3+ (10 + 60K3)s2+ (16 + 120(K3+K2))s+ 120 .
The state feedback gains
K2= 0.283 and K3= 0.15
place the poles at the desired locations. The plot of the roll output for a
unit step disturbance is shown in Figure AP11.7.
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628 CHAPTER 11 The Design of State Variable Feedback Systems
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
00.5 1 1.5 2 2.5 3 3.5 4
Time (secs)
Amplitude
FIGURE AP11.7
Roll angle response to a step disturbance.
AP11.8 The state equations are (using the parameters of P3.36 in Dorf and
Bishop)
˙
h= ˙x1=1
50[80θ50h] = x1+8
5x2
˙
θ= ˙x2=ω=x3
˙ω= ˙x3=Km
Jia=KmKb
JRa
ω+KmKa
JRa
vi=353
30 x3+25000
3vi.
In state variable form we have (without feedback)
˙
x=
18
50
0 0 1
0 0 353
30
x+
0
0
25000
3
vi.
(a) In this case we have vi=kh +ar =kx1+ar, where kand aare
the parameters to be determined and ris the reference input. With
the feedback of h(t) we have
˙
x=
18
50
0 0 1
25000
3k0353
30
x+
0
0
a25000
3
r .
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Advanced Problems 629
Since we only have one parameter to adjust, namely k, we will proba-
bly not be able to simultaneously meet both design specifications, In
fact with
k= 0.00056
we obtain the percent overshoot P.O. = 9.89%. The settling time cri-
terion cannot simultaneously be met—the best that can be obtained
is Ts7.5 seconds. In this case, we choose a= 0.00056 to make the
steady-state value of h(t) = 1.
(b) In this case we have vi=k1hk2θ+ar =k1x1k2x2+ar,
where k1,k2, and aare the parameters to be determined and ris the
reference input. Since we have two parameter to adjust, namely k1
and k2we will probably be able to simultaneously meet both design
specifications. In fact with
k1= 0.00056 and k2= 0.001
we obtain the percent overshoot P.O. = 4.35%. The settling time
criterion is easily met— Ts5 seconds. In this case, we choose a=
0.0012 to make the steady-state value of h(t) = 1.
AP11.9 (a) The state vector differential equation is
˙
x=
0 1 0 0
2 0 1 0
0 0 0 1
1 0 1 0
x+
0
0
0
1
u,
where x1=z,x2= ˙z,x3=yand x4= ˙y.
(b) The characteristic equation is
s4+ 3s2+ 1 = (s+j0.618)(sj0.618)(s+j1.618)(sj1.618) = 0 .
So, the system is oscillatory.
(c) Let u=kx4. Then characteristic equation is
s4+ks3+ 3s2+ 2ks + 1 = 0
which is stable if k > 0.
(d) Rewrite the characteristic equation as
1 + ks(s2+ 2)
s4+ 3s2+ 1 = 0 .
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630 CHAPTER 11 The Design of State Variable Feedback Systems
The root locus is shown in Figure AP11.9. A reasonable solution for
kis k= 1.35.
-3
-2
-1
0
1
2
3
-3 -2 -1 0 1 2 3
x
x
x
x
o
o
o
Real Axis
Imag Axis
FIGURE AP11.9
Root locus for 1 + ks(s2+2)
s4+3s2+1 = 0.
AP11.10 The state differential equation is
¨y=ky +αu
where kand αdepend on the system parameters, such as mass and length.
The transfer function is
y
u=α
s2k
which is unstable at the top of the arc. Since we can only use ˙yfor
feedback, we have
˙y
u=
s2k.
Let
Gc(s) = K1s+K2
s.
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Advanced Problems 631
Then
GGc(s) = α(K1s+K2)
(s2k)
and the closed-loop characteristic equation is
αK1s+αK2+s2k= 0
or
s2+αK1s+αK2k= 0 .
Select αK2k > 0 and αK1>0 for stability.
AP11.11 The state-space representation of the plant is
˙
x=Ax +Bu
y=Cx
where
A=
0 1
23
,B=
0
1
,and C=h1 0 i.
With the intermediate variables defined as
z=˙
xand w= ˙u
we have
˙e
˙
z
=
010
001
023
e
z
+
0
0
1
w
where
e=yr .
To meet the design specifications, we require the closed-loop poles to lie
to the left of the line in the complex plane defined by s=0.8. We choose
K2= [10 3]
and
Gc(s) = 8
s.
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632 CHAPTER 11 The Design of State Variable Feedback Systems
This places the closed-loop poles at s=2,2 and 2. The closed-loop
transfer function with the internal model controller is
T(s) = 8
s3+ 6s2+ 12s+ 8 .
The step response is shown on Figure AP11.11.
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0 2 4 6 8 10 12 14
Time (secs)
Amplitude
FIGURE AP11.11
Internal model controller step response.
AP11.12 The state-space representation of the plant is
˙
x=Ax +Bu
y=Cx
where
A=
0 1
23
,B=
0
1
,and C=h1 0 i.
With the intermediate variables defined as
z=¨
xand w= ¨u
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Advanced Problems 633
we have
˙e
¨e
˙
z
=
0 1 0 0
0 0 1 0
0 0 0 1
0 0 23
e
˙e
z
+
0
0
0
1
w
where e=yr.
0
1
2
3
4
5
6
0 1 2 3 4 5 6
Time (secs)
Amplitude
FIGURE AP11.12
Internal model controller ramp response.
To meet the design specifications, we require the closed-loop poles to lie
to the left of the line in the complex plane defined by s=0.67. We
choose
w=[K1K2K3]
e
˙e
z
=[16 32 22 5]
e
˙e
z
.
Then,
Gc(s) = K1+K2s
s2=16 + 32s
s2.
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634 CHAPTER 11 The Design of State Variable Feedback Systems
The closed-loop transfer function with the internal model controller is
T(s) = 32s+ 16
s4+ 8s3+ 24s2+ 32s+ 16 .
This places the closed-loop poles at s=2,2,2 and 2. The ramp
response is shown in Figure AP11.12.
AP11.13 The controllability matrix is
Pc=
53
1 18
and the observability matrix is
Po=
43
22 44
.
Computing the determinants yields
det Pc=87 6= 0 and det P0= 242 6= 0 ,
hence the system is controllable and observable. The controller gain ma-
trix
K=h3.02 6.11 i
places the closed-loop poles at the desired locations. Similarly, the ob-
server gain matrix
L=
2.38
1.16
places the observer poles at the desired locations.
AP11.14 The controllability matrix is
Pc=
0 0 4
0 4 12
412 24
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Advanced Problems 635
and the observability matrix is
Po=
29 2
16 415
120 29 41
.
Computing the determinants yields
det Pc=64 6= 0 and det P0= 10870 6= 0 ,
hence the system is controllable and observable. The controller gain ma-
trix
K=h0.5 1.25 0.5i
and the observer gain matrix
L=
57.43
16.11
104.43
yields the desired closed-loop system poles and observer poles, respec-
tively.
AP11.15 The state-variable representation of the system is
˙
x=
0 1
72
x+
0
1
u
y= [ 1 4 ]x+ [0]u .
The observability matrix is
P0=
1 4
28 7
,
and det P0= 105 6= 0, hence the system is observable. The observer gain
matrix
L=
7.18
6.29
places the observer poles at s1,2=10 ±10, as desired.
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636 CHAPTER 11 The Design of State Variable Feedback Systems
Design Problems
A state variable representation isCDP11.1
˙
x=
0 1
033.14
x+
0
0.827
va
y=h1 0 ix
where x1=xand x2= ˙x. Note that we are neglecting the motor induc-
tance and assuming that the position x(t) is the output. Assume that we
have available for feedback the angle θand angle rate ˙
θ(see CDP4.1), so
that
va=k1
rx1k2
rx2+au
where u(t) is the reference input (that is, the desired position x(t)), the
gains k1and k2and the scaling parameter aare to be determined. Recall
that
x=rθ = 0.03175θ .
With the feedback in the loop we have
˙
x=
0 1
26.03k133.14 26.03k2
x+
0
0.827a
u
y=h1 0 ix
Choosing k1= 50, k2= 1 and a= 1574.1 results in
P.O. = 1.1% and Ts= 0.11 second .
The closed-loop poles are s1,2=29.59 ±20.65j.
DP11.1 The governing differential equation is
¨y2000y=20i .
In state variable form, the system is described by
˙
x=
0 1
2000 0
x+
0
20
i .
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Design Problems 637
Consider the state feedback
i=k1x1k2x2+βr
where r(t) is the reference input and k1, k2and βare to be determined.
Then, the closed-loop system is
˙
x=
0 1
2000 20k120k2
x+
0
20β
r .
The characteristic equation is
s2+ 20k2s2000 + 20k1= 0 .
For stability, let 20k12000 >0. Select k1= 125. Then, ωn= 22.36
rad/sec, and
k2=2ζωn
20 .
Let ζ= 0.59 to meet 10% overshoot specification. Thus,
k2=2(0.59)(22.36)
20 = 1.32 .
The closed-loop transfer function is
T(s) = 20β
s2+ 26.4s+ 500 .
Choose β=25 so that
T(s) = 500
s2+ 26.4s+ 500 .
The feedback law is
i= 125x1+ 1.32x225r .
DP11.2 The automobile engine control system (see DP10.8 in Dorf and Bishop)
is modeled as
G(s) = 2esT
(0.21s+ 1)(4s+ 1) .
In this case, we will assume the delay is negligible. Therefore, T= 0. A
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638 CHAPTER 11 The Design of State Variable Feedback Systems
state variable representation of the system is
˙
x=
0 1
1.19 5.01
x+
0
1.19
r
y= [1 0]x.
Let
r(t) = k1x1k2x2+k3u
where u(t) is the command input. Using ITAE methods, our desired char-
acteristic polynomial is
q(s) = s2+ 1.4ωns+ω2
n= 0 .
Select ωn= 11.315 to obtain a settling time Ts<0.5 seconds. The char-
acteristic polynomial of the closed-loop system is
s2+ (5.01 + 1.19k2)s+ (1.19 + 1.19k1) = 0 .
Equating coefficients and solving for the gains yields
k1= 106.59 and k2= 9.235 .
Select k3= 107.59 to yield a zero steady-state error to a step input.
0
0.2
0.4
0.6
0.8
1
1.2
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7
Time (secs)
Amplitude
FIGURE DP11.2
The step response of the engine control system.
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Design Problems 639
DP11.3 The compensator is
˙
ˆ
x= [ABK LC]ˆ
x+Ly+Mr
˜u=Kˆ
x
where
ABK LC =
28.7 1
365.19 20
,M=
0
200
,
N= 363.64 ,K=h344.55 15.82 i,and L=
28.7
165.19
.
We selected the desired eigenvalues of ABK at p=10 ±10jand the
desired eigenvalues of ALC at q=20 ±10j. For initial conditions we
let x(0) = [1 1] and ˆ
x(0) = [0 0].
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
0
0.5
1
1.5
Time (s)
x1
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
−2
0
2
4
6
Time (s)
x2
Actual x1
Actual x2
Estimated x1
Estimated x2
FIGURE DP11.3
The step response showing the actual and estimated states.
DP11.4 The design specifications are
(a) Percent overshoot <20%
(b) Ts<1.5s, and
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640 CHAPTER 11 The Design of State Variable Feedback Systems
(c) steady-state error less than 20% of the input magnitude.
The state differential equation is
˙
x=Ax +Bu
y=Cx
where
A=
0 1 0
0σ1α1
gα2σ2
=
0 1 0
00.415 0.0111
9.81.43 0.0198
,
B=
0
n
g
=
0
6.27
9.8
and C=h1 0 0 i.
The transfer function is
θ(s)
δ(s)=ns +2α1g
s3+ (σ1+σ2)s2+ (σ1σ2α1α2)s+α1g
=6.27s+ 0.0154
s3+ 0.435s20.0077 + 0.109 .
Let u=K1x1K2x2K3x3. Then the closed-loop system matrix is
ABK =
0 1 0
nK1σ1nK2α1nK3
ggK1α2gK2σ2gK3
,
where K= [K1K2K3]. From the design specifications, we have the
desired roots at
s3+a2s2+a1s+ao=s3+36s2+225s+1350 = (s+30)(s+3+j6)(s+3j6) = 0 .
The actual characteristic equation is
s3+ (gK3+K2n+σ1+σ2)s2+ (α1α2α1gK2+K1nα2nK3
+gK3σ1+K22+σ1σ2)s
+α1gα1gK1+gK3n+σ2nK1= 0 .
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Design Problems 641
Comparing coefficients yields
0n g
n nσ2α1gα2n+gσ1
α1g+σ2n0gn
K1
K2
K3
=
a2σ1σ2
a1+α1α2σ1σ2
a0α1g
where
a2= 36
a1= 225
a0= 1350 .
The solution for Kis
K= [53.11 28.64 21.96] .
DP11.5 The controllability and observability matrices are
Pc=
0.05 0.04
0.001 0.001
and
P0=
1 0
0.8 0.02
,respectively.
Computing the determinants yields
det Pc=1.002e05 6= 0 and Po= 0.02 6= 0 ,
hence the system is controllable and observable. The feedback gain matrix
K= [ 3820 179620 ]
yields the desired closed-loop poles. The observer gain matrix
L=
120
180000
yields the desired observer poles. The integrated system is shown in Fig-
ure DP11.5.
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642 CHAPTER 11 The Design of State Variable Feedback Systems
+
-
System Model
x=Ax+Bu
.C
xy
Observer
Control Law
-K x=Ax+Bu+Ly
.
^ ^ ~y=y-Cx
^
~
C
x
^
u
A= -0.8 0.02
-0.02 0 B= 0.05
0.001
C= 1 0
K= 3820 -179620
L= 120
180000
FIGURE DP11.5
Integrated controller and observer.
DP11.6 (a) The characteristic equation associated with the system matrix is
q(s) = s2+ (12 + K2)s+ (36 + K1) = 0 ,
where we have assumed state feedback of the form
u=K1x1K2x2.
The deadbeat control characteristic equation is
s2+αωns+ω2
n= 0 ,
where α= 1.82 and we use ωn= 9.64 to meet the settling time
specification. Then, equating coefficients and solving for the gains
yields
K1= 56.93 and K2= 5.54 .
(b) Since the closed-loop poles are located at s1,2=8.77 ±4, we can
select the observer poles to be about ten times farther in the left-half
plane, or
s1,2=88,88 .
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Design Problems 643
Then the observer gains are
L=
164
5740
.
(c) The block diagram is shown in Figure DP11.6.
+
-
System Model
x=Ax+Bu
.C
xy
Observer
Control Law
-K x=Ax+Bu+Ly
.
^ ^ ~y=y-Cx
^
~
C
x
^
u
A= 0 1
-36 -12 B= 0
1
C= 1 0
K= 56.93 5.54
L= 164
5740
FIGURE DP11.6
Block diagram for integrated controller and observer.
DP11.7 The compensator is
˙
ˆ
x= [ALC]ˆ
x+Ly+Bu
u=Kˆ
x
where
ALC =
60 1 0
1095 0 1
3750 510
,
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644 CHAPTER 11 The Design of State Variable Feedback Systems
N= 4000 ,K=h3998 595 30 i,and L=
60
1095
3748
.
We selected the desired eigenvalues of ABK at p1,2=10 ±10j,
p3=20 and the desired eigenvalues of ALC at q1,2=20 ±10j,
q3=30. For initial conditions we let x(0) = [1 1 1] and ˆ
x(0) = [0 0 0].
The transfer function from rto yis
T(s) = 4000s3+ 2.8e05s2+ 6.8e06s+ 6e07
s6+ 110s5+ 5100s4+ 1.29e05s3+ 1.9e06s2+ 1.58e07s+ 6e07 .
The bandwidth is 11.7 rad/s.
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
0
0.5
1
1.5
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
−10
0
10
20
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
−200
−100
0
100
200
FIGURE DP11.7
The step response showing the actual and estimated states.
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Computer Problems 645
Computer Problems
CP11.1 The controllability and observablity matrices have nonzero determinants,
as shown in Figure CP11.1. Therefore, the system is observable and con-
trollable.
>>
A=[-6 2 0;4 0 7;-10 1 11]; b=[5;0;1]; c=[ 1 2 1]; d=[0];
sys = ss(A,b,c,d);
Co=ctrb(sys); dt_Co=det(Co)
Ob=obsv(sys); dt_Ob=det(Ob)
dt_Co =
-84933
dt_Ob =
-3.6030e+03
FIGURE CP11.1
Determining controllability and observability.
CP11.2 The system is controllable since the determinant of the controllability
matrix is nonzero , as shown in Figure CP11.2.
a=[0 1;-6 -5]; b=[0;6]; c=[1 0]; d=[0];
sys_ss = ss(a,b,c,d);
Pc=ctrb(sys_ss);
dt_Pc=det(Pc)
Ob=obsv(sys_ss);
dt_Ob=det(Ob)
sys_tf=tf(sys_ss)
dt_Pc =
-36
dt_Ob =
1
Transfer function:
6
-------------
s^2 + 5 s + 6
FIGURE CP11.2
M-file script to determine controllability and to compute equivalent transfer function model.
CP11.3 The gain matrix (computed as shown in Figure CP11.3) is K=h0.5 0.5i.
a=[0 1;-1 -2]; b=[1;1]; c=[1 -1]; d=[0];
p=[-1;-2];
K=acker(a,b,p)K =
0.5000 0.5000
FIGURE CP11.3
M-file script to place the closed-loop system poles using state feedback.
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646 CHAPTER 11 The Design of State Variable Feedback Systems
CP11.4 The constant velocity guided missile is not controllable since the control-
lablity matrix, Co, has a zero determinant, as shown in Figure CP11.4.
Using the tf function (see Figure CP11.4), we determine that the transfer
function is
G(s) = 5s
s5+ 0.5s4+ 0.1s3.
Cancelling common terms in the transfer function yields
G(s) = 5
s4+ 0.5s3+ 0.1s2.
Then, using the ss function, we determine a state-space representation of
G(s). As shown in Figure CP11.4, the state-space representation is
˙
x=Ax +Bu
y=Cx
A=[0 1 0 0 0;-0.1 -0.5 0 0 0;0.5 0 0 0 0;0 0 10 0 0;0.5 1 0 0 0];
b=[0;1;0;0;0];
c=[0 0 0 1 0];
d=[0];
sys_ss = ss(A,b,c,d);
% Part (a)
Co=ctrb(sys_ss); dt_Co=det(Co)
% Part (b)
sys_tf = tf(sys_ss)
sys_new = minreal(sys_tf);
sys_new_ss=ss(sys_new)
% Part (c)
Co_new=ctrb(sys_new_ss); dt_Co_new=det(Co_new)
% Part (d)
evalues=eig(sys_new_ss)
dt_Co_new =
32
evalues =
0
0
-0.2500 + 0.1936i
-0.2500 - 0.1936i
a =
x1 x2 x3 x4
x1 -0.50000 -0.10000 0 0
x2 1.00000 0 0 0
x3 0 1.00000 0 0
x4 0 0 2.00000 0
b =
u1
x1 2.00000
x2 0
x3 0
x4 0
c =
x1 x2 x3 x4
y1 0 0 0 1.25000
d =
u1
y1 0
Continuous-time system.
Transfer function:
5 s
-----------------------
s^5 + 0.5 s^4 + 0.1 s^3
dt_Co =
0
FIGURE CP11.4
Analysis of the constant velocity guided missile state-space model.
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Computer Problems 647
where
A=
0.50.100
1 0 0 0
0 1 0 0
0 0 2 0
B=
2
0
0
0
and C=h0 0 0 1.25 i.
The reduced system is controllable but not stable, since there are two poles
at the origin. Systems that are not controllable have too many states.
After eliminating unnecessary states, a controllable system of minimal
complexity (i.e. states) is obtained. In this case, the number of states is
reduced from five to four.
CP11.5 The eigenvalues of Aare
e1=2.0727
e2=0.2354
e3,4= 0.2761 ±0.2593j
The system is unstable since there are two eigenvalues in the right half-
plane, see Figure CP11.5. The characteristic polynomial is
A = [-0.0389 0.0271 0.0188 -0.4555; 0.0482 -1.0100 0.0019 -4.0208;
0.1024 0.3681 -0.7070 1.4200; 0 0 1 0];
b1 = [0.4422;3.5446;-6.0214;0];
b2 = [0.1291;-7.5922;4.4900;0];
% Part (a)
evalues = eig(A)
%part (b)
p = poly(A)
r = roots(p)
% Part (c)
Co1 = ctrb(A,b1); dt1 = det(Co1)
Co2 = ctrb(A,b2); dt2 = det(Co2)
>>
evalues =
0.2761 + 0.2593i
0.2761 - 0.2593i
-0.2354
-2.0727
p =
r =
1.0000 1.7559 -0.6431 0.0618 0.0700 -2.0727
0.2761 + 0.2593i
0.2761 - 0.2593i
-0.2354
dt1 =
-1.8451e+03
dt2 =
-90.6354
FIGURE CP11.5
Analysis of the VTOL aircraft model.
p(s) = s4+ 1.7559s30.6431s2+ 0.0618s+ 0.0700 .
The roots of the characteristic equation are the same as the eigenvalues.
Also, the system is controllable from either u1or u2. If the aircraft should
lose the control of the vertical motion through u1, then the control u2can
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648 CHAPTER 11 The Design of State Variable Feedback Systems
be used to control both vertical and horizontal motion, and vice versa.
CP11.6 The m-file script to analyze the translunar halo orbit problem is shown
in Figure CP11.6. The translunar equilibrium point is not a stable point
dt1 =
0
dt2 =
0
dt3 =
0
A=[0 0 0 1 0 0;0 0 0 0 1 0; 0 0 0 0 0 1;7.3809 0 0 0 2 0;
0 -2.1904 0 -2 0 0; 0 0 -3.1904 0 0 0];
c=[0 1 0 0 0 0];d=[0];
b1=[0;0;0;1;0;0];
b2=[0;0;0;0;1;0];
b3=[0;0;0;0;0;1];
sys_ss_1 = ss(A,b1,c,d);
sys_ss_2 = ss(A,b2,c,d);
sys_ss_3 = ss(A,b3,c,d);
% Part (a)
evalues=eig(A)
% Part (b)
Cb1=ctrb(sys_ss_1); dt1=det(Cb1)
% Part (c)
Cb2=ctrb(sys_ss_2); dt2=det(Cb2)
% Part (d)
Cb3=ctrb(sys_ss_3); dt3=det(Cb3)
% Part (e)
sys_tf = tf(sys_ss_2);
sys_tf=minreal(sys_tf)
% Part (f)
sys_ss=ss(sys_tf);
Co=ctrb(sys_ss); dt_Co=det(Co)
if dt_Co ~= 0
disp('System is completelly Controllable')
else
disp('System in uncontrollable')
end
% Part (g)
P = [-1+i; -1-i;-10;-10];
[A,B]=ssdata(sys_ss);
K = acker(A,B,P)
evalues =
2.1587
-2.1587
0 + 1.8626i
0 - 1.8626i
0 + 1.7862i
0 - 1.7862i
Transfer function:
s^2 - 7.381
----------------------------
s^4 - 1.19 s^2 - 16.17
dt_Co =
64
System is completelly Controllable
a =
x1 x2 x3 x4
x1 0 0.59525 0 2.02089
x2 2.00000 0 0 0
x3 0 2.00000 0 0
x4 0 0 2.00000 0
b =
u1
x1 1.00000
x2 0
x3 0
x4 0
c =
x1 x2 x3 x4
y1 0 0.50000 0 -0.92261
d =
u1
y1 0
FIGURE CP11.6
Analysis of the translunar satellite halo orbit.
as evidenced by the eigenvalues of Ain the right half-plane; the system is
not completely controllable from any uiindividually. The transfer function
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Computer Problems 649
from u2to ηis
T(s) = s44.191s223.55
s6+ 2s419.97s251.58 .
A careful analysis reveals that T(s) can be reduced by eliminating com-
mon factors. The common factors are s2+ 3.1834. The reduced transfer
function is
T(s) = s27.3815
s41.1837s216.2030 .
Using state feedback
u2=Kx
the gain matrix Kwhich places the desired poles (using Ackermann’s
formula) is
K=h22 71.56 60 27.02 i.
CP11.7 The m-file script to determine the initial state is shown in Figure CP11.7a.
Given three data points at t= 0,2,4, we construct the three equations
A=[0 1 0;0 0 1;-2 -4 -6];
b=[0;0;0]; c=[1 0 0]; d=[0];
sys=ss(A,b,c,d);
%
% Part (b)
v1=c*expm(0*A); v2=c*expm(2*A); v3=c*expm(4*A);
V=[v1;v2;v3]; Vi=inv(V);
n=[1;-0.0256;-0.2522];
x0=Vi*n
%
% Part (c)
t=[0:0.1:4]; u=0.0*t;
[y,x]=lsim(sys,u,t,x0');
plot(t,y,[0 2 4],[1;-0.0256;-0.2522],'*'), grid
xlabel('Time (sec)'), ylabel('y(t)')
title('Data points denoted by *')
FIGURE CP11.7
(a) Script to determine the initial state from three observations.
y(0) = 1 = Ce0Ax0
y(2) = 0.0256 = Ce2Ax0
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650 CHAPTER 11 The Design of State Variable Feedback Systems
y(4) = 0.2522 = Ce4Ax0
or, in matrix form
Ce0A
Ce2A
Ce4A
x0=
1
0.0256
0.2522
.
The problem is solvable if the matrix
Ce0A
Ce2A
Ce4A
is invertible. In this case, the inverse does exist and the solution is
x0=
1
1
1.9998
.
The simulation is shown in Figure CP11.7b.
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
00.5 1 1.5 2 2.5 3 3.5 4
*
*
*
Time (sec)
y(t)
Data points denoted by *
FIGURE CP11.7
CONTINUED: (b) System simulation using computed initial state.
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Computer Problems 651
CP11.8 Suppose we are given
A=
0 1
1 0
B=
0
1
and the feedback
u=Kx =[K1K2]x.
Solving HTP+PH =Ifor Pyields
p11 =K2
2+K2
1+ 3K1+ 2
2(K1+ 1)K2
p12 =1
2(K1+ 1)
p22 =K1+ 2
2(K1+ 1)K2
Then, with xoT= [1,0] we find that
J=xoTPxo=p11 .
Computing the partial of Jwith respect to K2yields
J
K2
=1
21
K1+ 1 K1+ 2
K2
2.
Setting
J
K2
= 0
and solving for K2, we find that
K2=q(K1+ 2)(K1+ 1) .
For a given value of K1, the value of K2that minimizes Jcan be computed
via the above equation. With K2given as above, we can compute Jto be
J=sK1+ 2
K1+ 1 .
A plot of Jversus K1(with K2equal to the minimizing value) is shown
in Figure CP11.8. As K1increases, the performance index Jdecreases.
However, we see that the rate of decrease slows considerably after K1>
20. Also, K2increases as K1increases. We want to keep both gains as
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652 CHAPTER 11 The Design of State Variable Feedback Systems
small as possible, while still having a small J. A reasonable selection is
K1= 20 and K2= 21.5.
1
1.1
1.2
1.3
1.4
1.5
0 5 10 15 20 25 30 35 40 45 50
K1
J
Performance index J versus K1
0
20
40
60
0 5 10 15 20 25 30 35 40 45 50
K1
K2
K2 versus K1
FIGURE CP11.8
Performance index as a function of K1and K2.
CP11.9 In this problem, A=1 and B= 1. Computing Qyields
Q= (1 + λ(k)2) = 1 + λk2.
Define
H=ABk =1k .
Solving
HTP+P H =Q
yields
p=1 + λk2
2(k+ 1) .
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Computer Problems 653
0.41
0.42
0.43
0.44
0.45
0.46
0.47
0.48
0.49
0.5
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
k
J/x0^2
0
0.5
1
1.5
2
2.5
0 1 2 3 4 5 6 7 8 9 10
lambda
k min
FIGURE CP11.9
Plot of J/x2
0versus kand the minimizing kversus λ.
The performance index is
J=x2
0pwhich implies J/x2
0=1 + λk2
2(k+ 1) .
The plot of J/x2
0versus kis shown in Figure CP11.9. The minimum value
is achieved when k= 0.41. To arrive at this result analytically, take the
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654 CHAPTER 11 The Design of State Variable Feedback Systems
partial of J/x2
0with respect to k, set the result to zero and solve for k:
J/x2
0
k = 0 when k2+ 2K1= 0 .
Solving for kyields k=1±p1 + 1. So, when λ= 1, k= 0.41. The
plot of kmin versus λis shown in Figure CP11.9.
CP11.10 The m-file is shown in Figure CP11.10.
A=[0 1;-18.7 -10.4]; B=[10.1; 24.6]; C=[1 0]; D=[0];
% Controller Gains
p=[-2;-2 ];
K=acker(A,B,p)
% Observer Gains
q=[-20+4*j;-20-4*j];
L = acker(A',C',q); L=L
>>
K =
-0.3081 -0.1337
L =
29.6000
89.4600
FIGURE CP11.10
Using the acker function to compute the controller gains and the observer gains.
CP11.11 The m-file is shown in Figure CP11.11(a). The compensator can be rep-
A=[0 1 0;0 0 1;-4.3 -1.7 -6.7]; B=[0;0;0.35]; C=[0 1 0]; D=[0];
% Controller Gains
p=[-1.4+1.4*j;-1.4-1.4*j;-2];
K=acker(A,B,p)
% Observer Gains
q=[-18+5*j;-18-5*j;-20];
L = acker(A',C', q); L=L'
% Simulation of closed-loop system with the observer
Ac=[A -B*K;L*C A-B*K-L*C];
Bc=[zeros(6,1)];
Cc=eye(6);
Dc=zeros(6,1);
sys=ss(Ac,Bc,Cc,Dc);
x0=[1;0;0;0.5;0.1;0.1]; t=[0:0.001:3.5];
[y,t]=initial(sys,x0,t);
subplot(311)
plot(t,y(:,1),t,y(:,4),'--'), grid
subplot(312)
plot(t,y(:,2),t,y(:,5),'--'), grid
subplot(313)
plot(t,y(:,3),t,y(:,6),'--'), grid
>>
K =
10.1143 22.3429 -5.4286
L =
1.0e+003 *
-1.6223
0.0493
0.7370
FIGURE CP11.11
(a) M-file.
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Computer Problems 655
resented as
˙
ˆ
x= (ABK LC)ˆ
x+Lyand u=Kˆ
x.
Since y=Cx, we can write
˙
ˆ
x= (ABK LC)ˆ
x+LCx .
Similarly, with
˙
x=Ax +Buand u = Kˆ
x
we obtain
˙
x=Ax BKˆ
x.
In matrix form, we have
˙
x
˙
ˆ
x
=
ABK
LC A BK LC
x
ˆ
x
,
with initial conditions
hx(0)Tˆ
x(0)TiT=h1 0 0 0.5 0.1 0.1iT.
The response of the system is shown in Figure CP11.11(b).
0 1 2 3 4
5
0
5
0 1 2 3 4
1
0
1
0 1 2 3 4
2
0
2
Estimated state (dashed line)
True state
(solid line)
x1
x2
x3
Time (sec)
FIGURE CP11.11
CONTINUED: (b) Response of system to an initial condition.
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656 CHAPTER 11 The Design of State Variable Feedback Systems
CP11.12 The Simulink block diagram is shown in Figure CP11.12.
FIGURE CP11.12
Simulink block diagram.
CP11.13 The m-file to design the compensator is shown in Figure CP11.13(a). The
Simulink simulation is shown in Figure CP11.13(b). The output shown
on the x-y graph depicts the state xof the system. The initial conditions
selected for the simulation are
x(0) =
1
0
0
0
and ˆ
x(0) =
0.5
0.1
0.1
0.1
.
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Computer Problems 657
0 2 4 6 8 10
?100
0
100
0 2 4 6 8 10
?2
0
2
0 2 4 6 8 10
?10
0
10
A=[0 1 0 0;0 0 1 0;0 0 0 1;-2 -5 -1 -13]; B=[0;0;0;1]; C=[1 0 0 0]; D=[0];
% Controller Gains
p=[-1.4+1.4*j;-1.4-1.4*j;-2+j;-2-j];
K=acker(A,B,p)
% Observer Gains
q=[-18+5*j;-18-5*j;-20;-20];
L = acker(A',C',q); L=L'
% Simulation of closed-loop system with the observer
Ac=[A -B*K;L*C A-B*K-L*C];
Bc=[zeros(8,1)];
Cc=eye(8);
Dc=zeros(8,1);
sys=ss(Ac,Bc,Cc,Dc);
x0=[1;0;0;0;0.5;0.1;0.1;0.1]; t=[0:0.001:10];
[y,t]=initial(sys,x0,t);
subplot(311)
plot(t,y(:,1),t,y(:,4),'--'), grid
subplot(312)
plot(t,y(:,2),t,y(:,5),'--'), grid
subplot(313)
plot(t,y(:,3),t,y(:,6),'--'), grid
>>
K =
17.6000 24.6800 19.1200 -6.2000
L =
63
1369
10495
1479
FIGURE CP11.13
(a) M-file to design the compensator, including the observer.
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658 CHAPTER 11 The Design of State Variable Feedback Systems
x=[A-BK-LC]x+Ly
.
^ ^
u=-Kx
^
FIGURE CP11.13
CONTINUED (b) The Simulink simulation.
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C H A P T E R 1 2
Robust Control Systems
Exercises
E12.1 The plant transfer function is
G(s) = 3
s+ 3 .
Try a PI controller, given by
Gc=K1+K2
s.
The ITAE characteristic equation is
s2+ 1.4ωns+ω2
n,
where ωn= 30. Then
K1= 13 and K2= 300 .
Without a prefilter, the closed-loop system is
Y(s)
R(s)=39s+ 900
s2+ 42s+ 900 ,
and with a prefilter, the closed-loop system is
Y(s)
R(s)=900
s2+ 42s+ 900 ,
where
Gp(s) = 23.07
s+ 23.07 .
The step response, with and without the prefilter, is shown in Figure E12.1.
659
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660 CHAPTER 12 Robust Control Systems
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5
0
0.2
0.4
0.6
0.8
1
1.2
1.4
Time (sec)
y(t)
Without prefilter
With prefilter
FIGURE E12.1
Step response: (a) w/o prefilter (solid line), and (b) w/prefilter (dashed line).
E12.2 The disturbance response is shown in Figure E12.2.
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5
−0.01
0
0.01
0.02
0.03
0.04
0.05
Time (sec)
y(t)
FIGURE E12.2
Disturbance response for system in E12.1.
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Exercises 661
E12.3 The closed-loop transfer function is
T(s) = 25
s2+bs + 25 ,
and the sensitivity function is
S(s) = s2+bs
s2+bs + 25 ,
where b= 8, nominally. The sensitivity of Tto changes in bis determined
to be
ST
b=T
b
b
T=bs
s2+bs + 25 .
The plot of T(s) and S(s) is shown in Figure E12.3, where b= 8.
10−1 100101102
−60
−50
−40
−30
−20
−10
0
10
Frequency (rad/sec)
Gain dB
20log|T|
20log|S|
FIGURE E12.3
Plot of T(s)and the sensitivity function S(s).
E12.4 The plant transfer function is
G(s) = 1
(s+ 20)(s+ 36) ,
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662 CHAPTER 12 Robust Control Systems
and the PID controller is given by
Gc(s) = K3(s+a)(s+b)
s.
Let a=20, b=500, and K3= 200. Then, the closed-loop system is
T(s) = 200s2+ 4000s+ 100000
s3+ 256s2+ 4720s+ 100000 .
The closed-loop poles are s1=-237.93 and s2,3=9.04 ±j18.5 and the
zeros are s1,2=10±j20. Therefore, there is an approximate cancellation
of the complex poles and zeros and the approximate system is
ˆ
T(s) = 238
s+ 238 .
The actual response and approximation are shown in Figure E12.4.
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5
0
0.2
0.4
0.6
0.8
1
1.2
1.4
Time (sec)
y(t)
approximation
actual
FIGURE E12.4
Step response for closed-loop actual and approximate transfer functions.
E12.5 The loop transfer function is
L(s) = Gc(s)G(s) = 10KD(s+KP/KD)
s(s+ 3)(s+ 10) .
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Exercises 663
Select KP/KD= 10. Then
L(s) = Gc(s)G(s) = 10KD
s(s+ 3) ,
and the closed-loop transfer function is
T(s) = 10KD
s2+ 3s+ 10KD
.
Let ζ= 0.69, which implies P.O. < 5%. Also, 2ζωn= 3, so ωn= 2.17.
Thus,
10KD=ω2
n= 4.72 .
Thus, the controller is Gc(s) = 0.47(s+ 10). The settling time is Ts= 2.8
s and the percent overshoot is P.O. = 4.6%. As Kincreases, the per-
cent overshoot increases from 0% to 16% and the settling time generally
decreases from 3.8 sec to 2.6 sec.
E12.6 The loop transfer function with the PID controller is
Gc(s)Gs(s) = KDs2+KPs+KI
s
1
(s+ 5)2.
The ITAE step response requires
s3+ 1.75ωns2+ 2.15ω2
ns+ω3
n=s3+ (10 + KD)s2+ (25 + KP)s+KI.
For n= 3 we estimate the normalized settling time to be
ωnTs8 seconds.
Thus, ωn6, and
KD= 0.5, KP= 52.4,and KI= 216.
The step response is shown in Figure E12.6. The transfer function from
the disturbance to the output is
Y(s)
Td(s)=G(s)
1 + Gc(s)G(s)=s
s3+ 10.5s2+ 77.4s+ 216 .
The disturbance response is shown in Figure E12.6. The system is effective
in reducing the effects of the disturbance, and the maximum output is
reduced by 1/100 for a step disturbance.
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664 CHAPTER 12 Robust Control Systems
0 0.5 1 1.5 2
0
0.2
0.4
0.6
0.8
1
1.2
1.4
TIme (s)
(a) Step response
y(t)
0 0.5 1 1.5 2
−2
0
2
4
6
8
10
12 x 10−3
TIme (s)
(b) Disturbance response
y(t)
FIGURE E12.6
(a) Step response: w/o prefilter (solid line) and w/prefilter (dashed line); and (b) distur-
bance response.
E12.7 The plant transfer function is
G(s) = 1
(s+ 4)2,
and the PID controller is
Gc(s) = K1s+K2+K3s2
s.
Using the ITAE criteria and selecting ωn= 10 yields
K3= 9.5K2= 1000 and K1= 199 .
The step response is shown in Figure E12.7. The disturbance response is
also shown in Figure E12.7. The maximum y(t) = 0.0041, so the system
is effective in rejecting the step disturbance.
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Exercises 665
0
0.2
0.4
0.6
0.8
1
1.2
1.4
00.5 1 1.5 2
Time (sec)
y(t)
(a) step response
-0.5
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
x10-3
00.5 1 1.5 2
Time (sec)
y(t)
(b) disturbance
FIGURE E12.7
(a) Step response: w/o prefilter (solid line) and w/prefilter (dashed line); and (b) distur-
bance response.
E12.8 The maximum ωn= 60. Then K1= 3600 and K2= 80. The maximum
control input is max |u(t)| ≈ 80. The plot of the step response and the
control input u(t) is shown in Figure E12.8.
0
0.2
0.4
0.6
0.8
1
1.2
0 0.05 0.1 0.15 0.2
Time (sec)
y(t)
(a) step response
-10
0
10
20
30
40
50
60
70
80
90
0 0.05 0.1 0.15 0.2
Time (sec)
u(t)
(b) control input u(t)
FIGURE E12.8
Step response w/o prefilter; and (b) control input u(t).
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666 CHAPTER 12 Robust Control Systems
E12.9 One possible PD controller is
Gc(s) = 27.6s+ 8.25s .
When
K= 1 ,
the system roots are
s1,2=3.2±j4.3
s3=9.5.
The step response is shown in Figure E12.9 for K= 0.5,1, and 1.5.
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2
0
0.2
0.4
0.6
0.8
1
1.2
1.4
Time (sec)
y(t)
K=1 (solid); K=0.5 (dashed); and K=1.5 (dotted)
FIGURE E12.9
Step response for K= 0.5,1, and 1.5.
E12.10 One possible PI controller is
Gc(s) = 2.2s+ 22
s.
When K= 1, the system roots are s1,2=1.31 ±j1.31, and s3=6.37.
The step response is shown in Figure E12.10.
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Exercises 667
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
0
0.2
0.4
0.6
0.8
1
1.2
1.4
Time (sec)
y(t)
K=1 (solid); K=0.5 (dashed); and K=1.5 (dotted)
FIGURE E12.10
Step response for K= 0.5,1, and 1.5.
E12.11 The plot is shown in Figure E12.11.
0 0.5 1 1.5 2 2.5 3 3.5 4
0
10
20
30
40
50
60
70
80
90
100
Time (sec)
P.O. (%)
FIGURE E12.11
Percent overshoot as a function of kin the interval 0.1k4.
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668 CHAPTER 12 Robust Control Systems
E12.12 The controllability matrix is
Pc=
c1c2
c2ac1bc2
and
det Pc=c2
2+ [bc1]c2+ac2
1.
For controllability we require det Pc6= 0, hence
c2
2+ [bc1]c2+ac2
16= 0
implies
c2
c16=b
2±q(b/2)2a
where (b/2)2a0. For real-valued c1and c2, if (b/2)2a < 0, all real
values of c1and c2are valid. Valid values of the constants are c1= 0,
c2= 10, a= 10, and b= 3. The step response is shown in Figure E12.12.
0 0.5 1 1.5 2 2.5 3 3.5 4
0
0.2
0.4
0.6
0.8
1
1.2
1.4
Step Response
Time (sec)
Amplitude
FIGURE E12.12
Step response with c1= 0,c2= 10,a= 10, and b= 3.
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Problems 669
Problems
P12.1 The closed-loop transfer function is
T(s) = 4(s+ 2)
s2+ 4s+ 8
and the sensitivity function is
S(s) = s2
s2+ 4s+ 8 .
The plot of 20 log |T|and 20 log |S|is shown in Figure P12.1. The band-
width is
ωB= 6.31 rad/sec .
Then
|ST
K|ωB= 0.98
|ST
K|ωB
2= 0.78
|ST
K|ωB
4= 0.30 .
20log|T|
20log|S|
10-1 100101
-60
-50
-40
-30
-20
-10
0
10
Frequency (rad/sec)
Gain dB
FIGURE P12.1
Plot of T(s)and the sensitivity function S(s).
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670 CHAPTER 12 Robust Control Systems
P12.2 (a) The loop transfer function is given by
Gc(s)G(s) = K
s(0.02s+ 1)(0.002s+ 1) .
When
K= 100 ,
the peak magnitude is
Mpω= 1.84 .
(b) The plot of 20 log |T|and 20 log |S|is shown in Figure P12.2a.
-120
-100
-80
-60
-40
-20
0
20
101102103104
20log|T|
20log|S|
Frequency (rad/sec)
Gain dB
FIGURE P12.2
(a) Plot of T(s)and the sensitivity function S(s).
(c) The bandwidth is
ωB= 117 rad/sec ,
and
|ST
K|ωB= 1.47
|ST
K|ωB
4= 0.39
|ST
K|ωB
2= 1.62 .
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Problems 671
(c) The disturbance response is shown in Figure P12.2b.
0
1
2
3
4
5
6
7
8
x10-7
0 0.05 0.1 0.15 0.2 0.25 0.3
Time (secs)
Amplitude
FIGURE P12.2
CONTINUED: (b) Disturbance response for K= 100.
P12.3 (a) The loop transfer function is
L(s) = Gc(s)G(s) = K(s4)(s1)
(s+ 0.02)(s+ 2)2.
The characteristic equation is
1 + Gc(s)G(s) = 1 + K(s4)(s1)
(s+ 0.02)(s+ 2)2= 0
or
s3+ (4.02 + K)s2+ (4.08 5K)s+ 0.08 + 4K= 0 .
Using Routh-Hurwitz we find that the system is stable for
4.6987 < K < 0.6947 .
(b) The steady-state error is
ess =1
1 + 50K.
Select K= 0.18 to obtain a steady-state error to a unit step of 0.1.
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672 CHAPTER 12 Robust Control Systems
(c,d) The plots of y(t) for
K= 0.18 (nominal)
K= 0.21 (+15%)
K= 0.15 (15%)
are shown in Figure P12.3.
0 5 10 15 20 25
−0.2
0
0.2
0.4
0.6
0.8
1
1.2
Time (sec)
y(t)
K=0.18 (solid) & K=0.21 (dashed) & K=0.15 (dotted)
FIGURE P12.3
Step input response for K= 0.18,K= 0.21 and K= 0.15.
P12.4 (a) The plant is given by
G=1
ss
25 + 1.
We desire P.O. < 10% and Ts<100 ms. Using a PD controller
Gc(s) = 100 + 2.2s ,
we determine that P.O. = 7%, Ts<100 ms and ess =A
100 for a ramp
input. The plot of y(t) is shown in Figure P12.4.
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Problems 673
(b) The sensitivity is
|Sr
K1|= 27.95
when K1= 1.
(c) The plot of y(t) when K1= 2 (the compensator Gc(s) is unchanged)
is shown in Figure P12.4.
(d) The disturbance response is shown in Figure P12.4.
0
0.2
0.4
0.6
0.8
1
1.2
0 0.05 0.1 0.15 0.2
Time (sec)
y(t)
(a) step response
0
0.002
0.004
0.006
0.008
0.01
0.012
0 0.05 0.1 0.15 0.2
Time (sec)
y(t)
(b) disturbance
FIGURE P12.4
(a) Step response: K1= 1 (solid line) and K1= 2 (dashed line); and (b) disturbance
response.
P12.5 (a) The plant is given by
G(s) = 1
s(s+p)
where p= 2, nominally. One solution is
Gc(s) = 18.7(s+ 2.9)
(s+ 5.4) .
Then,
T(s) = 18.7(s+ 2.9)
(s+ 3.41)(s+ 2 + 23j)(s+ 2 23j).
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674 CHAPTER 12 Robust Control Systems
(b,d) The step responses are shown in Figure P12.5 for p= 2 and p= 1.
(c,d) The disturbance responses are shown in Figure P12.5 for p= 2 and
p= 1.
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
0 5
Time (sec)
y(t)
(a) step response
-0.14
-0.12
-0.1
-0.08
-0.06
-0.04
-0.02
0
0 5
Time (sec)
y(t)
(b) disturbance
FIGURE P12.5
(a) Step response: p= 2 (solid line) and p= 1 (dashed line); and (b) disturbance response:
p= 2 (solid line) and p= 1 (dashed line).
P12.6 (a) The plant is given by
G(s) = 1
s(s2+ 4s+ 5) ,
and the PID controller is
Gc(s) = K(s+z)2
s.
When
z= 1.25
and
K= 4 ,
all roots are
s=1±j1.22 .
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Problems 675
Then, the closed-loop transfer function is
T(s) = 4(s+ 1.25)2
s4+ 4s3+ 9s2+ 10s+ 6.25 .
(b,c) The step responses with and without a prefilter are shown in Fig-
ure P12.6.
(d) The disturbance response is shown in Figure P12.6.
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
0 5 10
Time (sec)
y(t)
(a) step response
-0.1
-0.08
-0.06
-0.04
-0.02
0
0.02
0 5 10
Time (sec)
y(t)
(b) disturbance
FIGURE P12.6
(a) Step response: w/o prefilter (solid line) and w/prefilter (dashed line); and (b) distur-
bance response.
P12.7 (a) The loop transfer function is
Gc(s)G(s) = 10Ka(5s+ 500 + 0.0475s2)
s3.
When
Ka= 374.5,
the phase margin is
P.M. = 40o.
(b) The root locus is shown in Figure P12.7a.
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676 CHAPTER 12 Robust Control Systems
-150
-100
-50
0
50
100
150
-150 -100 -50 0 50 100 150
xxx
o
o
Real Axis
Imag Axis
*
*
*
FIGURE P12.7
(a) Root locus for 1 + Ka10(0.0475s2+5s+500)
s3= 0.
When
Ka= 374.5,
the roots are
s1=139.8
s2,3=19.1±j114.2.
(c) The transfer function from Td(s) to Y(s) is
Y(s)
Td(s)=s
s3+ 182s2+ 19150s+ 1915000 .
The maximum is
max |y(t)|= 0.0000389 .
(d) The step responses, with and without a prefilter, are shown in Fig-
ure P12.7b.
P12.8 The polynomial under investigation is
s3+ 3s2+ 3s+ 4 = 0 .
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Problems 677
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
0 0.05 0.1 0.15 0.2 0.25 0.3
Time (sec)
y(t)
FIGURE P12.7
CONTINUED: (b) Step response: w/o prefilter (solid line) and w/prefilter (dashed line).
From the uncertainty bounds on the coefficients, we define
α0= 4 β0= 5
α1= 1 β1= 4
α2= 2 β2= 4
Then, we must examine the four polynomials:
s3+ 2s2+ 4s+ 5 = 0
s3+ 4s2+s+ 4 = 0
s3+ 4s2+ 4s+ 4 = 0
s3+ 2s2+s+ 5 = 0
The fourth polynomial is not stable—therefore, the system is not stable
for the uncertain parameters.
P12.9 One possible PID controller is
Gc(s) = 0.058s2+ 2.17s+ 16.95
s.
A first-order Pade approximation was used in the design to account for
the delay system. The step input response is shown in Figure P12.9. A
prefilter should also be used with the PID controller. A suitable prefilter
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678 CHAPTER 12 Robust Control Systems
is
Gp(s) = K2
K3s2+K1s+K2
.
0
0.2
0.4
0.6
0.8
1
1.2
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5
Time (sec)
y(t)
FIGURE P12.9
Step response with the PID controller and prefilter.
P12.10 The PID controller is given by
Gc(s) = KDs2+KPs+KI
s.
Using the ITAE method, we desire the characteristic polynomial to be
q(s) = s3+ 1.75ωns2+ 2.15ω2
ns+ω3
n= 0 ,
where we select ωn= 4 to obtain a peak time of Tp= 1 second. Here we
use the approximation for ITAE third-order systems that ωnTp4 from
Figure 5.30(c) in Dorf and Bishop. The actual characteristic equation is
s3+ 25KDs2+ 25KPs+ 25KI= 0 .
Equating coefficients and solving for the gains yields
KP= 1.376 , KD= 0.28 ,and KI= 2.56 .
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Problems 679
The step response is shown in Figure P12.10, with the prefilter
Gp(s) = KI
KDs2+KPs+KI
.
0 0.5 1 1.5 2 2.5 3
0
0.2
0.4
0.6
0.8
1
1.2
1.4
Step Response
Time (sec)
Amplitude
FIGURE P12.10
Step response with the PID controller and prefilter.
P12.11 We will design for the case where K= 1 and p= 1. The design plant is
G(s) = 1
s(s+ 1)(s+ 4) .
The nominal plant is given by
G(s) = 2.5
s(s+ 2)(s+ 4) ,
and the PID controller is
Gc(s) = KDs2+KPs+KI
s.
Using the ITAE method, we desire the characteristic polynomial to be
q(s) = s4+ 2.1ωns3+ 3.4ω2
ns2+ 2.7ω3
ns+ω4
n= 0 ,
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680 CHAPTER 12 Robust Control Systems
where we select ωn= 2.38 to obtain a peak time around Tp= 3 seconds.
The actual characteristic equation (with the worst-case plant) is
s4+ 5s3+ (4 + KD)s2+KPs+KI= 0 .
Equating coefficients and solving for the gains yields KP= 36.40, KI=
32.08, and KD= 15.26. The step response is shown in Figure P12.11,
with the prefilter
Gp(s) = KI
KDs2+KPs+KI
.
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
0
0.2
0.4
0.6
0.8
1
1.2
1.4
Time (sec)
y(t)
Worst−case plant
Nominal plant
FIGURE P12.11
Step response with the prefilter: nominal plant (dashed line) & worst-case plant (solid line).
P12.12 The transfer function is
G(s) = C(sIA)1B=h2 0 i
s3
5s+K
0
1
=6
s2+Ks + 5 .
The sensitivity is
SG
K=G
K
K
G=Ks
s2+Ks +5.
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Advanced Problems 681
Advanced Problems
AP12.1 Let Gp(s) = 1. A viable PID controller is
Gc(s) = KP+KI
s+KDs=1000s2+ 3000s+ 100
s.
The loop transfer function is
Gc(s)G(s) = 1000s2+ 3000s+ 100)
s(50s2+ 1) .
We can check that Kv= 100, as desired. The step response is shown in
Figure AP12.1.
Step Response
Time (sec)
Amplitude
0 0.2 0.4 0.6 0.8 1 1.2
0
0.2
0.4
0.6
0.8
1
1.2
1.4
System: syscl
Peak amplitude: 1.1
Overshoot (%): 9.5
At time (sec): 0.234
FIGURE AP12.1
Step response with PID controller.
AP12.2 For all three controllers, choose K= 1 as the design value. Also, use
as the nominal points a= 2 and b= 5 for each design. ITAE methods
were employed in all designs, although this did not work well for the PI
controller.
(a) PI controller: Let
Gp(s) = 1 .
Not all specifications could be met simultaneously with a PI con-
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682 CHAPTER 12 Robust Control Systems
troller. The best over-all results are achieved when using a= 3 and
b= 4.5 as the design values. An acceptable PI controller is
Gc(s) = 1.2 + 3.96
s.
Controller P.O. TsTp|u(t)|max
PI 0% 2.29s n.a. 4.43
PD 4.6% 1.72s 1.26s 12.25
PID 1.97% 0.65s 0.47s 37.25
TABLE AP12.2 PI, PD, and PID controller performance summary.
The final design is based on root locus methods since the ITAE meth-
ods did not produce an effective controller. The closed-loop transfer
function is
T(s) = 1.2s+ 3.96
s3+ 3s2+ 5.7s+ 3.96 .
(b) PD controller: Let
Gp(s) = 12.25
7.25 + 2.9s.
The closed-loop transfer function is
T(s) = 7.25 + 2.9s
s2+ 4.9s+ 12.25 ,
where the PD controller (based on ITAE methods) is
Gc(s) = 7.25 + 2.9s .
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Advanced Problems 683
(c) PID controller: Let
Gp(s) = 1000
15.5s2+ 210s+ 1000 .
The closed-loop transfer function is
T(s) = 15.5s2+ 210s+ 1000
s3+ 17.5s2+ 215s+ 1000 .
And the PID controller (based on ITAE methods) is
Gc(s) = 15.5s2+ 210s+ 1000
s.
The performance of each controller is summarized in Table AP12.2.
AP12.3 (a) The PID controller is
Gc(s) = KDs2+KP
KDs+KI
KD
s.
Since we want P.O. < 4% and Ts<1s, we choose the dominant
closed-loop poles to have ωn= 6 and ζ= 0.8. Therefore, we place the
zeros at
s2+KP
KD
s+KI
KD
=s2+ 10s+ 36 .
Solving for the constants yields,
KP
KI
= 10 ,KI
KD
= 36 .
Then, using root locus methods, we choose KD= 91 to place the
roots near the zeros. The PID controller gains are computed to be
KP= 910, KI= 3276 and KD= 91.
(b) The loop transfer function is
Gc(s)G(s) = KDs2+KPs+KI
s2(s2+ 5s+ 4) .
The closed-loop system characteristic equation is
s3+ 5s2+ 4s+KDs2+KPs+KI= 0 .
Solving for the PID gains yields KP= 73.4, KI= 216 and KD= 5.5.
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684 CHAPTER 12 Robust Control Systems
Therefore, the controller is
Gc(s) = 5.5(s2+ 13.35s+ 39.3)
s.
Using the prefilter
Gp(s) = 39.3
s2+ 13.35s+ 39.3,
we obtain the closed-loop transfer function
T(s) = 216
s3+ 10.5s2+ 77.4s+ 216 .
The percent overshoot is P.O. 3.5% and the settling time is Ts
1.67 sec.
AP12.4 The PID controller is
Gc(s) = KDs2+KP
KDs+KI
KD
s.
The bounds 1 a2 and 4 b12 imply that 2 ωn3.46
and 0.5ζωn1. One solution is to place the PID controller zeros at
0 0.5 1 1.5 2 2.5 3
0
0.2
0.4
0.6
0.8
1
1.2
1.4
Time (sec)
Amplitude
FIGURE AP12.4
Family of step response with PID controller with nominal case (a, b) = (1.5,9) denoted by
the solid line.
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Advanced Problems 685
s=1±j8 (i.e. ζωn= 1 and ωn= 3). So,
s2+KP
KD
s+KI
KD
=s2+ 2ζωns+ω2
n=s2+ 4s+ 9 .
The nominal case for design is chosen to be a= 1.5 and b= 9. Using
root locus, we select KD= 2.1 to place the closed-loop characteristic
roots near the zeros. Then, the PID controller gains are computed to be
KP= 8.4, KI= 18.9, and KD= 2.1. The plot of the response to a step
input is shown in Figure AP12.4. The off-nominal cases shown in the
simulations are (a, b) = (1.2,4), (1.4,6), (1.6,10), and (1.8,12).
AP12.5 To obtain a phase margin of P.M. = 49.77o, select K= 1.5, b= 36 and
choose Gp(s) = 1. The PID controller is
Gc(s) = 1.5(s2+ 20s+ 36)
s.
When K1= 0.75, the phase margin is reduced to P.M. = 45.45o; and
when K1= 1.25, the phase margin is increased to P.M. = 52.75o.
AP12.6 With the settling time Ts= 1 and percent overshoot P.O. < 10% specifi-
cations, we target for dominant closed-loop poles with ωn= 10. Here we
estimate ωnTs10 associated with the ITAE performance. The closed-
loop transfer function is
T(s) = Gp(s)1.5(KDs2+KPs+KI)
(1 + 1.5KD)s2+ 1.5KPs+ 1.5KI
,
where we have neglected τ. Using the ITAE method, the desired charac-
teristic polynomial is
s2+2ωns+ω2
n=s2+1.5Kp
1 + 1.5KD
s+1.5KI
1 + 1.5KD
.
Let KD= 0.25. Then solving for the remaining PID gains yields KP=
12.96 and KI= 91.67. The pre-filter is
Gp(s) = 137.5
0.375s2+ 19.45s+ 137.5.
Then the closed-loop transfer function (with τ= 0.001) is
T(s) = 137.5
0.001s3+ 1.375s2+ 19.45s+ 137.5.
The transfer function from the disturbance to the output is
Y(s)/Td(s) = 1.5s
0.001s3+ 1.375s2+ 19.45s+ 137.5.
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686 CHAPTER 12 Robust Control Systems
The step input response and disturbance response are shown in Fig-
ure AP12.6.
0 0.2 0.4 0.6 0.8 1
0
0.2
0.4
0.6
0.8
1
1.2
1.4
(a)
Time (sec)
Amplitude
0 0.2 0.4 0.6 0.8 1
−0.01
0
0.01
0.02
0.03
0.04
0.05
(b)
Time (sec)
Amplitude
FIGURE AP12.6
(a) Input response; (b) Disturbance response.
AP12.7 The PI controller is given by
Gc(s) = KPs+KI
s.
We will also use the prefilter
Gp(s) = KI
KPs+KI
.
Using the ITAE method, we determine that
KP=2ωnand KI=ω2
n.
Let ωn= 2.2. Then KP= 3.11 and KI= 4.8. The step response and
control u(t) are shown in Figure AP12.7.
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Advanced Problems 687
0
0.2
0.4
0.6
0.8
1
1.2
0 5
Time (secs)
Amplitude
(a)
-0.2
0
0.2
0.4
0.6
0.8
1
1.2
0 5
Time (secs)
Amplitude
(b)
FIGURE AP12.7
(a) Input response; (b) Control history u(t).
AP12.8 (a) A suitable PD controller is given by
Gc(s) = 0.6 + 0.4s .
The percent overshoot is P.O. = 18.8% and the peak time is Tp= 2.4
sec.
(b) A suitable PI controller is given by
Gc(s) = 0.15 + 0.01
s.
The percent overshoot is P.O. = 23.7% and the peak time is Tp= 7.8
sec.
(c) A suitable PID controller is given by
Gc(s) = 0.6 + 0.01
s+ 0.4s .
The percent overshoot is P.O. = 19.9% and the peak time is Tp= 2.5
sec.
(d) The PD or PID controllers are the best choices.
AP12.9 A robust PID controller designed with ITAE methods will be a suitable
controller. From the settling time specification we select ωn= 10, where
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688 CHAPTER 12 Robust Control Systems
we have used ζ= 0.8. The worst case is
a= 1 and K= 2 .
The desired closed-loop transfer function is
T(s) = ω3
n
s3+ 1.75ωns2+ 2.15ω2
ns+ω3
n
and the actual characteristic equation is
q(s) = s3+ (2a+KKD)s2+ (a2+KKP)s+KKI.
Equating like terms, we find that
KP= 107 KI= 500 KD= 7.75 .
AP12.10 We use as the design plant
G(s) = s+ 2
s(s+ 3) .
0
0.2
0.4
0.6
0.8
1
1.2
1.4
0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1
Time(sec)
Amplitude
FIGURE AP12.10
Family of step responses with the design plant (p, q, r) = (3,0,2) denoted by the solid line.
Select
p1= 2 and z1= 3
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Advanced Problems 689
to cancel a design plant pole and zero. Then, choose p2= 0 to have
zero steady-state error to a unit step. The remaining variables Kand
z2are selected based on ITAE methods, where ωn= 100. A suitable
compensator is
Gc(s) = 141.42(s+ 3)(s+ 70.71)
s(s+ 2) .
A plot of the step responses for various values of p, q and ris shown in
Figure AP12.10.
AP12.11 A suitable compensator is
Gc(s) = 1000(s+ 1.8)(s+ 3.5)(s+ 5.5)
s(s+ 600) .
0
0.2
0.4
0.6
0.8
1
1.2
1.4
012345678910
Time (sec)
Amplitude
FIGURE AP12.11
Step responses with nominal plant (solid line) and off-nominal plant with all poles reduced
by 50% (dashed line).
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690 CHAPTER 12 Robust Control Systems
Design Problems
The plant model with parameters given in Table CDP2.1 in Dorf andCDP12.1
Bishop is given by:
θ(s)
Va(s)=26.035
s(s+ 33.142) ,
where we neglect the motor inductance Lmand where we switch off the
tachometer feedback (see Figure CDP4.1 in Dorf and Bishop). With a
PID controller ,the closed-loop system characteristic equation is
s3+ (33.142 + 26.035KD)s2+ 26.035KPs+ 26.035KI= 0 .
A suitable PID controller is
Gc(s) = 50 + s+0.1
s.
This PID controller places the closed-loop system poles to the left of
the ζωnline necessary to meet the settling time requirement. The step
response is shown below. The settling time is Ts= 0.12 second. In the
steady-state the error due to a step disturbance is zero.
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5
0
0.2
0.4
0.6
0.8
1
1.2
Time (secs)
Amplitude
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Design Problems 691
DP12.1 The closed-loop transfer function is
Y(s)
R(s)=KmGc(s)
s2+ (2 + KmK1)s+Gc(s)Km
.
(a) When Gc=K, we have
T(s) = 15K
s2+ (2 + 15K1)s+ 15K,
where Km= 15. Using ITAE criteria and ωn= 10, we determine that
K1= 0.81 and K= 6.67. For the disturbance, we have
Y(s)
TL(s)=1
s2+ 14.14s+ 100 .
The input and disturbance responses are shown in Figure DP12.1,
without prefilters.
0
0.2
0.4
0.6
0.8
1
1.2
0 0.5 1
Time (sec)
y(t)
(a) Step response
-0.012
-0.01
-0.008
-0.006
-0.004
-0.002
0
0 0.5 1
Time (sec)
y(t)
(b) Disturbance response
FIGURE DP12.1
(a) Step response: Gc(s) = K(solid line) and Gc(s) = KP+KDs(dashed line); and (b)
disturbance response (same for both compensators).
(b) When Gc=KP+KDs, we have
Y(s)
R(s)=15(KP+KDs)
s2+ (2 + 15K1+ 15KD)s+ 15KP
.
For ωn= 10 and with the ITAE criteria, we determine that (with
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692 CHAPTER 12 Robust Control Systems
KD= 0.1)
Y(s)
R(s)=15(6.67 + 0.1s)
s2+ 14.14s+ 100 .
DP12.2 The nominal plant is given by
G(s) = 1
s(s+ 5) .
The closed-loop transfer function is
T(s) = K(KDs2+KPs+KI)
s3+ (5 + KKD)s2+KKPs+KKI
.
Let
KP= 450 , KI= 750 ,and KD= 150 .
A family of responses is shown in Figure DP12.2 a for various values of
K. The percent overshoot for 0.1K2 is shown in Figure DP12.2b.
0 0.5 1 1.5 2 2.5 3 3.5 4
0
0.2
0.4
0.6
0.8
1
1.2
1.4
Time (s)
Step response
FIGURE DP12.2
(a) Family of step responses for various values of K.
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Design Problems 693
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2
0
1
2
3
4
5
6
7
8
9
K
Percent overshoot
FIGURE DP12.2
CONTINUED: (b) Percent overshoot for various values of K.
DP12.3 (a) The dexterous hand model is given by
G(s) = Km
s(s+ 5)(s+ 10) ,
where Km= 1, nominally. The PID controller is
Gc(s) = KD(s2+ 6s+ 18)
s.
The root locus is shown in Figure DP12.3a. If we select
KD= 90 ,
the roots are
s1,2=5.47 ±j6.6
s3,4=2.03 ±j4.23 .
Thus, all roots have
ζωn>4/3
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694 CHAPTER 12 Robust Control Systems
to meet the design specification
Ts<3 sec .
(b) The step responses for
Km= 1
and
Km= 1/2
are shown in Figure DP12.3b. When
K= 1/2,
an off-nominal value, the settling time specification is no longer sat-
isfied.
-20
-15
-10
-5
0
5
10
15
20
-20 -15 -10 -5 0 5 10 15 20
xxxx
o
o
Real Axis
Imag Axis
*
*
*
*
FIGURE DP12.3
(a) Root locus for 1 + KDs2+6s+18
s2(s+5)(s+10) = 0.
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Design Problems 695
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
Time (sec)
y(t)
FIGURE DP12.3
CONTINUED: (b) Step response (without prefilters): PID with K3= 90 and Km= 1
(solid line) and PID with K3= 90 and Km= 0.5(dashed line).
DP12.4 The nominal plant is
G(s) = 17640
s(s2+ 59.4s+ 1764) ,
and the PID controller is
Gc(s) = KI(τ1s+ 1)(τ2s+ 1)
s.
(a) Using ITAE methods, we determine that ωn= 28.29, KI= 36.28,
τ1+τ2= 0.0954 and τ1τ2= 0.00149. So,
Gc(s) = 36.28(0.00149s2+ 0.0954s+ 1)
s.
(b) The step response for the nominal plant and the PID controller is
shown in Figure DP12.4a, with and without a prefilter.
(c) The disturbance response is shown in Figure DP12.4b.
(d) The off-nominal plant is
G(s) = 16000
s(s2+ 40s+ 1600) .
The step response for the off-nominal plant is shown in Figure DP12.4a.
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696 CHAPTER 12 Robust Control Systems
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
00.5 1
Time (sec)
y(t)
(a) nominal plant
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
00.5 1
Time (sec)
y(t)
(b) o-nominal plant
FIGURE DP12.4
(a) Step response for (i) nominal plant: w/o prefilter (solid line) and w/prefilter (dashed
line); and (ii) for off-nominal plant: w/o prefilter (solid line) and w/prefilter (dashed line).
-0.05
0
0.05
0.1
0.15
0.2
0.25
0.3
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
Time (sec)
y(t)
disturbance response
FIGURE DP12.4
CONTINUED: (b) Disturbance response for the nominal plant.
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Design Problems 697
DP12.5 One possible solution is
Gc(s) = 0.08(0.01s+ 1)(0.99s+ 1)
s.
The phase margin with this controller is
P.M. = 45.5o.
The step response is shown in Figure DP12.5 for the nominal plant (with
and without a prefilter); the step response for the off-nominal plant is also
shown in Figure DP12.5. The prefilter is
Gp(s) = 1411
13.97s2+ 1411s+ 1411 .
0 10 20
0
0.2
0.4
0.6
0.8
1
1.2
1.4
Time (sec)
y(t)
(a) nominal plant
0 10 20
0
0.2
0.4
0.6
0.8
1
1.2
1.4
Time (sec)
y(t)
(b) o nominal plant
FIGURE DP12.5
(a) Step response for nominal plant: w/o prefilter (solid line) and w/prefilter (dashed line);
and (b) for off-nominal plant: w/o prefilter (solid line) and w/prefilter (dashed line).
DP12.6 Using ITAE methods, three controllers are designed for the nominal plant:
(i) PID controller:
Gc(s) = 0.225s2+ 0.535s+ 34.3
s
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698 CHAPTER 12 Robust Control Systems
(ii) PI controller:
Gc(s) = 0.9s+ 22.5
s
(iii) PD controller:
Gc(s) = 0.9s+ 22.5
The step responses for each controller is shown in Figure DP12.6. The
responses for the PID and PI controller are the same since the gains were
selected to obtain the same ITAE characteristic equation. An appropriate
prefilter is used in all cases.
0
0.2
0.4
0.6
0.8
1
1.2
0 0.5 1 1.5 2
Time (sec)
y(t)
(a) nominal plant
0
0.2
0.4
0.6
0.8
1
1.2
0 0.5 1 1.5 2
Time (sec)
y(t)
(b) o-nominal plant
FIGURE DP12.6
(a) Step response for nominal plant: PID (solid line); PI (dashed line); and PD (dotted
line); (b) for off-nominal plant: PID (solid line); PI (dashed line); and PD (dotted line).
DP12.7 The loop transfer function is
G(s) = KaKm
(0.5s+ 1)(τfs+ 1)s(s+ 1) =K
s(s+ 2)(s+ 1)
since τfis negligible. A suitable PID controller is
Gc(s) = KKD(s2+as +b)
s=300(s2+ 2.236s+ 2.5)
s.
The step response is shown in Figure DP12.7. The percent overshoot is
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Design Problems 699
P.O. = 4.6% and the settling time is Ts= 3.74 seconds.
0
0.2
0.4
0.6
0.8
1
1.2
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
Time (secs)
Amplitude
FIGURE DP12.7
Step response for the elevator position control.
DP12.8 The system transfer function is
Y(s) = G(s)Gc(s)Gp(s)
1 + G(s)Gc(s)R(s).
We are given
G(s) = esT where T = 1 second .
Using a second-order Pade approximation yields
G(s)s26s+ 12
s2+ 6s+ 12 .
Three controllers that meet the specifications are
Gc1(s) = 0.5
s(Integral controller)
Gc2(s) = 0.04s+ 0.4
s(PI controller)
Gc3(s) = 0.01s2+ 0.04s+ 0.4
s(PID controller) .
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700 CHAPTER 12 Robust Control Systems
In all cases, the steady-state error is zero.
Integral PI PID
P.O.(%) 4.05 0 0
Ts(sec) 6.03 6.12 6.02
Tp(sec) 4.75 N/A N/A
|V(t)|max (volts) 1.04 1 1
The prefilter Gp(s) = 1 is used in all designs. To compute the voltage,
the transfer function is
V(s) = Gp(s)Gc(s)
1 + Gc(s)G(s)R(s).
DP12.9 The space robot transfer function is
G(s) = 1
s(s+ 10) .
(a) Consider Gc(s) = K. Then
T(s) = Gc(s)G(s)
1 + Gc(s)G(s)=K
s2+ 10s+K.
We determine that K= 50.73 for ζ= 0.702. Thus, we expect P.O. <
4.5%. So,
Gc(s) = 50.73 .
(b) Consider the PD controller
Gc(s) = KP+KDs .
Then
T(s) = KP+KDs
s2+ (10 + KD)s+KP
.
Using the ITAE method, we compute
KP= 100 and KD= 4 .
Thus,
Gc(s) = 4s+ 100 ,
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Design Problems 701
and the prefilter is
Gp(s) = 100
4s+ 100 .
(c) Consider the PI controller
Gc(s) = KP+KI
s=KPs+KI
s.
Then,
T(s) = KPs+KI
s3+ 10s2+KPs+KI
.
Using the ITAE method, we have
ωn= 5.7KP= 70.2 and KI= 186.59 .
Thus,
Gc(s) = 70.2 + 186.59/s ,
and the prefilter is
Gp(s) = 186.59
70.2s+ 186.59 .
(d) Consider the PID controller
Gc(s) = KDs2+KPs+KI
s.
Then,
T(s) = KDs2+KPs+KI
s3+ 10s2+KDs2+KPs+KI
.
Using the ITAE method with ωn= 10, we have
KD= 7.5KP= 215 and KI= 1000 .
Thus,
Gc(s) = 7.5s2+ 215s+ 1000
s,
and the prefilter is
Gp(s) = 1000
7.5s2+ 215s+ 1000 .
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702 CHAPTER 12 Robust Control Systems
A summary of the performance is given in Table DP12.9.
Gc(s)P.O. tptsyss max|y(t)|
K4.5% 0.62 s 0.84 s 0 0.026
PD 5.2% 0.39 s 0.56s 0 0.010
PI 1.98% 0.81 s 1.32s 0 0.013
PID 1.98% 0.46 s 0.75 s 0 0.004
TABLE DP12.9 A summary of performance to a disturbance input.
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Computer Problems 703
Computer Problems
CP12.1 The closed-loop transfer function is
T(s) = 8K
s2+ 2s+ 8K,
and the sensitivity function, ST
K, is
S(s) = s2+s
s2+ 2s+ 8K.
The plot of T(s) and S(s) is shown in Figure CP12.1, where K= 10.
nt=[80]; dt=[1 2 80]; syst = tf(nt,dt);
ns=[1 2 0];ds=[1 2 80]; syss = tf(ns,ds);
w=logspace(-1,2,400);
[magt,phaset]=bode(syst,w);magtdB(1,:) = 20*log10(magt(1,1,:));
[mags,phases]=bode(syss,w); magsdB(1,:) = 20*log10(mags(1,1,:));
semilogx(w,magtdB,w,magsdB,'--')
legend('20log|T|','20log|S|')
xlabel('Frequency (rad/sec)')
ylabel('Gain dB')
grid
10−1 100101102
−60
−50
−40
−30
−20
−10
0
10
20
Frequency (rad/sec)
Gain dB
20log|T|
20log|S|
FIGURE CP12.1
Plot of T(s)and the sensitivity function S(s).
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704 CHAPTER 12 Robust Control Systems
CP12.2 A reasonable value of the gain K= 4. The family of step responses is
shown in Figure CP12.2.
p=[0.5:0.5:20]; K=4;
t=[0:0.01:1];
for i=1:length(p)
n=[K*p(i)]; d=[1 p(i)]; sys = tf(n,d);
sys_cl = feedback(sys,[1]);
y=step(sys_cl,t); Y(:,i)=y;
[y2,t2]=step(sys_cl);
S=stepinfo(y2,t2); Ts(i)=S.SettlingTime;
end
plot(t,Y) , xlabel('Time (sec)'), ylabel('Step response')
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
Time (sec)
Step response
FIGURE CP12.2
Family of step responses for 0.5< p < 20.
CP12.3 The closed-loop characteristic equation is
1 + KD
s2+as +b
Js3= 0
where
a=KP/KD
b=KI/KD.
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Computer Problems 705
We select
a= 1 and b= 2
to move the root locus into the left hand-plane (see Figure CP12.3a).
Then, we choose
KD= 71
from the root locus using the rlocfind function. The closed-loop Bode
plot in Figure CP12.3b verifies that the bandwidth
ωB<5 rad/sec.
Also, the phase margin is
P.M. = 45.7o,
which meets the design specification. The plot of phase margin versus J
is shown in Figure CP12.3c. We see that as Jincreases, the phase margin
decreases.
J=25; a=1; b=2;
ng=[1];dg=[J 0 0]; sysg=tf(ng,dg);
nc=[1 a b]; dc=[1 0]; sysc=tf(nc,dc);
sys=series(sysc,sysg); rlocus(sys)
−1.2 −1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4
−2
−1.5
−1
−0.5
0
0.5
1
1.5
2
Root Locus
Real Axis
Imaginary Axis
FIGURE CP12.3
(a) Root locus for 1 + KDs2+s+2
10s3= 0.
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706 CHAPTER 12 Robust Control Systems
PM =
45.7093
J=25; a=1; b=2; KD=71; KP=a*KD; KI=b*KD;
ng=[1]; dg=[J 0 0]; sysg=tf(ng,dg);
nc=[KD KP KI]; dc=[1 0]; sysc = tf(nc,dc);
sys=series(sysc,sysg); sys_cl = feedback(sys,[1]);
bode(sys_cl);
[GM,PM]=margin(sys); PM
−30
−20
−10
0
10
20
Magnitude (dB)
10 100101102
Frequency (rad/sec)
FIGURE CP12.3
CONTINUED: (b) Closed-loop Bode plot with ωB<5rad/sec.
Ji=[10:1:40];
for i=1:length(Ji)
numc=[KD KP KI]; denc=[Ji(i) 0 0 0]; sysc = tf(numc,denc);
[gm,pm]=margin(sysc);
Pm(i)=pm;
end
plot(Ji,Pm), grid
xlabel('J'), ylabel('Phase Margin (deg)')
0 5 10 15 20 25 30 35 40
−10
0
10
20
30
40
50
60
70
80
90
J
Phase Margin (deg)
FIGURE CP12.3
CONTINUED: (c) Phase margin versus J.
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Computer Problems 707
CP12.4 The closed-loop characteristic equation is
1 + K1
s2+bs +a= 0
where a= 8 and the nominal value of b= 4. The root locus is shown in
Figure CP12.4a.
ÈSelect a point in the graphics window
selected_point =
-2.0165 + 2.5426i
ans =
2.4659
K
clf, hold o
a=8; b=4;
num=[1]; den=[1 b a]; sys = tf(num,den);
rlocus(sys), hold on
zeta=0.59; wn=1.35;
x=[-10:0.1:-zeta*wn]; y=-(sqrt(1-zeta^2)/zeta)*x;
xc=[-10:0.1:-zeta*wn];c=sqrt(wn^2-xc.^2);
plot(x,y,':',x,-y,':',xc,c,':',xc,-c,':')
rlocnd(sys)
-4
-3
-2
-1
0
1
2
3
4
-4 -3 -2 -1 0 1 2 3 4
x
x
Real Axis
Imag Axis
+
+
FIGURE CP12.4
(a) Root locus for 1 + K1
s2+4s+8 .
The performance region is specified by
ζ= 0.59 and ωn= 1.35 ,
which derives from the design specifications
Ts<5 sec and P.O. < 10% .
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708 CHAPTER 12 Robust Control Systems
Using an m-file, the value of
K= 2.5
is selected with the rlocfind function. The step responses for b= 0,1,4
and b= 40 are shown in Figure CP12.4b. When b= 0, the system is
marginally stable; b= 1 results in a stable system with unsatisfactory
performance. The nominal case b= 4 is stable and all performance specs
are satisfied. When b= 40, the system is heavily damped: the percent
overshoot specification is satisfied, but the settling time is too long.
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0.5
01 2 3 4 5 6 7 8 9 10
Time (secs)
Amplitude
Time (secs)
Amplitude
Time (secs)
Amplitude
Time (secs)
Amplitude
b=0
b=1
b=40
b=4
FIGURE CP12.4
CONTINUED: (b) Step responses for b= 0,1,4and 40.
CP12.5 (a) An acceptable lead compensator (designed with root locus methods)
is
Gc(s) = Ks+a
s+b= 5s+ 0.3
s+ 2 .
The compensated root locus is shown in Figure CP12.5a, where
K= 5
is selected to place the closed-loop poles in the performance region.
(b) The step responses for ζ= 0,0.005,0.1 and 1 are shown in Fig-
ure CP12.5b.
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Computer Problems 709
-4
-3
-2
-1
0
1
2
3
4
-4 -3 -2 -1 0 1 2 3 4
xxx
x
x
o
o
o
Real Axis
Imag Axis
+
+
+
+
+
FIGURE CP12.5
(a) Compensated root locus.
(c) You would like the actual structural damping to be greater than the
design value, if it must be different at all.
0
0.2
0.4
0.6
0.8
1
1.2
1.4
02 4 6 8 10 12 14 16 18 20
Time (sec)
Amplitude
zeta=0,0.005 (solid); zeta=0.1 (dashed); zeta=1 (dotted)
FIGURE CP12.5
CONTINUED: (b) Step responses for ζ= 0,0.005,0.1and 1.
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710 CHAPTER 12 Robust Control Systems
CP12.6 The m-file script which computes the phase margin as a function of the
time delay (using the pade function) is shown in Figure CP12.6. The
maximum time delay (for stability) is td= 4.3 seconds.
time delay vector
K=5;
numg=K*[1]; deng=[1 10 2]; sysg = tf(numg,deng);
td=[0:0.1:5];
for i=1:length(td)
[ndelay,ddelay]=pade(td(i),2); sysd = tf(ndelay,ddelay);
sys = series(sysg,sysd);
[mag,phase,w]=bode(sys);
[gm,pm,w1,w2]=margin(mag,phase,w);
pmv(i)=pm;
end
plot(td,pmv), grid
xlabel('time delay [sec]')
ylabel('phase margin [deg]')
-20
0
20
40
60
80
100
120
00.5 1 1.5 2 2.5 3 3.5 4 4.5 5
time delay [sec]
phase margin [deg]
FIGURE CP12.6
Phase margin versus time delay.
CP12.7 The m-file script is shown in Figure CP12.7a. The steady-state error
(shown in Figure CP12.7b) is zero when
a= 0.5
and increases rapidly as aincreases past a= 0.5. The maximum initial
undershoot is shown in Figure CP12.7c. As aincreases, the initial under-
shoot increases linearly. The gain margin is shown in Figure CP12.7d. It
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Computer Problems 711
can be seen that as aincreases, the gain margin decreases very rapidly.
gain margin
negative unit step input
max initial undershoot
steady-state tracking error
a=[0.01:0.01:0.99];
t=[0:0.1:30];
for i=1:length(a)
num=a(i)*[1 -0.5]; den=[1 2 1]; sys_o = tf(num,den);
[mag,phase,w]=bode(sys_o);
[gm,pm,w1,w2]=margin(mag,phase,w);
gmv(i)=gm;
sys_cl = feedback(sys_o,[1]);
[y,x]=step(-sys_cl,t);
yf(i)=1-y(length(t));
ym(i)=-min(y)*100;
end
gure(1), plot(a,gmv), grid, xlabel('a'), ylabel('gm')
gure(2), plot(a,yf), grid, xlabel('a'), ylabel('steady-state error')
gure(3), plot(a,ym), grid, xlabel('a'), ylabel('maximum initial undershoot [%]')
FIGURE CP12.7
Script to generate all the plots.
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
a
steady−state error
FIGURE CP12.7
CONTINUED: (b) Steady-state tracking error.
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712 CHAPTER 12 Robust Control Systems
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
0
5
10
15
20
25
a
maximum initial undershoot [%]
FIGURE CP12.7
CONTINUED: (c) Maximum initial undershoot.
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
0
50
100
150
200
250
a
gm
FIGURE CP12.7
CONTINUED: (d) Gain margin.
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Computer Problems 713
CP12.8 The plant (balloon and canister dynamics plus motor) is given by
G(s) = 1
(s+ 2)(s+ 4)(s+ 10) ,
and the PID controller is
Gc(s) = KD(s2+as +b)
s.
Let a= 6. Then using the root locus methods, we determine that with
KD= 12.5 and b= 10
we have the roots
s1=8.4
s2=4.7
s3,4=1.43 ±j1.05 .
Thus, ζ= 0.8. The plot of y(t) is shown in Figure CP12.8. The percent
overshoot is less that 3%, as desired.
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
0
0.2
0.4
0.6
0.8
1
1.2
1.4
Time (sec)
y(t)
With prefilter
Without prefilter
FIGURE CP12.8
Simulation of the GRID device.
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C H A P T E R 1 3
Digital Control Systems
Exercises
E13.1 (a) Elevation contours on a map are discrete signals. (b) Temperature in
a room is a continuous signal. (c) A digital clock display is a discrete
signal. (d) The score of a basketball game is a discrete signal. (e) The
output of a loudspeaker is a continuous signal.
E13.2 (a) Using long-division we determine that
Y(z) = z1+ 3z2+ 7z3+ 15z4+···
Therefore, with
Y(z) =
X
k=0
y(kT )zk
we have
y(0) = 0 y(T) = 1 y(2T) = 3 y(3T) = 7 y(4T) = 15 .
(b) The exact solution is
y(kT ) = ekln 2 1.
E13.3 For the system response
y(kT ) = kT
where k0, we have
Y(z) = T z
(z1)2.
E13.4 The partial fraction expansion of Y(s) is
Y(s) = 5
s(s+ 2)(s+ 10) =0.25
s+0.0625
s+ 10 0.3125
s+ 2 .
714
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Exercises 715
Then, using Table 13.1 in Dorf and Bishop, we determine that
Y(z) = 0.25 z
z1+ 0.0625 z
ze10T0.3125 z
ze2T
= 0.25 z
z1+ 0.0625 z
z0.135 0.3125 z
z0.670 ,
where T= 0.1.
E13.5 The Space Shuttle and robot arm control block diagram is shown in Fig-
ure E13.5. The human operator uses information from the computer gen-
erated data display and visual sensory data from the TV monitor and by
looking out the window. He/she commands the robot arm via a joystick
command to the computer.
-
+Computer D/A Robot arm
& motors/gears
TV monitor
& window view
measurement
joint angle &
rate sensors
measurement
A/D
human
operator
data
display
digital analog
analog
digital
digital
ref.
joystick
command tip
position
FIGURE E13.5
The Space Shuttle/robot arm control block diagram.
E13.6 From Section 10.8 in Dorf and Bishop, we find that the design resulted
in the compensator
Gc(s) = 6.66s+ 1
66.6s+ 1 = 0.1s+ 0.15
s+ 0.015 .
Using the relationships
A=eaT , B =ebT ,and C1A
1B=Ka
b,
we compute
A=e0.15(0.001) = 0.99985 , B =e0.015(0.001) = 0.999985 ,and C= 0.1.
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716 CHAPTER 13 Digital Control Systems
Therefore,
D(z) = CzA
zB= 0.1z0.99985
z0.999985 .
E13.7 Using long-division, we determine that
Y(z) = 1 + 3.5z1+ 5.75z2+ 6.875z3+···
Therefore, with
Y(z) =
X
k=0
y(kT )zk
we have
y(0) = 1 y(T) = 3.5y(2T) = 5.75 y(3T) = 6.875 .
E13.8 The closed-loop system with
T(z) = z
z2+ 0.2z1.0
is unstable since one of the poles of the transfer function (z=1.1 and
z= 0.90) lies outside the unit circle in the z-plane.
E13.9 (a) Using long-division we determine that
Y(z) = z1+z2+z3+z4+···
Therefore, with
Y(z) =
X
k=0
y(kT )zk
we have
y(0) = 0 y(T) = 1 y(2T) = 1 y(3T) = 1 y(4T) = 1 .
(b) The exact solution is
y(kT ) = 1 δ(k)
where δ(k) = 1 when k= 0 and δ(k) = 0 when k6= 0.
E13.10 We compute T= 1.25.
(a) Using Figure 13.19 in Dorf and Bishop, we determine that Kτ = 0.8
which implies K= 100.
(b) Using Figure 13.21 in Dorf and Bishop, we determine that ess = 0.75.
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Exercises 717
(c) Using Figure 13.20 in Dorf and Bishop, we determine that Kτ = 0.7
which implies K= 88.
E13.11 (a) The transfer function (including the zero-order hold) is
Go(s)Gp(s) = 100(1 esT )
s(s2+ 100) .
Expanding into partial fractions yields
G(z) = (1 z1)Z1
ss
s2+ 100
= (1 z1)z
z1z(zcos 10T)
z22 cos 10T z + 1 .
When T= 0.05 we ha,ve
G(z) = 0.1224(z+ 1)
z21.7552z+ 1 .
(b) The system is marginally stable since the system poles, z=0.8776±
0.4794j, are on the unit circle.
(c) The impulse response and sinusoidal input response are shown in
Figure E13.11.
-0.5
0
0.5
02 4 6 8 10 12 14 16
No. of Samples
Amplitude
-40
-20
0
20
40
010 20 30 40 50 60 70 80 90 100
No. of Samples
Amplitude
FIGURE E13.11
Impulse and sinusoidal (natural frequency) input response.
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718 CHAPTER 13 Digital Control Systems
E13.12 The partial fraction expansion of X(s) is
X(s) = s+ 1
s2+ 5s+ 6 =2
s+ 3 1
s+ 2 .
Then, with T= 1, we have
X(z) = 2z
ze3z
ze2=2z
z0.0498 z
z1353 .
E13.13 The root locus is shown in Figure E13.13. For stability: 2.2< K < 5.8.
−2 −1.5 −1 −0.5 0 0.5 1 1.5 2
−2
−1.5
−1
−0.5
0
0.5
1
1.5
2
K=5.8
K=2.2
Root Locus
Real Axis
Imaginary Axis
FIGURE E13.13
Root locus with unit circle (dashed curve).
E13.14 Given Gp(s), we determine that (with K= 5)
G(z) = 5(1 e1)z
z(ze1).
The closed-loop characteristic equation is
z2+ 1.792z+ 0.368 = 0
and the system is unstable, since there is a pole at z=1.55. The
system is stable for
0< K < 4.32 .
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Exercises 719
E13.15 The transfer function G(z) is
G(z) = 0.1289z+ 0.02624
z20.3862z+ 0.006738 .
The sampling time is T= 1 s.
E13.16 The transfer function G(z) is
G(z) = 0.2759z+ 0.1982
z21.368z+ 0.3679 .
The sampling time is T= 0.5 s.
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720 CHAPTER 13 Digital Control Systems
Problems
P13.1 The plot of the input to the sampler and the output r(t) is shown in
Figure P13.1.
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2
Time (sec)
r(t), r*(t)
FIGURE P13.1
Plot of r(t) = sin(ωt)and r(t).
P13.2 The plot of the input and the output is shown in Figure P13.2.
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2
Time (sec)
r(t)
FIGURE P13.2
Plot of r(t) = sin(ωt)and output of sample and hold.
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Problems 721
P13.3 The transfer function
Y(z)/R(z) = G(z) = z
zeT.
The ramp input is represented by
R(z) = T z
(z1)2.
The output Y(z) = G(z)R(z) is obtained by long division:
Y(z) = T z1+T(2 + eT)z2Th(1 + 2eT)(2 + eT)2iz3
+TheT+ (1 + 2eT)(2 + eT)
(2 + eT)(1 + 2eT)(2 + eT)2iz4+···
P13.4 The transfer function
Y(s)/R(s) = 1esT
s(s+ 2) .
The partial fraction expansion (with T= 1) yields
G(z) = (1 z1)Z0.5
s0.5
s+ 2= (1 z1)0.5z
z10.5z
z0.1353
=0.4323
z0.1353 .
P13.5 The step input is
R(z) = z
z1.
Also,
T(z) = G(z)
1 + G(z)=0.6321
z+ 0.2643 .
So,
Y(z) = T(z)R(z) = 0.6321
z+ 0.2643
z
z1=0.6321z
z20.7357z0.2643 .
Using long-division we determine that
Y(z) = 0.6321z1+ 0.4650z2+ 0.5092z3+ 0.4975z4+ 0.5006z5+···
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722 CHAPTER 13 Digital Control Systems
Therefore, with
Y(z) =
X
k=0
y(kT )zk
we have y(0) = 0, y(T) = 0.6321, y(2T) = 0.4650, y(3T) = 0.5092,
y(4T) = 0.4975, and y(5T) = 0.5006.
P13.6 Using the final value theorem (see Table 13.1 in Dorf and Bishop), we
determine that (for a step input)
Yss = lim
z1(z1)Y(z) = lim
z1(z1) 0.6321
z+ 0.2643
z
z1=0.6321
1.2643 = 0.5.
And using the initial value theorem, we compute
Yo= lim
z→∞ Y(z) = lim
z→∞
0.6321
z+ 0.2643
z
z1= 0 .
P13.7 Using Figures 13.19 and 13.21 in Dorf and Bishop, we determine that
the performance specifications are satisfied when Kτ = 0.5 and T
τ= 2.
Computing Kand T(with τ= 0.5) yields K= 1 and T= 1.
P13.8 We can select K= 1 and r= 0.2. The step responses for the compensated
and uncompensated systems are shown in Figure P13.8.
0 10 20 30 40 50 60 70
0
0.2
0.4
0.6
0.8
1
1.2
No. of Samples
Amplitude
Uncompensated
Compensated
FIGURE P13.8
Plot of compensated and uncompensated systems.
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Problems 723
P13.9 Consider the compensator
Gc(s) = Ks+a
s+b.
Then, using Bode methods we can select a= 1, b= 4, and K= 1. The
compensated system phase margin is P.M. = 50oand the gain margin is
G.M. = 15dB. The crossover frequency is ωc= 2.15 rad/sec. Utilizing the
Gc(s)-to-D(z) method and selecting T= 0.01 second, we determine
D(z) = CzA
zB=z0.99
z0.96 .
We use the relationships
A=eaT , B =ebT ,and C1A
1B=Ka
b,
to compute A=e0.01 = 0.99, B=e0.04 = 0.96, and C= 1.
P13.10 (a) The transfer function G(z)D(z) is
G(z)D(z) = K0.0037z+ 0.0026
z21.368z+ 0.3679 .
(b) The closed-loop system characteristic equation is
1 + K0.0037z+ 0.0026
z21.368z+ 0.3679 = 0 .
(c) Using root locus methods, the maximum value of Kis found to be
Kmax = 239.
(d) Using Figure 13.19 in Dorf and Bishop for T = 1 and a maximum
overshoot of 0.3, we find that K= 75. (e) The closed-loop transfer
function (with K= 75) is
T(z) = 0.2759z+ 0.1982
z21.092z+ 0.5661 .
The step response is shown in Figure P13.10.
(f) The closed-loop poles with K= 119.5 are z= 0.4641 ±0.6843j. The
overshoot is 0.55.
(g) The step response is shown in Figure P13.10 (for K= 119.5).
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724 CHAPTER 13 Digital Control Systems
0
0.5
1
1.5
02 4 6 8 10 12 14 16
No. of Samples
Amplitude
K=75
0
0.5
1
1.5
2
0 2 4 6 8 10 12 14 16 18 20
No. of Samples
Amplitude
K=119.5
FIGURE P13.10
Step response for K= 75 and K= 119.5.
P13.11 (a) Consider the compensator
Gc(s) = Ks+a
s+b.
Then, using Bode methods we can select a= 0.7, b= 0.1, and K=
150. The compensated system overshoot and steady-state tracking
error (for a ramp input) are P.O. = 30% and ess <0.01.
(b) Utilizing the Gc(s)-to-D(z) method (with T= 0.1 second), we deter-
mine
D(z) = CzA
zB= 155.3z0.9324
z0.99 .
We use the relationships
A=eaT , B =ebT ,and C1A
1B=Ka
b,
to compute
A=e0.007 = 0.9324 , B =e0.01 = 0.99 ,and C= 155.3.
(c) The step response for the continuous system with Gc(s) in part(a)
and for the discrete system with D(z) in part (b) is shown in Fig-
ure P13.11a.
(d) Utilizing the Gc(s)-to-D(z) method (with T= 0.01 second), we de-
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Problems 725
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
0 2 4 6 8 10 12 14 16 18 20
No. of Samples
Amplitude
T=0.1 sec
FIGURE P13.11
(a) Step response for continuous and discrete systems (T=0.1s) in Parts (a) and (b).
termine
D(z) = CzA
zB= 150z0.993
z0.999 .
We use the relationships
A=eaT
B=ebT
C1A
1B=Ka
b
to compute
A=e0.07 = 0.993
B=e0.001 = 0.999
C= 150 .
The step response for the continuous system with Gc(s) in part(a)
and for the discrete system with D(z) in part (d) is shown in Fig-
ure P13.11b.
(e) The ramp response for the continuous system with Gc(s) in part(a)
and for the discrete system with D(z) in part (b) is shown in Fig-
ure P13.11c.
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726 CHAPTER 13 Digital Control Systems
0
0.2
0.4
0.6
0.8
1
1.2
1.4
0 20 40 60 80 100 120 140 160 180 200
No. of Samples
Amplitude
T=0.01 sec
FIGURE P13.11
CONTINUED: (b) Step response for continuous and discrete systems (T=0.01s) in Parts
(a) and (d).
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
0 2 4 6 8 10 12 14 16 18 20
No. of Samples
Amplitude
T=0.1 sec
Ramp input (dashed line)
FIGURE P13.11
CONTINUED: (c) Ramp response for continuous and discrete systems (T=0.1s) in Parts
(a) and (b).
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Problems 727
P13.12 The root locus is shown in Figure P13.12. For stability: 0 < K < 2.
Real Axis
Imaginary Axis
−2 −1.5 −1 −0.5 0 0.5 1 1.5 2
−2
−1.5
−1
−0.5
0
0.5
1
1.5
2
Unit circle (dashed line)
FIGURE P13.12
Root locus for 1 + Kz+0.5
z(z1) = 0.
P13.13 The root locus is shown in Figure P13.13. When K= 0.027, the char-
acteristic equation has two equal roots: z1,2= 0.7247 and z3= 0.2593.
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
-2 -1.5 -1 -0.5 0 0.5 1 1.5 2
xxxo o
Real Axis
Imag Axis
Unit circle (dashed line)
FIGURE P13.13
Root locus for 1 + Kz2+1.1206z0.0364
z31.7358z2+0.8711z0.1353 = 0.
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728 CHAPTER 13 Digital Control Systems
P13.14 The root locus is shown in Figure P13.14. When K= 9.5655 ×105, the
two real roots break away from the real axis at z= 0.99. For stability:
K < 9.7×105.
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
-2 -1.5 -1 -0.5 0 0.5 1 1.5 2
x
x
x
xo o
Real Axis
Imag Axis
Unit circle (dashed line)
FIGURE P13.14
Root locus for 1 + Kz3+10.3614z2+9.758z+0.8353
z43.7123z3+5.1644z23.195z+0.7408 = 0.
P13.15 Given
Gp(s) = 20
s5
and the sample and hold (T=0.1s) as shown in Figure 13.18 in Dorf and
Bishop, we determine that
G(z) = 2.595
z1.649 .
Then, with R(z) = z/(z1), we have
Y(z) = 2.595z
(z1)(z+ 0.9462) .
Therefore, Y(z) = 2.59z1+ 0.14z2+ 2.46z3+ 0.26z4+···.
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Problems 729
P13.16 Given Gp(s) and the sample and hold (T=1s) as shown in Figure 13.18
in Dorf and Bishop, we determine that
G(z) = 0.22775z+ 0.088984
z21.0498z+ 0.049787 .
Then, with R(z) = z/(z1), we have
Y(z) = 0.22775z+ 0.088984
z20.82203z+ 0.13877
z
z1.
The plot of y(kT ) is shown in Figure P13.16.
12345678
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
kT
y(kT)
FIGURE P13.16
Plot of y(kT )for a step input.
P13.17 The root locus is shown in Figure P13.17 for
1 + K0.39532z+ 0.30819
z21.4724z+ 0.47237 = 0 .
The limiting value of the gain for stability is K= 1.71.
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730 CHAPTER 13 Digital Control Systems
−5 −4 −3 −2 −1 0 1 2
−2
−1.5
−1
−0.5
0
0.5
1
1.5
2
Unit circle (dashed line)
Root Locus
Real Axis
Imaginary Axis
FIGURE P13.17
Root locus for 1 + K0.39532z+0.30819
z21.4724z+0.47237 = 0.
P13.18 The plot of the step responses for 0 T1.2 is shown in Figure P13.18.
The overshoot and settling time summary is given in Table P13.18.
T0 0.2 0.4 0.6 0.8 1.0 1.2
P.O. 16.3% 20.6% 25.6% 31.3% 36.9% 40.0% 51.0%
Ts8.1 8.4 8.8 11.4 14.4 16.0 19.2
TABLE P13.18 Performance summary.
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Problems 731
0 10 20 30 40 50 60 70 80 90 100
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
Amplitude
No. of Samples
FIGURE P13.18
Step responses for 0T1.2.
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732 CHAPTER 13 Digital Control Systems
Advanced Problems
AP13.1 Given the sample and hold with Gp(s), we determine that
G(z) = 10.5K(z0.9048)
(z1)2.
The root locus is shown in Figure AP13.1. For stability: 0 < K < 0.2.
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
-2 -1.5 -1 -0.5 0 0.5 1 1.5 2
xxo
Real Axis
Imag Axis
Unit circle (dashed line)
FIGURE AP13.1
Root locus for 1 + K10.5(z0.9048)
(z1)2= 0 with unit circle (dashed line).
AP13.2 The root locus is shown in Figure AP13.2a. The loop transfer function is
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
-2 -1.5 -1 -0.5 0 0.5 1 1.5 2
xxo
Real Axis
Imag Axis
Unit circle (dashed line)
FIGURE AP13.2
(a) Root locus for 1 + K0.0379z
(z1)(z0.368) = 0.
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Advanced Problems 733
G(z)D(z) = K0.0379z
(z1)(z0.368) .
For stability: Kmax = 72. We select K= 8.2. The step response is shown
in Figure AP13.2b.
0
0.2
0.4
0.6
0.8
1
1.2
0 1 2 3 4 5 6 7 8 9 10
No. of Samples
Amplitude
FIGURE AP13.2
CONTINUED: (b) Step response with K= 8.2.
AP13.3 The root locus is shown in Figure AP13.3a. The maximum gain for
−3 −2.5 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2
−2
−1.5
−1
−0.5
0
0.5
1
1.5
2
System: sysz
Gain: 5.99
Pole: 0.736 + 0.257i
Damping: 0.596
Overshoot (%): 9.74
Frequency (rad/sec): 8.36
Unit circle (dashed line)
Root Locus
Real Axis
Imaginary Axis
FIGURE AP13.3
(a) Root locus for 1 + K0.07441z+0.06095
z21.474z+0.6098 = 0.
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734 CHAPTER 13 Digital Control Systems
stability is Kmax = 44.3. We select K= 6. The step response is shown in
Figure AP13.3b.
0 0.2 0.4 0.6 0.8 1 1.2
0
0.2
0.4
0.6
0.8
1
1.2
1.4
Step Response
Time (sec)
Amplitude
FIGURE AP13.3
CONTINUED: (b) Step response with K= 6.
AP13.4 The loop transfer function is
G(z) = 10(1 eT)
zeT,
and the closed-loop transfer function is
T(z) = 10(1 eT)
z(11eT10) .
For stability, we require
|11eT10|<1.
Solving for Tyields
0< T < 0.2.
Selecting T= 0.1s provides a stable system with rapid response; the
settling time is Ts= 0.2s. The step response is shown in Figure AP13.4.
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Advanced Problems 735
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0 0.5 1 1.5 2 2.5 3 3.5 4
No. of Samples
Amplitude
FIGURE AP13.4
Step response with T= 0.1s.
AP13.5 The maximum gain for stability is Kmax = 63.15.
Root Locus
Real Axis
Imaginary Axis
−3 −2.5 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2
−2
−1.5
−1
−0.5
0
0.5
1
1.5
2
System: sysz
Gain: 63.2
Pole: 0.725 − 0.686i
Damping: 0.00308
Overshoot (%): 99
Frequency (rad/sec): 7.58
Unit circle (dashed line)
FIGURE AP13.5
Root locus for 1 + K0.004535z+0.004104
z21.741z+0.7408 = 0.
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736 CHAPTER 13 Digital Control Systems
Design Problems
The plant model with parameters given in Table CDP2.1 in Dorf andCDP13.1
Bishop is given by:
Gp(s) = 26.035
s(s+ 33.142) ,
where we neglect the motor inductance Lmand where we switch off the
tachometer feedback (see Figure CDP4.1 in Dorf and Bishop). Letting
G(z) = ZG()G()
we obtain
G(z) = 1.2875e05(z+ 0.989)
(z1)(z0.9674) .
A suitable controller is
D(z) = 20(z0.5)
z+ 0.25 .
The step response is shown below. The settling time is under 250 samples.
With each sample being 1 ms this means that Ts<250 ms, as desired.
Also, the percent overshoot is P.O. < 5%.
0 50 100 150 200 250 300
0
0.2
0.4
0.6
0.8
1
1.2
No. of Samples
Amplitude
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Design Problems 737
DP13.1 (a) Given the sample and hold with Gp(s), we determine that
KG(z) = K0.1228
z0.8465 .
The root locus is shown in Figure DP13.1a. For stablity: 0 K < 15.
-1.5
-1
-0.5
0
0.5
1
1.5
-1.5 -1 -0.5 0 0.5 1 1.5
x
Real Axis
Imag Axis
Unit circle (dashed line)
FIGURE DP13.1
(a) Root locus for 1 + K0.1228
z0.8465 = 0 with unit circle (dashed line).
(b) A suitable compensator is
Gc(s) = 15(s+ 0.5)
s+ 5 .
Utilizing the Gc(s)-to-D(z) method (with T= 0.5 second), we deter-
mine
D(z) = CzA
zB= 6.22z0.7788
z0.0821 .
We use the relationships
A=eaT , B =ebT ,and C1A
1B=Ka
b,
to compute
A=e0.5(0.5) = 0.7788 , B =e0.5(5) = 0.0821 ,and C= 6.22 .
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738 CHAPTER 13 Digital Control Systems
(c) The step response is shown in Figure DP13.1b.
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0 2 4 6 8 10 12 14 16 18 20
No. of Samples
Amplitude
FIGURE DP13.1
CONTINUED: (b) Closed-loop system step response.
DP13.2 With the sample and hold (T=10ms), we have
G(z) = 0.00044579z+ 0.00044453
z21.9136z+ 0.99154 .
A suitable compensator is
D(z) = Kz0.75
z+ 0.5,
where Kis determined so that ζof the system is 1/2. The root locus is
shown in Figure DP13.2. We choose K= 1400.
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Design Problems 739
−5 −4 −3 −2 −1 0 1 2
−1.5
−1
−0.5
0
0.5
1
1.5
Curve of constant zeta=0.707 (dashed line)
Root Locus
Real Axis
Imaginary Axis
FIGURE DP13.2
Root locus for 1 + Kz0.75
z+0.50.00044579z+0.00044453
z21.9136z+0.99154 = 0.
DP13.3 The root locus is shown in Figure DP13.3a.
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
-2 -1.5 -1 -0.5 0 0.5 1 1.5 2
xxo
Real Axis
Imag Axis
Curve of constant zeta=0.707 (dashed line)
FIGURE DP13.3
(a) Root locus for 1 + Kz+1
(z1)(z0.5) = 0.
The gain for ζ= 0.707 is K= 0.0627. The step response is shown in
Figure DP13.3b. The settling time is Ts= 14T= 1.4s and P.O. = 5%.
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740 CHAPTER 13 Digital Control Systems
0
0.2
0.4
0.6
0.8
1
1.2
02 4 6 8 10 12 14 16 18 20
No. of Samples
Amplitude
FIGURE DP13.3
CONTINUED: (b) Step response with K= 0.0627.
DP13.4 With the sample and hold (T=1s), we have
G(z) = 0.484(z+ 0.9672)
(z1)(z0.9048) .
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
-2 -1.5 -1 -0.5 0 0.5 1 1.5 2
xxxo o
Real Axis
Imag Axis
Curve of constant zeta=0.5 (dashed line)
FIGURE DP13.4
(a) Root locus for 1 + Kz0.88
z+0.5
0.484(z+0.9672)
(z1)(z0.9048) = 0.
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Design Problems 741
0
0.2
0.4
0.6
0.8
1
1.2
0 2 4 6 8 10 12 14 16
No. of Samples
Amplitude
FIGURE DP13.4
CONTINUED: (b) Step response for K= 12.5.
A suitable compensator is
D(z) = Kz0.88
z+ 0.5,
where Kis determined so that ζof the system is 0.5. The root locus
is shown in Figure DP13.4a. We choose K= 12.5. The step response
is shown in Figure DP13.4b. Also, Kv= 1, so the steady-state error
specification is satisfied.
DP13.5 Select T= 1 second. With the sample and hold, we have
G(z) = 0.2838z+ 0.1485
z21.135z+ 0.1353 .
The root locus is shown in Figure DP13.5. To meet the percent overshoot
specification, we choose Kso that ζof the system is 0.7. This results in
K= 1. The step response has an overshoot of P.O. = 4.6%. Also, from
Figure 13.21 in Dorf and Bishop, we determine that the steady-state error
to a ramp input is ess = 2 (since T= 2, and Kτ = 0.3).
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742 CHAPTER 13 Digital Control Systems
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
-2 -1.5 -1 -0.5 0 0.5 1 1.5 2
xxo
Real Axis
Imag Axis
Curve of constant zeta=0.7 (dashed line)
FIGURE DP13.5
Root locus for 1 + K0.2838z+0.1485
z21.135z+0.1353 = 0.
DP13.6 With the sample and hold at T= 1 , we have
G(z) = 0.298z+ 0.296
z21.98z+ 0.9802 .
Consider the digital controller
Dz) = Kz0.9
z+ 0.6.
The root locus is shown in Figure DP13.6. To meet the percent overshoot
specification, we choose Kso that ζof the system is greater than 0.52.
We select K= 2. The step response has an overshoot of P.O. = 11.9%
and the settling time is Ts= 17.8s.
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Design Problems 743
−5 −4 −3 −2 −1 0 1 2
−1
−0.8
−0.6
−0.4
−0.2
0
0.2
0.4
0.6
0.8
1
Root Locus
Real Axis
Imaginary Axis
FIGURE DP13.6
Root locus for 1 + Kz0.9
z+0.60.298z+0.296
z21.98z+0.9802 = 0.
Step Response
Time (sec)
Amplitude
0 5 10 15 20 25 30 35
0
0.2
0.4
0.6
0.8
1
1.2
1.4 System: syscl
Peak amplitude: 1.12
Overshoot (%): 11.9
At time (sec): 2 System: syscl
Settling Time (sec): 17.8
FIGURE DP13.6
CONTINUED: (b) Step response for K= 2.
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744 CHAPTER 13 Digital Control Systems
Computer Problems
CP13.1 The m-file script and unit step response are shown in Figure CP13.1.
num=[0.2145 0.1609]; den=[1 -0.75 0.125];
sysd = tf(num,den,1);
step(sysd,0:1:50)
0
0.2
0.4
0.6
0.8
1
1.2
05 10 15 20 25 30 35 40 45 50
No. of Samples
Amplitude
FIGURE CP13.1
Step response.
CP13.2 The m-file script utilizing the c2d function is shown in Figure CP13.2.
% Part (a)
num = [1]; den = [1 0]; T = 1;
sys = tf(num,den);
sys_d = c2d(sys,T,'zoh')
%
% Part (b)
num = [1 0]; den = [1 0 2]; T = 1;
sys = tf(num,den);
sys_d=c2d(sys,T,'zoh')
Transfer function:
1
-----
z - 1
Transfer function:
0.6985 z - 0.6985
------------------
z^2 - 0.3119 z + 1
FIGURE CP13.2
Script utilizing the c2d function for (a) and (b).
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from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying,
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Computer Problems 745
% Part (c)
num = [1 4]; den = [1 3]; T = 1;
sys = tf(num,den);
sys_d = c2d(sys,T,'zoh')
%
% Part (d)
num = [1]; den = [1 8 0]; T = 1;
sys = tf(num,den);
sys_d = c2d(sys,T,'zoh')
Transfer function:
z + 0.267
-----------
z - 0.04979
Transfer function:
0.1094 z + 0.01558
-------------------
z^2 - z + 0.0003355
FIGURE CP13.2
CONTINUED: Script utilizing the c2d function for (c) and (d).
CP13.3 The continuous system transfer function (with T= 0.1 sec) is
T(s) = 13.37s+ 563.1
s2+ 6.931s+ 567.2.
The step response using the dstep function is shown in Figure CP13.3a.
The contrinuous system step response is shown in Figure CP13.3b.
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
0 2 4 6 8 10 12 14
No. of Samples
Amplitude
FIGURE CP13.3
(a) Unit step response using the dstep function.
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from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying,
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746 CHAPTER 13 Digital Control Systems
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
0 0.2 0.4 0.6 0.8 1 1.2 1.4
*
*
*
*
*
*
**
*
*
*****
FIGURE CP13.3
CONTINUED: (b) Continuous system step response (* denote sampled-data step response).
CP13.4 The root locus in shown in Figure CP13.4. For stability: 0 < K < 2.45.
−2 −1.5 −1 −0.5 0 0.5 1 1.5 2
−2
−1.5
−1
−0.5
0
0.5
1
1.5
2
Root Locus
Real Axis
Imaginary Axis
FIGURE CP13.4
Root locus for 1 + Kz
z2z+0.45 = 0.
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Computer Problems 747
CP13.5 The root locus in shown in Figure CP13.5. For stability: 0 < K < .
-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
Real Axis
Imag Axis
FIGURE CP13.5
Root locus for 1 + K(z0.2)(z+1)
(z0.08)(z1) = 0
CP13.6 The root locus is shown in Figure CP13.6.
1.5 1 0.5 0 0.5 1 1.5
1.5
1
0.5
0
0.5
1
1.5
Root Locus
Real Axis
Imaginary Axis
FIGURE CP13.6
Root locus.
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748 CHAPTER 13 Digital Control Systems
We determine the range of Kfor stability is 0.4< K < 1.06.
ÈSelect a point in the graphics window
selected_point =
-0.8278 + 0.5202i
ans =
0.7444
Select a point in the graphics window
selected_point =
-0.9745 - 0.0072i
ans =
0.3481 Kmin
Kmax
% Part (a)
num=[1 4 4.25 ]; den=[1 -0.1 -1.5];
sys = tf(num,den);
rlocus(sys), hold on
xc=[-1:0.1:1];c=sqrt(1-xc.^2);
plot(xc,c,':',xc,-c,':')
hold off
%
% Part (b)
rlocfind(sys)
rlocfind(sys)
FIGURE CP13.6
CONTINUED: Using the rlocus and rlocfind functions.
CP13.7 Using root locus methods, we determine that an acceptable compensator
is
Gc(s) = 11.7s+ 6
s+ 20 .
With a zero-order hold and T= 0.02 sec, we find that
0
0.2
0.4
0.6
0.8
1
1.2
00.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
**
*
*
*
*
*
*
*
*
*
*
*
*
************************************
Time (sec)
Amplitude
FIGURE CP13.7
System step response (* denotes sampled-data response).
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from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying,
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Computer Problems 749
D(z) = 11.7z10.54
z0.6703 .
The closed-loop step response is shown in Figure CP13.7.
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