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TECHNICAL HANDBOOK
FOR
CATV SYSTEMS

by

KEN SIMONS

THIRD EDITION

JERROLD ELECTRONICS CORPORATION
HORSHAM, PENNSYLVANIA, U.S.A.

TECHNICAL HANDBOOK FOR CATV SYSTEMS
Third Edition, Revised and Enlarged, March 1968,
Fifth Printing, July 1974
Printed in the U.S.A.

Copyright© 1968 by the JERROLD ELECTRONICS CORP.
Horsham, Pennsylvania, U.S.A.
All rights reserved
U.D.C. 621.397.743 : 621.315.212
Jerrold Stock No. 436-001

Published by JERROLD ELECTRONICS CORPORATION
ENGINEERING LABORATORY, PUBLICATIONS DEPT., HORSHAM, PA.

II

CONTENTS
Pages
PREFACE ........ ................ ........ ........... ....... ........................ .................... ...... .... ............ VII
CHAPTER I

DECIBELS AND dBmV ..............................................................

1-4

The Need for dB Expressions. Decibels Start with Logarithms. The Bel
and the Decibel. The dBmV. Signal Voltage in CATV Systems. Voltage
Measurement of the TV Sound'Signal. Voltage Measurement of the TV
Picture Sigual. Field Strength Meter Calibration.

CHAPTER II

THE COMBINATION OF VOLTAGE,
CURRENT OR POWER ............................................................

5-12

Introduction. Voltage and Current Addition-DC. Voltage and Current
Addition-Phased AC. Equal Contributions. Voltage Addition-Two Unequal Contributions. Voltage Addition-Effect of Phase Variation. Power
Addition-Different Frequencies. Power Addition-Equal Powers. Power
Addition-Two Unequal Contributions. Power Splitting Relationships.

CHAPTER III

RANDOM NOISE IN CATV SYSTEMS .............................. 13-15

Fundamentals. Signal-to-Noise Ratio. Noise Figure. Noise Figure in a
Cascaded Amplifier System.

CHAPTER IV THE FUNDAMENTALS OF DISTORTION
IN CATV AMPLIFIERS .......................................................... 16-29
Summary. Introduction. Distortionless Amplification. Amplification with
Distortion. Second Order Distortion. "AC" and "DC." AC and DC Components. The Spectrum of a Composite Voltage. A Voltage with Two AC
Components. Second Order Distortion by Addition of Components. The
Spectrum of a Voltage with Second Order Distortion. Third Order Distortion. Third Order Distortion by Addition of Components. Spectrum of
a Voltage with Third Order Distortion. The Sum of Two Sinusoidal Voltages Having Slightly Different Frequencies. Two Sinusoidal Input Voltages
with Second Order Distortion. Why Second Order Distortion is Unimportant in Present CATV Systems. Two Sinusoidal Input Voltages with
Third Order Distortion. Cross-Modulation and Compression. Conclusion.

CHAPTER V A MATHEMATICAL ANALYSIS OF
DISTORTION AS IT OCCURS IN CATV
AMPLIFIERS ................ .... ............ ............... ... ......... ............... ...... 30-39
A Summary of the Decibel Expressions. Output Levels Expressed in
dBmV. Relative Distortion Expressed in Decibels. Introduction. Results of
Basic Mathematical Development. Decibel Expressions for the Levels of
the Output Components. Relative Distortion Expressed in Decibels. Conclusion. The Detailed Mathematical Development of The Distortion Components. Detailed Development of Self- and Cross-Compression (and
Expansion) with Unmodulated Input Signals. The Detailed Development
of the Relationship Between Cross-Modulation and the Triple Beat Ratio.

III

CHAPTER VI CROSS-MODULATION IN A CATV SYSTEM .................. 40-45
The Nature of Cross-Modulation. Expressing Cross-Modulation in Percent.
Expressing Cross-Modulation in dBs. Maximum Tolerable Cross-Modulation. The Change in Cross-Modulation with Number of Channels. The
Increase in Cross-Modulation in a Cascaded Amplifier System. The Variation of Cross-Modulation with Output Level. Reduction in Output in a
Cascaded Amplifier System.

CHAPTER VII

THE ARITHMETIC OF CASCADED
TRUNK LINE AMPLIFIER SYSTEMS

46-47

Definitions. Noise Relationships. Cross-Modulation Relationships. How
Noise and Cross-Modulation Limit System Length. Maximum Number of
Amplifiers. Optimum System Operating Level.

CHAPTER VIII

MEASURING NOISE, CROSS-MODULATION,
AND HUM MODULATION ................................................ 48-56

Measuring Noise. Noise Level Measurements with a Jerrold Field Strength
Meter Model 704-B. Noise Calibration of a Field Strength Meter. Measuring Cross-Modulation. The Basic Test for TV Cross-Modulation. A CrossModulation Test Set. Measuring Cross-Modulation with a 704-B and
Picture Signals. An example of Cross-Modulation Measurement with TV
Modulation. Using Sine-Wave Modulated Signals for Measuring CrossModulation. Relation Between Apparent Cross-Modulation and Modulation
Ratio. Determining RMS Voltage Corresponding to Peak with Sine-Wave
Modulation. Examples of Cross-Modulation Measurement with Sine-Wave
Modulation. Measuring Hum Modulation. Measurements with a 704-B.
Hum Modulation in Cascaded Amplifier Systems.

CHAPTER IX REDUCING THE EFFECTS OF
REFLECTION IN CATV FEEDERS

57-63

Improvements in Pressure Tap Design. ReRection Introduced into a Feeder
by One Tap at the Worst Frequency (216 MHz). Periodicity. Periodicity
in Cable. Perioaicity Problems with Pressure Taps. Using Periodicity to
Minimize ReRections. Directional Coupler Multi-taps. Summary.

CHAPTER X SWEEP FREQUENCY TESTING OF
COAXIAL CABLES ...................................................................... 64-69
Introduction. Transmission Loss Measurement. Impedance Testing. Structural Return Loss Testing. Comparison of the Three Results.

IV

Pages

CHAPTER XI

CHARTS AND TABLES ...................................................................... 71-104

INSTRUCTIONS ON THE USE OF CHARTS AND TABLES ..................................................

71

APPLICATION OF POWER VS. dB CHARTS AND TABLES .................................................. 71-73
Power Gain or Loss. Percent Power. Conversion of Power to dBm. Conversion of dBm to Power. Relation between dBm and dBmV. Noise Factor
and Noise Figure. System Noise Calculations. Power Addition. Power
Division. Application of Power Split Data to Reflection Problems.
GUIDE TO POWER VS dB CHARTS
To Convert between -dB and Power Ratio ................................................................ Chart
To Convert between +dB and Power Ratio ................................................................ Chart
Power Sum Calculation-Relation between Difference of Two dBmV
Levels and dB to be Added to Larger Level ................................................................ Chart
Power Split Calculation (Tap Loss and Line Loss) .................................................... Chart

PI
P2

74
75

P3
P4

76
77

GUIDE TO POWER VS. dB TABLES
Given Power Ratio ... Find -dB ..............................................................................Table P5
Given -dB ... Find Power Ratio .............................................................................. TabIe P6
Given Power Ratio ... Find +dB ..............................................................................Table P7
Given +dB ... Find Power Ratio ..............................................................................Table P8
Given Level Difference ... Find Total Level for Powers Adding ............................Table P9
Given Level Difference ... Find Total Level for Powers Dividing ........................Table PlO
APPLICATION OF VOLTAGE VS. dB CHARTS AND TABLES ..............................................
Voltage or Current Gain or Loss. Percent Voltage. Relation Between
dBmV and mV. Voltage or Cross-Modulation Addition.

78-79
80-81
82-83
84-85
86
86
87

GUIDE TO VOLTAGE VS. dB CHARTS
To Convert between -dB and Voltage Ratio or % Voltage; and to Convert
between -dBmV and Millivolts .................................................................................. Chart VI 88-89
To Convert between +dB and Voltage Ratio, and to Convert between
+dBmV and Millivolts ................................................................................................ Chart V2
90
To Convert between Signal or Cross-Modulation Level Difference and
Level for Voltage Sum .................................................................................................. Chart V3
91
GUIDE TO VOLTAGE VS. dB TABLES
Given Voltage Ratio ... Find -dB ............................................................................ Table
Given -dB ... Find Voltage Ratio ............................................................................ Table
Given Voltage Ratio ... Find +dB ............................................................................ Table
Given +dB ... Find Voltage Ratio ............................................................................ Table
Given Signal or Cross-Modulation Level Difference ... Find dBmV corresponding to Voltage Sum ......... ~ ................................................................................Table
To Convert between dBmV and Voltage, or Current, or Power ..............................Table

V4
V5
V6
V7

92-93
94-95
96-97
98-99

V8
V9

100
101

CONVERSION BETWEEN RETURN LOSS, REFLECTION, AND VSWR ............Table Tl

102

GUIDE TO TABLE ON CHANNEL VARIABLES ......................................................................

103

TV CHANNELS 2-83: THEIR CORRESPONDING PICTURE AND
SOUND CARRIER FREQUENCIES, HALF-WAVELENGTHS FOR
AIR, POLYETHYLENE AND FOAM DIELECTRIC, AND DI-POLE
ANTENNA FACTOR .................................................................................................. Table T2

104

APPENDIX I

Mathematical Development of Distortion Components .................................. 105-106

APPENDIX II

Mathematical Development of Self- and Cross-Compression
(and Expansion) with Unmodulated Input Signals ..........................................

APPENDIX III

107

Mathematical Development of Relationship between
Cross-Modulation and Triple Beat Ratio ............................................................ 107-109

V

VI

PREFACE
This book on many important facets of CATV systems is a compilation of numerous
papers and articles previously published and of lectures given by the author.
Ken Simons has been with CATV right from its inception in the early fifties and the
contents of this book reflect his deep devotion to and involvement in the problems
with which this fast-growing industry is confronted daily. Not only is the CATV industry
deeply indebted to him for his contributions to the state of the art and to the opening
up of a whole new area of industrial, educational, and consumer electronics markets
with the accompanying creation of many thousands of jobs, but the public at large has
greatly benefited by being able to view television broadcasts in areas where previously
without CATV systems they could not.

Editor
Jerrold Electronics Corporation
Engineering Publications Department

Philadelphia, April 1968

GIDEON RATH,

VII

VIII

CHAPTER I
DECIBELS AND dBmV

TABLE A

The Need for dB Expressions
As the television signal progresses through a CATV system
from the antenna through miles of cable and dozens of
amplifiers, it undergoes many changes in power.level before
it is finally delivered to the subscriber's receiver. The power
is very low at the beginning of the system, coming out of
tpe antenna with a power level in the neighborhood of
0.0000000001 watt and increasing to around 0.0001 watt
at the outputs of the distribution amplifiers. The wide range
of power levels and all those zeroes (especially when you
start doing calculations) make this a pretty cumbersome
method to work with. There's a much better one, however,
which utilizes elements known as decibels (abbreviated
"dB") and decibel-millivolts (or "dBmV"). Restating the
unwieldy quantities we used above in this method's terms,
the antenna power would be expressed simply as "-20
dBmV" and the amplifier outputs as "+40 dBmV." The
whole range between the two would then be expressed
by numbers between -20 and +40. There is nothing mysterious about dB's and dBmV's. They show how basic
mathematics is applied to create a simpler way of analyzing systems.

Number
10,000,000,000
1,000,000,000
100,000,000
10,000,000
and so on down to
10
1
0.1
0.01
down to very small
numbers like
0.000,000,000,1

in terms of
power of 10

Lorrithm
(to ase 10)

1010
109
108
107

10
9
8

10
100
10-1
10-2

1
0
-1

-2

10-10

-10

7

The Bel and the Decibel
When telephones were first put into use it was discovered
that the longer the wires between two phones, the weaker
the signal arriving at the receiving end became. The convenient and obvious method for expressing how much the
signal has weakened was to do so in terms of a length of
"standard" cable. Just as llO volts means a quantity llO
times a standard of one volt, the telephone people would
refer to a weakening of a signal as being some number of
times that which would occur in a mile of this standard
cable.

Decibels Start with Logarithms
Most secretaries would feel right at home with logarithms,
which are really just a shorthand method of writing large
or small numbers. You learned in school that the number
100 can be written as 10 x 10 or 102 • That doesn't save
much time. But if you have to write the number 100,000,000, or 10 multiplied by itself eight times, it's quicker
to write it as 108 • Or if you are faced with 0.00000001
(which is 1/100,000,000 or 1/108 ), it's just 10-8 •

As time went by, the telephone people further discovered
that 10 miles of the standard cable reduced the Signal
power by a factor of approximately 10 times. They set up
this amount of attenuation (decrease), namely 10 to 1
power loss, as a unit. They called it a "bel," inspired by the
name of their company's famous founder. Mathematically,
they defined the bel in logarithmic terms since it described
signal attenuation in logarithmic rather than simply linear
terms. The formula is:

All the numbers you put in this form, 102 , 10-8 or even 10 77
have one thing in common, the "base" number of 10. The
idea of logarithms is to use this fact to carry our shorthand
one step further. We don't bother writing the 10 at all. We
just write the exponent-the power to which 10 is raised to
get the number-and call it a logarithm or "log" for short.
The logarithm for our old friend 100,000,000 or 108 , is 8. To
remind us that it is 10 that is raised to the power of the
logarithm, the log is referred to as log to the base 10, abbreviated log10, So, logarithm to the base 10 of 100,000,000, written log10 100,000,000, is 8. Remembering that
1/100 or 0.01 = 10-2 then log10 0.01 = -2, and so on, as
you can see from the following table A of examples.

Loss (in bels)

input power

= log10 ---output power

When input power = lOx output power, this becomes a
loss (in bels)
log10 10
1; in other words, a loss of
1 bel corresponds to a reduction of 10: 1 in power.

=

1

=

Just as a farad was discovered to be too large for practical
use, so that most capacitors are measured in microfarads
or picofarads, the bel was found to be too clumsy. The
unit which came into use was the one-tenth-of-a-bel or
"decibel" (abbreviated "dB"). Adjusting our formula to
read in decibels, we find that:
Loss in decibels =

The dB can also be used to express gain, the opposite of
loss. The formula becomes:
output power
Gain (in dB) = 10 10glO - - - - input power '
or, for voltage:

input power
10 10gio - - - - output power

. .
Gam (m dB)

A decibel then was the attenuation caused by a mile of the
standard phone cable. The important feature to remember
about a dB is that it is an expression of a ratio between
two levels - an input and an output. We've . originally
stated this as a ratio between two power levels, but it can
be used for voltage or current levels as well. Assuming the
input and output impedances are the same, we can convert the formula directly: ~

and again, Similarly for current.
ThedBmV
Power levels at various points in a CATV system are of
primary importance, forming the technical basis of system
operation. The antenna output must be high enough to
provide noise-free pictures. The amplifier output must be
held to optimum level for minimum system noise and crossmod. To express levels (as contrasted with power ratios),
a standard reference level is needed so the dB ratio of the
power at any point in the system to this standard level can
be computed. Early in the history of CATV systems the
power corresponding to an RMS'" voltage of 1 millivolt
across 75 ohms was chosen as the reference. This is approximately the input required for a noise-free picture on
an ordinary receiver, so its use provides dB levels indicating
approximately how much attenuation is allowable between
the point in question and the receiver.

E2

Since Power

= R'
input power

(input voltage) 2

output power

( output voltage) 2

We know that:

= 2 loglo (that number) .
For example, loglo (27)2 = 210g1027.
loglo (any number) 2

so
10gio

(input voltage) 2
(output voltage) 2

and since loss in dB
1010glo

10 (210g10

The level at any point in the system expressed in dB's above
the 1 millivolt/75-ohm standard is said to be the level in
decibel-millivolts or dBmV. In oilier words: voltage level
(indBmV)

input voltage

= 2 10gio output voltage ;

=

=

input power
output power

then Loss in dB

output voltage
I
'
mputvotage

= 20 10giO .

voltage in millivolts at that point
20 loglo ----:---::-:---:--:-:--::-:---:-:-standard level (1 millivolt)

(input voltage) 2

= 10 10glO (output voltage) 2 ,

when the voltage is measured at the 75-ohm impedance
level.

=

input voltage

input voltage

output voltage

)

= 20 10glO ------,,.---

Simplified, this reads:

output voltage

=

and similarly for current ratios.

20 10gio (voltage in millivolts) at 75 ohms
dBmV
impedance;

How iliis works out for a range of voltage and power ratios
can be seen in table B:

Table C shows the dBmV levels in a 75-ohm system corresponding to decade multiples of 1 microvolt.

(The two voltages must be measured at the same impedance level. )

TABLE C

TABLE B
Input power
Output power

Loss in dB

Input voltage
Output voltage

10,000,000,000
100,000,000
1,000,000
10,000
100
1

100
80
60
40
20
0

100,000
10,000
1,000
100
10
1

'" RMS

2

RMS voltage
across 75 ohms

dBmV

1 volt
100 millivolts
10
"
"
1
100 microvolts
10
"
"
1

+60
+40
+20
0
-20
-40
-60

= root mean square; explained on page 4.

Tables D and E can be used to find the voltage for any
integral number of dBmV. Follow the applicable rule:

2. Divide by 100 (corresponding to the 40 dB added)
giving 1.12 microvolts.

For Positive dBmV:

TABLE E

1. Subtract from the given level that multiple of 20 dB
giving a level within the range of Table D.

For- dBmV

2. Multiply the voltage found opposite that level by the
multiple of ten corresponding to the dBs subtracted:
Multiply millivolts
from Table D by

If you subtract

20
40
60
80

dB
dB
dB
dB

10
100
1,000
10,000

For example: Given + 78 dBmV.
1. Subtract 60 dB giving +18 dBmV. Corresponding voltage is 7.94 millivolts.

dBmV

Microvolts

dBmV

Microvolts

0
-1
-2
-3
-4
-5
-6
-7
-8
-9
-10

1000
891
794
708
631
562
501
447
398
355
316

-10
-11
-12
-13
-14
-15
-16
-17
-18
-19
-20

316
282
251
224
200
178
159
141
126
112
100

Conversion Ta.bles

2. Multiply by 1,000 (corresponding to the 60 dB added).
Voltage corresponding to +78 dBmV is 7.94 X 1,000
= 7,940 millivolts or 7.94 volts.

Chapter XI contains Tables V3, V4, V5 and V6 which allow
more accurate conversion between dBmV and voltages than the
example Table E above. In addition, other tables and charts are
given for converting between decibel expressions and related
electrical quantities. How those tables were derived and how to
use them is explained in the following chapters.

TABLE D

Signal Voltage in CATV Systems

For + dBmV
dBmV

Millivolts

dBmV

Millivolts

0
+1
+2
+3
+4
+5
+6
+7
+8
+9
+10

1.00
1.12
1.26
1.41
1.59
1.78
2.00
2.24
2.51
2.82
3.16

+10
+11
+12
+13
+14
+15
+16
+17
+18
+19
+20

3.16
3.55
3.98
4.47
5.01
5.62
6.31
7.08
7.94
8.91
10.00

We know that when we speak of a picture level of, say,
+20 dBmV, we mean that the voltage is 20 dB above
one-millivolt-across-75-ohms, or 10 millivolts. But what
exactly do we mean by that?
An actual picture signal is a high frequency sinusoidal
carrier modulated with video information. Its peak amplitude varies all the time. Does our 10 millivolts refer to
average voltage, peak voltage, effective voltage or what?
A quick review of some fundamentals may help to understand. Consider the relationships in a simple dc circuit as
shown in Fig. 1:
IOmV+

I
I

For Negative dBmV:

1. Add to the given level that multiple of 20 dB giving a
level within the range of Table E.

+0.133mA
CURRENT 0

2. Divide the voltage found opposite that level by the
multiple of ten corresponding to the dBs added:
Uyou add

Divide microvolts
from Table E by

20 dB
40 dB

10
100

00.133mA

VOLTAGE 0 . - - - - - - - TIME-

t-.- - - - - - - -

I

10
millivolts

E-IR - O"MsX 75· IOmV
I 0
·\OmV

+1.33 t--..
microwatt.
, , - - - . - -......- POWER 0 ...._ _ _ _ _ _ __

75 allma

ALL

THE TIME

P'EI•.£oXI~~:
• 1.33 mlc.-ts

ALL

THE TIME

Figure 1.

In a dc circuit like this everything is easy, because all the
quantities are constant. The voltage is 10 millivolts (all
the time), the current is 0.133 mA(all the time) and the
power is constant. There is no difficulty agreeing on what
each quantity means in this situation.

For example: Given -59 dBmV.
1. Add 40 dB, giving -19 dBmV. Corresponding voltage
is 112 microvolts.

3

amplitude is constant, the sound signal is measured in
terms of the RMS voltage present, just like a c-w signal,
its effective voltage being 0.707 x the peak voltage.

When the voltage and current vary with time there is
need for explanation. What kind of voltage is meant? Consider a circuit, similar to that in Fig. 1, but with sine-wave
ac voltage and current as shown in Figure 2.

Voltage Measurement of the TV Picture Signal

r

The TV picture signal in a CATV system is amplitudemodulated. It shows up as a high-frequency sine-wave ac
voltage with constant frequency, but with amplitude varied
by the picture information. At the end of each horizontal
line it goes to full amplitude during the sync pulse, as
shown in Figure 4.

----A-'-:~TAGEO
_
I \ IOmV.,
75ahma
0 f--::T;::"IM=E'TV---f--->--

+IomvpVOLTAGE

-IOmV
+O.l33mA -

"RMS

~-CURRENT
~
O.I33mA
CURRENT 0 f--~V---+------'''-

E"",' Irma X R

-

2.66IAA A

Pav.· Ermaxl r _

-O.l33mA

: : . 1.3:~~-~

SYNC TIP
--14.14 mV

RMS VOLTAGE
CORRESPONDING
TDSYNC TIP
VOLTAGE I~
14.14 X 70.7%
'IOmV RMS

AVERAGE POWER
1.33microwatt.

Figure 2.

Since the voltage, current and power -are varying all the
time, they can be expressed or measured by a single
number only after there is agreement as to what that
number means in reference to the varying quantities. In
the case. of sine-wave ac, "RMS" voltage and current are
generally used. The RMS quantities are the effective quantities. An ac voltage of 10 millivolts RMS will have the
same "effect," that is, it will cause the same average power
loss in a resistor, as a 10 mV de voltage. -Although the
actual voltage in the above circuit reaches plus and minus
peaks of 14.14 millivolts, 10 millivolts describes its "effect."
The current goes to plus and minus 0.188 rnA, but its heating effect is the same as that of a dc current of 0.133 rnA.
The power reaches 2.66 microwatts but the average power
is 1.33 microw:atts. Using RMS quantities simplifies calculations by making the product of RMS volts times RMS
amps. equal to average watts.

Figure 4.

When the picture is viewed like this, with the _oscilloscope
timed to show one horizontal line, the individual cycles
of sine-wave variation are not seen because there are so
many of them. During one horizontal sync pulse .the voltage
of a Channel 6 TV signal, for example, .goes through more
than 400 cycles. As the camera scans, the part of the
signal between sync pulses constantly changes in amplitude to spell out the picture infomlation. The only part of
the signal that does not change in amplitude is the pedestal
and the sync pulse. For this reason, signal strength measurements for a modulated TV picture signal are always referred to the sync level. The statement "the pix level is
+20 dBmV" or "There is 10 millivolts of picture signal"
means that the amplitude of the ac voltage during the
sync pulse is the same as the amplitude of an unmodulated
100millivolt RMS signal. Another way of saying the same
thing is: "a modulated TV picture signal is measured in
terms of 0.707 times its instantaneous peak value" (which
peak occurs at the maximum of each cycle during the
sync pulse) .

Note that, for a constant amplitude sine-wave voltage or
current, the RMS value is always 0.707 x the peak voltage
or current.
Voltage Measurement of the TV Sound Signal
The TV sound signal is frequency-modulated. This means
that it shows up in the CATV system as a high-frequency
sine-wave ac voltage with constant amplitude, and with a
frequency that is varied above and below the center frequency at an audio rate. Figure 3 illustrates this. Since the

Field Strength Meter Calibration
Field Strength Meters for CATV are generally calibrated
so they read the RMS voltage and the dBmV level of the
signal to which they are tuned. In the calibration procedure of Jerrold Models 704-B and 727 the instruments
are adjusted to read correctly on an accurately measured
c-w signal. Since the detector which drives the indicating
meter responds only approximately to the peak of the
signal (even though the meter indicates RMS) the indica~
tion is slightly low when a TV picture carrier is tuned in.
This error is usually quite small, in the order of 1 dB or
less, depending on how close the detector comes. to responding to true peak voltage.

SOUND SIGNAL

Figure 3.

4

CHAPTER II
THE COMBINATION OF VOLTAGE, CURRENT, OR POWER
Introduction

presence of the second source increases the current flow
through the first, so the power output of the first is
doubled.

To analyze the operation of a CATV system it is essential
to understand how electrical quantities combine to produce a final result. The way in which noise and distortion
combine in a cascaded amplifier system to produce a net
effect at the output of the last amplifier determines the
requirements on each amplifier. How the signal "splits" or
"divides" in a customer tap or line splitter determines the
net effect on system levels.

When a third source is connected in series,as indicated
in Figure 8, the voltages add to produce 3 volts across the
load, resulting in a current of 30 rnA and in a power of
90 m W. In summary, as sources having equal voltages are
added in series, the power output into a common load in-

This understanding is approached by first considering a
few simple cases involving the addition of voltage, current
or power.

1-10 mA

!III

Voltage and Current Addition-DC
First consider the conditions that exist in the simple dc
circuit shown in Fig. 5. A constant voltage" dc source
maintains a I-volt drop across a 100~ohm resistive load.
The load draws a current of 10 mA and dissipates 10 mW,
which is the power output of the source.

CONSTANT
CURRENT
SOURCE

Similarly Fig. 6 illustrates a constant-currentU dc source.
This source maintains a current of 10 mA through a 100ohm load. The voltage drop is 1 volt, and the load dissipates the full output of the source, which also is 10 mW.

+

>100

P-IOmW
IV

~

c

OHMS

Ir
Figure 6.

1= 20mA

Now consider what happens when a second equal voltage
source is connected in series with the first (compare Fig.
7 with Fig. 5). It is not surprising to find that, with twice
the voltage, the IOO-ohm load draws twice the current,
20 mAo It is perhaps a little surprising, when you think
about it, that the power dissipated in the load, which is
the same as the power output of the two generators, is 40
mW, 4 times what it was with only one source. The

i
1.

4

r:
E2-IV

L...

~+

1- IOmA

EI"IV

!III

P - 40mW

2V

>

!

100
OHMS

Figure 7.
CONSTANT +
VOLTAGE -'-'SOURCE -

1 c::=>

P-IOmW

IV

<'100

">OHMS
c
4

1=30mA

I

Figure 5.

3V

• A constant-voltage source maintains the same output voltage
regardless of load. The internal resistance of a constantvoltage source is very low compared to the load resistance .
•• A constant-current source maintains the same output current
regardless of load. The internal resistance of a constantcurrent source is very high compared to the load resistance.

1

P=90mW

===>

Figure 8.

5

100
OHMS

creases as the square of the total voltage (P

E2

When ac voltage or current components have identical
frequency and phase, as in Fig. 11, they are said to be
phased. A general rule can be formulated for the addition
of phased voltage or current components:
When the ac voltages (or currents) from a number
of phased sources are combined, the total rms voltage (or current) is the sum of the individual rms
voltages (or currents).
Example: The output currents of six phased ac sources
are combined in a WOO-ohm load as shown in Fig. 12.
What is the total current, what power is dissipated in the
load, and how many dB is this power increased over that
dissipated by source # 1 only?
Solution: Since the sources are phased, the total rms current is simply the sum of the given currents: 1 + 4 + 11 +
3 + 6 + 5 = 30 rnA. The power dissipation is J2R=
30/1000 X 30/1000 X 1000
0.900 watts. The power
dissipation with source # 1 alone would be 1 rnA through
1000 ohms, or 1/1000 X 1/1000 X 1000 = 0.001 W.

= R)'

The situation is similar when two . currents are added.
When a second current source is added, causing a second
10-mA current to flow in the 100-ohm load (compare
Fig. 9 with Fig. 6), the total current of 20 rnA causes a
voltage drop of 2 volts, and a dissipation in the load of 40
mW. The presence of the second source increases the'
power output of the first.
Similarly, when the outputs of three constant current dc
sources are combined to feed a common load as illustrated
in Fig. 10, the current through the load is three times the
current from one source and the power dissipated is nine
times. the power that would be dissipated with a single
source and the same load. (P = I2R).

=

+

!

!,omA

1= 20m A

+--

TWO

10mA

COMPONENTS AND TDTAL

REACH EACH POINT IN THE
IDENTICAL TIMES

CONSTANT
CURRENT
SOURCE
#1

CONSTANT
CURRENT
SOURCE

P=40mW

=:::)

::

+141

100
OHMS

#2

e.o
-141
+141

ezo

~~~~-+~~~~

-1.41
+2.82

Figure 9.

TOTAL

e f-+-+-If+--+-+--+~

+ 1= 30mA +-

1~lomA
CONSTANT
CURRENT
SOURCE
#1

~lomA

~ 10mA
CONSTANT
CURRENT
SOURCE
#2

CONSTANT
CURRENT
SOURCE
#3

-2,82

•

~TIME

100
OHMS

==>

I

<

P=90mW
Figure 10.

IV
rms

Voltage and Current Addition-Phased AC
Next, what will happen when the outputs of two sinewave ac sources having identical phase and frequency
are added? Fig. 11 illustrates the situation: Since the two
voltages reach each point in the cycle at the same time,
their outputs add directly, and the total voltage is just
twice that of either one. Thus the power output of each is
doubled by the presence of the other, just as in the dc
case.

20mA rms

GEN #2

,ool

OHMS E TOTAL
40mW
GEN#I

2 V rms

J

Figure 11. Two Sinewave AC Sources, Same Frequency, Same Phase.

6

---"::>~TOTAL

PHASED
AC
CONSTANT
1'lmA
CURRENT
rms
SOURCES

't

I

til

4mA

1

mA

1

6mA

3mA

1 - 30 mA

1

5mA

;)

1000.n.)

#2

#3

#4

#S

#6

rv

rv

rv

rv

rv

==>

TOTAL

<

POWER

-EI
-0.9 WATTS

E=
30X1000=
1000
30 VOLTS

J

Figure 12. Addition of Phased Currents.

Table F can be used; so. the dB increase corresponding to.
45 contributio.ns is found o.PPo.site number 45 in the table,
,33.06 dB.
TABLE F
Voltage o.r Current Addition, Phased So.urces,
Equal Contribution from Each So.urce

The Po.wer ratio. is 0.900 divided by 0.001 o.r 900. From
Table P7 in Chapter XI we find that the dB number fo.r
a Po.wer ratio. o.f 9 would be 9.54 dB; adding 20 dB
co.rresponding to. a Po.wer ratio o.f 100, the so.lutio.n is
29.54 dB.
One step in the calculation o.f the dB expressio.n could be
eliminated by simply using the current ratio.. The to.tal current is 30 rnA, an increase o.f 30 times o.ver 1 rnA fro.m
source # 1. From Table V6 we get that a current ratio. of
30 corresponds to 29.54 dB.

dB
dB
Corresponding
Number
CorresPo.nding
Number
to This
of Equal
o.fEqual
to This
Ratio.
Contributions
Contributions
Ratio

General rule for finding the dB increase for a combination
o.f phased so.urces:
To find the dB increase when the outputs of phased
voltage or current sources are combined, add the voltages (or currents) and use the voltage-dB table to find
dB corresponding to the ratio of the total to the original voltage or current. This kind of combination is
called "Voltage (or current) addition."
Equal Co.ntributio.ns
When a co.mmo.n o.utput is o.btained by Co.mbining the
current o.r vo.ltage o.utputs o.f a number o.f phased so.urces,
and each so.urce co.ntributes equally to. that output, the dB
increase in o.utput fo.r any 'given number o.f so.urces can
be easily tabulated. The dB increase is simply the vo.ltage
o.r current ratio. expressed in decibels, and since the ratio.
is equal to. the number o.f so.urces, this is simply that
number expressed in dB.
Fo.r example; two. equal phased so.urces in series generate
an output 6.02 dB abo.ve a single source, since 20 Io.g102
=6.02.
Table F has been prepared to sho.W these dB ratio.s fo.r
numbers o.f equal co.ntributio.ns from I, to 50 so.urces.
Example: By ho.W much will the o.utput of a system increase when 45 equal, phased so.urces in series are used as
compared with the o.utput o.f o.ne such so.urce?
So.lution: Since vo.ltage additio.n is applicable in this case,

7

1
2
3
4
5

0
6.02
9.54
12.04
14.00

26
27
28
29
30

28.30
28.62
28.94
29.24
29.54

6
7
8
9
10

15.56
16.90
18.06
19.09
20.00

31
32
34
35

29.82
30.10
30.36
30.62
30.88

11
12
13
14
15

20.82
21.58
22.28
22.86
23.52

36
37
38
39
40

31.12
31.36
31.60
31.82
32.04

16
17
18
19
20

24.08
24.60
25.10
25.58
26.02

41
42
43
44
45

32.26
32.46
32.66
32.86
33.06

21
22
23
24
25

26.44
26.84
27.24
27.60
27.96

46
47
48
49

33.26
33.44
33.62
33.80
34.00

33

50

Voltage Addition-Two Unequal Contributions

We want a table arranged so that, when given Lh - L,.,
we can find L t - L h.
By definition: Lh
20 loglo eh and Ls
20 loglo e.;
hence: Lh - L. = 20 10glO eh - 20 10glO e.
e.
-2010g1o eh
e.
so given Lh - L. we can find --'-;
eh
Now, since: et = e. + eh and L t = 20 loglo (es + eh);
hence: L t - Lh = 20 10glO (es + eh) - 20 10glO eh

Consider a case where the outputs of tWo unequal voltage
or current sources are phased so that they add directly on
a voltage basis. If the levels of the two are given in
dBmV, and the total is wanted in dBmV, one way to find
the answer would be to convert each level to an equivalent
rms voltage, add the two voltages, and convert the total
back to dBmV.

=

Example: Two phased generators supply a 75-ohm load.
The :first alone develops a level of +35 dBmV, the second
alone develops +43 dBmV. What is the level with both
connected? (Fig. 13).

= 20 10glO

e.+eh )
( ~

=

= 20 loglo

(eh
e+.1) .

e
Thus knowing --.!., we can simply add 1 to that number
eh
and then find the corresponding number of dB, representing L t - L h.

LEVEL IS

+35dBmV

e.

e.

=

Example: let ~ - L. be 8 dB, then 0.398 and eh
eh
1
1.398; then L t - Lh = 20 loglo 1.398
2.. 91 dB.

=

+43dBmV

OUTPUTS OF A AND B
ARE IN PHASE

75

=

With this result the solution for the example in Fig. 13 is
greatly simplified. Given two sources, having levels of
+35 and +43 dBmV, and assuming voltage addition, the
difference in level between the higher and the lower level
is 43-35 = 8 dB. The level of the total is found by adding
2.91 dB (Lt - Lh from above) to the higher level. 43 +
2.91 = 45.91 dBmV.

LEVEL IS

OHMS

+

A table has been worked out showing L t - ~ for values
of Lb - L. between 0 and 40 dB in 0.1 dB .steps. This
table is given as TABLE V8 in chapter XI; an example
will illustrate its use.

WHAT IS
TOTAL
LEVEL?

Example: A source delivers an output into a 75-ohm load
at a level of -12 dBmV. A second source, phased with the
first and connected to an equal load, supplies an output
at a level of -18 dBmV. What will he the level when both
sources are connected to the same load?

Figure 13.

Solution: In this case, voltage addition applies, thus Table
V8 can be used. The difference in the levels of the two
sources is 6 dB. Opposite 6 dB in the table we find 3.53 dB,
the increase in the total level as compared with the le~el
of the higher one of the two sources. The higher level is
-12 dBmV (it is the smaller numher, because both levels
are negative). The resulting output level is found by adding 3.53 algebraically -12 + 3.53 = - 8.47 dBmV.

Solution: From the dBmV-voltage table V7 find the rms
voltages corresponding to the two levels: +35 dBmV corresponds to 56.23 mY; +43 dBmV corresponds to 141.3
mY. The total voltage is 56.23 + 141.3 = 197.53 mY. In
Table V6 we find the corresponding level is about +45.9
dBmV.
This calculation can be simplified by the use of a table
which allows making the calculation in dB language without converting to voltage. The table relates the difference
in dB between the higher level and the lower to the difference between the total and the higher. The preparation
of the table can be understood from the following:

Voltage Addition-Effect of Phase Variation When two voltage sources have the same frequency, but
different phase, their outputs reach various points in the
cycle at different times so they do not add directly, and
the rms voltage of the total is not equal to the sum of the
two rms voltages. This is illustrated in Fig. 14 which shows
each of two equal voltages and the total, for various phase
relations between the two. When the phase angle is 0, the
total voltage is twice either one. For small phase angles
the total is only slightly reduced. When the second voltage
lags the first by 90° the total voltage is 3 dB higher than
either one (i.e. it is reduced 3 dB from the in-phase case)

Let:
Ls represent the lower level in dBm V, es the corresponding
voltage in millivolts.
Lh represent the higher level in dBmV, eh the corresponding voltage in millivolts.
L t represcnt the total level in dBmV, e t the corresponding
voltage in millivolts.-

8

e,

AN)

to determine the total. In some cases, when there are a
great many contributions, it is reasonable to assume that
Power Addition applies, as described below.

e2

IN PHASE

Power Addition-Different Frequencies

~

e.

e2 LAGS e, ~
90"

e"a;ro7

2. Since (total current)2 times 1000 = 208 mW, the total
0.208
current is ~
. - - = 14.4 rnA rms.
1000

100
OHMS

20mW

12MHz

3. The dB ratio can be calculated by taking the ratio of

No. 1 current to the total (14.4) and looking up the
corresponding number of dB in the current ratio-dB
table V6 (23.2 dB). Alternatively take the ratio of the
power output of No. 1 source (1 mW) to the total

Figure 16. Sum of Two Voltages of Different Frequency.

TOTAL

rms

rms

--......;:>~

CURREN T
OUTPUT S

l'

limA

#1

-#2

'f

4mA

#3

'f

II mA

'f

3mA

-#4

#5

CURRENT

14.4 mA

l'

SmA

-#S

l

5mA

~

>1000

=208
mW
Figure 17. Addition of Current of DiHerent Frequencies.

10

rms

OHMS VOLTAGE

<
TOTAL
POWER

TOTAL

=

14.4V

J

TABLE G

power (208 mW) and find the corresponding dB in
the Power-dB chart P7. The power ratio would be 208.
Divide by 100 to find dB corresponding to 2.08 (3.201)
and add 20 dB (because you had to divide by 100)
getting 23.2 dB.

dB
Increase
Number
Corresponding
of Equal
to This
Contributions Power Ratio

Power Addition-Equal Contributions
When the outputs of a number of equal sources are to be
combined, and the conditions indicate power addition, the
calculation can be simplified by using a table. The table is
prepared by noting that the power ratio is equal to the
number of sources. Thus the dB ratio of the output level is
10 10glO times the number of sources. Using this relationship, Table G has been prepared to show the increase in
output level fOiJ: 1 to 50 equal sources combining on a
power basis.
Example: 32 sources all having equal outputs and different frequencies combine in a common load. How much is
the total output increased over that which would· be supplied by a single source? Table G opposite 32 sources
shows an increase of 15.05 dB.
Power Addition-Two Unequal Contributions
Just as it was possible to simplify calculations by preparing a table for two unequal contributions in the case of
Voltage Addition, a table can be developed for the case
of Power Addition relating total level to input levels with
two unequal contributions. The preparation of the table
can be understood from the following:
Let L. represent the lower level in dBmV, p. the corresponding power ratio.
Let Lh represent the higher level in dBmV, Ph the corresponding power ratio.

We want a table arranged so that, when given Lh - L.,
we can find L t - L h.

=

=

-1010g1o

=

(!: );

=

= Ps + Ph and L t = 1010g1o (Ps + Pb),
Lh = 10 10glO (Ps + Ph) - 10 10glO Ph
h
= 10 10glO p.+P
= 10 loglo (-PP+ .1) ;
Ph

26
27
28
29
30

14.15
14.31
14.47
14.62
14.77

6
7
8
9
10

7.78
8.45
9.03
9.54
10.00

31
32
33
34
35

14.91
15.05
15.18
15.31
15.44

11
12
13
14
15

10.41
10.79
11.14
11.43
11.76

36
37
38
39
40

15.56
15.68
15.80
15.91
16.02

16
17
18
19
20

12.04
12.30
12.55
12.79
13.01

41
42
43
44
45

16.13
16.23
16.33
16.43
16.53

21
22
23
24
25

13.22
13.42
13.62
13.80
13.98

46
47
48
49
50

16.63
16.72
16.81
16.90
17.00

Power Splitting Relationships

-

In the analysis of CATV systems a situation is frequently
encountered, particularly with regard to taps and splitters,
where the power from a given source divides unequally
between two loads. Given the dB ratio between the power
from the source and the power in one load, we want to
find the dB ratio between the source power and the power
in the other load. This is another case of Power Addition,
but the relationship is a little different from that described
above, and another table is needed. Its preparation ean be
understood from the following:

b

P
Thus knowing ~, we can simply add 1 to that number
Ph
and then find the corresponding number of dB, representing L t - L h.

p.

p.

=

Example: let Lb - L. be 12 dB, then 0.063 and Ph
Ph
+ 1 1.063; then L t - Lh
10 10glO 1.063
0.265 dB.

=

0
3.01
4.77
6.02
7.00

Solution: The level difference is 11 dB, so the amount to
be added to the higher level (from TABLE P9) is 0.33 dB.
The combined output level is +64.33 dB or about 36
milliwatts.

so given Lh - L. we can find Ph

=

1
2
3
4
5

Example: The output of a channel 6 strip amplifier is combined with the output of a channel 10 amplifier in a common load. Channel 6 level is +54 dBmV, channel 10 level
is +64 dBmV. What is the total power output?

p.

Now, since: P t
hence: L t

to This

Power Ratio

Based on the above, a table has been worked out showing,
for the case of Power Addition, L t - Lh for values of
Lh - Ls between 0 and 20 dB in 0.1 dB steps. This table
is given as TABLE P9 in Chapter XI. An example will
illustrate its use.

Let L t represent the total level in dBmV, P t the corre;\.
.
spon d109 power ratIo.

10 10glO Ph and Ls
10 10glO p.,
By definition: Lh
hence: Lh - Ls
10 loglo Ph - 10 10glO p.

Number
of Equal
Contributions

dB
Increase
Corresponding

=

11

Let L. be the level of the lower one of the outputs, and
p. the corresponding power ratio
Ph the corresponding power ratio

Ph
p.
we first have to find - - 1, then invert to -'-, add 1, and
p.
Ph
then find the corresponding number of dB, representing
Lt- L h •

L t be the level of the input, and
P t the corresponding power ratio

Example: let L t

Lh be the level of the higher one of the outputs, and

We want a table arranged so that, when given L t
we can find L t - L h •

=

-

Lo,

=

=

=

=

= 10 loglo (

p.- + Ph)
Pb
.

= 10 loglo (P-Ph + 1
a

p.

+1=

39.81 'and

t -

A table PIO in Chapter XI has been .prepared for the case
of Power Splitting Relationships. An example will illustrate its use:

=

= 10 10glO( Ph:' P.) = 10 10glO(~ + 1 )
Ph
Thus given L t - L. we can find -;
,.
p.
but since L t - Lb
10log1o (P. + Pb) - 10 loglo Ph

Ph
L. be 16 dB, then -,-

-Ph = 38.81; inverted -p. = 0.026 and -p. + 1 = 1.026;
~
~
~
then L
Lh = 10 10glO 1.026 = 0.1104 dB.

=

By definition: Lh
10 10glO Ph and L.
10 loglo p.
and since: Pt
Ph + p., Lt
10 loglo (Ph + p.),
hence: L t - L. 10 loglo Ph + p. - 10 loglo p.

-

Example: A directional coupler divides energy between
two output terminals. Assuming it has no internal power
loss, and the loss to the tap output is 18 dB, what is the
line loss?

=

Solution: In the table opposite 18 dB (Lt - Lo) we find
0.069 dB (L t - ~). So the line loss is 0.069 dB.

)

12

CHAPTER III
RANDOM NOISE IN CATV SYSTEMS
Fundamentals

Thus the noise input into 7::; ohms is 1.1 microvolts RMS,
or -59 dBmV. This is the basic noise level, the minimum
that will exist in any part of a 75-ohm system.

In a CATV system, the lowest levels that can be allowed
at antenna output terminals, at repeater inputs, or at the
customer's set, without producing snowy pictures, are
determined by thermal noise. An understanding of this
noise, where it comes from and how strong it is, helps to
understand system limitations.

Sigllal-to-Noise Ratio
In order to avoid snowy pictures, the signal, at any point
in a system, must be sufficiently strong to override the
noise.

Any resistor or source which looks resistive over the band
in use (including antennas, amplifiers, or long cables)
generates a thermal noise signal. In the case of a resistor
this noise is due to the random motion of electrons, and its
strength can be calculated.

This relationship is expressed by the "Signal-toNoise Ratio," which is the difference between the signallevel, measured in dBmV, and the noise level, also
measured in dBmV; both levels being measured at the
same point in the system.
Although the CATV industry has not reached agreement
on the noise which can be tolerated in a picture signal,
careful tests have been made by other organizations and
much can be learned by considering the results.

R

75
ohms

Barstow and Christopher, of the Bell Telephone Laboratories, have published the results of careful studies on the
subject. (1), (2) Their tests are summarized in terms of the
signal-to-noise ratio which is rated "noise just perceptible"
by the average judgment of a group of trained observers.
The result which applies most nearly to the CATV situation is mentioned by Carson(3) referring ·to a picture
viewed at eight times picture height with Hat noise having
a 4.2 MHz bandwidth. The given ratio, 39 dB, refers to the
video signal (after detection) and corresponds approximately to an r-f ratio of 43 dB.

Figure 18.

If, as in Fig. 18, a sensitive high-impedance voltmeter
(which generated no noise itself) could be connected
across a 75-ohm resistor (or resistive source) it would
measure an open-circuit noise voltage calculated by:
en =y4RBk
where en is the RMS noise voltage
R is resistance in ohms
B is the bandwidth of the voltmeter in MHz
k isa constant approximately equal to
40 x 10- 16 at room temperature (68 0 F)

For comparison, consider the results of another series of
tests conducted by the Television Allocations Study Organization (TASO) and published in their report to the FCC
in 1959. Their ratings, corrected for a 4 MHz bandwidth
instead of the 6 MHz they used, are shown below:

A reasonable bandwidth for TV is 4 MHz. Assuming this
bandwidth, the open-circuited noise voltage for a 75-ohm
resistor is:
en = y 4 x 75 x 4 x 40 x 10
=
=

Y4.87 x 10 -

TASO Picture Rating
1. Excellent (no perceptible snow)
2. Fine (snow just perceptible)
3. Passable (snow definitely perceptible but
not objectionable)
4. Marginal (snow somewhat objectionable)

12

2.2 microvolts RMS.

If, as in Fig. 19, this source were counected to a 75-ohm
load (which had no noise in itself) it would deliver half
this voltage to the load:

45 dB
35 dB
29 dB
25 dB

(1) J. M. Barstow, and H. N. Christopher, "Measurement of

2.2,.,.V

Random. Video Interference to Monochrome and Color
TV," A.I.E.E. Transactions, Part I, Communications and
Electronics, Vol. 63, Nov. 1962, pp. 313-320.
(2) J. M. Barstow and H. N. Christopher, "The Measurement
of Random Monochrome Video Interference," A.I.E.E.
Transactions, Part I, Communications and Electronics, Vol.
73, Jan. 1954, pp. 735-741.
(3) D. N. Carson, "CATV Amplifiers: Figure of Merit and the
Coefficient System" 1966 IEEE International Convention
Record, Part I, Wire and Data Communication, March'

75-ohm

75-ohm
resistor
generating
noise

SIN Ratio

16

LOAD

1966, pp. 87-97.

Figure 19.

13

When it is decided how much noise is tolerable, the levels
required 'in a system can be specified. With a signal-tonoise ratio of 43 dB, for example, the minimum signal
level that would be required at the input to the first amplifier (if thermal noise were the only problem) would be
-59 + 43
-16 dBmV. Actual levels (to achieve this
signal-to-noise ratio) must be quite a bit higher because
of the noise that is contributed by the amplifiers.

Noise Figure in a Cascaded AmpliJier System
General:
The trunk-line of a CATV system often consists of a series
of nearly identical amplifiers equally spaced along a
coaxial cable. Determining the increase in noise due to
each amplifier in such a system helps to understand trunk
line operation.

=

Two Amplifiers:
Consider two amplifiers having equal gain and noise figure,
separated by a length of cable whose loss equals one amplifier's gain as shown in Fig. 21.

Noise Figure
When a 75-ohm resistor is connected to the input of an
amplifier having known gain, the noise output of the
amplifier is NOT, as might be expected, the input. noise
(-59 dBmV) increased by the amplifier gain.

-59dBmV

CABLE LOSS=GdB

750

N.F.=FdB

750 ~

AMPLIFIER
GAIN-40 dB

L....-N_.F._=_Fd_B_~

75n

Figure 21.

The noise level at the first amplifier's output is thermal
input noise (-59 dBmV) plus Gain (dB) plus Noise
Figure (dB). The cable attenuates this noise back down
to -59 + F, so that, in effect, there are two equal noise
sources at the input of the second amplifier: the output
of the first amplifier, attenuated by the· cable, and the
thermal noise at that point, including 'the equivalent noise
of the second amplifier. Since the sum of two equal powers
is 3 dB higher than either, the noise output from the
second amplifier is increased 3 dB over that of the first. It
follows that the noise figure of two identical amplifiers in
cascade is 3 dB higher than the noise figure of each one.
(See Note 1 on next page) .

NOISE OUTPUT
-9 dBmV

-

GAIN-GdB

Noise --59+G+F
Noise=-59+F+3dB
Noise=-59+G+F+3dB----------'

Consider, for example, an amplifier (Fig. 20) whose gain
is known (from measurement with a signal) to be 40 dB.
The measured noise output, with the input terminated, is
-9 dBmV. What is the Noise Figure?
NOISE INPUT
-59 dBmV

GAIN=GdB

!

LJ

----..r---------~

Since the amplifier always generates some internal noise,
noise output is always greater than it would be from a
noiseless amplifier having the same gain. This increase in
noise output, expressed in dB, is called the "Noise Figure"
of the Amplifier.

..

~~----------~Figure 20.

If the amplifier had no internal n~ise, the noise output
would be the input noise (dBmV) plus the amplifier gain
(dB).

More Than Two Amplifiers:
By extending this logic it can be seen that, when a system
is extended from two amplifiers to four amplifiers, the
Noise Figure is again increased 3 dB and in general:

Noise output (no amp. noise) = -59 +40 = -19 dBmV.
The measured noise is -9 dBmV, showing that the
amplifier adds 10 dB to the noise output, and that the
amplifier Noise Figure is 10 dB.

When identical amplifiers, connected by identical
cable lengths whose individual losses equal one am.plifier's gain, are cascaded, the system noise figure
increases 3 dB each time the number of cascaded
amplifiers is doubled.

The Noise Figure of an amplifier or system is the
difference between the measured output noise level
(in dBmV) with a terminated input, and the thermal
noise (-59 dBmV) plus the gain (in dB) of the
amplifier.

or stated mathematically:

=

Fm Fl + 10 10glOm
where F m is system noise figure, F 1 is amplifier noise figure
and m is the number of cascaded amplifiers. The quantity
"10 loglo m" is called the cascade factor (C).
Since any increase in noise figure decreases the signal-tonoise ratio, it follows that:
With a noise-free input signal, the system signal-tonoise ratio decreases 8 dB each time the number of cascaded amplifiers is doubled (See Note 2 on 'next
page).

If an amplifier contributed no noise, the signal and the
noise going through it would be amplified equally and the
signal-to-noise ratio would be unchanged. Since the amplifier output contains added noise, as indicated by the
Noise Figure, it follows that the -Output Signal-to-Noise
Ratio (in dB) is decreased as compared to the Input Signalto-Noise Ratio. If the input noise is at the minunum level
( -59 dBmV), that is if the input signal is noise free, the
output Signal-to-Noise ratio (in dB) .is found by subtracting
the Noise Figure from the input Signal-to-Noise ratio.
14

Table H can be used to find system noise figure or signalto-noise ratio when these quantities are known for the
individual amplifier.

Where: fI> f2 and f3 are the noise factor 'of the first,
second and third device, respectively.
gl and g2 are ~he power gain of the first and
second device, respectively.
(3) The power "gain" of an attenuating device is the ratio
output power
.
, a number less than one. The noise facmputpower
tor of an attenuating device is the reciprocal of its
power gain.
To apply these facts to the analysis of a cascade of two
amplifiers assume that:
Device # 1 is an amplifier. fl fa, g1 ga
Device #2 is a length of cable whose loss is equal to
the amplifier's gain

To find the system noise figure for a given number of
identical amplifiers in cascade: To the noise figure of an
individual amplifier ADD the cascade factor found in the
table opposite the number of amplifiers in cascade.
To find the system signal-to-noise ratio for a given number
of identical amplifiers in cascade: From the signal-to-noise
ratio at the output of the first amplifier SUBTRACT the
cascade factor found in the table opposite the number of
amplifiers in cascade (see Note 2 below).

TABLE
No of
Amps. in
Cascade

Cascade
Factor (C)

=

H
No of
Amps. in
Cascade

Cascade
Factor (C)

1
2
3
4
5

0
3.01
4.77
6.02
7.00

26
27
28
29
30

14.15
14.31
14.47
14.62
14.77

6
7
8
9
10

7.78
8.45
9.03
9.54
10.00

31
32
33
34
35

14.91
15.05
15.18
15.31
15.44

11
12
13
14
15

10.41
10.79
11.14
11.43
11.76

36
37
38
39
40

15.56
15.68
15.80
15.91
16.02

16
17
18
19
20

12.04
12.30
12.55
12.79
13.01

41
42
43
44
45

16.13
16.23
16.33
16.43
16.53

21
22
23
24
25

13.22
13.42
13.62
13.80
13.98

46
47
48
49
50

16.63
16.72
16.81
16.90
17.00

so g1g2

= ~~2 = 1,

g2

=

= -ga1

Device #3 is an amplifier identical to #1 so f3
g3 = ga

= fa,

Substituting these values in equation (2)
f123
fa

f2 - 1

f3 - 1

ga - 1

fa -.;-1

= f1 + -~~.
- + - - = fa + --- + - - =
L
1

+1-

1
-

+ fa g.

1, so f123

= 2 fa -

1
ga

which says that the noise factor of two identical amplifiers
in cascade is equal to twice the noise factor of one amplifier; less the reciprocal of its power gain. With noise factors
of about 10 X (10 dB) and power gains of about 100 X
(20 dB) it is apparent that the ·effect of the second term is
negligible, so f123 = 2 fa or, in logarithmic terms, the noise
figure of two identical amplifiers in cascade is 3 dB higher
than the noise figure of each.
Note 2:
The "system signal-to-noise ratio" (in dB) is the difference
(in dB) between the system's noise output level (indBmV)
with the input terminated, and the operating signal level at
the output terminals (in dBmV) .
Where the input signal is not noise-free, the noise it contains can be taken into account to find a total effective
signal-to-noise ratio. This can be done for usual conditions
by assuming that the total noise power output is the sum of
two components: the noise power input increased by the
system gain, and the noise power output of the system with
its input terminated. In dB terms this is done by the method
outlined on p. 72, par. 8, or as follows:

Note 1:
The analysis on the preceding· page for the case of two
amplifiers in cascade is not very precise. An accurate
analysis can be made based on three established facts:

1. Take the difference between the SIN ratio (in dB) of
the input signal at the first cascaded amplifier and the
SIN ratio (also in dB) of the cascaded amplifier system.
2. Use chart P3 or table P9 to find the number of dBs to
be subtracted from the smaller of the two ratios to find
the total effective SIN ratio.

(1) Noise factor (f) is related to noise figure (F) by:
F 10 loglof (see page 72, pal'. 6)

=

(2) The noise factor for three devices connected in succession (output of the first to input of the second, etc.) is:
f2 - 1
fa-l
f123 fl + - - - + - - g1
g1~

=
=

=

Example: System SIN ratio
43 dB; Input SIN ratio
47 dB; Difference 47 - 43 4 dB; From table
P9 opposite 4 dB find 1.46 dB. Thus total effective SIN ratio is: 43 - 1.46 = 41.4 dB.

=

15

=

CHAPTER IV
mE FUNDAMENTALS OF DISTORTION IN CATV AMPLIFIERS
Summary

interfering one. These harmonics and beats can combine
with the carrier on the channel in use to cause "herringbone" patterns in the picture. A· study of this type of
distortion will help in understanding how CATV amplifiers
can be operated to avoid these problems.

This chapter presents an elementary introduction to the
nature and effects of non-linear distortion in CATV
amplifiers. Two forms of distortion are important: second
order distortion which results, in extreme cases, in the
compression of one peak of a sine-wave and the expansion
of the other; and third order distortion which results, in
the extreme case, in compression (or expansion) of both
peaks. It, is shown that the effects of second order distortion may be analyzed by an equation involving a linear
term and a term proportional to the input voltage squared,
while the effects of third order distortion can be analyzed
by an equation including a linear term and a term involving the input voltage cubed.

Distortionless Amplification
Perhaps the simplest way to describe amplifier distortion
is to say what it is NOT. A distortionless amplifier would
be one which increased' the amplitude (voltage swing) of
the input signal without changing its waveform. Suppose,
for example, an amplifier could be built so that the output
voltage, at each instant, was exactly 10 times the input
voltage. A graph showing the output voltage plotted against
the input voltage would be a straight line, as illustrated in
Fig. 22. Such a graph is called the "transfer characteristic"
or "input-output curve," for the amplifier. A transfer characteristic which is a straight line is called a "linear transfer
characteristic."

In present CATV systems, second order distortion is generally not considered because, as will be shown, it results in
distortion products at frequencies which are either rums,
differences, or second harmonics of the carriers present and,
with the standard frequency assignments, these products
fall outside of the channels used.

Mathematically, the performance of this amplifier would
10 ein; where eont
be described by the equation: eont
is the instantaneous output voltage, and el n is the instantaneous input voltage. Calculating for particular voltages
would give a table:

=

Thus, the limiting factor in determining permissible ,output
levels in most CATV amplifiers is third order distortion
since it results in cross-modulation between channels, and
beats between carriers which fall inside the channels being
used.

eont (= 10 eln )

0

0
-0.2
-0.4
-0.6
-0.8
-1.0

Introduction
Distortion in sound reproducing equipment is familiar to
anyone who has heard a worn-out juke box, or an overloaded public address system. This harsh, unpleasant sound
presents the essential nature of all distortion: What comes
out of the system is different from what went in! In CATV,
distortion does not show up in the same way, but it is
present, and it places restrictions on amplifier operation
which must be understood if a system is to be intelligently
planned and operated.

eont (=

eln

-4
-6
-8
-10

eln )

0
+2
+4
+6
+8
+10

0
+0.2
+0.4
+0.6
+0.8
+1.0

-2

l{)

This is the table from which the characteristic of Fig. 22 is
plotted.
+10
+8

The amplifiers used in CATV have only one intended
function: to raise the signal levels. The other things
they do, the differences they generate between the outgoing signals and the incoming signals, are distortion. What
forms does this distortion take? Several effects properly
called distortion, such as the addition of noise to the signal,
hum. modulation and variations in amplifier frequency
response are NOT the subject of this chapter. Here, we are
concerned with only one kind of distortion: effects due to
the same causes that create "harmonic distortion" in audio
amplifiers.

V

+6
OUTPUT
+4

1.0

This distortion is due entirely to non-linearity in the transfer
characteristics of the transistors. Its worst' aspect is crossmodulation, crossing over of the modulation from one
channel to another, which causes "windshield wiper" effects
in the picture. Other effects include harmonics, where an
unwanted signal is generated at a frequency which is some
multiple of the frequency of a wanted one; and beats,
where two or more wanted signals combine to generate an

-.8

~

[7

1/

-'~NPu~4 -.~

~

~

~

0

-2

+.2

~

+.4 +.6
VOLTAGE

-4

VOLTAGE
-6
-8
-10

Figure 22. A Linear Transfer Characteristic

16

~

+.8

+I.o

The way in which such a linear transfer characteristic
results in an undistorted output is shown in Fig. 23. A plot
of the sinusoidal input voltage against time is illustrated
[Fig. 23(a)]. If, at each point along the time scale, the
instantaneous input voltage is projected downward to the
transfer characteristic [Fig, 23 (b)], the corresponding output voltage is found. Projecting this to the right, and
plotting against the same time scale, constructs graphically
the waveform of the output voltage [Fig. 23 (c) ]. For example, when the input is 0.75 volts and decreasing (point
"A"), the output is 7.5 volts and decreasing (point "B").
Since the output voltage at any time is simply ten times
the input voltage, the output duplicates the input waveform.

When a varying voltage is applied. to an amplifier with a
characteristic of this sort, the output voltage has a different
waveform than the input voltage. Consider the examples
shown in Figure 26. Figure 26 (a) illustrates the output
voltage waveform obtained when a sinusoidal voltage with
a voltage swing between + 1 and -1 volt is applied to the
amplifier whose characteristic is illustrated in Figure 25.
Since the transfer characteristic is symmetrical, both peaks
of the output voltage are flattened by the non-linearity,
giving the waveform illustrated.
A 1.0 volt peak-to-peak sinusoidal voltage applied to the
input of the same amplifier and biased at -0.5 volts so
INPUT
VOLTAGE

o

-I

Distortionless amplification does not require that the input
be a pure sinusoidal voltage. It is achieved when the waveform of the output voltage precisely duplicates that of the
input, regardless of what that waveform may be. Fig. 24,
for example, presents a diagram similar to Fig. 23 except
with a pyramidal input, showing how an identically-shaped
pyramidal output results.

TIME

t

,
I

,

Unfortunately, amplifiers that can be built using real-life
transistors do not have a linear relationship between the
input voltage and output voltage. Figure 25 illustrates a
non-linear transfer characteristic which might be found in
a real amplifier. As the input voltage swings either way
from 0, the output changes along a curve which produces
less and less change in output voltage as the input swings
further and further from O. If, in Fig. 25, the output
would continue to increase almost linearly, as it does between 0 and 4 V, it would reach about 20 V for a I-V
input instead of 10 V as shown.

./

....

........... I,

I

-....:.....

I

....... .,/

, .>

'.

,
I

''1I
~

I
I

I

+10

/~

-I

/

/

/

+10

,,,'

/1L_---/

I

OUTPUT
VOLTAGE

/

-

or---~~----+-----~-

_ _ +1 _ _ -

_ _ __

/
-10

-10

TRANSFER
CHARACTERISTIC

Figure 24. Distortionless Amplification.

INPUT
VOLTAGE

o

.,........,

././
./

I

Amplification with Distortion

-I

'-

+1

+1

+10
~

+8
A

/

OUTPUT +6
VOLTAGE

eout =IOe in

I

+4

/'

v

~

V

+21(
1.0 -.8
+10

-.6 -.4 -.2
INPUT

+10

(C)

(b)

OUTPUT
VOLTAGEOr-______~~______~-

-I

-10

-10

TRANSFER
CHARACTERISTIC

~

Y

/

V

/

10

o

II

0

+.2 +.4 +.6 +.8 +1. o

-2

VOLTAGE

-4
-6
-8
-10

Figure 25. A Non-linear Transfer Characteristic.

figure 23. Distortionless Amplification.

17

a gain of ten times and no distortion. Since all practical
amplifiers cause distortion, a sensible question is: "Can
the transfer characteristic of a practical amplifier be expressed in some simple mathematical way which will allow
analysis of the distortion generated?" The answer is that
the transfer characteristic of a practical amplifier can be
approximated by a simple mathematical expression and the
subject of what follows is how this is done.

(0)
-10

First, consider an amplifier which generates the kind of
distortion illustrated in Figure 26 (b). The transfer characteristic causing this kind of distortion can be approximated
by an equation having the form "( eout) equals (some
number X ein) plus (some other number X etn2 )."

( b)

The following may help to understand how this works.
Consider first the curve that results when e 2 is plotted
against e; the numbers are tabulated below.

+2

(c)

e

e2

e

e2

-1.0
-0.8
-0.6
-0.4
-0.2
0

+1.0
+0.64
+0.36
+0.16
+0.04
0

+1.0
+0.8
+0.6
+0.4
+0.2
0

+1.0
+0.64
+0.36
+0.16
+0.04
0

The corresponding curve is plotted in Figure 27. Notice
that it is symmetrical around the vertical axis, curving up
smoothly for both positive and negative magnitudes of e.

Figure 26. Output Voltage -Waveforms, Non-linear Characteristic.

that it varies between 0 and -1 volt produces an output
varying between 0 and -10 volts with the waveform illustrated in Figure 26 (b). The lower peak is flattened because the transfer characteristic bends over at -1 volt input; the upper peak is faithfully reproduced because the
characteristic is very nearly a straight line near O.

Next consider an example of what happens when such a
curve is added to a linear transfer characteristic. The output voltage is separated into two parts:
for the linear part:
el = 10 eln
for the "squared" part: e2 = 5 eln2
and for the total:
eout = el + e2
= 10 eln + 5 eln 2

Reducing the amplitude of the input voltage to 0.2 volts
peak-to-peak and biasing it at 0 so that it varies between
+0.1 and -0.1 volt gives the output voltage shown in
Fig. 26 (c). Because the signal varies along a nearly linear
part of the characteristic, this is almost an undistorted
reproduction of the sinusoidal input.

The transfer characteristic described by an equation having
a linear term (one involving "e") and a "squared" term
(one involving "e 2") is known as a "square-law" transfer
characteristic.

It should be clear from these examples that the nature as

well as the degree of distortion is dependent not only on
the shape of the transfer characteristic of the amplifier but
also on the amplitude of the input signal and on the operating point (bias). Two very different and significant kinds
of distortion are illustrated: one where the peaks are flattened symmetrically [Fig. 26(a)] and the other where only
one peak is flattened [Fig. 26 (b) ]. In what follows these
two cases will be explored more thoroughly.
Second Order Distortion
In the section on distortionless amplification, it was shown
that a linear transfer characteristic could be expressed in
very simple mathematical terms. The equation "eout =
10 eln" says very clearly that the amplifier in question has

-1.0

+--e

Figure 27.

18

o

e+-+

e2 Plotted Against e.

+1.0

INPUT
VOLTAGE
-I'--_ _O~--+.!.I

The numbers are tabulated below:
eln

10 ein

-1.0
-0.8
-0.6
-0.4
-0.2
0
+0.2
+0.4
+0.6
+0.8
+1.0

-10
- 8
- 6
- 4
- 2
0
+ 2
+ 4
+ 6
+ 8
+10

ein2
+1.0
+0.64
+0.36
+0.16
+0.04
0
+0.04
+0.16
+0.36
+0.64
+1.00

10 el n + 5 ein2

5 ein2

5.0
4.8
- 4.2
- 3.2
- 1.8
0
+ 2.2
+ 4.8
+ 7.8
+11.2
+15.0

+5.0
+3.2
+1.8
+0.8
+0.2
0
+0.2
+0.8
+1.8
+3.2
+5.0

-

(0)

+10

-10
TRANSFER
CHARACTERISTIC

-10

Figure 29. Second Order Distortion.

+1

-I

that "a square-law transfer characteristic (or a characteristic having second-order curvature) causes second-order
distortion of the output." Observe that not only is the
upper peak of the output voltage stretched by the action
of the second-order distortion and the lower peak flattened,
but also the entire curve is shifted upward so that its
average is above O.

eout =

2
--IOe in +5e in

/
+10

/

It has been shown that one way to study the effects of

/

second-order distortion mathematically is to use a squarelaw equation. There is a second approach which is also
very useful. This involves the addition of DC and AC
component voltages to produce a distorted total. Before
this is presented it might be well to review the meaning
of the three terms "DC," "AC" and "component."

/

V

+1

-I

"AC" and "DC"
The idea of "DC" is familiar. A DC voltage is typified by
the voltage between the terminals of an ideal battery
[Fig. 30(a)].

-10

It does not vary either in magnitude or polarity. The plot
of a DC voltage is a straight line parallel to the time axis
[Fig.30(b)].

Figure 28. Square-law Transfer Characteristic.

Fig. 28 shows the two curves plotted separately (a and b)
and the total ( c ). Notice the similarity between this
total curve (the plot of a simple mathematical equation)
and the lower half of a particular non-linear transfer characteristic (Fig. 25).

"AC" is almost equally familiar. To qualify as "AC" a
characteristic must vary above and below zero in such a
way that its average is zero. A periodic "AC" voltage is
one that goes through identical cycles of change over and
over again, and its average over anyone cycle is zero.
Another way of saying this is: "the area enclosed by the
plot is the same on the positive half-cycles as that on the
negative ones."

Fig. 29 illustrates graphically how the introduction of a
sinusoidal voltage into an amplifier having a square-law
transfer characteristic results in an output of the one-peakstretched, one-peak-flattened variety. Since this kind of
distortion results from the addition to the linear characteristic of a quantity involving e2 , it is called "second
order" distortion. In these terms it is said that Fig. 29 shows

Figures 31 (a), (b) and (c) all represent periodic AC
voltages, since in each case the variation is repeated in
identical cycles and the average is zero. To see how the

19

+

(a)

14

THE TOTAL VOLTAGE

+20

I

e

4-AVE.=+IOV

(0)
0

I

14

0

TIME-+ 100
MICRO -SECONDS

(c)

THE AC COMPONENT

(b)

200

A~ l
~

~

VOLTS

( b)

J

TOTAL

DC-=L;-

0

-10

+10

(d)

THE DC COMPONENT
VOLTS + 1 0 1 - - - - - - - - - - - - - -

o~~------------------------------

TIME--+

OL---__________________________
Figure 30. DC Voltage.
Figure 32. A Voltage with AC and DC Components.

average is zero, imagine a DC voltmeter reading each of
these voltages. Assuming the change to be faster than the
needle can follow, it would be pushed equally in both
directions, so would read zero.
VOLTAGE 15 AC WHEN AREA

AC and DC Components

"X" = AREA "y"

+2
(a)

TIME-

-2

( b)
AREA

Or-----~-------r--~X~_+------,__

How then is a voltage described which varies periodically
but in such a way that its average is not zero? Figure
32 (a), for example, illustrates a voltage which has a sinusoidal waveform, and an average of +10 volts. It is conveniently described in terms of its parts or "components." This
waveform could be obtained by connecting a 10-volt battery in series with an AC source generating a 10-volt peak
sinusoidal AC voltage [Fig. 32(b)]. Although it is actually
generated in some other way (it might, for example, occur
at the collector of a transister amplifier), it is still convenient to describe it in terms of these components. In these
terms it is called a "composite AC and DC voltage, its AC
component being a 10-volt peak voltage with sinusoidal
waveform, and its DC component +10 volts." By breaking
this complex voltage into two simpler components, it is
made easier to talk about and to measure. The DC component voltage can be measured with a DC voltmeter, and
the AC component with an AC voltmeter.
The Spectrum of a Composite Voltage:

-10

When a periodic varying voltage contains several components like this, it can be conveniently analyzed by plotting
its "spectrum." A spectrum is simply a graph which plots,
in the vertical direction, the peak voltage or amplitude of
each component and, in the horozinotal direction, the frequency at which each of these qomponents exists. Its
importance rests on the fact that "spectrum analyzers" are
available which plot these diagrams automatically, providing tremendously useful tools for distortion analysis.
The spectrum of a sinusoidal voltage is a single spike
showing the amplitude and frequency of that voltage.

Figure 31. Three AC Voltages with Different Waveforms.

20

Fig. 33 shows the spectrum of the composite voltage
whose waveform is plotted in Fig. 32 ( a). The spectrum
contains the same information as the time plot. It says that
this voltage consists of two components, a DC component
of 10 volts (represented by the 10-volt spike at 0 frequency) and a sinusoidal component of 10 volts peak
amplitude at a frequency of 10 kHz (represented by the
10-volt spike at 10 kHz).

A Voltage with Two AC Components
This technique of representing a varying voltage as the
sum of several components has very wide application.
Fig. 34 illustrates a second situation where it is useful.
Fig. 34 (a) shows a composite voltage which quite obviously contains a high-frequency variation (causing the
rapid oscillation) and a low-frequency one (causing the
slow oscillation). It is the sum of two equal sinusoidal
voltages; one having a frequency of 100 kHz (one cycle
in 10 microseconds) shown in Fig. 34 ( c ) ; the other
higher frequency component, Fig. 34 (d), completes 10
cycles in 10 microseconds so its frequency is ten times
higher, or 1 MHz.

10

Fig. 35 shows the spectrum of this composite voltage.
The spectrum indicates two 5-volt sinusoidal components,
one at 0.1 and one at 1.0 MHz.

t!,

~
g

"

C(

'"

5 --

Q.

4
0

'0

PEAK
VOLTAGE

FREQUENCY-

- - _.

3

i
I

i
I

2

Figure 33. Spectrum of a Composite Voltage.

I
-- --

+10

(0) THE

o

~

'-

o 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

TOTAL VOLTAGE

FREQUENCY MHz
Figure 35. Spectrum of Voltage with Two AC Components.

+5

Second Order Distortion by Addition of Components
VOLTS

0

The use of sinusoidal components to represent a nonsinusoidal varying voltage has its most important application in the study of distortion. Earlier we discussed the'
distortion produced when a sinusoidal input is applied to
an amplifier with second order distortion.

(b)

TIME"

-5

L~{l
HI~
J
~

(C)rHE LOW FREQUENCY COMP.

TOTAL

The waveform resulting. from this distortion is plotted in
Fig. 29 (c). An identically distorted waveform can be
formed by adding sinusoidal components (or, conversely
such a waveform can be separated into sinusoidal components) . Fig. 36 ill;ustrates the process.

O.
microseconds

This diagram shows how a distorted output can be generated by adding three components: the FUNDAMENTAL
component, a sinusoidal voltage having a frequency of
1 MHz (1 cycle in 1 microsecond) in. this example; the
SECOND HARMONIC compnnent, a sinusoidal voltage
having twice this frequency, 2 MHz (2 cycles in 1 microsecond); and a positive DC component.

'V

0

-5
+5

(d) THE

HLGH FREQUENCY COMPONENT

0

Notice first that the total voltage has a waveform identical
to that shown in Fig. 29 (c) (the one produced when a
sinusoidal voltage is passed through an amplifier with
a sqmue-law characteristic). Now see how the three components add in Fig. 36: At 0 time on the diagram, the

Figure 34. A Voltage with Two Sinusoidal Components.

21

+15

a 2.5-volt DC component, a 10-volt peak I-MHz component (the fundamental) and a 2.5-volt 2-MHz component
(the second harmonic). Some points shouJd be noted in
regard to these diagrams. Each spike on a spectrum diagram ALWAYS represents a pure sinusoidal component.
If a periodic v9ltage is non-sinusoidal, its spectrum shows
more than one component. When a DC component exists,
it does not appear on the spectrum displayed by the usual
spectrum analyzer. The input circuits of most analyzers
respond only.to the AC components.

(d)

:rOTAL.FUND.+2NO HARMONIC+OC

(0)
FUNDAMENTAL COMPONENT(TlOV PEAK)

+10

(e)
DC COMPONENT (+2.5V)

+5

Third Order Distortion
VOLTAGE 0 ~-.,L--+--~r--~r---:;~--+--~---T
0.75
I.O....SEC

In a previous section it has been shown that the kind of
non-linearity which results in the "one-peak-Hattened" kind
of distortion can be expressed by a simple square-law
mathematical equation. In very much the same way, the·
kind of distortion which results in both peaks being flattened can be expressed by a cube-law equation. This
equation has the form: (eont) = (some number x eln ) (some other number x ein3 ). It approximates the transfer
characteristic which results in a waveform with both peaks
flattened, as illustrated in Fig. 26 (a) .

TIME ( ....SEC)

I

-5

'/

I
I
,

, .... _ / I

I

...

-10

Figure 36. Addition of Second Order Components.

fundamental component is 0, the second harmonic is at
its negativ,e peak (-2.5 volts) and the DC component is
at +2.5 volts. Adding the three gives the total voltage
which is O. At 0.25 microseconds the fundamental has
gone through one-quarter cycle to its positive maximum
( + 10 volts), the second harmonic component has gone
through one-half cycle to its positive maximum (+2.5
volts) so the three add to produce the stretched peak of
the total (+15
2.5 + 2.5 + 10). At 0.75 microseconds
the second harmonic and-the DC are at +2.5 volts so they
subtract from the -10 volt peak of the fundamental to
flatten the peak of the total (-5 = -10 + 2.5 +2.5).

Consider the curve that results when e 3 is plotted against e.
The numbers are tabulated here.

=

Figure 36
principle:
produced
sinusoidal

illustrates one case of a very important general
Any non-sinusoidal periodic waveform can be
by an appropriate combination of DC and
components.

e

e3

e

e3

-1.0
-0.8
-0.6
-0.4
-0.2
0

-1.000
-0.512
-0.216
-0.064
-0.008
0

+1.0
+0.8
+0.6
+0.4
+0.2
0

+1.000
+0.512
+0.216
+0.064
+0.008
0

The corresponding curve is plotted in Fig. 38. It is "skew
symmetrical"; that is, the curve for negative magnitudes
of e has the same shape as for positive magnitudes, but is
upside down and has opposite direction.
+1.0

The Spectrum of a Voltage with Second Order Distortion
The distorted voltage of Fig. 29(c) and Fig. 36(d) can
also be represented by the spectrum diagram shown in
Fig. 37. This shows the three components that make it up:
10.0

6,0

w

C>

./

;'!

~~~~~~:~:(IIOO%)

...J

0

e+-+

.."

EiO >
W

Q.

4.0

JDC

component{25%
/

,

0

o

I,
1MHz

2nd Harmonic component
(25%)

1\
2M Hz

3MHz

-1.0

FREQUENCY

Figure 38.

Figure 37. Spectrum Analysis Showing 2nd Order Distortion.

22

e3 PloHed Against e.

+1.0

When this curve is added to a linear transfer characteristic,
it affects both extremes in the same way, since the linear
part and the "cubed" part go positive together and negative together. Consider an example:
Let the linear part of the characteristic be: e1 = 10 ein
and the "cubed" part be: es = 3 ein3 •
To get a curve which flattens the peaks, the cubed part is
subtracted from the linear part, so the total is:
eout = e1 - es = 10 eln - 3 elns
In this example, the cubed tenn is subtracted from the
linear tenn to give a characteristic which flattens the peaks.
In an amplifier with such a characteristic, the gain decreases as the input level is increased. (This is called
"compression.") It is also possible to design amplifiers
whose gain increases as the input level is increased. Such
a characteristic is approximated by adding the cubed tenn
to the linear tenn. In this case both peaks are stretched
(this is called "expansion").
+10

For the above example of compression the numbers are
tabulated here.
eln

10eln

6;n3

3 eins

10 ein - 3 elns

-1.0
-0.8
-0.6
-0.4
-0.2
0
+0.2
+0.4
+0.6
+0.8
+1.0

-10
-8
-6
-4
-2
0
+ 2
+4
+ 6
+ 8
+10

-1.000
-0.512
-0.216
-0.064
-0.008
0
+0.008
+0.064
+0.216
+0.512
+1.000

-3.000
-1.536
-0.648
-0.192
-0.024
0
+0.024
+0.192
+0.648
+1.536
+3.000

-7.000
-6.464
-5.352
-3.808
-1.976
0
+1.976
+3.808
+5.352
+6.464
+7.000

-10
Figure 39. Cube Law Transfer Characteristic.

Fig. 39 shows the two component curves plotted separately
(a and b) and the total (c). Notice the similarity between
this total curve, the plot of a simple equation, and the
non-linear transfer characteristic shown in Fig. 25.

INPUT
VOLTAGE
-I

o

+1

Fig. 40 illustrates graphically the way in which the introduction of a sinusoidal voltage into an amplifier having a
"cube-law" transfer characteristic results in an output oj:
the ''both-peaks-flattened'' variety. Since this kind of distortion results when a signal is passed through a transfer
characteristic having an equation with a tenn containing
e 3, it is called "third order" distortion. In these tenns it is
said that Figure 40 shows that "a cube-law transfer characteristic (or a characteristic having third order curvature)
causes third order distortion of the output."
Third Order Distortion by Addition of Components
In the foregoing it was found possible to duplicate the
effects of second order distortion by adding sinusoidal
components. In a similar way, the effects of third order
distortion can be obtained. Figure 41 illustrates the Jlddition of a 10-volt peak, I-megahertz fundamental component (a) and a I-volt peak, 3-megahertz third hannonic
component (b) to produce a distorted total (c) having the
same wavefonn as that generated by the cube-law equation
illustrated in Fig. 40 (c). Because of the 3: 1 frequency

TRANSFER
CHARACTERISTIC

Figure 40. Third Order Distortion.

23

concerned with what happens in an amplifier when more
than one sinusoidal voltage is introduced into it. Although
the picture carrier on each channel is not a constant-amplitude sine-wave (since it is modulated with the picture information), by temporarily pretending that it is, we can
learn a great deal about the nature of distortion in this
case.

+10.0

( b ) 3RD

HARMONIC (10%)

L- ....

The first question then is: What is the waveform resulting
when two sine-waves having slightly different frequencies
are added? The two television carriers are said to have
SLIGHTLY different frequencies because in general their
frequency separation is small compared with their frequencies. (In contrast to the case illustrated in Fig. 34
where one frequency is ten times the other.)

,;'

TIME-+

To answer this question it is helpful first to consider the
way in which two sinusoidal voltages add when both have
the same frequency and amplitude, but have various phase
relationships. Fig. 43 illustrates several cases showing
each voltage separately (a and b) and the resulting total
voltage (c).

-10.0

Figure 41. Addition of Third Order Components.

relationship, the third harmonic voltage is opposite in
phase to the fundamental at its positive peak, with the
result that the peak of the total is Hattened and is again
opposite in phase at its negative peak, so the peak of the
total is also Hattened at that time.

Spectrum of a Voltage with Third Order Distortion
Figure 42 illustrates the spectrum of this distorted voltage.
Since the distorted waveform is duplicated by the sum of
two components, the spectrum shows only these two: a
10-volt-peak fundamental component at 1 MHz and a
I-volt-peak third harmonic at 3 MHz.
INDICATED PHASE
RELATION IS THAT OF
VOLTAGE (b) WITH

~c

RESPECT TO VOlTAGE(a)

.--5FUNDAMENTAL (100%)
U>

Figure 43. The Addition of Two Sinusoidal Voltages.

-'
0

>

I
....
C

::::>

E\.O

When both voltages are sinusoidal, and the frequencies are
identical, and the voltages are exactly in phase, the two
reach their peaks at the same instant and at that time they
add directly (e.g. 1.0 + 1.0 2.0) so the peak voltage of
the total is the sum of the two components (shown as the
0 condition).

..J

0.

::E

'"
4,0

I

.!

=

2P

0

~RD HARMONI C

0

o

IMHz

When there is a 90 phase difference between the two, the,
total reaches its maximum at a time when each of the
components is at 0.7 of peak, so the peak voltage of the
total is reduced to 0.7 of the sum of the peak voltages of
the components [e.g. +0.7 + 0.7 = 1.4 (the +90 and
-90 conditions)]. When the two voltages have opposite
phase (180 out of phase), they are equal and opposite at
all times, and the total is 0 (the 180 condition).
0

(10"/0)

,
2MHz
FREQUENCY

3MHz

Figure 42. Spectrum Analysis Showing 3rd Order Distortion
of a Sinusoidal Voltage.

0

0

0

0

The Sum of Two Sinusoidal Voltages Having Slightly
Different Frequencies

Next consider two sinusoidal voltages having slightly different frequencies. Figures 44 (a and b) illustrate the waveforms of two particular voltages. Each is sinusoidal, with
a peak amplitude of 2 volts. One has a frequency of 5 MHz,
a time-per-cycle of 1/5 microsecond; the other has a fre-

Since a major objective of this chapter is to explain the distortion that occurs in broadband amplifiers when many
"channels" are handled simultaneously, it is necessarily
24

quency of 6 MHz, and a time-per-cycle of 1/6 f-tsec. Thus,
the former completes 5 cycles in a microsecond while the
latter is completing 6 cycles.

+4

Superimposing the two waveforms on each other [Fig.
44(c)] shows clearly a highly significant fact: the phase
relation between them is changing constantly. Initially they
are in phase (both at positive peak). After microsecond
the 5-MHz voltage has gone through H4 cycles and is 0,
going negative, while the other has gone through three
half-cycles and is at its negative peak. They differ in phase
by 90°. After l~ microsecond the 5-MHz one is at its negative peak, while the 6-MHz one is at its positive peak, and
they are 180° out of phase. As time goes on, they go
through all possible phase relations, coming back to the
"in phase" condition once each microsecond. It is true in
general that when signals have different frequencies their
phase relation changes constantly.

+2

(0 )

*

VOLTS

0
Il SE C

-2
-4
-TIME
TOTAL PLOTTED ALONE

+4
"-

"-

I

..,.

I/

+2
/

~

//

V

(b)
VOLTS

0
1.0 II- SEC

VOLTS

(0 )

-2
-4

~

0

Y

'-.

./

"-

114

112

314

WAVEFORM OF A 6 MHz SINUSOIDAL VOLTAGE

+2

Figure 45. The Sum of Two Equal 5 MHz and 6 MHz Voltages.

(b)

VOLTS

peaks reduce on successive cycles reaching 0 after l~ microsecond when the two components are 180° out of phase,
and building up again to a 4-voIt maximum peak after
one microsecond when they come back in phase again.
Care must be exercised in using the term "peak" in reference to a varying voltage with a waveform like this. The
basic meaning of the word "peak" for any periodic waveform is "the highest voltage reached at any point in the
cycle." In this sense the peak voltage of this waveform is
4 volts, and it is reached once each microsecond. In another
sense this voltage reaches a 4-volt positive peak at
on
the time scale, then goes to a slightly lower (-3.5 volt)
negative peak, then to a 3-volt positive peak and so on.
lf two lines are drawn through these peaks, as illusb'ated
in Fig. 45 (b), they are said to outline the "envelope" of
the waveform.

(c)
-2

o

V4

112

314

°

I.OII-sec

TlME_
Figure 44. Waveform of a 5 MHz and of a 6 MHz
Sinusoidal Voltage.

The two kinds of peak voltage can be distinguished by
calling the former the "peak of the envelope" and the latter
the "high-frequency" peak. In reference to Fig. 45(b), it
would be correct to say that the envelope peak voltage is
four volts, and the envelope frequency is 1 MHz. It could
be said further that the high frequency peak voltage varies
from a maximum of 4 volts down to a minimum of zero.

Now what happens when these two voltages are added?
The total follows the principles illustrated in Fig. 43.
When the two components are in phase, they add to produce a maximum peak voltage, when they are 180° out
they cancel, and in between the peak amplitude changes
from one condition toward the other. The resulting waveform is illustrated in Fig. 45 (a), showing the two component voltages and the total superimposed, and Fig. 45 (b) ,
showing the total alone. The total voltage reaches a 4-volt
maximum peak initially when the two are in phase, the

This sum of two particular sinusoidal voltages demonstrates
several characteristics common to all sums of two such
voltages without regard to their frequencies. One characteristic is the variation in the amplitude of the high fre-

25

Two Sinusoidal Input Voltages with Second Order
Distortion

2.0

It has been shown that, when two sinusoidal components
are fed into a distortionless amplifier, the output contains
only the two original components, or in other words, the
waveform of the output is the same as that of the input.

I.6

"'"~
..J

0

I. 2>

"'"

Fig. 47 illustrates again the waveform and spectrum in this
case, showing how the high-frequency peak voltage varies
at the difference frequency (f2-f1) as the phase relation
between the components changes.

'"
Q.

8

0

,'--' ,

5

MHz

6

MHz

FREQUENCY MHz

Figure 46. Spectrum of the Sum of Two Sinusoidal Voltages.
VOLTSO~HHT+~~+HHrl++HHT+~~~Hrl--

quency peak. For the sum of two equal voltages with ANY
frequencies, the amplitude of the high frequency peak
varies from maximum to 0 and back to maximum at a
frequency which is the difference of the frequencies of the
two components. In other words, the envelope frequency
is the difference of the frequencies of the two component
voltages. In this example the envelope frequency is 1 MHz,
the difference between 6 MHz and 5 MHz.

(0)

IDr-----~--------------------

Next consider the spectrum diagram of the voltage whose
waveform is illustrated in Fig. 45 (b). This is shown in
Fig. 46. It shows two components, one at 5 MHz and one
at 6 MHz, each having an amplitude of two volts. There is
no component at 1 MHz, nor at any frequencies other than
5 and 6 MHz. This should not be surprising since this
voltage is initially defined as the sum of two, and only two,
sinusoidal components.

(b)
PEAK

VOLTS

In discussing AC voltage it was clear that saying "the DC
component is zero," does not mean "there is no voltage
present." Any AC voltage averages zero, and thus has no
DC component, when the area above zero and that below
are equal over one cycle (see Fig. 31). Similar reasoning
applied to the waveform of Fig. 45 (b) shows how it can
be true that "the amplitude of the high frequency peaks
varies at a I-MHz rate" and yet "there is no I-MHz component present." Since the peaks above zero have higher
amplitudes at the same time as those below zero and lower
amplitudes at the same time, they cancel each other and
there is no average variation at the I-MHz frequency.

O~--~.v~--------

o

_________

fl fa

FREQUENCYFigure 47. The Sum of Two Sinusoidal Voltages Undlstorted.

Next consider what happens when two sinusoidal voltages
are added and introduced into an amplifier with second
order distortion. Fig. 48 ( a) shows a plot of the resulting
distorted waveform.
Since the output waveform has a decidedly different shape
from that of the input [compare Fig. 48 ( a ) and Fig.
47 ( a) ], it is clear that there must be components at frequencies other than the two original ones. Fig. 48 (b)
illustrates the five new frequency components that are
added to the output voltage by second order distortion.
Since the positive peaks in the output are stretched, and
the negative peaks flattened, there is a general shift in level
in the positive direction, and there must be a corresponding
positive DC component. Since the peaks above 0 no longer
average out with the peaks below 0, there is also a component at the difference frequency (f2-f1)' For a similar

The statement "this voltage has components only at 5 MHz
and 6 MHz" means that the waveform shown is duplicated
precisely by adding together equal 5 and 6-MHz sinusoidal
voltages. As long as this waveform is duplicated WITHOUT ANY DISTORTION no additional components are
necessary to reproduce it. A general prfuciple relating to
amplifiers can be stated:

Only when an amplifier distorts does the output signal
contain components at frequencies differing from the
frequencies of the input signal components.

26

TIME-

(0)

-10

Pr.

.....

+
VOLTS 0 ~H+++1J.¥\-l+I-++H1+1-I-H+fl4f+Hrt+tt-

DIFFERENCE

BEA~

~~
0

50

BEATS AND HARMONICS

"
-AAANtiNlAAA
160

150

z60

FREQUENCY MH z

1.0

Figure 49. Spectrum of 12 CW Signals with Appreciable
2nd Order Distortion.

( b)

PEAK

VOLTS

to the problem of second order distortion. The usual amplifier specification states the noise figure, gain and crossmodulation but does not mention sum or difference frequency beats or second harmonics. The reason for this is
related to the standard channel frequency assignments
established by the FCC. If one takes any pair of picture
carrier frequencies in the standard 12-channel assignments,
their sum or difference does not fall in any of those channels. Similarly, with one minor exception (channel 6 sound
carrier), the second harmonics of all low band carriers fall
between the two bands. Figure 49 shows the spectrum
obtained when 12 CW signals on the normal picture
carrier frequencies were introduced into a CATV amplifier
at levels somewhat higher than normal operating level.

DC component
not shown by
analyzer _ _

H

2fl\

I -f2

/2f2

1l1

"

fI~

fH
I 2

FREQUENCYFigure 48. The Sum of Two Sinusoidal Voltages with
Second Order Distortion.

reason, there is a component at a frequency which is the
sum of the frequencies of the two original signals (f1+f2)'
And, of course, each of the original signals generates a
second harmonic (at 2f1 and 2f2)' Thus the spectrum of
the output signal looks like Fig. 48(b) with components
at the two original frequencies as well as at the five new
ones.

This shows how the spurious signals resulting from second
order distortion fall below and between the bands, but not
within the channel limits. Since this is true, second order
distortion has no bad effects on an amplifier carrying up
to twelve standard TV channels, and normally is not considered in this case.

An important conclusion can be drawn from this one example: Whenever more than one sinusoidal voltage (that
is when more than one signal) is introduced into an amplifier which has second order distortion, the output will
include signals at certain frequencies differing from those
of the input signals. There will be a DC component, a shift
in the average collector current of the distorting stage
(which does not show up in the output when AC coupling
is used), a component at a frequency which is the difference of the two original frequencies, a component at a
frequency which is the sum of the original frequencies,
and components at twice each of the original frequencies.

Two Sinusoidal Input Voltages with Third Order
Distortion
Figure 5O(a) illustrates the appearance of the output voltage of an amplifier having third order distortion when the
sum of two sinusoidal voltages [similar to that shown in
Figure 47(a)] is introduced into the input. The flattening
of the larger vertical peaks is clearly evident. A spectrum
diagram showing the frequency components in the output
is shown in Figure 50 (b). In addition to the two original
sinusoidal components (at f1 and f2) spurious signals occur
at the following frequencies:

When the original signals are modulated with picture information, each of these spurious signals will carry the
modulation of both of the original signals from which it
comes.
Why Second Order Distortion Is Unimportant in
Present CATV Systems
Anyone who has worked with CATV equipment in the past
recognizes the fact that very little attention has been paid
27

2f1-f2

This falls below f1 at a spacing corresponding
to the frequency difference between f1 and f2.

2f2-f1

This falls above f2 at a spacing corresponding
to the frequency difference between £1 and £2-

(<<)

(h)

(c)

(d)

JO-~II

Input , - - - Millivolts .. _ _ __

(0 )
+

Z---,---

VOLTSO~~~++THHH;+++HH~++rrHH;;-

0_-'-_ _
JDD----

r.---Output
Millivolts

N---_

fD---AO---,--

TIME-

0 - - ' -_ _
f~

1.0

Figure 51. Spectra of Input and Output Signals Showing
Effects of 3d Order Distortion.

(b)
effects. The upper spectrum diagrams illustrate the input
signal components in an amplifier which has severe third
order distortion, the lower diagrams illustrate the resulting
output signal components. The amplifier voltage gain, for
small signal input, is 10 times. Thus, as illustrated in
Flgure 51 (a), an input of 2 millivolts gives an output of
approximately 20 millivolts.

VOLTS

o

-

l 'J

711 ,

3fl \

-

1I/3f2
,/ \

2:fr+fz

The transfer characteristic of this amplifier can be approximated by an equation where a "cubed" term is subtracted
from a "linear" term, with a shape resembling Fig. 39 (c) .
With such a characteristic the effective gain decreases as
the input signal amplitude increases. Thus, as shown in
Figure 51 (b), increasing the input signal of this amplifier
to 10 millivolts results in an output of 90 millivolts, rather
than 100 millivolts which would be obtained if the gain
were not reduced by the effects of third order distortion.
This effect, the reduction in gain at a single frequency as
the signal amplitude increases, is called compression 'a:rld
results in the distortion of the modulation envelope on any
modulated signal going through such an amplifier. When
this effect occurs in an amplifier carrying a single TV-modulated signal, it results in a flattening of the sync peaks
which is called "sync compression."

2fz.,.fl

FREQUENCY
Figure 50. The Sum of Two Sinusoidal Voltages
with Third Order Distortion.

3f1 and These are the third harmonics and the spacing
3f2
between is three times the spacing between
fl and f 2 •
2fl +f2

This falls above 3f1 at a spacing corresponding
to the frequency difference between fl and f 2 •

2f2 +f1

This falls below 3f2 at a spacing corresponding
to the same difference.

Figure 51(c) shows what happens when a signal is introduced at low level on another frequency. Several effects
can be seen: The output level on the new frequency is
somewhat below the 20-millivolt point it would reach if
the strong signal were not present; the strong signal output
is slightly reduced by the presence of the new signal [compare with (b)], and a spurious component at 2f1 -f2 can
be seen.

Cross-Modulation and Compression
The spurious slgnals generated by third order distortion
can give trouble in any multi-channel system since it is
possible for them to fall within some of the channels. In
present CATV systems they do not generally cause as
much trouble as another effect of, third order distortion,
"cross-modulation." This is one of the two important aspects of third order distortion which do not result in
components at new frequencies. (The other being "compression"). Each of these effects represents a change in
gain at the channel frequencies rather than the generation
of new frequency components. Figure 51 illustrates these

As shown in Figure 51 ( d), increasing the second input
signal to full amplitude results in a further reduction in
gain so that both output signals at the orlginal frequencies
are below 60 millivolts and the spurious signals increase
in amplitude. The most significant effect here is that the
gain on each channel is reduced not only by an increase
in level on that channel but also by the increase in level on
the other channel. This results in a transfer of any varia-

28

tion, or modulation, on one carrier to any other carriers
going through the same amplifier. This transfer is called
cross-modulation and represents the worst effect of nonlinearity in present-day CATV amplifiers .

1.0f------.------------

.-?CW SIGNAL

(0)

This effect is further illustrated in Fig. 52. Figure 52(a)
shows the output signal obtained when a sinusoidal input
is applied to an amplifier with a small amount of thirl
order distortion. Figure 52 (b) shows what happens when
a second signal, fully modulated, is fed through the same
amplifier, simultaneously with the original CW signal. The
output includes the modulated signal (which shows up in
the frequency spectrum as a carrier with smaller sidebands
on each side), the output at the frequency of the original
CW signal, and two spurious sideband components which
show up adjacent to the CW signal frequency as a result
of third order distortion. It is cleat how this distortion
results in a transfer of modulation from one signal to the
other.

O~------J~---------------------

o

FREQUENCY-

1.0

Conclusion
MODULATED
SIGNAL

(b)

This chapter has attempted to describe all of the effects
which result from the simplest kinds of non-linearity,
second order and third order distortion, in amplifiers of
the type used for CATV systems. It has shown that second
order effects are generally unimportant with present-day
frequency assignments and that, of all the third order
effects, cross-modulation is the most important, representing the factor which limits the output level at which the
amplifiers in these systems may be operated without degrading the picture quality at the receivers served by the
system.

CROSS-

o
o

111.,7 MOD.
FREQUENCYFigure 52. Spectrum Showing Cross-Mod.

29

CHAPTER V

A MATHEMATICAL ANALYSIS OF DISTORTION

AS IT OCCURS IN CATV AMPLIFIERS
LIST OF SYMBOLS
(in the order of their use)
A,B,C

Amplitudes of each of three sinusoidal input voltages.
Corresponding frequencies.

a, b, c
Constants characterizing the first, second and third order distortion of the amplifier.
The total instantaneous input and output voltages.
First Order output levels, in dBmV.
A decibel constant characterizing second order distortion defined
by:

~

K,

= 20 log" ( ~ )'

Expressed in dBmV

K2 is always a negative number since k2

<< k12.

Sum or difference beat level, in dBmV.
Second harmonic levels, in dBmV.
A decibel constant characterizing third order distortion, defined
by:
ks
Ks = 20 logio

3
2

v'2

s Expressed in dBmV

( ~12)

Ks is always a negative number since k3

30

«

ki S.

Labe

Triple beat levels, in dBmV.

L2ab, L 2ae, etc.

Intermodulation Component levels, in dBmV.

Lsa, Lsb, Lse

Third Harmonic levels, in dBmV.

Rab/a, etc.

The relative level of the sum or difference beat between signals
at a and b referred to the level of a, in dB. SimUar symbols are
used for other pairs of signals and other reference signals.

R2a/a, R2b/b, R2e/e

Relative level of second harmonic, referred in each case to· the
output level at the fundamental frequency, in dB.

Rabe/a, Rabe/b, Rabe/e

Relative level of the triple beat, referred to each of the three
output signals, in dB.

R2ab/a, etc.

Relative level of each of the Intermodulation components referred
to the indicated output. In this example the relative level of the
component at 2fa ± fb with reference to the level at fa, in dB.

Rsa/a, RSb/b, Rse/e

Relative level of each third harmonic component, referred to the
output level at the fundamental frequency, in dB.

XM, xm

XM is the NCTA standard cross-modulation ratio expressed in
decibel form, xm is the same ratio in voltage form.
XM = 20 log 10 xm.

M,m

m is the conventional modulation factor: m= 6ma>:
6mu

+

emfn
emfn

(100 m = % modulation).
M is the modulation factor in decibel form: M = 20 loglo m.
n

n is the total number of input signals applied to an amplifier,
including the test signal.

ea(D)

This is the amplitude of the output voltage at fa including the
effects of third order distortion. (As contrasted with the undistorted output = k 1A.)
The voltage ratio of the triple beat output component amplitude
to the first order output at fa.
k
When A = B = C, rabe = 3/2 k: A2.
Rabe/a = 20 loglo rabe'

31

A SUMMARY OF THE DECIBEL EXPRESSIONS
OUTPUT LEVELS EXPRESSED IN dBmV
First Order Output Component Levels:

Second Order Distortion Component Levels:

Sum and Difference Beats:

Lbe = K2

+ La + Lb (at fa ± fb)
+ Lb + Le (atfb ± fe)

Lae = K2

+ La + Le

Lab = K2

(at fa ± fe)

Second Harmonics:
L2a

= K2 + 2La + 2Lb -

6, at 2fb

= K2 + 2Le -

6, at 2fe

L 2b = K2
L2e

6, at 2fa

Third Order Distortion Components:

Triple Beat Component Levels:

Intermodulation Component Levels:
L 2ab = Ka

+ 2La + Lb - 6dB

+ 2La + Le L2ba = Ka + 2Lb + La L2ae = Ka

(at 2fa ± f b)

6dB (at 2fa ± fe)

6dB (at 2fb ± fa)
L 2be = Ka + 2Lb + Le - 6dB (at 2fb ± fe)
L 2ea = Ka + 2Le + La - 6dB (at 2fe ± fa)
L 2eb = Ka + 2Le + Lb - 6db (at 2fc ± fb)

Third Harmonic Component Levels:
Laa

= Ka + SLa -

Lab

= Ka + SLb -

15.5 (at Sfb)

+ SLc -

15.5 (at Sfc)

Lac = Ka

32

15.5 (at Sfa )

BELATIVE DISTORTION EXPRESSED IN DECIBELS
Relative Second Qrder Distortion:
Relative Sum or Difference Beat

Rab/a = K2 + Lb
R be/b K2 + Le
Rae/a = K2 + Le
Rab/b = K2 + La
R be/e = K2 + Lb
R ae/e = K2 + La

=

(at fa ±
(atfb ±
(at fa ±
(at fa ±
(at fb ±
(atfa ±

f b)
fe)
fe)
fb)
fe)
fe)

Relative Second Harmonic:

= K2 + La -

6 (at 2fa)

+ Lb R 2e/e = K2 + Le -

6 (at 2fb)

R2a/a

R2b/b = K2

6 (at 2fe)

Relative Third Order Distortion:
Relative Triple Beat:

+ Lb + Le (atfa ± fb ± fe)
Rabe/b = Ka + La + Le (at fa ± fb ± fe)
Rabe/e = Ka + La + Lb (at fa ± fb ± fe)
Rabe/a = Ka

Relative Intermodulation:

R2ab / a = Ka+ La
R2ab / b

+ Lb -

= Ka + 2La -

6dB (at 2fa ± f b)

6dB

+ La + Le - 6dB
R2ae / e = Ka + 2La - 6dB
R2be/b = Ka + Lb + Le - 6dB
R2be/c = Ka + 2Lb - 6dB
R2ba/a = Ka + 2Lb - 6dB
R2ba / b = Ka + La + Lb - 6dB
R2ca/a = Ka + 2Lc - 6dB
R2ae / a = Ka

(at 2fa ± f b)
(at 2fa ± fe)
(at 2fa ± fe)
(at 2fb ± fe)
(at 2fb ± fc)
(at 2fb ± fa)
(at 2fe ± fa)

= Ka + Lc + La - 6dB (at 2fe ± fa)
R2eb/b = Kg + 2Le - 6dB
(at 2fe ± f b)
R2cb/e = Ka + Lc + Lb - 6dB (at 2fc ± fb)

R2ea/e

Relative Third Harmonic:

= Ka + 2La Rab/b = Ka + 2Lb Rae/c = Ka + 2Le Rania

15.5dB (at 3fa)
15.5dB (at 3fb)
15.5dB (at 3fc)

NCTA Cross-Modulation
XM

= Ka + 2L + 20Iog1o (0 -

1)

Cross-Modulation in Terms of "M"
M

= Ka + 2L + 20Iog1o (0 33

1) - 6dB

INTRODUCTION:

pedances and with sinusoidal input voltages at each of
three frequencies:

The preceding chapter "The Fundamentals of Distortion in
CATV Amplifiers" develops, chiefly from a graphical standpoint, the general nature of distortion as it occurs in
CATV amplifiers. It illustrates the spectra of spurious
signals generated by second and third order curvature in
the transfer characteristic, and describes the compression
and cross-modulation effects resulting from third order
curvature. In this chapter the quantitive relationships between the various distortion products will be developed
mathematically, and a consistent way of expressing them in
the convenient decibel "language" of CATV will be shown.

el n = A cos a

+ B cos b + C cos c

(Note that A cos a and A sin a are both sinusoidal voltages.
Their waveforms are identical except for a 90° phase
difference. The cosine form is used throughout this analysis because it results in simpler expressions.)

A, Band C are the amplitudes of each voltage respectively, measured in millivolts.

It must be borne in mind in considering what follows that
some of the basic assumptions necessary to permit easy
mathematical development are not completely justified in
reality. For example, the assumption that the gain and
distortion coefficients (kl> k2' and kg) are constant for all
input signal frequencies is distinctly at variance with the
measured performance of practical amplifiers. These discrepancies, while very significant, do not nullify the usefulness of the relationships developed. However, extreme
care must be exercised when drawing conclusions concerning real amplifiers from the mathematical considerations.

Assume further that the transfer characteristic of the
amplifier can be accurately represented at all frequencies
by a power series containing three terms:

el n is the instantaneous input voltage as described above,
kl> k2 and k3 are complex numbers describing the gain,
phase shift and distortion properties of the amplifier.
Although, in a real amplifier, any phase angle may be
associated with each of these constants, in this example the
phase angle for kl and k2 will be assumed to be 0°, and
that for k3 either 0° (indicated by a + sign) or 180°
(indicated by a - sign).

RESULTS OF THE BASIC MATHEMATICAL
DEVELOPMENT:
Assume an amplifier with 75-ohm input and output im-

With these conditions the output will contain the following sinusoidal components (see Appendix I, p. 105 for derivation):
First Order Components:

Components identical with the input
signals except with increased amplitude.

klein =

+ klA cosa
+ klB cosb

+ k1C cosc

Three First Order Output Components
These represent the result of linear
amplificatiop.. k1 is the small-signal
voltage gain of the amplifier.

Second Order Distortion Components:

Three DC Components due to second
order distortion.

These represent a shift in average level.

Components at frequencies which are
combinations of input signal frequencies.

+ k2AB cos(a ±

b)
c)
+ k2BC cos(b ± c)

+ k2 AC cos(a ±

Six Sum and Difference Beat Components.

+ k2A~ cos 2a
Components at twice the input signal
frequencies.

+
+

2
k2B2 cos 2b
2
k2C2 cos 2c
2
34

Three Second Harmonic Components.

Third Order Distortion Components:
The first three groups of Third order components contain components at frequencies differing from the frequencies of
the input signals.
kSelnS

Components at three times each input
frequency.

Components at frequencies which are
combinations of input signal frequencies.

=

+ 1/4ksAs cos 3a
+ 1/4ksB3 cos 3b
+ 1/4kaCS cos 3c

Three Third Harmonic Components.

+ 3/4ksA2B cos(2a ± b)
+ 3/4ksA2C cos(2a ± c)
+ 3/4ksB2A cos(2b ± a)
+ 3/4ksB2C cos(2b ± c)
+ 3/4ksC2A cos(2c ± a)
+ 3/4ksC2B cos(2c ± b)

Twelve Intermodulation (2a ± b beat)
Components, and

+ 3/2ksABC cos(a ±

Four Triple (abc) Beat Components.

b ± c)

The last two groups contain components at the frequencies of the input signals. When ks is positive, these components
add to the first order output causing an increase in gain which is called expansion. When ka is negative, they subtract
causing a decrease in gain which is called compression.
Components at the frequency of each
input signal with amplitude determined
by the input voltage of that signal
cubed.
Components at the frequency of each
input signal with amplitude determined
by the input voltage at that frequency,
and the square of the input voltage at
one other frequency.

+ 3/4ksAs cosa
+ 3/4ksBs cosb
+ 3/4ksCs cosc

Three components causing Self-compression when k$ is negative, Self-expansion when ks is positive.

+ 3/2ksAB2 cosa
+ 3/2ksAC2 cosa
+ 3/2ksBA2 cosb
+ 3/2ksBC2 cosb
+ 3/2ksCA2 cosc
+ 3/2ksCB2 cosc

Six components causing Cross-compression when ks is negative, Cross-expansion when ks is positive.

The corresponding first-order RMS output voltage is

Self-compression and self-expansion describe a condition
where the gain at the frequency of one of the input signals
is decreased or increased as a result of increasing the input
voltage of that signal.

kl~

v'2

obtained by multiplying the input voltage by kl> the

small signal voltage gain. Since the level in dBmV is defined as 20 IOglO (RMS voltage in millivolts) the first
order output levels, expressed in dBmV are:

Cross-compression and cross-expansion describe a condition
where the gain at the frequency of one of the input signals
is decreased or increased as a result of increasing the
voltage of another input signal.

DEcmEL EXPRESSIONS FOR THE LEVELS
OF THE OUTPUT COMPONENTS:
Decibel expressions for the output levels of each of the AC
distortion components will now be formulated, and for
CATV purposes they are most conveniently expressed in
dBmV, the decibel ratio of the RMS voltage of the component to one millivolt RMS. The dBmV is an expression
-48.75 dBm.
of power level, OdBmV

=

First Order Output Component Levels:
Since the instantaneous input voltage at fa is A cosa,
where A is measured in millivolts across 75 ohms, the peak
input voltage is A, and the RMS input voltage is

Second Order Distortion Component Levels:
Sum and Difference Beats:

Expressions for second order distortion levels can be obtained in the simplest form by carefully defining the constant "K2" in the decibel equation. Its definition is accom-

~, both in millivolts.
35

So the second harmonic output levels expressed in dBmV
are:

plished in the following manner: The voltage expression
for the amplitude of a sum beat is "k2AB", so "K2 + La +
Lb" is chosen as a convenient decibel expression for its
level. With this choice, since the level of a component is
20 10g10 of the RMS voltage of that component:
K2 + La + Lb

+ 2La L 2b = K2 + 2~ -

6, at 2fa

+ 2Le -

6, at 2fe

L2a = K2

= 20 log 10 ( k2AB)
Y2

L2e - K2

6, at2fb

Substituting previously defined quantities for La and Lb:
Third Order Distortion Components:
First, a quantity "Ks" is defined to provide a simple expression for each Triple Beat component:
:3/2 kgABC)
Assume: Kg + La + Lb + Le = 20 10g10 (
v'2

Kg

3
~ ABC )
= 20 10g10 ( "'2.
V2.

- 20 loglo

or

k1

v'2

B - 20 10g1o

- 20 loglo

~
V2

A

k1

v'2 C

With this definition of K2 Sum and Difference beat levels
in dBmV are expressed as follows:

~
Lab = K2

+ La + Lb (at fa ±
Lbe = ~ + Lb + Le (atfb ±
Lac

fb)

or

fe)

-J2

Ks

= 20 10g10 3/2 (~]) 3

= K2 + La + Le (at fa ± fe)
With this definition of Kg, the triple beat levels are given by:

Second Harmonics:

Triple Beat Component Levels:

I

The amplitude of the second harmonic output at 2fa is
k2A2
-2-·
The corresponding RMS voltage is

~

v'2.

Labe

= Ks + La + Lb + Le (at fa ± fb ± fe)

A2
2
Intermodulation Component Levels:

so the level is: L2a .= 20log10

(~. ~2)

Since the coefficient of each of the Intermodulation Components is 3/4, which is 1/2 of the coefficient of the Triple
Beat components, 20 loglo 1/2
-6 dB is added to obtain the decibel expressions for the intermodulation components:

=

Dividing the quantity in parentheses into convenient
factors:

L 2ab

= Ks + 2La + Lb = Kg + 2La + Le L 2ba = Ks + 2Lb + La L 2be = Ks + 2Lb + Le -

6dB (at 2fa ± fb)

L2ae

6dB (at 2fa ± fe)

+ 2Le + La -

6dB (at 2fe ± fa)

L2ea = Kg

6dB (at 2fb ± fa)
6dB (at 2fb ± fe)

L2eb = Ks + 2Le + Lb - 6db (at2fe ± fb)

36

Third Harmonic Component Levels:

Relative Second Harmonic:

The coefBcient of each Third Harmonic Component is 1/4,
1
which is - the coefficient of each Triple Beat,. component.
6
1
To express this in dB, 20 10glo '6
15.5dB is added to

Similarly the second harmonic of the output at fa relative
to the fundamental is given by:
R2a/a

=-

=L2a -

La

= K2 + 2La -

6 - La

=K2 +La -6

each Third Harmonic Component level:
R2a/ a = K2

+ 3La -

15.5 (at 3fa)

= Ks + 3~ -

15K (at 3.fb)

Lac = Ks + 3Le -

15.5 (at 3fe)

Lsa = Ks
LSb

so

+ La -

=

Expansion/compression

Relative Third Order Distortion:

It is pointless to express the levels of the expansion/compression terms in dBmV. They are considered later in this
chapter.

Relative Triple Beat:

RELATIVE DISTORTION EXPRESSED IN DECmELS:

and similarly for other terms, so:

In the measurement and specification of distortion, it is
common practice to deal with relative distortion, i.e. to
relate the amplitude of the distortion component to the
amplitude of the undistorted output component. Harmonic
distortion, for example, is commonly described in terms of
"percent harmonic" which is 100 times the ratio of the harmonic component amplitude to the fundamental component
amplitude.

= Ks + ~ + Le (atfa ± fb ± fe)
Rabe/ b = Ks + La + Le (at fa ± fb ± fe)
Rabe/ e = Ks + La + Lb (at fa ± fb ± fe)
Rabe/a

When the levels of the undistorted (first order) components and of the various distortion components are expressed in decibel terms it is easy to arrive at a statement
of relative distortion. The relation, in decibels, between
any given distortion component, expressed in dBmV, and
a given first order output, also in dBmV, is simply the difference (in dB) between the two levels.

Relative Intermodulation:

+ La + Lb - 6dB
R2ab/ b = Ks + 2La - 6dB
R2ae/ a = Ks + La + Le - 6dB
R2ae/ e = Ks + 2La - 6dB
R2be/ b = Ks + Lb + Le - 6dB
R2be/ c = Ks + 2Lb - 6dB
R2ba / a = Ks + 2Lb '- 6dB
R2ba / b = Ks + La + Lb - 6dB
R2ea / a = Kg + 2Le - 6dB
R2ea/ e = Ks + Le + La - 6dB
R2eb / b = Kg + 2Le - 6dB
R2eb/ e = Ks + Le + Lb - 6dB
R2ab / a = Ks

Relative Second Order Distortion:
Relative Sum or Difference Beat
The level of a particular sum beat component is, for
example:
Lab = K2 + La + Lb (atfa + fb)
and the level of the first order output at fa is La, so the
relative sum beat is the difference of the two levels:

Rab/a

= Lab -

La
=K2 +Lb

= K2 + La + Lb -

6 (at 2fa)

R2b/ b = K2 + Lb - 6 (at2fb)
R2e/ e K2 + Le - 6 (at 2fe)

La

Where 

it follows that: percent mod. = 100 X

p-p variation

,

-l-

2 X ave. level
!

p-p variation
=50x-----unmodulated level

:

In this case the 'scope has been adjusted so that the unmodulated level corresponds to 10 divisions, so:
100 m = 50 X

= 5

I
,

,

-i-

Figure 72.

p-p variation (in div.)
10 div.

Fig. 72 shows the trace with the input set for normal
amplifier output, and 'scope gain is set so the average deflection is 10 divisions.

X p-p variation (div.)

55

!--r-r

II

I

~

/

indicated, with a deflection of about 0.7 div. p-p is about
0.5 x 0.7 or about 0.35%, which is quite good.

I

,

I

I

Expressing Hum Modulation in dB

----'-.

.J

~

.. '"

It is often convenient, particularly when dealing with
cascaded systems, to express hum modulation in dB rather
than as a percent. The percent figures from measurement
are converted to dB as follows:

~

Hum Modulation (dB reI. to 100%) =
%modulation
20 10glO - - - - 100
For example:

Figure 73.

Fig. 73 shows the 'scope trace centered, with no change
in gain. The peak-to-peak deflection is 2 divisions, indicating modulation of 5 x 2 divisions or 10%: This degree of
hum mQdulation would be excessive in an amplifier intended for CATV use; the photo was made while using an
amplifier whose hum had been increased above normal by
removing a filter capacitor.

Hum Modulation
(dB reI. to 100%)

%

Modulation

-6

50.0
10.0
1.0
0.1

-20
-40
-60
(See curve p. 49).

Hum Modulation in Cascaded Amplifier Systems

-

When each one of a chain of cascaded amplifiers produces
a certain amount of hum modulation, the system hum modulation will increase directly with the number of amplifiers,
unless the phase of the modulation from some of the
amplifiers is different from that of others. To be safe, it is
good practice to calculate system hum modulation based
on .the "worst case" where they are all in phase.

r- I!"""
I
~

i

In this worst case, the percent hum modulation is doubled
each time the number of amplifiers is doubled; or in dB
terms, the hum modulation (in dB reI. to 100%) increases
6 dB each time the number of amplifiers is doubled (i.e.
the negative dB number gets smaller). To aid in system
calculations, Table L shows the increase in hum modulation for up to 50 cascaded amplifiers.

I

Figure 74.

Fig. 74 illustrates the hum modulation with a high-grade
commercial signal generator connected directly to the
input of a 704B. All of the steps have been followed, and
the scope gain increased lOx, so the percent modulation

TABLE L
INCREASE OF HUM MODULATION IN A CASCADED AMPLIFIER SYSTEM
WORST CASE:
ASSUMING IDENTICAL AMPLIFIERS, AND PHASED HUM
To get system
To get system
To get system
To get system
To get system
mod. (dB)
mod. (dB)
mod. (dB)
mod. (dB)
mod. (dB)
No. of
ADD TO
No. of
ADD TO
No. of
ADD TO
ADD TO
No. of
ADD TO
No. of
Cascaded
Cascaded dB MOD.
dB MOD.
dB MOD.
Cascaded
dB MOD.
dB MOD.
Cascaded
Cascaded
Amplifiers FOR 1 AMP. Amplifiers FOR 1 AMP. Amplifiers FOR 1 AMP. Amplifiers FOR 1 AMP. Amplifiers FOR 1 AMP.
1
2
3
4
5

0
6.02
9.54
12.04
14.00

11
12
13
14
15

20.82
21.58
22.28
22.86
23.52

21
22
23
24
25

26.44
26.84
27.24
27.60
27.96

31
32
33
34
35

29.82
30.10
30.36
30.62
30.88

41
42
43
44
45

32.26
32.46
32.66
32.86
33.06

6
7
8
9
10

15.56
16.90
18.06
19.09
20.00

16
17
18
19
20

24.08
24.60
25.10
25.58
26.02

26
27
28
29
30

28.30
28.62
28.94
29.24
29.54

36
37
38
39
40

31.12
31.36
31.60
31.82
32.04

46
47
48
49
50

33.26
33.44

56

3~.62

33.80
34.00

CHAPTER IX
REDUCING THE EFFECTS OF REFLECTION IN CATV FEEDERS
Introduction
The presstrre type tap is a convenient and economical way
of connecting the customer house drop cable to the feeder
cable in a CATV system. Because of this convenience the
large majority of taps in use today are of this type. Hundreds of thousands of them are providing satisfactory service in systems all over the country. Unfortunately there is
a penalty attached to the pressure tap's convenience. Because it must tap into the feeder cable with no opportunity
for series compensation, the pressure tap inevitably introduces reflections into that cable. This chapter shows how
these reflections can be minimized by careful tap design,
and how their adverse effect on the transmission of picture
signals can be greatly reduced by grouping taps in optimum
arrangements.

capacitive tap had high efficiency (i.e., the loss on the
feeder "",,;;w,l\;jljH~HI'AI'e'I.r' 26dB
REFLEI:I10N

AT
INPUT

(a)
2BdB---+~--r~~~~~~~~~­

Figure 83. Reflection vs. Frequency3 Reels of CATV Cable Showing Effects of Periodicity.

~I~~~~~-=~~~~-~~~I

174 MHz

216MHz

29dB-------------------30
(b)

37dB

(a) NONE

:~

--j

2?:J. .:;,

=r-v:

TRANSMISSION
1ST TAP

3OdB------------------

(b) I TAP

:Ir 1
".;ww'\

Ic}
Ic)

=r

V

•

JI-TO
V ~

,

23d~~
2 TAPS ' \
,

174 MHz

(d) 4 TAPS

V

TRANSMISSION

'a~v:T08THTAP
V

V

V

I

216MHz

J\MJ\J\

Figure 8!i. 16 8MT Taps on %" Feeder, 800' Long, 50 Foot Spacing,
High-Band Performance.

The problem is reduced somewhat when taps are installed
at irregular intervals so that there is no repetitive pattern.
Fig. 86 shows the reflections and responses that resulted
when the same 16 taps were installed on the same feeder
but spaced completely at random. The reflection pattern
is no longer regular and shows reduced amplitude. Transmission variations are improved to a little more than 1 dB
in the worst case. This still represents a situation somewhat short of one that would guarantee excellent picture
transmission.

(e) 16 TAPS

Figure 84. Addition of Reflections, Pressure Taps Spaced at
50' Intervals Along V2" Foam Cable.

60

poorer transformer taps (CMT) were substituted for the
better ones (BMT). This increases the peak reflection from
16 dB (16%) to 11 dB (28%), md increases the variation
from a little over one dB to more than 3 dB.

,., . . M~ '~~~.
I
174NHz

Using Periodicity to Minimize Reflections

I
21SMHz

The patterns obtained with simplified reflection conditions
(Fig. 81, for example) show narrow peaks with relatively
broad areas between where the reflections were low. This
effect can be used to reduce reflections in the TV bands
by installing taps in periodic arrays with peaks outside of
these bands. If the spacing between taps is made 36 inches
(for foam-insulated cable) the reflection spike will be at
135 MHz, where this spacing is one-half wavelength and
where the reflection does no harm. At 67 and at 201 MHz,
where the spacing is one-quarter and three-quarters of a
wavelength respectively, the reflections from successive
taps will cancel, causing a minimum effect for the TV
channels between 54 and 88 and between 174 and 216
MHz.

LOSS
TO 1ST

(b)

TAP

33

LOSS
TO

29dB

8TH TAP

(e)

Fig. 88 illustrates this. Fig. 88 (a) shows a plot of reflection vs. frequency for a single transformer tap (BMT25).
It reaches a maximum of close to 28 dB (over 4%) at
216 MHz. When two of these taps are attached to the
line 36 inches apart, their reflections cancel at the center
of the low band and the center of the high band, as illustrated in Fig. 88 (b). The net effect is that the two taps
cause somewhat less reflection than one! An even more
dramatic effect is obtained when three taps, Fig. 88 (c)

Figure 86. 16 BMT Taps on 1/2" Feeder 800' Long, Hi·Band Performance.
There was one tap in each 50' cable section but spacing was rando·m.

The importance of optimizing the design of the individual
tap for minimum reflection is illustrated by Fig. 87. This
shows a situation identical with Fig. 86, except that the

-lldB

,

ONE TAP
209

14

\1

(0)

=-1 .-:::: 28 dB
00

~

J

(0)

28dB
00 1
174MHz

~.

V

~

1<-36"---->1

~

-0--0REFL.B:TION
AT
INPUT

~

28dB

(b)

00

I
216MHz

LOSS
TO

--4:-----+--~t_-r~_!J. 28dB
(e) _ _~"::::::;::::""",~ _ _ _""":---,,,::-.L. 00

1ST TAP

27dB
28

(e)
29

LOSS
TO
8TH TAP

30

(d)

31

Figure 87. 16 CMT Taps on 112" Feeder 800' Long High Band
Performance. There was one tap in each 50' interval
but spacing was random.

---'t---?'-----4---JI- 28dB

Figure 88. Reflection vs. FrequentyBMT Pressure Taps on 1/2" Cable.

61

and four taps, Fig. 88 (d), are connected. Whereas four
of these taps could cause as low as 16 dB return loss (16%
reflection) if they were installed so that their. reflections
would add in phase, by scientific grouping they can be
made to give less reflection within the TV bands than one
tap alone.

to a minimum of 24 dB. Transmission variations across any
one TV channel were reduced from about 2 dB down to
about 0.5 dB. With no increase in equipment cost, tap
grouping substantially reduces the possibility of picture
degradation due to reflections and response variations in
the feeder.

When four taps are to be installed at a given location, a
fairly common situation in a CATV system, several arrangements ure possible. Fig. 89 shows the reflection plots for
some of them. The arrangement shown in Fig. 89 (a)
probably is the physically most convenient, in that the installer has to reach out only 18" to either side of the pole.
The electrical performance, however, is poor, showing
excessive reflection on channels 6, 7 and 13. Either of
the other arrangements shown is good; the one shown in
Fig. 89 (c) with taps close together at the center, and
the other two spaced 36" on either side, seems to be the
best from both the electrical and the mechanical point of
view.

This technique is quite evidently useless where signals are
distributed covering the entire spectrum between the low
frequencies and the upper end of the band. Since the
reflection patterns achieved by grouping of taps improve
conditions within· the two bands at the expense of the
region between the bands, the grouping method is useles/l
where this region too has signals.

• •

R

I
174MHz

I
216MHz

:BI-ftVi~r..------rc;A;,JJ\-------:-' E~:;:~ON

(ll)

Ib) 33dB "iI..,v'V"'vrV"V....,"'~,.,.~==;;;Ji~T~I~ION
34

- - - - - - - - - - - - - - INPUT THRU
1ST TAB

(0)
30dB

Ie)
--~r-;:------\-....!J-- 28dB

31

~

-,..

32

TRANSMISSION
LOSS WINPUT THRU
~
BTH TAP

---~'-------~~--~
28dB

Id)

29

.,.......

30
(b)

28dB _ _ _

Ie)

29
30

TRANSMISSION
LOSS;:;;:;;;;:; INPUT THRU
12TH TAP

TRA NSMISSION
LOSSINPUT THRU
16TH TAP

=

Figure 90. 16 BMT Taps on V2" Feeder 800' Long. Grouped for
Minimum Reflection-High-Band.

(e)

--~---+-----.lt---+- 28dB

Directional Coupler Multi-Taps
It is interesting to compare the performance of pressure
taps under these optimum conditions with the results obtained when using a tapping device having, electrically,
the best possible characteristics. This device is the directional coupler multi-tap. Where low-reflection performance
is d(:lsired throughout the entire spectrum, the directional
coupler multi-tap is the best device and has three other
important advantages over the pressure tap:

Figure 89. Reflection vs. Frequency4 Pressure Taps in Various Possible Arrangements.

To show the improvement that can be obtained by this
simple technique, an 800-foot feeder was equipped with
the same 16 BMT taps used in the earlier examples. They
were installed in four groups of four, each group arranged
in the pattern illustrated in Fig. 88 (d). Fig. 90 shows
the result. (Note that the vertical scale of Fig. 90 (a) is
doubled to exaggerate the reflections.) Grouping of the
taps in this way increased the feeder return loss as compared with the conditions of Fig. 86 from as low as 16 dB

1. Directivity: It is more sensitive to waves corning down
the feeder than to waves traveling back up the feeder,
and thus discriminates against reflections or spurious
signals corning from taps or receivers further down the

line.
62

2. Lossless Backmatch: With a transformer tap approximately half the energy tapped off the line is lost in the
backmatch resistor. With a directional coupler none
of this energy is lost, the reverse termination acts only
to absorb energy reflected from the receiver. Thus the
efficiency of a directional coupler (which determines
the line loss for a given tap loss) can be very high.

\0-- 200' ---01
~~~~
i'lII'l

H

H

IN-fH

~~~~
.2

~~~~
#3

~~~~

2BdB~

(0)

00
174 MHz

(b)

(e)

20 dB

~

""2
•

~TR~ON~~i~~ON
LOSS

25 dB

1. Installation: To install a multi-tap the feeder cable
(d)

20 dB
21 d B -

TO
""I
COUPLER

TO

23 dB
24 dB

must be cut. This takes time and interrupts transmission in systems already in operation, so the pressure
tap is more convenient.

INPUT

216 MHz

21 dB

The multi-tap has three disadvantages as compared with
pressure taps:

REFLECTION

AT
i

I

3. Multiple Outputs:. This means that fewer units are
needed with correspondingly fewer possibilities of reflection. The Jerrold Starline series of couplers (e.g.
Models DCM) have four outputs, so only one-fourth
the number of units is required as compared with pressure taps.

h

*4

-- =

-

TO

::::

.... 3
COUPLER

22 dB

2. Pre-loading: A complete multi-tap must be installed

even if only one house drop cable is to be connected
initially; thus the use of multi-taps requires more advance planning and investment.

(e)

.... 4

25 dB - - - - - - 26 dB

3. House-drops: Although the multi-tap is usually located
at a point on the feeder cable where the average
house-drop can be kept as short as practical, some of
the house drop cables will be longer than they would
be if pressure taps were used.
To allow comparison of directional coupler multi-tap performance with that of the foregoing pressure tap arrangements, the 800t feeder was equipped with four DCM units
at 200' intervals. The reflection and response characteristics are shown in Fig. 91. It can be seen that the coupler
has slightly higher return loss (28 dB vs. 24 dB) as compared with the best arrangement of pressure taps (Fig. 90)
and less response variation (about 0.3 dB vs. 0.5 dB)
but their performance is quite comparable. The coupler
requires no care in regard to spacing, works equally well
across the entire spectrum, and has the other advantages
listed above.

TO

24 dB

~

COUPLER

Figure 91. Four OeM Multitaps on Y2" Feeder 800' Long
Hi-Band Performance.

Summary
This chapter has presented a technique for minimizing the
reflections from pressure taps by careful grouping. While
the best results are obtained with the better transformer
type of taps (BMT), the same improvement will be experienced with any pressure taps. The directional coupler
multi-tap is shown to have slightly better performance
than the best that can be obtained from pressure taps.
With its other advantages this suggests the use of the
coupler for situations where the'very best performance is
desired and where the necessity of cutting the cable is not
too great a deterrent; in other situations the use of pressure taps may be quite adequate.

63

CHAPTER X
SWEEP FREQUENCY TESTING OF COAXIAL CABLE

Introduction

Investigation showed that this effect was due to periodic
discontinuities. Something in the manufacture of the cable
produced variations in characteristic impedance recurring
at precisely spaced intervals throughout the length of the
cable. Due to this precise spacing, many reflections, precisely phased at a certain frequency, arrived back at the
input end of the cable, causing this severe distortion of the
transmission characteristic.

Sweep testing is essential for coaxial cables used in ETV
and CATV distribution systems. This chapter compares
three basic methods: measurement of transmission loss,
measurement of input impedance, and measurement of
reflection coefficient.
The technical requirements for flexible coaxial cable were
organized in Military Specification JAN-C-17, originally
issued in 1944. This specification and its subsequent revisions spell out in detail the requirements for physical
construction and a number of electrical parameters, including attenuation and dielectric strength of the cable.
Regarding the characteristic impedance, JAN-C-17 specified the nominal impedance which was determined by a
calculation involving the total measured capacitance of a
reel of cable, and the delay factor measured on a short
sample. For cables of relatively short lengths, this specification was adequate; but the advent of CATV systems,
where TV signals are transmitted through many miles of
cable, uncovered the need for an additional specification.

To prevent the recurrence of this problem, a program of
factory tests was begun. The original test method involved
measuring the transmission loss through each reel of cable
over the frequency bands then in use. A reel of cable was
rejected if the loss in these bands dipped more than 0.25
dB below the smoothed attenuation characteristic.
After this transmission loss measurement method had been
used for several years, it became evident that a more sensitive test was needed. It was found that a measurement
of the input impedance at each end of a reel of cable gave
a more sensitive indication of the existence of periodic
reflections. Experience with the impedance measurement
method showed two major defects: it was difficult to
arrive at an accurate calibration, and the measured deviation was a critical function of cable length. Removing two
or three feet from the end of the cable would change the
entire pattern.

The problem first came to light in our organization about
twelve years ago when one of our field engineers returned
to the laboratory a reel of cable which, he claimed, would
not pass TV channel 6.

To overcome these defects, a test method was developed
employing a bridge; this method allowed observation and
measurement of the reflections from the cable end, eliminating uncertainty and allowing easier calibration. This
reflection measurement method has been used by a number
of cable manufacturers during the past five years and has
provided a satisfactory way of controlling periodic defects.

Figure 92 shows the measured attenuation of this reel of
cable, indicating an attenuation spike 50 dB deep at 87
MHz!

30

'"
"

.::-- r-.....
....

40

'. r---.
....

---

.

...

m 50

~

"

""'....

M~:UREO ~OSS
NOMINAL LOSS

"t--. r....

z

o

~ 60

. Vi'..
....

. r-. ....

........

.... ~

j

Z

ILl

~

The relative merits of the three methods of sweep frequency cable testing can perhaps best be developed by
describing each method in some detail and comparing
results.
Transmission Loss Measurement
Figure 93 shows a diagram of the equipment used in the
sweep frequency technique for measuring cable loss vs.
frequency. A wide-band sweep frequency transmission
measuring set is connected alternately to the cable under
test and to a variable standard attenuator. This arrangement provides an attenuation reference line on the oscilloscope against which the loss of the cable can be compared. "
For accurate measurement it is essential that the cable
face a well-matched impedance at each end. 10-dB fixed
attenuators are used to establish this condition.

70

60

90

t

Il

IDO

14!O

"'~O

UO

1110

Z )0

Z.!O

FREQUENCY MHz

Lo.. ver.u. frequency (or 2.000 feet of defective RGll/U
cable
Figure 92.

"See the author's article SWEEP MEASU~MENTS MAKING THE TOUGH ONES EASY, in ELECTRONICS, Vol. 36, No. 16,
April 19, 1963.
.

64

With the high end-to-end attenuation present on this reel,
the single shield allowed sufficient coupling to produce
ripples in the frequency characteristic.
Impedance Testing
A more sensitive test, free from this coupling problem, is
obtained by using the technique diagrammed in Figure 96.

WIDE BAND
SWEEP
FREQUENCY
GENERATOR

Figure 93 •.Equipment Connection for Transmission Loss Test.

VARIABLE
ATTENUATOR

Figure 94 illustrates the loss characteristic of a particular
reel of cable measured with this technique. The frequency
range was chosen to include a major defect at 137 MHz.
The rapid change in attenuation with frequency makes
accurate measurement of the dip at 137 MHz difficult.

-----

Ih
....

T
....

VARIABLE
TERMINATION

TO SCOPE
VERTICAL
INPUT

Figure 96. Equipment Connection for Impedance Test.

The output of a wide-band generator is fed through a
bridging detector to one end of the cable under test, with
the other end accurately terminated. The detector measures variations in the input voltage as a function of frequency. With a well-matched source (assured by the
10-dB attenuator) the input voltage varies almost directly
with the magnitude of the cable's input impedance. An
impedance plot made by this technique for the same reel
of cable is illustrated in Figure 97 (compare with Figure
95).

39dB

2dB

41dB

_ _--80ohma

130MHz

140MHz

150MHz

==::::=:'_~::'-

160MHz

__~W~7C:::::~=::"'-="'::::-~-

700hml

Figure 94. Transmission Loss Measurement, 100' Reel of RG59/U.

The measurement is made easier by inserting an equalizer
in series with the cable so that the average loss is flat and
the irregularity is more clearly displayed and measured,
as shown in Figure 95. One of the defects of the transmission loss measurement method appears on this plot.

130MHz

140MHz

150MHz

I60MHz

Figure 91. Impedance Test on 1000' Sample of RG59/U.

Calibration was obtained by substituting a precise 75-ohm
terminator for the cable end and varying the attenuator
above and below 10 dB by an amount corresponding to the
indicated impedance levels.

2dB

VARIABLE
STANDARD

75A

,..---{\,:)j--r,..,...""4 TERMINATION

75 OHMS
tlO%
INSERTION LOSS FROM "IN" TO ·OUT"ON dB)
=12.5 + RETURN LOSS OF CABLE
RETURN IDSS
130MHz

I40MHz

150MHz

= 20 lDG lO (REFLECTION COEFFICIENT)

I60MHz

Figure 95. Transmission Loss Measurement 1000' Reel RG59/U
Equalized.

Figure 98. Variable Bridge for Cable Reflection Testing.

65

Structural Return Loss Testing
The bridge used for reflection testing is diagrammed in
Figure 98. When the variable standard arm of the bridge
is adjusted to equal the average characteristic impedance
of the cable, the bridge acts as a directional coupler with
directivity in excess of 50 dB and a constant insertion loss
of about 12.5 dB. The bridge is connected into a test
system, as shown in Figure 99.

--------i-------~--+_----------------~2.2%
33dS RL
REFLECTiON

WIDE BAND
SWEEP
FREQUENCY
GENERATOR

CABLE
UNDER
TEST

130MHz

I
150MHz

140MHz

I
160MHz

Figure 101. Same as Figure 100 with Gain Increased. and
Reference Changed.

VARIABLE
TERMINATION

Comparison of Three Methods
Figure 99. Equipment Connection for Reflection Test.

Although the defect plotted in Figures 97 and 100 showed
up clearly in all three tests, it should be noted that it was
a particularly bad defect, i.e., one that would cause picture
distortion if it fell within a television channel. A defect
which is about the worst that can be tolerated in a cable
television system is illustrated in Figures 102, 103, and
104.

The variable attenuator generates a reference trace which
is set to cross the cable trace at peaks of the reflection
characteristic. Since the measurement is made in dBs, the
results are most conveniently expressed in these terms. The
reflection coefficient expressed in dB is the "return loss,"
and the return loss characteristic of cable, due to periodic
variations in its structure, has become known as the "structural return loss."

Figure 102, illustrating the transmission loss measurement
of such a defect, shows the difficulty of this method: using
all the scale expansion available, and equalizing the transmission characteristic, the O.I-dB variation is difficult to
discern and impossible to measure accurately.

Figure 100 illustrates a structural return loss plot with the
characteristics of the same cable defect as shown on the
curves in figures 95 and 97.

~

O.ldSr

-----

---I
I dE

-------------r~------------------------ladB
RETURN LOSS

- - - - - -__J

(12.6% REFLECTION'

Figure 102. Measurement of a Cable Defect by Transmission
Loss Method.

Figure 103 shows a great improvement in sensitivity obtained by impedance measurement, but also illustrates the
weakness of this method in that four different measurements were obtained, depending critically on small variations in the point at which the cable was connected to the
detector. The reading on this particular defect varied
from 4.4% up to 12.3%, depending on the length of the
connection.
DO

RL

Figure 104 shows the advantage of the return loss bridge
method, which gives a high degree of sensitivity with
essentially the same reading, regardless of the point of
connection (compare with Figure 103).

=:~~L __________--"::=~==""",:::::;:::".,L~ (0% REFLECTiON)

13DMHz

140MHz

150MHz

I
160MHz

Figure 100. Reflection Test on RG59jU Sample.

This comparison is further illustrated by measurements
made on a reel of good CATV trunk line cable. Figure 105
illustrates the return loss characteristic taken over the
entire TV spectrum, showing excellent structural return
loss characteristics.

The sensitivity of this method is illustrated by Figure 101,
which is similar to Figure 100, but shows a 15-dB increase
in sensitivity. Return loss variations as low as 50 dB (0.3%
reflection) can be clearly displayed.

66

/------~

76'5011""(+4'1%)

AVERAGE!!
730bml

~

~

O~IGINAL

INPUT
iMPEDANCE

\ 30dB

~--

_

68.801l"".(-5.B%)

J(

B20h"" (+12.3%)

_ _ /""<

_________

~

~

I/B WAVELENGTH
ADDED

------~--r_----------_i~~It--------t_----t_--~ 40dB

-

I

71.2011 ... ,(+4.4%)

_

I

(

50MHz
1/4 WAIiELENCITIt
ADDED

(OOMHz

150MHz

200MHz

Figure 105. Reflection Test on Sample Reel of CATV Trunk Cable.

Figures 106 and 107 show transmission loss measurements
near the worst defect. Note that the transmission loss
variation at this poin~ can hardly be seen. None of the
variations below this level could be ~een or measured by
this method.

66.2 oh... , (-9.3%)

Figure 103. Sa'me Defect as Figure 101 Measured by the
Impedance fest Method.
8dB

26 dB

o
9dB
--'-----------~----------

26dB

----60MHz
I

I

50 MHz

----------------------

1/1 >"ADDED

----------------------~--------------------26dB

IOd8

Figure 106. Transmission Loss Test on Same Reel as Figure 105
Near Worst Defect.

~ldBt

1/4 hADDED

--------------------w------------26dB

--~/

1

3/8 h ADDED

Figure 104. Same Defect as Figures 101 and 103 Measured by Ibe
Reflection Test Method.
SOMHz

60MHz

Figure 107. Same as Figure 106 Except Transmission Line Equalized.

67

Figure 108 illustrates an impedance test of this worst defect, and Figures 109 and 110 show return loss tests in this
same frequency range.
_____________________________________________

reflection test method, using a return loss bridge adjusted
to the average impedance of the cable under test, provides a high degree of sensitivity, ease of calibration, and
freedom from ambiguous readings.

80~ms

Adjusting the Return Loss Bridge and the
Cable Termination:
The bridge used for measuring Structural Return Loss of
cable (for example the Jerrold Model RLBV-4H-7F, with
"F" fittings) used for 75-ohm cable, is provided with a
resistance and a capacitance adjustment which allow
matching the bridge to the average characteristic impedance of each particular reel of cable. This allows distinguishing defects in cable due to non-standard characteristic impedance from those due to periodic discontinuities.
In measuring cable, it is most important that these adjustments be set correctly before reading Return Loss. The
following shows the result of correct adjustment, as well
as the result when one or the other adjustment is wrong.

____________________--,---______________________ 70 ohms
60MHz

50MHz

Figure IDS. Impedance Test on Same Reel as Figure 105
Near Worst Defect.

-----r---~ ,,,.

I~
I

Figure III illustrates the trace obtained on a reel of highgrade CATV cable with the bridge and the far-end
terminator correctly adjusted. The frequency range is from
50 to 220 MHz and the reference line at the top is set at
30 dB return loss. It is essential to set the bridge so that
the minimums in the trace touch the base line as they do
here. Erroneous readings are obtained if the bridge is adjusted for minimum readings on the "spikes", without
considering the base line.

I~n

IA

'VV"v'

\~

I
50MHz

60MHz
TRACES ILLUSTRATING ADJUSTMENT Of BRIDGE TO MATCH
CABLE'S AVERAGE IMPEDANCE

Figure 109. Reflection Test on Same Reel Near Wont Defect.

____------------------------rtr-~------------40dB
il
II

I
\

'!

Ii

Figure Ill. Correct Settings-Minimums Reach Base Line At All
Frequencies.

'\

Figure ll2 shows the effects of misadjusting the far-end
termination. With a long reel of cable this has a major
effect only at the low-frequency end of the trace, producing a "fuzzy" trace due to the rapid oscillations in impedance caused by reflections from the far end.

I

50MHz

60MHz

Figure 11 O. Same as Figure 109 Except Gain Increased,
Reference Shifted.

In summary, three methods that have been used to determine the existence of electrical problems due to periodic
discontinuities in cable have been described. The transmission measurement method suffers from low sensitivity
and the need for equalization. The impedance measurement method presents difficulties in calibration and is
ambiguous because of variations peculiar to the point of
connection between cable and test set. In contrast, the

Figure 112. Termination Misadjusted-Makes Trace Fuzzy,
Especially At Low End.

68

Figure 113 shows a trace with the far end correctly terminated, but with the resistance balance on the bidge
slightly offset from the average impedance for this reel.
The effect is a trace raised off the base line across the
entire frequency range.

Figure 114 shows a trace with far-end termination and resistance balance set correctly, but with a slight misadjustment of the bridge capacitance balance. This has its
greatest effect at the high-frequency end, causing a rising
characteristic. This is very similar to the effect produced
when the connector between the bridge and the cable end
has a characteristic impedance which differs greatly from
the average impedance of the reel.

Figure 113. Resistance Balance Set Wrong-Capacitance Balance and
Termination Set Correct. (NOTE: Minimums Do Not Reach Base Line.)

Figure 114. Capacitance Balance Set Wrong. Curve Tips Up At High End.
This Can Also Be Caused By Bad Connection Between Cable and Bridge.

Note: Error Due to Detector Envelope Response
To show a flicker-free display on an oscilloscope the horizontal scan of the usual sweepfrequency generator has a frequency of 25 to 60 Hz. When a complex return loss pattern,
such as the one shown in Figure 105 is scanned 60 times per second, severe requirements
are placed on the detector if its detected output is to follow accurately the rapid changes
in r-f input. Since many detectors in common use are not fast enough, errors in the measurement of Structural Return Loss may result when scanning rapidly over wide bands.
To test for this trouble, or to be sure of an accurate measurement, reduce the sweepwidth to a few MHz and tune the sweep's center frequency to center each major spike
in turn on the 'scope (so that it looks like Figure 109, for example). When the measurement at this setting shows a poorer Return Loss than the measurement with wide sweep,
the latter is in error due to detector trouble and should be disregarded.

69

70

CHAPTER XI
CHARTS AND TABLES

INSTRUCTIONS ON THE USE OF CHARTS
AND TABLES

APPLICATIONS OF THE POWER vs. dB CHARTS
AND TABLES

General

1. Power Gain or Loss The direct application is to
cases where dBs are related to power gai~ or loss. Example: The input to an amplifier is 0.1 milliwatt. Its gain
is 32 dB, what is the output power? From the Chart (P2)
32 dB corresponds to a power gain of about 1600. Thus
the output is 1600 x 0.1 mW or 160 milliwatts.

The charts are intended for use where approximate
answers are sufficient, and to determine the decimal
point (for ratios) or the prefix (for dBs) when the given
ratio is outside the range of the tables. The charts generally cover the range from -60 to +60 dB with an
accuracy of about 0.5 dB, while the tables cover the range
-10 to +10 dB for power ratios, and -20 to +20 dB
for voltage ratios, with an accuracy of 0.001 dB.

2. Percent Power The tables relating power ratio to
-dBs can be used for percent by making use of the fact
that:
a smaller power, expressed in percent of a larger one

Use of Charts

= 100 x the power ratio

Each chart may be entered either horizontally (when you
know the ratio) or vertically (when you know the dBs).
For example, assume you know the power ratio is 4200 and
want the corresponding dBs. Use chart P2, covering power
ratios between 1 and 1,000,000. Notice that 4200 shows
on the right-hand scale, and follOwing the indication of
the arrows, the bottom scale is used with this for dBs.
Moving across the 4200 line you will see that it intersects
the slanting line above 36.5 dB on the bottom scale. So
36.5 dB is the answer.

smaller power)
( 1
argerpower

Example: When the ratio of two powers is 0.119, the
smaller is 11.9% of the larger (11.9 = 100 X
0.119).
Example: The loss of a certain directional coupler from input to tap is 17.2 dB. What percent of the input
power reaches the tap? From chart PI and table
P6 relating power ratio and - dB, -17.2 dB
corresponds to a power ratio of 0.0190. So the
power reaching the tap is 1.9% (0.019 X 100)
of the input.

As an example of the reverse conversion, suppose you are
given -14 dB, to find the power ratio. Use chart PI.
-14 dB occurs on the bottom scale, and according to the
arrows the left-hand scale goes with this. The 14 dB
line intersects the slanting line at a power ratio of 0.039,
the answer.

3. Conversion of Power to dBm The dBm (abbreviation for "dB above one milliwatt") is commonly used to
express power levels in electronics (other than in CATV).
It is defined by: Power level in dBm
dBs corresponding
to the ratio of the given power to the standard power of 1
milliwatt.

=

Use of Tables
Each table is always entered from the left. Find the first
two significant figures of the given number in the lefthand column, then locate the answer under the column
headed by the third significant figure. As an example,
assume a given power ration of 4250. Use table P7 (power
ratios between 1.0 and 10) to find the dBs corresponding to
4.25. Find 4.2 in the left-hand column, follow this line to
the right to find the answer 6.284 dB in the 0.05 column.
From the chart, as described above, 36.5 dB corresponds
approximately with a power ratio of 4200, so the accurate
dB !lumber for a ratio of 4250 is found by adding 30 to
that found for 4.25, giving 36.284 dB.

To find dBm corresponding to a given power:
a. Convert power to milliwatts
b. Consider this as a power ratio, and find dBm in
table or chart relating ratio and dB.
Example: Express a power of 7.75 watts in dBm.
a. 7.75 watts is 7750 milliwatts (mW

=W

X 1000).

b. From Chart P2, a power ratio of 7750 corresponds to 38.9 dB. So 7.75 watts corresponds
to +38.9 dBm.

When the given number is in dB, another table is used.
Given -14.15, for example, the approximate ratio from
chart PI, is 0.039 (as illustrated above). To find the
accurate ratio, look up -4.15 dB in table P6. On the 4.10
line under the 0.05 column find a ratio of 0.3846. Moving
the decimal point to agree with the chart answer gives the
accurate power ratio 0.03846..

4. Conversion of dBm to Power The power ratio corresponding to the given dBm is the power in milliwatts.
Example: What power is represented by -16.2 dBm?
From Chart PI, -16.2 dB corresponds to a ratio of 0.024.
So the answer is 0.024 mW.

7I

5. Relation between dBm and dBmV. Since dBm and
dBmV both express power levels in dBs, they are related
by a constant factor. To find this factor, we have to calculate first the power corresponding to 1 millivolt across 75
ohms: .

b. Find the combined noise factor fl.
formula:

= 4.27 + 7.96 = 12.23
c. To find the combined noise figure
we convert this
ratio to dBs. Using chart P2, table P7, we find a power
ratio of 12.33 corresponds to 10.9 dB.

Then we divide this into one milliwatt to find the power
ratio:

8. Power Addition A problem that occurs in studying
noise in CATV systems is to find the power level corresponding to the sum of the powers represented by two
given levels. One way to solve this is to use the Power vs.
dB tables:

1
1000
1

by applying the

.
23.44 - 1
22.44
fi 2 = 4.27 +
= 4.27 + - - =
•
2.82
2.82

1
E2
(1000)2
1
Power = = ----=
watts·
'
R
75
75 x 106
.

Power Ratio =

2

75 X 10 6
- - - = 75,000.
1000

75 X 106

Example: The noise at a certain point in a system is due to
two components, one at +7.31 dBmV, the other at +12.73
dBmV. What is the level of the total noise?

From Chart P2 and Table P7, we find that the number of
dBs corresponding to a power ratio of 75,000 is 48.75 dB.

a. Find the power ratios (to 0 dBmV) corresponding to
the given levels from Chart P2, Table P8.

From this follow the rules:

a. To convert dBmV to dBm: subtract 48.75 dB.

b. Add the ratios to find the power ratio of the total.

b. To convert dBm to dBmV: add 48.75 dB.

c. Find corresponding dBmV.
6. Noise Factor and Noise Figure. In some noise problems
it is necessary to find the Noi~e Factor of a device. This is
simply the power ratio corresponding to the Noise Figure.
As Noise Figure is expressed in dB, and Noise Factor is a
power ratio, we can use the Power vs. dB charts and tables
to find one when the other is known.

In this case: +7.S1 dB corresponds to a power ratio of
5.S8S, and +12.73 dB corresponds to a power ratio of
18.75. The sum of the power ratios is 24.1S, and (from
Chart P2, Table P7) the corresponding total noise level is
13.82dBmV.

Example: An amplifier has a measured noise figure of 8.S
dB; what is the noise factor?

A somewhat faster way to solve problems involving power
addition, is to use the chart PS or table P9 relating the
difference between two dBmV levels to the dB to be added
to the larger to find the level corresponding to the power
sum.

From table P8 we find the noise factor is 6.761.
7. System Noise Calculations-when one amplifier follows
another with no attentuation between them. When the output of one amplifier is connected directly to the input of
another, the noise factor of the combination is found from
the formula:
f 2 -1
fl.2=fl+~

Example: Assume the same problem as above, where we
were given two noise contributions of +7.S1 dBmV and
+ 12. 7S dBm V to find the level corresponding to the total
noise. The dB difference is 12.7S - 7.S1 == 5.52 dB. On
Chart PS, opposite 5:5 dB difference, we find 1.1 dB.
Adding 1.1 to 12.73, we obtain the answer: 13.8 dBmV
( compare with solution above) .

where: fl is the noise factor of the first amplifier,
f2 is the noise factor of the second amplifier,
G1 is the power gain of the first amplifier.

When negative levels are involved, remember that the
higher level is the one with the smaller number of negative dBs: To find the level corresponding to the power sum
of -17 and -20 dBmV, opposite the difference (S dB)
on chart PS, find 1.8 dB. Add this to the higher level, which
is -17 dBmV: -17 + 1.8 = -15.2.

This formula applies only to noise factors and power gain,
so when noise figures and dB gain are given, they must be
converted to the corresponding ratios before the formula
can be used.

9. Power Division Another application of the "power
sum" idea allows simplified analysis of splitters and taps
in CATV systems. A splitter divides energy between two
outgoing cables; a directional coupler or tap diverts a
small part of the feeder's energy to the house drop, passing
the major part down the feeder. Each of these devices is
designed to have as little internal power loss as possible,
so their performance can be estimated by assuming that
there is no power loss, all of the energy dividing between
the output. Figure 115 is a diagram of such a device.

Example: An amplifier with a noise figure F 1 of 6.3 dB and
gain of 4.5 dB is followed by a second amplifier with a
noise figure F 2 of 13.7 dB; what is the noise figure Fl. 2 of
the combination?
a. Find the power ratios from chart P2, table P8:
Fl (power ratio corresponding to 6.3 dB) = 4.27
F2 (power ratio corresponding to IS.7 dB) = 2S.44
G1 (power ratio corresponding to 4.5 dB)
2.82

=

72

..

OUTPUT" A"
INPUT

-

LOSSLESS
DIVIDER
(TAP OR
(SPLITTER)

A commonly used measure of reflection in CATV equipment and cables is the "return loss", the dB ratio of input
power to reflected power. When the device is loss less, the
sum of the output power and the reflected power must
equal the input power. Thus the "power-split" curve can

--

OUTPUT "B"

be used to find the insertion loss (Pin) when the return
(Pout)
(Pin) .
loss - - - IS known.
(Pretl)

Figure 115.

Assuming no internal loss:
Input Power = Output Power "A"

Example: A filter has the impedance match shown in Figure 118:

+ Output Power "B".

What is its insertion loss, if there are no internal losses?

When this relation is established, chart P4 or table PlO
can be used to find the loss from input to one output, when
the loss from the input to the other output is known.
Example: A directional coupler has 12 dB tap loss; what
would the line loss be if the coupler had no internal losses?
On chart P4, or in table PlO, opposite 12 dB tap loss we
find a line loss of 0.28 dB.

RETURN

LOSS
16dB I---::::oo-_:-----_+_

These relations assume no internal loss, but one case with
loss can be handled readily. When a non-directional tap
includes a resistor to provide back-match of the tap output,
the tap output energy divides almost equally between this
resistor and the tap output (see Figure 116).

~ ~------~~---

FREQUENCY - - - - .

Figure 118.

Loss at minimums, where return loss is infinite, is O.
Loss at maximums, where return loss is 16 dB, is O.ll -dB
(from P4 or PlO) so the transmission curve is approximately as shown in Figure 119.

OUTPUT

INPUT

LOSS
dB
Figure 116.

Figure 119.

--n

In this situation the line loss can be found by first subtracting 3 dB from the actual tap loss, and then looking
up the corresponding line loss on chart P4.

Example: A cable has the return loss spike shown in Figure 120.
20 dB
FREQUENCY

A l6-dB back-matched tap, for example, would have the
same loss (0.22 dB) as a lossless l3-dB tap.

MATCH

10. Application of Power-Split Data to ReSection Problems
When a lossless device having mismatch is introduced between a previously matched source and load (see Figure
117) the power-split curve on chart P4 can be used to find
the resulting insertion loss.

Figure 120.

What insertion loss is caused? (assuming it is due only to
the mismatch at the input terminal).
With a 20-dB return loss spike the insertion loss (from
PlO) is 0.04 dB as shown in Figure 121.

Pout

Pin

-----.

P refl

LOSSLESS
DEVICE

-----.

~ 0.04 dB

LOAD

V

t

Figuren7.

Figure 121.

73

LOSS

CHART Pl
dB

-30

1.0

R~

0.8

-35

-40

-45

-50

-55

__

-60
1 X 10-3
8

0.6
0.5

6

0.4

4

0.3

3

0.2

2

0.1

IX 10- 4

5X10- 4

0.08

8

0.06
0.05

6

0

~

0.04

4

0::

0.03

3

0::

w

5X 10-5

3:
a. 0.02

~

0::
0::

W

3:
a.

0

0

2

IX10- 5

0.01
0.008

8

0.006
0.005

5 X10-6

6

Ii

0.004

4

0.003

3

0.002

2

0.001

0

o

-5

-10

-15
dB

74

-20

-25

IX10- 6

-30

CHART P2

o

5

dB
15

10

20

25

',000

800,000

800
600
500
400
300

30
1,000.000

-

600.000
500,000

~-

400,000
300,000

200

200.000

!J

100

100,000

80

80,000

~

60
50

60,000
50,000

0::

40

40,000

0::
UJ

30

30,000

a. 20

20,000

0

0==

I

III 10,000
8,000

10
8
6
5
4
3

--

6,000
5,000
4,000
3,000

2,000

2

I

30

35

40

50

45

dB
75

55

1,000
60

0

~
0::
0::
UJ

0==

a.

CHART P3

o

...J

5

POWER (OR NOISE) ADDITION
dB DIFFERENCE BETWEEN LEVELS
25
20
15
10

30

35

3.0

3.0

2.0

2.0

1.0

1.0

0.6

0.6

0.4

0.4

0.2

0.2

L&.I

>
L&.I

...J

0:::

~

C)

::I:

~
fi)

\

0.1

0.1

c

c

«

m 0.06

0.06

"t:J

0.04

0.04

0.02

0.02

,
\

0.01

0.01

a006

0.006
0.004 0

5

10
15
20
25
dB DIFFERENCE BETWEEN LEVELS
76

30

350 .004

CHART P4

4

6

POWER SPLIT CHART
dB LOSS FROM INPUT TO IIBII
8
10
12

14

16

3.0

3.0

2.0

2.0
OUTPUT "A"
INPUT

LOSSLESS
DIVIDER
(TAP OR
SPLITTER)

OUTPUT "B"

- -

1.0

.

-"-

1.0

0.8

0.8

0.6

0.6

zO.4

0.4

«
=

~

I-

::;:)

a.
~

o

a: 0.3

0.3

u.

en
en
o
...J

m

0.2

0.2

0.1

0.1

"0

0.08

4

6

8
10
12
dB LOSS FROM INPUT TO "B"

77

14

16

0.08

TABLE P5

GIVEN: POWER RATIO (0.1 to 1.0); FIND -dB (-10 to 0)

For power ratios less than 0.1:

of 10

Add

Multiple

-dBs

oflO

-dBs

10
10'
10"
10"

-10
-20
-30
-40

10·
10·
10 7
10·

-50
-60
-70
-80

Multiple

1. Multiply given ratio by a multiple of 10 as required to give a number
between 0.1 and 1.0.
2. Look up -dB corr«:;sponding to this number in Table P5.
3. Add to the result -dBs corresponding to the multiple of 10 used in
step 1.

Add

=

Example: Given power ratio 0.0000076 (7.6 x 10- 6 ). Multiply by 10- 5 to get 0.76. From Table P5, -dB corresponding
to 0.76 is -1.192. Adding -50 dB (corresponding to the 10- 0 multiplier used) we get -1.192 + (-50)
-51.192 dB.

=

NOTE: The minus signs in front of all dB numbers in this table have been omitted!

.001

.002

.003

.004

.005

.006

.007

.00B

.009

10.000
9.586
9.208
8.861
8.539

9.957
9.547
9.172
8.827
8.508

9.914
9.508
9.136
8.794
8.577

9.872
9.469
9.101
8.761
8.447

9.830
9.431
9.066
8.729
8.416

9.788
9.393
9.031
8.697
8.386

9.747
9.355
8.996
8.665
8.356

9.706
9.318
8.962
8.633
8.327

9.666
9.281
8.928
8.601
8.297

9.626
9.245
8.894
8.570
8.268

8.239
7.959
7.696
7.447
7.212

8.210
7.932
7.670
7.423
7.190

8.182
7.905
7.645
7.399
7.167

8.153
7.878
7.620
7.375
7.144

8.125
7.852
7.595
7.352
7.122

8.097
7.825
7.570
7.328
7.100

8.069
7.799
7.545
7.305
7.077

8.041
7.773
7.520
7.282
7.055

8.013
7.747
7.496
7.258
7.033

7.986
7.721
7.471
7.235
7.011

6.990
6.778
6.576
6.383
6.198

6.968
6.757
6.556
6.364
6.180

6.946
6.737
6.536
6.345
6.162

6.925
6.716
6.517
6.326
6.144

6.904
6.696
6.498
6.308
6.126

6.882
6.676
6.478
6.289
6.108

6.861
6.655
6.459
6.271
6.091

6.840
6.635
6.440
6.253
6.073

6.819
6.615
6.421
6.234
6.055

6.799
6.596
6.402
6.216
6.038

6.021
5.850
5.686
5.528
5.376

6.003
5.834
5.670
5.513
5.361

5.986
5.817
5.654
5.498
5.346

5.969
5.800
5.638
5.482
5.331

5.952
5.784
5.622
5.467
5.317

5.935
5.768
5.607
5.452
5.302

5.918
5.751
5.591
5.436
5.287

5.901
5.735
5.575
5.421
5.272

5.884
5.719
5.560
5.406
5.258

5.867
5.702
5.544
5.391
5.243

5.229
5.086
4.949
4.815
4.685

5.214
5.072
4.935
4.802
4.672

5.200
5.058
4.921
4.789
4.660

5.186
5.045
4.908
4.776
4.647

5.171
5.031
4.895
4.763
4.634

5.157
5.017
4.881
4.750
4.622

5.143
5.003
4.868
4.737
4.609

5.129
4.984
4.855
4.724
4.597

5.114
4.976
4.841
4.711
4.584

5.100
4.962
4.828
4.698
4.572

4.559
4.437
4.318
4.202
4.089

4.547
4.425
4.306
4.191
4.078

4.535
4.413
4.295
4.179
4.067

4.522
4.401
4.283
4.168
4.056

4.510
4.389
4.271
4.157
4.045

4.498
4.377
4.260
4.145
4.034

4.486
4.365
4.248
4.134
4.023

4.473
4.353
4.237
4.123
4.012

4.461
4.342
4.225
4.112
4.001

4.449
4.330
4.214
4.101
3.990

3.979
3.872
3.768
3.665
3.565

3.969
3.862
3.757
3.655
3.556

3.958
3.851
3.747
3.645
3.546

3.947
3.840
3.737
3.635
3.536

3.936
3.830
3.726
3.625
3.526

3.925
3.820
3.716
3.615
3.516

3.915
3.809
3.706
3.605
3.507

3.904
3.799
3.696
3.595
3.497

3.893
3.788
3.686
3.585
3.487

3.883
3.778
3.675
3.575
3.478

.000

78

TABLE P5 continued

.000

.001

.002

.003

.004

.005

.006

.007

.008

.009

3.468
3.372
3.279
3.188
3.098

3.458
3.363
3.270
3.179
3.089

3.449
3.354
3.261
3.170
3.080

3.439
3.344
3.251
3.072

3.429
3.335
3.242
3.152
3.063

3.420
3.325
3.233
3.143
3.054

3.410
3.316
3.224
3.134
3.045

3.401
3.307
3.215
3.125
9.036

3.391
3.298
3.206
3.116
3.028

3.382
3.288
3.197
3.107
3.019

3.010
2.924
2.840
2.757
2.676

3.002
2.916
2.832
2.749
2.668

2.993
2.907
2.823
2.741
2.660

2.984
2.899
2.815
2.733
2.652

2.933
2.890
2.765
2.684
2.604

2.976
2.882
2.807
2.725
2.644

2.967
2.874
2.798
2.716
2.636

2.958
2.865
2.790
2.708
2.628

2.950
2.857
2.782
2.700
2.620

2.941
2.848
2.774
2.692
2.612

2.596
2.518
2.441
2.366
2.291

2.588
2.510
2.434
2.358
2.284

2.581
2.503
2.426
2.351
2.277

2.573
2.495
2.418
2.343
2.269

2.565
2.487
2.411
2.336
2.262

2.557
2.480
2.403
2.328
2.255

2.549
2.472
2.396
2.321
2.248

2.541
2.464
2.388
2.314
2.240

2.534
2.457
2.381
2.306
2.233

2.526
2.449
2.373
2.299
2.226

2.218
2.147
2.076
2.007
1.938

2.211
2.140
2.069
2.000
1.931

2.204
2.132
2.062
1.993
1.925

2.197
2.125
2.055
1.986
1.918

2.190
2.118
2.048
1.979
1.911

2.182
2.111
2.041
1.972
1.904

2.175
2.104
2.034
1.965
1.897

2.168
2.097
2.027
1.959
1.891

2.161
2.090
2.020
1.952
1.884

2.154
2.083
2.013
1.945
1.878

1.871
1.805
1.739
1.675
1.612

1.864
1.798
1.733
1.669
1.605

1.858
1.791
1.726
1.662
1.599

1.851
1.785
1.720
1.656
1.593

1.844
1.778
1.713
1.649
1.586

1.838
1.772
1.707
1.643
1.580

1.831
1.765
1.701
1.637
1.574

1.824
1.759
1.694
1.630
1.568

1.818
1.752
1.688
1.624
1.561

1.811
1.746
1.681
1.618
1.555

1.549
1.487
1.427
1.367
1.308

1.543
1.481
1.421
1.361
1.302

1.537
1.475
1.415
1.355
1.296

1.530
1.469
1.409
1.349
1.290

1.524
1.463
1.403
1.343
1.284

1.518
1.457
1.397
1.337
1.278

1.512
1.451
1.391
1.331
1.273

1.506
1.445
1.385
1.325
1.267

1.500
1.439
1.379
1.319
1.261

1.494
1.433
1.373
1.314
1.255

1.249
1.192
1.135
1.079
1.024

1.244
1.186
1.129
1.073
1.018

1.238
1.180
1.124
1.068
1.013

1.232
1.175
1.118
1.062
1.007

1.226
1.169
1.113
1.057
1.002

1.221
1.163
1.107
1.051
.996

1.215
1.158
1.101
1.046
.991

1.209
1.152
1.096
1.040
.985

1.203
1.146
1.090
1.035
.980

1.198
1.141
1.085
1.029
.975

.969
.915
.862
.809
.757

.964
.910
.857
.804
.752

.958
.904
.851
.799
.747

.953
.899
.846
.794
.742

.947
.894
.841
.788
.737

.942
.888
.835
.783
.731

.937
.883
.830
.778
.726

.931
.878
.825
.773
.721

.926
.872
.820
.768
.716

.921
.867
.814
.762
.711

.706
.655
.605
.555
.506

.701
.650
.600
.550
.501

.696
.645
.595
.545
.496

.691
.640
.590
.540
.491

.685
.635
.585
.535
.487

.680
.630
.580
.531
.482

.675
.625
.575
.526
.477

.670
.620
.570
.521
.472

.665
.615
.565
.516
.467

.660
.610
.560
.511
.462

.458
.410
.362
.315
.269

.453
0405
.357
.311
.264

.448
0400
.353
.306
.259

.443
.395
.348
.301
.255

.438
.391
.343
.297
.250

.434
.386
.339
.292
.246

.429
.381
.334
.287
.241

.424
.376
.329
.283
.237

.419
.372
.325
.278
.232

.414
.367
.320
.273
.227

.223
.177
.132
.088
.044

.218
.173
.128
.083
.039

.214
.168
.123
.079
.035

.209
.164
.119
.074
.031

.205
.159
.114
.070
.026

.200
.155
.110
.066
.022

.195
.150
.106
.061
.017

.191
.146
.101
.057
.013

.186
.141
.097
.052
.009

.182
.137
.092
.048
.004

3~161

79

TABLE P6

GIVEN: -dB (0 to -10); FIND: POWER RATIO (1.0 to 0.1)

For-dB below -10:

FordBs
Added

Multiply
By .

FordBs
Added

Multiply
By

10
20
30
40

0.1
0.01
0.001
0.0001

50
60
70
80

10-"
10-"
10-'
10- 8

1. Add dBs as required to give a number between 0 and -10.
2. Look up power ratio corresponding to this number of dBs in Table P6.
3. Multiply the power ratio found by the number corresponding to dBs
added in step 1.

Example: Given -37.21 dB. Add 30 dB to get -7.21 dB. From Table P6, ratio corresponding to -7.21 dB is 0.1901.
Corresponding to the 30 dBs added, multiply 0.1901 x 0.001 to get power ratio of 0.0001901 or 1.901 x 10- 4 •
0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

1.0000
.9772
.9550
.9333
.9120

.9977
.9750
.9528
.93Il
.9099

.9954
.9727
.9506
.9290
.9078

.9931
.9705
.9484
.9268
.9057

.9908
.9683
.9462
.9247
.9036

.9886
.9661
.9441
.9226
.9016

.9&63
.9638
.9419
.9204
.8995

.9840
.9616
.9397
.9183
.8974

.9817
.9594
.9376
.9162
.8954

.9795
.9572
.9354
.9141
.8933

.8913
.8710
.85Il
.8318
.8128

.8892
.8690
.8492
.8299
.8IlO

.8872
.8670
.8472
.8279
.8091

.8851
.8650
.8453
.8260
.8072

.8831
.8630
.8433
.8241
.8054

.8810
.8610
.8414
.8222
.8035

.8790
.8590
.8395
.8204
.8017

.8770
.8570
.8375
.8185
.7998

.8750
.8551
.8356
.8166
.7980

.8730
.8531
.8337
.8147
.7962

.7943
.7762
.7586
.7413
.7244

.7925
.7745
.7568
.7396
.7228

.7907
.7727
.7551
.7379
.72Il

.7889
.7709
.7534
.7362
.7194

.7870
.7691
.7516
.7345
.7178

.7852
.7674
.7499
.7328
.7161

.7834
.7656
.7482
.73Il
.7145

.7816
.7638
.7464
.7295
.7129

.7798
.7621
.7447
.7278
.7Il2

.7780
.7603
.7430
.7261
.7096

.7079
.6918
.6761
.6607
.6457

.7063
.6902
.6745
.6592
.6442

.7047
.6887
.6730
.6577
.6427

.7031
.6871
.6714
.6561
.6412

.7015
.6855
.6699
.6546
.6397

.6998
.6839
.6683
.6531
.6383

.6982
.6823
.6668
.6516
.6368

.6966
.6808
.6653
.6501
.6353

.6950
.6792
.6637
.6486
.6339

.6934
.6776
.6622
.6471
.6324

.6310
.6166
.6026
.5888
.5754

.6295
.6152
.6012
.5875
.5741

.6281
.6138
.5998
.5861
.5728

.6266
.6124
.5984
.5848
.5715

.6252
.6109
.5970
.5834
.5702

.6237
.6095
.5957
- .5821
.5689

.6223
.6081
.5943
.5808
.5675

.6209
.6067
.. 5929
.5794
.5662

.6194
.6053
.5916
.5781
.5649

.6180
.6039
.5902
.5768
.5636

.5623
.5495
.5370
.5248
.5129

.5610
.5483
.5358
.5236
.5Il7

.5598
.5470
.5346
.5224
.5105

.5585
.5458
.5333
.5212
.5093

.5572
.5445
.5321
.5200
.5082

.5559
.5433
.5309 .
.5188
.5070

.5546
.542(}
.5297
.5176
.5058

.5534
.5408
.5284
.5164
.5047

.5521
.5395
.5272
.5152
.5035

.5508
.5383
.5260
.5140
.5023

.5012
.4898
.4786
.4677
.4571

.5000
.4887
.4775
.4667
.4560

.4989
.4875
.4764
.4656
.4550

.4977
.4864
.4753
.4645
.4539

.4966
.4853
.4742
.4634
.4529

.4955
.4842
.4732
.4624
.4519·

.4943
.4831
.4721
.4613
.4508

.4932
.4819
.4710
.4603
.4498

.4920
.4808
.4699
.4592
.4487

.4909
.4797
.4688
.4581
.4477

.4467
.4365
.4266
.4169
.4074

.4457
.4355
.4256
.4159
.4064

.4446
.4345
.4246
.4150
.4055

.4436
.4335
.4236
.4140
.4046

.4426
.4325
.4227
.4130
.4036

.4416
.4315
.4217
.4121
.4027

.4406
.4305
.4207
.4Ill
.4018

.4395
.4295
.4198
.4102
.4009

.4385
;4285
.4188
.4093
.3999

.4375
.4276
.4178
.4083
.3990

.3981
.3890
.3802
.3715
.3631

.3972
.3882
.3793
.3707
.3622

.3963
.3873
.3784
.3698
.3614

.3954
.3864
.3776
.3690
.3606

.3945
.3855
.3767
.3681
.3597

.3936
.3846
.3758
.3673
.3589

.3926
.3837
.3750
.3664
.3581

.3917
.3828
.3741
.3656
.3573

.3908
.3819
.3733
.3648
.3565

.3899
.38Il
.3724
.3639
.3556

0.00

80

TABLE P6 continued

0.00

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

.3548
.3467
.3388
.3311
.3236

.3540
.3459
.3381
.3304
.3228

.3532
.3451
.3373
.3296
,3221

.3524
.3443
.3365
.3289
.3214

.3516
.3436
.3357
.3281
.3206

.3508
.3428
.3350
.3273
.3199

.3499
.3420
.3342
.3266
.3192

.3491
.3412
.3334
.3258
.3184

.3483
.3404
.3327
.3251
.3177

.3475
.3396
.3319
.32'43
.3170

.3162
.3090
.3020
.2951
.2884

.3155
.3083
.3013
.2944
.2877

.3148
.3076
.3006
.2938
.2871

.3141
.3069
.2999
.2931
.2864

.3133
.3062
.2992
.2924
.2858

.3126
.3055
.2985
.2917
.2851

.3119
.3048
.2979
.2911
.2844

.3112
.3041
.2972
.2904
.2838

.3105
.3034
.2965
.2897
.2831

.3097
.3027
.2958
.2891
.2825

.2818
.2754
.2692
.2630
.2570

.2812
.2748
.2685
.2624
.2564

.2805
.2742
.2679
.2618
.2559

.2799
.2735
.2673
.2612
.2553

.2793
.2729
.2667
.2606
.2547

.2786
.2723
.2661
.2600
.2541

.2780
.2716
.2655
.2594
.2535

.2773
.2710
.2649.2588
.2529

.2767
.2704
.2642
.2582
.2523

.2761
.2698
.2636
.2576
.2518

.2512
.2455
.2399
.2344
.2291

.2506
.2449
.2393
.2339
.2286

.2500
.2443
.2388
.2333
.2280

.2495
.2438
.2382
.2328
.2275

.2489
.2432
.2377
.2323
.2270

.2483
.2427
.2371
.2317
.2265

.2477
.2421
.2366
.2312
.2259

.2472
.24l5
.2360
.2307
.2254

.2466
.2410
.2355
.2301
:2249

.2460
.2404
.2350
.2296
.2244

.2239
.2188
.2138
.2089
.2042

.2234
.2183
.2133
.2084
.2037

.2228
.2178
.2128
.2080
.2032

.2223
.2173
.2123
.2075
.2028

.2218
.2168
.2118
.2070
.2023

.2213
.2163
.2113
.2065
.2018

.2208
.2158
.2109
.2061
.2014

.2203
.2153
.2104
.2056
.2009

.2198
.2148
.2099
.2051
.2004

.2193
.2143
.2094
.2046
.2000

.1995
.1950
.1905
.1862
.1820

.1991
.1945
.1901
.1858
.1816

.1986
.1941
.1897
.1854
.1811

.1982
.1936
.1892
.1849
.1807

.1977
.1932
.1888
.1845
.1803

.1972
.1928
.1884
.1841
.1799

.1968
.1923
.1879
.1837
.1795

.1963
.1919
.1875
.1832
.1791

.1959
.1914
.1871
.1828
.1786

.1954
.1910
.1866
.1824
.1782

.1778
.1738
.1698
.1660
.1622

.1774
.1734
.1694
.1656
.1618

.1770
.1730
.1690
.1652
.1614

.1766
.1726
.1687
.1648
.1611

.1762
.1722
.1683
.1644
.1607

.1758
.1718
.1679
.1641
.1603

.1754
.1714
.1675
.1637
.1600

.1750
.1710
.1671
.1633
.1596

.1746
.1706
.1667
.1629
.1592

.1742
.1702
.1663
.1626
.1589

.1585
.1549
.1514
.1479
.1445

.1581
.1545
.1510
.1476
.1442

.1578
.1542
.1507
.1472
.1439

.1574
.1538
.1503
.1469
.1435

.1570
.1535
.1500
.1466
.1432

.1567
.1531
.1496
.1462
.1429

.1563
.1528
.1493
.1459
.1426

.1560
.1524
.1489
.1455
.1422

.1556
.1521
.1486
.1452
.1419

.1552
.1517
.1483
.1449
.1416

.1413
.1380
.1349
.1318
.1288

.1409
.1377
.1346
.1315
.1285

.1406
.1374
.1343
.1312
.1282

.1403
.1371
.1340
.1309
.1279

.1400
.1368
.1337
.1306
.1276

.1396
.1365
.1334
.1303
.1274

.1393
.1361
.1330
.1300
.1271

.1390
.1358
.1327
.1297
.1268

.1387
.1355
.1324
.1294
.1265

.1384
.1352
.1321
.1291
.1262

.1259
.1230
.1202
.1175
.1148

.1256
.1227
.1199
.1172
.1146

.1253
.1225
.1197
.1169
'.1143

.1250
.1222
.1194
.1167
.1140

.1247
.1219
.1191
.1164
.1138

.1245
.1216
.1189
.1161
.1135

.1242
.1213
.1186
.1159
.1132

.1239
.1211
.1183
.1156
.1130

.1236
.1208
.1180
.1153
.1127

.1233
.1205
.1178
.1151
.1125

.1122
.1096
.1072
.1047
.1023

.1119
.1094
.1069
.1045
.1021

.1117
.1091
.1067
.1042
.1019

.1114
.1089
.1064
.1040
.1016

.1112
.1086
.1062
.1038
.1014

.1109
.1084
.1059
.1035
.1012

.1107
.1081
.1057
.1033
.1009

.1104
.1079
.1054
.1030
.1007

.1102
.1076
.1052
.1028
.1005

.1099
.1074
.1050
.1026
.1002

.1000

.0998

.0995

.0993

.0991

.0989

.0987

.0984

.0982

.0980

81

TABLE P7

GIVEN: POWER RATIO (1.0 to 10.0); FIND: +dB (0 to +10)

For power ratios greater than 10:

Multiple

oflO

Add
dBs

Multiple

oflO

Add
dBs

10
10'
10'
10'

10
20
30
40

10·
10·
10'
10 8

50
60
70
80

1. Divide by the multiple of 10 required to give a number between 1.0
and 10.0.
2. Look up + dB corresponding to this number in Table P7.
3. Add to the result +dBs corresponding to the multiple of 10 used in
step 1.

Example: Given power ratio = 397,000 (3.97 x 105 ). Divide by 105 to get 3.97. From Table P7, dBs corresponding to 3.97
is 5.988. Adding 50 dB (corresponding to the 105 multiplier used) we get 5.988 + 50 = 55.988 dB •

.00

.01

.02

.03

.04

.05

.06

.07

.08

.09

.000
.414
.792
1.139
1.461

.043
.453
.828
1.173
1.492

.086
.492
.864
1.206
1.523

.128
.531
.899
1.239
1.553

.170
.569
.934
1.271
1.584

.212
.607
.969
1.303
1.614

.253
.645
1.004
1.335
1.644

.. 294
.682
1.038
1.367
1.673

.334
.719
1.072
1.399
1.703

.374
.755
1.106
1.430
1.732

1.761
2.041
2.304
2.553
2.788

1.790
2.068
2.330
2.577
2.810

1.818
2.095
2.355
2.601
2.833

1.847
2.122
2.380
2.625
2.856

1.875
2.148
2.405
2.648
2.878

1.903
2.175
2.430
2.672
2.900

1.931
2.201
2.455
2.695
2.923

1.959
2.227
2.480
2.718
2.945

1.987
2.253
2.504
2.742
2.967

2.014
2.279
2.529
2.765
2.989

3.010
3.222
3.424
3.617
3.802

3.032
3.243
3.444
3.636
3.820

3.054
3.263
3.464
3.655
3.838

3.075
3.284
3.483
3.674
3.856

3.096
3.304
3.502
3.692
3.874

3.118
3.324
3.522
3.711
3.892

3.139
3.345
3.541
3.729
3.!109

3.160
3.365
3.560
3.747
3.927

3.181
3.385
3.579
3.766
3.945

3.201
3.404
3.598
3.784
3.962

3.979
4.150
4.314
4.472
4.624

3.997
4.166
4.330
4.487
4.639

4.014
4.183
4.346
4.502
4.654

4.031
4.200
4.362
4.518
4.669

4.048
4.216
4.378
4.533
4.683

4.065
4.232
4.393
4.548
4.698

4.082
4.249
4.409
4.564
4.713

4.099
4.265
4.425
4.579
4.728

4.116
4.281
4.440
4.594
4.742

4.133
4.298
4.456
4.609
4.757

4.771
4.914
5.051
5.185
5.315

4.786
4.928
5.065
5.198
5.328

4.800
4.942
5.079
5.211
5.340

4.814
4.955
5.092
5.224
5.353

4.829
4.969
5.105
5.237
5.366

4.843
4.983
5.119
5.250
5.378

4.857
4.997
5.132
5.263
5.391

4.871
5.011
5.145
5.276
5.403

4.886
5.024
5.159
5.289
5.416

4.900
5.038
5.172
5.302
5.428

5.441
5.563
5.682
5.798
5.911

5.453
5.575
5.694
5.809
5.922

5.465
5.587
5.705
5.821
5.933

5.478
5.599
5.717
5.832
5.944

5.490
5.611
5.729
5.843
5.955

5.502
5.623
5.740
5.855
5.966

5.514
5.635
5.752
5.866
5.977

5.527
5.647
5.763
5.877
5.988

5.539
5.658
5.775
5.888
5.999

5.551
5.670
5.786
5.899
6.010

6.021
6.128
6.232
6.335
6.435

6.031
6.138
6.243
6.345
6.444

6.042
6.149
6.253
6.355
6.454

6.053
6.160
6.263
6.365
6.464

6.064
6.170
6.274
6.375
6.474

6.075
6.180
6.284
6.385
6.484

6.085
6.191
6.294
6.395
6.493

6.096
6.201
6.304
6.405
6.503

6.107
6.212
6.314
6.415
6.513

6.117
6.222
6.325
6.425
6.522

6.532
6.628
6.721
6.812
6.902

6.542
6.637
6.730
6.821
6.911

6.551
6.646
6.739
6.830
6.920

6.561
6.656
6.749
6.839
6.928

6.571
6.665
6.758
6.848
6.937

6.580
6.675
6.767
6.857
6.946

6.590
6.684
6.776
6.866
6.955

6.599
6.693
6.785
6.875
6.964

6.609
6.702
6.794
6.884
6.972

6.618
6.712
6.803
6.893
6.981

82

TABLE P1 continued

.00

.01

.02

.03

.04

.05

.06

.07

.08

.09

6.990
7.076
7.160
7.243
7.324

6.998
7.084
7.168
7.251
7.332

7.007
7.093
7.177
7.259
7.340

7.016
7.lDl
7.185
7.267
7.348

7.024
7.110
7.193
7.275
7.356

7.033
7.118
7.202
7.284
7.364

7.042
7.126
7.2lD
7.292
7.372

7.050
7.135
7.218
7.300
7.380

7.059
7.143
7.226
7.308
7.388

7.067
7.152
7.235
7.316
7.396

7.404
7.482
7.559
7.634
7.709

7.412
7.490
7.566
7.642
7.716

7.419
7.497
7.574
7.649
7.723

7.427
7.505
7.582
7.657
7.731

7.435
7.513
7.589
7.664
7.738

7.443
7.520
7.597
7.672
7.745

7.451
7.528
7.604
7.679
7.752

7.459
7.536
7.612
7.686
7.760

7.466
7.543
7.619
7.694
7.767

7.474
7.551
7.627
7.701
7.774

7.782
7.853
7.924
7.993
8.062

7.789
7.860
7.931
8.000
8.069

7.79€
7.868
7.938
8.007
8.075

7.803
7.875
7.945
8.014
8.082

7.810
7.882
7.952
8.021
8.089

7.818
7.889
7.959
8.028
8.096

7.825
7.896
7.966
8.035
8.102

7.832
7.903
7.973
8.04).
8.109

7.839
7.910
7.980
8.048
8.116

7.846
7.917
7.987
8.055
8.122

8.129
8.195
8.261
8.325
8.388

8.136
8.202
8.267
8.331
8.395

8.142
8.209
8.274
8.338
8.401

8.149
8.215
8.280
8.344
8.407

8.156
8.222
8.287
8.351
8.414

8.162
8.228
8.293
8.357
8.420

8.169
8.235
8.299
8.363
8.426

8.176
8.241
8.306
8.370
8.432

8.182
8.248
8.312
8.376
8.439

8.189
8.254
8.319
8.382
8.445

8.451
8.513
8.573
8.633
8.692

8.457
8.519
8.579
8.639
8.698

8.463
8.525
8.585
8.645
8.704

8.470
8.531
8.591
8.651
8.710

8.476
8.537
8.657
8.716

8.482
8.543
8.603
8.663
8.722

8.488
8.549
8.609
8.669
8.727

8.494
8.555
8.615
8.675
8.733

8.500
8.561
8.621
8.681
8.739

8.506
8.567
8.627
8.686
8.745

8.751
8.808
8.865
8.921
8.976

8.756
8.814
8.871
8.927
8.982

8.762
8.820
8.876
8.932
8.987

8.768
8.825
8.882
8.938
8.993

8.774
8.831
8.887
8.943
8.998

8.779
8.837
8.893
8.949
9.004

8.785
8.842
8.899
8.954
9.009

8.791
8.848
8.904
8.960
9.015

8.797
8.854
8.910
8.965
9.020

8.802
8.859
8.915
8.971
9.025

9.031
9.085
9.138
9.191
9.243

9.036
9.090
9.143
9.196
9.248

9.042
9.096
9.149
9.201
9.253

9.047
9.lDl
9.154
9.206
9.258

9.053
9.lD6
9.159
9.212
9.263

9.058
9.112
9.165
9.217
9.269

9.063
9.117
9.170
9.222
9.274

9.069
9.122
9.175
9.227
9.279

9.074
9.128
9.180
9.232
9.284

9.079
9.133
9.186
9.238
9.289

9.294
9.345
9.395
9.445
9.494

9.299
9.350
9.400
9.450
9.499

9.304
9.355
9.405
9.455
9.504

9.309
9.360
9.410
9.460
9.509

9.315
9.365
9.415
9.465
9.513

9.320
9.370
9.420
9.469
9.518

9.325
9.375
9.425
9.474
.9.523

9.330
9.380
9.430
9.479
9.528

9.335
9.385
9.435
9.484
9.533

9.340
9.390
9.440
9.489
9.538

9.542
9.590
9.638
9.685
9.731

9.547
9.595
9.643
9.689
9.736

9.552
9.600
9.647
9.694
9.741

9.557
9.605
9.652
9.699
9.745

9.562
9.609
9.657
9.703
9.750

9.566
9.614
9.661
9.708
9.754

9.571
9.619
9.666
9.713
9.759

9.576
9.624
9.671
9.717
9.763

9.581
9.628
9.675
9.722
9.768

9.586
9.633
9.680
9.727
9.773

9.777
9.823
9.868
9.912
9.956

9.782
9.827
9.872
9.917
9.961

9.786
9.832
9.877
9.921
9.965

9.791
9.836
9.881
9.926
9.969

9.795
9.841
9.886
9.930
9.974

9.800
9.845
9.890
9.934
9.978

9.805
9.850
9.894
9.939
9.983

9.809
9.854
9.899
9.943
9.987

9.814
9.859
9.903
9.948
9.991

9.818
9.863
9.908
9.952
9.996

10.000

10.004

10.009

10.013

10.017

lD.022

10.026

10.030

lD.035

lD.039

0

~.597

83

TABLE P8

GIVEN +dB (0 to +10); FIND: POWER RATIO (1.0 to 10.0)
For dBs Multiply For dBs Multiply
By
Subtracted
Subtracted
By

For +dBs greater than +10:
1. Subtract dBs as required to give a number between 0 and 10.

10
20
30
40

2. Look up power ratio corresponding to this number of dBs in Table P8.
3. Multiply the power ratio found by the number corresponding to the
dBs subtracted in step 1.

10

10 2
10 3
10'

50
60
70

BO

10 5
10·
10 1
10 8

Example: Given 76.21 dB. Subtract 70 dB to get 6.21 dB. From Table P8, power ratio for 6.21 dB is 4.178. Corresponding
to the 70 dBs subtracted, multiply 4.178 X 107 to get power ratio of 41,780,000.
0.00

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

1.000
1.023
1.047
1.072
1.096
1.122
1.148
1.175
1.202
1.230
1.259
1.288
1.318
1.349
1.380
1.413
1.445
1.479
1.514
1.549
1.585
1.622
1.660
1.698
1.738
1.778
1.820
1.862
1.905
1.950
1.995
2.042
2.089
2.138
2.188
2.239
2.291
2.344
2.399
2.455
2.512
2.570
2.630
2.692
2.754

1.002
1.026
1.050
1.074
1.099
1.125
1.151
1.178
1.205
1.233
1.262
1.291
1.321
1.352
1.384
1.416
1.449
1.483
1.517
1.552
1.589
1.626
1.663
1.702
1.742
1.782
1.824
1.866
1.910
1.954
2.000
2.046
2.094
2.143
2.193
2.244
2.296
2.350
2.404
2.460
2.518
2.576
2.636
2.698
2.761

1.005
1.028
1.052
1.076
1.102
1.127
1.153
1.180
1.208
1.236
1.265
1.294
1.324
1.355
1.387
1.419
1.452
1.486
1.521
1.556
1.592
1.629
1.667
1.706
1.746
1.786
1.828
1.871
1.914
1.959
2.004
2.051
2.099
2.148
2.198
2.249
2.301
2.355
2.410
2.466
2.523
2.582
2.642
2.704
2.767

1.007
1.030
1.054
1.079
1.104
1.130
1.156
1.183
1.211
1.239
1.268
1.297
1.327
1.358
1.390
1.422
1.455
1.489
1.524
1.560
1.596
1.633
1.671
1.710
1.750
1.791
1.832
1.875
1.919
1.963
2.009
2.056
2.104
2.153
2.203
2.254
2.307
2.360
2.415
2.472
2.529
2.588
2.649
2.710
2.773

1.009
1.033
1.057
1.081
1.107
1.132
1.159
1.186
1.213
1.242
1.271
1.300
1.330
1.361
1.393
1.426
1.459
1.493
1.528
1.563
1.600
1.637
1.675
1.714
1.754
1.795
1.837
1.879
1.923
1.968
2.014
2.061
2.109
2.158
2.208
2.259
2.312
2.366
2.421
2.477
2.535
2.594
2.655
2.716
2.780

1.012
1.035
1.059
1.084
1.109
1.135
1.161
1.189
1.216
1.245
1.274
1.303
1.334
1.365
1.396
1.429
1.462
1.496
1.531
1.567
1.603
1.641
1.679
1.718
1.758
1.799
1.841
1.884
1.928
1.972
2.018
2.065
2.113
2.163
2.213
2.265
2.317
2.371
2.427
2.483

1.014
1.038
1.062
1.086
1.112
1.138
1.164
1.191
1.219
1.247
1.276
1.306
1.337
1.368
1.400
1.432
1.466
1.500
1.535
1.570
1.607
1.644
1.683
1.722
1.762
1.803
1.845
1.888
1.932
1.977
2.023
2.070
2.118
lU68
2.218
2.270
2.323
2.377
2.432
2.489

1.016
1.040
1.064
1.089
1.114
1.140
1.167
1.194
1.222
1.250
1.279
1.309
1.340
1.371
1.403
1.435
1.469
1.503
1.538
1.574
1.611
1.648
1.687
1.726
1.766
1.807
1.849
1.892
1.936
1.982
2.028
2.075
2.123
2.173
2.223
2.275
2.328
2.382
2.438
2.495

1.019
1.042
1.067
1.091
1.117
1.143
1.169
1.197
1.225
1.253
1.282
1.312
1.343
1.374
1.406
1.439
1.472
1.507
1.542
1.578
1.614
1.652
1.690
1.730
1.770
1.811
1.854
1.897
1.941
1.986
2.032
2.080
2.128
2.178
2.228
2.280
2.333
2.388
2.443
2.500

1.021
1.045
1.069
1.094
1.119
1.146
1.172
1.199
1.227
1.256
1.285
1.315
1.346
1.377
1.409

2.541
2.600
2.661
2.723
2.786

2.547
2.606
2.667
2.729
2.793

2.553
2.612
2.673
2.735
2.799

2.559
2.618
2.679
2.742
2.805

84

~.442

1.476
1.510
1.545
1.581
1.618
1.656
1.694
1.734
1.774
1.816
1.858
1.901
1.945
1.991
2.037
2.084
2.133
2.183
2.234
2.286
2.339
2.393
2.449
2.506
2.564
2.624
2.685
2.748
2.812

TABLE P8 continued

0.00

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

2.818
2.884
2.951
3.020
3.090

2.825
2.891
2.958
3.027
3.097

2.831
2.897
2.965
3.034
3.105

2.838
2.904
2.972
3.041
3.112

2.844
2.911
2.979
3.048
3.119

2.851
2.917
2.985
3.055
3.126

2.858
2.924
2.992
3.062
3.133

2.864
2.931
2.999
3.069
3.141

2.871
2.938
3.006
3.076
3.148

2.877
2.944
3.013
3.083
3.155

3.162
3.236
3.311
3.388
3.467

3.170
3.243
3.319
3.396
3.475

3.177
3.251
3.327
3.404
3.483

3.184
3.258
3.334
3.412
3.491

3.192
3.266
3.342
3.420
3.499

3.199
3.273
3.350
3.428
3.508

3.206
3.281
3.357
3.436
3.516

3.214
3.289
3.365
3.443
3.524

3.221
3.296
3.373
3.451
3.532

3.228
3.304
3.381
3.459
3.540

3.548
3.631
3.715
3.802
3.890

3.556
3.639
3.724
3.811
3.899

3.565
3.648
3.733
3.819
3.908

3.573
3.656
3.741
3.828
3.917

3.581
3.664
3.750
3.837
3.926

3.589
3.673
3.758
3.846
3.936

3.597
3.681
3.767
3.855
3.945

3.606
3.690
3.776
3.864
3.954

3.614
3.698
3.784
3.873
3.963

3.622
3.707
3.793
3.882
3.972

3.981
4.074
4.169
4.266
4.365

3.990
4.083
4.178
4.276
4.375

3.999
4.093
4.188
4.285
4.385

4.009
4.102
4.198
4.295
4.395

4.018
4.111
4.207
4.305
4.406

4.027
4.121
4.217
4.315
4.416

4.036
4.130
4.227
4.325
4.426

4.046
4.140
4.236
4.335
4.436

4.055
4.150
4.246
4.345
4.446

0.064
4.159
4.256
4.355
4.457

4.467
4.571
4.677
4.786
4.898

4.477
4.581
4.688
4.797
4.909

4:487
4.592
4.699
4.808
4.920

4.498
4.603
4.710
4.819
4.932

4.508
4.613
4.721
4.831
4.943

4.519
4.624
4.732
4.842
4.955

4.529
4.634
4.742
4.853
4.966

4.539
4.645
4.753
4.864
4.977

4.550
4.656
4.764
4.875
4.989

4.560
4.667
4.775
4.887
5.000

5.012
5.129
5.248
5.370
5.495

5.023
5.140
5.260
5.383
5.508

5.035
5.152
5.272
5.395
5.521

5.047
5.164
5.284
5.408
5.534

5.058
5.176
5.297
5.420
5.546

5.070
5.188
5.309
5.433
5.559

5.082
5.200
5.321
5.445
5.572

5.093
5.212
5.333
5.458
5.585

5.105
5.224
5.346
5.470
5.598

5.117
5.236
5.35S·
5.483
5.610

5.623
5.754
5.888
6.026
6.166

5.636
5.768
5.902
6.039
6.180

5.649
5.781
5.916
6.053
6.194

5.662
5.794
5.929
6.067
6.209

5.675
5.808
5.943
6.081
6.223

5.689
5.821
5.957
6.095
6.237

5.702
5.834
5.970
6.109
6.252

5.715
5.848
5.984
6.124
6.266

5.728
5.861
5.998
6.138
6.281

5.741
5.875
6.012
6.152
6.295

6.310
6.457
6.607
6.761
6.918

6.324
6.471
6.622
6.776
6.934

6.339
6.486
6.637
6.792
6.950

6.353
6.501
6.653
6.808
6.966

6.368
6.516
6.668
6.823
6.982

6.383
6.531
6.683
6.839
6.998

6:397
6.546
6.699
6.855
7.015

6.412
6.561
6.714
6.871
7.031

6.427
6.577
6.730
6.887
7.047

6.442
6.592
6.745
6.902
7.063

7.079
7.244
7.413
7.586
7.762

7.096
7.261
7.430
7.603
7.780

7.112
7.278
7.447
7.621
7.798

7.129
7.295
7.464
7.638
7.816

7.145
7.311
7.482
7.656
7.834

7.161
7.328
7.499
7.674
7.852

7.178
7.345
7.516
7.691
7.870

1.194
7.362
7.534
7.709
7.889

7.211
7.379
7.551
7.727
7.907

7.228
7.396
7.568
7.745
7.925

7.943
8.128
8.318
8.511
8.710

7.962
8.147
8.337
8.531
8.730

7.980
8.166
8.356
8.551
8.750

7.998
8.185
8.375
8.570
8.770

8.017
8.204
8.395
8.590
8.790

8.035
8.222
8.414
8.610
8.810

8.054
8.241
8.433
8.630
8.831

8.072
8.260
8.453
8.650
8.851

8.091
8.279
8.472
8.670
8.872

8.110
8.299
8.492
8.690
8.892

8.913
9.120
9.333
9.550
9.772

8.933
9.141
9.354
9.572
9.795

8.954
9.162
9.376
9.594
9.817

8.974
9.183
9.397
9.616
9.840

8.995
9.204
9.419
9.638
9.863

9.016
9.226
9.441
9.661
9.886

9.036
9.247
9.462
9.683
9.908

9.057
9.268
9.484
9.705
9.931

9.078
9.290
9.506
9.727
9.954

9.099
9.311
9.528
9.750
9.977

10.000

10.023

10.046

10.070

10.093

10.115

10.140

10.163

10.186

10.210

85

TABLE

P9

POWER ADDITION

TWO SIGNAL OR NOISE LEVELS COMBINED ON A POWER BASIS.
GIVEN: dB DIFFERENCE BETWEEN THE TWO LEVELS;
FIND: dBs TO ADD TO THE HIGHER LEVEL TO GET TOTAL LEVEL
IN dBs.

0.0

0.1

3.01

2.96

2.91

2.54
2.13
1.76
1.46
1.19

2.50
2.09
1.73
1,43
1.17

2,45
2.05
1.70
1,40
1.15

0.97
0.79
0.64
0.51
0,41

0.95
0.77
0.62
0.50
0,40

0.93
0.76
0.61
O,4g
0.39

0.33
0.26
0.21
0.17
0.13

0.32
0.26
0.21
0.16
0.13

0.10
0.08
0.07
0.05
0.04

0.10
0.08
0.07
0.05
0.04

TABLE PIO
POWER DIVISION

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

2.86

2.82

2.77

2.72

2.67

2.63

2.58

2,41
2.01
1.67
1.37
1.12

2.37
1.97
1.64
1.35
1.10

2.33
1.94
1.60
1.32
1.08

2.28
1.90
1.57
1.29
1.06

2.24
1.87
1.54
1.27
1.04

2.20
1.83
1.51
1.24
1.01

2.16
1.80
1,48
1.22
0.99

0.91
0.74
0.60
0,48
0.39

0.90
0.72
0.59
0,47
0.38

0.88
0.71
0.57
0,46
0.37

0.86
0.70
0.56
0,45
0.36

0.84
0.68
0.55
0,44
0.35

0.82
0.67
0.54
0,43
0.35

0.81
0.65
0.53
0,42
0.34

0.31
0.25
0.20
0.16
0.13

0.31
0.25
0.20
0.16
0.13

0.30
0.24
0.19
0.15
0.12

0.30
0.24
0.19
0.15
0.12

0.29
0.23
0.19
0.15
0.12

0.28
0.23
0.18
0.15
0.12

0.28
0.22
0.18
0.14

0.27
0.22
0.17
0.14

O.II

O.II

0.10
0.08
0.07
0.05
0.04

0.10
0.08
0.06
0.05
0.04

0.10
0.08
0.06
0.05
0.04

0.09
0.08
0.06
0.05
0.04

0.09
0.07
0.06
0.05
0.04

0.09
0.07
0.06
0.05
0.04

0.09
0.07
0.06
0.05
0.04

0.09
0.07
0.06
0.04
0.03

A SIGNAL IS DIVIDED SO THAT THE SUM OF THE TWO OUTPUT
POWERS EQUALS THE INPUT POWER.
GIVEN: dB DIFFERENCE BETWEEN INPUT LEVEL AND LOWER OUTPUT LEVEL;
FIND: dB DIFFERENCE BETWEEN INPUT LEVEL AND HIGHER OUTPUT LEVEL.

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

3.02
2.20
1.65

2.92
2.14
1.61

2.83
2.08
1.56

2.74
2.02
1.52

2.65
1.96
1,48

2.57
1.90
1.44

2,49
1.85
1,40

2,42
1.80
1.36

2.34
1.75
1.33

2.27
1.70
1.29

1.26
0.97
0.75
0.58
0.46

1.22
0.94
0.73
0.57
0,45

1.19
0.92
0.71
0.56
0.44

1.16
0.89
0.70
0.54
0,43

1.13
0.87
0.68
0.53
0.42

1.10
0.85
0.66
0.52
0,41

1.07
0.83
0.65
0.50
0.40

1.04
0.81
0.63
0,49
0.39

1.02
0.79
0.61
0,48
0.38

0.99
0.77
0.60
0,47
0.37

0.36
0.28
0.22
0.18
0.14

0 ..35
0.28
0.22
0.17
0.14

0.34
0.27
0.21
0.17
0.13

0.33
0.26
0.21
0.16
0.13

0.33
0.26
0.20
0.16
0.13

0.32
0.25
0.20
0.16
0.12

0.31
0.25
0.19
0.15
0.12

0.30
0.24
0.19
0.15
0.12

0.30
0.23
0.18
0.15
0.12

0.29
0.23
0.18
0.14

O.II

O.II

O.II

0.09
0.07
0.06
0.04

0.09
0.07
0.05
0.04

0.08
0.07
0.05
0.04

0.10
0.08
0.06
0.05
0.04

0.10
0.08
0.06
0.05
0.04

0.10
0.08
0.06
0.05
0.04

0.10
0.08
0.06
0.05
0.04

0.09
0.07
0.06
0.05
0.04

0.09
0.07
0.06
0.05
0.04

0.09
0.07
0.06
0.04
0.04

86

O.II

VOLTAGE vs dB CHARTS AND TABLES
Applications of the Voltage vs. dB Charts
and Tables

microvolts )
.
1000
dB corresponding to a voltage ratio of 0.0332 is -29.577
dB, so the answer is -29.577 dB.
volts is 0.0332 millivolts ( millivolts

1. Voltage or Current Gain or Loss: The direct application
of these tables is to cases where the dBs are related to gain
or loss of voltage or current.
Example: The current gain at 100 MHz of a certain transistor is rated as 15 times. What is this in dB? From Chart
V2 and Table V6 a current ratio of 15 eorresponds to
23.52 dB.

=

Example of a reverse case: Find the voltage corresponding
to +17.21 dBmV. From chart V2 and table V7, 17.21 dB
corresponds to a voltage ratio of 7.261, so the answer is
7.261 millivolts.

2. Percent Voltage: Chart VI has scales allowing direct
conversion of percent voltage to dBs. The tables relating
voltage (or current) ratio to dBs can be used for percent by
making use of the relation:

4. Voltage {or Cross-Modulation} Addition When two
levels (or cross-modulation components) are given in dB
form and it is desired to find the dB expression corresponding to their voltage sum (or total cross-mod.) either of
two procedures may be used:

a smaller voltage (or current), expressed as percent of
smaller voltage)
a larger = 1 ()() times the voltage ratio ( 1
1
arger va tage

a. The direct approach is to change each one to a voltage
(or % cross-mod.) expression, add the voltages (or % crossmod. figures) and find the dB expression corresponding
to the total.

Example: When the ratio of two voltages is 0.275, the
smaller is 27.5% of the larger (27.5 = 100 x 0.275).

For example: At its normal operating level a trunk line system of amplifiers generates -58 dB cross-modulation. It is
followed by a distribution system generating -53 dB of
cross-mod. What is the total cross-mod. in dB? Using chart
VI and table V5 the % cross-mod. figures are found. For
-58 dB: 0.1259%. For -53 dB: 0.2239%. The total
% cross mod. is 0.1259 + 0.2239
0.3498% and the dB
figure corresponding to 0.35% (from chart VI and table
V4) is -49.119 dB.

Example of the use of the tables: -52 dB is sometimes
considered the limit of tolerable cross-modulation. What is
the corresponding percent cross-modulation? Using Chart
VI and Table V5, the voltage ratio corresponding to
-52dB is 0.0025. So the answer is 100 x 0.0025
0.25%.

=

=

3. Relation between dBmV and m V: The voltage ratio
between any given voltage and one millivolt is that voltage
. expressed in millivolts. Thus the voltage vs. db charts and
tables can be used directly to find dBmV when voltage
is known or vice versa.

b. Chart V3 and Table V8 allow a quicker 'solution To
solve the problem above, first take the difference between
the given dB levels: -58 - (-53)
-5 dB. Opposite
5 dB in table V8 find 3.874 dB to be added to the higher
level, which is -53 dB. -53 + 3.874 = -49.126 dB.

=

Example: What dBmV level corresponds to 33.2 microvolts
across 75 ohms? Use Chart VI and Table V4. 33.2 micro-

87

CHART VI (VOLTAGE RATIO)

_.
dB (OR dBmV)

. 10

o

-10

8

-20

-30

-40

-50

-60
1.0
0.8

6

X10- 4

4

0.6
0.5
0.4

3

0.3

2

0.2

0.1

->
.
--

0.04

-e
.
-

0.03

tia:::

8

0.08

E
o

6

0.06
0.05

o

4

•

~ X10- 5 3

a:::

~
:.J

0

2

l&.I

1&.1
(!)

>

(!)

2

0.02

o
>

c(

~

0

>

--

I

8
6

0.01
0.008
0.006
0.005
0.004

Dm

Hi

0.003
0.002

2

I
-60

-70

-80

-90

-100

dB (OR dBmV)
88

-110

0.001
-120

CHART Vl (PERCENT OF MAX. VOLTAGE)

o

-10

-20

dB
-30

-40

-50

0.1

-60
100

0.08

80

0.06
0.05

60
50

•

0.04
0.03

30

0.02

20

0.01
UJ
(!)

-

~~

0.008

c:(

~

0

>
:IE
::::>
:E

x

c:(

:E

40

0.006
0.005

10
UJ

8

(!)

6

0

5

0.004

4

~

~

>
2

::::>

2

X

0.003

3

c:(

2

u.

u.

0

c:(

0

0.002

2

z

~

z

UJ

0

UJ
0

UJ

a:

UJ

a:

D-

0.001
0.0008
0.0006
0.0005

-

D-

0.8
0.6
0.5

0.0004

0.4

0.0003

0.3

0.0002

0.2
l)

0.0001
-60

-70

-80

-90
dB
89

-100

-110

0.1
-120

CHART V2

1,000

o

10

dB(or dBmV)
30
40

20

E

60
1,000,000

800

800,000

600
500

600,000
500,000

400

400,000

300

300,000

200

200,000

I

100

>

50

100,000

80

80,000

60
50

60,000
50,000

->

E

-

-

40

~

30

30,000

0
I
1/

10

~~

8
6
5

>

-

10.000
8,000
6,000
5,000

~~

4

4,000

3

3,000

2

2,000

1 II
60

70

80

90
dB (or dBmV)
90

100

110

1,000
120

CHART V3

o

5

VOLTAGE (OR CROSS-MOD) ADDITION
DIF FERENCE BETWEEN TWO LEVELS (d B)
10
15
20
25

30

35

4

3

.J

UJ

2

~

.J

0:

~

-

(!)

X

o

F-

1.0

a
0.8
a
a

UJ
~

w 0.6

"0

0 .4

0.3

0.2

O.l~~~~~~~~~~~~~~~~~~~~~~~~~

o

5

10

15

20

25

DIFFERENCE BETWEEN TWO LEVELS (dB)
91

30

35

TABLE V4

GIVEN: VOLTAGE RATIO (or MILLIVOLTS) (0.1 to 1.0);
FIND: -dB (or -dBmV) (-20 to 0)

For voltage ratios less than 0.1:
1. Multiply given ratio by a multiple of 10 as required to give a number between 0.1 and 1.0.
2. Look up -dB corresponding to this number in Table V4.
3. Add to the result -dBs corresponding to the multiple of 10 used in step 1.

For Multiple of 10

Add-dBs

10
100
1000
10,000

-20
-40
-60
-80

EXAMPLE: Given voltage ratio = 0.000172. Multiplying by 1000 we get 0.172. From Table V4, 0.172 corresponds to
-15.289 dB. Adding -60 dB corresponding to the multiple used, we get -15.289 + (-60) = -75.289 dB.
NOTE: The minus signs in front of all dB numbers in this table have been omitted!

.000

.001

.002

.003

.004

.005

.006

.01}7

.008

.009

20.000
19.172
18.416
17.721
17.077

19.914
19.094
18.344
17.655
17.016

19.828
19.016
18.273
17.589
16.954

19.743
18.938
18.202
17.523
16.893

19.659
18.862
18.132
17.458
16.833

19.576
18.786
18.062
17.393
16.773

19.494
18.711
17.993
17.329
16.713

19.412
18.636
17.924
17.266
16.654

19.332
18.562
17.856
17.202
16.595

19.251
18.439
17.788 !
17.140
16.536

16.478
15.918
15.391
14.895
14.425

16.420
15.863
15.340
14.846
14.379

16.363
15.810
15.289
14.799
14.334

16.306
15.756
15.239
14.751
14.289

16.250
15.703
15.189
'14.704
14.244

16.193
15.650
15.139
14.657
14.199

16.138
15.598
15.090
14.610
14.155

16.082
15.546
15.041
14.563
14.111

16.027
15.494
14.992
14.517
14.067

15.972
15.442
14.943
14.471
14.023

13.979
13.556
13.152
12.765
12.396

13.936
13.514
13.112
12.728
12.360

13.893
13.473
13.073
12.690
12.324

13.850
13.432
13.034
12.653
12.288

13.807
13.392
12.992
12.616
12.252

13.765
13.351
12.956
12.579
12.217

13.723
13.311
12.918
12.542
12.181

13.681
13.271
12.879
12.505
12.146

13.639
13.231
12.841
12.468
12.111

13.597
13.191
12.803
12.432
12.076

12.041
11.701
11.373
11.057
10.752

12.007
11.667
11.341
11.026
10.722

11.972
11.634
11.309
10.995
10.692

11.938
11.601
11.277
10.964
10.663

11.903
11.568
11.245
10.934
10.633

11.869
11.535
11.213
10.903
10.604

11.835
11.502
11.182
10.873
10.574

11.801
11.470
11.150
10.842
10.545

11.768
11.437
11.119
10.812
10.516

11.734
11.405
11.088
10.782
10.487

10.458
10.173
9.897
9.630
9.370

10.429
10.145
9.870
9.603
9.345

10.400
10.117
9.843
9.577
9.319

10.371
10.089
9.816
9.551
9.294

10.343
10.061
9.789
9.525
9.269

10.314
10.034
9.762
9.499
9.244

10.286
10.006
9.736
9.473
9.218

10.257
9.979
9.709
9.447
9.193

10.229
9.951
9.683
9.422
9.169

10.201
9.924
9.656
9.396
9.143

9.119
8.874
8.636
8.404
8.179

9.094
8.850
8.603
8.383
8.156

9.069
8.826
8.589
8.359
8.134

9.045
8.802
8.566
8.336
8.112

9.020
8.778
8.543
8.313
8.090

8.995
8.754
8.519
8.291
8.068

8.971
8.730
8.496
8.268
8.046

8.947
8.707
8.473
8.246
8.024

8.922
8.683
8.450
8.223
8.002

8.898
8.659
8.427
8.201
7.991

7.959
7.744
7.535
7.331
7.131

7.937
7.723
7.514
7.310
7.111

7.915
7.702
7.494
7.290
7.092

7.894
7.681
7.473
7.270
7.072

7.872
7.660
7.453
7.250
7.052

7.851
7.639
7.432
7.230
7.033

7.829
7.618
7.412
7.210
7.013

7.808
7.597
7.391
7.190
6.994

7.787
7.576
7.371
7.171
6.974

7.766
7.556
7.351
7.151
6.955

6.936
6.745
6.558
6.375
6.196

6.916
6.726
6.540
6.357
6.178

6.897
6.707
6.521
6.339
6.161

6.878
6.688
6.503
6.321
6.143

6.859
6.670
6.484
6.303
6.125

6.840
6.651
6.466
6.285
6.108

6.821
6.632
6.448
6.267
6.090

6.802
6.614
6.430
6.249
6.073

6.783
6.595
6.411
6.232
6.055

6.764
6.577
6.393
6.212
6.038

92

TABLE V4 continued

.000

.001

.002

.003

.004

.005

.006

.007

.008

.009

6.021
5.849
5.680
5.514
5.352

6.003
5.832
5.663
5.498
5.336

5.986
5.815
5.647
5.482
5.320

5.969
5.798
5.630
5.465
5.304

5.951
5.781
5.613
5.449
5.288

5.934
5.764
5.597
5.433
5.272

5.917
5.747
5.580
5.417
5.256

5.900
5.730
5.564
5.401
5.240

5.883
5.713
5.547
5.384
5.224

5.866
5.697
5.531
5.368
5.209

5.193
5.036
4.883
4.731
4.583

5.177
5.021
4.867
4.716
4.568

5.161
5.005
4.852
4.702
4.554

5.145
4.990
4.837
4.687
4.539

5.130
4.974
4.822
4.672
4.524

5.114
4.959
4.807
4.657
4.510

5.098
4.944
4.792
4.642
4.495

5.083
4.928
4.776
4.627
4.481

5.067
4.913
4.761
4.612
4.466

5.052
4.898
4.746
4.598
4.451

4.437
4.293
4.152
4.013
3.876

4.423
4.279
4.138
3.999
3.863

4.408
4.265
4.124
3.986
3.849

4.394
4.251
4.110
3.972
3.836

4.379
4.237
4.096
3.958
3.822

4.365
4.222
4.082
3.945
3.809

4.351
4.208
4.069
3.931
3.795

4.336
4.194
4.055
3.917
3.782

4.322
4.180
4.041
3.904
3.768

4.308
4.166
4.027
3.890
3.755

3.742
3.609
3.479
3.350
3.223

3.728
3.596
3.466
3.337
3.210

3.715
3.583
3.453
3.324
3.198

3.702
3.570
3.440
3.312
3.185

3.688
3.557
3.427
3.299
3.173

3.675
3.544
3.414
3.286
3.160

3.662
3.531
3.401
3.274
3.148

3.649
3.517
3.388
3.261
3.135

3.635
3.504
3.375
3.248
3.123

3.622
3.491
3.363
3.236
3.110

3.098
2.975
2.853
2.734
2.615

3.086
2.963
2.841
2.722
2.604

3.073
2.950
2.829
2.710
2.592

3.061
2.938
2.817
2.698
2.580

3.049
2.926
2.805
2.868
2.569

3.036
2.914
2.793
2.674
2.557

3.024
2.902
2.781
2.662
2.545

3.012
2.890
2.769
2.651
2.534

2.999
2.878
2.757
2.639
2.522

2.987
2.865
2.745
2.627
2.510

2.499
2.384
2.270
2.047

2.487
2.372
2.259
2.147
2.036

2.476
2.361
2.248
2.136
2.025

2.464
2.350
2.236
2.125
2.015

2.453
2.338
2.225
2.114
2.004

2.441
2.327
2.214
2.103
1.993

2.430
2.315
2.203
2.092
1.982

2.418
2.304
2.192
2.081
1.971

2.407
2.293
2.180
2.069
1.960

2.395
2.281
2.169
2.058
1.949

1.938
1.830
1.724
1.618
1.514

1.927
1.820
1.713
1.608
1.504

1.917
1.809
1.703
1.598
1.494

1.906
1.798
1.692
1.587
1.483

1.894
1.788
1.681
1.577
1.473

1.884
1.777
1.671
1.566
1.463

1.873
1.766
1.660
1.556
1.453

1.863
1.756
1.650
1.545
1.442

1.852
1.745
1.639
1.535
1.432

1.841
1.734
1.629
1.525
1.422

1.412
1.310
1.210
1.110
1.012

1.401
1.300
1.200
1.100
1.002

1.391
1.290
1.190
1.091
.993

1.381
1.280
1.180
1.081
.983

1.371
1.270
1.170
1.071
.973

1.361
1.260
1.160
1.061
.964

1.351
1.250
1.150
1.051
.954

1.340
1.240
1.140
1.042
.944

1.330
1.230
1.130
1.032
.934

1.320
1.220
1.120
1.022
.925

.915
.819
.724
.630
.537

.906
.810
.715
.621
.528

.896
.800
.705
.612
.519

.886
.791
.696
.602
.510

.877
.781
.687
.593
.501

.867
.772
.677
.584
.491

.857
.762
.668
.574
.482

.848
.753
.658
.565
.473

.838
.743
.649
.556
.464

.829
.734
.640
.547
.455

.446
.355
.265
.175
.087

.436
.346
.256
.167
.079

.427
.336
.247
.158
.070

.418
.327
.238
.149
.061

.409
.318
.229
.140
.052

.400
.309
.220
.131
.044

.391
.300
.211
.122
.035

.382
.291
.202
.114
.026

.373
.282
.193
.105
.017

.364
.274
.184
.096
.009

2~158

93

TABLE V5

·GIVEN: -dB (or -dBmV) (O to -20);
FIND: VOLTAGE RATIO (or mV) (LO to 0.1)

For - db below -20:
For dBs Added

1. Add dBs (in multiples of 20) as required to give a number between 0 and
-20.

Multiply By

20
40
60
80

2. Look up voltage ratio corresponding to this number in table V5.
3. Multiply the voltage ratio found by the number corresponding to the dBs
added in step 1.
.

0.1
0.01
0.001
0.0001

EXAMPLE: Given -53.22 dB. Adding 40 dB we get -13.22 dB. From Table V5, voltage ratio corresponding to -13.22 dB
is 0.2183. Multiplying by 0.01 (corresponding to 40 dB added) we get 0.2183 x 0.01 = 0.002183.

0.00

0.02

0.04

0.06

0.08

0.00

0.02

0.04

0.06

0.08

1.0000
.9886
.9772
.9661
.9550

.9977
.9863
.9750
.9638
.9528

.9954
.9840
.9727
.9616
.9506

.9931
.9817
.9705
.9594
.9484

.9908
.9795
.9683
.9572
.9462

.6310
.6237
.6166
.6095
.6026

.6295
.6223
.6152
.6081
.6012

.6281
.6209
.6139
.6067
.5998

.6266
.6194
.6124
.6053
.5984

.6252
.6180
.6109
.6039
.5970

.9441
.9333
.9226
.9120
.9016

.9-;1:19
.9311
.9204
.9099
.8995

.9397
.9290
.9183
.9078
.8974

.9376
.9268
.9162
.9057
.8954

.9354
.9247
.9141
.9.036
.8933

.5957
.5888
.5821
.5754
.5689

.5943
.5875
.5808
.5741
.5675

.5929
.5861
.5794
.5728
.5662

.5916
.5848
.5781
.5715
.5649

.5902
.5834
.5768
.5702
.5636

.8913
.8810
.8710
.8610
.8511

.8892
.8790
.8690
.8590
.8492

.8872
.8770
.8670
.8570
.8472

.8851
.8750
.8650
.8551
.8453

.8831
.8730
.8630
.8531
.8433

.5623
.5559
.5495
.5433
.5370

.5610
..5546
.5483
.5420
.5358

.5598
.5534
.5470
.5408
.5346

.5585
.5521
.5458
.5395
.5333

.5572
.5508
.5445
.5383
.5321

.8414
.8318
.8222
.8128
.8035

.8395
.8299
'.8204
.8110
.8017

.8375
.8279
.8185
.8091
.7998

.8356
.8260
.8166
.8072
.7980

.8337
.8241
.8147
.8054
.7962

.5309
.5248
.5188
.5129
.5070

.5297
.5236
.5176
.5117
.5058

.5284
.5224
.5164
.5105
.5047

.5272
.5212
.5152
.5093
.5035

.5260
.5200
.5140
.5082
.5023

.7943
.7852
.7762
.7674
.7586

.7925
.7834
.7745
.7656
.7568

.7907
.7816
.7727
.7638
.7551

.7889
.7798
.7709
.7621
.7534

.7870
.7780
.7691
.7603
.7516

.5012
.4955
.4898
.4842
.4786

.5000
.4943
.4887
.4831
.4775

.4989
.4932
.4875
.4819
.4764

.4977
.4920
.4864
.4808
.4753

.4966
.4909
.4853
.4797
.4742

.7499
.7413
.7328
.7244
.7161

.7482
.7396
.7311
.7228
.7145

.7464
.7379
.7295
.7211
.7129

.7447
.7362
.7278
.7194
.7112

.7430
.7345
.7261
.7178
.7096

.4732
:4677
.4624
.4571
.4519

.4721
.4667
.4613
.4560
.4508

.4710
.4656
.4603
.4550
.4498

.4699
.4645
.4592
.4539
.4487

.4688
.4634
.4581
.4529
.4477

.7079
.6998
.6918
.6839
.6761

.7063
.6982
.6902
.6823
.6745

.7047
.6966
.6887
.6808
.6730

.7031
.6950
.6871
.6792
.6714

.7015
.6934
.6855
.6776
.6699

.4467
.4416
.4365
.4315
.4266

.4457
.4406
.4355
.4305
.4256

.4446
.4395
.4345
.4295
.4246

.4436
.4385
.4335
.4285
.4236

.4426
.4375
.4325
.4276
.4227

.6683
.6607
.6531
.6457
.6383

.0068
.6592
.6516
.6442
.6368

.6653
.6577
.6501
.6427
.6353

.6637
.6561
.6486
.6412
.6339

.6622
.6546
.6471
.6397
.6324

.4217
.4169
.4121
.4074
.4027

.4207
.4159
.4111
.4064
.4018

.4198
.4150
.4102
.4055
.4009

.4188
.4140
.4093
.4046
.3999

.4178
.4130
.4083
.4036
.3990

94

TABLE V5 continued

0.00

0.02

0.04

0.06

0.08

0.00

0.02

0.04

0.06

0.08

.3981
.3936
.3890
.3846
.3802

.3972
.3926
.3882
.3837
.3793

.3963
.3917
.3873
.3828
.3784

.3954
.3908
.3864
.3819
.3776

.3945
.3899
.3855
.3811
.3767

,.1995
.1972
.1950
.1928
.1905

.1991
.1968
.1945
.1923
.1901

.1986
.1963
.1941
.1919
.1897

.1982
.1959
.1936
.1914
.1892

.1977
.1954
.1932
.1910
.1888

.3758
.3715
.3673
.3631
.3589

.3750
.3707
.3664
.3622
.3581

.3741
.3698
.3656
.3614
.3573

.3733
.3690
.3648
.3606
.3565

.3724
.3681
.3639
.3597
.3556

.1884
.1862
.1841
.1820
.1799

.1879
.1858
.1837
.1816
.1795

.1875
.1854
.1832
.1811
.1791

.1871
.1849
.1828
.1807
.1786

.1866
.1845
.1824
.1803
.1782

.3548
.3508
.3467
.3428
.3388

.3540
.3499
.3459
.3420
.3381

.3532
.3491
.3451
.3412·
.3373

.3524
.3483
.3443
.3404
.3365

.3516
.3475
.3436
.3396
.3357

.1778
.1758
.1738
.1718
.1698

.1774
.1754
.1734
.1714
.1694

.1770
.1750
.1730
.1710
.1690

.1766
.1746
.1726
.1706
.1687

.1762
.1742
.1722
.1702
.1683

.3350
.3311
.3273
, .3236
.3199

.3342
.3304
.3266
.3228
.3192

.3334
.3296
.3258
.3221
.3184

.3327
.3289
.3251
.3214
.3177

.3319
.3281
.3243
.3206
.3170

.1679
.1660
.1641
.1622
.1603

.1675
.1656
.1637
.1618
.1600

.1671
.1652
.1633
.1614
.1596

.1667
.1648
.1629
.1611
.1592

.1663
.1644
.1626
.1607
.1589

.3162
.3126
.3090
.3055
.3020

.3155
.3119
.3083
.3048
.3013

.3148
.3112
.3076
.3041
.3006

.3141
.3105
.3069
.3034
.2999

.3133
.3097
.3062
.3027
.2992

.1585
.1567
.1549
.1531
.1514

.1581
.1563
.1545
.1528
.1510

.1578
.1560
.1542
.1524
.1507

.1574
.1556
.1538
.1521
·.1503

.1570
.1552
.1535
.1517
.1500

.2985
.2951
.2917
.2884
.2851

.2979
.2944
.2911
.2877
.2844

.2972
.2938
.2904
.2871
.2838

.2965
.2931
.2897
.2864
.2831

.2958
.2924
.2891
.2858
.2825

.1496
.1479
.1462
.1445
.1429

.1493
.1476
.1459
.1442
.1426

.1489
.1472
.1455
.1439
.1422

.1486
.1469
.1452
.1435
.1419

.1483
.1466
.1449
.1432
.1416

.2818
.2786
.2754
.2723
.2692

.2812
.2780
.2748
.2716
.2685

.2805
.2773
.2742
.2710
.2679

.2799
.2767
.2735
.2704
.2673

.2793
.2761
.2729
.2698
.2667

.1413
.1396
.1380
.1365
.1349

.1409
.1393
.1377
.1361
.1346

.1406
.1390
.1374
.1358
.1343

.1403
.1387
.1371
.1355
.1340

.1400
.1384
.1368
.1352
.1337

.2661
.2630
.2600
.2570
.2541

.2655
.2624
.2594
.2564
.2535

.2649
.2618
.2588
.2559
.2529

.2642
.2612
.2582
.2553
.2523

.2636
.2606
.2576
.2547
.2518

.1334
.1318
.1303
.1288
.1274

.1330
.1315
.1300
.1285
.1271

.1327
.1312
.1297
.1282
.1268

.1324
.1309
.1294
.1279
.1265

.1321
.1306
.1291
.1276
.1262

.2512
.2483
.2455
.2427
.2399

.2506
.2477
.2449
.2421
.2393

.2500
.2472
.2443
.2415
.2388

.2495
.2466
.2438
.2410
.2382

.2489
.2460
.2432
.2404
.2377

.1259
.1245
.1230
.1216
.1202

.1256
.1242
.1227
.1213
.1199

.1253
.1239
.1225
.1211
.1197

.1250
.1236
.1222
.1208
.1194

.1247
.1233
.1219
.1205
.1191

.2371
.2344
.2317
.2291
.2265

.2366
.2339
.2312
.2286
.2259

.2360
.2333
.2307
.2280
.2254

.2355
.2328
.2301
.2275
.2249

.2350
.2323
.2296
.2270
.2244

.1189
.1175
.1161
.1148
.1135

.1186
.1172
.1159
.1146
.1132

.1183
.1169
.1156
.1143
.1130

.1180
.1167
.1153
.1140
.1127

.1178
.1164
.1151
.1138
.1125

.2239
.2213
.2188
.2163
.2138

.2234
.2208
.2183
.2158
:2133

.2228
.2203
.2178
.2153
.2128

.2223
.2198
.2173
.2148
.2123

.2218
.2193
.2168
.2143
.2118 .

.1122
.1109
.1096
.1084
.1072

.1119
.1107
.1094
.1081
.1069

.1117
.1104
.1091
.1079
.1067

.1114
.1102
.1089
.1076
.1064

.1112
.1099
.1086
.1074
.1062

.2113
.2089
.2065
.2042
.2018

.2109
.2084
.2061
.2037
.2014

.2104
.2080
.2056
.2032
.2009

.2099
.2075
.2051
.2028
.2004

.2094
.2070
.2046
.2023
.2000

.1059
.1047
.1035
.1023
.1012

.1057
.1045
.1033
.1021
.1009

.1054
.1042
.1030
.1019
.1007

.1052
.1040
.1028
.1016
.1005

.1050
.1038
.1026
.1014
.1002

.1000

.0998

.0995

.0993

.0991

95

TABLE V6

GIVEN: VOLTAGE RATIO (or mV) (1.0 to 10.0);
FIND: +dB (or +dBmV) (0 to +20)

For voltage ratio greater than 10:
1. Divide by the multiple of 10 required to give a number between 1.0 and
10.0.
2. Look up +dB corresponding to this number in table V6.
3. Add to the result the number of dBs corresponding to the multiple of 10
used in step 1.

For Multiple of 10

Add dBs

10
100
1000
10,000

20
40
60
80

EXAMPLE: Given voltage ratio = 65,200. Dividing by 10,000 we get 6.52. From table V6, dBs corresponding to 6.52 is
16.285. Adding the number of dBs according to divisor used in step 3 We get 16,285 + 80 = +96.285 dB .

.00

.01

.02

.03

.04

•05

.06

.07

.08

.09

.000
.828
1.584
2.279
2.923

.086
.906
1.656
2.345
2.984

.172
.984
1.727
2.411
3.046

.257
1.062
1.798
2.477
3.107

.341
1.138
1.868
2.542
3.167

.424
1.214
1.938
2.607
3.227

.506
1.289
1.007
2.671
3.287

.588
1.364
2.076
2.734
3.346

.668
1.438
2.144
2.798
3.405

.749
1.511
2.212
2.860
3.464

3.522
4.082
4.609
5.105
5.575

3.580
4.137
4.660
5.154
5.621

3.637
4.190
4.711
5.201
5.666

3.694
4.244
4.761
5.249
5.711

3.750
4.297
4.811
5.296
5.756

3.807
4.350
4.861
5.343
5.801

3.862
4.402
4.910
5.390
5.845

3.918
4.454
4.959
5.437
5.889

3.973
4.506
5.008
5.483
5.933

4.028
4.558
5.057
5.529
5.977

6.021
6.444
6.848
7.235
7.604

6.064
6.486
6.888
7.272
7.640

6.107
6.527
6.927
7.310
7.676

6.150
6.568
6.966
7.347
7.712

6.193
6.608
7.008
7.384
7.748

6.235
6.649
7.044
7.421
7.783

6.277
6.689
7.082
7.458
7.819

6.319
6.729
7.121
7.495
7.854

6.361
6.769
7.159
7.532
7.889

6.403
6.809
7.197
7.568
7.924

7.959
8.299
8.627
8.943
9.248

7.993
8.333
8.659
8.974
9.278

8.028
8.366
8.691
9.005
9.308

8.062
8.399
8.723
9.036
9.337

8.097
8.432
8.755
9.066
9.367

8.131.
8.465
8.787
9.097
9.396

8.165
8.498
8.818
9.127
9.426

8.199
8.530
8.850
9.158
9.455

8.232
8.563
8.881
9.188
9.484

8.266
8.595
8.912
9.218
9.513

9.542
9.827
10.103
10.370
10.630

9.571
9.855
10.130
10.397
10.655

9.600
9.883
10.157
10.423
10.681

9.629
9.911
10.184
10.449
10.706

9.657
9.939
10.211
10.475
10.731

9.686
9.966
10.238
10.501
10.756

9;714
9.994
10.264
10.527
10.782

9.743
10.021
10.291
10.553
10.807

9.771
10.049
10.317
10.578
10.832

9.799
10.076
10.344
10.604
10.857

10.881
11.126
11.364
11.596
11.821

10.906
11.150
11.387
11.618
11.844

10.931
11.174
11.411
11.641
11.866

10.955
11.198
11.434
11.664
11.888

10.980
11.222
11.457
11.687
11.910

11.005
11.246
11.481
11.709
11.932

11.029
11.270
11.504
11.732
11.954

11.053
11.293
11.527
11.754
11.976

11.078
11.317
11.550
11.777
11.998

11.102
11.341
11.573
11.799
12.019

12.041
12.256
12.465
12.669
12.869

12.063
12.277
12.486
12.690
12.889

12.085
12.298
12.506
12.710
12.908

12.106
12.319
12.527
12.730
12.928

12.128
12.340
12.547
12.750
12.948

12.149
12.361
12.568
12.770
12.967

12.171
12.382
12.588
12.790
12.987

12.192
12.403
12.609
12.810
13.006

12.213
12.424
12.629
12.829
13.026

12.234
12.444
12.649
12.849
13.045

13.064
13.255
13.442
13.625
13.804

13.084
13.274
13.460
13.643
13.822

p.103
13.293
13.479
13.661
13.839

13.122
13.312
13.497
13.679
13.857

13.141
13.330
13.516
13.697
13.875

13.160
13.349
13.534
13.715
13.892

13.179
13.368
13.552
13.733
13.910

13.198
13.386
13.570
13.751
13.927

13.217
13.405
13.589
13.768
13.945

13.236
13.423
13.607
13.786
13.962

I

96

TABLE V6 continued

.00

.01

.02

.03

.04

.05

.06

.07

.08

.09

13.979
14.151
14.320
14.486
14.648

13.997
14.168
14.337
14.502
14.664

14.014
14.185
14.353
14.518
14.680

14.031
14.202
14.370
14.535
14.696

14.049
14.219
14.387
14.551
14.712

14.066
14.236
14.403
14.567
14.728

14.083
14.253
14.420
14.583
14.744

14.100
14.270
14.436
14.599
14.760

14.117
14.287
14.453
14.616
14.776

14.134
14.303
14.469
14.632
14.791

14.807
14.964
15.117
15.269
15.417

14.823
14.979
15.133
15.284
15.432

14.839
14.995
15.148
15.298
15.446

14.855
15.010
15.163
15.313
15.461

14.870
15.026
15.178
15.328
15.476

14.886
15.041
15.193
15.343
15.490

14.902
15.056
15.208
15.358
15.505

14.917
15.072
15.224
15.373
15.519

14.933
15.087
1$.239
15.388
15.534

14.948
15.102
15.254
15.402
15.549

15.563
15.707
15.848
15.987
16.124

15.577
15.721
15.862
16.001
16.137

15.592
15.735
15.876
16.014
16.151

15.606
15.749
15.890
16.028
16.164

15.621
15.763
15.904
16.042
16.178

15.635
15.778
15.918
16.055
16.191

15.649
15.792
15.931
16.069
16.205

15.664
15.806
15.945
16.083
16.218

15.678
15.820
15.959
16.096
16.232

15.692
15.834
15.973
16.110
16.245

16.258
16.391
16.521
16.650
16.777

16.272
16.404
16.534
16.663
16.790

16.285
16.417
16.547
16.676
16.802

16.298
16.430
16.560
16.688
16.815

16.312
16.443
16.573
16.701
16.827

16.325
16.456
16.586
16.714
16.840

16.338
16.469
16.599
16.726
16.852

16.351
16.483
16.612
16.739
16.865

16.365
16.496
16.625
16.752
16.877

16.378
16.509
16.637
16.764
16.890

16.902
17.025
17.147
17.266
17.385

16.914
17.037
17.159
17.278
17.396

16.927
17.050
17.171
17.290
17.408

16.939
17.062
17.183
17.302
17.420

16.951
17.074
17.195
17.314
17.431

16.964
17.086
17.207
17.326
17.443

16.976
17.098
17.219
17.338
17.455

16.988
17.110
17.231
17.349
17.466

17.001
17.122
17.243
17.361
17.478

17.013
17.135
17.255
17.373
17.490

17.501
17.616
17.730
17.842
17.953

17.513
17.628
17.741
17.853
17.964

17.524
17.639
17.752
17.864
17.975

17.536
17.650
17.764
17.875
17.985

17.547
17.662
17.775
17.886
17.996

17.559
17.673
17.786
17.897
18.007

17.570
17.685
17.797
17.908
18.018

17.582
17.696
17.808
17.919
18.029

17.593
17.707
17.820
17.931
18.040

17.605
17.719
17.831
17.942
18.051

18.062
18.170
18.276
18.382
18.486

18.073
18.180
18.287
18.392
18.496

18.083
18.191
18.297
18.402
18.506

18.094
18.202
18.308
18.413
18.517

18.105
18.212
18.319
18.423
18.527

18.116
18.223
18.329
18.434
18.537

18.127
18.234
18.340
18.444
18.547

18.137
18.244
18.350
18.455
18.558

18.148
18.255
18.361
18.465
18.568

18.159
18.266
18.371
18.475
18.578

18.588
18.690
18.790
18.890
18.988

18.599
18.700
18.800
18.900
18.998

18.609
18.710
18.810
18.909
19.007

18.619
18.720
18.820
18.919
19.017

18.629
18.730
18.830
18.929
19.027

18.639
18.740
18.840
18.939
19.036

18.649
18.750
18.850
18.949
19.046

18.660
18.760
18.860
18.958
19.056

18.670
18.770
18.870
18.968
19.066

18.680
18.780
18.880
18.978
19.075

19.085
19.181
19.276
19.370
19.463

19.094
19.190
19.285
19.379
19.472

19.104
19.200
19.295
19.388
19.481

19.114
19.209
19.304
19.398
19.490

19.123
19.219
19.313
19.407
19.499

19.133
19.228
19.323
19.416
19.509

19.143
19.238
19.332
19.426
19.518

19.152
19.247
19.342
19.435
19.527

19.162
19.257
19.351
19.444
19.536

19.171
19.226
19.360
19.453
19.545

19.554
19.645
19.735
19.825
19.913

19.564
19.654
19.744
19.833
19.921

19.573
19.664
19.753
19.842
19.930

19.582
19.673
19.762
19.851
19.939

19.591
19.682
19.771
19.860
19.948

19.600
19.691
19.780
19.869
19.956

19.609
19.700
19.789
19.878
19.965

19.618
19.709
19.798
19.886
19.974

19.627
19.718
19.807
19.895
19.983

19.636
19.726
19.816
19.904
19.991

20.000

20.009

20.017

20.026

20.035

20.043

20.052

20.061

20.069

20.078

97

TABLE V7

GIVEN: +dB (or +dBmV) (0 to +20);
FIND: VOLTAGE RATIO (or mV) (1.0 to 10.0)

For +dBs greater than +20:
1. Subtract the multiple of 20 dB required to give a number between 0 and
+20.

For dBs Subtracted

Multiply By

20
40
60
80

10
100
1000
10,000

2. Look up the voltage ratio corresponding to this number in table V7.
3. Multiply the ratio found by the number corresponding to the dBs subtracted in step 1.

EXAMPLE: Given +49.72 dB. Subtracting 40 dB we get 9.72 dB. From table V7 the voltage ratio corresponding to 9.72 dB
is 3.062. Multiplying by 100 (corresponding to 40 dB subtracted in step 1) we get a voltage ratio of 3.062 x
100 = 306.2.

0.00

0.02

0.04

0.06

1.000
1.012
1.023
1.035
1.047

1.002
1.014
1.026
1.038
1.050

1.005
1.016
1.028
1.040
1.052

1.007
1.019
1.030
1.042
1.054

1.059
1.072
1.084
1.096
1.109

1.062
1.074
1.086
1.099
1.112

1.064
1.076
1.089
1.102
1.114

1.122
1.135
1.148
1.161
1.175

1.125
1.138
1.151
1.164
1.178

1.189
1.202
1.216
1.230
1.245

.0.08

0.00

0.02

0.04

0.06

0.08

1.009
1.021
1.033
1.045
1.057

1.585
1.603
1.622
1.641
1.660

1.589
1.607
1.626
1.644
1.663

1.592
1.611
1.629
1.648
1.667

1.596
1.614
1.633
1.652
1.671

1.600
1.618
1.637
1.656
1.675

1.067
1.079
1.091
1.104
1.117

1.069
1.081
1.094
1.107
1.119

1.679
1.698
1.718
1.738
1.758

1.683
1.702
1.722
1.742
1.762

1.687
1.706
1.726
1.746
1.766

1.690
1.710
1.730
1.750
1.770

1.694
1.714
1.734
1.754
1.774

1.127
1.140
1.153
1.167
1.180

1.130
1.143
1.156
1.169
1.183

1.132
1.146
1.159
1.172
1.186

1.778
1.799
1.820
1.841
1.862

1.782
1.803
1.824
1.845
1.866

1.786
1.807
1.828
1.849
1.871

1.791
1.811
1.832
1.854
1.875

1.795
1.816
1.837
1.858
1.879

1.191
L205
1.219
1.233
1.247

1.194
1.208
1.222
1.236
1.250

1.197
l.211
1.225
1.239
1.253

1.199
1.213
1.227
1.242
1.256

1.884
1.905
1.928
1.950
1.972

1.888
1.910
1.932
1.954
1.977

1.892
1.914
1.936
1.959
1.982

1.897
1.919
1.941
1.963
1.986

1.901
1.923
1.945
1.968
1.991

1.259
1.274
1.288
1.303
1.318

1.262
1.276
1.291
1.306
1.321

1.265
1.279
1.294
1.309
1.324

1.268
1.282
1.297
1.312
1.327

1.271
1.285
1.300
1.315
1.330

1.995
2.018
2.042
2.065
2.089

2.000
2.023
2.046
2.070
2.094

2.004
2.028
2.051
2.075
2.099

2.009
2.032
2.056
2.080
2.104

2.014
2.037
2.061
2.084
1.109

1.334
1.349
1.365
1.380
1.396

1.337
1.352
1.368
1.384
1.400

1.340
1.355
1.371
1.387
1.403

1.343
1.358
1.374
1.390
1.406

1.346
1.361
1.377
1.393
1.409

2.113
2.138
2.163
2.188
2.213

2.118
2.143
2.168
2.193
2.218

2.123
2.148
2.173
2.198
2.223

2.128
2.153
2.178
2.203
2.228

2.133
2.158
2.183
2.208
2.234

1.413
1.429
1.445
1.462
1.479

1.416
1.432
1.449
1.466
1.483

1.419
1.435
1.452
1.469
1.486

1.422
1.439
1.455
1.472
1.489

1.426
1.442
1.459
1.476
1.493

2.239
2.265
2.291
2.317
2.344

2.244
2.270
2.296
2.323
2.350

2.249
2.275
2.301
2.328
2.355

2.254
2.280
2.307
2.333
2.360

2.259
2.286
2.312
2.339
2.366

1.496
1.514
1.531
1.549
1.567

1.500
1.517
1.535
1.552
1.570

1.503
1.521
1.538
1.556
1.574

1.507
1.524
1.542
1.560
1.578

1.510
1.528
1.545
1.563
1.581

2.371
2.399
2.427
2.455
2.483

2.377
2.404
2.432
2.460
2.489

2.382
2.410
2.438
2.466
2.495

2.388
2.415
2.443
2.472
2.500

2.393
2.421
2.449
2.477
2.506

98

TABLE V7 continued

0.00

0.02

0.04

0.06

0.08

0.00

0.02

0.04

0.06

0.08

2.512
2.541
2.570
2.600
2.630

2.518
2.547
2.576
2.606
2.636

2.523
2.553
2.582
2.612
2.642

2.529
2.559
2.588
2.618
2,649

2.535
2.564
2.594
2.624
2.655

5.012
5.070
5.129
5.188
5.248

5.023
5.082
5.140
5.200
5.260

5;035
5.093
5:152
5.212
5.272

5.047
5.105
'5.164
5.224
5.284

5.058
5.117
5.176
5.236
5.297

2.661
2.692
2.723
2.754
2.786

2.667
2.698
2.729
2.761
2.793

2.673
2.704
2.735
2.767
2.799

2.679
2.710
2.742
2.773
2.805

2.685
2.716
2.748
2.780
2.812

5.309
5.370
5.433
5.495
5.559

5.321
5.383
5.445
5.508
5.572

5.333
5.395
5.458
. 5.521
5.585

5.346
5.408
5.470
5.534
5.598

5.358
5.420
5.483
5.546
5.610

2.818
2.851
2.884
2.917
2.951

2.825
2.858
2.891
2.924
2.958

2.831
2.864
2.897
2.931
2.965

2.838
2.871
2.904
2.938
2.972

2.844
2.877
2.911
2.944
2.979

5.623
5.689
5.754
5.821
5.888

5.636
5.702
5.768
5.834
5.902

5.649
5.715
5.781
5.848
5.916

5.662
5.728
5.794
5.861
5.929

5.675
5.741
5.808
5.875
5.943

2.985
3.020
3.055
3.090
3.126

2.992
3.027
3.062
3.097
3.133

2.999
3.034
3.069
3.105
3.141

3.006
3.041
3.076
3.112
3.148

3.013
3.048
3.083
3.119
3.155

5.957
6.026
6.095
6.166
6.237

5.970
6.039
6.109
6.180
6.252

5.984
6.053
6.124
6.194
6.266

5.998
6.067
6.138
6.209
6.281

6.012
6.081
6.152
6.223
6.295

3.162
3.199
3.236
3.273
3.311

03.170
3.206
3.243
3.281
3.319

3.177
3.214
3.251
3.289
3.327

3.184
3.221
3.258
3.296
3.334

3.192
3.228
3.266
3.304
3.342

6.310
6.383
6.457
6.531
6.607

6.324
6.397
6.471
6.546
6.622

6.339
6.412
6.486
6.561
6.637

6.353
6.427
6.501
6.577
6.653

6.368
6.442
6.516
6.592
6.668

3.350
3.388
3.428
3.467
3.508

3.357
3.396
3.436
3.475
3.516

3.365
3.404
3.443
3.483
3.524

3.373
3.412
3.451
3.491
3.532

3.381
3.420
3.459
3.499
3.540

6.683
6.761
6.839
6.918
6.998

6.699
6.776
6.855
6.934
7.015

6.714
6.792
6.871
6.950
7.031

6.730
6.808
6.887
6.966
7.047

6.745
6.823
6.902
6.982
7.063

3.548
3.589
3.631
3.673
3.715

3.556
3.597
3.639
3.681
3.724

3.565
3.606
3.648
3.690
3.733

3.573
3.614
3.656
3.698
3.741

3.581
3.622
3.664
3.707
3.750

7.079
7.161
7.244
7.328
7.413

7.096
7.178
7.261
7.345
7.430

7.112
7.194
7.278
7.362
7.447

7.129
7.211
7.295
7.379
7.464

7.145
7.228
7.311
7.396
7.482

3.758
3.802
2.846
3.890
3.936

3.767
3.811
3.855
3.899
3.945

3.776
3.819
3.964
3.908
3.954

3.784
3.828
3.873
3.917
3.963

3.793
3.837
3.882
3.926
3.972

7.499
7.586
7.674
7.762
7.852

7.516
7.603
7.691
7.780
7.870

7.534
7.621
7.709
7.798
7.889

7.551
7.638
7.727
7.816
7.907

7.568
7.656
7.745
7.834
7.925

3.981
4.027
4.074
4.121
4.169

3.990
4.036
4.083
4.130
4.178

3.999
4.046
4.093
4.140
4.188

4.009
4.055
4.102
4.150
4.198

4.018
4.064
4.111
4.159
4.207

7.943
8.035
8.128
8.222
8.318

7.962
8.054
8.147
8.241
8.337

7.980
8.072
8.166
8.260
8.356

7.998
8.091
8.185
8.279
8.375

8.017
8.110
8.204
8.299
8.395

4.217
4.266
4.315
4.365
4.416

4.227
4.276
4.325
4.375
4.426

4.236
4.285
4.335
4.385
4.436

4.246
4.295
4.345
4.395
4.446

4.256
4.305
4.355
4.406
4.457

8.414
8.511
8.610
8.710
8.810

8.433
8.531
8.630
8.730
8.831

8.453
8.551
8.650
8.750
8.851

8.472
8.570
8.670
8.770
8.872

8.492
8.590
8.690
8.790
8.892

4.467
4.519
4.571
4.624
4.677

4.477
4.529
4.581
4.634
4.688

4.487
4.539
4.592
4.645
4.699

4.498
4.550
4.603
4.656
4.710

4.508
4.560
4.613
4.667
4.721

8.913
9.016
9.120
9.226
9.333

8.933
9.036
9.141
9.247
9.354

8.954
9.057
9.162
9.268
9.376

8.974
9.078
9.183
9.290
9.397

8.995
8.099
9.204
9.311
9.419

4.732
4.786
4.842
4.898
4.955

4.742
4.797
4.853
4.909
4.966

4.753
4.808
4.864
4.920
4.977

4.764
4.819
4.875
4.932
4.989

4.775
4.831
4.887
4.943
5.000

9.441
9.550
9.661
9.772
9.886

9.462
9.572
9.683
9.795
9.908

9.484
9.594
9.705
9.817
9.931

9.506
9.616
9.727
9.840
9.954

9.528
9.638
9.750
9.863
9.977

10.000

10.023

10.047

10.072

10.096

99

TABLE VB
VOLTAGE ADDITION

TWO SIGNALS, OR CROSS-MOD. COMPONENTS, COMBINED ON A
VOLTAGE BASIS.
GIVEN: dB DIFFERENCE BETWEEN THE TWO LEVELS;
FIND: dBs TO ADD TO HIGHER LEVEL TO GET TOTAL LEVEL IN dBs.

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

6.02

5.97

5.92

5.87

5.82

5.77

5.73

5.68

5.63

5.58

5.53
5.08
4.65
4.25
3.88

5.49
5.03
4.61
4.21
3.84

5.44
4.99
4.57
4.17
3.80

5.39
4.94
4.53
4.13
3.77

5.35
4.90
4.49
4.10
3.73

5.30
4.86
4.45
4.06
3.70

5.26
4.82
4.41
4.02
3.66

5.21
4.78
4.37
3.98
3.63

5.17
4.73
4.33
3.95
3.60

5.12
4.69
4.29
3.91
3.56

3.53
3.21
2.91
2.64
2.39

3.50
3.18
2.88
2.61
2.36

3.46
3.15
2.86
2.59
2.34

3.43
3.12
2.83
2.56
2.32

3.40
.'3.09
2.80
2.53
2.29

3.36
3.06
2.77
2.51
2.27

3.33
3.03
2.74
2.48
2.25

3.30
3.00
2.72
2.46
2.22

3.27
2.97
2.69
2.44
2.20

3.24
2.94
2.66
2.41
2.18

2.16
1.95
1.75
1.58
1.42

2.13
1.93
1.74
1.56
1.41

2.11
1.91
1.72
1.55
1.39

2.09
1.89
1.70
1.53
1.38

2.07
1.87
1.68
1.51
1.36

2.05
1.85
1.67
1.50
1.35

2.03
1.83
1.65
1.48
1.33

2.01
1.81
1.63
1.47
1.32

1.99
1.79
1.61
1.45
1.31

1.97
1.77
1.60
1.44
1.29

1.28
1.15
1.03
0.92
0.83

1.26
1.14
1.02
0.91
0.82

1.25
1.12
1.01
0.90
0.81

1.24
1.11
1.00
0.89
0.80

1.22
1.10
0.99
0.88
0.79

1.21
1.09
0.98
0.87
0.78

1.20
1.08
0.96
0.86
0.77

1.19
1.06
0.95
0.86
0.77

1.17
1.05
0.94
0.85
0.76

1.16
1.04
0.93
0.84
0.75

0.74
0.66
0.59
0.53
0.48

0.73
0.66
0.59
0.53
0.47

0.73
0.65
0.58
0.52
0.46

0.72
0.64
0.57
0.51
0.46

0.71
0.64
0.57
0.51
0.45

0.70
0.63
0.56
0.50
0.45

0.69
0.62
0.56
0.50
0.44

0.69
0.61
0.55
0.49
0.44

0.68
0.61
0.54
0.49
0.43

0.67
0.60
0.54
0.48
0.43

0.42
0.38
0.34
0.30
0.27

0.42
0.38
9. 34
0.30
0.27

0.42
0.37
0.33
0.30
0.26

0.41
0.37
0.33
0.29
0.26

0.41
0.36
0.32
0.29
0.26

0.40
0.36
0.32
0.29
0.26

0.40
0.35
0.32
0.28
0.25

0.39
0.35
0.31
0.28
0.25

0.39
0.35
0.31
0.28
0.25

0.38
0.34
0.31
0.27
0.24

0.24
0.22
0.19
0.17
0.15

0.24
0.21
0.19
0.17
0.15

0.24
0.21
0.19
0.17
0.15

0.23
0.21
0.19
0.17
0.15

0.23
0.21
0.18
0.16
0.15

0.23
0.20
0.18
0.16
0.14

0.23
0.20
0.18
0.16
0.14

0.22
0.20
0.18
0.16
0.14

0.22
0.20
0.18
0.16
0.14

0.22
0.19
0.17
0.15
0.14

0.14
0.12
0.11
0.10
0.09

0.14
0.12
0.11
0.10
0.09

0.13
0.12
0.11
0.09
0.08

0.13
0.12
0.10
0.09
0.08

0.13
0.12
0.10
0.09
0.08

0.13
0.12
0.10
0.09
0.08

0.13
0.11
0.10
0.09
0.08

0.13
0.11
0.10
0.09
0.08

0.12
0.11
0.10
0.09
0.08

0.12
0.11
0.10
0.09
0.08

100

TABLE V9

CONVERSION BETWEEN dBmV, RMS VOLTAGE, RMS CURRENT,
AND POWER, at 75 OHMS

Abbreviations Used:
Voltage
p.V ...... microvolts •..... 10' 6volts
m V .••... millivolts ....... 10-3 volts

RMSVoltage

RMS Current

Current
nA ..... nanoamperes ..... 10. 9 amp.
p.A ..... microamperes .... 10.6 amp.
rnA ..... milliamperes ..... 10. 3 amp.

Power

fW ..... femtowatts ..... 10.15 watts
pW ...•. picowatts ...... 10.12 watts
nW .•... nanowatts .•.•. 10.9 watts
p.W . •... microwatts ..... 10.6 watts
mW ..... milliwatts ...••. 10.3 watts

Power

RMS Voltage

RMS Current

Power

1.00 p.V
L12p.V
1.26 p.V
1.41p.V
1.59p.V

13.30 nA
15.00nA
16.80nA
18.80nA
21.10 nA

13.30 fW
16.80 fW
21.10fW
26.60 fW
33.50 fW

1000.00 p,V
1.12 mV
1.26 mV
1.41 mV
1.59 mV

13.30p.A
15.00 pA
16.80 p,A
18.80 p,A
21.10 p.A

13.30nW
16.80nW
21.lOnW
26.60nW
33.50nW

1.78 p.V
2.00/LV
2.24/LV
2.51/LV
2.82/LV

23.70nA
26.60 nA
29.90nA
33.50nA
37.60nA

42.20fW
53.10fW
66.80fW
84.10 fW
106.00fW

1.78 mV
2.00mV
2.24mV
2.51 mV
2.82mV

23.70 p.A
26.60 p,A
29.90 p,A
33.50 p,A
37.60 p,A

42.20nW
53.lOnW
66.80nW
84.lOnW
106.00nW

3.16/LV
3.55/LV
3.98/LV
4.47/LV
5.02/LV

42.20nA
47.30nA
53.lOnA
59.60nA
66.80nA

133.00fW
168.00fW
211.00fW
266.00fW
335.00 fW

3.16mV
3.55mV
3.98mV
4.47 mV
5.01 mV

42.20 p,A
47.30 p,A
53.10 p,A
59.60 p,A
66.80 p,A

133.00nW
168.00nW
211.00nW
266.00nW
335.00nW

5.62/LV
6.31p.V
7.08/LV
7.94/LV
8.91/LV

75.00nA
84.lOnA
94.40 nA
106.00nA
119.00nA

422.00 fW
531.00fW
668.00fW
841.00fW
1.06 pW

5.62mV
6.31mV
7.08mV
7.98mV
8.91 mV

75.00 p,A
84.1Dp.A
94.40 p,A
106.00 p,A
119.00 p,A

422.00nW
531.00 nW
668.00nW
841.00nW
1.06p.W

10.00 /LV
11.20p.V
12.60/LV
14.10 /LV
15.90/LV

133.00nA
150.00nA
168.00nA
188.00nA
211.00 nA

1.33 pW
1.68pW
2.11 pW
2.66pW
3.35pW

10.00mV
11.20mV
12.60mV
14.lOmV
15.90mV

133.00p.A
150.00 p,A
168.00 p,A
188.00 p,A
211.00 p.A

1.33 p,W
1.68 p,W
2.11p.W
2.66 p,W
3.35p,W

17.80/LV
20.00p.V
22.40/LV
25.10p.V
28.20/LV

237.00nA
266.00nA
299.00nA
335.00nA
376.00nA

4.22pW
5.31 pW
6.68pW
8.41 pW
1O.60pW

17.80mV
20.00mV
22.40 mV
25.10mV
28.20mV

237.00 p.A
266.00 p,A
299.00 p,A
335.00 p.A
376.00p.A

4.22p.W
5.31 p,W
6.68 p,W
8.41 p,W
10.60p.W

31.60 p.V
35.50/LV
39.80/LV
44.70/LV
50.10 /LV

422.00nA
473.00nA
531.00 nA
596.00nA
668.00nA

13.30pW
16.80pW
21.l0pW
26.60pW
33.50pW

31.60mV
35.50mV
39.80mV
44.70mV
50.lOmV

422.00 p,A
473.00p,A
531.00 p,A
596.00 p,A
668.00 p.A

13.30p,W
16.80p,W
21.10 pW
26.60 p,W
33.50p.W

56.20/LV
63.10 /LV
70.80p.V
79.40 p.V
89.1Op.V

750.00nA
841.OOnA
944.00nA
1.06 pA
L19p.A

42.20pW
53.lOpW
66.80pW
84.10pW
106.00pW

56.20mV
63.10mV
70.80mV
79.40 mV
89.lOmV

750.00 p,A
841.00 p.A
944.00 p.A
1.06 rnA
1.19 rnA

42.20p,W
53.10 p,W
66.80.p,W
84.10 p,W
106.00p.W

100.00 p,V
112.00 p,V
126.00 p,V
141.00 p,V
159.00p,V

1.33p,A
1.50 p.A
1.68 p,A
1.88 p,A
2.11p.A

133.00pW
168.00pW
211.00pW
266.00pW
335.00pW

100.00mV
112.00mV
126.00mV
141.00mV
159.00 mV

1.33 rnA
1.50 rnA
1.68 rnA
1.88 rnA
2.11 rnA

133.00p.W
168.00p.W
211.00 p,W
266.00p.W
335.00 p,W

178.00p.V
200.00 p,V
224.00p,V
251.00 p,V
282.00p.V

2.37 p,A
2.66p.A
2.99 p,A
3.35 p,A
3.76p.A

422.00pW
531.00pW
668.00pW
841.00 pW
1.06 nW

178.00mV
200.00mV
224.00mV
251.00mV
282.00mV

2.37 rnA
2.66 rnA
2.99 rnA
3.35 rnA
3.76 rnA

422.00 p,W
531.00 p,W
668.00 p,W
841.00 p,W
1.06mW

316.00p.V
355.00 p,V
398.00 p,V
447.00 p,V
501.OOp.V

4.22 p,A
4.73 p,A
5.31 p,A
5.96p.A
6.68p.A

1.33nW
1.68nW
2.11nW
2.66nW
3.35nW

316.00mV
355.00mV
398.00mV
447.00mV
501.00mV

4.22 rnA
4.73 rnA
5.31 rnA
5.96mA
6.68mA

1.33mW
1.68mW
2.11mW
2.66mW
3.35mW

562.00 p,V
631.00 p,V
708.00 p,V
794.00 p.V
881.00 p.V

7.50 p,A
8.41p,A
9.44p.A
10.60 p,A
11.90p.A

4.22nW
5.31nW
6.68nW
8.41 nW
10.60nW

562.00mV
631.00mV
708.00mV
794.00mV
891.00mV

7.50mA
8.41 mA
9.44mA
10.60 rnA
11.90mA

4.22mW
5.31mW
6.68mW
8.41 mW
1O.60mW

1000.00mV

13.30mA

13~30mW

101

TABLET!

RETURN LOSS, REFLECTION COEFFICIENT, AND VSWR

EXAMPLE: AssUme that a discontinuity exists in a transmission line, producing a reHected wave whose amplitude is "p"
times that of the main wave: Er = p Em;
where Er is the rms voltage of the reHected wave
and Em is the rms voltage of the main wave.

The reHection coefficient expressed in percent

.

= 100 P, and expressed as a return loss in- dB = 20 log

10

P.

At points along the line where the reHected wave and the main wave are in phase, they add to produce a voltage maximum:
Em + E. = Em (1 + p).

Emax

=

At points where they are 180 0 out of phase, they subtract to produce a voltage minimum: E m1n
Then the V8WR " 8" ( ratio 0 f maximum to minimum voltage) is:

8

Emu Em + Er
= -= Em-E.
E
m1n

and in terms of S, p
Return
Loss
dB down

0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
5.5
6.0
6.5
7.0
7.5
8.0
8.5
9.0
9.5
10.0
10.5
11.0
11.5
12.0
12.5
13.0
13.5
14.0
14.5
15.0
15.5
16.0
16.5
17.0
17.5
18.0
18.5
19.0
19.5
20.0

Reflection
Coefficient

%

100.00
94.4
89.1
84.1
79.4
75.0
70.8
66.8
63.1
59.6
56.2
53.1
50.1
47.3
44.7
42.2
39.8
37.6
35.5
~3.5

31.6
29.9
28.2
26.6
25.1
23.7
22.4
21.1
20.0
18.8
17.8
16.8
15.9
15.0
14.1
13.3
12.6
11.9
11.2
10.6
10.0

VSWR

Return
Loss
dB down

Reflection
Coefficient

%

VSWR

== Em - E. = Em (1 -

Em (1 + p)
Em(l-p)
8-1

1+p
= --,
I-p

=--.
8+1

Return
Loss
dB down

Reflection
Coeffic ient

%

VSWR

00

34.7
17.3
11.6
8.71
7.00
·5.85
5.02
4.42
3.95
3.57
3.26
3.01
2.80
2.62
2.46
2.32
2.21
2.10
2.01
1.92
1.85
1.79
1.72
1.67
1.62
1.58
1.53
1.50
1.46
1.43
1.40
1.38
1.35
1.33
1.31
1.29
1.27
1.25
1.233
1.222

20.5
21.0
21.5
22.0
22.5
23.0
23.5
24.0
24.5
25.0
25.5
26.0
26.5
27.0
27.5
28.0
28.5
29.0
29.5
30.0
30.5
31.0
31.5
32.0
32.5
33.0
33.5 .
34.0
34.5
35.0
35.5
36.0
36.5
37.0
37.5
38.0
38.5
39.0
39.5
40.0.

9.44
8.91
8.41
7.94
7.50
7.08
6.68
6.31
5.96
5.62
5.31
5.01
4.73
4.47
4.22
3.98
3.76
3.55
3.35
3.16
2.99
2.82
2.66
2.51
2.37
2.24
2.11
2.00
1.88
1.78
1.68
1.59
1.50
1.41
1.33
1.26
1.19
1.12
1.06
1.00
102

1.208
1.196
1.184
1.172
1.162
1.152
1.143
1.135
1.127
1.119
t.112
1.105
1.099
1.094
1.088
1.083
1.078
1.074
1.069
1.065
1.062
1.058
1.055
1.051
1.049
1.046
1.043
1.041
1.038
1.036
1.034
1.032
1.030
1.029
1.027
1.026
1.024
1.023
1.021
1.020

40.5
41.0
41.5
42.0
42.5
43.0
43.5
44.0
44.5
45.0
45.5
46.0
46.5
47.0
47.5
48.0
48:5
49.0
49.5
50.0
50.5
51.0
51.5
52.0
52.5
53.0
53.5
54.0
54.5
55.0
55.5
56.0
56.5
57.0
57.5
58.0
58.5
59.0
59.5
60.0

.944
.891
.841
.794
.750
.708
.668
.631
.596
.562
.531
.501
.473
.447
.422
.398
.376
.355
.335
.316
.299
.282
.266
.251
.237
.224
.211
.200
.188
.178
.168
.159
.150
.141
.133
.126
.119
.112
.106
.100

1.0191
1.0180
1.0170
1.0160
1.0151
1.0143
1.0134
1.0127
1.0120
1.0113
1.0107
1.0101
1.0095
1.0090
. 1.0085
1.0080
1.0075
1.0071
1.0067
1.0063
1.0060
1.0057
1.0053
1.0050
1.0048
1.0045
1.0042
1.0040
1.0038
1.0035
1.0034
1.0032
1.0030
1.0028
1.0027
1.0025
1.0024
1.0022
·1.0021
1.0020

p).

TABLE T2. CHANNEL VARIABLES-INTRODUCTION'

or, if the voltage is known:

*WAVELENGTH IN AIR: The velocity of radio waves-

E

in air is the same as the velocity of light: 300,000,000
(3 X 108 ) meters per second. Since one meter equals 39.37
inches, the velocity of radio waves, expressed in inches per
second, is 39.37 X 3 X 108 = 1.181 X 10 10 inches per
second.

= 1.181f X 10

MHz

microvolts per meter

where:
V is the output voltage in millivolts rms
E is the field strength in microvolts per meter
f MHz is the signal frequency in MHz

One wavelength is the distance a wave travels while one
cycle is completed. If the cycle is completed in one second,
the wave has a frequency of one Hertz (1 Hz). Hence the
wavelength "X" for 1 Hz equals the distance the wave
travels in one second, which is 1.181 X 1010 inches, and
for a freqency of f Hz :
X

= 20.7 V X f

For greatest convenience in CATV work, output level
should be expressed in dBmV (dB above one millivolt)
and the field strength level in dBu (dB above one microvoJt per meter) .
In the above expression the field strength level in dBu is
20 log 10 E

10

inches.

Hz

so: field strength level (dBu)

In CATV, frequency is more conveniently expressed in
mega-Hertz (MHz), 1 MHz being 106 Hz.
Thus X

=

1.181 X 10 10

11,810
= --inches.
f

f Hz X 106

wavelength -

X
2

20 log 10 V is the level in dBmV, so if we let
20 log 10 (20.7 X f MHz )
dipole antenna factor, then

=

and the half

MHz

=

field strength level (dBu)
dipole output (dBmV) + dipole antenna factor.

5905

= -inches.
f
lIHz

Examples:

lfz wavelength in other dielectrics: A dielectric other than
air has the effect of slowing the wave down, so one wavelength becomes shorter:
1
Wavelength in dielectric wavelength in air X

1. A field strength meter using a half-wave dipole on channel12 indicates an antenna output level of -18 dBmV.

-ve

=

From table T2: for channel 12, dipole antenna
factor
72.54 db. So field strength level ill
72.54-18 54.54 dBu.

=

where e is the dielectric constant of the material.
For solid polyethylene, e
and

= 2.26 so

*wavelength = 0.665f

X 5900"
MHz

*wavelength = 0.81f

X 5900
MHz

2. The field strength level at a given location on channel 6 is predicted to be + 12 dBu at an elevation of
30 feet. What is the expected output level from an
antenna array having a gain of 15 dB, mounted at an
elevation of 300 feet?

1
va
= 0.665

3920

=-inches;
f
= 1.51 so

4779
= -f

Three factors are involved in determining the output level:

1
va
= 0.81

a. The height is increased lOX, from 30 to 300 feet.
Assuming linear increase of field strength with
height, this is handled by adding 20 dB.

inches.

MHz

b. The antenna gain adds 15 dB.

THE DIPOLE ANTENNA FACTOR is convenient when
a field strength meter, calibrated in dBmV, is used with a
dipole antenna to measure field strength. The factor is the
number of dB which must be added to the output (in
dBmV) of a
'wavelength dipole to obtain the field
strength in dBu (dB above one microvolt per meter).

=

c. Since field strength level
dipole output + dipole
antenna factor from the table: factor
64.72 for
channel 6;

*

The expected output level in dBmV is thus:
+12 +20 +15 - 64.72 = -17.72 dBmV

When a 75-ohm half-wave dipole antenna is exposed to
an r-f field, the rms voltage delivered to a 75-ohm load is:

=

5.58 X

yI75

1000

E
X -flUIz

=

so: 64.72 dB is subtracted.

The factor is listed on Table T2 for each TV channel. It is
derived as follows:

V

=

MHz

for the usual foamed polyethylene e
and

= 20 log 10 E

= 20 log 10 V + 20 log 10 (20.7 X fMHz )

E

= 0.0483 - mV rms
fill
Hz

.. Rounded off.

103

TABLE T2
.. V2
..'"
z
CD

0

Ii
C
C

'"

.c

(.)

'':
(.)

I!!"
=.!!! ...

t;t::z:

ii:~:::E

TV CHANNEL FREQUENCIES AND CORRESPONDING DI·POLES

Wave Length, Inches
Type of Dielectric

"CI

§~
~:::E

CD

Foam

Solid

CD

0

cm

z

.!! ..

Ii

cc"CI

C
C

00

Air

..

'"CC

Il.-

"E
CD ..
"CD

.a".:
N
u ... :z:

'"

(.)

V2 Wave Length, Inches

,u
.-0"'"

.c

(.)

'"

ii:~::a:

=:z:
~:::E

CD

cm

cc"CI

Type of Dielectric

.!! ..

"CI

COol

'"CC

00

Air

Foam

Solid

Il.-

,u
.-0"""

2
3
4
5

55.25
61.25
67.25
77.25

59.75
65.75
71.75
81.75

103.8
93.8
85.7
74.8

84.0
76.0
69.4
60.6

68.5
61.9
56.6
49.4

61.2
62.1
62.9
64.1

43
44
45
46

645.25
651.25
657.25
663.25

649.75
655.75
661.75
667 ..75

9.2
9.1
9.0
8.9

7.4
7.4
7.3
7.2

6.1
6.0
5.9
5.9

82.5
82.6
82.6
82.7

6
7
8
9

83.25
175.25
181.25
187.25

87.75
179.75
185.75
191.75

69.5
33.4
32.8
31.5

56.3
27.0
25.6
25.5

45.9
22.0
21.6
19.8

64.7
71.2
71.5
71.8

47
48
49
50

669.25
675.25
681.25
687.25

673.75
679.75
685.75
691.75

8.8
8.7
8.7
8.6

7.2
7.1
7.1
7.0

5.8
5.7
5.7
5.6

82.7
82.8
82.9
83.0

10
11
12
13

193.25
199.25
205.25
211.25

197.75
203.75
209.75
215.75

30.0
29.4
28.5
27.7

24.5
23.8
23.0
22.5

20.0
19.4
18.8
18.2

72.0
72.3
72.5
72.8

51
52
'53
54

693.25
699.25
705.25
711.25

697.75
703.75
709.75
715.75

8.5
8.4
8.4
8.3

6.9
6.8
6.8
6.7

5.6
5.5
5.5
5.5

83.1
83.1
83.2
83.3

14
15
16
17

471.25
477.25
483.25
489.25

475.75
481.75
487.75
493.75

12.5
12.4
12.2
12.0

10.1
10.0
9.9
9.7

8~2

8.1
8.0
7.9

79.8
79.9
80.0
80.1

55
56
57
58

717.25
723.25
729.25
735.25

721.75
727.75
733.75
739.75

8.2
8.1
8.0
8.0

6.6
6.5
6.5
6.5

5.4
5.3
5.3
5.3

83.3
83.4
83.5
83.6

18
19
20
21

495.25
501.25
507.25
513.25

499.75
505.75
511.75
517.75

11.9
11.8
11.6
11.5

9.6
9.5
9.4
9.3

7.8
7.7
7.6
7.5

80.2
80.3
80.4
80.5

59
60
61
62

741.25
747.25
753.25
759.25

745.75
751.75
757.75
763.75

7.9
7.8
7.8
7.7

6.4
6.3
6.3
6.2

5.2
5.2
5.2
5.1

83.6
83.7
83.8
83.9

22
23
24
25

519.25
525.25
531.25
537.25

523.75
529.75
535.75
541.75

11.3
11.2
11.1
11.0

9.2
9.1
9.0
8.9

7.4
7.4
7.3
7.2

80.6
80.7
80.8
80.9

63
64
65
66

765.25
771.25
777.25
783.25

769.75
775.75
781.75
787.75

7.7
7.6
7.5
7.5

6.2
6.2
6.1
6.1

5.1
5.0
4.9
4.9

84.0
84.1
84.1
84.2

26
27
28
29

543.25
549.25
555.25
561.25

547.75
553.75
559.75
565.75

10.8
10.7
10.6
10.5

8.7
8.7
8.6
8.5

7.1
7.1
7.0
6.9

81.0
81.1
81.2
81.3

67
68
69
70

789.25
795.25
801.25
807.25

793.75
799.75
805.75
811.75

7.4
7.4
7.3
7.3

6.0
6.0
5.9
5.9

4.9
4.9
4.8
4.8

84.3
84.4
84.4
84.5

30
31.
32
33

567.25
573.25
579.25
585.25

571.75
577.75
583.75
589.75

10.4
10.3
10.2
10.1

8.4
8.3
8.3
8.2

6.9
6.8
6.7
6.6

81.4
81.5
81.6
81.7

71
72
73
74

813.25
819.25
825.25
831.25

817.75
823.75
829.75
835.75

7.2
7.2
7.1
7.0

5.8
5.8
5.7
5.7

4.7
4.7
4.6
4.6

84.5
84.6
84.6
84.7

34
35
36
37

591.25
597.25
603.25
609.25

595.75
601.75
607.75
613.75

10.0
9.9
9.8
9.7

8.1
8.0
7.9
7.8

6.6
6.5
6.4
6.3

81.8
81.9
82.0
82.1

75
76
77
78

837.25
843.25
849.25
855.25

841.75
847.75
853.75
859.75

7.0
6.9
6.9
6.8

5.7
5.6
5.6
5.5

4.6
4.6
4.6
4.5

84.8
84.8
84.9
84.9

38
39
40
41

615.25
621.25
627.25
633.25

519.75
625.75
631.75
637.75

9.6
9.5
9.4
9.3

7.7
7.7
7.6
7.5

6.3
6.2
6.2
6.1

82.2
82.3
82.3
82.4

79
80
81
82

861.25
867.25
873.25
879.25

865.75
871.75
877.75
883.75

6.8
6.8
6.7
6.7

5.5
5.5
5.4
5.4

4.5
4.5
4.4
4.4

85.0
85.0
85.1
85.2

42

639.25

643.75

9.2

7.4

6.1

82.4

83

885.25

889.75

6.6

5.3

4.4

85.3

NOTE: Propagation velocity in cable with polyfoam dielectric is calculated at 81% of that in free space; solid polyethylene at 66%.

CABLE ATTENUATION
To convert nominal cable attenuation given in dB/100 ft. at 70°F into dB/loo
meters at 21 ·C: Multiply by 3.281.
104

Appendix I

THE DETAILED MATIlEMATICAL DEVELOPMENT

OF
THE DISTORTION COMPONENTS

Let el n = A cosa + B cosb + C cosc
where a > b > c
and eout = klein + k2eln2 + kSelns

k2A2 cos2 a =

First Order Components:

k B2
k2B2cos2b = _2_

+ ++
k A2

k A2

2

By direct multiplication:
klein = klA cosa + klB cosb + klC cosc

~C2

k2C2COS2C= -2-

cos 2a

k B2
+ _2_

cos 2b

+

cos 2c

+

2
kC2

Second Order Components:

k2 e l n 2 =
= k2 (A cosa + B cosb + C cosc) 2

= k2A2cos 2a + k2B 2cos2b

Substituting the other three expressions from above into
eq. (1) gives the sum and difference beat components:

+ k 2C2COS2C

+ 2k~ cosa cosb + 2k2AC cosa cosc + 2k2BC cosb cose

+ b) + k2AB cos
2k2AC cosa cosc = k 2AC cos (a + c) + k2 AC cos
2k2BC cosb cosc = k2BC cos (b + c) + k 2 BC cos

To reduce these expressions to ones containing identifiable
frequencies, expand each term in accordance with the
applicable one of the following two relationships (eq. 1 or
2): (These can be found in any elementary trigonometry
text, or on pages 1041-2 of the ITT "Reference Data for
Radio Engineers" Fourth Edition.)

2k2AB cosa cosb = k2AB cos (a

(a - b)
(a - c)
(b - c)

Third Order Components:
equation 1 cos x cos y

1

="2 cos

(x

1

+ y) + "2 cos

kselna = ka(A cosa + B cosb

(x - y)

=

equation 2

1

1

COS2 X="2+"2 cos2x

+ C coscl)a
kaAacos3a + kaBscossb + kaCscosac
+ 3kaA2B cos2 a cosb + 3ksA2C cos2a cosc
+ 3kaB2C cos2b cosc + 3kaB2A cos2b cosa
+ 3ksC2A cos 2c cos a + 3kaC2B cos2c cosb
+ 6ksABC cosa cosb cosc

Three trigonometric relationships are needed to allow expansion of the terms of the types "cosax", "cos2x cosy", and
"cos x cos y cos z".

Substituting the appropriate expressions from above into
eq. (2) gives the DC and second harmonic components:

105

1
1
= 2" cos (x + y) cosz + 2" cos (x - y) cosz

From the ITT handbook (loc. cit.) p. 1041:
cos 3x = -3cosx + 4cosax
4cosax = 3cosx + cos 3x

=!

[.!. cos (x + y + z) +

2 2

3

1

+

cos 3 x = "4 cosx + "4 cos 3x

giving eq. 3

~[!.cos

2 2

~ cos

2

(x + y - z) ]

(x - y + z) +! cos (x - y - z)]
2

giving eq. 5
cos 2x cosy =

(~+ ~ cos 2x)
1

cosy (substituting from eq. 2)
1
1
cosx cosy cosz = "4 cos (x + y + z) + "4 cos (x + y - z)

1

= 2" cosy + 2" cos 2x cosy
1

1

1

+ "4 cos (x - y + z) + "4 (x - y - z)

1

but: cos 2x cosy = 2" cos (2x + y) + 2" cos (2x - y)
Applying eq. 3 to the appropriate terms gives the selfexpansion/compression and third harmonic components:

giving eq. 4

I I I
cos 2x cosy = 2" cosy +"4 cos (2x + y) + "4 cos (2x - y)

cosx cosy cosz = (cosx cosy) cosz

=

[~ cos

(x + y) +

~ cos (x -

y)] cosz

Applying eg. 3 to the appropriate terms gives the cross-expansion/compression and intermodulation components:
3
3
3
3kaA2B cos2a cosb = 2"kaA2B cosb + "4kaA2B cos (2a + b) + "4kaA2B cos (2a - b)
3kaA2C cos 2a cosc

3
= 2"kaA2C
cosc +

3
3
laA2C cos (2a + c) + "4kaA2C cos (2a - c)

3
3
3
3kaB2C cos2b cosc = 2"kaB2C cosc + "4kaB2C cos (2b + c) + "4kaB2C cos (2b - c)
3
3
3
3kaB2A cos 2b cosa = 2"kaB2A cosa + "4kaB2A cos (2b + a) + "4kaB2A cos (2b - a)
3
3
3
3kaC2A cos 2c cosa = 2"k aC2A cosa + "4kaC2A cos (2c + a) + "4kaC2A cos (2c - a)
3
3
3
3kaC2B cos 2c cosb = 2"kaC2B cosb + "4kaC2B cos (2c + b) + "4kaC2B cos (2c - b)

Using eq. 5 gives the triple beat components:
3
3
6k aABC cosa cosb cosc = 2"kaABC cos (a + b + c) + 2"kaABC cos (a + b - c)

3
3
+ 2"kaABC cos (a - b + c) + "2kaABC cos (a - b - c)
106

Appendix II
THE DETAILED MATHEMATICAL DEVELOPMENT OF SELF- AND CROSS-COMPRESSION
(AND EXPANSION) WITH UNMODULATED INPUT SIGNALS
The output voltage at fa, for example, including the effects
of distortion is:

and the ratio of this component to the first order output is:

ea(D) = k1A + 3/4 ksAs + 3/2 ksAB2 + 3/2 ksAC2

eabe = 3/2 ka As = 3/2 ka A2
k1A
k1A
kl

A measure of the degree of Cross-compression (or -expansion) is found by dividing this output by that which would
occur if thert: were no distortion:

Because this expression is frequently used in what follows,
it is abbreviated r abc (the ratio of the triple beat output
voltage to the first order output voltage).

ea(D) _ k1A + 3/4 ksAs + 3/2 ksAB2 + 3/2 ksAC2
k1A k1A

eabe
/ ka 2
rabe = k1A = 3 2 kl A .
(Note that rabe may be either positive or negative depending on the sign of ks.)

= 1 + 3/4 ka A2 + 3/2 ks B2 + 3/2 ka C2
kl
kl
kl

This is identical with the term preceding the parentheses
in equation A. Substituting in that equation:

Although the foregoing has only been concerned with conditions involving three CW input signals, in this case it is
instructive to discover the relationship with any number
of input signals. Assume all input amplitudes = A. Then:

equation
"B"

ea(D) = 1 + 3/4 ks A2 + 3/2 ks A2 + 3/2 ks A2 ++ etc.
~A

~

~

~

k
=1+ k:A2(3/4+3/2+3/2+ .... etc.)

ea(D)
k1A = 1 + rabe (n1
- 2')

Which shows the relationship between the degree of expansion (or compression) and the ratio of Triple beat to
first order output.

k
= 1 + 3/2 k: A2 (1/2 + 1 + 1 + .... etc.)

Compression or expansion, in dB is expressed
When there is one input signal the quantity in parentheses
= ~, with two it = 1~, with three it = 2~ and in general:
equation
"c" 20 Iog 10 eaID)
k1A -- 20 IoglO [1 + rabe ( n equation
"A"

~(~) =
1

1 + 3/2 ks A2 (n
kl

-!.)
2

where n is the
total number of
input signals.

I)J

2'

The tables for Compression and Expansion were computed
by selecting progressive values for Rabe/a (= Labe - La)
and calculating the corresponding value for rabc from the
relationship: Rabe/a= 20 loglo rabc'

To allow comparison between the various effects of third
order distortion, a common measure of this distortion is
needed. The triple beat component is chosen. Its amplitude
is given by:
e"be = 3/2 ks ABC;

The value for rabc was then inserted in equation B, with a
- sign for compression, or a + sign for expansion. With a
chosen number of inputs (n) the ratio of ek:~) was then
calculated. Finally this ratio was expressed in decibel form
by the use of equation C.

where A = B = C, eabe = 3/2 ksAs.

Appendix III
THE DETAILED MATHEMATICAL DEVELOPMENT OF THE RELATIONSHIP BETWEEN
CROSS-MODULATION AND THE TRIPLE BEAT RATIO
1. NCTA Standard Cross-Modulation

amplitude with no interfering signals [Section IV (E)].

The NCTA Standard "Amplifier Distortion Characteristics" requires that the interfering signals be modulated
approximately 100% with a 15.75 KHz square-wave [Section IV(D)], and that modulation on the desired carrier
shall be expressed in terms of the "cross-modulation ratio,"
the ratio of the peak-to-peak variation in the amplitude of
the desired carrier with interfering signals applied, to its

For the purposes of this analysis the effect of 100% squarewave modulation is achieved by assuming that the amplitudes of the interfering signals vary periodically between
two extreme conditions. During the "ON" half of the modulation cycle the amplitudes Of the interfering signals are
given by: Bmnx = B, Cmax = C, etc. During the "OFF" half
of the modulation cycle, these amplitudes are given by:
107

Bmin = 0, Cmin = 0, etc.

This is accurate within 5% when:

When third-order distortiori exists, the output voltage at
fa during the "ON" half of the modulation cycle is found
by summing all the components at that frequency including the first order output (klA), the self-compression/expansion component (3/4 ksA3) and the crosscompression/expansion components (3/2 ksAB2, 3/2 ksAC2,
and so on for any number of interfering signals). So:

rabc
rabc
1 + - ~ 1.05, or ~ 0.05 when rabe ~ 0.1

eON = klA

+

3/4 ksA3

+

3/2 k3AB2

+

3/2 k3AC2

2

A decibel expression for this relationship can be obtained
by defining the NCTA decibel ratio:
XM
20 logloxm

=

and the relative Triple Beat (in dB) :

++

Rabe/ a = 20 loglo rabe

etc. (including the effects of compression and cross-compression) and during the "OFF" half of the modulation
cycle the amplitudes of the interfering signals are zero so
the output at fa is:

then, for equal input levels (La =. Lb = Le = L):
XM = Rabc / a + 2010g1o (n - 1)
Since Rabe/a = K3 + 2L,
XM = K3 + 2L + 20 loglo (n - 1)

+ 3/4 k3 A3

eOFF = klA

2

(including the effect of compression only).

with an accuracy of ±0.5dB (equivalent to ±5% when
Rabe / a ~ - 20dB (rabe ~ 0.1).

The NCTA "cross-modulation ratio" (xm) is defined as
the ratio of the variation on the output at fa, to its amplitude with no other inputs:

2. Cross-Modulation Measured in Terms of "m," the Conventional Modulation Factor

+ 3/4 k3A3 + 3/2 k3AB2 + 3/2 k3AC2 + + etc.) k,A + 3/4 k3A3
3/2 k3AB2 + 3/2 k3AC2 .. + .. etc.
k,A + 3/4 k3A3

(k,A

(k,A

+ 3/4 k3A3)

While cross-modulation measurements in terms of the
NCTA "cross-modulation ratio" are commonly used in
CATV, it may sometimes be convenient to express the
modulation transferred to the desired carrier in terms of
the "modulation factor" defined by:

This can be simplified by assuming equal input amplitudes
at all frequencies: (A = B = C = etc.)

. )
m = e max - em in (100m = percent mo d u IatlOn
emax + emin

where n is the total number of input signals including the one at fa.

Assuming the same 100% square-wave modulation on the
interfering signals as in section 1 of Appendix III, and
assuming that kg is positive, the maximum amplitude of
the output signal at fa occurs during the "ON" period in
the modulation cycle:

To get this into a more convenient form, divide the numerator and the denominator by k]A:

em.. =
(n-1) 3/2 kg A2
kl
xm = ------,----"----;-

1

+ ..!2 (3/2

+ 3/4 k.A3 + 3/2 k3AB2 + 3/2 k3AC2 + .. , + .. etc.

And the minimum amplitude occurs during the "OFF"
period
emin = klA + 3/4 k3A3

~
A2 )
kl

In this case "m" is given by:

In Appendix II the Triple Beat Ratio was defined as a
convenient measure of third-order distortion:
rabe = 3/2

k,A

~: A2

m = emax - emin =
emax + eml n
_
-

Substituting this in the above equation:

+ k,A + 3/4 k3A3 + 3/2 k3AB2 + 3/2 k3AC2 + .. etc.
+ k,A + 3/4 k3A3 + 3/2 k3AB2 + 3/2 k3AC2 + .. etc.
3/2 k3AB2 + 3/2 1eAC2 + ... + etc.
2k,A + 3/2 k3A3 + 3/2 k3AB2 + 3/2 kaAC2 + ... + etc.

-k,A - 3/4 k3 A3
k,A + 3/4 k3A3

(n - 1) rabc

xm=~----

1

+ rabc

If we had assumed initially that k3 was negative e max and

2

that emin would have been interchanged in the expressions
above, we would have obtained the same final result, except preceded by a - sign, indicating negative crossmodulation.

which expresses the NCTA Cross-Modulation Ratio in
terms of the Triple Beat Ratio.

r~e is usually

and

<<

1, so 1

+ r~e

The expression can be simplified by considering the special
case where all input signals have the same amplitude
(A = B = C = etc.). Then
.

= 1

xm=(n-1)ra bc

m =
108

(n - 1) 3/2 k sA3
2klA + n(3/2 k3A3)

where n is the total number
of input signals.

rna

Dividing numerator and denominator byk1A:

0- 9(3/2~A2)
2 + n ( 3/2
Since 3/2

= m within 5%

nrabe
when 1 + - - ~ 1.05,

2

~ A2 )

nrabe

or-2-~0.05

~: A2 = rabe as previously defined:

0.1
Whenrabe ~­
n

The following table shows the limiting value of rabe and
Rabe/a (= 20log101"abe) for various number of inputs:
'n
rabe
.tI.abe/a

(n - 1) rabe
m=
2 + n rabe

2
5
This relates the modulation factor to the triple-beat ratio
for any number of input signals.
Since n rabe is very much smaller than 2 in most practical
cases, it is convenient to consider the approximate relationship: .
rna

=

7
12
20

The use of the approximation rna

where rna is the approximate modulation factor.

(n - 1) rabe
2

-26.OdB
-34.0
-36.9
-41.6
-46.0

= -n2--I rabe

allows a

simple decibel expression for cross-modulation:
Let M = 20 loglOm, then:
n-l
M 20 logloma 20 loglo -2- rabe

=

To determine the degree of error involved in this approximation under various conditions, take the ratio:

=

= Rabe/a - 6 + 20log1o (n - 1)

rna _ (n - 1) rabe (2 + n rabe)
2 X (n - 1) rabe

m-

=

However for equal input levels: Rabe/a
Ks + 2L,
hence: M
Ks + 2L + 20 loglO (n -1) - 6 dB.

=

= 2 + n rabe = 1 + n rabe
2

0.050
0.020
0.014
0.008
0.005

2

109

BM- 7/74

Printed in U.S.A ... (fJ"" 0

32531



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