# Alan V. Oppenheim, Ronald W. Schafer Discrete Time Signal Processing Solution Manual (1996)

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1

Solutions - Chapter 2
Discrete-Time Signals and Systems

3

= g[nl*l
Stable: Let 1*11 $M (a) T(:z:[n]) 2.1. • then IT[*] I :;; lg[n]IM. So, it is stable if lg[n]l is bounded. • Causal: y,[n] = g[n]z 1 [n] and 112[n] = g[n]:z:2[n], so if :z:,[n] = :z:2[n] for all n < no, then y,[n] = 112[n] for all n no, so tbis is not memoryless. • Stable: IT(z[n])l :;; I:::.~ ... lz[k]l$ L::.!..~ :z:[k]M $l2no +liM for lz[n]l$ M, so it is
stable.
• Not Causal: T(:z:[n]) depends on future values of :z:[n], so it is not causal.

4

• Linear:
n+no

L

T(az1 [n] + bz2[n]) =

azt[k]

+ bz2[k]

t=n-no

n+no

=

a

L

n+no

L

:tt[k] +b

:t2[k]

= aT(z,[n]) + bT(:t 2[n])

This is linear.
• TI:

a+no

T(z[n- no]

=

L:

z[k- nol

.t=n-ne
n

=
=

L:

:t[k]

t=n-no

11!n- no]

This is TI.
• Not memoryless: The values of 11[n] depend on 2no other values of :t, not memoryless.
(d) T(:t[n])

= z[n- no]

• Stable: IT(z[n])l = [z[n- no]!~ M if [z[n] ~ M, so stable.
• Causality: If no ~ 0, this is causal, otherwise it is not causal.
• Linear:
T(az,[n]

+ bz•[n]) = az,[n- no]+ bx.[n- no]
= aT(:t,[n]) + bT(x.[n])

This is linear.
• TI: T(:t[n- nd] = :t[n- no- nd] = 11[n- n•l· This is TI.
• Not memoryless: Unless no = 0, this is not memoryless.
(e) T(:t[n]) e•l•l
• Stable: jz[n]l ~ M, ]T(x[n])l = [e•l•lj ~ el•l•ll ~eM, this is stable.
• Causal: It doesn't use future values of z[n], so it causal.
• Not linear:

=

T(az 1 [n] + b:t 2[n])

=

eu•I•J+i>z•l•l

=

eAZt(n)eb,[n]

# aT(:t 1 [n]) + bT(z.[n])
This is not linear.
• TI: T(:t[n -no]) e•l•-nol 11[n - no], so this is TI.
• Memory]ess: 11[n] depends on the
value of :t only, so it is memoryless.

=

=

=

n••

(f) T(:t[n]) az[n] + b
• Stable: IT(z[n])l = [az(n] + bl ~ t>[MI + [b(, wbicb is stable for finite a aDd b.
• Causal: This doesn't use future values of z[n], so it is causal.
• Not linear:

T(e:t 1 [n] + d:t 2 [n])
This is not linear.

=

acz 1 [n] + t>d:t•[n] + b
# eT(:t 1 [n]) + d7'(:t2(n])

5
• Tl: T(:z:(n -noll = c:z:(n - no] + b = y(n -no]. It is Tl.
• Memoryless: y(n] depends on the n'h value of :z:(n] only, so it is memoryless.
(g) T(:z:(n])

= :z:(-n]

• Stable: IT(:z:(n])l :>l:z:(-n]l $M, so it is stable. • Not causal: For n < 0, it depends on the future value of :z:(n], so it is not causal. • Linear: T(az 1 (n] + b:r 2(n]) = c:z:,[-n] + b:r2:z:[-n] = aT(z 1 (n]) + bT(z2[n]) This is linear. • Not Tl: T(z(n - no]) This is not Tl. • Not memoryless: For n memoryless. (h) T(z(n]) # = z( -n- no] # y(n- no] = z[-n +no] 0, it depends on a value of z other than the n'h value, so it is not = z[n] + u(n + 1] • Stable: IT(z(n])l$ M + 3 for n;::: -1 and IT(:z:(n])l $M for n < -I, so it is stable. • Causal: Since it doesn't use future values of z(n], it is causal. • Not linear: T(az,(n] + b:r2(n]) = a:r,[n] + b:r2(n] + 3u(n + 1] # aT(:z: 1 (n]) + bT(r2(n]) This is not linear. • Not Tl: T(:r(n- no] = :r(n -no]+ 3u(n + 1] = y(n- no] = r(n- no]+ 3u(n- no+ 1] This is not TI. • Memoryless: y(n] depends on then"' value of :r only, so this is memoryless. 2.2. For an LTI system, the output is obtained from the convolution of the input with the impulse response of the system: 00 y(n] = L h(k]z(n - k] b-oo (a) Since h[k] # 0, for (No$ n $N 1 ), N, y(n] = L h[k]z(n - k] t=No The input, r(n] # 0, for (N2$ n $N3), so r[n- k] # 0, for N2$ (n- k) ::; N 3

6
Note that the minimum value of (n - l:) is N2. Thus, the lower bound on n, which occurs for
k =No is
Using a similar argument,
N•

= N1 +N,.

Therefore, the output is nonzero for
(No +N2):,; n:,; (N, + N,).
(b) H :r[n] # 0, for some n 0 :,; n:,; (no+ N- 1), and h[n] # 0, for some n, :,; n:,; (n, + M -1), the
results of part (a) imply that the output is nonzero for:
(n0 + n,) :,; n :,; (n0 + n 1 + M + N- 2)
So the output sequence is M + N- 1 samples long. This is an important quality of the convolution
for finite length sequences as we shall see in Chapter 8.
2.3. We desire the step response

to

a system whose impulse response is
h[n]

=4-nu[-n],

The convolution sum:

for 0 < 4 < 1.

00

L

y[n] =

h[k]:r[n - k]

.k=-~

The step response results when the input is the unit step:
:r[n]

= u[n] = { 0,1•

forn2:0
for n < 0

Substitution into the convolution sum yields
00

y[n] =

L

4-•u[-k]u[n- k]

b-oo

For n :,; 0:
00

y[n]

=

I:

4-·

1=-CX)
00

=

I: 4•
b-n

=

4-n
1-4

Forn>O:
0

y[n]

=

I:

..
=
=

•=-em
r:4•
"="
1
1-4

4-·

7
2.4. The difference equation:

3

1

y(n]- 411!n- 1] + Sy[n- 2]

= 2l:(n- 1]

To solve, we take the Fourier transform of both sides.

The system function is given by:

The impulse response (for :r[n]

= 6[n]) is the inverse Fourier transform of H(eiw).
H(eiw)-8
+
8
- 1 + le-Jw
1- le-Jw
•
2

Thus,

h[n]
2.5.

= -8( 41)"u[n] + 8( 21)"u[n].

::::

y [>1)

(a) The homogeneous difference equation:

y[n]- 5y[n- 1] + 6y[n- 2]

=0

Taking the Z-transform,
1- sz-l +6z-• = 0

(1- 2z- 1 )(1- 3z- 1 ) = 0.
The homogeneous solution is of the form

ll•[n] = A 1 (2)" + A,(3)".
(b) We take the z-transform of both sides:
Y(z)[1- sz- 1 + 6z- 2 ]

= 2z- 1 X(z)

Thus, the system function is
H(z)

Y(z)

=

X(z)

=

1

=

-2
1 - 2z

Sz- 1 + 6z- 2

2

1

+::---;:--,.
1 - 3z-l '

where the region of convergence is outside the outermost pole, ~ the system is causal. Hence
the ROC is jzj > 3. Taking the uiverse z-transform, the impulse response is

h(n) = -2(2)"u(n] + 2(3)"u(n].

8
(c) Let z[n]

= u[n] (unit step), then
X(z) = 1

1
z

1

and

Y(z)

= X(z) · H(z)

=

(1- z

1 )(1-

2z ')(I- 3z- 1f

Partial fraction expansion yields

1
Y(z) = 1 -z '

4

3

-:--7=+
l-2z 1
l-3z ,.

The inverse transform yields:

y[n] = u[n] - 4(2)"u[n] + 3(3)"u[n].
2.6.

(ar The difference equation:

1
y[n] - 211[n- 1]

= z[n] + 2z[n- 1] + z[n- 2]

Taking the Fourier transform of both sides,

Y(eiw)[l- !e-jw]
2

= X(.,.iw)[l + 2e-)w +.-;:a...].

Hence, the frequency response is

=

1 + 2e-;... +.-;:a...
1- !e-iw

(b) A system with frequency response:
H(eiw)

cross multiplying,

Y(eiw)[l

+ ~·-jw + ~e-i2wJ = X(eiw)[l - ~e-iw + e-i3w],

and the inverse transform gives

1
3
1
y[n] + v[n- 1] + 11[n- 2] = z[n] - z[n- 1] + z[n - 3].
4
2
2
2.7. z[n] is periodic with period N if z[n] = :z:[n + NJ for some integer N.
(a) :z:[n] is periodic with period 12:

.;< fn) = .;< f )(n+N) =.;< fn+. .k)
~ 2?rk =iN, for integers k,N
Making k

= 1 and N = 12 shows that z[n] has period 12.

9
(b) :r[n] is periodic with period 8:

= eJ'( 'i' )(n+N) = eJ'( 'i'n+2d)

eJ'( 'i'nJ

=

3

= : N, for integers k, N

27Tk

=

= ~k,for integers k,N

N

The smallest k for which both k and N are integers are is 3, resulting in the period N being 8.
(c) :r[n]

= [sin(..n/5)]/(7Tn) is not periodic because the denominator term is linear inn.

(d) We will show that :t[n] is not periodic. Suppose that :t[n] is periodic for some period N:

=,i!?,)(n+N) =,i!?,n+2••J
= 2..-k = ~N,for integers k,N

,i!?,n)

=

N = 2v'2k, for some.integers k, N

There is no integer k for which N is an integer. Hence :r[n] is not periodic.

2.8. We take the Fourier transform of both h[n] and :t[n], and then use the fact that convolution in the time
domain is the same as multiplication in the frequency domain.
H(e'w)
Y(e"w)

5

=
=

1 +le-i""'
2

H(e"w)X(e"w)
1
5
1 +le-i~>~ 1- fe
2
3
2
1 + ~e jw + 1 -

=
=
y[n]

=

ie

1

2(3)nu[n]

jw

J·w

1

+ 3(-2)nu[n]

2.9. (a) First the frequency response:
Y(eJ'w) _ ~.-iwy(eJ'w)

6

H(.,JW)

=

=

+ .!, 0 -2iwy(e"w) = .!..-2jw X(e'w)
6

3

Y(eiw}
X(eiw)
le-2jw
3

1 - !e-iw + le

•

•

2jw

Now we take the ;,;...,rse Fourier transform to find the impulse response:

H(eJ'W)

h(n]

=

10
For the step response s[n]:
00

s[n]

=

L

=

L:

h[k]u[n - A:]

•

h[kJ

1:=-oo

=

-2

1- (1/3)•+•
1 - (1/2)n+'
1- 1/3 u[n] + 2 1- 1/2 u[n]

= (1 + (~)"- 2(~)")u[n]
(b) The homogeneous solution ll>[n] solves the difference equation when :[n] = 0. It is in the form
· ll>[n] = 2: A(c)", where the c's solve the quadrMic equation

5
6

1
6

c"--c+-=0
So for c = 1/2 and c = 1/3, the general form for the homogeneous solution is:
ll>[n]

=A,(~)"+ A,(~)"

(c) The total solution is the sum of the homogeneous and particular solutions, with the particular
solution being the impulse response found in part (a):

y[n]

=

!l>[n] + y9 [n]

=

A,(~)"+ A2 (~ )" + -2(~ )"u[n] + 2(~ )"u[n]

Now we use the constraint y[O] = y[1] = 1 to solve for A1 and A,:

y[O]
y[1]
A1 +A2
A,f2+ A,/3

= A 1 + A, - 2 + 2 = 1
= A,/2 + A,f3- 2/3 + 1 = 1
= 1
= 2/3

With A, = 2 and A,= -1 solving the simultaneous equations, we find that the impulse response
is
1
1
1
1
lf[n] = 2(-)"- (- )" + -2(- )"u[n] + 2(- )"u[n]
2
3
3
2

2.10. (a)

y[n]

=
=

h[n]• z[n]
00

L

a•u[-k- 1]u[n- A:]

lo=-oo

=

=

{ t_, ..
L a•,
.......
a•

n::; -1

lo=-oo

{

n> -1

n$-1 1-1/a' 1/a n > -1 1- 1/a' 11 (b) First, let us define u[n] = 2nu(-n- 1]. Tben, from part (a), we know that w(n] = u[n] ov(n] ={ 2'* 1 1, ' n < -1 n; _1 Now, y(n] = = = u(n - 4] • u(n] w{n- 4] { 2n-3 n$3
1,
n>3
0

(c) Given the same definitions for u(n] and w(n] from part(b), we use the fact that h(n] = 2n- 1 u(-(n1) - 1] = u(n- 1] to reduce our work:

y[n]

=
=
=
=

r(n] • h(n]
r[n]• u(n- 1]
w{n -1]
{ 2n, n$0 1, n>O (d) Again, we use u(n] and w[n] to help us. y[n] = = = = = r[n] • h[n] (u[n]- u[n- 10]) • u[n] w[n]- w[n - 10] (2n+lu[-(n + 1)] + u[n])- (2n- 9 u[-(n- 9)] + u[n- 10]) 2(n+l) - 2(n-1). n$ -2
1- 2(n-l),
-1 $n$8
{ 0,
n~9

2.11. First we re-write :r[n] as a sum of complex e:rponentials:

Since comple¥ exponentials are eigenfunctions of LTI systems,

y[n]

=

H(.J•I<).J"'i'- H(e-i•l')e-iwn/4
2j

Evaluating the frequency response at w =

±7:/4:

=

H(eif)

We get:
y(n] =

=

2../2£-i•l•ei•n/4 _ 2,.T2.J•I (iv).
(b) If the system were LTI, the output should he in the form of A(1/2)", since (1/2)" would have been
an eigenfunction of the system. Since this is not true, the system cannot be LTI. ==> (i).
(c) Given the ioformation, the system may be LTI, but does not have to be. For example, for any
input other than the given one, the system may output 0, making this system non-LTI. ==> (iii).
If it were LTI, its system function can be found by using the DTIT:

=
=
=

H(ei"')

h(n]
2.15.

Y(ei"')
X(e;"')
1
1-

le-;~o~

2

1

(2')"u(n]

(a) No. Consider the following input/outputs:

:z:,(n]
:t2(n]

=6(n]

=6(n- 1]

=(41 )"u(n]
.
1
112[n] = (4)"- 1u(n]

==> y.(n]
==>

Even though :z: 2[n] = :z:1(n- 1], 112[n] "# y1[n- 1] = (i)"- 1u[n- 1]
(b) No. Consider the input/output pair :z:2(n] and 112[n] above. :z: 2(n] = 0 for n < 1, but y2(0] # 0.
(c) Yes. Since h[n] is stable and multiplication with u[n] will not cause any sequences to become
unbounded, the entire system is stable.
1!.16. (a) The homogeneous solution y,[n] solves the difference equation when :z:(n] = 0. It is in the form
Y•(n] = 2: A(c)", where the c's solve the quadratic equation
1
1
c'- -c+- = 0

4

8

So for c = 1/2 and c = -1/4, the general form for the homogeneous solution is:

y,(n]

= A1 ( _!2 )" + A2(_ _!4 )"

(b) Taking the z-transform of both sides, we find that
Y(z)(1- .!.-•- .!z- 2 )
4
8

= 3X(z)

and therefore

H(z)

=
=
=
=

Y(z)

X(z)

3
1 - 1/4z 1 - 1/8z-2
3
(1 + 1j4z-1)(1-1/2z ')
1
2
:--~,.,-...,. + ~-;-:~...,.
1 + 1/4z 1 1 - 1/2z 1

14
The causal impulse response corresponds to assuming that the region of convergence extends outside
the outermost pole, making

=((-1/4)" + 2(1/2)")u[n]

hc[n]

The anti-causal impulse response corresponds to assuming that the region of convergence is inside
the innermost pole, making

h..e[n]

= -((-1/4)" + 2(1/2)")u[-n -1]

(c) h<{n] is absolutely SUIDDlable, while h..,[n] grows without bounds.
(d)

Y(z) = X(z)H(z)
1
1
= 1- !z-1 · (1 + lz-')(1- !z-•)
1/3
2
2/3
= 1 + 1/4z-l + 1- 1/2z-• + 1- 1/2z-l
1 1
( )"u[n] + 4(n + 1)("21 )"+lu[n + 1] + 2 (1 )"u[n]
y[n] =

32

34

2.17. (a) We have

r(n]

= { ~:

forOSnSM
otherwise

Taking the Fourier transform
R(eJw)

=

M

I>-jw•
n=O

=

1- .-jw(M+l)

(b) We have

- { i{1 +cos( 'if),

[J
wnWe note that,

Thus,
W(_,..)

O,

for05n5M
otherwise

15
(c)

IR(ei ro )I

2.18. h[n] is causal if h[n]

=0 for n < 0. Hence, (a) and (b) are causal, while (c), (d), and (e) are not.

2.19. h[n] is stable if it is absolutely summable.
(a) Not stable because h[n] goes to oo as n goes to oo.
(b) Stable, because h[n] is non...ero only for

0::; n::; 9.

(c) Stable.
_,

L: Jh!nll = L:
n

(d) Not stable. Notice tbat

00

3" = 2::<113>"

n=-oo

= 112 < oo

n=l

L:• Jsin(.-n/3)1 =2h
n=O

•

and summing Jh[n]l over all positive n therefore grows to oo.
(e) Stable. Notice that Jh[n]l is upperbounded by (3/4)1nl, which is absolutely summahle.
(f) Stable.

-51

16
2.21. For an arbitrary linear system, we have

Let x[n]

y[n]

= T{x[n]},

y[n]

= T{x[n]}

y 1 [n]

= T{x1[n]}

= 0 for all n.

For some arbitrary x 1 [n], we have

Using the linearity of the system:
T{x[n] + x,[n]}

Since x[n] is zero for all n,

=
=

T{x[n]} + T{x 1 (n]}
y[n] + !11 [n]

+ x,[n]} = T{x,[n]} = y 1 [n]

T{x[n]
Hence, y[n] must also be zero (or all n.

2.22. We use the graphical approach to compute the convolution:

y{n]

=

x{n]• h[n]
00

= L

x[k]h[n - k]

k=-oo

(a) y{n] = x{n] • h{n]

y[n]

.
(h) y[n]

= o[n- 1]• h[n] = h{n- 1]

. .rr ..
0

2

°

3

= x[n]• h{n]
5

0

~2
(c) y[n]

= x[n]• h[n]
s
4

3

IT

s s

s

4

4
l 3

3 '
2 2

t4
3
2

'I

O I 2 3 4 S 6 7 8 9 10 II 12 13 14 15 16 17 18 19 20 n

(d) y[n] = x{n] • h{n]

17

3

3

0123456

2.23. The ideal delay system:

= T{z[n]} = z[n- n

11[n]

n

0]

Using the definition of linearity:

=

T{u 1 [n] + bo: 2 [n]}

az,[n- n.J + bo:2 [n- n.]
oy 1 [n] + by2 [n]

=

So, the ideal delay system is LINEAR.
The moving average system:

L"'·

1

= Tz[n] = M

y[n]

l

+

M
2

+ 1 k=--MI

z[n- k]

by linearity;

T{a2:,[n] + bx,[n]}

=

L"''

1

M, + M, + 1

(az,[n] + bo:,[n])

lt=-Ml

l

=
=

l

lrl2

M, + M2 + 1

L

~-~

M2

az,[n] + M1 + M 2 + 1

oy1 [n] + i>y,[n]

Conclude, the moving average is LINEAR.
2.24. The response of the system to a delayed step:

Jl[n]

=

z[n] • h[n]
00

L

=

z[k]h[n - k]

1=-oo
00

L

=

u[k - 4]h[n - k]

1=-oo
00

y[n]

= L h[n -

k]

t=4

EVllluating the above Slll1lmation:
For n < 4:
For n = 4:
For n = 5:
For.n 6:
For n = 7:
For n = 8:
For n ~ 9:

=

=

y[n] 0
11[n] = h[OJ = 1
11[n] = h[1] + h[O] = 2
11[n] = h{2] + h[l] + h[O] = 3
11[n] = h[3] + h[2] + h[1] + h[OJ = 4
y[n] h[4] + h(3] + h{2] + h[l] + h[O] = 2
y[n] = h(5] + h{4] + h[3] + h(2] + h[l] + h[O]

=

L

k=-M1

=0

ln,[n]

18
2.25. The output is obtained from the convolution sum:

= o:[n]• h[n]
= L o:[k]h[n -

y[n]

00

k]

'==-co
00

L

=

o:[k]u[n-k]

1:=-CIO

The convolution may be broken into five regions over the range of n:

= O,for n < 0

y[n]
n

:La•
.,..,

=

y[n]

1 - a=<>

1_

=

1-a

N,

y[n]

=
=
y[n]

1_

~ a<'=-N2)

I:=N2
(N1+t)
4

4

(N1+1) _

1

=

1_

+

1-a

2_

, for N 1 < n < N2

n

= L alt. +
k=O

0 (N1+ll

_a

4 (n+l)

1-a
4

(n+t)

, (or N2 $N1 N1+N2 b:::O k=N2 :La• + L n::; (N1 + N2) a<•-N,) N, = :La• + L N,a"' N, = 2La• .,.., = 2· ( 1 _ a(N,+l)) _ 1 4 , forn > (N1 +N2 ) 2.26. Recall that an eigenfnnction of a system is an input sigDal which appears at the output of the systel scaled by a complex constant. 19 (a) z[n] = 5"u[n]: 00 L: = 11[n] h[k]z[n- k] k=-oo 00 }: h[k]5<•-•lu[n - k] = t=-oo n = L: 5" h[kJ5-• t=-oo Becuase the summation depends on n, z[n] is NOT AN EIGENFUNCTION. (b) z[n] = e'""'": 00 L: y[n] = h[k]ei""'<•-•1 1::=-oo 00 = eJ1wn L: h[k]e-;:z..• .t=-oo = e'""'" · H(ei""') YES, EIGENFUNCTION. (c) eiwn + ei""'•: 00 00 L y[n] = h[k]eiw(•-•1 + L h[k]ei""'(•-•1 lt=-oo = L 00 ,;wn L 00 h[k]e-;wt + eJ:Z..n h[k]e-i:Z..t Since the input cannot be extracted from the above expression, the sum of complex exponentials is NOT AN EIGENFUNCTION. (Although, separately the inputs are eigenfunctions. In general, complex exponential signals are always eigenfunctions of LTI systems.) (d) z[n] = 5": 00 y[n] = L: h[k]5<•-•l i:=-oo 00 = 5" }: h[k]s-• .11:=-oc YES, EIGENFUNCTION. (e) z[n] = s•ei""'": 00 y[n] = L: h[k]s<•-•1 ,;:z..<•-•1 b-oo = sneJ':L.In ..L: t=-oo YES, EIGENFUNCTION. h[k]s-•.-1""'• 20 2.27. • System A: 1 = (2)n :~:(n] This input is an eigenfunction of an LTI system. That is, if the system is linear, the output will be a replica of the input, scaled by a complex constant. Since y(n] (t)", System A is NOT LTL = • System B: :~:(n] =e-i•18 u[n] The Fourie< transform of :~:(n] is L 00 X(e-iw) = ~/ 8 u(n]e-;wn --00 = 00 L:e-j(w-l)n n-o = The output is y[n] 1 1- ,-;(w f). = 2:z:[n], thus Y(eiW) = 2 1- ,-;(w-f) . Therefore, the frequency response of the system is Hence, the system is a linear amplliie<. We conclude that System B is LTI, and unique. • System C: Since :~:(n] be given by = ei•/8 is an eigenfunction of an LTI system, we would expect the output to y(n] = -yei•l•, where 7 is some complex constant, if System C were indeed LTI. The given output, y(n] indicates that this is so. Hence, System C is LTI. However, it is not 11llique, since the only constraint is that 2.28. z(n] is periodic with period N if :~:(n] (a) = :~:(n + N] for some intege< N. :~:(n] is periodic with period 5: ei<'f•l = ei<'fH•+NJ = .;<'f•+•••l = Making k 2 2.. k = ; N,for integers k,N = 1 and N = 5 shows that :~:(n] has period 5. (b) :~:(n] is periodic with period 38. Since the sin function has period of 27r: z[n + 38] = sin(r(n + 38)/19} = sin(rn/19 + 21r} = z(n] = 2ei•l•, 21 (c) This is not periodic because the linear term n is not periodic. (d) This is again not periodic. e;w is periodic over period 2.-, so we have to find k, N such that :r[n + N] = .;(n+N) = .;(n+2d) Since we can make k and N integers a.t the same time, :r[n] is not periodic. 2.29. (a) (b) (c) (d) (e) e 2.30. I e ., (a) Since cos(.-n) only takes on values of +1 or -1, this transformation outputs the current value of :r[n] multiplied by eitber ±1. T(:r[n]) (-l)n:r:[n]. = • Hence, it is stable, because it doesn't change the magnitude of :t[n] and hence satisfies boundedin fbounded-<>ut stability. • It is causal, because each output depends only on the current value of :r:[n]. • It is linear. Let 111[n] T(:r:,[n]) cos(.-n):r: 1 [n], and y,[n] T(:r:,[n]) cos(lm):t,[n]. Now = • = = = T(a:r: 1 [n] + b:r:2 [n]) =cos(.-n)(a:r:,[n] + b:r:2 [n]) = ay1 [n] + i>y,[n] It is not time-inwriant. If 11[n] = T(:r[n)) = (-1)n:r:[n], then T(:r:[n- 1)) = (-1)n:r:[n - 1] f. y[n- 1]. (b) This transformation simply "samples" :r[n] at location which can be expressed as k'. • 'l'be system is stable, siDce if :r{nj is bounded, :r[n2] is also bounded. • It is not cansal. For eumple, T.o:{4] :r:[16]. • It is linear. Let llt[n] T(:,[n)) :r,(n2 j, and 112[nj = T(:r:,[n)) :r:2(n2 ]. Now = = = = =a:r:,(n2] + b:r:,(n2 ]) =ay1 (n] + t>y,[n] It is not time-invariant. If y[n] = T(:r[n]) = :r[n 2], then T(:r[n- 1]) = :r[n 2 T(a:r:1 (nj + b:r:,[n]) • 1] f. y[n- 1]. 22 (c) First notice that 00 L6[n- k] ...., =u[n] So T(x[n]) = :z:[n]u[n]. This transformation is therefore stable, causal, linear, but not timeinvariant. To see that it is not time invariant, notice that T(6[n]) = 6[n], but T(6[n + 1]) = 0. (d) T(:r[n]) = I;:.,._, :r[k] • This is not stable. For example, T(u[n]) = oo for all n i. 1. • It is not causal, since it sums /OMJJard in time. • It is linear, since 00 00 2: k=n-1 00 2: o:r,[kJ + b:z:,[kJ =a :z:,[kJ l:=n-1 .t=n-1 • It is time-invariant. Let 2: :z:,[kl + b 00 L y[n] = T(x[n]) = :r[k], .t=n-1 then T(:r[n- no]) L"" = x[k] = y[n- no] i:=n-no-1 2.31. (a) The homogeneous solution !lh[n] solves the difference equation when x[n] = 0. It is in the form y,[n] =I; A(c)", wbere the c's solve tbe quadratic equation 1 15 2 5 C+ - c - - =0 So for c = 1/3 and c = -2/5, the general form for the homogeneous solution is: Y•[n] =A,(!)"+ A,(-!)" 3 5 (b) We use the z-transform, and use different ROCs to generate the causal and anti-causal impulses responses: 1 _ 5/11 + 6/11 - (1- iz- 1 )(1 + !z ')- 1-lz ' 1 + ~z-1 H(z) _ hc[n] = : ( ~ )"u[n] + : 1 (-~ )"u[n] 1 h [n] = .. -~(!)"u[-n -1]- .!.(-!)"u[-n -1] 113 115 (c) Since h,(n] is causal, and the two exponential bases in hc[n] are both less than 1, it is absolutely summable. hoc(n] grows without bounds as n approaches -oo. (d) Y(z) = = = y[n] = X(z)H(z) 1 1 1-Jz-1 (1- !z-')(1 + iz-1) 27/20 -25/44 55/12 + + 1- 1/3z-l 1 + 2/5z-1 1 3/5z- 1 ~(!)"u[n]+ 55 (-!)''u[n]+ 27 (~)"u[n] 44 3 12 5 20 5 23 2.32. We first re-write the system function H(&w): Let lit [n] =.:[n]• g[n], then .:[n] 1m = cos(2) = et•n/2 2 G(ei•/2)&•n/2 llt[n] = + e-j'fln/2 + G(e-i•/2)e-i•n/2 2 Evaluating the frequency response at w = :!:7r /2: Therefore, and y[n] 2.33. Since H(e-iw) = ef•l•yt[n] = &f•l• cos(~n~) 2 2 = H"(e.>w), we can apply the results of Example 2.13 from the text, To find H(ei'f), we use the fact that H(eiw) is periodic over 211", so H(ei'f) Therefore, 311" 2 = H(e-ii) = &'f ~ 211" 3 11[n] = cos(-n +- + -) 2.34. (a) Notice that r[n] 4 311" 1111" = cos(-n + -12) 2 =:.o[n - 2] + 2%0 [n- 4] + :.o[n- 6] Since the system is LTI, 11[n] = 110(n - 2] + 2l!O(n - 4] + 110[n - 6], and we get sequence shown here: 2 -2 24 (b) Since !lo[n] 2.35. (a) Notice that zl[n] = -l:l:o[n + 1] + zo[n -1] = :J:o[n]• (-o[n + 1] + o[n- 1]), h[n] = -6[n + 1] + 6[n- 1] = z 2[n] + z 3[n + 4], so if T{·} is linear, T{z [n]} = T{z,[n]}+T{z3[n+4]} 1 = y,[n] + Y•[n + 4] From Fig P2.4, the above equality is not true. Hence, the system is NOT LINEAR. (b) To find the impulse response of the system, we note that 6[n] = z 3[n + 4] Therefore, = T{6[n]} = y3 [n+4] 36[n + 6].+ 26[n + 5] (c) Since the system is known to be time-invariant and not linear, we cannot use choices such as: 6[n] = zl[n]- :r2[n] and 6[n] = 1 2:r 2 [n + 1] to determine the impulse response. With the given information, we can only use shifted inputs. 2.36. (a) Suppose we form the impulse: 6[n] 1 1 = 2"'' [n]- 2:r2[n] + :r,[n] Since the system is linear, L{6[n]} • = 21 yl[n]- 21y2[n] + 113[n] A shifted impulse results when: The response to the shifted impulse Since, L{o[n]} ""L{6[n -1]} The system is NOT TIME INVARIANT. (b) An impulse may be formed: 6[n] 1 1 = 2"'' [n]- z,[n] + zs[n] 2 since the system is linear, L{6[n]} = = 1 1 2y,[n]- 2113[n] + 1/s[n] h[n] 25 from the figure, J11 (n] = -6(n + 1] + U[n] + U[n- 1] + 6(n- 3] 112[n] = -6(n + 1] + 6(n]- U(n -1]- 6(n ~ 3] lls(n] =U(n + 2] + 6(n + 1]- U(n] + U(n - 2] Combining: . h(n] = U(n + 2] + 6{n + 1]- U(n] + U[n- 1] +U(n- 2] +6[n- 3] 3 2 2 o 1 2 n -2 2.37. For an LTI system, we use the convolution equation to obtain the output: 00 y(n] = L :r(n - k]h(kJ k=-oo Let n = m+ N: 00 y(m + NJ = L = L :r(m + N- k]h(k] 00 :r((m- k) + N]h(kJ t=-oo Since :r(n] is periodic, :r(n] = :r(n + rN] for any integer r. Hence, 00 Jl(m + N] = = L :r(m - k]h(k] t=-oo y(m] So, the output must also be periodic with period N. 2.38. (a) The homogeneous solution to the second order difference equation, 3 1 y(n]- 41/(n - 1] + sllln - 2] = 2:r(n - 1], is obtained by setting the input (forcing term) to zero. 3 1 Jl(n]- 4y(n - 1] + sy(n - 2] = 0 Solving, 3 -1 1- 4-z 1 -2 + 8-z = 0• 26 (1- ~z- 1 )(1- ~z- 1 ) = 0, ll•{n] =A,(~)"+ A,(~)", 2 and the hom(lgeneous solution takes the form 4 for the constants A1 and A2 . (b) Substituting the intial conditions, 11•{-1] = A,(~)- 1 + A,(~)- 1 = 1, and Y•{O] =A, +A, = 0. We have 2A 1 +4A, =1 At +A2 =0 · Solving, A1 = -1/2 A2 = 1/2. and (c) Homogeneous equation: 1 y(n]- y(n- 1] + -y[n- 2] 4 =0 Solving, 1 - z- 1 + ~z- 2 4 = 0' 1 -1) (1- 1 z-1)( 1- 2z 2 = 0, and the homogeneous solution takes the form ll•(n) =A,(~)". Invoking the intial conditions, we have 11•(-1] = 2At = 1 =0 11•{0] =A, Evident from the above contradiction, the initial conditions cannot be met. (d) The homogeneous difference equation: 1 11[n]- y(n- 1] + y[n- 2] 4 =0 Suppose the homogeneous solution is of the form ll•{n] =A,(~)" +nB,(~)", substituting into the difference equation: 1 llh(n]- !lh{n- 1] + 411•[n- 2] = 0 A,(~)" +nB,(~)"- A 1 (~)n-l- (n -1)B1 (!)n-l 2 2 2 2 !A,(!)"- 2 +!(n-2)Bt(~)"- 2 4 2 4 2 = 0. 27 (e) Using the solution from part (d): and the initial conditions 11•[-1] = 1 and 11•[0] = 0, we solve for A1 and B 1 : A,= 0 B 1 = -1/2. 2.39. (a) For z 1 [n] =6[n], y 1 [0] = 1 111[1] = ay[O] =a For z 2[n] = 6[n- 1], y,[O] y,[1] = = 1 ay[O] + z2[1] = a+ 1 # y,[O] Even though z 2[n] = z 1 [n- 1], y,[n] # y,[n- 1]. Hence the system is NOT TIME INVARIANT. (b) A linear system has the property that T{az,[n] + bz 2[n]} = aT{z,[n]} + bT{z.[n]}} Hence, if the input is doubled, the output must also double at each value of n. Because y[O] 1, always, the system is NOT LINEAR. = (c) Let %3 = az.(n] + tlz.[n]. For n?: 0: ll>[n] • = :2:3[n] + ay3[n- 1] = az,[n] + ,8z2[n] + a(z.[n- 1] + y,[n- 2]) = a :l::>•z,[n- k] + ,8 n-1 n-1 .=0 L a•z.[n- k] 1=0 = o(h[n]• z,[n]) + ,9(h[n]• z 2 [n]) = Qlll[n] + ,8y,[n]. For n < 0: Y>[n] = .,- 1 (y,[n + 1]- z,[n]) = -a .. .. •=-1 J=-1 L: .;•z,[n- kJ- .e L: ..•.,.tn- kJ = oy1 (n] + ,8y,(n]. Forn=O: Y>[n] =Y>(n] =112[n] = 0. Conclude, Y>(n] = oy,[n] + PY.(n], for all n. Therefore, the system is LINEAR. The system is >till NOT TIME INVARIANT. 28 2.40. For the input = = :t[n] cos(wn)u[nj (-l)"u{nj, the output is y[n] "' = L (j/2)•u[k](-1)(n-llu[n- k] f:=-ao = " (-1)" I;U/2) 1 {-W 1 11=0 = " {-1)" I;H/2) 1 "c-( 1=0 = For large n, (-j/2)(n+l) -t -j /2)(n+l)) 1+j/2 {- 1) 0. Thus, the steady-state response becomes y[n] = = (-1)" 1 + j/2 cos{wn) l+j/2" 2.41. The input sequence, :t{n] "' = I: o[n + 16k}, k=-oo has the Fourier representation X (eiw) = L"' L"' .S(n + 16k]e-iwn n=- oo b:: -oo Therefore, the frequency representation of the input is also a periodic impulse train. There are 16 frequency impulses in the range -.- ~ w ~ 1r. We sketch the magnitudes of X(&w) and H(eiw): IH(e -.}!) -fl! ~ -1! 16 16 16 IX(ei (I) )I 0 1! "" 16 16 ~ 16 (I) 29 From the sketch, we observe that the LTI system is a lowpass filter which removes all but three of tbe frequency impulses. To these, it multiplies a phase factor .-;..,. The Fourier transform of the output is 1 1 .... 2"> -o(w) + -•-'" o(w- 16 16 16 1 2"> + 16e' o(w + 16 = Y(ei"') ·u Thus the output sequence is 1 1 21m 3" tl(n] =-+-cos(-+-). 16 8 16 8 2.42. (a) From the figure, = = 11[n] (z(n] + z(n]• h1 (n]) • h2(n] (z(n]• (o[n] + h1 [n])) • h2[n]. Let h(n] be the impulse response of the overall system, y(n] = z(n]• h[n]. Comparing with the above expression, (o[n] + h.(n]) • l>,(n] = h(n] = = h,[n] + h.(n]• h2[n] a"u(n] + p<•-•lu[n- 1]. (b) Taking the Fourier transform of h(n] from part (a), 00 H(ei"') = }: h[n]e-;"'" n=-oc 00 00 = }: o"u(n]e-:;"'" + P 00 L 00 0 ne-i"'" t = (n- +p L 0 (t-l)e-;'"'t, l=O n=O where we have used a)[1- ae-;"'1 = X(ei"')[1 + 13•-;"'1 taking the inverse Fourier transform, we have y[n1- ay[n- 11 =:[n1 + j3:(n- 1). (d) From part (a): h[n1 =0, for n < 0. This implies that the systtm is CAUSAL. U the system is stable, its Fourier transform existo. Therefore, the condition for stability is the same as the condition imposed on the frequency response of part (b). That is, STABLE, if Ia! 2.43. For (-1 )1..=• .. :[n)e-;"'n!w=O --oo = ,..._.., L :[n1 = (b) L = = .. 6 L :(n1e-;•n __L.., :r(nj(-1)" --00 00 = =2 < 1. 31 (c) Because z[n] is symmetric about n = 2 this signal has linear phase. X(.,;"')= A(w)e-;:z.. A(w) is a zero phase (real) function of w. Hence, LX(.,;"') = -2w, -.-$ w $.- (d) i: forn=O: X(.,;"')dw =2.-z[O] =4.- (e) Let y[n] be the unknown sequence. Then Y(e-i"') X(e-i"') = L z[n].,;"'" n L z[-n]e-i"'n n L y[n]e-i"'" = = = n Hence y[n) = z[-n). •I • o -4 -I (f) We have determined that: X(.,;"')= A(w)e-i:Z.. = = XR(ei"') "-{X(.,;"')} A(w) cos(2w) = !A(w) (.,;:z.. +e-;:z..) 2 Taking the inverse transform, we have I 1 1 1 2a[n + 2] + 2a[n- 2] = 2z[n + 4] + 2"'[n] 2 • • •-112 -4 I I . I 0 I . . I 112T •••• 4 -112 32 2.45. Let :z:[nJ = o[n), then X(ei"') =1 The output of the ideal lowpass filter: W(ei"') = X(ei"')H(ei"') = H(ei"') The multiplier: causes a shih in the frequency domain: The overall output: y[n] =e-;•nw[n] + w[n] Y(ei"') = H(eilw-•l) + H(ei"') Noting that: Y(ei"') 2.46. = 1, thus y[nJ = o[n]. (a) We first perform a partial-fraction expansion of X(ei"'): X(ei"') = :[n) = = = 1-a2 {1-"" '"')(1- at'"') 1 aei"' . + . 1- ae'"' anu(n] + a-nu[-n- 1] 1-M-'~~~ a in\ (b) 1 , [ -• X(ei"')cos(w)dw 2 = - 1 j" X(ei"') e 21< -· jw + e-jw dw 2 [ X(ei"')ei"'dw+ 2_ [ X(ei"')e-'"'dw) = !2 ( 2_ 2'1' -· 211' -· 1 = 2(:z:[n -1) + :z;{n + 1J) 4 \n-1\ + 4 \n+ll) = !( 2 2.47. (a) y[nJ h{n] = = = = :z:(nJ + 2:z:[n- 1) + :(n - 2] :z:[n) • h[nJ :[n] • {6(nJ + 26[n - 1) + o[n- 2)) 6!nJ + 26[n - 1) + .S[n - 2] (b) Yes. h[n] is finite-length and absolutely summable. 33 (c) H(~w) = 1 + 2e-jw = 2e-Jw(cos(w) + 1) + .-2jw = 2e-jw(,!_~w + 1 + .!..-jw) 2 2 (d) IH(~w)l = 2(cos(w) + 1) LH(~w) = -w p Magnitude It It -It -It -It (e) h.[n] = _.!._ 2?r = _.!._ = 2.48. 1 1 1 <2~r> H,{~w)~w dw H(e'(w+•)~wndw 2"' <211'> _.!._ 2~ H(~(w)e'(w-•l•tJw <2•> 0 -j•n...!... 1 H(~(w)~wndw = = = 2w <211'> -1"h[n] 6[n]- 26[n- 1] + 6[n- 2] s[n] = 1 +cos(=) = 1 + ( -1)" (a) Notice that S(~w) = 27f L 6(w- h) • s[n] ..ttJJ (b) Since y[n] = :[n]s[n], = _.!._ f' S(.,.)X(~(w-Bl)dw 271' -· Ol 34 = = _!._ 27r 1• S(~ )X(ei<~-•>)dw 1 -'JI' X(d~) + X(ei<~-•l) Y(eiw) contains copies of X(ei"') replicated at intervals of.-. (c) Since w[n] = y[n] + (1/2)(y[n + 1] + y[n- 1]), W(ei"') = Y(ei"') +.!. (ei"'Y(ei"') + .-;"'Y(ei"')) 2 = Y(ei"')(1 + cos(w)) (d) The following figure shows X(d"'), Y(d"'), and W(d"') for a< 2 and a> 2. Notice that X(ei"') = { 1, lwl$ wfa,
•0, wfo $lwl 2::" So, for a > 2, Y(ei"') contains two noiH>verlapping replications of X(d"'), whereas for a < 2, "aliasing" occurs. When there is aliasing, W(ei"') is not at all close to X(d"'). Hence, a must be greater than 2 for w[n] to be "close" to :r[n]. X. Y. and W 1o< a<2 2r-----------------. X.Y,andWforl>2 6r------------------, 1.5 .{1 iS 0.5 oLJ~------------~-J -2 0 2 " 3~====~~====rl 2.5 { 2 2:: - 1.5~ oL---~--------L----" .. 6.--------------------. -2 0 2 ~: oL---~~------~U---~ -2 -2 2 . 0 2 10 0~~~------------~~ -2 0 2 ., 2.49. 0~~-U~------~U---~ 0 • (a) We start by interpreting each clue. (i) The system is causal implies h[n] = 0 for n$0.

(ii) The Fourier transform is conjugate symmetric implies h[n] is real.
(iii) The DTFT of the sequence h(n + 1] is real implies h(n + 1] is even.

From the above observations, we deduce that h(n] has length 3, therefore it has finite duration.

35

(b) From part (a) we know that h[n] is length 3 with even symmetry around h[1]. Let h[O] = h[2] =a
and h[1] = b, from (iv) and using Parseval's theorem, we have
2c2 +b2

= 2.

From (v), we also have
2a -b= 0.
Solving the above equations, we get
1

= v'3
2
= v'3

h[O]
h[1]

1

= v'3

h[2]
or
h[O]
h[1]

=

h[2]

2.50.

1·

= -v'l
2
= -v'l
1

- y'l"

(a) Carrying out the convolution sum, we get the following sequence q[n]:
4

3

4

3

3

3
q[n]

1

1

1

1

n
012345678910
(b) Again carrying out the convolution sum, we get the following sequence r[n]:

3

3

r[n]

444444

44
0

1

2

n
3

4

5

6

7

8

9

10

11 12 13l14 .15 16

-12

- 16
-20

-8

-4

36
(c) Let a[n]

= v[-n] and b[n] = w[-n], then:
+oo

a[n] • b[n]

=

L

=

L

a[.I:Jb[n- .1:]

+oo

v[-k]w{k- n]

1:=-oo

+oo

=

L

=

q[-n].

v[r]w[-n- rJ where

r = -k

We thus conclude that q[-n] = v[ -n] • w[ -n].

2.51. For (-1

:r,[n- k]h[k]

••-oo
+co

L ,-iwo(•-•>:r[n- no- k]h[k]
= 1=-co
'#

y[n- no].

We thus conclude that system S is not time in..ariant.
(c) Since the magnitude of .->won is always bounded by 1 and h[n] is stable, a bounded input :t[n]
will always produce a bounded input to the stable LTI system and therefore the output y[n] will
be bounded. We thus conclude that system S is stable.
(d) We can rewrite y[n] as:

y[n]

(e-iwo•z[n])
= h[n]•
+co
L .,-;... (n-•>:r[n = r--oo

k]h[k]

+co

= L

.,-jwon.,Jwo•:z:[n- k]h[k]

k=-oo

=

e-;won

+oo

ei-•:z:[n- k]h[k].
E
•=-oo

System C should therefore be a multiplication by ,-;won.
2.57. (a) H 1 (eiw) corresponds to a &equency shifted version of H(eJw), specifically:

H,(.,Jw) = H(.,J(w-•lj.
We thus have:

39

H,(~"') = { 0 , lwl < 0.8.1 '

0.8" $lwl$

"·

This is a highpass filter.

.--

1

.....1.-.....L---.....L---.....L-.....L-...

- .. -o.s..

o.s..

0

w

1(

(b) H2(e;"') corresponds to a frequency modulated version of H(~"'), speciDcally:
H2(~"')

=

H(~"') • (o(w- O.Sw)

+ o(w + O.S1r))

where

lwl $". We thus have: lwl < 0.3lr 0.3"$ Jwl $0. 7.0.7" < lwl$ "·
This is a bandpass filter.
H.(~"')

1

•

w
-1r

-0. 7.- -0.3lr 0

0.3..

0. 7w "

(c) H 3 (e;"') corresponds to a periodic convolution of H,,(e'"') with another lowpass filter, specifically:

where H(eJ"') is given by:

H(~"')
Carrying out the convolution, -

={ 01

, Jwl < O.l?r
, O.b $lwl$ ..

get:

0.1 , lwl < O.l?r
-~ +0 • 15 ' 0.1.- $lwl$ 0.3..
0
0.3.. < lwl $... H,(~"') = { •• 40 2.58. Note that X(dw) is real, and Y(eiw) is given by: Y(eiw) = { -jX(~) +jX(e'w) w[n] = z[n] + jy[n], therefore: O[n] = - ....L ..., y[n - k] . 41 (b) By carrying out the convolution, we get: -1 h[n] = { -~ , n =0,n=2 , n =1 o.w. 2.61. The system is not stable, any bounded input that excites the zero input response will result in an unbounded output. The solution to the difference equation is given by: y[n] = y,;,[n] + y..,.[n] where y,;.[n] is the zero input response and y,.,.[n] is the zero state response, the response to zero initial conditions: y,,,[n] y..,.[n] = a(~)n where a is a constant determined by the initial condition. 1 = (l)nu[n]• z[n]. Ail example of a bounded input that results in an unbounded output is: z[n] = J[n + 1]. The output is unbounded and given by: 1 1 1 y[n] = (-t+'u[n+ 1]- -(-)n. 2 2 2 2.62. The definition of causality implies that the output of a causal LTI system may only be derived from past and present inputs. The convolution sum: 00 y[n] = 'E h!l:]z[n- 1:] k=-ao -1 = 'E 00 'E h[k]z[n- k] h[k]z[n- k] + k=-ao Note that the first summation represents a weighted sum of future values of the input. Thus, if the system is causal, _, 'E h[k]z[n- k] = 0. b-oo = 0 for n < 0. Using reverse logic, we can show that if h[n] = 0 for n < 0, This can only be guaranteed if h[k] 00 y[n] = 'E h!l:Jz[n- k]. .... Since the convolution snm specifies that the input is formed from past and present input values, the system is, by definition, causal. 2.63. The system could be LTI. A possible impulse response is: h[n] 2.64. Let the input be z[n] = o[n -1], if the system is causal then the output, 11[n], should be zero for n < 1. Let's e-valuate y[O]: 11[0] = 2~ = = 1:• Y(~)dw ..!._ ~+~ e-1'-~e-1'-~12 dw 271' -· 2 - 3w -# 0. This proves that the system is not causal 2.65. z,[n] is even-symmetric around n = 1.5, furthermore since I;zl[n] < 0 and we want A1(0) ~ 0, we need to include a .- factor in the phase. An appropriate choice for 81 (w) is therefore: 81 (w) 0 -j z2[n] is odd-symmetric around n = 3, therefore: Bo(w) ... =-3w + 2 lwl < "· e.(w) " 0 -j _, 2.66. (a) E(~"') F(~"') G(~"') Y(e'"') = = = = = = = H 1 (~"')X(~"') E(e-i"') H,(e-i"')X(e-i"') H 1 (~"')F(~"') H 1 (~"')H 1 (e-i"')X(e-i"') G(e-1"') H 1 (e-i"')H 1 (e'"')X(~"'). (b) Since: We get: (c) Taking the inverse transform of H (e'"'), we get: h[n] 2.67. = h,[-n]• h 1 [n]. (a) Using the properties of the Fourier transform and the fact that ( -1)" V(~"') W(~"') Y(ei"') H(e'"') is thns given by: (b) = = = = = X{~(w+<)) H,(~"')V(~"') H,(e'"')X(~ = for z[n] real, z[n] = z•[n], so L: ns-oo .,·[nJ·-;wn 00 x·<·-jw> = L: z[n1.-jwn = X(ei"'). Tbus, the Fourier tr~orm of a real input is conjugate symmetric. X(ei"') = XR(eiw) + jXr(~w) X•(e-;w) = XR(<-;w)- jXr(e;w) 48 From property 7, X(ei"') =X"(e-i"') for o:[n] real. Thus, XR(ei"') We may infer + jXJ(ei"') = XR(e-i"')- iXI(e-i"'). property 8: XR(ei"') = XR(e-i"') property 9: X1(ei"') = -X1 (e-i"') = iX(ei"')ieiLX(<'") X"(e-'"'l = IX(e-1"')1•-;Lx(.-;.) X(ei"') From property 7: So, property 10: property 11: IX(ei"')! LX(ei"') = IX(e-iw)l = -LX(e-i"'). 2. 7 4. Theorem 1: 00 L 00 (a:r,[n] + b:r2 [n])e-i"'n = 00 L az,[nje·iwn + n.=-oc n=-oo n=-oo L Theorem 2: = L: 00 o:[n- n·l··jwn n=:-oo = :E o:[l]e-jw(t-n,) l=-CJO 00 = ~41lft4 L: x[l]e- 1"'' l=-oo = ei"'"' X(ei"') Theorem 3: 00 L z[n]eJ"'o"'e-jwn = L 00 :t[nje·i(w-.,.)n n.=-ac Theorem 4: n.=-oo Theorem 5: L 00 n.=-OCl n:&[nje·jwn l=-oo b:r,[n]e·iwn 2. 75. The output of an LTI system is obtained by the convolution sum, 00 y[n] = L 1=-oo z[kjh[n - kj. Taking the Fourier transform, Y(eiw) = = ~f;oo 1: 1: 00 = (1: z[k)h[n- k)) e-i~ 00 Cf;oo·h[n- kje-iwn) z[kje-i"> (_too h{n- k}e-iw(n->)) z[kj 00 Hence, 2. 76. The Modulation theorem: the time-domain representation, v[n] = __!_, (' d8 (' dw X (ei 1 )W(e'(w-B>)e->w• (21r) '-· L. = 2... f' d8 X(ei1 )w[n]ei'• 21f )_ .. = z[n]w[n] 2. 77. (a) The Fourier transform of y•[-n] is Y"(e'"), and X(ei")Y(e'") forms a transform pair with z[n] • y[n]. So and g[n] = z[n] • y"[-n) form a transform pair. (b) 00 L (z[n] •11"[-nj) e-jwn --co 00 = 00 L L z[k)y"[k- n)e-i"" a=-cob-cc forn=O: 1 [ . X(ei")Y"(.,;w)dw = 221' -oc f: t=-oo z[kjy"[k) 50 (c) Using the result from part (b): :(n] y"(n] sin(rn/4) = 2m = sin(lm/6) 511'11 We recogn.i2e eacll sequence to be a pulse in the frequency domain: X(ei"') -+-------L---L---L------~~ w " _, Substituting into Eq. (P2.77-1): 00 2: --= :(n]y"(n] = 2_ (' X(ei"')Y"(e'"')dw 271" = 2_ 2.. J_w (] 2 1 = 60 2. 78. X (e'"') is given by: (a) 1/.(n] = { :(n], n even 0, nodd = ~ (1 + eiwn) :(n] 51 which transforms to (b) Y•[n] = :r[2n] Y.(~w) = ~ [x(~'l') + X(~!'!'+•ll] = Y,(~'t) (c) _{ • :r[n/2], n even [l y,n- 0, Y.(~W) nodd = X(~""') ~·~ --t/.~:.._---~-.1-.~C-----~-· i -j 2.79. 0 w (a) 00 +.(-N,-w) +;(N,w) = L :r[n-N):r'[n+N)~wn. ,..._.., .. = (L =-oo :r[n + N):r'[n- N]e-,wnr 52 00 = L (o:(n + N]:r*(n- N]e-jwn)• n=-oo 00 L = o:*(n + N]:r(n- N]e-iwn n.=-oo = +.(-N,-w). (b) 00 +.(N,w) L = Aan+Nu[n + N]Aan-Nu[n- N]e-;wn n=-oo L 00 = A2 a2ne-jvln nEN 00 L = A2 (a•e-iw)n n=N = A2 (a•e-iw)N 1- a2e-Jw 0 2Ne-iwN = A•-"--':-~ 1- c2e-iw. (c) X(e-i<•+(w/2))) = L :r(n]e-j(v+w/2)n. X*(e-i(•-(w/2))) = L :r*(n]ei(v-w/2)n_ 00 00 --co Let S = .L 2Jf J" -1f X(ei(•+(w/2H)X*(ei<•-(w/ 2 ll)ei2 •N dv ' then·. s = = = f" L -1 00 27r -· n=-oo .. L L l,.e o co :r(n]:r*(k]e-;•<•t•> 2 A=-oo.t=-oo _!_ :r*(k]e-i<•-wl2l•e-i2•N dv 1:=-oo f: f: 2.- n=-oo >=-co f" ..;<•-n+2N) dv -tr :r(n]:r*(k]e-;-';+» 2sin(r(k- n + 2N)) k - n + 2N 00 = " = L = L . o:(n]e-'<•+w/2)n ( [ .... sh-:aNl £.J z n]:r* n- 2N)e-' ' 00 :r(n + N):r*(n- N)e-iwn +s(N,w). We thus conclude that: +.(N,w) = 1 2 1f f" X(ei<•+(w/2ll)X*(ei<•-!w/2HjeJ2•N dv. -· 53 2.80. w(n] = o:[n] + y(n] The mean of w(n]: = m, E{ w[n]} E{o:[n] + y(n]} E{o:(n]} + E{y(n]} = = = m.+m, The variance of w[n]: a!, E{(w[n]- m,) 2 } = E{ w'[n]} - m! = E{(o:(n] + y(n])2 } - m! E{.,'(n]} + 2E{o:[n]l![n]} + E{y'[n]}- m~- 2m.m,- m~ = = If :z:[n] and y[n] are uncorrelated: a! = -_ . E{o:2 (n]}- m! + E{y2 [n]} - m! . a2+u2 2.81. Let e[n] he a white Doise sequeDce and E{s(n]e(m]} = 0 for all nand m. E{y[n]y(n + m]} = = E{s(n]e[n]s[n + m]e(n + m]} E{s[n]s[n + m]e[n]e(n + m]} SiDce s[n] is uncorrelated with e[n]: E{y(n]y[n + m]} = = E{ s(n]s(n + m]} E{ e(n]e[n + m]} cr!a;o[m] 2.82. (a) 1/>•• (m] = = E(o:(n]o:[n + m]) E((s(n] + e(n])(s(n + m] + e[n + m])) = E(s[n]s(n + m]) + E(e[n]e[n + m]) + E(s(n]e(n + m]) + E(e[n]s[n + m]) = 1/>,.[m] + tl>.. (m] + 2E(e[n])E(s(n]) siDce s(n] and e(n] are iDdepeDdeDt and statiOIWY· = tj>,,(m] + 1/>•• [m] where we assumed e[n] has zero mean. Taking the Fourier transform of the above equation, we get: ••• (ei"') = t.,(ei"') + ••• (ei"'). {b) tl>•• (m] = = = = E(o:[n]e(n + m]) E((s(n] + e[n])e[n + m]) E(s(n])E(e(n]) + tl>ce(m] siDce s(n] and e(n] are indepe11dmt and statiOIWY· tl>.._(m] where - assumed e(n] has zero mean. Taking the Fourier transform of the above equatiOD, - ••• (ei"') get: =••• (ei"'). 54 (c) ~. [m] = E(:z:[n]s[n + m]) = E((s[n] + e[n])s[n + m]) = ,P.,[m] + E(e[n])E(s[n]) since s[n] and e[nJ are independent and stationary. = ,P., [m] where we assumed e[n] has zero mean. Taking the Fourier transform of the above equation, we get: +.,(.;w) = +.,(.;w). 2.83. (Throughout this problem, we will usume lal < 1.) (a) IP••[m] = h[m]• h[-m]. Taking the Fourier transform, we get: +.. (.;w) = B(dw)H(e-jw) 1 1 Taking the Inverse Fourier transform, we get: alnl ,p.. [m] = -•. 1-a (b) Using part (a), we get: IB(.;"')i' = = H(.;w)B'(dw) B(dw)B(e-J"') since h[n] is real = •••(eJw) 1 1 = (1- ae->w) (1- aeJ"') 1 1 = --.< . + 1- 1oe''"'. ). 1-a 1- ae_,., - (c) Using Parseval's theorem: = L lh[n]i• L lal 2"u[n] +oc = :E<1al 2 )" ft=n[m] = = = = = = E(y[n]y[n + m]) E((z[n]- z[n- 1])(z[n + m]- z[n + m- 1])) E(z[n]z[n + mj)- E(z[n]z[n + m- 1])- E(z[n- 1]z[n + mj) +E(z[n -1]z[n +m -1]) t/>00 [m]- tl>•• [m- 1]- t/>.,[m + 1] + t/>•• [m] 21/>•• [m]- t/>•• [m- 1]- tl>•• [m + 1] 2.. [O]: tl>.. [o] = 2<1!. (c) The noise power increased by going through the first-bad:ward-dilference system. This tells us that the first backward di1ference amplilies the noise of a signal. 56 .. 2.85. (a) E{z(n]y(n]} = ..,_.., h(n]z(n - E{z(n] E .. L = .. k]} h(k]E{z(n]z(n- k]} 1:=-CXI = E h!kJ.. !kl >=-co Because z(n] is a real, statioua:y white noise process: •• (n] = a!o(n]. Therefore, = a! E{z(n]y(n]} .. L h[k]o[k] = a!h[O]. (b) The variance of the output: u! = E{(y[n]- ,.)2 } = E{y2 (n]} - m;. When a zero-mean random process is input to a determistic LTI system, the output is also zeromean: .. y(n] = z(n]• h[n] L = ..L Taking the expected value of both sides: ""' = .... = 0, z[k]h[n- k]. E{z[n]}h[n- k] ifm. = 0. So, u! = = = E{y2 [n]} {...,.i;.., h[m]z(n- m] >=f;oo h{k]z[n- k]} ..E L.. h[m]h[k]E{z[n- m]z[n- k]} .. L h[m]h[k]o[m - k] a!E a! L h'[m]. E .. .. I'JUII'-oct=-oo = ,.. • = m=-oo&=-oc m,.;:-oo 57 2.86. Using the solution to problem 2.85: (a) 00 ~ =~ :E h~[kJ this statement is TRUE, because o:[n] is a white noise sequence. (b) Since y[n] is not a white noise sequence, this statement is FALSE. (c) Let h, [n] = h2[n] = anu[n] bnu[n]. These systems are cascaded: = h[n] h1 [n] • hz[n] ...., n = :Ea•bn-•, n;<:O bn ( 1 - (a/W+') u[n] 1 -(a/b) = w[n] = o:[n] • h[n]. Since o:[n] is zero-mean, mw = 0 also. o-~ = E{ w2 [n]} 00 = 0"~ :E h•[kJ. •=<> 2.87. (a) x[n] is a stationary white noise process. 00 L y[n] = h[k]:r[n - k], n ?: 0 b-oo {.~.., h[k]o:[n - E{y[n]} = E k]} , n?:O n = L h[k]E{z[n- k]} 11:=-oo = { m, ~~-oo h[k], 0, n?:O n,.[n,, n2] = E{11[n,)11[n2]}. = E {J.., n, = h[k]z[n 1 - k] .I;.., h[m]z[n2 - m]} , no L L h[A:)h[m)E{z[n, - A:)z[n2 - m]} 1:--oo m=-oo ,., = "2 L L t=-oom=-co h[k)h[m]t/>u[n 1 - k,n2 - m]. n ?: 0 58 (c) co L = m, k=-oo h[k] .t=-oo = m.,. lim n1,ft2-oo ~ ~ ~ L.J h[k]h[m]t/>•• [n 1 - k, n2 - m] t=-oo m=-oo =~ ~ ~ L..J h[k]h[m]t/>.. [k, m]. i=-oo m=-oo (d) h[n] =anu[n] co = E{y[n]} m, L anu[n] m. = 1-a 2.88. (a) No, the system is not linear. In the expression of y[n], we have nonlinear terms such as :r2 [n] and divisions by :r[n], :r[n- 1] and :r[n + 1]. (b) Yes, the system is shift invariant. If we shift the input by no, m,[n] shifts by no as well as ~[n] and cr:!n], therefore y[n] shifts by no and the system is thus shift invariant. (c) If :r[n] is bounded, m.[n] is bounded so is u~[n] and U:!n]. ~ a result, y[n] is bounded and therefore the system is stable. (d) No, the system is not causal. Values of the output at time n depend on values ofthe input at time n +I (through u~[n] and m,[n]). Since present values of the ouput depend of future values of the input, the system cannot be causal. (e) Wben cr![n] is very large, u;[n] is zero, therefore: = y[n] m,[n] l n+l = 3 L: :r[k] i:=n-1 which is the average of the previous, present and next value of the input. When cr![n] is very small (approximately zero), then: y[n] = :r[n]. y[n] makes sense for these extreme cases, because in very small noise power, the ouput is equal to the input since the noise is negligible. On the other band, in very large noise power, the input is too noisy and so the output is an average ofthe input. 2.89. (a) E{ :r[n]:r[n]} = t/>.,[0]. (b) +n(ei"') = = X(ei"')X"(~w) = +-(ei"')IH(~)I 2 = ~ W(~W)H(~W)W"(~w)H"(~w) 1 •t-cos(w)+ 1/4" 59 (c) t/>u(n] 2.90. = t/>-(n]• h(n]• h(-n] = = cr; (G)" Grn u(n)• u(-n]) cr;,p,,[n]. (a) •• (n] = E{z(p)v(p- t]}, therefore: 00 00 = :L :L tf>.,(n] h(r)h(m]E_{z(p)v(p- (n + t- s)]} r=-oom=-oo = h[-n)• h(n)• t/>•• [n]. +,,(eiw) = JH(e'w)i"+•• (e'w). (b) No, consider z(n] white and v(n) = -z[n) tl>•• (n) = -~6(n] + •• (e'w) =-cr;. Noting that JH(eiw)l' is positive, Hence, the cross power spectrum can be negative. 2.91. (a) Since f[n] = e(n]- e(n- 1], +u(e'w) is given by: +IJ(eiw) • ..• ·- = = = = HI(e'w)H,(e-iw)+,,(e'w) (1- e-iw)(l - e'w)a~ a~(2 _ e'w _ e-iw) ~(2- 2 cos(w)) . 60 (b) 2 1 (b) = - z[-G)"u(-n-1)] (1)" 2 .-· = - L:(2zl" -1 00 L: n=-oo = 2z -1- 2z n=l 1 1 = 1- !z lzl < 2 1 (c) Z[G) n u[-nJ] = ioo (2z)" = 1! 2z I lzl < 2 (d) Z[c5[n]J = z0 = I all z (e) I• I > 0 Z[c5[n- 1]] = ,-• (f) Z[c5[n + 1]] = z+ 1 0~ lzl < oo (g) [( 21)" (u[n]- u[n- ]= ~ 1)" = 1 (2:)-l 9 • Z 10]) 3.2. :r[n] = { n :r[n] <* n u[n) <* n N, O n u[n] <* -zdz lzl > 1 dz 1-z-1 z-1 (l _ z 1 )2. lzl > 1 -N-1 z[n- no] <* therefore I• I> 0 X(z) · .-... =:> (n- N)u[n- N] <* (1z z-1)2 lzl >I 64 3.3. (a) z.(n] x.cz> = alnl -1 = 00 L a-"'z-" + La"z-" ta=-ao = 0 < lal 2 Partial fractions: one pole-+ inspection, :z:(n] = (-i)nu[n] Long division: 1 1+iz- 1 - jz-1 + ... + iz-2 1 1 + !z-1 - !z-t -jz-1 - iz-2 + lz 2 • + ~z-2 + jz-3 = :z:(n] =( -D n u(n] 1 2 66 (b) 1 1 X(z) = 1 1 + •• Partial Fractions: one pole-+ inspection, :r[n) 1•1 < 2 1 = -( -!Jnu[-n- 1] Long division: 2z 1 +1 lz2 - + 8z3 + ... 4z2 1 1 + 2z - 2z - 2z - 4z2 + 4z2 +4z' + Sz' -D => :r[n) =- ( n u[-n- 1) (c) 1- lz-1 1 2 X(z)- - 1 + ~z 1 + !z-2 X(z) :r[n) • • Partial Fractions: -3 1•1 > 2 4 1 1•1 > 2 = 1 + t• 1 + 1 + !• = [-3(-~r +4(-~rJ ..[nJ Long division: 1 + (-~- + (- 136 + 1)z-2 !l·- 1 + ... 1 1 C-l-!l•(-~- 1 + tH- !l·-· !Jz-1 H+W+!ll• => z[n) = [-3(( -n -n n+ ( 2 n-2] u[n) (d) X(z) = 1- lz-1 2 1-lz • Partial Fractions: X(z) z[n) = 1-lz-1 1- L- • 2 2 1 1•1 > 2 1 1 = "1-+--2....1 .~1 1•1> 2 = ( -~ru[nJ 67 Long division: see part (i) above. (e) X(z) Partial Fractions: = 1z - 1 -a (1Z-1 lzl > la- 11 1 X(z) =-a- a- ( 1 - a•) 1- a-lz lzl > la- 11 1 :(n] = -a6[n]- (1- a2 )a-<•+>lu[n] Long division: -a.+z-1 ·-'-•) - ( --;y-z I 1 1 + ... - a.z-1 (a- 1 - a)z 1 =3.7. -· x[n] = -a6[n]- (1- a2 )a-<•+>lu[n] (a) x[n] = .. r-n- 11 + -1 r G u[n] 1 1 2 < 1•1 < 1 X(z) = 1- z 1 + 1- lz-1 2 Now to find H(z) we simply use H(z) Y(z) H(z) = X(z) H(z) causal~ ROC =Y(z)/X(z); i.e., . (1- z- 1)(1- !z-1) -lz- 1 = (1- !z 1)(1 + z-1) -!z-' = 1- z- 1 1 + z-l 1•1 > 1. (b) Since one of the poles of X(z), which limited the ROC of X(z) to be less than 1, is cancelled by the zero of H(z), the ROC of Y(z) is the region in the z-plane that satisfi.. the remaining two constraints lzl > and lzl > 1. Hence Y(z) converg.. on 1•1 > 1. (c) ! 1 Y(z) = 1 - -lz-• 2 Therefore, y[n] I + 3 1 + z-' Iz I > 1 I = --3I(1)" -2 u[n] + -(-l)"u[n] 3 3.8. The causal system has system function 1-.-1 H (z) - --.'---:- 1 + lz-1 • and the input is :[n] = (!)" u[n] + u[-n- 1). Therefore the z-transform of the input is X (z) = 1 1 -jz-1 - 1- z-1 = -::--r---7:--;:---~ 1- jz-1 (1 ""- jz-1)(1- z-1) 1 3 < 1•1 < 1 68 (a) h[n] causal ~ u[n -1] ( 43)n u[n]- (- 3)n-l 4 h[n] = (b) _!z-1 Y(z) = _.!. = 13 1- lz-1 + 3 Therefore the output is y[n] = (c) .!. 13 1 + !z-1 • must be f < lzl which includes the unit circle. 1 + z- 1 2 H(z)--~-..,.,. - (1- ~z-')(1 + tz-') - (1- ~z-1) (a) h[n] causal~ ROC outside (b) ROC includes (c) 1•1 = 1 i < 1•1 8(-i3)n u[n] -138(1)n 3 u[n] + 13 For h[n] to be causal the ROC of H(z) · h[n] absolutely summable. 3.9. 3 = (1 - i•_ 13)( 1 + {z-') X(z)H(z) 1 (1 + tz-1) 1•1 = ~ ~ lzl > !· =>-stable. y[n] = - 1 ( - 1)" u[n]- 4(2)"u[-n- 1] 4 3 = Y(z) = X(z) = 3 -1 1 + lz- 1 + ! 3 1- 2z-1 • 1 + .-1 (1 + t• 1)(1- 2z- 1 1 ) Y(z) _ (1- jz- 1) H(z) - (1- 2z-1) 4 < I• I< 2 1•1 < 2 z[n] = -(2)"u[-n- 1] + ~(2)"- 1 u[-n] 2 (d) h[n]=2G)" u[n]3.10. (-Dn u[n] (a) z[n] = = G)" u[n _ 101 + G)" u[n _ 101 Gr Gr -[( Gr (ir) u[n] + + u[n] (u[nJ-u!n The last tern> is finite length and cooverges everywhere except at z Therefore, ROC outside largest pole < 1•1- i -nn] = 0. Therefore, 69 (b) :z:[n] = { 1, -10.$ ." .$10 0, othefWISO Finite length but has positive and negative powers at z in its X (z ). Therefore the ROC is 0 < 1•1 <(X). (c) :z:[n] :z:[-n] G) •u[n] cr· 2 = ... 2"u[-n] = C)-· 2 u[-n] X(1/z) 1 => ROC is lzl > 2 u[-n] => ROC is lzl < 2 (d) :z:[n] (D =[ •+<- (e''/ 3 )"] u[n- 1] z[n] is right-sided, so its ROC extends outward from the outermost pole e1•1 3 • But since it is non-zero at n = -1, the ROC does not include oo. So the ROC is 1 < lzl < oo. (e) :z:[n] = u[n + 10]- u[n + 5] = { 1, -10 .$ n .$-6 01 otherwise z[n] is finite-length and has only positive powers of z in its X(z). So the ROC is lzl < oo. (f) :z:[n] = (21)n-1 u[n] + (2 + 3j)"- u[-A- 1] 2 z[n] is two-sided, with two poles. Its ROC is the ring between the twn poles: ! < lzl < j.;. ! < 1•1 < j,-.'31 j, or 3.11. 00 :z:[n] causal => X(z) = L:z:[n]z-• n=O · which means this summation will include no positive powers of z. This means that the closed form of X(z) must converge at z oo, i.e., z oo must he in the ROC of X(z), or Jim,~ 00 X(z)-! oo. = = (a) Jim ( 1- .-1)2 •~oo(1- !z-1) =1 could be causal 70 (b) (z- 1) 2 lim •~oo (z -ll -co - could not be causal (c) lim (• 1)5 -. <-+oo(z- f)' =0 could be causal (d) (z- l4 ) 6 =oo <-+oo(z lim 3.12. -!)• could 110t be causal (a) 1-l.-1 Xl(z) = 1 + ~z 1 The pole is at -2, and the zero is at 1/2. (b) 1- lz-1 x.(z) = (1 + ~z-1 )3(1- jz-1) The poles are at -1/2 and 2/3, and the zero is at 1/3. Since :r2 [n] is causal, the ROC is extends from the outermost pole: 1•1 > 2/3. (c) X3(z) 1+z- 1 -2z-• = 1- •13• 1 +. 2 The poles are at 3/2 a.od 2/3, a.od the zeros are at 1 and -2. Since :r 3 [n] is absolutely summable, the ROC must include the Ullit circle: 2/3 < lzl < 3/2. 3/2 71 3.13. G(z) = = sin(z- 1 )(1 1 + 3z-2 + 2z-4 ) z-3 (z- - - z-5 z-7 + -5! - -)(1 + 3z 7! 3! -2 + 2z -4 ) = Lg[n]z-" ft g[ll] is simply the coefficient in front of z- 11 in this power series expansion of G(z ): g[llJ 3 2 -w1 + 9!11!. = 3.14. B(z) 1 = 1- !z- 2 • = (1- 1 !z- 1 )(1 + !• ') 0.5 0.5 + 1 1-lzI+lz 2 2 = Taking the inverse z. Transform: 1 1 I I h{n] = -(-)"u[n] + -(-- )"u[n] 2 2 2 2 So, I A,= 2; 0:2 1 = --; 2 3.15. Using long division, we get 1- = B(z) I -10 iOfiZ 1 - ~=-l Taking the inverse z-transform, h[nJ={ (l)", n=0,1,2, ... ,9 0, otherwise Since h[n] is 0 for n < 0, the system is causal. 3.16. (a) To determine H(z), we first find X(z) md Y(z): X(z) = 1 1- fZ 1 I 1- 2z I 1 3 < 1•1 < 2 72 5 1- !z-1 = Y(z) 5 l-iz I -Jz-1 = lzl > (1 - }z-1 )(1 - fZ I)' 2 3 Now H(z) = Y(z) X(z) = 1- ~z- 1 3 lzl > 2 3 The pole-zero plot of H(z) is plotted below. H(z) (b) Taking the inverse z..transform of H(z), we get h[n] (c) Since H(z) = Y(z) X(z) • we can write Y(z)(lwhose inverse z-transform leads = 1- ~z1- 3 z 1 , I ~z- 1 ) = X(z)(l- 2z- 1 ), to 2 y[n]- 311[n- 1] = :t[n]- 2z[n- 1] (d) The system is stable because the ROC includes the unit circle. It is also c:a.usaJ since the impulse response h[n] = 0 for n < 0. 3.17. We solve this problem by finding the system function H(z) ofthe system, and then" looking at the different impulse responses whith can result &om our thoice of the ROC. Taking the z..transform of the difference equation, we get Y(z)(l- iz- 1 + z- 2 ) =X(z)(l- z-1 ), and thus H(z) = Y(z) X(z) 1- z- 1 = 1- ~z-1 + z-2 73 1- z- 1 = = 1 )(1-iz- 1 ) (1-2z + 2/3 1 1- 2z 1/3 1- ~z lithe ROC is (a) izl < (b) i: t < izi < 2: (c) izl > 2: 1 1 = 32 2nu[n] + J(2)nu[n] h[n] = (d) izl > 2 or izi < h[O] = l. ~: 3.18. (a) H(z) = = 1 + 2'-1 + z- 2 1 )(1(1 + !z- -2+ z-•) 1+ ! ~ ~z ! 1 + ""1__...%-_..,.1 Taking the inverse 2>-transform: h[n] 8 = -2o[n] + 31 (- 21 )nu[n] + 3u[n]. (b) We use the eigenfunction property of the input: y[n] = H(~•l 2 )z[n], where = 1 -2+ 1+ 1 • 3 + 3 fe-J•/2 1- e-i•/ 2 1 = -2+-1--1 l·+-L-1. - -2j = Putting it together, -~-·· ,+f 'j) +J 74 3.19. The ROC(Y(z)) includes the intersection of ROC(H(z)) and ROC(X(z)). (a) 1 Y(z) = (1 + ~z-1)(1- tz-1) l· The intersection of ROCs of H(z) and X(z) is 1•1 > ~· So the ROC of Y(z) is 1•1 > (b) The ROC of Y(z) is exactly the .intersection of ROCs of H(z) and X(z): < 1•1 < 2. (c) ! Y(z) = (1The ROC is 3.20~ i• I 1)(1 + l• 1) 1•1 > f· 1n both cases, the ROC of H(z) has to he chosen such that ROC(Y(z)) includes the intersection of ROC(H(z)) and ROC(X(z)). (a) 1- 4~z- 1 - 1 + lz-1 H(z)- 3 The ROC is 1•1 > ~­ (b) 1 H(z) = 1- !z-1 • 3 •2 1. (a) The ROC is y[n] 1•1 > t· = y[n] = n <0 0 n " L:;:r[k}h[n- k) =I;an-t =a" A:=O .b:O N-1 y[n] = N-1 1- a- lal 1•1 > 0 n=O Therefore, 1- z-N 1 Y(z)- (1 - az-1 )(1- z-1) - (1- az-1 )(1- z 1) (1- az Now, z-N 1 }(I - z-•) 1•1 > lal So = 1- Qn+l 1 - a u[n] - 1 _ 4 n-N+l = u[n - N] n <0 0 y[n] 1- a 1-a•+l OSnSN-1 1-G 4 n+l ( 1-0-N) •-1 n?. N 3.22. (a) ~ y[n] = I: h[k]:r[n - k] = = H(z)X(z) i=-oc (b) Y(z) • y[n] = 3 1 1 + lz-1 1 z-1 3 l ! + • 1 + !z-1 1- z-1 = 1r u[n] + -u[n] 9 -3 ( -4 3 4 = ~ ( 1 + ~ (- n") u[n] = 49 ( 1- ( -3l+1) u[n] 3.23. (a) H(z) = = h[n] 1- lz-2 2 (1-lz- 1}(1-lz 1) 5+ !z-1 = -4+ 1 _ lz-1 + lz-2 2 7 -4+ 1- lz 1 1lz-1 2 • = -40[n]- 2 • • G) n u[n] + 7 (D n u[n] 76 (b) 3 1 1 y(n]- -y(n- 1] + -11[n- 2] = z(n]- -z(n- 2] 4 8 2 3.24. The plots of the sequences are shown below. (a) Let a(n] = co L 6(n - 4k], i=-oo Then co A(z) =L z-•n l=-oo (b) b(n] B(z) 1•1 > 1 z-• 1- .... •.. cu ••• ·~~~~===~~~~ ·10-4 ....... ~0 • 10 i r. n . c .. ... Z" • ... .. . ... .. l 0 n • • • • 3.25. X (z) Obtain a proper fraction: = (z - z• a)(z - b) z' = -z~2--"""(a_::.+_,.b,...)z_+__,ab 77 1 z2 - (4 + b)z + ab z2 z2 X(z) - (4 + b)z + ab (4+ b)z - ab = ( + b) z - ab 1+ a ~-~~-~ L = 1+ ~ Z-4 z[n] = = .S[n] + .S[n] + 42 _ L ..!.=!... z-b =1+ (•+>!•-•• •-• z-4 + (o+O)o--oo 1 ( 4 2z-l =1+ 4-b 1-<>z b 4n- 1 u(n- 1]- 4 •-• z-b 1 - b'z-1 ) ;-'--7-::;1-bz-l ,.. 4 _ b bn-lu[n- 1] (-1-) (4n+l - bn+l )u(n- 1] 4-b 3.26. (a) :r(n] is right-sided and 1 - !.z- 1 X(z) = 1 + Long division: 1 1+ ~z- !1 1 + igZ -2 lz ' + ... 1 + lz • Therefore, z(n] 2 = 2(-l)nu(n]- .S(n] (b) 4 !. Poles at and-~. :r(n] stable,=> Therefore, izl > ! :r(n] = 4 G) =>causal. n ..[n]- 4 ( -n n ..[n] (c) X(z) = Therefore, ln(1- 4z) 1 izl < 4 78 (d) X(z) = 1- 11 z- 3 (z( > (3) -l => causal 3 By long division: 1 + jz-3 3 1- tz1 1 - + lz-• + ... iz-3 + !z 3 3 + j.z-3 - iz-6 + lz • • = 3.27. n z(n]= { = 0,3,6, ... otherwise (a) X(z) 1 1 !z 1)2(1- 2z 1)(1- 3z 1) 2 < [z[ < 2 =(l+lz-2)2 i s +dis-~+~ (l+lz 1) (1-2z- 1) (1-3z = (I+ 1) Therefore, (-1)"+1 S8 (-1)" 1568 2700 1 z(n] = ~ (n+1) 2 u(n+l]+ ( )2 5 35 2 u(n]+ (3S) 2 (2)"u(-n-l]- (3S) 2 (3)"u(-n-l] (h) -1 X(z) = e' Therefore, z(n] z- 2 z- 3 z- 4 = 1 + z- 1 + 2! + 3! + T! + ... = n!1 u(n]. (c) 3 X(z) 2 z - 2z =- = z2 + 2z + =--=--=z-2 1-2z Therefore, z(n] 3.28. [z[ < 2 = o(n + 2] + U(n + 1]- 2(2)"u(-n- 1] (a) nz(n] ~ d -z dz X(z) z(n- no]~ z-no X(z) 3z-• X(z) = (1- tz-1)2 = 12z _2 [ d ( -z dz 1 1- tz z(n] is left-sided. Therefore, X(z) corresponds to: 1)n-2 :r(n] = -12(n- 2) ( ;( u[-n + 1] 1)] 79 (b) X(z) = sin(z) = L.. t=O (-1). (2k + 1)! ROC includes izl = 1 z'l+l Therefore, Which is stable. (c) X(z) = z7 - 2 1 7 _ z- 7 = z - _ z- 7 1 1 X(z) .. .. 11- :E = z7- L lzl > 1 z-7n n=O Therefore, :t[nJ = o[n + o[n- 7kJ n=O 3.29. X(z) = e' +e'i• z~0 0 X(z) = .. 1 .. 1 (1)" 1 .. 1 L --;z" + L--; - = L (- )··-· + L ,.-· = n=o n. n. z a=-oo n . n. 1 :t[n] = -1I'+ o[n] n . n=O n=() 3.30. 1 X(z) = log2 (- - z) 2 izl < 1 2 (a.) 1 (I) 00 (2 )' -I 1 1 X(z) =log(!- 2z) =-"-•-. = ".l..J -l -2 .l..J • =- ".l..J -(2z)-' -l s=l Therefore, :t[n] = ;; 1 t=-oo l z-' t=-oo 2 u[-n- 1] C)" (b) n:t[n] .,. n:t[n] = Gf :t[n] = ;;~cr 2 u[-n-1] 1 -z!._log(l-2z)=-z(--)(-2)=z- 1 dz 1- 2z u[-n-1] ( 1 )• 1 - !z-1 2 - 1 1•1 < 2 80 ll.31. (a) X(z) = = X(z) = :z:(n] cnu(n] + bnu(n] + enu(-n- 1] 1 1 1 + 1-az 1 1-bz-1 1-ez 1- 2ez- 1 + (be+ ac- cb)z-• (1- cz 1)(1- bz 1)(1- ez-1) lei < lbl < lei lbl < lzl < lei lbl < izl < lei Poles: a, b, c, Zeros: z1, z2, ex> where z1 and z2 are roots of numerator quadratic. pole-zero pi t {a) (b) = = :z:(n] :,[n] : 2 (n] = nz.(n] = :(n] = nzo(n] = = X(z) n 2 cnu(n] anu[n] ¢> nanu[n] X,(z) ¢> =1 X2(z) 1 az- 1 izl >a az- 1 d = -z dzX1(z) = (1-az ') 2 izi >" 1 ) d d ( czn 2 c nu[n J ¢> -z dz X2 (z) = -z dz ( 1 _ cz- 1)• -az- 1(1 + az- 1) 1•1 >" (1 - az 1)3 (c) :z:(n] Therefore, X(z) {cos ;2n) u(n]- en• {cos ~n) u[n- 1] = .n• (cos ~n) (u[n]- u(n- 1]) =o(n] = en• = 1 for alllzl. 3.32. From the pole-zero diagram 3 lzl > 4 y(n] = ~ = = :z:[-(n- 3)] z- 3 z- 1 Y (z) = z -• X (z - 1 ) = .,-,----,,..-....,..,.--,.--,..,. (z 2- z-1 + iHz-1 + ~) :z:(-n + 3] 8/3 z(2- 2z + z2)(~ + z) lzl >a 81 Poles at 0,-~,1 ± j, zeros at oo z[n] causal =0> z[-n + 3] is left-sided =0> ROC is 0 < jzj < 4/3. -4/3 3.33. From pole-zero diagram X(z) = z2 + 11 z-2 (a) y[n] 4z2+1 = (-21)n z[n] =0> Y(z) =X(2z) =2z- 2 1 zeros poles Y(z) (b) w[n] W(z) =cos(";') z[n] = ~(ei•n/ 2 + e-i•nl )z[n] 2 2 z) + !x(ei•12 z) = !x(-iz) + !x(iz) = !x(e-i•l 2 2 2 2 1(-z +1) + -21(-r+1) z -1 = 2(z 2 + 2 W(z) = -2 poles at ±ti zeros at ±1 2 1 -jz- 2 jz- 1 2 1) i 82 Y(z) 3.34. 1 2 H(z)= 3-7z- +5z- =S+-;---;;-1~ 1- 2~z '+ z-2 1 2z-1 3 1- !z-1 h[n] stable => h[n] = SD[n]- 2nu[-n- 1]- 3 G) n u[n] (a) n y[n] = h[n] • z[n] = L h[l:] n - I: 2. = -2n+l n 1 = 1- •• 1- .-• 1 z 1•1 1 zl=z-1 z- 1 - z-4 (1- z-<)(1- z- 1 ) = 1- Z = u[n- 1]- L6[n- 4- 41:] ~ .-1 I .-- .. -1-z-< bO >1 1•1 > 1 2 < 1•1 < 2 83 3.36. z[n] = u[n] ¢> X(z) = 1 1 1•1 > 1 _ .- 1 1) n-1 (1)n+l 4z 11[n] = ( u[n + 1] = 4 u[n + 1] ¢> Y(z) = _ !z 1 2 2 1 1 1•1 > 2 (a) H(z) = Y(z) = 4z(1 ~ z- 1) X(Z) 1- :;z 1 1•1 > 2 1 (b) 4z t-lz-1 2 4 t-lz-1 2 1 H(z) = h[n] = 1 4Gr+ u[n+1J-4Gru[nJ = 40[n+1J-2Gru[nJ (c) The ROC of H (z) includes lz I = 1 1•1 >-2 => stable. (d) From part (b) we see that h[n] starts at n = -1 =>not causal 3.37. 1 X(z) l 3 4 = 1-lz-1 + 1-2z-1 2 bas poles at z =! and z = 2. Since the unit circle is in the region of oonvergence X(z) and z[n] have both a causal and an anticausal part. The causal part is "outside" the pole at The anticausal part is "inside" the pole at 2, therefore, z[OJ is the sum of the two parts · !. l lz 1 1 z[O] = &-+OO lim 1 - 1 z-l +lim J=-. = - +0= z-o z - 2 3 3 2 3.38. Y(z)- - (1- z- 1 + z-2 ~z- 1 )(1 2 - -1 + lz-1) . 1- z- 1•1 > 1 84 Therefore using a contour C that lies outside of jzj y[l] = = = 3.39. = 1 we get 1 j 2(z + l)z"dz 27rj !c (z- !Hz+ ~)(z- 1) 2(i +I)(!) 2(-t + 1)(-f) H-ll 18 2 5 +6=2 2(1 + 1)(1) (1-i)(I+!l + -s-- (a) X (z) z10 = .,.-,..,..,.-....,.,,.,..:::..,-.,.,-;;-;-....-;-;--;-;(z- iHz- ~)'O(z + ~)2(z + j)(z + ll = 1. Therefore, the ROC is ! < jzj < ~= E[residues of X(z)z-• inside C], where Cis contour in ROC (say the unit circle). Stable=> ROC includes jzj (b) :r[-8] :r[8] = E [residues of (z - First order pole at z z 3) ( 1)( 3) 2 z - 2 10 (z + 2 2 z + 25 = ! is only one inside the unit circle. )( 7 ) inside unit circle] z+2 Therefore 1 :r[-8] = 3.40. 1 = 96 (a) After writing the following equalities: V(z) W(z) = = X(z)- W(z) V(z)H(z) + E(z) we solve for W(z): • H(z) W(z) 1 = l+ H(z) X(z) + 1 + H(z) E(z) (b) H1(z) - = H(z) _, ~ - l+H(z)-1+ 1 :~~~ = 1 =z-1 1- z-1 1 + 1-r' ·-· (c) H(z) is not stable due to its pole at z 3.41. (a) Yes, and =1, but H 1 (z) and H2{z) are. h[nJ is BffiO stable if its ROC includes the unit circle. Hence, tbe system is stable if rmin < 1 > 1. Tm.cz (b) Let's consider the system step by step. = = (i) First, v[n] a-•x[nJ. By t.aking the z-transform of both sides, V{z) X(oz). (ii) Second, v[n] is <ered to get w(n]. So W(z) H(z)V(z) H(z)X(az). (iii) Finally, y(n] o"w[n]. In the z-transform domain, Y(z) = W(zfo) = H(zfo)X(z). = = = In conclusion, the system is LTI, with system function G(z) = H(zfo) and g(n] = o"h[n]. 85 (c) The ROC of G(z) is ar,.,. to he stable. 3.42. < 1•1 < ar.,... We want r,.,. < 1/a and r=• > 1/a for the system - (a) h(n] is the response of the system when o:(n] = o(n]. Hence, 10 h(n] + La .... 1 h[n- .1:] = 10 and still have it compatible with the difference equation for S? Note that tbe difference equation can describe systems up to order 10. If we choose .. h[n] = (0.9)" cos( n)u[n], 4 we only need a second order difference equation: The z-transform of h(n] can be found from the z-transform table: 1 o.o H(z)- 7> - (1-0.9ei•/•z 1)(1-0.9.-J•/ 4 86 • .... 1 l L 1 -... _, ~00511.52 (b) h(n] = • "' 4 u(n- 1] (43)n u(n]- 2 (3)n-l (c) 3 y(n]- 4y(n- 1] = z(n]- 2z(n- 1] (d) The system is stable because the ROC includes the unit circle. It is also causal since h(nj = 0 for n < 0. 3.44. (a) _! X(z) The ROC is i .z = 1- 13 + I i 3 1- 2z- 1 < lzl < 2. (b) The following figure shows the pole-zero plot of Y(z). Since X(z) has poles at 0.5 and 2, the poles at 1 and -0.5 are due to H(z). Since H(z) is causal, its ROC is lzl > 1. The ROC of Y(z) must contain the intersection of the ROC of X(z) and the ROC of H(z). Hence the ROC of Y(z) is 1 < 1•1 < 2. 2 -- -u'-:_,,_-4.5,.--+o-;;...,--,,-1:-:;•-•'"' (c) H(z) = = Y(z) X(z) llz-t (1 z-l)(l+ ,.-i)(l-2z-i) (l-12z-Ij(l 2z-l) 87 = (1 + z- 1 )(1- !z-1 ) (1- z 1)(1- !z-1) -i 2 = 1+ 3 1-z 1 + -1-+_,lf'-z---:-1 2 Taking the inverse z-transform, we find 2 2 1 h[nj = 6[nj + 3u[n]- 3(-2)"u[n] (d) Since H(z) has a pole on the unit circle, the system is not stable. 3.45. (a) ny[nj dY(z) -z~ Y(z) (b) To apply the results of part (a), we let z[n] W(z) = z[n] = X(z) = -I z- 1X(z)dz = u[n- 1], and w[n] = y[n]. I I z-1 z- 1 _ z- 1 dz 1 = - = - = = - 1-1 - + - -1d z dz 1 z(z- 1) z z -1 ln(z) -ln(z- 1) 3.46. (a) Since y[n] is stable, its ROC contains the unit-circle. Hence, Y(z) converges for~ < lzl < 2. • (b) Since the ROC is a ring on the z-plane, y[n] is a two-sided sequence. (c) z[n] is stable, so its ROC contains the unit-circle. Also, it has a zero at oo so the ROC includes oo. ROC: lzl > ~· (d) Since the ROC of z[n] includes oo, X(z) contains no positive powers of z, and so z[n] = 0 for n < 0. Therefore z[n] is causal. (e) z[OJ = = = X(z)l•=~ tz- 1) A(11 (1 + ~z- )(1 0 !z 1)1·=~ (f) H(z) has zeros at -.75 and 0, and poles at 2 and oo. Its ROC is lzl < 2. 88 l I 2 - -u'-:_,---,.._.,.,..---,,f-..,.:-:--,,:--.,.,,s,---:-'2 .... (g) Since the ROC of h[n] includes 0, H(z) contains no negative powers of z, which implies that h[n] = 0 for n > 0. Therefore h[n] is anti-causal. 3.47. (a) 00 = L:;z[n]z-n X(z) n=O = X(oo) 00 lim L:;z[n]z-n •~oo = z[O] n=O Therefore, X(oo) = z[O] # 0 and finite by assumption. Thus, X(z) has neither a pole nor a :zero at z = oo. (b) Suppose X(z) has finite numbers of poles and zeros in the finite z-plane. Then the most general form for X(z) is · M oo IT<•-c•) X(z) = L:;z[n]z-n = KzL~•;:;:''---- IT (z- d•) n=O Kzl where K is a constant and M and N are finite positive integers and Lis a finite positive or negative integer representing the net number of poles (L < 0) or zeros (L > 0) at z = 0. Clearly, since X(oo) = z[O] # 0 and < oo we must have L + M = N; i.e., the total number of :zeros in the finite z-plane must equal the total number of poles in the finite z-plane. 3.48. X( ) z = P(z) Q(z) where P(z) and Q(z) are polynomials in z. Sequence is absolutely swnmable ~ROC contains 1•1 = 1 and roots of Q(z) inside 1•1 = 1. These conditions do not necessarily imply that z[n] is causal. A shift of a causal sequence would only add more zeros at z 0 to P(z ). For example, consider = X(z) z' = --, .. - . z = --~ ~ z[n] z- i 1 1•1>2 =z· = (D 1 1 1- iz-1 n+l u[n + 1] => right-sided but non-caosal. 89 3.49. o[n] + ..O(n- N) lal < 1 X(z) = = X(z) = logX(z) = log(1 + az-N) = az-N- z[n] 1 +az-N a2z-2N 2 +4 sz-3N 3 - ... Therefore, :[nJ = z: 00 (-1)"+1 k a••rn - kNJ 1<=1 3.50. (a) :t[n] =:t[-n] =* X(z) =X G) Therefore, i.e., 1/zc is also a zero of X(z). (b) :r[n] real= :r[n] = :r"[n] Therefore X(zc) i.e., 3.51. = X(z) = X"(z") = 0 = X(z,j) z0 is also a zero and by part (a) so is 1/zf,. (a) Z[:r"[n]] = n=~oo X"[n]z-n = 00 ( 00 n~oo z[n](z")-n ) " = X"(z") (b) 00 Z[:r[-n]] =L 00 z[-n]z-n n=-oo =L z[n](z- 1)-n = X(z- 1) n.=-oo (c) Z[&{z(n]}] = Z [ z[n] ~ :r"[n]] =~ [X(z) + X"(z")] (d) Z[Im{:r[n]}] = Z ["'[n] ;;"'"[n]] = ;j [X(z)- X"(z")] 3.52. z1(n) = (-Wz(n) '* X1(z) = 00 L (-1)n:t(n)z-n = X(-z) --oo The poles and zeros are rotated 180 degrees about the origin. 90 N-l 3.53. (a) -l 9• (w} _ -tan . (Zm{X(e'w})) 7U{X(eiw)} - l::r(n]sin(nw} = ~ ( } _ -;:::=<>;:=---w - N-l tanvz 2: :r(n] coo(nw) n=O N-1 N-l 2: :r(n]cos(wn) =- 2: :r(n]sin(nw} tanB.(w) N-l tanB.(w}:t(OJ + 2: :r(n](tan B.(w} cos(nw) + sin(nw)) = 0 =• N-1 1 tanB.(w,) + :r(O] ~ :r(n](tan9.(wt) cos(nwt) + sin(nw,)) for N- 1 values of w, in the range 0 (b) :r(n] < Wt < "· = 6(n] + U(n -1] + 36[n- 2] = X(z} e.(w} =0 = 1 + 2z- 1 + 3z-2 = tan-• ( -2sin(w)- 3sin(2w) ) 1 + 2 cos(w) + 3 cos(2w) Consider the values 8: ( i) = 5: tane.G) tan - e. c;) and 9s ( 23Jf) = 5 ;, which give the equations + :r;od:t(lJ(tane.G)cos~+sin~) :r(2] (tan e. G) cos .. + sin,.) J = o + :0 [:r(1] (tan B. c;) cos 2; + :t(2] (tan B. + c;) 4 cos ; 2 +sin ; ) +sin~)] = 0 1 1 + :r(O] (:r(1] · 1 + :r(2]· -1} = 0 ~ + :r;O] (:r(1] + 2~ + ~) +:t(2] c~- ~) = 0 :r(O] + :r(1]- :t(2] = 0 } -:r(O] + 2:<(1]- :r[2] = 0 = { :r(1] = 2:r(O] :t(2] = 3:r(OJ Therefore :r[n] = :r(0](6(n] + U[n- 1] + 36(n- 2]} where :r(O] is undetermmed. 3.54. :r(n] = 0 for n < 0 implies: 00 lim X(z) ~-oo 00 = .z-oo~ lim ":r(n]z-• = :r(O] + lim ":r[n]z-• = :r(O] ;~-oo ~ n.-o n.=l . 91 For the case :r:[n] = 0 for n > 0, ... 0 •lim ....o X(z) a.ss. =.-o lim "~ :r:(n]z-" =:r:(O] +lim ,....., L :r:(-n]z" = :r:(O] ,..._.., n=1 (&) ... es.(n] = 00 2: :r:[A:J:r:(n + A:] = 2: :r:(-i;]:r:(n - A:] = :r:[-n]• :r:[n] C.. (z) = X(z- 1)X(z) = X(•)X(z- 1) X(z) has ROC: rll < 1•1 < r£ ud therefore X(z- 1) has ROC: r£ 1 < 1•1 has ROC: maz{r£ 1,rll) < 1•1 < miD[rji 1,r£) (b) :r:(n] = a"u[n] is st&ble if 1<>1 < 1. In this case 1 X(z) = 1- oz-1 lol < rji 1. Therefore c•• (z) 1 < 1•1 ud X(z-1) = -1-az - Therefore c•• (z) 1 -oz-1 1 = 1- az- 1 1- az = (1- az- 1)11- a 1z-1) = 1- az 1 - l - a-1z-1 1.!.s ~ lal < 1:; < Ia- 1 I This implies th&t es.[n] = Thus, in SUIDI!l&rY, the poles 1 is lal < izl < la- 1· &re &t _!._, [a"u[n] + a-"u(-n- 11) 1-a a and a- 1 ; the zeros &re &t 0 and""; and tbe ROC of c•• (z) a (c) Cleuly, :r1[n] = :r[-n) will have the same &UtocorrelatioD function. For ex.ample, 1 X1(z)=-1-oz 1 1 1•1 < ,..- 11=- c.,•• (z) = 1 -GZ 1 -G.Z -1 (d) Also, any del&yed w:rsioll of :r[n] will have the same autoconela.tioD function; implies Xo(z) .-.. = 1 _ oz- 1 z-,. z• lol < 1•1 =- c...,(z) = 1 - ... - 1 1- QZ =c•• (z) e.c., :r:,[n] = :r[n- m] =c•• (•) 92 S.56. In order to be a z·transform, X(z) must be analytic in some aDDular recK>n afthe z-plane. To determine if X(z) = z• is a~?a~rtic we examine the existence ol X'(z) by the Canchy Riemann conditions. If X(z) = X(:r +ill)= u(.,,v) + iu(%,11) then for the derivative to exist at z, we must have In our cue, X("'+ ill)=,.- ill and thus, 8u a, =-1 unless"' and 11 are zero. Thus, X'(z) exists only at z = 0. X(z) is not aulytic anywhere. Therefore, 8u 8,. = 1 ¥ %(n] = ~if X(z)z•-Jdz does not exist. S.57. If X(z) has a pole at z = zo then A(z) can be expressed as a Taylor's series about z A(z) = A(zo) + L"" n=l where A(zo) = 0. = zo. -A•(zo) -(z- Zo)• n. 1 Thus Res (X(z) at z = zo] = X(z)(z- zo)lz=ze =· = B(z) A(z) B(z)(z- Zo) ~ A•(Zo) (z- Zo)• ....4-t, n! = B(z) A'(Zo) + I I zsze •=ze ~ A•(Zo) (z- Zo)•-1 ~ ....2 n! I•=,. = B(Zo) A'(zo) ! 93 Solutions - Chapter 4 Sampling of Continuous-Time Signals 9 4.1. %[n] = Ze(nT) = sin ( 21r(100)11 ~) = sin {in) 4.2. The discrete-time seque~>ce z[n] = coa( 4'"' ) results by samplillg the coatiDuoWI-time sigDal Ze(t) = ms(o.,t). Since w = nT aDd T = 1/1000 seconds, tbe signal &eque~>cy could be: n.. = -.4 . 1000 = 25(hr or possibly: n. = (2" + "4>. 1000 = 22so... I 4.3. (a) Since z[n] = z.(nT), ..n 3 = T = 40()/)r.nT l 12000 (b) No. For example, since cos( " n) 77f 3 = cos( 3n), T can be 7/12000. 4.4. (a) Letting T = 1/100 gives :r[n] = Ze(nT) = sin ( 2().-n = sin(";) 1 ~) + cos ( 401rn 1 ~) +cnsC;') (b) No, anotber choice is T = 11/100: z(n] 4.5. A plot of H(eJw) appears below. = Ze(nT) (201m;~) + COl ( 40.-n 1~) = sin = m(ll;")+ms(22;") = liD(";) +cos C;') 96 . • (I) (a) =0, , 101 2: 2.. · 5000 The Nyquist rate is 2 times the highest frequency. ,. T .:,.. sec. This avoids all aliasing in 10 the C/D CODverter. z,(t) = (b) 1 = 10kHz = TO1 = 10,0000< 8 T ."' o, = '· (c) = 2.- · 625rad/sec 625Hz 1 T = 20kHz ,.."' = TO1 8 o, '· = 20,000°' = 2.- · 1250rad/sec = 1250Hz 4.6. (a) The Fourier transform of the filter impulse respcmse • H,(jO) = = L 1. hc(t)e-m• dt 010 a-a.te-;nt dt 1 = a+ ;n So, we take the magnitude llfc(jO)I 1/a 0 97 (b) Sampling the filter impulse response in (a), the cliscrete-time fii,J ia described by .. hd(n] = Te-•"T u[n] = L;Te-•"T.-jwn Hd(eiw) ....., T = 1 - e-.Te Taking the magnitude of this response . IH.(e'W)I = jfll T (1- 2e-aT cos(w) + e-2oT)i . Note that the frequency response of the cliscrete-time filter is periodic, with period 2... !%<• jCD)I I -2n -4lt (c) The minimum occurs at w = ... 21t 0 47t (J) The corresponding value of the frequency response magnitude is = T (1 + 2e-•T + e-2•T) i T 1 + e aT· 11\i(e jro)l T ----------------------- 112 1/a 2Ja 3/a 4. 7. The continuous-time signal contains an attenuated replica of the original signal with a delay of.,.•. Zc(t) =Sc(t) + OSc(t- 'rd) (a) Tailing the Fourier transform of the analog signal: Xc{iO) =Sc(iO) · (1 + ae-fr•n) Note that Xc{jO) is zero for 101 > ..fT. Sampling the continuous-time signal yields the discret< time sequence, z[n]. The Fourier transform of the sequence is X{.;w) 1 ~ jw .br = T f... Sc(T+1T) .,._.., + f f: ~-00 Sc(~ + ; 2; )e-j••(f+'f. Wo ..... = 21r/3. (h) Wo/3 1( /5 37t /5. (c) Since L > M, there is 110 cbance of aliasing. Hence Wo ....... = 1t . s = ..,._..... = •. 19. The nyquist sampliDg property must be satisfied: T S 1t /0.,. 4.20. (a) The Nyquist sampliDg property must be satisfied: T S ff/0., ===> F, 2: 2000. (b) We'd have to sample 10 that X(eiw) lies between Jwl < 1t/2. So F, 2:4000. 4.21. (a) Keeping ill mmd that after sampliDg,"' = OT, the Fourier transform of :r[n] is X (ei"') (b) A straigbt-forward application of the Nyquist criterion would lead to an incorrect conclusion that the sampliDg rate is at least twice the III&Jimum frequency of :z:,(l), or 202 • However, since the spectrum is bandpass, we only need to ensure that the replicatioiiS ill &equency which occur as a result of sampliDg do not overlap with the OficiDal. (See the following figure of X,UO).) Therefore, we only need to ensure n2 - -2r T < n, = 2ft T<- .0.0 i03 (c) The block diagram alODg with the frequerocy response of h(t) is sbawn here: . ~~·· h(t) convert x(n] sequenc::e to!mpulse llaiD lD • 4.22. x(t) (a) w = !lT, ~ -It LiZ I, It CD (b) To recover simply filter out the undesired pam of X(eiw). x(n] • I -21tff I xrr Bandpass Filter rI fiT (c) T <21r - -n. 4.23. In tbe frequency domain, we have X,._(t) I 21tff -n 104 s 0 (1) = 0, 101 ~ ;, Therefore, sinCe we ue sampllng this s.(t) at the Nyquilt frequacy s(n) will be fuJI band and nnaljaoed. s[n) = z.(nT,) 1/c(t) is a band-limi1ed interpolation of z(n) at a di!"erent period. Since DO aliasing occun at z(n), the spectrum of llc(t) will be a frequency axis sc.aliDg of the spectrum of z.(t) for T, > T2 or T, < T2. As we show ill the ~. 1/c(t) = ~:"'• (~:~) 4. 24. The Fourier transform of lie ( t) is sketched below for each case. 105 (a) I -lt x5 X tal (b) I (I) -2 lt X 5 X tal 21< X 5 X tal O (c) 2 II X S X 1al O (d) (I) 4.25. (a) z,(l) = : t ~ 1. Such a I}'SUm would rault iD z[n] 6[n] &Dd 11c(t) = liDc(r/5). Now suppose we delay the iDput to be z.(t - 2). Now z[n] = 0 &Dd lf,(t) = 0. = 4.H. We can aaalyze the system iD the frequency domaiD: X(e jm) X(e2j!ll tz 1 H1(ej(l> ) yl (e j(l> ) Xce2jm IHt (e j(l> ) '2 Y,(.;o-) is X(ri")H,(ei"') dOWilS&D1pled by 2: Y1 (ei") = ~ { X(e';..f2)H, (ei"''J + X(e(2i(w-2 + H,(eil't-•l)} 4.30. X,(jO) = 0 Y(jO) = IOIX.(jO), 101 ~ 4000lr 101 !> 2000lr 1000.. ~ SiDce only half the frequency band of X,(jO) is needed, we can alias everythiDg past 0 1/3000 s. T = Now that T is set, figure out H (ei") band edges. w1 =0 w, = O,T 1T B(ei") = '* w, =2" · 500 · ..,\,. '*"-'1=-"3 2w '* w, = 2" · 1000 · ~ '*""=3 { '···I 0.- '···I < '• i• -< ,_ - T O~lwl < f,1f = o, llr(t) = In discrete-time, we want . 1 -~ z,(r)dr . 101> f = 1 H.(jO) = 1"0 = 20001r. Hence, 109 IH(e jm )I 2Km 1t arg(H(e j(l) )) (I) -2x -lt 4.32. (a) The highest frequency is •IT = 1t x 10000. ··1-------.. _____.......:_ Mlr-~--~~--~--r-~--~~--~-, i -----~ r; r: 100 1\.. 0 \ ~ _.., (b) -· -u __ -...cy,..,.._ -u ~·~ 0 G.2 '" .. . . t)0.4 (c) To filter the 60Hz out, 1 3ft ""'= T!l =- · 2.--60 =-10,000 250 4.33. ll(n] = r(n) Y(eiw) = X(~) • X(eiw) therefore, Y(~) will occupy twice the frequency bud thai X(~) does if DO U Y(ei"') "I 0, Since w = _nT, -• < w < •, tha X(~) "I 0, -j < w < j ud so X(jSl) = 0, ,. 2 ~ T · 2•(1000) T :5 a!jasjng 1 4000 Occurs. 101 ~ 2•(1000). 110 4.S4. (a) Since there is DO aliasiDg involved in this process,"" may choose T to be uy value. Choose T = 1 for simplicitf· X.(jll) 0, !Ill 2: "fT. Since Y.(jll) H.(jll)X.(jll), Y.,(jll) = 0, !Ill 2: 1t fT. Therefore, there trill be DO aliasing problems in going &om 1/e(t) to 11[n]. Recall the relat.iODShip w OT. We caD simply use this in our system CDDversioD: = = = .-,.,1• H(ei"') = H(jll) = .-;art• e-;n;z, ; T=l Note that the choice ofT ud therefore H(jll) is Dot unique. {b) COS e; ft- i) = = i(ei(\'-R-f) + .-j('fR-t)] !.-l(•/4)e/(h/2)R + !e~(•/4) 0 -;( .. /2)R 2 2 Since H(e"") is aD LTI system,"" caD fiDd the respoDSe to each oftbe two eigeDfunctioDS separately. !l(n] = ~ 0 -i(•/4) H { e-fC.. /2)) eJC .. /2)R + ~ei(•/4) H { 0 -;(5•/2)) 0 -;(5•/2)R Since H(e'"') is defiDed for 0 5 lwl 5 S1r /2 => 51f /2 - 2" "/2. Therefore, = ~·-i(•/4) H y(n] = " we must evaluate the frequeDcy at the basebaDd, i.e., {ei<5•/2)) eJ("/2)R + ~ei(•/4) H = ~ (eii<5•/2)R-(•/2)} +e-i{( .. /2)R-(•/2)})' = cos (5Jt n- ~) 2 2 . (e-1(5•/2)) 0 -j(5•/2)n . y[n) n 0 ·I ) ( 4.35. The frequeDcy respoDSe H(e'"') "' H.(jOfT). FiDdiDg that e.unJ "' UOJ 2 + !unl + 3· H(ei"') = ; 4.36. (a) Since OT "' w, (2Jt · 100)T "' ~ => T 1 (10jw) 2 + 4{10jw) +3 1 -100w' + 3 + 40jw = .:., ! 111 = iJ, there will be no aliasing. (b) The downsamp1er bas M 2. Since o:[n) is bandlimited to frequency axis simply expa~~ds by a fador of 2. For llc(t) = .:.(t) Y.(in) = x.(in). Therefore ~ 2". lOOT' ~ = •· nr * The r 4.37. h both syatems, the speech was fikered first so that the subsequent sampling result.s ill no aliasing. Therefore, going •[n) to •l[n) basically requires rb•ngi~~& the sampling rate by & fador of 31dizf5kllz = 3/5. This is cloae .mh the following syatem: I Digital=1tl31-· LPFI --.o..; ~ S I-_,.,..;. cutoff Igain"3 . 4.38. X.(iO) is drawn below. • z.(t) is sampled at sampling period T, so there is no aliasing ill z[n). .AA'A. rr -Jt • (I) Inserting L - 1 zeros between samples compresses the frequency axis. ~. -wi L The wJL · 1D filter H(ei'-') removes frequency components between"/ L and"· .I\ -~·;·) &T -.1 L The multiplication by ( -1 )• shifts the - • _,. wJL 1\ ct of the frequency band from 0 to .-. " 1. (j) 112 The D/C conversion maps the raDge to .. to the ranp - ..IT to ..fT. - .. •U9. (a) lnl > (RL -1) h[n] = 0, Therefore, for causal system delay by RL - 1 samples. (b) GeDeral interpolator condition: = = 11(0] h{kL] 1 0, k =%1, :!:2, ... (c) (RL->l y(n] L = AL-1 h[k]v[n- II:]= h(O]v(n] + t=-(RL-1) L h[n](v[n- k] + v[n + kj) ksl This requires only RL-1 multiplies, (assuming h(O] = 1.) (d) n+(ltL-1) y(n] = L v[k]h[n - A:) k-on-(ltL-1) If n = mL (m an integer), then .., don't bave any multiplications since h(O] = 1 and the other non-zero samples of v(A:] hit at tbe zeroo h(n]. Otherwise the impulse response spans 2RL - 1 samples of v{n], but only 2R of tb- are DOn-zero. Therefore, there are 2R multiplies. 4.40. Split H(e!w} into a lowpas$ and a delay.

x,

(t)

C/D

x[n)

-

tL

I

H...,<~"'

)

w[n)

e

-j.,

v[n)

tL

I

ldcal Upsampler with gain
Then we allalyze the system as follows:

:[n] = :r.(nT)
1

T

DO aliasing assumed

tu{n] = r=c

rate chan&e

T) ,
£"'< ( nrT - £

v(n] = 111(n - 1] = 1
y(n]

=

v[nL] =

L"'< (nT -

i)

delay at higher rate

~t)

113

4.41. (a) See figures below.
(b) From part(a), -

see that

Y.(jO) = X.(j(O-

~)) + X.(j(O + ~ ))

Therefore,

V\f\IJ\N
-Slt/4

4.42.

-31114 -1114

1114

31114

51114 "'

(a) The Nyquist criteriOD states that :<0 (1) can be ncooaed as long as

1
T2~ ~ 2 X 2r(250) ~ T :S 500.

In this case, T = 1/500, so the Nyquist cri1eriOD is satisfied, and :<0 (1) can be recovered.
(b) Yes. A delay ill t.ime does DOt c:bange the bandwidth ol the sipal. Hence, y.(t) has the sam
bandwidth and same Nyqw.t sampling rate as :r.(t).

(c) Consider first the follcrwing a::pressiolll far X(ei"') and Y(ei"'):

X(ei"') =
Y(ei"') =

~X.(jO) Inc, = s!ox.(jSOG-.I)
~Y.(jO) lnaf = ~.-;ll/UMMl x.(jO) Ins,

= _1_._,...,. x.(j500.1J
500
= .-;..,• X(ei"')

114
Hence, we let

B(ei") = { 2eo, -;.. •

Jwl < j

otherwise

Then, iD the following figure,

""' < j
otherwise

y[n)

x[n]

H

(d) Yes, &om our analysis abow,

H2(ei") = .-jw/2
4.43.

(a) Notice tim that

X.(jO) =

{

F.(jO)IB..,(iO)Ie-;0', 101 ~ 4001r
E.(iO)JB..,(iO)Ie-;n', 4001r ~ 101 :S 8001r
0,

otherwise

For the given T = 1/800, there is no aliasing &om the C/D conversion. Hence, the equivalent CT
transfer function H.(jO) can be written as

B (jO)

= { B(ei"')l.,=nT.

c

FUrthermore, since Y.(iO)

0,

101 ~ wfT
otherwtSe

= H.(jO)X.(iO), the desired tranfer function is

Combining the two previous equatioiiS, we find

H(ei") = {

ei(OOO...)',

0,

lwl :S 1r /2
lwl :S r

7
- 400lr

= -T->
2..-

1200lt

The miDimum • il 1200lt. F« dlia choice, we get
H(ei'") = {

t1 S •

~l-l', ~/i ~·

U4. (a) See the following figure:

X(e

i'l)

(I)

(b) For this to be true, H(ei"') Deeds to filter out X(ei"') for •/3
Furthermore, we WUlt

•/2
T
2

:5 lwl :5 r. Hence let wo

= r/3.

=2..-(1000) =- T2 = 1/6000

(c) MatchiDg the followiDgligure of S(ei"') with the figure for R.,(jO), and remembering that 0 = wfT,
we get T3
(2r/3)/(2000r) 1/3000.

=

=

S(e.il l
YJ[n].,, f 3 1----+l•l E t~~l
y,[n].,i f 3 1----+l•i Ejz)l t
Yo[n]•!

(c) Yes, z[n] caD be recoDStrncted from 112[n] and W•[n]as demonstrated by the following figure:
x[n]

H 3 (z)

H 4 (z)

wJn]

w.[n)

~2

'2

y. (nl
.

y, (n)

t2

v, [n)

t2

•. [n]

•In1

H 4 (z)

s(n]

117

hi \he following discussion, It\ "'• [n] deno\e the even samples
samples ot :z:(n]:

= { 0,:z:(n],
hi the fignre, .,[n]

n

of :z:[n],

and z.(n] denote the odd

eftll

n odd

~inJ,

n even

=

{

_

{ :z:[n], n eftll
0,
nodd

nodd

,.[n] -

= :z:.[n]
Furthermore, i\ can be verified using \be IDFT \bat the impulse respoiiSe /4(n] corresponding w
H,(ei"') is
h.(n] = { -2/(jrn), n odd .
o\hennse
0,
Notice in particular that every otner sample of the impulse respoiiSe /4[n] is zero. Also, from the
form of H,(ei"'), it is clear that H,(e'"')H,(e'"') = 1 a.nd hence h,[n] • /4[n] 6[n].
Therefore,

=

v,[n]

n even
n odd

=

{ ll<(n/2],

=

{ tu4 (n],

=
=

{ (:r: • h.)[n], neven
0,
nodd

0,

0,

n even
n odd

:z:0 (n] • h.[n]

where the last equali~ follows from the fact that h.[n] is non-zero only in the odd samples.
Now, s[n] = v,[n]•h.[n] = :z:0 (n]•h.(n]•h.[n] = .rrlnl. aDd since :[n] = :.[n]+z.[n], •[n]+v3[n]

:[n].
4.47. Sampling random processes
¢ •••• (r)

= E(:z:,(t)z;(t + r)).,. P•••• (n) =

L:

¢ •••• (r)e-;'" dT

(a)

¢ .. [m]

= E(z[n]:r:"[n + m]) = E(:z:,(nT)z;(nT + mT))

=

¢ •••• (mT),

i.e., sampled antocovariance

(b) Since ¢u(m] is a sampled ¢ •••• (T)

P..(w)

=T1

~
2rk)
L.J P.... ("'
T+T

K=-oo

(c) If

P•••• = 0, for
then

P.. (w) =

lwl 2: r

~P•••• {f),

lwl

$r = 118 .(.48. {a) ..... (.,.) = E(z.(t)z.(t + .,.) = 4>•• (m) = E(z.(nT)z.(nT + mT)) E(z(n]z(n + m)) = 4>.,.,(mT) (b) 2Tr) -L 1 ... P.... ("' P..(w) = T T+T no. Therefore, we require that " ~ (e) For the JpeCtrum of F"JC P3.&-2 it is dear \hat if T will be wbi1e, as shown ill the figure above. = II; then the discrete-time power spectrum pxx ( m) I I ' I ' I I I I I '' I I ') I I ' I -411: I '' I -211: I I 'v I I ' '' '' I I I I '< I I I I '< '' '' I ' ' I '' ' 211: I I ' '' ' It '' (I) = (d) For white discrete-time signal:> 4>,.[m) = 0, m # 0 but 4>•• [m] 4>•••• (mT). Therefore, auy analog signal whose autocorrelation func:t.iou bas zeros equally spaced at illtenals of T will yield a white discrete-time leqllellce is sampled with sampling period T. For example, for Fig P3.&-l: 4>.,.,(.,.) ifT= 4.49. sill CloT =--:;;-- * ~ sill ClomT 4>•• (m) = rmT sill'"" ••• [m] = rm/0. = o, (a) Coasider the followillg plots. S)'Sial>l: SySiem2: -~ . /K. ;:vi\%\. - • m~ -&tm m 'b • 119 Jls(t) = !12(1): Convolution is a linear process. Aliasing is a linear process. Periodic convolution is equivalent to convolution followed by aliasing. This is clearly not YsUO). YsUO) is an aliased lls(t) 'I r(t): System 2 at S&ep 1 shows version of x.UO) (b) Now, x;uo). -1$(211)

.1t

'b

1$(211) 0 30(2>1) il -10(2Jr) 0 10(2Jr) b (c) :(t) = A cos(30lrt) 3 1 :3(1) = 4Acos(30.-t) + Acos(3. 30.-t), 4 v[n] = ~A cos (~.-n) +~A cos (~.-n) v[n] = :3[n] Jl[n] = z(n] We can see here that sometimes aliasing won't be destructive. When aliased sections do not overlap they can be reconstruc:1ed. (d) This is the inverse to part (c). Since multiplication in time corresponds to convolution in frequency, a signalz2 (t) has at most two times the bandwidth of :(t). Therefore, : 112 will have at least the bandwidth of z(t). If we run our signal through a box that will raise it to the 1/M power, then the sampling rate can be decreased by a factor of M. l 4.50. (a) :,[n) = :.(n) • h..h(n] h..,.[n] = { 1, 0 ~ n ~ L - 1 H (ei"') .,. = sin(wL/2) -i!L-1>-o/2 sin(w/2) e 0, else (b) The impuloe response h,.,. (n] corresponds to the convolution of two rectangular sequences, as shown below. A' • -L L •n II I = IlL· • - ..I.:l 2 ..I.:l 2 II I . *. n - ..I.:l 2 ..I.:l 2 •n 120 H (ei"') _ .!_ (siD(wL/2)) •- L sill(w/2) (c) The frequency respoue of -...order-hold is out- Tbe baDd...;dth of+ •• (.,;"') i.s no larger than the baDd.,.;dth of X(ei"'). Therefore, the outputs of the sys\ems will be the same if H2(ei"') is an ideallowpau 6her ...;th a c:utolf of wf L. 4.52. The idea here i.s to uploit the fact that every other sample supplied to h(n] ill Fig 3.27-1 is zero. That is, 111 (n] = h[n] • w[n] = __L...., w[n - k]h[k] lltll(n] + bou(n- 1] + cw(n- 2] + dw(n- 3] + ew[n- 4] az(n/2] + <=[(n/2)- 1] + e((n/2)- 2], n even O.:((n/2) - (1/2)] + dz((n/2) - (3/2)], n odd = ={ w 1 (n] = { ht(n/2] • z(n/2], 0, n ewn n odd = { h 1 (0]z(n/2] + h,[1]%{(n/2) -1] + h 1 (2j%((n/2)- 2], newn 0, w,(n] = = n odd h,(n/2] • z[n/2], n even { 0, "odd { h2(0]z(n/2] + h2(1]:r[(n/2)- 1] + h,[2]:r[(nf2) - 2], n even 0, nodd 121 Comparing w,[n], tuo[n) with Jl[n) above: w[n) tall give even samples if h 1 [0) = o, h 1 [1) = c,h,[2) =e. Similacly, w,(n] if h,(n) deLays tuo(n) by ODe sample, i.e., h 3 (0) = O,h3 (1) = O,h,[2) = 0. Thus tuo(n) tall give the odd samples = { h,(O)z[(n- 1)/2) + hz[1)z((n- 1)/2- 1) + hz(2]z((n- 1)/2- 2), ~ h 2 (0) =b, 11,(1) =d, • 1C/2 n eveo n~ 11,[2) = 0 4.53. Sketches appear below. (a) F"ust, X(ei"') is plotted. The lowpass filter cuts otr at j. ftl2 I (I) The downsampler expands the &equocy axis. Since Ro(eiw) is bandlimited to if, no aliasing occurs. ~ It It (I) The upsampler compresses the frequocy axis by a factor of 2. .~ -It It (I) The lowpass filter cuts of at j ~ Yo(eiw) = Ro(ei"') as sketched above. (b) Go(ei"') = l (X(eJW)B0 (ei"') + X(en"'+•l)B0 (en-•l)) 122 (c) Yo(.,;w) == ~Ho(.,;w) ( X(.,;w)Ho(.,;w) + X(eA-"~.,;r..-+"'1}) Y,(.,;w) == ~H,(.,;w) (.x(.,;w)H,(.,;w) +X(~)B.~)) Y(.,;w) == Yo(.,;w)- Y,(.,;w) == ~x(.,;wl [.Hl<.,;wl- S:<.,;wl] +~X(ei<-l) [s.,(.,;w}Ho(ei<.-t,_...,...,..{ei"'->)] The ali Mill& terms always c:aDcei. Y(ei"') Is propor&ioDaJ • X(«>") if il4(e"'}- JP.(.,...)\ is a COI!St&llt. C1111 be~ ./2 2 e l>./2 ede = ~L"' 12 2 = 1:;.1 1"''2 e de= -1>/2 = E(e[m)e[n]) = { e' 3 1>/2 1:;. L~>.; 2 =0 1:;. 2 =12 E(e[m])E(e[n]), m E(e2 [n]l, m ¥- n =n t;.2 r[n,m) =r[n- m) =l26[n- m) (b) SNR = ..-! =1~2 2..-! C7~ (c) Let e,[n) be~. ou'put noise. e,[n) = L h[k)e[n • k) The variance oC :[n) is .,..;ghted similarly so ~e SNR does no' change. SNRout =12~.• 125 (d) ffn] = :[n]e[n] E(/[n]) = E(z[n)e[n]) = E(z[n])E(e[n]) = 0 "' =E(/2 [n]) =E(:2 [n)e2 (n]) =E(:2 [nj)E(e2 [n]) =11!11! rr[n, m) =E(z(n):[m)e[n)e[m)) = E(z[n):[m)) · E(e[n)e(m)) (e) (f) Using ~e results of pan (c). .. 2 "c.- Again,~ ariaDce 2 ( I 1 ) --~.r. -I-a -1-o 4 4 ol :[n] is weighted by the same factor, so the SNR does not change. SNRout = 12 32· 4.58. Fim, notice that since 1/e(t) = : 1 (t):,(t), Ye(iO) = /,;(X 1 (j0) • X,(jO)), and so Ye(jO) = 0 for JOJ;::: 1l1r/2 x 10'. Hence the Nyquist rate T = 1/55000s. Choose System A and B such that w1[n) = a:r 1 (nT) ud w,[n] = hzz(nT). For System A, we need to resample such that M T T = T, = x, [n) 2 10-· 10 1/55000 = li X ·I I ·I I ·I I m n L =II file= IIIli M = 10 ! I [n) For System B, we need to resample such that M T 2 x 10-< 1 T = T, = 1/55000 = 11 System C is simply the identity system. 4.59. The speech is first sampled at 44.1 klh, and we wish to resample it so that ~e samplillg rate is at S kHz. There are 110 •li asing effects aaywhere in the sys~em. Hmce L 44.1 441 -=--M 8 80 We simply make L = 441, M =80, and "'• = "/441. 126 4.60. n,, and n, has to be chosen such that (a) The region lfll :$; n,

maps to

lwl

:$; fr/4: ., fl,T=4"==>0,=441r (b) No aJiasiDg ocaus ill the region l!ll :$; n, darillc sampliDc:

21r

T -n, = n, = n, = 2r(4·44l -44r = 308r

4.61. (a)
V(z)
U(z)
Y(z)

= H 1 (z)(X(z)- Y(z))

=

H,(z)(V(z)- Y(z))
= U(z) + E(z)

=
Substituting H 1 (z)

H 1 (z)H2 (z)
X
1
E( )
1 + H,(z)(1 + H,(z)) (z) + 1 + H 2 (z)(1 + H,(z)) z

=1/(1- z-

1)

and H2 (z)

=z- /(1- z1

1 ),

we fiDd

H.,(z) = .. -•
H.,(z) = (1- z- 1 ) 2

Hence the difference equation is y[n] = :r[n - 1] + f[n], where
f[n] = e[n]- 2e[n- 1] + e[n - 2].
(b)

P11 (ei")

= .,:IH,,(ei"JI'2

= ..:1(1- .-;...) 12

=
=
=
=
The total noise power

cry

..:(1- .-;")'(1 - e'")'
.,:(2- 2cos(w)) 2
o-:(4 sin2 (w/2)) 2
16o-! sin4 (w/2)

is the autocorrelation of f[n] evaluated at 0:

o-} = E[(e[n]- 2e[n- 1] +e[n- 2]) 2 ]

= E[e2 [n]] + E[-2e 2[n- 1]] + E[e[n- 2]']
= fia!,
where we have used linearity of expectatioDS, and the fact that sincu[n] is white, E[e[n]e[n-.1:]]
for k ;,! 0.

ee
~--~~~~----·
-1t

K Cll

=0

127

(c) Since X(ei") is bancllimited, o:[n]• loo[n] = o:[n]. Bence,
w[n] = y[n]• ha[n] = (o:[n- 1] + /[n]) • h3[n] = o:[n- 1] + g[n],
where g[n] is the quantization noise in the region

lwl < 1l/M.

(d) For a small angle o:, lin:"' o:. Therefore,

u:

1•'.,,.,"'
2 1•'"'

= b1

"'

1

,.. w/11

,.:(21inwf2) 4 d..>
J'!(2w/2)• ..,

~"'"I·'"'
-•1"'

= 2.. 5

-·-

= ~"·
sM•

=

(e) X,(;O) must be suflic:iently haDdlimited that X(ei")
X,(;OT) is zero for
= for 101 > "/MT.
Assuming that is satisfied, v.[n] = o:[Mn- 1] = :,(MTn- T).
Downsampling does Dot change the variance of the noise, and hence

x.unl o

a! =a!·

=

P11 (ei" 1"')
= 1oo~ sin4 (wf2M)

P.. (ei")

I

It Ill

4.62. (a) (i) The transfer function from o:[n] to y.[n] is
H sr (z ) =

.-•

r::?f

1+

z-t

1-.r-1

= Z -1

Hence y.[n] = o:[n- 1].
(ii) The transfer function from e[n] to lf,[n] is

H,,(z)

=

1
1+

_,

,!.-t

= 1-z-1

So

P,.(w) = P,(w)H,,(ei")H.,e-;w
= ,.:(1- .-;..)(1- ei")
= ,.:(2- 2cos(w))

(h) (i) o:[n] contributes cmly to 111 [n], but DOt Wz[n]. Therefore

= o:[n - 1]
r,[n] = o:[n- 2}

ll~a[n]

lwl > tr/M.

Hence

128
(ii) In pan(a), the cillference equation desaibing the sigma-delta noise-shaper is

y(nJ

=:(n- 1] + e[nJ- e[n- 1].

So here we apply the diJference equation to both &igm •-delta modulators:
lllc(nJ = e1[nJ- e1(n- 1]
,_[nJ = e1(n- 1] + e2[n]- e2[n- 1]
r 0 [nJ = lllc[n- 1]- (,_[nJ- "-["- 1])
= -e2[nJ + 2e2(n- 1]- e.[n- 2]
H ...(z) = -(1- z-1)2
P•• (w) = cr!(2- 2cosw)2

!

129

Solutions - Chapter 5
Transform Analysis of Linear Time-Invariant Systems

131

5.1.

_{

11[n 1-

1, 0 $n$ 10,

0, otherwise

Therefore,
Y(ei"') =

.-J.., ~sm•t"'

This Y(ei"') is full band. Therefore, since Y(~") = X(~")H(~"), the Oll!y possible z}n) ud "'" that
could produce 11(n] is z[n] = lf[n] &Dd "'• = 11.

5.2. We haft 11[n- 1] -Jflf[n] + 11[n + 1]

=z[n] or z- Y(z)- !IY(z) + zY(z) =X(z). So,
1

1

=

H(z)

z-1

-If+ z
z

= <•- !H•- 3)
_!
!
= __J_+_L.

•-!

z-3

(a)

1 zeroatz=-

Re

3

• (b)

Stable

"* ROC is l $1•1$ 3. Therefore,
h[n]

= --81(1)ft-l
-3 u[n- 1]- 9
-(3)"- u[-n]
8
1

5.3.
1

11[n- 1] + 311[n- 2] = z[n]
z- 1 Y(z)

H(z)

+ ~z-2 Y(z) =
Y(z)

X(z)

1

= X(z) =
z-1 + l·-·
%
H(z) =
1 + !z-1

132

f < lzl, h[n] = (-l)n+ u{n + 1] ~ uswer (a)
ii) ! > 1•1.
i)

1

h{n] = - (

-n

n+l u[-n- 2]

=

-(-D(-~fu[-n-2]

=

3 - 3 u[-n- 2] ~ aaswer (d)

1( 1).

5.4. (a)

z(n] = Gf u(n] + (2).u(-n -1]
1
z
X(z)= 1-lz-1- z-2'

1

2 4

•

4
u(n- 1)
(3)n u(n]- 2 (3)•1

h(n] = 4
(c)

H(z) - Y(z) - 1 - 2z-1
- X(z) - 1- {z-1
Y(z)-

~z- 1 Y(z) = X(z)- 2z-1X(z)

11[n] - ~11[n - 1) =z(n] - 2z[n - 1]
4

3

1•1 > 4

i33

i•

(d) The ROC is outside JzJ =
which includes the unit circle. Therefore the system is stable. The
h[n] we found in part (b) tells us the system is also uusal.

5.5.

11[n]
Y(z) =

=G) •u[n) + (U" u[n) +u[n)
1

I

+

-lz-1

1
1- iz-'

I
I - z-1'

+--

JzJ > I

%[n) = u[n]
1

= I_ z-l,

X(z)

.
lzl

>I

Y(z)
3 -ljz-1 + Jz- 2
H(z) - - -:--r-.-:-+--.
- X(z) - 1- f.z-• + fiz-2'

lzl

>

I

3

(a) Cross multiplying &lid equating z-• with a delay in ti!J>e:

11[n]-

7

12

y[n- I]+

I

12

y[n- 2] = 3%[n]-

19

2

6 :r[n- 1] + J%[n- 2]

(b) Using partial fractions on H(z) we get:

H(z) =
So,

z- 1
I
z- 1
1
+
1 1 +I,
1 - 3 z 1 1- iz-1
1 - ~z-1
1 - 4 zI

I

1•1 > 3

3
u[n- I]+ (I)"
4 u[n]- (1)"u[n- I]+ 6[nJ
(I)" u[nJ- (l)n-1
4
1

h[n] =

3

(c) Since tbe ROC of H(z) includeslzl =I tbe system is stable.

5.6. (a)

%[n]
X(z)

= I--f!z

1

+

= -3I(I)"
2 u[nJ- 34(2)"u[-n- I]

1 = (I- fz-1)(1I
2z- 1)'

I- 2z-1

1

2 < fzl < 2

(b)

1 _7:7.,-:-2-::-~
Y(z) - -;-:--,-".
- (1- !z-1)(1- 2z-1)
This has tbe same poles as tbe input, then!Core tbe ROC is still

! < Jzl < 2.

(c)

Y(z)

H(z) = X(z) = 1- z

_2

.,. h[n] = 6[n]- cf(n- 2]

134
5. 7. (a)

y(n] = (2

or

:z(n]

+3

= 5u(n] *

= 1 _5z-

X(z)

(-i)") u(n] *

Y(z)

=1 _

Y(z)
1-z-1
B(z) = X(z) = (1-lz-1)(1 +

1•1 > 1

1,

lz-

1

Jz-1)'

+ 1+

i.-

3

1,

1•1 > i

3

l•l > i

lm

Re

(b)
7
1-z-1
-1
H(z) = (1- !z-1)(1 + tz-1) = (11) + (1 + iz-1)'

!•

h(n] = -

3

lzl > i

2(1)" u(n] + 57(-.3)" u(n]

5 2

{c)

H z _ Y(z) _
1 - z- 1
( ) - X(z) - 1 + !z-1- lz •

•

•

Y{z) +

•

~z- 1 Y(z)- ~z- 2 Y(z) = X(z)- z- 1X(z)

1
3
u[n] + -y(n- 1]- -u(n- 2] = z[n] - :z(n- 1]
4

8

5.8. (a)

Therefore,

y(n]

=

3

Y(z)

=

3
2z- 1Y(z) + z-2 Y(z)

2y[n- 1] + y[n- 2) + z(n- 1]

Y(z)
z- 1
- X(z)- 1- Jz 1 - z- 2

H(z)---

-

+ z- 1X(z)

z- 1
(1- 2z- 1)(1 + ~z-1)'

lzl > 2

lm

1 zeroalz•-

Re

2

!

135
(b)

1•1 > 2

h{n] =
(c) Use ROC

~ [(2)"- (

-D ")

u[n]

of! < 1•1 < 2 lillce the ROC m- i.Dclude lzl = 1 for a stable system.
h[n]

=- 52(2)"u[-n- 1]- 52 ( -21)" u[n]

5.9.

5
y[n- 1]- 211[n] + y[n + 1] = z[n]
z- 1 Y(z)H(z)

~Y(z) + zY(z) = X(z)
Y(z)

= X(z)
=

1-

.-•

~z-1
2

+ z-2

.-•

= (1- 2z-1)(1- !z 1)

=

a
1- 2z-•

1 zeroatz~-

Three regiOIIS of couverseuce:
(a)

1•1 < ~:
h[n]

2
=--(2)"u[-n1] + -2(1)"
- u[-n- 1]
3
3 2
2

2(1)" u[n]

h[n] = --(2)"u[-n- 1]- - 3
3 2
!Deludes 1•1

= 1, so this is stable.

136
(c)

I• I> 2:
h(n)

2
= l(2)"u(n)J2(1)"
2 u[n]

ROC outside of largest pole, so this is c:aasal.
5.10. Figure P5.16 shows two zeros and three poles inside the unit circle. SiDc:e the number of poles must
equal the number of zeros, there must be an additioul zero at z = oo.

H (z) is c:aasal,

10

the ROC lies outside the largest pole and includes tbe unit circle. Therefore, the

system is also stable.

=

The inverse sywtem switches poles and zeros. The inverse system could have a ROC that includes 1•1 1,
m.akiDg it stable. However, the zero at z oo of H(z) is a pole for H;(z), so the system H;(z) c:aDDot
be causal

=

5.11. (a) It connot be determined. The ROC micht or might not include the unit circle.
(b) It connot be determined. The ROC micht or might not include z = oo.
(c) FaJ.e. Given that the system is causal, we know that the ROC must be outside the outermost pole.
Since the outermost pole is outside the unit circle, th~ ROC will not include the unit circle, and
thus the system is not stable.
(d) True. H the system is stable, the ROC must include the unit circle. Because there are poles both
inside and outside the unit circle, any ROC including the unit circle must be a ring. A ring-shaped
ROC means that we have a two-sided system.
5.12.

(a) Yes. The poles z

= :!:j(0.9) are inside the unit circle so the system is stable.

(b) First, factor H(z) into two parts. The first should be minimum phase aad therefore have all its
poles and zeros inside the unit circle. The second part should contain the rtm•i•ing poles and

zeros.

H(z)

=

1 +0.2z- 1
1 + O.Slz 2

1- 9z-•

minimum phase

poles & zeros
outside Wlil circle

I

All pass systems have poles and zeros that occur in conjugate reciprocal pairs. H we include the
factor (1 - iz- 2 ) in both parts of the equation above the first part will remain minimum phase
and the second will become allpass.
H(z)

=

(1 + 0.2z- 1 )(1- tz- 2 ) . 1- 9z- 2
I+ 0.81z-2
1

-lz-2

= H 1 (z)H,.(z)
5.13. An a.ride: Technically, this problem is not well defined, since a pole/zero plot does not uniquely
determine a system. That is, ma.ny system functillns caa have the same pole/zero plot. For example,
consider the systems

= .-•

H.(z)
Hz(z) = 2z- 1
Both of these systems have tbe same pole/zero plot, namely a pole at zero and a zero at infinity.
Clearly, the system H 1 (z) is allpass, as it passes all frequencies with unity gain (it is simply a unit
delay). However, one could ask whether H 2 (z) is allpass. Looking at the standard definition of an

137
allpaas system provided iD this chapter, the amwer would be
&eque~~cies with •nitll gain.

jH(ei"')j

DO,

Iince the system does not pass all

de&DitioD of &D allpaas system would be a system for wbic:b the system magDitude respoue

=o, where a is a real coDSt&Dt. Such a system would pass all &equccies, ud scale the output

by a CODSt&Dt factor a. Ill a practical setting, this deliDitiOD of &D allpass system is satisfactDry. Uncle.
this defiDitiOD, both systems H 1(z) &Dd H 2(z) woaJd be CODSidered allpaas.
For this problem, it is assumed that DODe of the poles or seros sbowD iD the pole/sero plots are scaled,
so this issue of 'UiDg the proper defiDitiOD of u allpass system does Dot apply. The st&Ddard defiDition
of &D allpaas system is Died.
Solution:

(a) Yes, the system is allpaas, siDce it is of the appropriate form.
{b) No, the system is not allpaas, Iince tbe sero does Dot occur at the coajucate reciprocallocatioa of
the pole.
(c) Yes, the system is allpaas, Iince it is of the appropriate form.
{d) Yes, the system is allpaas. This system coasists of &D allpass system iD cascade with a pole at sero.
The pole at sero is simply a delay, ud does aot c:b&Dge the magDitude spectrum.
5.14.

(a) By the symmetry of z 1[n] we kaow it bas liDear phase. The symmetry is arouad n
continuous phase of X 1 (e''") is arg[X 1 (~w)] = -5.1. Thus,

.

grd[X,(e'w)]

= 5 so the

d
. }
d
= - dw {arg[XI(e'w)] = - dw { -5.1} = 5
4

3
2

34.56789

n

(b) By the symmetry of z2[n] we kaow it bas IIDear phase. The symmetry is arouad n
kaow the phase of X 2 (~w) is arg[X2 (~] -w/2. Thus,

=

.
grd[X,(e'w)]

d { arg[X,(e'w)]
. } =- dw
d { -2
"'}
=- dw

= 21

312

314

T •••

•••
-2
5.15. (a) h[n] is symmetric about n

-1

0

1

2

3

=1.
H(~)

= 2 + e-,;.. + 2e-2jw
= e-"'c~ + 1 + 2e-,;..)
= (1 + 4cosw)e-iw

n

= 1/2 so we

138

=

=

A(w) 1 + 4a~~w, a= 1, IJ 0
Generalized Linear phase but Dot Linear Phase Iince A(w) is Dot always positive.
(b) This sequence bas no even or odd symmetry, 10 it does not possess &eneralised linear phase.
(c) h(n] is symmetric about n = 1.

=
=

H(ei'-)

=

1 + :se-Jw + e-2Jw
.-Jw(el" + 3 + .-;...)
(3 + 2 coow)e-Jw

A(w) = 3 + 2 a

(a) Cross multiplying and taking the inverse transform

Jl(n]- OJI(n- 1]

1
=z[n]- ..-z(n1]

(b) Sinoe H(z) is causal, we know that the ROC is
Wlit circle. So, H(z) is stable for Ia! < 1..

(c) a=!

lzl

>a. For stability, the ROC must include the

.
lzf > 112
Re

2

142
(d)
1
B(z) = 1-""

o-1 z-l
1- u-1'

1

lzl

>"

..

h{n) = (cs)•u[n)- !(cs)•- 1u(n- 1)
(e)

.
1- ..-1.-jw
B(e"") = B(z)l- = 1 ...-.;..
1
IB(..;..)I 2 = H(..;..)B"(ei") = 1 - .. - ·~;..
1 - ... -...

COIW)

2

! (" + 1- 2cscosw) l
.. 1 + ... - 2cs cosw
1

=

.

=
5.23.

(a) Type I:
Jl/2

-

A(w) ='E cs(n]coswn

cosO= 1, cos .. = -1, so there are no restrictioas.
Type II:

1

(JI+l)/2

A(w) =

L

b(n)cosw (n-

2)

-1

cooO = 1, coo (n.. - j) = 0. So H(ei•) = 0.

Type ill:

Jl/2

A(w)

=

L c(n) siDwn

sinO= 0, sinn.- = 0, so B(ei") = B(ei•) = 0.
Type IV:

1

(JI+l)/2

A(w) =

L

d(n)sillw ( n-

-=1

sinO= 0, sill (n.. - j) ~ 0, so just B(ei") = 0.
(b)
Lowpus
Bandpass
High pass
Bandstop
Dilferenti&tor

Type I

TypeD

y
y

y
y

y

N
N
N

y
y

Type ill
N

y
N
N
N

1 - .. -•..;...
1 - ......

1 + :!f - 1
" cos"' i
( 1 + tJ"2 - 2a

=

IB(ei'-)1

•

Type IV
N

y
y

N
y

2)

143
5.24. (a) Taking the z-trausform ol both sides and reananging

-l

B(z)- Y(z) _
+ z-•
- X(z) - 1- tz-2
Since the poles and zeros {2 poles at z = ±1/2, 2 zeros at z = ±2} occur ill conjugate reciprocal
pairs the system is allpus. This property is easy to """""iu since, as ill the system above,
the c:oellicients ol the n111Df!rator aad denomilla&or z..polynomials get reversed (and ill general
conjugated).
(b) It is a property ol aiJpus systems that the output energy is equal to the illput energy. Here i.s the
proof.

-

..
= :E
-..
= ..!.. [
= 2~ £:
= 2~ L:

N-l

L l11[n]l

2

l11[n]l 2
2

(by Parseoa.l's Theorem)

IY(eiw)l dw

2• -·

=
=
=

..

L

2

IH(,,w)X(eiw)l dw
IX(,,wJI' dw

[z[n]l 2

2

!IH(eiw)l =I since h[n] is allpass)

(by Parseval's theorem)

"-'= -oo
N-l

-

L [z[n]l

2

5

5.25. The statement is false. A non-causal system
example, collSider the non-ausal system

C&ll

indeed have a positive collStallt group delay. For

h[n] = 6[n + 1] + 6[n] + 46[n - 1] + 6[n - 2] + 6[n - 3]
This system has the frequency respcmse

IH(eiw)l

= ei"" + 1 + 4e-;w + e-jl.wl + e-jlf.J
= .-JW(.,i,._ + ,,w + 4 + .-jw + .-j2w)
= .-1w(4 + 2cos(w) + 2cos(2w))
= 4 + 2cos(w) + 2 cos(2w)

LB(e'w)

= -w

H(eiw)

grd[B(,,w)]

5.26.

(a) A labeled

pol~zero

=

1

diagram appears below.

lm

1 zeroatz=-

Re

144
The table ol common z-transfarm pain ci...,. us
1

.
) [ ]
{rlin""')z(r ,. llD""'n
u n ._. 1- (2r cos""')z-1 + r'z-2'

lzl > r

which enables us to derift /a(n).

la{n)

.
B(z)

=1

= (-.-1 - ) (r"lin""'n)u(n)

liD""'

rz- 1
{2r c:cs""')z-1 + r'z-•

=

rz- 1
(1- rz-1)2'

lzl > r

Again, using a table lookup ci...,. us

la(n) = nr"u[n)
lm

1 zeroatz=-

Re

5. 27. Making use of some DTFT properties can aide in the solution of this problem. First, note that

/a2 (n)
/a2 (n)

= (-1)"h1 [n)

=

e-i•"h.[n]

Using the DTFT property that states that modulation in tbe time domain corresponds to a shift in the
&equeocy domain,
B2(~)

=

H,(ei(w+•lj

Consequently, B2 (~) is simply B 1 (ei"') shifted by ... Tbe ideal low pass &Iter has now become the
ideal high pass &Iter, as shown below.

1

-rd2

0

It Q)

1

-It

-rd2

0

It (I)

145
5.28. (a)

A

1

1•1 > 2 h(n] causal

H(z)- (1-}z-1)(1 + lz-1)'

H(1)

=6~A

=4

(b)
H(z)

h(n]

1

4

=

(1-lz-1)(1 + lz-1)'

=

1 -lz-1

=

-12cr
u(n] +- -1r u(n]
15 2
5
3

1•1> 2

+

1 + lz-1

8(

(c) (i)

:r:(n]

1

= u(n]- 2u(n -1] #

Y(z)

=
=

y(n]

X(z) =

1
1- lz_ 2. - 1 ,
1

X(z)H(z)
1-2lz-1
_
____

4

.

_:- ~·-')(l+!z- 1 )'

, -·

1•1 >I

1•1 >I

4

=

(1- z- 1)(1 +

=

1 - z- 1 + 1 +

=

3u[n] + (

iz- 1 )

3

1
lz-1
3

-D"

u(n]

(ii)

:r:(t) =50+ 10cos(201rt) + 30cos(40rt)

T=

1
40

t=nT

,.

:r:(n] = 50+ lOcos n "- 30c:os7n

2

=

50+ s.,i(n•/2) + Se-i(../2) + ts.,i•• + 15e-in•

Using the eigenfwlction property:

11[n]

= 50H(ei0 )+s.:i<"•l2 >H(ei<•l2>)+5e-i) = 7 (H)H(e'") = 4, H(e-••) 4

=

y(n]

iH.

=1 -

4

l - 1 ....,
1 e-JW - 1 e-J
H(e-i<•/21) = 7 (H)+

=300 + 24v'2 cos (in -

tan - 1

iH.

G)) + 120 cos

...n

I46
5.29.

2I

=

H(z)

(I- !z-1 )(I- 2z-1 )(I- 4z 1)
I
28
46
- I - 2.- 1 + I - u-1
I - lz
1
2

=

SiDce we bow the leq1leiiCe is not stable, the ROC must not iDdude lzl = 1, and since it is two-sided,
the ROC must be a riDe- Tllia ._,. Clllly 011e poaible choice: the ROC is 2 < lzl < 4.
(a)

h[n] =

or

u[n]- 28(2)"u[n]- 48(4)"u[-n -11

(b)
H 1(z)

28

1

= 1- !z-1 - 1- 2z-1
H2(z) =

48

1

_4z

1

5.30. (a)

H(z)

= (z + ~~z- !l = z-(M-2) ( 1- ~z-2)

1

M

M-2

M-1

n

-1/4

(b)

111[n] = z[n - (M - 2)]-

v[n]
Let v[n]

=

1
4zln
- M]
1

w[2n] = z[2n- (M- 2)]- 4z[2n- M]

= z[2n],
1

y[n] = v[n- (M- 2)/2]- 4v[n- (M/2)]
Therefore,

1

g[n] = 6{n- (M- 2)/2]- 46[n- (M/2)],

M nen

G(z) _ z- 1
1

1

1 < 1•1

_ z- 1,

<3

1 -1)- u[n)
= 54(1)"
2 u[n]- '5(3)"u[-n

(b) ROC includes z = oo so h{n] is caasal. Since both h[n) and :[n] are 0 for n
isalsoOforn e-•

= H1~z) = 1- e!loz-•

Hable and causal,

1•1 < .-•

not causal

or Rabie

148
(c) Only the causal ho(n] is stable, therefore only it

h[n 1=

be used to recover •[n].

e-on, n = 0,8,16, ...

{ 0,

otherwise

(d)

z[n]• ho[n] = 6(n]- .-aoo[n- 8]
+ e-ao(6(n- 8]- .-ao.s(n- 16])
+ e- 1ao(6(n -16]- e-ao6(n- 32]) + · · ·
= 6(n]
5.U.

=G)" u(n] +G)" u(n]

h(n]
(a)

H(z) =
Since h(n], z(n]

1

2- lz- 1

1

1- 12 z-1

+ 1- jz-' =

1-

1

•
p-•
+ 11 z-2 , lzl > -2

= 0 for n < 0 we caD assume initial rest conditions.
5
1
5
y(n] = 611[n- 1]- 6y[n- 2] + 2>:[n]- 6>:[n- 1]

(b)

h,[n] = { h,[n], n < 100
100
0

n;

(c)
1

H(z) =

~~:~ = ~ h[m]z-m,

1

N- 10

+1

m=O

N-1

11[n] =

L

.......

h(m]z(n- m]

(d) For IIR., we have 4 multiplies md 3 adds per ontput point. This cnes us a total of 4N multiplies
md 3N adds. So, IIR grows with order N. For FIR. we have N multiplies md N - 1 adds for the
n" output point, so this collfiguration bas order /(l.
·
5.35. (a)

=oo :o- pole at ~<•1•1
20Jo&,oiH(~<'-I•I)I = -oo :o- aero at ~(h/51
20Jo&10 1H(~<•IIl)l

I!Monmo:e at "' .. 'f :o- pole Wide l1l1it clzcle here.
Since tbe impuloe respOiiiH is real, t.be poles aDd seroo must be ill CODjupte pairs. The remaining
2 zeros are at aero (the number of poles al-yo equalo the number of oeroo).

149
lm

Re

(b) Since H(z) has poles,

wet:- h{n] is IIR.

(c) Since h{n] is causal and IIR, it cannot be symmetric, and thus c:annot have lineae phase.
{d)

Since there is a pole at lzJ = 1, the ROC dOe. Dot include the unit circle.

This means the system

is DOt stable.

5.36. (a)
H(z)

=
=

lz-

1)(1 + 0.9z- 1)
(1- 2z- 1 )(1 +
(1- z-1)(1 +0.7;z-1)(1- 0.7;z-1)
1 - O.&z-1 - 2.35z- 2 - 0.9z-•
1 - z- 1 + 0.49z- 2 - 0.49z- 3
Y(z)

=

X(z)

Cross multiplying and taking the inverse z-transform gives

y[n]- y(n- 1] + 0.49y[n- 2]- 0.49y{n- 3] = :{n]- o.&:[n- 1] - 2.35:r[n- 2j- 0.9:r[n- 3]
(b)

lm

Re

Note that since h[n] is causal, ROC is
(c)

lzl > 1.

150

I H(el"') I
~r------------,,-------------,

l

f

0

0

-It

1t

(I)

(d) (i) The system is not stable since the ROC does not include 1•1 = 1.
(ii) Because h(nJ is not stable, h(nJ does not approach a coDStant as n -+ ex>.
(iii) We can see peaks at"'= ±j in the graph of IH(e1w)l shoWD in part (c), so this is false.·
(iv) Swapping poles and zeros gives:
lm

Re

There is a ROC that includes the unit circle (0.9 < lzl < 2). However, this stable system
would be two sided, so - must conclude the statement is false.
5.37.

tz-

1)(1- !z) 6 (1- !z-1)(1- !z- 1)(1- sz- 1)
(1- !z-1)(1X(z)=
(1-tzl
• =s
(1-~z- 1 )
ft

[

a :r n

J

~

X( -1 )-~(1-!az-1)(1-faz-1)(1-5az-')

a

z - 5

·

(1

6az-')

A minimum phase sequence has all poles and zeros inside the unit circle.

la/21 < 1 •
la/41 < 1 •

lal < 2
lal < 4

l5al<1 •

lal< 5
1
lal<6

16al<1 •

1

I

!t
'

151
Therefore, a•:[n] is real aDd minimum phase iff a is real aDd ]a]

< l·

5.38. (a) The eaUA1 systems haw conjugate zero pairs inside or outside the UDit circle. Therefore

=

H(z)
H,(z)

(1-0.!If:IO·hz- 1 )(1-0.9e-JO·••z- 1 )(1-1.~·••z- 1 )(1-1.25e-;o.a.z-•)

=

Hz(z)

=

Hs(z)

=

(0.9) 2 (1.25)2 (1- (10/9~-hz- 1 )(1- (10/9)e-JO·'•z- 1 ) •
(1- o.se>O·.... - 1 )(1- o.se-;o.a. .,- 1 )
(0.9)2 (1- (10/9~·a.z- 1 )(1- (10/9)e-;o.a.z- 1 )(1-1.~·••z- 1 ) •
(1 - i.25e-JO.k .-•)
(1.25)2 (1- O.!if:IO·'•z- 1 )(1- 0.9e-;o·•• z- 1 )(1 - o.se'O·'• z- 1 ) •
(1- o.ae-;o·•·

.-•>

Hz(z) baa all its zeros outside the UDit circle, aDd is a maximum phase sequence. Hs (z) has all its
zeros imide the UDit circle, aDd thus is a minimum pbaae sequence.
(b)

H(z)
h(n]
H1 (z)
h 1 (n]
Hz(•)
h.(n]
H•(•)
hs(n]

1 + 2.5788z-• + 3.4975z-• + 2.5074z- 3 + 1.2656z4
6(n] + 2.57886(n- 1] + 3.49756(n- 2) + 2.50746(n- 3) + 1.26566(n- 4]
1.2656 + 2.5074z- 1 + 3.4975z-• + 2.5788z-• + z 4
1.26566(n) + 2.50746(n- 1) + 3.49756(n- 2] + 2.57886(n- 3] + 6[n- 4]
= 0.81 + 2.1945z-• + 3.3906z- 2 + 2.8917z-• + 1.5625z-•
0.816(n) + 2.19456(n -1) + 3.39066(n- 2) + 2.89176(n- 3) + 1.56256[n- 4]
1.5625 + 2.8917·-· + 3.3906·-· + 2.1945·-· + 0.8lz 4
= 1.56256(n] + 2.89176(n -1) + 3.39066(n- 2] + U9456(n- 3] + 0.816(n- 4)
=
=

=
=

=
=

(c)

n E(n) E,(n) E,(n) E,(n)
1.0
1.6
0.7
2.4
0
7.7
5.5
10.8
7.9
1
17.0
22.3
20.1
2 19.9
27.1
25.3
26.8
3 26.2
27.8
27.8
27.8
4 27.8
27.8
27.8
27.8
5 27.8
26.2

216.1 27.. %7..

%7.. %7,..
20.1

1!1.11

012345n

012345n

27.8 27.1

%7.1

17

s..s
0.7

%7.8 %7.8

EJ.~

EJ.~

f

012345n

l.

10

~

012345n

152
The plot ol E,[n] COITeSpODds ~ the miDimWII phase sequence..
5.S9. All zeros iDside the llllit circle meaDS the sequence is miDimWII phase. Since
Ill

Ill

- -

L lh.u,.[n]l ~ L lh!nJI"
2

is true for all M, we can use M
phase leqlleJlCle.

= 0 lAd just compute 112[0]. The largest result will he the minimWII

The aunr is F.

5.40.
(i) A zero phase sequence bas all its poles &Dd zeros in conjupt.e reciprocal pain. Generalized
linear phase sySiemS ue zero phase sySiemS with additiOD&i poles or oeroo at z = 0, oo, 1 or
-1.
(ii) A stable system's ROC includes the UDit circle.
(a) The poles aze DOt in conjugau reciprocal pain, 10 this does DOt have zero or generalized linear
phase. H;(z) bas a pole at z = 0 lAd perhaps z =co. Therefore, the ROC is 0 < 1•1 < oo, which
meaDS the iDvene is stable. If the ROC includes z = co, the inverse will also he causal.
(b) Since the poles aze not conjllple reciprocal pairs, this does DOt have rero or generalized linear
phase either. H;(z) bas poles illside thellllit circle, 10 ROC is 1•1 > ~match the ROC of H(z).
Therefore, the inverse is both stable <d causal.

J

(c) The oeroo occur in conjugate reciprocal pain, 10 this is a zero phase system. The inverse has poles
both inside and outside the UDit circle. Therefore, a stable non-causal inverse emu.
(d) The zerOs occur in conjupt.e reciprocal pairs, 10 this is a zero phase system. Since the poles of the
inverse system are 011 the llllit circle.& stable invene does 110t exist.
5.41. Convolving two symmetric sequences yields another symmetric sequence. A symmetric sequence convolved with an antisymmetric sequence gives an &Dtisymmetric sequence. If you convolve two antisym·
metric Sequences, you will get a symmetric sequence.

A: h,[n]• ho[n]• h3[n] = (h1 [n]• ho[n]) • h3[n]

h.[n]• ho[n] is symmetric about n = 3, (-1 :5 n :57)
(h.[n]• ho[n]) • h•[n] is antisymmetric about n = 3, (-3 :5 n :59}
Thus, system A has generalized linear phase

B: (h,[n]• ho[n]) + h•[n]

h.[n] • h2 [n] is symmetric about n = 3, 11 we noted above. Adding h3 [n] ~this sequence will desuoy
all symmtery, 10 this does not have pneralized linear phase.
5.42.

(a)

= 1, !wl <,.
= --, lwl < ..

A(ei")
f(w)

153

A(J'>)

1

•••

•••

• ••

•••

-·

&

(I)

-all ........... .

(b)

-·

0

•

01

..!_ (" .-;-~""dw = siD1r(n- a)

h(n] =

21r /_.

r(n- a)

l

11=3

•••

•

0

(l"'

•••

b

•••

•

2

l

2

•

3

I

1

3

l

3.25

2

0
6

•

4

II

3.5

0

(l:

1

1

3

e
6

n

5

y

•••
n

4

I

6

T

5

'i'

4

.. .

e
5

6

6

•••
n

(c) U a is"" integer, then h(n] is symmetric about about the point n =a. U a = ':,where M is odd,
which is not a point of the sequence. For a in general, h(n] will
not be symmetric.

If,

5.43. Type I: 5)'1DIIletric:, M Even, Odd

-

Lezlcth

N

H(ei"')

=

L h[n]e-;..n

(N-2)/2

=

L

N

h[n]e--

+

(N-2)/2

=

L
"""'

L

h[nje-;..n + h[M/2]•-;w(N/2)

.-(JI+2)/2
(N-2)/2
h[n]e-;wn

+

L
......

h[M- m]e-jw(N-,.J

+ h[M/2]•-;..(N/2)

154
=

(JI-2)/2

e-iw(JI/2)
(

L
......,

h(m)el"((JI/2)-"') +

(JI-2)/2

L
.......

h(m)e-i"((JI/2)-M) + h(M/2)

=

e-iw(JI/2)

(:~

=

e-MM/2l

(~2/I((M/2)-n)cos""'+h(M/2))

12

Let

2h(m) cosw((M/2)- m) + h(M/2))

ca{n) ={ h(M/2),

n

=0

2/o[(M/2)-n), n= 1, ... ,M/2

Thea

B(ei") = e-,;..(Jt/2)
ud.., have

...L
Jl/2

...

o(n) cos""'

Jl/2

.A(w) =

L o(n) cos(wn),

M

o=T, 13=0

Type U: Symmetric, M Odd, EYeD Lencth

...
Jl

H(el")

=

L ll(n)e-i""

(JI-1)/2

=

L

.....
(JI-1)/2

=

L

.....

Jl

+

L

(JI-1)/2

-

h(n]e-1"'"

ll(n)e-;""
..-(J1+1)/2
(JI-1)/2
ll[n)e-;wn +
h[M- m)e-iw(JI-m)

L

= .-,iw(Jt/2)

.?;,

(

ll(m)e"'((JI/2)-,.) +

l;,

(JI-1)/2

ll(m)e-i"'((M/2)-ml

2

= e-MM/2) (Jt,?;/ 2h(m) c:osw((M/2)- m))
2

= e-iw(JI/2)
Let

(Jt~/ 2h[(M + 1)/2- n) cosw(n- (1/2)))

6{n) = 2h[(M + 'tJ/2- n),

Thea

B(e"')

= e-MM/2l

n = 1, ... , (M + 1)/2

(JI+l)/2

L

6(n) cosw(n- (1/2))

-1

ud.., have

(JI+1)/2
.A(w) =

L

-1

b[n) cosw(n- (1/2)),

M

a=2• 13=0

)

)

155
Type ffi: Alltisymmetric, M Even, Odd Length
B(ei"')

= L"' h(n)rder. When the allpass filter llips a pole or zero outside the unit
circle, one is left in the conjugate reciprocal Jocati011, giving us linear phase.
(b) We know that h[n] is lqth 8 aad theftfore has 7 zeros. Since it is an even length generalized
linear phue filter with real c:oeflidents and odd syuunetry it must he a Type IV filter. It therefore
has the property that its zeros come in coajugate reciprocal pain stated mathematically as B (z) =
B(1/z"). The zero at z
-2 implies a zero at z
the zero at z O.Sei(•/4 1 implies
zeros at z = 0.&-i(•/4 1, z = 1.25ei<•14 l and z 1.25e-i(•/4 l. Becouae it is a IV filter, it also must
have a zero at z = 1. Putting all this toðer gi...,. us

=

=

=-!.while

=

B(z) = (1 + 2z- 1 )(1 + o.sz- 1 )(1- o.Sei<•14 lz- 1 )(1- o.s.-i<•/41.-• l.
(1-1.25ei<•l4 lz- 1 )(1-1.25e-i(•/4 lz- 1 )(1 _ .-•)
5.47. The input :t[n] in the frequency domain loob like

!

159
X(a~'")

·~(10lt)

(10lt)

5

-Q.Sit -Q.4lt

0

0.4lt O.Sn

while the c:ortespODdiDg output 11[n]looks like

Y(el'")
10e-j10..

-It

-Q.31t

0

0.31t

It

(I)

It

(I)

Therefore, the filter must be

H(el'")

-It

-Q.3lt

0

0.31t

In the time domain this is
h(n]

= 2 sin(0.3r(n- 10)]
r(n- 10)

5.48. (a)

Property
Stable

Applies?
No

IIR

Yes

FIR

No

For a stable, causal system, all poles must be
inside the unit circle.
The system has poles at locatioiiS other thaD

z=Oorz=oo.

rnr systems caD ollly have poles at • -

0 or

z- co.

Minimum
Phase
Allpass

No

Generalized Linear PIWe

No

Positive Grouo Delay for all w

No

No

Minimum ptwe . , . . - have all poles and zeros
loeated inside tbe unit circle.
Allpua systems have poles and zeros in conjugate
reciprocal pain.
The causal geaeralised linear p!We systems
presented in this chapter are FIR.
This system is not in the appropriate form.

160
(b)
Property

Applies?

St&ble

Yes

IIR
FIR

No
Yes
No

Minimum
Phase

Allpus

No

GeDeralizecl LiDear Phase

Yes

Posiu.e Group Delay for all w

Yes

Collllllellts

The ROC for this system fwlctiOII,
1•1 > 0, COIIWDs the llllit circle.
(Note there is 7th order pole at z = 0).
The system has poles ODiy at z = o.
The .,._ has poleo ODiy at • - o.
By debiUoa., a Jllillimum phase system mast
ha.e all its poles uad ...,. located
iMU the IIIIi& cirde.
Noce lhat the 1er01 011 the llllit circle will
C&UI4! the mapitude opectrum to drop r.ero at
cerWD &equeDCies. Clearly, tbis syslem il
DOt allpus.
This illhe polefzero plot of a type n FIR
liDear phase system.
This syslem il causal uad liDear phase.
CoDJeqUeDdy, its group delay il a positift

.

(c)
Property
Stable

Applies?

IIR

Yes

CoiiUDellts
All poles ue iDSide the Wlit circle. Since
the system is causal, the ROC includes the
Wlit circle.
The system bas poles at locatioDS other than

Fffi

No

• = 0 or z = oc.
FIR IJS'ellll cua ollly ha.e poles at z - 0 or

Minimum
Phase
All pass

No
Yes

Generalized LiDear Phase

No

Positive Group Delay for all w

Yes

Yes

z = oo.
Minimum phase systems haft all poles uad zeros
located inside the wUt circle.
The poles inside the llllit circle have
WiiespODdiDg leiOIIocated at CODjupte ·
reciprocallocatioas.
The causal generalizedlillear phase systems
preseated in this chapter are FIR.
Stable allpass systems have positift group delay
for all ....

5.49. (a) Yu. By the region of convergence we know there are DO poles at z = oc uad it therefore mast be
causal. Another way to see this is to use lollC divisiOil to write H,(z) u
1

H,(z)

=1

=:-I =
-s

1 + .-• +

,-s + ,-• + z-< ,1•1 > 0

(b) h 1 [n] is a causal rectaJ1CU1ar puhe of ~el~Cth 5. If we COII'IO!ft h,[n] with uaolher causal recW1gUlar
pulse of length N we will pt a triUlplar pulse of lellClh N + 5 - 1 "' N + 4. The trWigular pulse
is symmetric aro1111d its apa uad thus has liDear phase. To make the triuacDlar pulse g[n] haft at
least 9 11011zer0 aamples we CUI c:hoc.e N 5 « let hs[n] h, [n].
Proof:

=

=

161

=

[1-.-;... ]2
1- e

Jw

= r·-jwS/2 (&wS/2 _ .-jwS/2)]
e-i~<~/2

=

2

(eiw/2 _ e-jc.>/2)

sin2 (Sw/2) -;w,
r(n- 4.3)

. 0,

Proof: Although the group delay is constant ( grd (H(e1'")]

=
=
=
=

h[n]

H(eiw)
e-;4.3w

M

lwl < w,
otherwise

= 4.3) the resulting M is not an integer.

±h(M-n]
±eJMw H(e-iw)
±e:i(M+4.3)~o~

1

[wl

0.

164
lm

4lh ordlf pole

(c) A sketch of the pole-zero plot for Y(z) is shown below. Note that the ROC for Y(z) is [zl

> !·

lm

5.55.

•
•
•
•

Since z[n] is real the poles & zeros come in complex conjugate pairs.
From ( 1) we know there are no poles except at zero or infinity.
From (3) and the fact that z[n] is finite we know that the signal has generalized linear phase.
From (3) and (4) we have a = 2. This and the fact that there are no poles in the finite plane
except the five at zero (deduced from (1) and (2)) tells us the form of X(z) must be
X(z)

=z[-1]z + z[O] + z[1]z- 1 + z[2]z- 2 + z[3]z- 3 + z[4)z 4

+ z[S)z-•

=

The phase changes by .. at w = 0 and .. so there must be a zero on the unit circle at z ±1. The
zero at z 1 tells us ~>:[n] 0. Tb,e zero at z -1 tells us :[(-1)":r:[n) 0.
We can also conclude z[n] must be a Type m filter since the length of z[n) is odd and there is a
zero at both z ±1. :r:[n] must therefore be antisymmetric around n 2 and z[2) 0.
• Fr~m (5) and Parseval's theorem we have L: [:r:[n][ 2 = 28.
• From (6)

=

=

=

=

=

=

y[OJ =
=

11[1]

=

2_
2'JI'

2_
27r

1•

Y(eiw)dw

-r

=

=4

:r:[n]• u[n) In=<> = :r:[-1) + :r:[O)

1•

Y(eiw)ei"'dw

=6

-r

= :r:[n] ou[n) [..,., = z[-1) + :r:[O] + :r:[1]
• The conclusion from (7) that :[(-1)":r:[n) = 0 we already derived earlier.
• Since the DTIT {z.[n]} 1U {X(...... )} we have

=

z[5] + :r:[-5)
2

=

3

-2

z[S] = -3+z[-SJ
:t(S]
-3

=

Summarizing the above we have the following (dependent) equations

!

165
(1)
(2)
{3)
(4)
(5)
(6)
(7)
(8)
(9)
(10)

=

:t(-1] + :t(O] + :[1] + :t[2] + :[3] + :[4] + :(5] 0
-:t(-1] + :t(O]- :t(l] + :t(2]- :(3] + :t(4]- :t(S] = 0
:t[2] = 0
:t(-1] = -:t(S]
:t[O] -:t[4]
:t[1] -:t[3]
:t[-1] 2 + z[0] 2 + :t[1] 2 + :[2] 2 + :[3] 2 + z[4]' + :t[5]'
:t(-1] + :t(O] = 4
:t[-1] + :t(O] + :t(1] = 6
:t[5] = -3

=
=

=28

:t[n] is easily obtained from solving the equations in the following order: (3},(10),(4),(8},(5),(9), and (6).

3

JC(nj

2

3

-1

0

1

5

4

n

2
-2
-3

5.56. (a) The LTI system 5 2 is characterized as a Jowpass filter.
The •·transform of h 1 [n] is found below.
· y[n]- y[n- 1] +

~y[n- 2] =

4
1

:t(n]

Y(z)- Y(z)z- 1 + 4Y(z)z- 2

=

X(z)

(1- z- + ~z- 2 )

=

X(z)

Y(z)

1

1

1

= (1- z-• + tz-') = (1- !z-')'
This system function has a second order pole at z = i· (There is also a second order zero at z = 0).
H,(z)

Evaluating this pole-zero plot on the unit circle yields a low pass filter, as the second otder pole
boosts the low frequencies.
Since

=

H 2 (ei"')
H 2 (z)

H 1 (-ei"')

= H 1 (-z)

If we replace all references to z in H 1 (z) with

-z, we will get H 2 (z).
1

H,(z)

= (I+ tz-' )2
-!·

Consequently, H 2 (z) has two poles at z =
(There is also a second order zero at z = 0).
Evaluating this pole-zero plot on the unit circle yields a high pass filter, as the second order pole
now boosts the high frequencies.

166
(b) Tbe LTI system 5 3 is cbara.cterized as a highpass filter. H3 (~} is the inverse system of H 1 (eJw},
since H,(ei"')H,(eJw) 1. Consequently, H 3 (z}H1 (z) = 1.
~ shown in part (a), H, (z) has a second order pole at z
~. and a second order zero at z
0.
Thus, H,(z) has a second order zero at z =
and a second order pole at z = 0. Evaluating this
pole-zero plot on the unit circle yields a high pass filter, as the second order zero attenuates the
low frequencies.
S3 is a minimum phase filter, since its poles and zeros are located inside the unit circle. However,
because the zeros of S. do uot occur in conjugate reciprocal pairs,
cannot be classified as one
of the four types of FIR filters with generalized linear phase.
(c) First, we compute the system function H4 (z}

=

=

!,

=

s.

11[n] + a 1y[n- 1] + a 2 v[n- 2] = Poz(n]
Y(z)

+ a 1 Y(z}z-• + a2Y(z}z-2 =

PoX(z)

f3o

H4 (z) =

1 + o 1 z- 1 + a 2 z- 2
is a stable and noncausal LTI system. Therefore, its poles must be located 01dsidt the unit
circle, and its ROC must be an interior region that includes the unit circle. We place a second
order pole at z = 2, whicb is the (conjugate} recip>;OC&llocation of the second order pole of H, (z)
at z =
This gives

s.

!-

(1-2z-1)2

=

Po

(1 - 4z- 1 + 4z- 2}

In order for
an appropriate value of Po must be found. Consider the case when z

= 1. Then,

=

I(1- z- 11+ ~z 2) I

=

1(1-:+tll

= 4
The values a 1
solution.

= -4, a 2 = 4, and Po = 4 satisfy the criteria. Note that Po = -4 also is a valid

(d) If h,[n]• h 1 (n] is FIR, then the poles of H 1 (z) must be cancelled by zeros of Hs(z). Thus, we
expect a second order zero of H,(z) at z =
Therefore, H,(z) will have the term (1- !z- 1 }2 •
In order for the filter ho(n] to be zero phase, it must satisfy the symmetry property h,[n] 11$[-n], whicb means that H,(z) = H 0(z). For this property to be satisfied, we need two more zeros located at z = 2. In addition, we want these zeros to correspond to a noncausal sequence. Therefore, H,(z) will also have the term (1- ~z) 2 • Combining tbese two results, l· = 167 Taking the inverse :<-transform yields hs[n] 5.57. 33 = !o[n2] - ~o[n- 1] + o[n]- ~o[n + 1] + !o[n + 2] 4 4 16 4 4 (a) 1 . 1 . :t[n] = s[n] cos won= s[n]e-'won + s[n]e-Jwon 2 X(eiw) y[n] = = 2 = !s(eJ!w-wol) + !s(eio) (b) This time, Y(eiw) = H(e-'w)X(e'w) = !.-J-•e-Jw••S(eio) s[n - n•] cos(won - won• - t/>o) = t/>o + won 4 then y[n] = s[n- n,]cos(won- ~) for narrowband s[n]. (c) d . d - dw arg[H(e-'w)J =- dw[-t/>o- wn,] = Tgr = Tph = _.!.arg[H(e-'w)] = _.!_[-1/>o -wn,] = t/>o- n• w w n4 w y[n] = s[n- Tgr(wo}]cos[wo(n- rph(wo))) (d) The effect would he the same as the following: (i) Bandlimit interpolate the composite signal to aC-T signal with some rate T. (ii) Delay the envelope by T · Tgr, and delay the c:.arrier by T · Tph. (iii) Sample to a D-T signal at rate T 5.58. (a) . m. = 0 ~ c;..(m] =r .,(m] ~ r .,(z) = +,..(z) .;•• [m] = y[n] • y[-n] = :t[n] • :t[-n] • h[n] • h[-n] 168 N M .1=1 ..... 11!nJ = :L "•ll!n - tl + :L o•=ln - tJ, N 11o =1 M Y(z) = L"•Y(z)z-• + X(z) + Lb•X(z)z-• t=l i=l So, Or equivalently, n c.z- c.z) r ..,(z) = A u!::.";;;.''---------n(1- d.z- )(1- d••> M (1- 1 )(1- 2 1 t=l (b) To "whiten" the signaly(n] we need a system: H,.(z)H,.(z- 1 ) = H(z)!(z ') Therefore, n N 1 (1- d.z- )(1- H,.(z)H,.(z- 1 ) d·•> = ":";"1' - - - - - - - . . , . - IT (1- c.z- )(1- c,z) 1 .1=1 The poles of H,.(z) are the zeros of H(z) aDd the zeros of H,.(z) are the poles of H(z). \_Ve musl now decide which N of the 2N zeros of H,.(z)H,.(z- 1 ) to associate with H,.(z). The remaininG N zeros and M poles will he reciprocals and will he usoci•ted with H,.(z- 1). In order for H,.(z) to he stable, we must chose all its poles inside the nnit circle. Thus for a pair c., c; 1 we chose the one which is inside the nnit circle. 169 {c) There is no real constraint on the zeros of H,.(z), so we can select either d• or 4, 1 • Thus, it is not unique. 5.59. (a) 1 .11-1 H(ei"') . H;(e'"') 1-e-Jw = 1.M .-,., -;wM =~ .-;..n = - e . ~ 1-e-'"' ..... .,. h;[n] = Loo o[n- kM]- o[n- kM- 1] •=<> 1 1 1 ••• 2M+1 1 n 0 -1 -1 -1 h;[n] has infinite length, so we can never get a result without infinite sums. Therefore, it is not a real time filter. We can use the transform approach but we must have all the input data aVailable to do this. {b) The proposed system is a windowed version of h;[n]: h,(n]• h2(n] = h;[n]p[n] Where n] Pl = { 1, 0, 0 ~ n ~ qM othennse :(n]• h(n]• h;(n]p{n] = w(n] Therefore, if :[n] is shorter than qM points, we can recover it by looking at w(n] in the range 0 ~ n ~ qM -1. {c) q h 1 [n] =L •=<> 6(n- kM].,. H 1 (z) = 1-z-qM -M 1-z Thus, 1 1-z-M H 2 {z) = H(z) 1- z tM Note that 1-z-<~M bas M zeros and qM poles. Since H2(z) is causal, there are no poles at z = oo. If H(z) bas P poles and Z zeros: Z+M$P+qM

170
5.60. (a)
1

a- z- 1

cz-1

=z--=-= =-~
a
a
.u- 1

H(z)

lm
, poleatz=-

Re

.

H(e'")

1/a

.
1
= e'"-= cosw + j sinw--a1
a

arg[H(ei")] =

tan-1 ( sinw I)
cosw--

•

{b)

lm

Re

.

G(z)

G(ei")-

z

1

- 1 - OL!-;~.~

arg[G(ei")]

1

= -z- a = .,........;=--...,
1- az-1

=

= I - a cosw1+ jasinw

-tan- 1 ( 1 ~:~w)

= tan-1

=
5.61. (a) Because hl[n], h2 [n] are
be inside the unit circle.

minimWll

(a::: 1)

arg{H(ei")J

phase sequences, all pales and zeros of their .transforms must
hi[n] • h,[n] ++ H 1 (z)Hz(z)

Since H 1 (z) and H,(z) have all their pales and zeros inside the unit circle, their product will alsa.
(b)

hi[n] + h,(n] ++ H1 (z) + H 2 (z)

xi[n] =

(21)n u[n] +-+ 1iz-' = Xl(z)
1

_

171

(1)n u(n]+---+ 21z 1 = x,(z)

z,[n] = 2 -2

1- 0

= 0 and a pole at z =

Both of these are minimum phase, with a zero at z

X1(z) + X,(z)

= 1- 31z
0

~.

1

This is minimum phase, with the same pole and zero as X1(z) and X,(z).

(1)n
2 u(n]+---+ 1 -lz- 1 = X1(z)
1)n
-6
z,[n] = -6 (-3 u(n] ........
1 -1 = x,(z)
6

z 1(n] = 6

1- JZ

X1(z) has a pole at z =!and a zero at z = 0. X 2 (z) has a pole at z =land a zero at z

z-1

= (1- 1~ _1)( 1 -

X1(z) + X2 (z)
This has zeros at z
5.62.

= 0, oo an9 poles at z

= 0.

2

1

3z

1)

= ~~l· Therefore, it is not minimum phase.

(a)

(1)n u[n] + -(2)"u[-n- I]

r(n] =-4 3

R(z)

=

=
=

4
3

2

~

1- ~z

~

1 - 2z-1
-2z-l
1

(1 - !z-1 )(1 - 2z-1)

(1-

1
)(1- ~z)'

!z

1

ROC:

! < lzl < 2

lm
1

zeroatz=Re

2

(b)

r(n] = h(n] • h[-n] ~ R(z) = H(z)H(z-1)
.
1
R(z)
1

= -;:--,-....;,..,,..-..,...,.
(1- !z- )(1- !z)

We have two choices &om H(z). Suu:e h[n] is minimum phase we need the one which has the pole
at z ·= ~, which is inside the llllit circle.
±1
1- 1

H(z) = (

h[n]

,z

1 ),

=±G)

ROC: lzl
n

u[n]

>!

172

5.63. (a) Maximum phase systems are of the form
Jl

II• z- 1 )(1 -

0.&-iO·•• z- 1 )(1 - (5/6)~.?. z- 1 )(1 - (S/6)e-JO.?• z- 1)
1

Hc(z)

= (1.44)(1- O.Sei0-3•z-1 )(1- O.Se-iO.l•z-1 )(1- (S/6)ei0· 7 •z

1)(1- (S/6)e i 0 ·1 • z-1)
(z- 1 - (5/6)e-i0 ?o)(z-•- (S/6)ei0·?o)
G(z) = H •• (z) = (1- (5/6)ei0.1•z-1)(1- (S/6)e ,o.?oz 1)

Ae
41h order pole

Ae
C#'l order zero

5.65.

H(z) = Hmin(z)

z- 1 -a
,
1 -az 1

Jal < 1

Thus,
lim Hmin(z) = lim

.&-+oo

1- az- 1

.1-+oo

h,.;,[O]

z

1

-a

H(z)

= - !h[O]
a

Therefore, Jhm;n[OJI > Jh[OJI since Jal < 1. This process can be repeated if more than one allpass system
is cascaded. In each case, the factor for each will be larger than unity in the limit.
·
5.66. (a) We use the allpass principle and place a pole at z

H(z)

= •• and a zero at z = .J,.
.
••

174
(b)

H(z) = Q(z)z- 1 - z0Q(z)

h(n]

= q(n- 1]- z;q(n]

Hmin(z) = Q(z)- z•Q(z)z- 1

h...in[n] = q(nJ- ••q[n- 1]
(c)
n

€

=

L

n

lh...u.[m]l 2

L

-

lh[m]l'

~

"'=0
n

= L

(lq[m]l 2

-

••q[m- ljq*[mJ- z;q*[m -ljq[mJ + l••l 2 lq[m- 1]1')

n

- L (lq[m- 1]1

2

-

z0q*[m- 1]q[m]- ••q[m- 1]q*(m] + l••l 2 lq[m]l 2 )

n

L (lq(m]l' -lq[m- 1]1')

=

(1 -

=

(1 -l••l 2 )lq[n]l 2

1••12 )

(d)
€

= (1 -l••l')lq[n]( 2 ~ 0

Then

n

n

L

L

lh...u.[m]l' -

m=O

lh[m]l 2 ~ 0

m=O

n

L

Vn since I•• I< 1

n

jh[m]l' $L lh...in[m]j 2 Vn m=O m=D 5.67. (a) x[n] is real, minimum phase and x[n] = 0 for n < 0. Consider the system: ~·1--+i H,.;.(z) x(n} H..,(z) f-+ylnl x[n] is the impulse response of a minimum phase system. y[n] is the impulse response of a system which has the same frequency response magnitude as that of x[n] but it is DOt minimUIII phase. Therefore, the equation applies. n n :E l:r:[kJI' ~ :E jy[kJI .... .... 2 Since h•• [n] is causal and :r:[n] is causal, y[n] is also causal, and these sums are meaningful. 175 (b) As discussed in the book, the group delay for a rational allpass system is always positive. That is, Therefore, filtering a signal z[n] by such a system will delay the energy in the output y[n]. H we require that z[n] is causal, then 11[nJ will be causal as well, and the equation n n bO .... :L lziA:JI• ::: :L ly[A:JI' applies to the system. 5.68. (a) = = g[n] r[n] z[n]• h[n] g[-n] • h[n] s[n] = r[-n] = g[n] • h[-n] = z[n]• (h[n]• h[-n]) = hJ[n] = H,(ei"') h[n]• h[ -n] H(e'"')H"(ei"') = IH(ei"')l2 Since H 1 (ei"') is real, it is zero phase. (b) g[n] = z[n]• h[n] r[n] = :r[-n]• h[n] !f[n] = g[n] + r[-n] = :r[n]• h[n] + :r[n] • h[-n] = z[n] • (h[n] + h[-n]) h2[n] = h[n] + h[-n] H 2 (ei"') = H(e'"') + H"(ei"') = 21U{H(ei"')} H2(ei"') is real, so it is also zero phase. (c) ..J2 ······ 31t/4 1 ······....-------. 0 fl/4 " (I) -..J2 ····················· 0 3fl/4 " (I) In general, method A is preferable since method B causes a magnitude distortion which is a function of the (possibly non-linear) phase of h[n]. 176 5.69. False. Consider 1 H(z) = (1- 1 ~z-1 )(1- 2z-l) ~z- 1 + z - 1- 2 This system function bas poles at z = 1/2 and z = 2. However, as the foDowing shows it is a generalized linear phase fiher. H(.,;w) = 1 ;>w l- • 2eJw+e . • eiw = = 5. 70. . e'"'-i+e ( . JIM 1 ).,;w 2cosw-! (a) Since h[n] is a real causal linear phase filter the zeros must occur in sets of 4. Thus, if z1 is a zero of H(z) then zi,lfz, and 1/zi must also be zeros. We can use this to find 4 zeros of H(z) from the given information. z,, magnitude magnitude magnitude magnitude zi' 1/z., 1/zi, = = = = 0.5, 0.5, 2, 2, angle angle angle angle = = = = 153 degrees 207 degrees 207 degrees 153 degrees (b) There are 24 zeros so the length of h(n] is 25. Since it is a linear phase filter it bas a delay of (L- 1)/2 = (25- 1)/2 = 12 samples. That corresponds to a time delay of (o.5 ..,:1e) (12 samples)= 6 ms (c) The zero locations used to create the foDowing plot were estinlated from the figure using a ruler "!'d a protractor. Estimate of Continuous-Time Magnitude Response 0 m -20 :!:!. ..,.,:::> . 'E0> ::! (0.0444)lr/T ( 0.128)1=I = liM. i:=l = z-1- d• 1:=1 z- 1 (z- d•) z I - d• IT >=I (z-1- ••")(z-1 2 ••) ••H•- et) z- (z(z-1- e;)(z I - ••) z- d, liM, (z- e,)(z- et) 1- d•z A:=l (1- e0z)(1- •••) H(1/z), 1 lzl E II. which in the time domain is h;[n] = h[ -n] 5.74. We can model g[n] as g[n] = z[n] + ao[n - no] Now send the corrupted signal g[n] through a highpass filter hap/ [n] with a cutof of we = "/2. 179 112 • •• -1 ••• 1 n 0 The highpa.ss filter completely filters out the lowpass signal :r[n]. The output y[n] is = = = y[n] (z[n] + ao[n- no]). h,,,[n] ah,,1[n- no] a(-1)(n-no)sin~(n-no) .-(n- no) y[n] looks similar to the picture of h,,1 [n] above except that it is scaled by a and shifted to no. Thus, 2y[no] a= z[n] = g[n] - 2y[no]o[n - no] (a) When no is odd, y[nJ = 0 at all odd values of n except n :r[n] from g(n]: = no. This leads to a procedure to lind · • Filter g(n] with the highpass filter described above. • Find the only nonzero value at an odd index in the output y[n]. This value is y[no]. • z[n] = g[n]- 2y[no]o[n- no] (b) The only time three cousecutive nonzero samples occur in y[n] is at n = no. The procedure to lind z[n] is • Filter g[n] with the highpass filter described above. • Look for three cousecutive nonzero output samples. The middle value is y[no]. • z[n] = g[n] - 2y[no]o[n- no] 5. 75. Looking at the z-transform of the FIR filter, H(z) = L"" = h[nV" n=O N-1 L h[N- 1 - n]z-n n=O Substituting m =N - i - n into the summation gives 0 H(z) = L m=N-l h(m]zm-N+l 180 N-1 = L h[m]z"'z-N+ 1 -=<> = = N-1 z-N+l L h[m]z"' .....0 z-N+l H(z- 1 ) Thus, for such a filter, H(l/z) = zN- 1H(z) Ji zo is a zero of H(z), then H(zo) = 0, and H(l/zo) Consequently, even-symmetric linear phase = zf- 1H(zo) = o Fm filters have zeros that are reciprocal images. 181 Solutions - Chapter 6 Structures for Discrete-Time Systems 183 6.1. We proceed by obtaining the transfer functions for each of the networks. For network 1, Y(z) = 2rcos8z- 1Y(z)- r 2 z- 2Y(z) + X(z) or H,(z) = Y(z) = X(z) 1 1-2rcos8z '+r•z 2 For network 2, define W(z) as in the figure below: W(z) z[n] -rsin8 rsin8 rcos8 y[n] .-• then W(z) = X(z)- rsin8z- 1Y(z) + rcos8z- 1 W(z) and Eliminate W (z) to get Y(z) H.(z) rsin8z-• 1 +r 2 z = X(z) = 1- 2rcos8z- 2 Hence the two networks have the same poles. 6.2. The only input to the y[n] node is a unity branch connection from the z[n] node. The rest of the network does not a1Iect the input-output relationship. The difference equation is y[n] z[n]. = 6.3. 1 2+ !z4 H(z)- 1 + lz-1 _ lz 2 • • System (d) is recognizable as a transposed direct form ll implementation of H(z). 6.4. (a} From the flow graph, we have: Y(z) =2X(z) + (~X(z)- ~Y(z) + ~Y(z}z- 1 }z-'- 184 That is: Y(z)(1 + ~z- 1 - ~z- 2 ) = X(z)(2 + ~z- 1 ). The system function is thus given by: H(z) = Y(z) = X(z) 1+ 2 + lz- 1 lz 1 -lz •· (b) To get the difference equation, we just inverse Z-transform the equation in a. We get: 1 3 1 11[n] + -y[n- 1]- -y[n- 2] = 2:[n] + -z(n- 1]. 4 8 4 6.5. The flow graph for this system is drawn below. y[n] z[n] 1 1 (a) w[n] = z[n] + 3w[n- 1] + w(n- 2] y[n] =w[n] + 11[n- 1] + 2y[n- 2] (b) = X(z) + 3z-1W(z) + z- 2W(z) Y(z) = W(z) + z- 1Y(z) + 2z- 2Y(z) W(z) So Y(z) X(z) = H(z) 1 = (1- z-1 - 2z-2)(1- 3z-1 - z-2) = 1 1- u- 1 + 7z • + 2z-•. (c) Adds and multiplies are circled above: 4 real adds and 2 real multiplies per output point. (d) It is not possible to rednce the number of storage registers. Note that implementing H(z) above in the canonical direct form n (minimum storage registers) also requires 4 registers. 6.6. The impulse responses of each system are shown below. 185 4 4 3 3 1 1 1 (a) 1 (b) 0 -1 -1 -2 -2 3 3 3 2 2 2 2 1 1 1 1 (c) (d) 0 -1 -1 -1 6.7. We have _! H(z) -1 + z-2 = 1-• iz 1 . 2 Therefore the direct form II is given by: -1/4 •-~-::::-~---· y[n] x[n] ...___,.- - - - . 1 1/4 6.8. By looking at the graph, we get: y[n] = 2y[n - 2] + 3x[n - 1] + x[n - 2]. 6.9. The signal flow graph for the system is: x(n] w,(n] w1[n] y(n] z-1 2 -1 w,(n] w3(n] z-1 4 186 (a) First we need to determine the transfer function. We have = :[nj - w3[nj + 4w4 [n- 1] = IDt [nj = ID2[n-1j = 2w3[nj = 102[nj + :[n- 1] + w,[nj. w.(nj "'>[nj "'>[nj w4 [nj y[nj Taking the Z -transform of the above equations, rearranging and substituting terms, we get: + .-• - s.-• = 1 + 13z. + z- 1 - Sz-2 1 H(z) The difference equation is thus given by: y[nj + y[n- 1]- 8y[n- 2] = :[nj + 3:t[n- 1] + :[n- 2]- S:[n- 3]. The intpulse response is the response to an impulse, therefore: h[nj + h[n- 1]- 8h[n- 2] = J[n] + 3J[n- 1] + 6[n- 2]- 8J[n- 3]. From the above equation, we have: =1 = 3- h[O] = 2. h[OJ h[1] (b) From part (a) we have: y[nj + y[n- 1]- Sy[n- 2] 6.10. = :t[n] + 3:t[n- 1] + :[n - 2] - S:[n - 3]. (a) w[nj v[nj y[nJ = = = 1 2 y[nj + :t[nj 1 2y[n] + 2:t[nj + w[n- 1] v[n- 1] + :t(n]. (b) Using the Z-transform of the difference equations in part (a), we get the transfer function: _ Y(z) _ 1 + 2z- 1 + z- 2 H( z ) - X( ) 1 2· z 1- •1 .- 1 - •• We can rewrite it as : H(z) = (1 + .-1)(1 + z-1) 1 )(1- z- 1 ) (1 + !• We thus get the following cascade form: -1/2 187 (c) The system function has poles at z the system is not stable. = -! and z = 1. 6.11. (a) H(z) can be rewritten as: H(z) = z- 1 - Since the second pole is on the unit circle, 6z-• + 8z-• 1 1- 2 z 1 . We thus get the following direct from ll ftow graph : :[n] y[n] rz-1 1/2 rz-1 -6 ~-1 8 (b) To get the trausposed form, we just reverse the arrows and exchange the input and the ouput. The graph can then be redrawn as: :[n] y[n] f-1 f-1 1/2 -6 ~-1 8 6.12. We define the intermediate variables w,[n], w.[n] and w3 [n] as follows: -1 w1[n] 2 :[n] ---r-------11'--'-----ll[n] 188 We thus have the following relationships: = -z[n] + w:o[n] + w,[n] w,[n] = z[n- 1] + 2w,[n] w,(n] = w,[n- 1] + y[n- 1] ll(n] 2w,(n]. w,fn] = Z -transforming the above equations and rearranging and grouping terms, we get: H(z) = Y(z) = -2 + 6z- 1 + 2z-2 • X(z) 1- Sz 1 Taking the inverse Z -transform, we get the following difference equation: 6.13. 1- lz-2 B(z) = 1- 1 z _: 4 1 -2 · 8z The direct form I inlplementation is: y[n] z[n] z-1 z-1 1/4 z-1 z-1 1/8 -1/2 6.14. H(z)- 1+ §.z-1 6 - 1- + lz-2 iz-1- iiz-2. The direct form II implementation is: y[n] z[n] z-1 1/2 5/6 z-1 1/2 1/6 189 6.15. H(z) = 1 7 -1 + 1 -2 - •• •• . l+z l+lz 2 2 To get the transposed direct form II implementation, we first get the direct form II: y[n] z[n] .-1 -1 -7/6 .-1 1/6 -1/2 Now, we reverse the arrows and exchange the role of the input and the ouput to get the transposed direct form II: y[n] :r[n] .-1 -1 -7/6 .-1 1/6 -I . ....' 6.16. (a) We just reverse the arrows and reverse the role of the input and the output, we get: y[n] :r[n] .-1 .-1 -2 -1/2 .-1 1/4 3 190 (b) The original system is the cascade of two tr&DSposed direct form n structures, therefore the system function is given by: H(z) = ( 1 - 2z-1 + 3z-2 )(1- !z-1). 1- !z-2 2 • Tbe transposed graph, on the other band, is the cascade of two direct form n structures, therefore the system function is given by: H(z) = (1- !z-1)( 1- 2z-1 + 3z-• ). 2 1- !z-2 Tbis confirms tbat both graphs have the same system function H(z). 6.17. H(z) =1- ~z- 1 + ~z-2 + z-3 . (a) Direct form implementation of this system: (b) l!ansposed direct form implementation of the system: 6.18. The flow graph is just a cascade of two tr&DSposed direct form U structures, the system function is thus given by: Which can be rewritten as: 191 1) (1+2z- 1)(1-lz3 H(z)- (1 + ~z-1 - jz-2)(1- az-1) In order to implement this system function with a second-order direct form n signal flow graph, a pole-zero cancellation has to occur, this happens if a a -2 or a 0. U a the overall system function is: · 1 + 2z- 1 = l, = _ H< z ) - 1 + iz 1 1 3 - az = = f, 2" U a = -2 , the overall system function is: H(z) = And finally if a 1 - 2-z- 1 1 1 + iz 3 3 1- iz 2. = 0, the overall system function is: 1 1 H(z) = (1 + 2z- )(1 - ~z- ) 1 + lz-1- !z-2 • • 6.19. Using partial fraction expansion, the system function can be rewritten as: H(z) = 1 -8 1 - JZ + 3Z 1 -1 + 1 2 -1 + 9. Now we can draw the Bow graph that implements this system as a parallel combination of first-order transposed direct form n sections: 9 :r(n] -8 y(n] z-1 1/3 z-1 -2/3 6.20. The transfer function can he rewritten as: {1 + 2z- 1 + !z-2 ) H(z)• - (1 + ~z-2)(1- !z 1 + z-2) which can be implemented as the following cascade of second-order transposed direct form n sections: 192 z[n] y[n] z-1 z-1 5/2 2 z-1 z-1 -1 -1/4 5/4 6.21. h[n] l Y(z) =e'w. "u[n] ,.._.. H(z) = 1- eJ.,.z-1 . = --. X(z) 0 = = e'w•y[n- 1] + :t[n]. Let y[n] llr[n] + jy;[n]. Then llr[n] + jy;[n] 1] + jy;[n- 1]) + :r[n]. Separate the real and imaginary parts: So y[n] y,[n] y;[n] = = = (cosWo +; sinWQ)(y,[n- :r[n] + coswoy.[n- 1]- sinWQy;[n -1] sinWQy,[n- 1] + COSWQy;[n- 1]. :r[n] !lr[n] -sinwo COSWo Jl;[n] 6.22. (1 H(z) = (I - + z-1)2 tz-1 )(1 - !z-1 )" 1 1 + z- ) H(z)"' ( 1- tz-1 ( 1 + :z- 1 1 -!z-1 ) · 193 :r[n] y[n] • • 1 • 1/4 F • J H(z) = • FI J 1/2 • ( 1 + z-1 ) ( 1+ z-1 ) 1- iz-1 1 -lz-1 . :r[n] y[n] • • • l LI l 1/2 • I !·-: 1/4 • Plus 12 systems of this form: :r[n] y[n] 1/4 1/2 with the three types of 1st-order systems taken in various orders. 6.23. Causal LTI system with system function: H(z)- (1- ~z-1 1- lz-1 5 + lz-')(1 + tz-1). (a) (i) Direct form I. so bo . 1 1 = 1, bt =--and At=5 4 5 , a2 1 = -, 03 = --. 24 12 194 y[n] z[n] z-1 z-1 -1/5 1/4 z-1 -5/24 z-1 -1/12 (ii) Direct form IT. y[ n] z[n] 1/4 .-1 -1/5 .-1 -5/24 .-1 -1/12 (iii) Cascade form using first and second order direct form IT sectious. 195 1 1 + !z-2). 3 So -! , bo1 =I , b11 = 1>,1 = 0, bu.= I, b12 = 0, ~ = 0 and au=-~, a21 = 0, au= i, on= -l· :r:[n] v[n] .-1 .-1 -1/4 1/2 -1/5 -1/3 (iv) Parallel form using first and second order direct form II sections. We can rewrite the transfer function as: H(z) 27 98 .J!.. -1 1uz = I+ffi!z 1 + 1 _m!z-1 - iz-2 · • So eot = t':s •- - .!!. ""U2-125' It - .. 11- -i1 I .. -0 ,_21 - 'eu =0' 36 e12--125' and I n _,a2 -3· .,.12- 1 22- 1 z-1 196 27/125 z-l -1/4 (n] y[n] z-• 1/2 -36/125 z-• 1/3 (v) Transposed direct form II We take the direct form II derived in part (ii) and reverse the arrows as well as exchange the input and output. Then redrawing the flow graph, we get: w 1 (n] [n] y[n] z-• to:z(n] -1/5 "'•(n] 1/4 z-• -5/24 z-• -1/12 (b) To get the difference equation for the flow graph of part (v) in (a), we first define the intermediate variables: w,[n] , w,[n] and w3 [n] . We have: 197 = {1) w1 [n) z[n) + w,[n- 1) 1 1 (2) w,[n) = :(ll[n) + ws[n- 1)- sz[n] 5 1 (3) V13[n) = - 24y[n)- 12 11[n- 1] (4} y[n) = w 1 [n]. Combining the above equations, we get: 1 5 . 1 y[n]- 4y[n -1] + 24y[n- 2] + 12y[n- 3] = z[n)- s1 * -1). Taking the Z-transform of this equation and combining terms, we get the following transfer function: 1- lz-1 H(z)- • - 1- lz-l + .l..z-2 + .l..z-3 4 24 12 which is equal to the initial transfer function. 6.24. (a} 1 H(z) = 1- az 1 y[n] z[n] • • c y[n] HT(z} = z[n) + cy[n- 1) 1 = 1 -cz = H(z} 1 (b) 1 -1 H(z) = 1 + iz 1- !• 1 y[n) z[n) • • 1/2 1/4 198 1 1 y[n] = :t[n] + :~:[n- 1] + y[n- 1] 4 HT(z) = 2 1 + lz-1 1- ,z 1 = H(z) 1 (c) H(z) =" + bz- 1 + cz- 2 y[n] • :r[n] • y[n] = 4:t[n] + bz[n - 1] + c:r[n - 2] HT(z) = "+ bz- 1 + cz- 2 = H(z) (d) H rsinllz- 1 (z) = 1- 2rcosllz 1 + r 2 z 2 y[n] .-1 w[n] -rsinll rsinll rcosll :r[n] 199 v = X +z-1U u = rcosiiV- rsiniiY w = rsiniiV + rcosllz- 1 W y = .-1w y Hr(z) =X = = = rsinez- 1 1- 2r cosBz-1 + r2z-2 H(z).· 6.25. (a) H(z) 1 - llz-1 • + •~z-2 - !z-3 8 (b) y[n] = 9 9 11 7 2z[n] + 8't:[n- 1] + Sz[n- 2] + g-z[n- 3] + Sz[n - 4] 11 5 7 + gll[n- 1] - 411[n- 2] + By[n- 3]. (c) Use Direct Form II: 2 z[n] .-1 11/8 9/8 .-1 -5/4 9/8 .-1 7/8 11/8 .-1 7/8 6.26. (a) We can rearrange H(z) this way: (1 + .-1)2 - 1- lz-1 + z-2 H (z) - ---'-:,...;...~'--:; 2 y[ n] 200 :z:[n] u[n] 0.2 w[n] v[n] y[n] z-1 2 z-1 2 2 2 z-1 -1 z-1 z-1 z-1 -1 1/2 z-1 z-1 -7/8 -1/2 The solution is not unique; the order of the denominator 2ud-order sections may be rearranged. (b) 1 u[n] = :z:[n] + 2:z:[n- 1] + :r[n- 2] + u[n- 1]- u[n- 2] v[n] = u[n]- v[n- 1]- 2"[n- 2] w[n] = v[n] + 2v[n- 1] + v[n- 2] y[n] = 2 1 7 w[n] + 2w[n- 1] + w[n- 2] + 2y[n- 1]- gll[n- 2]. 1 -"If = -· T 2• .. "' (b) For H,(z) H(-z), replace each z- 1 by -z-1 . Alternatively, replace each coeflicient of an odd-delayed variable by its negative. 201 (c) [n] II[n] .-• 2 -1 .-• 2 -1 .-• -2 1 .-• -2 1 6.28. z[n] y[n] b " .-• a w[n] -1 (a) y[n] = z[n] + abw[n] + bw[n- 1] + aby[n] w[n] = -y[n]. Eliminate w[n]: y[n] = z[n]- ablf[n]- by[n - 1] + aby[n] y[n] = :[n]- by[n- 1] So: (b) 1 H(z) = 1 + 6z-1. 202 z(n] y(n] • • -b 6.29. (a) z[n] • I y(n] • (b) From it follows that 7 ~n-n L-az R=0 = 1 -az • -· 1-az 1" (c) z[n] • 1.. a I.-: l . !-·: • y(n] (d) (i) (c) has the most storage: 9 vs. 7. (ii) (a) has the most arithmetic: 7 adds + 7 multiplies per sample, vs. 2 multiplies+ 2 adds per sample. 6.30. (a) 203 ...·-r-----------~-==-~'=-·~-----------. • - .... . •• ·-r-----------~-=-='~-=·-~-----------, . ... . ... .. . • • • • (b) (c) H(~w) = ..!_ 0 - ;7w [sin((1Sw)/2) _!e-Ns sin((1Sw)/2) _! eHr sin((1Sw)/2) ] . 15 sin(w/2) 2 sin((w- ~;)/2) 2 sin((w + (2.. )/15)/2) H(~w) = 1- e -jlSw 15 [ 1 . 1- .-,.., ~ -i~ + 20 . 1- e>fte-iw + 1 e'·~ ] -1--....--"';if;;-e_-j,-w When no = 15/2, . 1 r~, (1- = 15 .-jlSw) eif - e-if !~-- ..;"" (1- .-jlSw) - ei""-s2J•'u' e-r·-fh/l.S) - 204 !ei""+ (1- e-ilW)] eiw+C:I;/15) = e-;•+!2;'UJ _.!._ [·-;...7(~"'¥- .-;...¥) 15 2j sin 'f - i·-;...7.-;fr(~"'¥- .-;...¥) 2j sin ( w-(2;/1$))
!e-jw7 ~fr (~¥w

_ .-;¥"')]

2j sin ( w+(2; /15))

=

e-;w7 [sin(15w/2) _ !e-;fr sin{15w/2) _

15

sin(w /2)

sin ( w-(2; /IS))

!~n sin(15w/2)]
sin ( w+<2; /15))

- ·-

....... ...

..

When

no= 0,
B(ei"')

e-;w7 [sin{15w/2)

= -15-

sin(w /2)

!e-h'< sin(15w/2)

+ sin { w-("; /IS))

+

!~r. sin(15w/2J]
sin ( tot+(2;/15))

.....,.,_ .....

The system will have genesalized linear phase if the impulse sesponse has even symmetry (note it
cannot have odd symmetry), or alt.ematively, if the frequency sesponse can be expressed as:
B(ei"')

=.-;•7 A.(ei"')

205

where A.(ei"') is a real, even, periodic function in w. We thus conclude that the system will have
generalized linear phase for no = Jtk, where k is an odd integer.
(d) Rewrite H(z) as
1
cos (:!!c + ~) z= 1-15z- - + cos~"
1-z-1
1-2cos 2•z-l+z 2
15

H(z)

[

15

1

]

"

•

15

1/15

:tnJ

y"fin]
z-1

z-15

-1

Q

/3

")

-1

where "
6.31.

= cos(2?rno/15), /3 = - cos(2lr(Ro + 1)/15), and 1 = 2 cos(2" /15).

(a)

G

u[n]

-----.----•

:z:[n] •

l: Fl

y[n]

I

-1

(c)

z- 1 - !

.-1

z- 1 - 2

H(z)=(1-lz 1)(1-2z- 1 ).
3

This can he implemented as the cascade of the Bow graph in part (b) with the foUowing Bow graph:

I,
.
l: l
.-1

:r[n] •

-1
However the above Bow graph can be redrawn as:

-1

:r[n] •

. ,-·j

• y[n]

.-1

.-1
I

:

l' I

• y[n]

Now cascading tbe above Bow graph with the one from part (b) and grouping the delay element
we get the foUowiog system with two multipliers and three delays:

-1

!

-:_]-:_:-f_~3--·.'___.. _-. . ~:
-·] :J·~~. . . . . . .

:z:[n]-·

-I

--·y[n]

211
6.36. (a) Transpose

= reverse arrows direction and reverse the inputfoutput, we get:
w,(n]
:r(n] •

_,:,,~
!·-:
l :,
•

W3(n]

"'' (nJ.

•

l' .

z-l

(b) From part (a), we have:

=

(1) w 1 (n]
(2) 1t12(n] = :r(n] + w 1 (n -1]

2:r(n] + w,(n]

1

(3) w,(n] = - y(n] + 2y(n- 1]

2

(4) y(n]

= w,(n] + y(n- 1]

Taking the Z -transform of the above equations, substituting and rearranging terms, we get:
(1- !,-t- 2z- 2 )Y(z)

2

=. (2z- 1 + 1)X(z).

Finally, inverse Z-transforming, we get the following difference equation:
1

y(n]- 2y(n- 1]- 2y[n- 2] = :r(n] + 2z(n - 1].

(c) From part (b), the system function is given by:

H(z)
It has poles at

z=

=

~ + 2z-l

1- 2z- 1

8
-..:...,= and z

-

•.

2z

8
= - -..::....,=

1 - -133
1 + -133
which are outside the unit circle, therefore the system is NOT BIBO stable.
(d)

y(2] = :r(2] + 2z(1] + !Y(1] + 2y(OJ
y(Oj = :r(Oj = 1
y(1] :r[1] + 2z(O] + fy[O] ! + 2 + !

=

Therefore,

6.37.

(a)

=

1
3
19
y[2] = - + 1 + - + 2 = -.
4
2
4

=3

• y(n]

212

H[O]

.-•

...
ii[IJ

.-•
z,
1/N

:r[n]

y[n]

'
'
'
''

'
'
''
'

'

'

H[N- 1]

(b) Note that the •• 's are the zeros of (1- z-N). Then write H(z) over a common denominator:

H(z)

=
N-1

=

}:

H; II
-[

k=O

j

n:-',;,1<1- ••• -• >

N-1

....

(1- z;z-').

·-·

Therefore, H(z) is the sum of polynomials in z- 1 with degree :S N -1. Hence, the system impulse
response has length :S N.
(c)

z-1

[roi:/N)1- 2cos(21rk/N)z 1 + z- 2

21rk!N]]
•

= o also.
k[OJ/16

[n]

y[n]

.-•

-z-16

"

lik[IJI

.-•
{3

Q

.-•
'Y

where u
6.38.

= cos(Q(1]), o = -cos(Q(1]- (2•"/16)), {3 = 2cos(2.. /16), and 'Y = -1.

(a)

xTnJ
h(O]

h(1]

h(2]

h[10]
M:!

M (N

+ 1) multiplies per output sample; M N

215
(b)

-

z-1

'f"n]

M:1

M:1

M:1

h[Oj

h[10]

h[1]

vTnJ
N + 1 multiplies per output sample; N adds per output sample. The number of computations has
been reduced by a factor of M in both adds and multiplies.
(c)

M:1
X

[n]

z-•

-

1111n]

1/"!!
I

The total computation can not be reduced because to compute the value of any given output
sample, the previous output value must be known.

216
(d)

n
I

M:1

.-1

.-I

718

112

nu
M:1

.-1
112

7,8
I

w
M:1

.-1

.-1
1/2
I

7I 8

rIV>
M:1

.-1
718

112

Only direct form IT (ii) can be implemented more efliciently by commuting opefatiODS with the
doWDSalllplers.

6.39. Since each section is 3.4cm long, it takes
3.4cm

-10-·

3.4~ · 1o< -

sec

to traverse one section. Since the sampling rate is 20ldh (T, = 0.5 · 10--.sec), it takes two sampling
intervals to traverse a section. The entire system is linear and so the forward going and backward going

217
waves add at a boundary. Let

A. -A•

Otn=

(from A• into A.); then a••

An+A~r:

= -a•• and we get:
y(n]

•

•

x(n]

,...., = 1

6.40. (a) For rounding:
Ps(e)

1/A

-A/2

A/2

(b) For truncation:
Ps(•)

1/A

-A

218

.!..[ ede =.!.. e']"

m, = A

A 2 _6

-6

=~
2

~=.!..[ e•de_A• = ~]o _A• =A•.
'

A.

4

-6

6.41. Since the system is linear, y[n] is the sum

3A _6

L

=

12

or the outputs due to :z:1[n] and :z:2[n]. Therefore

00

y[n]

4

00

h1 [k]:z: 1 [n- k] +

L

h2[k]:z:2[n- 1.:]

= !lt[n] + 112[n].
The correlation between y,[n] and 112[n] is

E{y,(m]112[n]}

=

E

Lt..,
00

=

h,[l]z 1 [m -l] ·

.t..,

h2[k]:r2[n-

k]}

00

L L

h,[l]h2[k]E{x,[m -l]:r2[n- k]}

t=-oo 1=-oo

If :z:,[n] and :z:2[n] are uncorrelated, E{:r 1 [m -l]:z:2[n- k]}
y,[n] and 112[n] are uncorrelated.

6.42.

= 0; hence, E{y,[m]y2[n]} = 0.

(a) The linear noise model for each system is drawn below.

Therefore,

a

b,

(2a 2 )
(c)

x[n]

•

y[n]

bo

. .-·j

:

b,.

I

I

(3<7')

I

'

l·-:

•

a

(b) Clearly (a) and (c) are different. Thus the answer is either (a) and (b) or (b) and (c). H we take
(b) apart, we get

220

bo
z-1

z-1

a

b1

=

•

I:·-l r F I

•

a

b1

We see that the noise all goes through the poles. Note that the lei' source sees a system function
(1 - az- 1 )- 1 while the 2<1 2 source sees z- 1 /(1- az- 1 ). However, the delay (z- 1 ) does not affect
the average power. Hence, the answer is (b) and (c).
(c) For network (c),

or using the frequency domain formula,

..-} =
=

3u'-l-f
1
. _I_ dz
27rj
1- az- 1- az z
3u'-l-f
dz
21rj
(z- a){l- az)
1

3u'
For network (a),

H(z)

= bo + b,z-1
} - ClZ-l

h{n] = bo.S(n] + (bo + b, )a"u(n]
a
• Time domain calculation:

221

=

2'(b'+'(n]
ft

1
= -21fJ
/H(z)H(z- 1) dz
z

=

L

(residues of

H(z)H(z- 1 )
z

H(z)~(z-') inside unit circle).

(bo + b,z- 1 )(bo + b,z) z
(z:...a)(1-az);
(boz + b,)(bo + b,z)
z(z- a)(I- az)

=
=

residue (z

-b,bo
= 0) = a

2
.d (
) (boa+ b,)(bo + b,a)
~a+ I?, a+ b,bo + b1 boa
res1 ue z =a =
a( 1 - a')
=
o( 1 _ c')
.

~a+

=

-

b,bo + b,boa 2

a(1 - a 2 )
~

=
=
6.43.

I?, a+ b,bo + b,boa2

+ 1?, + 2bob,a
1 -a 2

b' +(abo+ b,)'
1- a2

o

(a)
I

y(n] = y(n- I]+

4

1

y(n]

(I)'

n

I
2,

n

~

0

II-U}n+l

= 2L 4 = 2
i=O

!
•

For large n, y[n] = (I/2)/(3/4) = 2/3.
(b) Working from the diHerence equation and qoa.ntizating after multiplication, it is easy to see that,
in the quantized case, y(OJ I/2 and y(n] 5/8 for n ~ I. In the unquantized case, the output
monotonically approaches 2/3.

=

=

ll
32

y[n] unqoa.ntized

----J---~--~L---J---~----L------

0

I

2

3

4

5

n

222

!

!

i
5

5

!

•I: ~-

--l...-.,-f_.l..__i..L--l-i___._j_ . _ _ j
0

1

3

2

4

5

(c) The system diagram is direct form II:

X(~w)

1

_

2

- 1 + e-Jt..~

So

=

which implies that y(n] (1/2)(1/4)", which approaches 0 as n grows large.
To find the quantized output (working from the difference equation): y(O] 1/2, y(1]
y(n] 0 for n ;:=: 2.

=

=

1

2

1

fi

_,
128

...!...
512

1

2048

=1/8, and

y(n] unqua.ntized

n
0

1

!

!

2

3

4

5

•
y(n] quantized

n
0

2

1

3

4

5

6.44. (a.) To check for stability, we look at the poles loca.tion. The poles are Joca.ted at
z "' 0.52 + 0.84j and z "' 0.52 - 0.84j.
Note that

[z[ 2

"'

0.976 < 1.

The poles are inside the unit circle, therefore the system function is stable.

223

(b) If the coefficients are rounded to the nearest tenth, we have
1.04 -+ 1.0 and 0.98 -+ 1.0.
Now the poles are at

z= 1 -;-13 andz= 1 -;-13_
2

2

Note that now,

[z[• = 1.
The poles are on the unit circle, therefore the system is not stable.
6.45. The flow graphs for networks 1 and 2 respectively are:

· ·z-·D-a• · n.-·
a"'-J

z[n] ----~--w-i,ll-[n_J_ _ _
w..[n_J_ __.._ _ _ _ y[n]

z[n]

.-•
•

~

I

1 1·:

·

z-•

r 1·: r 1·: r I·:

(a) For Network 1, we have:

w1 [n] = z[n]- a8 z[n- 8]
w,[n] = ay[n- 1] + w,[n]
y[n] = w,[n]
Taking the Z -transform of the above equations and combining terms, we get:

Y(z}(1- az- 1 ) = (1- a•z-1 )X(z)
That is:

H(z) = 1 -a • z -· .
1-az- 1

For Network 2, we have:

y[n)

=z[n) + az[n- 1) + a z[n- 2) + ... + a :r[n -7).
2

7

•

y[n]

224
Taking the Z -transform, we get:

= (1 + az- 1 + aZ z- 2 + ... + a7z-7)X(z).

Y(z)
So:

(b) Network 1:

z{n]

Network 2:

z{n]

.-•
1

.,

.-•
a'

••

z-•

z-1

a•

z-•

a•

.-•

z-•

a•

a1

y{n]

(c) The nodes are circled on the fignres in part (b).
(d) In order to avoid overflow in the system, each node in the network must he constrained to have
a magnitude less than 1. That is if w.{n] denotes the value of the ktb node variable and h.{n]
denotes the impulse response from the input z{n] to the node nriahle w•{n] , a sullicient condition
for Jw.{n]J < 1 is
1

z...,. < ~-co Jh>{mJr

In this problem, we need to make sure overflow does not occur in each node, i.e. we need to take
the tighter bound on z,.... For network 1, the impulse response from wo{n] to y{n] is anu{n],
therefore the condition to avoid overllow from that node to the output is

w,...<1-laJ.
Where we assumed tlw JaJ < 1. The transfer function from z{n] to w 1!nJ is 1- a•z-•, therefore
to avoid overflow at that node we need:

w1 {n] < zm.. (l- a 8 ) < 1-JaJ.

225
We thus conclude that to avoid overllow in network 1, we need:
1-[a[
:z:m4%< - 1
••

-a

Now, for network 2, the transfer function from input to output is given by 6[n] +acl[n -1] +a2 6[n2] + ... + a7 o[n -7], therefore to avoid averllow, we need:
1

:.,.... < 1+ Ia[+a2 + ... +,..
,_ 7·
1
(e) For network 1, the total noise power is ,~tr For network 2, the total noise power is 7~. For
network 1 to have less noise power than network 2, we need

~,,<7~.

1- a

That is:

5

[al < 7·
The largest value of

[al such that the noise in network 1 is less than network 2 is therefore

~.

227

Solutions - Chapter 7
Filter Design Techniques

229
7.1. Using the partial fraction technique, we see

H.(s)

=

s+a
(s + a) 2 + b2

=

0.5 _ +
0.5 _
s +a+ ]b s +a- Jb

Now we can use the Laplace transform pair
1

e-01 u(t) +---+ - s+a
to get

h,(t) = ~ ( .-(o+jb)l

+ .-(o-jb)l) u(t).

(a) Therefore,
h 1 [n]

H,(z)

=

jzi

1

(b) Since
s 0 (t)

=

'
1

.

H (s)

-~

h,(T)dT +---+ - '-

> .-•T

= Sc(s)

we get

S

s+a

_

_ A,

,(s)- •(s+a+jb)(s+a-jb)- s

A;

A2

+ s+a+jb + s+a-jb

where

At=

a2

a
+til'

0.5
a+ jb

A,=---

Though the system h2 [n] is related by step invariance to h,(t), the signals 2 [n] is related to s,(t) by
impulse invariance. Therefore, we know the poles ofthe partial fraction expansion of S,(s) above
must transform as Zl = e•• T, and we can find

s, (z) = 1 _Az1 1 + 1 _ e

A2

(a+jb)T z-1

•·
+ .,--..,.-.,....,=--;1 _ e-<• ,.}T z-1

Now, since the relationship between the step response and the impulse response is
n

s,[n]

=

L

oo

h,[k]

=

L

S,(z)

h,[k]u[n- k]

= 1H,(z)
z

= h2 [n]• u[n]

1

We can finally solve for H,(z)

H,(z)

= S 2 (z)(1- z- 1 )
=

1- .-•

1- .-•

A, + A, -:-1 -:'::T.~=--:e-(•+i•)T z 1 + A •:z -:---==r==.,-"'"
1 - e-<•-i6)T -1'
2

izl > .-•T

where A1 and A2 are as given above.

!

230
(c)

t
[1 _
2

h,[k) =

s,[n) =

1=-oo

t(

.-(o+jb)>T + .-(o-jb)•T)

1:=0

.-(o+jb)(n+l)T
1 _ e (o+jb)T

1

=

~

+

1_

.-(o-jb)(n+l)T]
1 _ e-<•-ib)T
u[n]

l

= [Bt + B,e-C•+jb)Tn + B;e-(•-ib)Tn u[n)
where
e-.2) + (1- e-<>·2)

=

(1+;_..)

(W.)
(• + 1)
a(1 + e-<>·•) + (1- e-<>·•)

(s+·~)- (1+!-•• ) (.+·~)

Since the bilinear transform does not introduce any ambiguity, the representation is unique.

233

7.5. (a) We must use the minimum specifications!

6 = O.Ql

= 0.05.-

l!.w

=

A

M

-20log10 6 = 40

A-8

+ 1 = 2.2BS.<:l.w + 1 = 90.2 -+ 91

= 0.5842(A- 21)0 ·4 + 0.07886(A- 21) = 3.395
Since it is a linear phase filter with order 90, it has a delay of 90/2 = 4S samples.
{3

(b)
(c)

Hd(ei"')
1

-0.62511

-11

h.i[n]

0

-0.311

0.311

0.62511

11

(I)

= sin(.625.-(n- 45))- sin(.3.-(n- 45))
.-(n- 45)

7.6. (a) The Kaiser formulas say that a discontinuity of height 1 produces a peak error of 6. If a filter has
a discontinuity of a different height the peak error should be scaled appropriately. This filter can
be thought of as the sum of two filters. This first is a lowpass filter with a discontinuity of 1 and
a peak error of 6. The second is a highpass filter with a discontinuity of 2 and a peak error of 26.
ln the region 0.37r::; lwl::; 0.4757r, the two peak errors add but must be less or equal to than 0.06.

c! + 26 ::; 0.06
6,.... = 0.02

A= -20log(0.02)
{3

= 0.5842(33.9794-

21) 0 "4

=33.9794

+ 0.07886(33.9794- 21) = 2.65

(b) The transition width can be
l:.w

= 0.37r - 0.27r
=

ar

l!.w

We must choose the smallest transition width so l!.w,....
Mis

=

0.525lf - 0.475lf

= 0.05.- rad. The corresponding value of

33.9794-8

M

= 2.285(0.05") = 72.38 -+ 73

7.7. Using the relation w =fiT, the passband cutoff &eqnency, w9 , and the stopband cutotr frequency, w.,
are found to he

w,

=
=
"'• =
=

20'(1000)10-'
2r(l100)10-'

234

Therefore, the specifications for the discrete-time frequency response H•(ei") are
0.99 $IH•(ei")l$

1.01,

IH.(ei"ll $O.Gl, 0 :Siwl$

0.201<

0.22 .. $'"''$ ..

7.8. Optimal Type I filters must have either L + 2 or L + 3 alternations. The filter is 9 samples long so its
order is 8 and L = M /2 = 4. Thus, to he optimal, the filter must have either 6 or 7 alternations.
Filter 1: 6 alternations
Filter 2: 7 alternations
Meets optimal conditions
Meets optimal conditions
7.9. Using the relation w

=nT, the cutoff frequency w, for the resulting discrete-time filter is
We

=

flcT

=
=

[2.. (1000)][0.0002]

7.10. Using the bilinear transform frequency mapping eqnation,
w,

=

2tan- 1

Cl;T)

=

2 tan -•

C"(2000)(~.4 x

w-•))

7.11. Using the relation w

= !lT,

n, =
=
=
=

w,
T
1 1
which has the impulse response

h[n]

Td (u[n] + u[n- 1J)
= 2"

238
(b) The difference equation is

y(n] =

r. (:z:(n] + :z:(n- 1]) + y(n- 1]
2

This system is not implementable since it has a pole on the unit circle and is therefore not stable.
(c) Since this system is not stable, it does not strictly have a frequency response. However, if we ignore
this mathematical subtlety we get

= r. (l+e-j"')
=·

=

2

1-e '"'

2

e;../2- .-;..12

r. (d--'2 + .-;..'•)
T•

j cot(w/2)

2

and since the Laplace transform evaluated along the j!l axis is the continous..time Fourier transform
we also have

I H(ei"') I

T(2

0

-It

!tl2

It

Ol

0

-It

L H (j.Q)
c

LH(ei"')
!tl2

0

lt

0

!tl2
It

lt
-It

!tl2

Ol

-lt

0

0

-!t12

-!t12

1n general, we see that we will not be able to approximate the high frequencies, but we can
approximate the lower frequencies if we

chooser.= 4/'lf.

(d) Applying the bilinear transform yields

G(z)

JzJ > 1
which has the impulse response

g(n]

239

(e) This system does not strictly have a frequency response either, due to the pole on the unit circle.
However, ignoring this fact again we get

=

G(.,;w)

=

;d [! ~ :=;:]
2
Tc~

=
=

G(jO)

(.,;w/2 _ e-iw/2)
eiw/2

+e

jw/2

i. tan(w/2)
2"

jO

~ I G(ei"') I

0

-II

1112

II

"'

0

-II

1112

LG (jO)

LG(ei"')
1112

c

-II

7112

-II

0

II

1112

"'

0

II

0

-1112

Again, we see that we will not be able to approximate the high frequencies, but we can approximate
the lower frequencies if we chooser.= 4/7f.
(f) If the same value of T• is used for each bilinear transform, then the two gystems are inverses of
each other, since then

240

10lt

-0.11t

-lt

0.11t

"

(J)

Then, to get the overall system response we scale the frequency axis by T and bandlimit the result
according to the equation

IH..-,(jfl)l = {

,(&wT)I, 11"11
11"11

OIH'

<;.
>;.

n
(b) Using the frequency mapping relationships of the bilinear transform,

0 = :.tan(~),
0
w = 2tan- ( ~·) '
1

we get

10lt

--

0.981t

-lt

Then, to get the overall system response we scale the frequency axis by T and bandlimit the result
according to the equation
H

I ....

(jfl)l - { IH,(&wT)I, 101 < f
0,
101 >;.

241

10lt

-98001t

7o24o (a) Expanding the sum

to

see things more clearly, we get

A

r

=

L (s- ••o)• + Ge(s)

bt

A- + ( A )2+000+( A. ) + GeS
( )
= •-.so
s-.ao
s-so r
2

1

~o)•

Now multiplying both sides by (• -

we get

(•- •ol' He(•) = A1(•- •o)•-l + A•(•- ~or-• + 000+A.+ (•- •ol'Ge(s)
Evaluating both sides of the equal sign at s

= so gives us

= (s- •ol' He(•) J..,,,
Note that (s- so)•Ge(s) =0 when s = 'o because Ge(•) bas at most one pole at s =SoAr

Similarly, by taking the first derivative and evaluating at s

~ [(s- sol' He(s)] =

r

.

L:-1
1
(k _ 1)! ·-"'u(t) -+ (• + a)•,
we get

"IU{s} > -a

= so we get

242

7.25. (a) Answer: Only the bilinear transform design will guarantee that a minimum phase discrete-time
filter is created &om a minimum phase continuous-time filter. For the following explanations
remember that a discrete-time minimum phase system bas all its poles and zeros inside the
unit circle.
Impulse Invaria.nce: Impulse invariance maps left-half 6-plane poles to the interior of the z-plane
unit circle. However, left-half s-plane zeros will not necusarilN be mapped inside the z-plane
unit circle. Consider:

H(z)

H we define Poly.(z) =

nf.,
{I-e•;T•z- 1), we can note that all the roots of Poly.(z) are
j"¢Jc

inside the unit circle. Since the numerator of H(z) is a sum of A•Poly.(z) terms, we see
that there are no guarantees that the roots of the numerator polynomial are inside the unit
circle. In other words, the sum of minimum phase filters is not necessarily minimum phase.
By considering the specific example of
s+ 10

= (s + 1)(s + 2)'

He(s)

and using T = 1, we can show that a minimum phase filter is transformed into a non-minimum
phase discrete time filter.
Bilinear 'Iransform: The bilinear transform maps a pole or zero at s = So to a pole or zero
Thus,
(respectively) at Zo =

..

,'+!••.
-

1

+f..

l.zol = 11- -.•o I
Since He( s) is minimum phase, all the poles of He(s) are located in the left half of the s-plane.
Therefore, a pole s 0 = a+ jrl must have a < 0. Using the relation for 1 0 , we get

l.zol =

(1 +
(1-

fa)• + (tfl)2
fa)• + (f0)2

< I
Thus, all poles· and zeros will be inside the z-plane unit circle and the discreu-time filter will
be minimum phase as well.
(b) Answer: Only tbe bilinear transform design will result in an allpass filter.
Impulse Invariance: In the impulse invariance design we have

H(e'w)

=.~..,He (j (;. + 2;:))

The aliasing terms can destroy the all pass nature of the continuous-time filter.

243
Bilinear Transform: The bilinear transform only warps the in!quency axis. The magnitude
response is not affected. Therefore, an all pass filter will map to an all pass filter.
(c) Answer: Only the bilinear transform will guarantee

H(ei"')l .. =<~

= H,(jO)In=<~

Impulse lnvariance: Since impulse invariance may result in aliasing, we see that
H(ei")

= H,(jO)

if and only if

or equivalently

which is generally not the case.
Bilinear Transform: Since, under the bilinear transformation, 0

= 0 maps tow= 0,

H(ei 0 ) = H,(jO)
for all H,(s).
(d) Answer: Only the bilinear transform design is guaranteed to create a bandstop filter from a
bandstop filter.
If H, (s) is a bandstop filter, the bilinear transform will preserve this because it just warps the
frequency axis; however aliasing (in the impulse invariance technique) can fill in the stop band.
(e) Answer: The property bolds under the bilinear transform, but not under impulse invariance.
Impulse Invariance: Impulse invariance may result in aliasing. Since the order of aliasing and
multiplication are not interchangeable, the desired identity does not hold. Consider Do, (s) =
H.,(s) .-•T/2.
Bilinear Transform: By the bilinear transform,

=

H(z)

(2(1-z-'))

= H, T• 1 + z-1

G: :=:))

-

H" (:.

=

H,(z)H,(z)

H" (:.

G: :=:))

(f) Answer: The property holds for both impulse invariance and the bilinear transform.
Impulse Invariance:

H(ei"') =

=

.fao
f

1:=-cc

H,

(j (;. + ~: k))

He~(;(;.+ ~k)) +

= H 1 (ei"') + H2 (e'"')

f H.,(;(;.+ ~k))

~=-CCI

244

Bilinear Transform:

H\~)

=

H.\~R\-+·~'0)

H.,(;. G::=:)) +H.,(:. G::=:))

=
= H1 (z) + Hz(z)

(g) Answer: Only the bilinear transform will result in the desired relationship.
Impulse Invariance: By impulse invariance,

H, (e'w)

= ·

f: H., (; (;. + 2;k))
f:
+ 2;•k))
4

A:=-oo

Hz (e'w)

=

H., (; (;.

11:=-oo

We can clearly see that due to the aliasing, the phase relationship is not guaranteed to be
maintained.
Bilinear Transform: By the bilinear transform,
H 1 (e'w)

= He, (; :. tan(w/2))

Hz(e'w) = He, (; :. tan(w/2))
therefore,

7.26.

(a) Since

and we desire

we see that

requires

f: H. (i~k) =

·--.....

0.

(b) Since the bilinear transform maps ll = 0 tow = 0, the condition will hold for any choice of H.Uil).
7.27.

245

(a)

h1 [n] = h[2n]
H, (e'w)

=

L""

h[2n]e'wn

n=-oo

=

I:

h[n]e"i"

f.

~ [h[n] + (-lth[n]]e'"f

n even

=

n.=-oo

I
¥ ) + 2H
I
( e>--r·•'")
= 2H(e'

H, (ei"')

1/2

0

-w/2

-lt

7[

(1)

7[

(1)

lt

(1)

(b)

H2(dw) =

I:

h[n/2]e-Jwn

n even

=

"" h[n]e-Jw2n

I:

n.=-oc

= H (e'>w)
H (ei"')
2

-

-

1

-x-7lt!8

7lt!8

-lt/8 0 lt!8

(c)

1

-lt

-3lt!4

0

3lt!4

246
7.28. (a) We have

•

=

;n =

1- z- 1
1 + z-1
1- e-jw

1 +e

jw

eil!ll/2 _ e-Jw/2

= eJw/2 + e j,.;/2
n = tan(~)
n. =tan (1'-) .......... w,. = 2tan- 1 (0,)
(b)

•

=

jO

=

=
n =
=
n. =

1 + z- 1

1- z- 1
1 + e-jw
1- e-iw
+ e-iw/2
ei,.,/2 _ e-iw/2
eJ~i~~/2

-cot
tan

G)

("'~w)

-w)

tan ( w "'2

(c)

=> w:P2

=wPl +

7r

(d)

The even powers of z do not get changed by this transformation, while the coefficients of the odd
powers of z change sign.

Thus, replace A, C, 2 with -A, -C, -2.
7.29.

(a) Substituting Z

= ei' and z = &w we get,
ei'

=
=

(b)

_.,..,
ei<""'+•)
8-1f
w=-2-

247
H, (ei"')

ec

ec

A

0

-1112

-It

It

(I)

(c)

+-+

h[n]

B(,;')

h 1 [n] +-+ B ( ,;<2w+•l)
In the frequency domain, we first shift by
can write that as
h [n]
1

1r

and then we upsample by 2. In the time domain, we

= { (-l)nl2 h[n/2],
O,

for n even
for n odd

(d) In general, a filter
H(z)

= bo + ~z-l + b,z-• + · · · + bM-IZM-l + bMz-M
Qo

will transform under Hl(z) =
H 1 ( z)

+ O.tz-l + 0.2Z

2

+ · · · + O.N-lZN

l

+ O.Nz-N

H(-z2 ) to

bo- b,z-2 + ~>,z-• +.--- b.v_,z>M-2 + bMz-•M
= GoO.tz-2 + a2z-4. + ... - 0.M-1Z2N-2 + O.Nz-2N

~--"-'.::..__..-':-=-::-.-.:,--_..:;.:;c::..:.;:....,o-:;..:,..:=.:.=

where we are assuming here that M and N are even. All the delay terms increase by a factor of
two, and the sign of the coellicient in front of any odd delay term is negated.
The given difference equations therefore become

=
=
=

g(n]
f[n]
y[n]

z[n] + a,g[n- 2]- b!/[n- 4]
-a.g[n- 2]- b,f[n- 2]
cd[n] + c2g[n - 2]

To avoid any possible confusion please note that the b, and
not the same b, and a, sbown above for the general case.

a,

in these difference equations are

7.30. We are given·
H(z)

= He(s) I -/3[=]
1+·-·
I=

where a is a nonuro integer and {3 is a real number.
(a) It is true for {3
Proof:

> 0.

•
S

+ $Z-a: s-{3 {3-. z-" zo z-o] = {3 [1l+z = {3- {Jz-o = -Pz- sz-o = z- ({3 + s) {3-. = fJ+• 0 4 - 0 {3 +. = fJ-. ! 248 The poles •• of a stable, causal, continuous-time filter satisfy the condition 1U {s} < 0. We want these poles to map to the points in the z-plane such that Jz•l < I. With a > 0 it is also true that if lz•l < 1 then lzfl < 1. Letting •• = u + jw we see that z• < < I.B+a + iOI < (,8 + a) 2 + 0 2 < 2<7,8 < lz•l lzfl 1 1 1.8- u- iOI (,8- a) 2 + 0 2 -2<7,8 But since the continuous-time filter is stable we have &{ ••} < 0 or a < 0. That leads to -,8 < .8 This can only be true if ,8 > 0. (b) It is true for ,8 < 0. The proof is similar to the last proof except now we have jz0 j > !. (c) We have z• = ~~:Ln 2 Jz 1 = 1 jzj = 1 Hence, the jO axis of the s-plane is mapped to the unit circle of z-plane. (d) First, find the mapping between 0 and w. 1- e-i'lw 1 + e-i2w eJw- e-il,, = j{l = e~w n = = w + e-iw lan(w) tan- 1(0) Therefore, 1- 6, { '"'' :s ~} u { 34" :s IH(e'~JI :S 1 + 6,, < '"'' M (b) Since we only have control over e[n] for 0$ m S M, we get that
for 0$n SM. (c) w(n] which is a rectangular window. = { ~: OSnSM, otherwise. <' is minimiud if h(n] = h..(n] 250 7.33. (a) Hd(~w) =(I - 2u{w)J~(rf2-rw) = 1, i- 'TW, IHd(~w)l for-" < w < 11" 'lw -j -Tw, :----JW2 -It 0 -W2 r-------.: (b) A Hilbert transformer of this nature requires the filter to have a zero at z = 0 which introduces the 180° phase diHerence at that point. A zero at z = 0 means that the sum of the filter coefficients equals zero. Thus, only Types ill and IV fulfill the requirements. (c) '1· For the -windowed Fm system to be linear phase it must be antisymmetric about Since the ideal Hilbert transformer h..[n] is symmetric about n = T we should choose T = (d) The delay is M/2 = 21/2 = 10.5 samples. It is therefore a Type IV system. Notice the mandatory '1· zero at w = 0. 1 0!---------------------~It 251 (e) The delay is M /2 = 20/2 zeros at w = 0 and ,.. . = 10 samples. It is therefore a Type ill system. Notice the mandatory lH(ei"')J 1 0~--------------------~ " 7.34. (a) It is weU known that convolving two rectangular windows results in a triangular window. Specifically, to get the (M + 1) point Bartlett window forM even, we can convolve the foUowing rectangular windows. r,lnJ = { n=O, ... ,~ -1 0, otherwise [k, ••In] = rt(n- 1] Using the known transform of a rectangular window we have WR,(eiw) = WR,(eiw) = Ws(eiw) = [i"sin(wM/4),-iw(lf-!) sin(w/2) VM {2 sin(wM/4) ,-,w( 'f+!) VM sin(w/2) WR, (eiw)WR,(eiw) = 2_ (sin (wM/4)) M sin(w/2) 2 -jwJI/2 e Note: The Bartlett window as defined in the text is zero at n = 0 and n = M. These points are included in the M + 1 points. For M odd, the Bartlett window is the convolution of r 3 1n] ••In] { .[b. = { .[b. = n-O M-l - , ... ,--y-- 0, otherwise nM-1 - 1, ... ,-20, otherwise In the frequency domain we have w•. (eiw) = ·~ W •(ei"') = Ws(ei"') = - {2sin(w(M +1)/4) -iw(¥) sin(w /2) e VM {2sin(w(M -1)/4) -jw(J¥+I) sin(wf2) e VM WR,(ei"')WR.(eiw) = 2_ (sinlw(M + 1)/2]) (sinlw(M- 1)/2]) -iwM/ 2 M sin(w/2) sin(w/2) e 252 (b) w[n] = [A+ Bcoo Ccoo (~n)] WR[n] = {21rA6(w) +Blr [6 (w + !;) +6 (w- !;) ] +C+ (w + ;;) + 6 (w- ;;)]} W(.;w) ~ { sin(w(M + 1)/2)) 21r sin(w/2) where C:) + @ .-jw/11/'} denotes periodic convolution. (c) For the Hanning window A Wftanning[n] Wftanning(.;w) =0.5, B = -0.5, and. C =0. c:)] w,[n] = [o.5- 0.5coo = = 0.5WR(.;w)- 0.25WR(.;w) ® [6 (w + !;) + 6(w- !;) l 0.5WR(.;w)- 0.25 [wR(.; 50 21$ A $50 A< 21 we can solve for A in the following manner: 1. We know P = 3.68. Therefore, from the formulas above, we see that 2. If we assume A > 50 we find, 3.68 = 21. 0.1102(A- 8.7) = A A~ 42.1 But, this contradicts our assumption that A > 50. Thus, 21$ A $50. 3. With 21 :5 A :5 50 we find, 3.68 A = = 0.5842(A- 21) 0·' + 0.07886(A- 21) 42.4256 With A, we can now calculate 6. 6 = 10-A/20 = 10-42.425<1/20 = 0.0076 The discontinuity of I in the first passband creates a ripple of 6. The discontinuity of 1/2 in the second passband creates a ripple of 6/2. The total ripple is 36/2 = 0.0114 and we therefore have 6, = 0. = 6, = 0.0114 Now using the relationship between M, A, and t:.w M t:.w = = A-8 2.285t:.w 42.4256-8 2 _28S(4S) =0.3139"' O.llr Putting it all together with the information about Hd(eiw) we arrive at our final answer. 0.5 o+--------r~~=-==~~~----------r--0 ! 254 7.36. (a) Since H(ei0 ) ¥0 and H(ei•) ¥0, this must be a Type I filter. (b) With the weighting in the stopband equal to 1, the weighting in the passband is ~W(Ol) 1.6f---------. 1 00~------------~~----~~--------------~----(1) 0.4lt 0.58lt It (c) IE(Ol)l (d) An optimal (in the Parks-McClellan sense) Type I lowpa.ss filter can have either L + 2 or L + 3 alternations. The second case is true only when an alternation occurs at all band edges. Since this filter does not have an alternation at w it should only have L + 2 alternations. From the figure, we see that there are 7 alternations so L 5. Thus, tbe filter length is 2L + 1 11 samples long. = .. = = (e) Since the filter is 11 samples long, it has a delay of 5 samples. (f) Note the zeroes off the unit circle are implied by the dips in tbe frequency response at the indicated frequencies. 0 Re 0 101h order pole 7.37. (a) The most straightforward way to find hc(n] is to recognize tbat H•(ei") is simply tbe (periodic) convolution of two ideallowpa.ss filters witb cutoff frequency "'• = ../4. That is, where 255 Therefore, in the time domain, hd[n) is (h1p 1 [n)) 2, or 2 = ('in:/4)) sin2 (.-n/4) = (b) h[n) must have even symmetry aronnd (N- 1)/2. h[n) is a type-1 FIR generalized linear phase system, since N is an odd integer, and H(eiw) ¥- 0 for"'= 0. Type-1 FIR generalized linear phase systems have even symmetry aronnd (N- 1)/2. (c) Shifting the filter hd[n) by (N - 1)/2 and applying a rectangular window will result in a causal h[n) that minimi:r.es the integral squared error •· Consequently, sin• ("(n- N-1 l] 2 h[n) = w[nJ = { 1, where • w[n) .-•(n-N21)2 0$ n$N - 1 0, othei"W!Se (d) The integral squared error < can be reformulated, using Parseval's theorem, to 00 • = L Ja[n)- hd[n)J 2 -oo Since an= [I { hd[n) 0, -(N-1)/2-1 < L = _!!=1. < n 2 otherwise ' < - L Ja[n)- hd[n)J + Ja[n)- hd[nll' + -(N-1)/2 L -oo 2 (N-1)/2 2 -(N-ll/2-1 = N-1 "" Ja[n)- h•[n)J :E (N-1)/2+1 2 00 2 Jhd[n)J + 0 + L Jhd[n)J 2 (N-1)/2+1 By symmetry, •=2 L"" Jhd[n)J 2 (N-1)/2+1 7.38. (a) A Type-llowpass filter that is optimal in the Parks-McClellan can have either L + 2 or L + 3 alternations. The second case is true only when an alternation OCCUIS at all bal!d edges. Since this filter does not have an alternation at "' 0 it only has L + 2 alternations. From the figure we see there are 9 alternations so L = 7. Thus, M = 2L = 2(7) = 14. = 256 (b) We have hHp(n] HHP(ei"') = -ei'"h£p(n] = -HLP(ei(w-•l) -A.(ei-time filter would not have a constant group delay. The bilinear transformation maps the entire jfl axis in the s-plane to one revolution of the unit circle in the z..plane. Consequently, the linear phase of the continuous-time filter will get nonlinearly warped via the bilinar tranSform, resulting in a nonlinear phase for the discrete-time filter. Thus, the group delay of the discrete-time filter will not be a constant. 7.42. (a) Using the fact that H,(s) = }:\:\ and cross multiplying we get H ( ) = Y;,(s) ' s X,(s) (s + c)Y;,(s) dy,(t) + cy,(t) dt = A s+c AX,(s) = Azo(t) = (b) dy,(t) dl. I t=nT y,(nT) - y,(nT- T) T = [Az,(l)- cy,(tlJJt=nT = Az,(nT)- cy,(nT) ::: Az,(nT)- cy,(nT) 258 (c) y[n] -y[n- 1] Az[n] - cy[n] = T (c+ f) y[n]- fll[n -1] (c +f) c+ +- +•-• Az[n] = H( ) z AX(z) = = Y(z) = X(z) Y(z)- fY(z)z-• A (d) H,(s) J ,_,_, = ~---.- = = s~c~ -=-.•-·-• A ~+c H(z) (e) First solve for z and then substitute s 1- .-• =T % = 1. 1-sT = "+ jn to get z = = If we let 8 s 1 1- (u + ;n)T ei--·us> 1 V(1- u)• + (nT) 2 = tan-• ( J':.) we see that = 1 J(l- u)2 + (nT) cos(8) 1- (7 2 1 ( .• ..) = 2(1- <7) e' + .-, and thus the s-plane maps to the z-plane in the following manner z = [2(1 1- u) (ei' + .-;•)] ei' = 1 + 1 e'"211 2(1- <7) = 2(1 1 2(1- <7) <7) + 1 2(1- <7) To find where the ;n axis of the s-plane maps, we let • 1 z =- 2 ei>--·ce.-> =;n, i.e., u = 0 and find 1 _;2 ... "'(0T) + 2-~ 259 Therefore, the jO-uis maps to a circle of radius 1/2 that is centered at 1/2 in the z-plane. We also see that_ the region u < 0, i.e., the left half of the s-plane, maps to the interior of this circle. s-plane z-plane If the continuous-time system is .table, it. poles are in the left half s-plane. As shown above, these poles map to the interior of the unit circle and so the discrete-time system will also be stable. The stability is independent of T. Since the jO-uis does not map to the unit circle, the ~ete-time frequency response will not be a faithful reproduction of the continuous-time frequency response. As T gets smaller, i.e., as we oversample more, a larger portion of the jO-uis gets mapped to the region close to the unit circle at w = 0. Although the frequency range becomes more compressed the shape of the two responses will look more similar. Thus, as T decreases we improve our approximation. (f) Substituting for the first derivative in the differential equation obtained in part (a) we get y,(nT + T) - yo(nT) T ( T) = = Ax[nJ H,(s) 1,-•-• _...,..... + cy, n y[n + ~- y[n] + cy[n] Y(z) H(z) =X( z ) = s = z = = A = 1 T+c Ax,(nT) z-1 T 1+sT I+ CT +jOT To find where the jO axis of the s-plane maps, we let s = jO, i.e., " = 0 and find z =I +jOT Therefore, the jO-uis lies on the line ~{z} = 1. We also see that the region u < 0, i.e., the left half of the s=plane, maps to the left of this line. s-plane z-plane ! 260 If the continuous-time system is stable, its poles are in tbe left half •-plane. As shown above, these poles can map to a point outside the unit circle and 10 the discrete-time system will not necessarily be stable. There are cases where -varying T can tum au oostable oystem into a stable system, but it is not true for the general case. Since the jO-axis does not map to the unit circle, the discrete-time frequency respoDOe will not be a faithful reproduction of the continuous-time frequency response. However, as T gets smaller our approximation gets better for tbe same reasons outlined for the first backward difierence. 7.43. (a} Just doing the integration reveal$

AT
/.nT-T

ile (T)dT + Ye(nT- T) =

lle(T}I:t-T + 1/e(nT- T) = lle(nT)

Using the area in the trapezoidal region to replace tbe integral above, we get

lle(nT)

=

Ye (T)dT + 1/e(nT- T)

/.AT
nT-T

"" [il. (nT)+ 'ile (nT- T)] ~ + 1/e(nT- T)
(b} Solving for i; < ( nT) in the difierential equation we get

il. (nT) = Az.(nT)- cy.(nT)
Substituting this into the answer &om part (a} yields

T

y.(nT) = (Az.(nT) - cy.(nT) + Aze(nT- T)- cy.(nT- T)] 2

+ y,(nT- T)

(c) The difierence equation is

y[n]

= (Az(nj- cy[n] + A.z[n- 1]- cy[n- 1]} T2 + y[n- 1]

y[nJ(1+c~) -y[n-1] (1-c~) =A~(z[n]+z[n-1]}
Therefore,
Y(z)

[1 + c~]- Y(z)z- [1- c~] = A~X(z) (1 + z1

Af(1 + z- 1 )
H(z) = X(z) = 1 + cf- z 1 + z-14
Y(z)

(d)

=

=

1- z

1

+ cf(l + z

H(z)

7.44.

+.(jO)
+(z)

=

=

H.(jO)H.(-jO)
H(z)H(z- 1 )

1}

1

]

261
(a} (i} Since He(•} has poles at ••• H.( -•} has poles at -••·
(ii} The material in this chapter shows that under impulse invariance

A•

T•A•

- +---+ 1- ~ltT,.z-1 .
8 - Sj;

Thus, going from step 1 to step 2 means that the autocorrelation of the discrete-time system
is a sampled version of the autocorrelation of the continuous-time system.
(iii} Since +(z) = H(z)H(z- 1) we can choose the poles and zeros of H(z) to be all the poles inside
the unit circle, and that choice leaves all the poles and zeros outside the unit circle for H(z- 1 ).
Consider the following example using h.(t) = ,-a•u(t).
H.(s)

1
=s+Q

+.(s)

+(z}

and

H.(-s)

=
=

H.(s)H.(-•i

=

s+o s-o

1
=-s+o

[.~a] [-s~a]
1/2o _ 1/2o

=

if a> 0, then

h[n] =
(b) Since IH.(jn)l 2

=+.(in)

l::.

and +(ei"')

T.¢.(mT.),

= H(ei"')H(e-;..} = IH(ei"')i', we see that since ¢[m] =
.

+(ei"')
Therefore, if +.(in)

(1- e->aT•) (e-oT•f u[n]

=

J;,., •. (j (;: + 2;./c)).

=:: 0 for 1n1 2! f;, then +(ei"') =::+.{if.;)

and IH(ei"')i'

=:: jH. (ii;) j' ·

(c) No. We could always cascade H(z) with an allpass filter. The new filter is di1ferent, but has the
same autocorrelation.

262
7.45.

(a) Since the two 8ow diagrams are equivalent we have

z-1

=

z =

.

z-1-a
1-az 1

z-a

1-az

z-a
(z-a)
1- -

=---

1-az

H(z) = H,,(Z)Iz= t--a
·-· = H,,
(b) Let Z

- -

QZ

= ei8 and z = ei"'. Then

z-1 =

z- 1 -a
1 az- 1
e-iw- Q
1- ae-Jw

e-;e- oe-i'e-iw
,-;"'(1 + ae-;8 )

=

e-jw- Q

=

e-il +

0

e-;• +a
1 + ae i 6
e-i8

+a

1 + aei 1
1 + aei 6

=

1+

=

,-;• + 2a + a'ei'
1 + 2ocos8 + a 2

ae-i8 ·

Using Euler's formula,
e-jw

=
=

Noting that -w

cose- j sin9 + 2a + a 2 cos9 + ja2 sinO
1 + 2ocose + a 2
2o + (1 +a') cose + j[(a'- 1) sinO]
1+2ocos9+ci'

= tan-• [ Zt:/]'
-w
w

=

tan-• [

= tan-• [

(a'- 1) sine

2o + (1

+ a 2 ) cose

]

2

(1- a ) sine ]
2a + (1 + oZ) case

This relationship is plotted in the figure below for a = 0, ±0.5.
Warping of the frequency scale, LPF to LPF
(1)

263
Although a warping of the frequency seale is evident in the figure, (except when o = 0, which
= .-• ), if the original system bas a piecewise-constant lowpass frequency
corresponds to
response with cutofF frequency 89 , then the transformed system will likewise have a similar lowpass
response with cutofF frequency w9 determined by the choice of o.

z-•

w _
9

tan-• [

(1 -a') sin(8,) )
2o + (1 + o2) cos(89 )

-

(c)
xjnl----r-------r---ytn]
xjn] ---+--~.---l----.l----+l y[n]
1

H(z)

r

H.,(Z)

- o.p Looking at tbe 6ow graph for H(z) we see a feedback loop with no delay. This effectively makes
the current output, y[n], a function of itself. Hence, there is no computable difference equation.
(d) Ye.<, the 6ow graph manipulation would lead to a computable difference equation. The 6owgraph
of an FIR filter has a path without delays leading from input to output, but this does not present
any problems in terms of computation. Below is an example.

bo

xjnJ---~--=--,----Yfn]

iz-1

The transformation would destroy the linear phase of the FIR filter since the mapping between 8
and w is nonlinear. The only exception is the special case when o = 0, i.e., when 8 = w.
Since there are feedback terms in the transformed filter, it must be an IIR filter. It therefore has
an infinitely long impulse response.
·
(e) Since the two 6ow diagrams are equiva.lent we have

=

z

Z

=

z-a
z-1-az

H(z)
Letting Z

=&

8

and z

z-• -a

z-•

-1

1- az- 1

=z

az
-z- a

-11 -

z--o)
= H, 9 (Z)Iz=• t-.a·-· = H,9 (z
1QZ

= ei"' we have,
= eiwei""-~

1-aeJ"'

· ei"" - a

1 - ae-;v,
. .
.
1 - aeJw 1 ae-J111

=

eJW

=

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

· ei"' - 2o + a'e-;"'
1- 2ocosw+oZ

264
Using Euler's formula,

ei' = .,;--(1 +a')cosw- 2a+ j(l- a')sinw
l-2acosw+a'
Noting that 8

= w + tan-• [ ~H

],

8 ="'+tan _ 1

[

(1- a') sinw ]
(1 +a') cosw- 2a

(I)"

2!t .......... .

-It

rc/2

It

9

-It

............ -211
We see from the plot of w ..,rsus 8 that a lowpass filter will not always transform into a lowpass
filter. Take, for example, the case when the originallowpass filter has a cutoff of 8 "f2. With
a = 0 it would transform into an allpass filter.

=

7.46.

(a) Since

=

y[n]

(2z[n]- h[n]• z[n]) • h[n]

= (2h[n]- h[n]• h[n]) • z[n]
the new transfer function is
g(n]

=2h(n]- h[n]• h(n]

(i) It is FIR since the convolution of two finite length sequences results in a finite length sequence.
(ii) Note that the term h[n] • h[n] is symmetric since it is the convolution of two symmetric
sequences. Therefore, g(n] must he symmetric since it is the ciliference of two symmetric
sequences.
(b) The frequency response for G(&~) is

G(&w)

= 2H(ei~) -

H(.,;~)H(ei~)

265

G(ei"')

1~ '----------...1
1~

1~

1

-f.1L----------

-2S2

~

H(ei"')

0

\

H(ei"')

As shown above, if the passband of H(eiw) is the region [1- 6,1 +6,], then the passband of G(&w)
is in the region (I - ~, 1] which is a smaller band. However, the stop band gets bigger since it
maps to (-26,- 6i, 262- 5i].

Thus,

=
B =
c =
D =
A

If 61

(1-6?}
1
-26, -6~
26,

-6i

« 1 and 0:, « 1 then,
Maximum passband approximation error "" 0
Maximum stopband approximation error "" 20:,

(c) Since

y(n]

=

(3z[n] - 2z(n] • h(n])• h(n] • h(n]

= (3h(n] • h(n] - 2h[n] • h(n] • h(n]) • z[n]
the new transfer function is

hsharp(n]

= 3h(n] • h[n] -

2h[n] • h(n] • h[n]

and so

The new tolerance specifications can he found in a similar manner to the last section. We get,

A = I-36?-26~
B
1
0
D = 36i+Ul

=
c =

If 61

«

1 and 0:,

«

1 then,
Maximum passband approximation error "" 0
Maximum stopband approximation error "" 0

266
(d) The order of the impulse response h(n] isM. Since it is linear phase it must therefore have a delay
of
samples. To convert the two systems we must add a delay in the lower leg of each network
to match the delay that was added by the first filter.

"f

l(nl------;f,...------1~'---~-n)- T - h [ n _ J__,t--y(n]
2z......

l(n]---,tr---i

h(n)

IJ

j

2

h[n)

H.__hl_nJ_.~y(n]

3z....,

The restrictions on the filter that carry over from part (a) are that it have
(i) Even symmetry
(ii) Odd Length
Hence, Type I Fffi filters can be used.
The length of h[n] is 2L+ l. Since the term that is longest in the twicing system's impulse response
is the h[n] • h[n] term, the length of g[n] is 4L + 1. Since the term that is longest in the sharpening
system's impulse response is the h[n] • h[n] • h[n] term, the length of h5 harp[n] is 6L + 1.
7.47. We know that any system whose frequency response is of the form
L

A.(~"')= L:a•(cos(w))•
1=0

can have at most L - 1 local maxima and minim• in the open interval 0 < w < .- since it is in the form
of a polynomial of degree L.

If we include all endpoints of the approximation region

{0!:: lwl !:: wp} U{w, !:: lwl !:: 1f}
then we see we can have at most L

+ 3 alteruation frequencies.

If the transition hand has two of the local minim• or m•xim• of A.(ei"'), then only L- 3 can be in the
approximation bands. Even with all four endpoints of the approximation region as alternation points,
we can only have a maximum of L + 1 alteruation points. This does not satisfy the optimality condition
of the Alternation Theorem which requires ill least L + 2 alternation points. It follows tbal the transition
band cannot have more than two local minim• or maxim• of A.(ei"') either.
If the transition band only has one of the local minim• or m•xima of A.(ei"'), then the error will not
alternate between w7 and w, and they cannot both be alteruation frequencies. In this case, only L- 2
of the local minima or maxima of A. (ei"') are in the approximalion bands. If we add the maximum of
three band edges to the total count of alternation frequencies we get L + 1, which is again too low.
· Therefore, the transition band cannot have any local minima or m•xim• and must he monotonic.
7.48. (a) A.(ei"') has 7 alternations of the error. If the approximati'>D bands are of eqnaliength and the
weighting function is unity in both bands, why would the stopband have 1 extra al&ernation than
the passband? The answer is thal, lf it were an optimal filter, it would not. The optimal filter for
this set of specifications shonld have the ICmt number of alternations in each hand and therefore
requires an even number of alternations. Since the optimal approximation is unique, the one shown
in the figure cannot be optimal.

267

(b) A polynomial of degree L can have at most L- 1 local minima or maxima in an open interval.
Since A. (~"') has t4ree local extrema in the interval from 0 < w < .-, we know L ~ 4.
Note that the optimal filter is half wave anti-symmetric if you lower its frequency response by one
half, i.e.,

A.,.(~"') =

=

-A.,. (~<•-wl)

where A• .,(~"') H.,.(e;"') - 1/2. Another way of saying this is to say that the optimal filter is
anti-symmetric around w .-/2 after lowering the response by 1/2. This property holds because the
optimal filter has symmetric bands with the same number of alternations. Plugging in A.,.(e-i"')
H..,(ei"')- 1/2 into the above expression gives

=

=

H.,.(ei"') -1/2
H..,(e-i"')

h.,.[n]

= -[H.., (~<•-..)) -1/2]
= -H.,. ( &<•-wl) + 1
= -( -Wh.,.[-n] + 6[n]

This condition implies that

h..,[n]

={

h.,.[-n], n odd
neven,n-:/0

0,
0.5,

A sample plot of h.,.[n] appears below, for L

n=O

= 6.

112

6

n

=

=

Note that because h.,.[n] 0 for n even, n -1 0, a plot of h..,[n] for L 5 would have the same
nonzero samples, and therefore be equivalent. So the optimal filter with L 6 is really the same
filter as the case of L 5, just as the optimal filter with L 4 is the same filter as the case with
L=3.
We know the filter non-optimal filter has 7 alternations. The optimal filter should be able to meet
the same specifications, but with a lower order. From part (a), we know the number of alternations
must be even. Thus, the optimal filter for these specifications will have 6 alternations.
An optimal lowpass filter has either L + 2 or L + 3 alternations which means L = 4 or L = 3.
However, we showed above that these are really the same filter. Since the optimal fi.lter has L = 4,
the fi.lter shown in the problem cannot have L = 4.
Putting it all together we find L > 4 for the fi.lter shown in the figure.

=

7.49. (a)

=

=

268

(b) The delay of tbe linear phase system is 51/2
order 51. Therefore, the total delay is

= 25.5 samples since it is a linear phase system of

--

H(ei=)

Delay

=

HoUO)

25.5T + O.ST
26T
= 2.6 ms

=

(c) H(dflT) should cancel the effects of H0 (jO). However, to cancel the effects ofthe delay introduced
by Ho(jfl) would require a noncausal filter which is not practical in this situation. Using the
relation w = nT,
H•(dw)

={

sin

~'f),

0,

lwl :S 0.2..
0.4.. :S lwl :S "

To obtain equiripple behavior in H~fl), we need to weight the error so that the ripples grow
with H•(e'w). Then when we multiply by H0 (jfl) the ripples will be decreased to an equal size.
Therefore, we need
sin('f)
""' :S 0.2..
W(w)=
{
1,
0.4.- :S lwl :S .-

T"'
W(m)

11---

oo~~~~~----------~-0.2lt 0.41t
It
"'

H(ei"')

1

It

"'

(d) lf Hr(ifl) is also sloping across the band, 101
Ho(jfl) and compensate as in part (c), i.e.,

H•(dw) = {

0

0.2lt

"'

< 7r/T, we would combine its effects with those of

sin~~) IHr ~If) I'

""' < 0.2.,

0,

O.b :S ""' :S "

This would take care of the distortion due to IH.(jfl)l but not of any phase distortion. The
weighting function will change in a similar manner.

269

7.50.

(a) To avoid aliasing, we require

M $" "'• So the maximum allowable decimation factor is " Mmax=- "'• (b) V(e"") 11------.. 0.9lt/100 n/100 Y(e"") 1/1001-------------------, " (I) " (I) (c) 11-----. 00 0.9lt/100 (1)$1
W1(fi'>}

'~I
00

.~

0.45ll

5()(1)$1 " (I) 270 V2(#>) ,~1 .~ 00 0.4511: O.SK " Y(#>) \ >0001 00 0.911 II Cll (d) After the first decimation by 50 is performed, W1 (&w) should look like the following: w,<#i Since we allow aliasing to occur in the transition bands, we have SOw,, w,, (e) Using 6,$
$1.55,. 0.031,. = 0.01, 6, =0.001, Aw = 0.001.- we get N = -10log10 {0.01 x 0.001)- 13 + 1 2.324(0.00l?r) :::: 5069 In general, the number of multiplies required to compute a single output sample is just N. For a linear phase 6lter, however, the symmetry in the coefficients allow us to cut the number of multiplies (roughly) in half if implementing the 6lter with a difference equation. The following is an example of how this is accomplished using the simple Type I linear phase 6lter h[n] = 0.256[n] + 6[n- 1] + 0.256[n- 2]. y[n] = 0.25z[n] + :r[n- 1] + 0.25:r[n- 2] (2 multiplies) = :r(n- 1] + 0.25(:r[n] + :r(n- 2]) (1 multiply) The procedure is similar for the other types of linear phase 6lters. Thus, we need 2535 multiplies to compute each sample of the output. (f) We have N, = -10log10 (0.01 x 0.001) -13 2.324(0.031 .. - 0.009.-) + I 271 "' N• = 232 -10log,.(O.Ol X 0.001)- 13 2.324(0.51<- 0.45.. ) "' 103 +1 H we again use linear phase filters we find 116 0 52 0 multiplies to get multiplies to get multiplies to get multiplies to get each each each each sample of vt [n] sample or [n] from v, [n] sample of v,[n] from w,[n] sample of y[n] from v,[n] w, The total number of multiplies is 168. (g) We have N, = "' N• = "' -1o log 10 (o.oo5 x o.001)- 13 2.324(0.022.-) 251 +1 -10log,.(0.005 X 0.001)- 13 2.324(0.051<) 111 +1 Therefore, we have a total of 126 + 56 = 182 multiplies per output point. (h) No. Since 6, « I we have ~ < 6, which means the stopband ripple is getting smaller. Thus, we could actually increase the specifications. (i) Performing a similar analysis on the other possibilities yields M, M• 50 2 25 . 4 20 5 10 10 5 20 4 25 2 50 Thus, the choice M 1 = 25 and M 2 example. 7.51. Multiplies per output 182 156 172 291 557 693 1375 = 4 yields the minimum number of multiplications for this (a) + .S,6[n- no] h,[n] = h[n] H, (eiw) = = H(eiw) + .S,e-;.,.. A,(eiw)e-;"""' + .s,,-;w.. = (A.(eiw) + .S.] 0 -;wne H,(eiw) H,(eiw) is real since A,(eiw) is real and 6, is real. It is nonnegative since A,(eiw) that H,(eiw) is an even function or w and is a zero-phase filter. 2: -6,. Note (b) H 3 (&w) is a zero-phase filter with real coeflicients. Thus, a zero at z• implies there must also be zeros at z<, 1/z•, and 1/zt· In addition, a zero on the unit circle must be a double zero because 272 both its value and its derivative is zero. Note that this last property is true for H,(&w) but not for A.(eiw). We can write H 3 (z) as H,(z) = H,(z)H,(l/z) where H,(z) contains all the complex conjugate zero pairs inside the unit circle and H,(I/z) contains the corresponding complex conjugate zero pairs outside the unit circle. We factor one of the double zeros on the nnit circle and its complex conjugate zero into H,(z). The other pair on the nnit circle goes into H 2 (1/z). Since H,(z) has its zeros on or inside the unit circle it is minimum phase ( - allow minimum phase systems to have zeros on the unit circle in this problem). Since the zeros occur in complex conjugate pairs, h2 [n] is real. (c) wher~ c = A,(e-'w) S ,/l+lh+lh~y't-lh+lh. Since 1- 61$ Ae{ei'"') :S 1 + 01 in the passband and -62

:5

62 in the stopband, we have
wE passband

wE stopband

Therefore,

The original filter h[n] has order M. Therefore, h1 [n] also has order M, but h2 [n] has order M/2
due to the spectral factorization. Since h,.;.[n] has the same order as h2 [n]we lind that the length
of h.un[n] is M /2 + 1.
(d) No. If we remove the linear phase constraint, then the zeros of H 3 (z) ue not distributed in
conjugate reciprocal quads. It then becomes impossible to express

where H 2 (z) is a minimum phase filter.
No. It will not work with a Type n linear phase filter. In this case no= M/2 is not an integer.
7.52.

(a)
M

H(eiw)

=

L h[n]e-.iwn

.....

273

(M-1)/2

= L

M

(M-1)/2

L

(M-1)/2

L

h[n]e-iwn +

.....

h[M- m]e-iwM .;wm

m=O

(M-1)/2

= .•-jwM/2

L

.-jwM/2

= .-jwM/2

(M-1)/2

L

h[n]eiw(M/2-n) +

.....

[

=

h[n]e-jwn

n.=(M+l)/2

n=O

=

L

h[n]e-jwn +

.....

]
h[n]e-jw(M/2-n)

(M-1)/2

L

2h[n]cosw(M/2- n)

.....

(.11+1)/2

L

2h[M{'- n]cosw(n- !l

=1

Then
(M+1)/2
H(eiw)

L

= .-jwM/2

=

where b[n] 2h[(M + 1)/2- n] for n
(b) Using the trigonometric identity

b(n]cosw(n- 1/2)

= 1, ... , (M + 1)/2.

we get

--.M->

cos(w/2)

L

n=O

ii[n] coswn

¥

¥-

n=O

n=O

= 21"L.. b[n]cosw(n+ !l + 21"L..b[n]cosw(n- 21 )
¥

¥

n=l

n=)

= 21"L.,.b[n-1]cosw(n-!)+ 1"L.,.b[n]cosw(n- 12 )
2
11-M 1
+ 2b[O]coswf2- 2b[+JcoswM/2

¥

= ~L

(b[n] +ii{n

-11) cosw(n- !l + }ii[O]cosw/2- !ii[M:[-I]coswM/

=1

Since this last expression must equal

¥

L

b(n] cosw(n-})

we can just matcb up the multipliers in front of the cosine \ermS of the two expressions. We get

b[1] + 2ii{Oj

b{n]=

n=1
2
ii{n] + ii[n - 1] 2 < n < .!!=!
2
1
2
11 1
ii[ 2 1
n - ll±l

2

•

-

2

274
(c) Consider

W(w) (Hd(eiw)- A(eiw)]

~ b(n]cosw ( n- D]

=

W(w) [ Hd(eiw)-

=

W(w) [ Hd(eiw)- cos(w/2)

=

W(w) [H4(eiw)- tb[n]coswn]

~ b(n] coswn]

where we have defined

M-1

L =
Hd(eiw)

=

W(w)

=

We also see that
!pin

•fn)

2
H.(eiW)

cos(w/2)
W(w)cos(w/2)

{ma,cO} =min {max{·}}
~nJ

wEF

wEF

(d) Type III filters:
M

H(eiw)

=

L: h\n]e-jwn
n=O

Jl/2-1

=

L

M

M/2-1

L

M/2-1
h(n]e-jwn-

....0

=

h(n]e-jwn

n=M/2+1

n.=O

=

L

h(n]e-jwn + 0 +

L

h(m]e-Jw(M-m)

m=O

M/2-1
0 -jwM/2

L

h(n] (e-jw(n-M/2) _ eiw(n-M/2))

n=O
M/2-1

=

.-jwM/2

=

.-iwM/2

L

(-2j)h(n] sinw(n- M/2)

M/2

L

2jh(M/2-m] sinwm

=1

Then

M/2

H(eiw)

=.-jwM/2 L

c(n] sinwn

=

where c(n] 2jh(M /2- n] for n = 1, ... , M /2.
If we Follow a similar analysis as the one in part (b) we get
.Y-1
sinw

2: .:(nJcoswn =

=0

~

.Y-1

.Y.-1

n=O

n=O

2: C!nJsinw(n + 11- ~ 2: .:(nJsinw(n -1)

275

llf

=

llf

2 I:c[n -l]sinwn- 2'LC!n]sinwn
n=l

+

~

=

n=l

~C[O]sinw+ ~c[¥JsinwM/2

t

(C{n- 1]- c[n])sinwn +

~C[O] sinw + ~c[¥JsinwM/2

n=1

Matching terms we get

2C{OJ- c[IJ

n=l

2

C[n - I]- C[n]

c[n]=

2

cr¥- 11

n --

2

M

2

In a manner similar to that of part (c) we can find
M

L =
if.(ei~)

Hd(ei~)

=
=
=

W(w)

i'
Type

--1
2
sinw
W(w) sinw

F

ry filters:
M

H(ei~)

= L

h[nje-i•.n

n=O

=

(M-1)/2

L

h[n]e-Jwn +
n=(Jl+l)/2

J\==0

=
=

(M-1)/2

L

(M-1)/2

h[nje-i"'n-

.-j~M/2

(M-1)/2

L

L

h[mje-J"'(M-m)

h[nj ( ,-J~(n-M/2)

_ eJw(n-M/2))

n=O
(M-1)/2

=

.-jwM/2

L

(-2j)h[n]sinw(n- M/2)

n=O

(M+1)/2

=

.-iwM/2

L

2jh((M + 1)/2- m]sinw(m- 1/2)

m=1

Then

(M+l)/2

H(ei"')

=.-JwM/> L

-1

d[n] sinw(n- 1/2)

276

where d[n]

= 2jh((M + 1)/2- n] for n = 1, ... , (M + 1)/2. We C&I. find
2d[o]- dill

n=l

2

d[n]

=

d[n- 1]- d[n]

2, [ ei"' +2·-'"']
= Qo + 01COSW

cosw

=

w

=

2

cos II- <>o

cos

"'(cos II- <>o)
' for ~cos~~""\ :S 1
"'

_1

(e)

!

278

I

·1·:.~

..8

a>

:··

0

_,

-~-~------~----~
cos(ro)
1

0

It

The picture above shows the mapping for a 0 somewhere between 0 and 1. The top right plot is
the mapping of
ms8 ao + (1-ao)msw

=

We see that as ao increases, the transformation pushes the new passband further towards .-. The
new filter is not generally an optimal filter since we lose ripples or alternations while keeping L
fixed. (Note that some of the original filter does not map anywhere in the new filter).
(f) In a similar manner, this choice of ao will cause the new passband to decrease with decreasing Qo·
7.54.

(a) Let Dt(z) be the z-transform of fl.(tl{:r[nj}. Then
Do(z)

=

D 1 (z)

=

Z {ti0 {:o:[n)}} = X(z)
Z {ti 1 {:o:[n]}} = (z- z- 1 )X(z)

D 2 (z)

=

Z{ti2 {:o:[n]}}=(z-z- 1 ) 2 X(z)

Dt(z)

=

Z {Ll.•{:o:[n]}} = (z- z- 1 )• X(z)

279
(b) By taking the transform of both sides of the continuous-time difterential equation one gets (assuming initial rest conditions)
N

M

:~:>•••Y(s)

= 2),s' X(s)
r=O

A:=O

Solving for He(s)

Similarly,
N

M

L at(z- z-l )•Y(z) =L b,(z- z....

1

)' X(z)

r=O

=> m(z) = z- z- 1
(c) First, map the continuous-time cutoff frequency into discrete-time and then make the sketch.

s = z- z- 1
j!l

=.!"'- .-;~

eiw- e-jw

0 =

j

w=-..6

2sin(w) =I

H(ei"')
1

-lt

->t/6 0

ltl6

7.55. (a) Using DTFT properties,

h1 [n]

=

h[-n]

Ht(ei~)

=

H(e-;~)

Since H(ei~) is symmetric about w = 0, H(e-;~) = H(ei"'). Thus, H 1 (ei~) = H(e-i-') = H(ei~).
H(e'~) is optimal in the minimax sense, so H1 (ei~) is optimal in minimax sense as well.

280

Bd(ei"')

={ ~:

W(ei"') = {

~fo,,

O$w$wp

w.

:s w :s

1r

0$w$wp
w. :S w :S r

(b) Using DTIT properties,
h2(n]
B2(ei"')

= (-1)nh(n]

=
=

(e-i•)nh(n]
B(ei!w+•l)

B2(ei"') is a high pass filter obtained by shifting B(ei"') by,. along the frequency axis. B2(e'"')
satisfies the alternation thereom, and is therefore optimal in the minimax sense.

:S 1r -w.
r-w,:Sw:SK

0$w 0$ w

:s 71"- "'·

w-w,:Sw511'

(c) Using DTIT properties,
h 3[n]
B3(e'"')

= h[n] • h(n]
= B(ei"')B(ei"')

ln the passband, B 3(e'"') alternates about 1 + Of with a maximal error of 26~. ln the stopband,
B 3(ei"') alternates al2out ~/2 with a maximal error of ~/2. At first glance, it may appear that
H3(eJ"') is optimal. However, this is not the case. Counting alternations, we find that the original
filter H (ei"') has 8 alternations.
We know that since H(ei") is optimal, it must have at least L + 2 alternations. It is also possible
that H (eJ"') has L + 3 alternations, if it corresponds to the extraripple case. So L is either 5 or 6
for this filter. Consequently, the filter length of h[n], denoted as N, is either 11 or 13.
The filter h3[n] is the convolution of two length N sequences. Therefore, the length .of h3(n], denoted as N', is 2N- 1. Since N is either 11 or 13, N' must be either 21 or 25. It follows that
the polynomial order for h 3 (n], denoted as L', is either 10 or 12. For h 3 (n] to be optimal in the
mininlax sense, it must have at least L' + 2 alternations. Tbns, hs(n] must abihit at least 12 alternations, for the non-extraripple case, or at least 14 alternations in the extraripple case to be optimal.
A simple counting ofthe alternations in H.(eJ"') reveals that there are 11 alternations, consisting
0. These are too few to satisfy
of the 8 alternations that were in H(eJ"') plus 3 where H(ei"')
either the non~raripple case or the extraripple case. As a result, this filter is not optimal in the

=

mmwmax JeDSe.

(d)

/4(n]
B 4 (ei"')

= h[n] - K 6[n]
= B(ei"') - K

This filter is simply H(ei"') shifted down by K along the Ht(eJ"') axis. Consequently, this filter
satisfies the alternation theorem, and is optimal in the mjnimax sense.

281

O$w$w,
w. :S w :S

1r

O$w$w,

w. ::; w::;

7(

(e) hs(n] is h(n] upsampled by a factor of 2. In tbe frequency domain, upsampllng by a factor of 2 will
cause the frequency axis to get scaled by a factor of 1/2. Consequently, Hs(&w) will be a bandstop
filter that satisfies the alternation theorem, with twice as many alternations as H (&w). This filter
is optimal in the minimax sense.

0$w$ w,/2
w,/2 $w 5 "-w,/2 .- -w,/2$ w $.0 5w 5 w,/2 w,/2$w$.. -w,/2 .- -w,/2$ w $.- 7 .56. We have an odd length causal linear phase filter with values from n either a Type I or Type ill filter. = 0, ... , 24. It must therefore be ' (a) True. We know either Type ill Type I h(m] = h(24- m] or h(m] = -h(24- m] for -oo < m < oo since the filter has linear phase. Substituting m = n h(n + 12] = h(12- n] or h(n + 12] + 12 we get = -h(l2- n] (b) Fwe. Since the filter is linear phase it either has zeros both inside and outside the unit circle or it has zeros only on the unit circle. If the filter has zeros both inside and outside the unit circle, its inverse has poles both inside and outside the unit circle. The only region of convergence that would correspond to a stable inverse would be the ring that includes the unit circle. The inverse would therefore be two-sided and not causal. If the filter only has zeros on the unit circle, its inverse has poles on the unit circle and is therefore unstable. (c) Insufficient /nformatioTL If it is a Type ill filter it would have a zero at z = -1 but if it is a Type I filter this is not necessarily true. (d) True. To minimize the maximum weighted approximation error is the goal of the Parks-McClellan algorithm. (e) True. The filter is Fm so there are no feedback paths in the signal flow graph. (f) True. The filter has linear phase and arg (H(eJw)) where {J ={J- 12w = 0," for a Type I filter or {J = ../2, 3r/2 for a Type ill filter. grd (H(ei"')) = = -! 12 > 0 {arg (H(ei"')]} Tbe group delay is 282 7.57. (a) The desired tolerance scheme is (b) W(w) = !· ~ j; (or~) (or 1) (or f.) ~:: :~ ·(or 1) O:S!wi:Swl .... :Siwl$ "'>
"'•:Siwi:S ..

(c) From the Alternation Theorem, the minimum number of alternations is L + 2.
(d) The trigonometric polynomial (of degree L) can have at most L- 1 points of local minima or
maxima in the open interval between 0 and ... If these are all alternation points and, in addition,
all the band edges are alternatinn points, we find the maximum number of alternations is

L-1+6=L+5
(e) If M = 14, then L = M/2 = 7. The maximum number of alternations is therefore 7 + 5 = 12.

T!!Picai E(w) looks like :
E(oo)

o

0

.,

.,

(f) ~ will be shown in part (g), the 3 band case can have maxima and minima in the transition
regions. It follows that we do not have to have an extremal frequency at w4 • Therefore, if we
started with an optin>al maximal ripple filter and jnst slid "'• over we may move a local minin>um
or maximum into the transition region, bnt there will still be enongh alternations left to satisfy
the alternation theorem. Thus, the maximum approximation error does not have to decrease.

(g) (i) If a point in the transition region has a local minimum or maximum then there is the possibility
that the surrounding points of maximum error do not alternate. Thus, we might lower the
number of alternations by two. B~, if...., started with L + 5 alternations this reduction
does not drop the number of alternations below the lower limit of L + 2 set by' the Alternation
Theorem. Therefore, local maxima and minin3a of A.(eiw) can occur in the transition regions.
Note that this is not true in the 2 band case.
(ii) If a point in the approximation bands is a local minimum or maximum, the surrounding points
of maximum error do not alternate. Thus, a local minimum or maximum in the approximation
bands implies that the total number of alternations is reduced by two. However, if we started
with L + 5 alternations this reduction does not drop the number of alternations below the
lower limit of L + 2 set by the Alternation Theorem. Therefore, we can have a local maximum
or minimum in the approximation bands. Note that in the 2-band case we drop &om L + 3 to
L + 1 which violates the Alternation Theorem.

283

7.58.

(a) In order for condition 3 to hold, G(z- 1) must he an allpass system, since

z-t =
.-jB =
=

G(z- 1)
G(e-"")

IG<·-jW)jeJLG(•-'"l

Clearly, !G(e-1wJI must equal unity to map the unit circle of the Z-plane onto the unit circle of
the z-plane.
(b) Consider one allpass term in the product, and note that a• is real.

The inside of the unit circle of the Z-plane is

0 $IZI I > 1 1•1 > z-1 z•-1 I lzi' I The inside of the unit circle of the Z-plane maps to the outside of the unit circle of the z-plane. This is not the desired result. However, if (1- ail > 0, then I < 1 < 1•1 < z-lz•-1 I lzi' 1 The inside of the unit circle of the Z-plane maps to the inside of the unit circle of the z-plane. This is the desired result. Thus, for condition 2 to be satisfied, 1- ai > 0 < 1 Ia•! < 1 ja,j 2 This condition holds for the general case as well since the general case is just a product of the simpler allpass terms. ! 284 (c) First, it is shown that G(z- 1) produces the desired mapping for some value of a. Starting with G(z- 1 }, z-• = e-;• e-il- ae-;•e-;w e-.;..(1 + ae-i1 ) e-;161 = = = = = = z-1 -a 1-az e-jf.J I -a 1- oe-Jw e-;.., -a e-;• + 0 e-i• +a 1 + ae-i' e-il +a 1 +aei' 1 + ae ;• 1 + aei1 ,-;• + 2a + a 2 ei' 1+2acos8+a2 Using Euler's formula, = = cosS- j sinS+ 2a + a2 cosO+ ja2 sinS 1 + 2acos8 + a2 2a + (1 + a2) cosO+ j((a2 - 1) sinD] 1 + 2acos8 + a2 Noting that -w = tan- 1 [ ~~Jl, -w w 2 = tan-• [ (a - 1} sinD ] 2a+(1 +a2 )cos8 = tan-' [ (1- a ) sinO ] 2a+(1+a2 )cos8 2 This relationship is plotted in the figure below for dilferent values of a. Although a warping of the frequency scale is evident in the figure, (except when a = 0, which corresponds to z-l = z-l }, if the original system has a piecewise-constant lowpass frequency response with cutoff frequency e,, then the transformed system will likewise have a similar lowpass response with cutoff frequency "'• determined by the choice of a. Warping of the frequency scale, LPF to LPF (I) 285 Next, an equation for a is found in terms of z-1 e. and "'•· Starting with G(z- 1 ), z-1-a 1-az 1 = e-;w,.- o 1 ae-iw, e-;w,.- Q = e-il, - e-;w,. - 1) e-il,. _ e-;w,. e i(l,.+w,.) - 1 o(e-i(B,.~,.) " = = e-i(l,.+w,.)/2(e-i(l,.-w,.)/2 _ ei(B,.-w,.)/2) e j(l,.+~,o~,.)f2(e j{B,.+w,.)/2 _ ei(B,.+w,.}/2) -2i sin((e,- "'•)/2] -2j sin((e. + "'•)/2] sin((e,- w,)/2] sin((6, + "'•)/2] = = (d) Using the equation for w found in part c, with e, = "/2, a2] w,=tan- 1 [1~ (i} -1 = "'• tan [1- (-0.2679) 2( -0.2679} -1 ( = = tan 2 ] 0.9282 ) -0.5358 2tr/3 (ii) = "'• = [I- (0)2] 2(0) tan- 1 (oo) tan-1 = 1f/2 (iii) "'• = = = -1 tan 2 [1- (0.4142) ] 2(0.4142) tan- 1 (1) 1f/4 (e) The first-order allpass system G(z-1) =- z-1 +a 1 +az- 1 satisfies the criteria that the unit circle in the Z-plane maps to the unit circle in the z-plane, and that e = 0 maps tow="· Next, a is found in terms of e, and "'•· z-1 = z- 1 +a 1 + QZ l 286 = e-iw,. +0 1 + ae-Jw., -e-;B, - oe-i(w,.+l) a(1 + e-i(w,+B,)) = a = = = -e-;1,. _ e-;w,. e-il, + e-iw,. 1 + e j(w,.+l,.) e-i(w,.+l,.)/2(e-i(-w,.+l,.}/2 + e-j(w,.-1.. )/2) e-i(w,.+l,)/2(ei("',.+l,.}/2 + e-i(~oo~,.+l,.)/2) cos[(wp- Bp)/2] cos[(wp + Bp)/2] (f) First, an equation for w is found in terms of 8 and a. z-• = = z- 1 +a 1 +az 1 e-Jw + Q 1 +ae ;w -e-i' - ae-i(w+B) = e-jw + ae-i') = -e-i'- a .-iw(l = = = = = e-il +Q +0 1 + ae-1 8 e-il +o 1 + ae-i' .-jl 1 + oei' 1 + oei8 + 2a + ci'ei' 1 + 2acos8 + a 2 cos8- j sin8 + 2a + a 2 ccs8 + ja 2 sinS 1 + 2acosS + a 2 cosS + 2a + ci' cosS + j(- sinS+ a 2 sinS) 1 + 2acos8 +a> Therefore, 2 -w + 7r w (a -1)sin8 ] = tan _ 1 [ 2a+(1+a 2 )cos8 = tan-• [ (1-ci')sin8 ] +w 2a + (1 + a 2 )cos8 Note that this lowpass to highpass expression is the similar to the lowpass to lowpass expression for w found in part (c). The only dilference is the additive " term, which shifts the lowpass filter into a highpass filter. The frequency warping is plotted below. 287 Warping of the frequency scale, LPF to HPF (J) For B, = "/2, this becomes (i) _1 w, = tan = tan = = Sr/3 [I- (-0.2679) 2 ] 2( -0.2679) 0.9282 ) -1 ( -0.5358 +" + 71" 2tr/3+tr The right edge of the low pass filter gets warped to 57f /3, which is equivalent to frequency response of this filter appears below. 1 ... -It " (ii) w, = = = = tan -1 tan- 1 The ... 0 -ltl3 -1r /3. r~- (O)'J 2(0) (oo) (I) + .. + .- .-f2+tr 37r/2 The right edge of the low pass filter gets warped to 3tr /2, which is eqaiYaleot to -ir /2. The frequency response of this filter appears below. 288 1 ••• • •• 0 -7112 -lt lt (I) (iii) "'• = = = = _, rl- (0.4142) 2 tan 2(0.4142) + 1 " _, (0.8284) tan 0.8284 + .. 7r/4+tr 57f/4 The right edge ofthe low pass filter gets warped to 57rf4. The frequency response of this filter appears below. 1 ••• ••• -lt -31t/4 0 3lli4 lt (I) 289 Solutions - Chapter 8 The Discrete-Time Fourier Transform 291 8.1. We sample a. periodic continuous-time signal with a sampling rate: n•. 1 6 F: = - = - = - H3 z ' 2.. T 10(a) The sampled signal is given by: :o[n] = :o,(nT) Expressed as a Discrete Fourier Series: L' :o[n] = "•ei'f•n 1:-=-• We note that, in accordance with the discussion of Section 8.1, the sampled signal is represented by the summation of harmonically-related complex exponentials. The fundamental frequency of this set of exponentials is 2.. /N, where N = 6. Therefore, the sequence :t[n] is periodic with period 6. (b) For any bandlimited continuous-time signal, the Nyquist Criterion may be stated from Eq. (4.14b) as: F,?. 2FN, where F, is the sampling rate (Hz), and FN corresponds to the highest frequency component in the signal (also Hz). As evident by the finite Fourier series representation of :r,(t), this continuous-time signal is, indeed, bandlimited with a maximum frequency of Fn Hz. Therfore, by sampling at a rate of F, ._, Hz, the Nyquist Criterion is violated, and aliasing 10 results. = ,:., = (c) We use the analysis equation of Eq. (8.11): N-1 X[k] =L x[n]e-i?,Hn n=O From part (a), x[n] is periodic with N = 6. Substitution yields: X[k] = t (t = a,.ei'fmn) ,-;'f•n tn=-8 n=O L• L• a,.ei(2•/l)(m-•)n A=Om=-t We reverse the order of the summations, and use the orthognnality relationship from Example 8.1: X[k]=6 • 00 m=-1 r=-oc La,. L o[m-k+rNJ Taking the infinite summation to the outside, we recognize the convolution between a, and shifted impulses (Recall = 0 for JmJ > 9). Thus, a,. 00 X[k] =6 L "•-•· Note that from X[k], the aliasing is apparent. · =-oo 292 8.2. (a) Using the analysis equation of Eq. (8.11) N-1 X[k] =L f[n]W~n Since %[n] is also periodic with period 3N, SN-1 X,[k] = L %(n]W3";J n=O N-1 2N-1 L f[n]W:;";J + L = JN-1 f[n]W:;";J n=2N n=N n=O L f[n]w;.-: + Performing a change of variables in tbe second and third summations of X3-[k], X,[k] N-1 N-1 n=O n=O = L z[n]w;.-: + w;: L Since f[n] is periodic with period N, and N-1 N]w;.-: + Wf~N L z[n + 2N]w;.-: f[n + n=O = w~tln, W,tR x,[kJ = (~+·-;•·(tl +•-;•·(>t') N-1 z::: z[nJw~tnJ n=G = (l+e-;b(ll+e-jb("fl)X[k] = { 3X[k/3J, k = 3l otherwise 0, (b) Using N = 2 and %[n] as in Fig P8.2-l: N-1 = L X[kj f(n]W~n = %(0] + %(1]e-;.. = 1+2(-1). = {3, -1, k=O k = 1 Observe, from Fig. P8.2-1, that i(n] is also periodic with period 3N = 6: 3N-1 L i(n]w;.-: = L• %(n]t-jfk X3 (k] = n=G n=G = (1 + .-;'f• +e-i¥>)(1 + 2(-l)l) = (1 + .-;'f• + .-;'f•)X[k/3] = ( ~3. ::~ 0, lc = 1,2,4,5. 293 8.3. (a) The DFS coefficients will be real if i[n] is even. Only signal B can be even (i.e., is[n] = %8 [-n]; if the origin is selected as the midpoint of either the nonzero block, or the zero block). (b) The DFS coefficients will be imaginary if x[n] is even. None of the sequences in Fig P8.3-1 can be odd. (c) We use the analysis equation, Eq. (8.11) and the closed form expression for a geometric series. Assuming unit amplitudes and discarding DFS points which are zero: XA[k] • = L;ei'fkn = = =<> 1- eif.U: 1- eit• 1-(-1)• .•• 1- .... = 0, k = ±2, ±4, ... 2 L;ei'f•· Xs[kj = =<> = 1- eJtlk 1- eit.t 3 Xc[kj = :E e~>t>n- :E 7ei>t•· =• n=O 3 L = (eit•• _ e~f•ln+4)) n=O = = (1 - ei.. ) 1 - ..,:• 1-eJ'i.C 0, k = ±2, ±4, 8. 4. A periodic sequence is constructed from the sequence: as follows: ~ i[n] = L z[n + rN], lal < 1 ~-oc (a) The Fourier transform of :r[n]: __L.,. z[n]e-j.m ~ X(eiw) = = = .. L:a"e-;..,n ..... 1 1- ae-P.,' (b) The DFS of i[n]: N-1 X[kj = L n=O x[n]w~· lal < 1 294 oo N-1 L L = N-l z{n+rNJW~n oo L L = <>n+rNu{n+rNJW~n n=O r=-oo N-1 oo = LLQn+rNw~ ft=O r=O Rearranging the summations gives: ,lal < 1 ,lal < 1 8.5. (a) = o[n] L o[n]W~". = n=O = 1 z[n] N-1 X[kj 0 ::; k ::; (N - 1) (b) = z[n] c5[n- no], o::; no :5 (N -1) N-1 X{k] = L o(n - no]W~n' 0$ k $(N -1) n=O = w•no N (c) z[n] = { 1, n even 0, nodd N-1 X[kj = Ezfn]W~n, 0$ k $(N -1) n=O (N/2)-1 = E W]./'"' n=O X[kj 1- e-i2•" = 1- e = { N/2, k=O,N/2 0, otherwise j(••fN) 295 (d) :r:(n] = { 1, 0::; n::; ((N/2)- 1) 0, N/2$ n $(N- 1) N-1 X[.i:J = L 0$ k $(N - 1) ..... :r:[n]Wg>, (N/2)-1 ....I: = W'" N l 1- e-id = 1- e X[k] = i(2d)fN k=O N/2, 1- e 0, ~2•k/NJ' k odd k even, 0::; k 5 (N- 1) (e) :r:[n] = X[k] = {an, 0 ::; n$ (N - 1)
otherwise
0,
N-1

X[k]

L anwfy",

O$k$(N-1)

n=

X(ei"') =

=

1 _ e-i{w-wo)N
1- e

j(w We)

e-i(w-Wo)(N/2)
e-j(w-wo)/2

X(ei"')

=

(sin [(w- Wo)(Nf2)])
sin [(w- Wo)/2]

-i(w-... )((N-1)/2)

e

(sinf(w -Wo)(N/2)])
sin ((w- Wo)/2]

(b) N-point DFT:
N-1

X(k]

= L

=•

:r:(nJWtn,

0 $k$ (N- 1)

296
N-1

L eJwonw;t

=

n=O

1_

=

.-;((2••/N)-wo)N

1- e

j((blc/N) w 0 )

. ((2••
"]
-;('N'--.... )(!!?)sm
N" -Wo ) T

=

sin[(~- w,) /2]

e
Note that X[k] = X(&w)lw=(2d)/N
(c) Suppose Wo = (2do)/N, where ko is an integer:

X[k]

1-

=

1_

=

e-i(A:-to)b

e-i(•-.. )(2•)/N

sin.-(k- ko)

-j(>•/NH•-.. )((N-1)/2)

sin(.-(k- ko)/N)

•

8. 7. We have a six-point uniform sequence, :z:[n], which is nonzero for 0 :S n :S 5. We sample the Z-transform
of :z:[n] at four equally-spaced points on the unit circle.

X[k]

= X(z)I,=•"·•J•>

We seek the sequence :1 [n] which is the inverse DFT of X[k]. Recall the definition of the Z-transform:
00

X(z) =

L

:z:[n]z-n

n=-oo

Since x[n] is zero for all n outside 0 :S n :S 5, we may replace the infinite summation with a finite
summation. Funhermore, after substituting z = ei( 2 ·d"/ 4 ), we obtain
X[k]

• :z:[nJWt",

=L

0 :S k :S 4

n=O

Note that we have taken a 4-point DFT, as specified by the sampling of the Z-transform; however, the
original sequence was of length 6. As a result, we can expect some aliasing when we return to the time
domain via the inverse DFT.
Performing the DFT,
X[k]

= Jl1• + Wt + w;• + w;• + Wf" + W;",

0 :S k :S 4

Taking the inverse DFT by inspection, we note that there are six impulses (one for each value of n
above). However,

and
W4•• -- ......
"'"

w•• - w•
4.

4'

-

so two points are aliased. Tbe resulting time-domain signal is

2

2

:z:,[n]
1

•

l l

-1 0

1

1

II
2

3

•4 •5 •6

n

297
8.8. Fourier transform of z[n] = (1/2)nu[n]:
00

X(eJw)

L

=

z[n]e->wn

n=-CICI

~G)" .-jwn

=

1

=
Now, sample the frequency spectra of z[n]:
Y[k]

We have the

l~pt

= X(eJw)iw=2d/1D•

0$k$ 9

DFT:

Y[k]

I

=

1_

=

L• y[n1w,•;

y[n]

= 1- '

n

302

= 6 + swt + 4W;' + 3w.f• +

X2[kj

•

= Li>[nJWt"

=

Y1 [k]

2w:• + w~·

=
=

.....

w;>

.X1 [k]X2 [k]
6W;' + sw:> + 4wt• + 3w~• + 2wt• +

w;•

Noting that w;• = &"fml = 1 = W¥•, we use the synthesis equation of Eq. (8.12) to construct
y,[n]. Tbe result is identical to the sequence depicted above.
(b) The DFS of the signal illustrated in Fig. P8.21-2.is given by:

Xs[k] =

L• ia[n]Wf"

.....

= 1+Wt•
Therefore:

Y,[kJ

=
=

.X,[k].X.[kJ
.X,[kJ + w:> X,[kJ

Since the DFS is linear, the inverse DFS of Y2 [k] is given by:

!i2[nj = :t,[n] + %1[n - 4].
8.22. For a finite-length sequence z[n], with length equal toN, the periodic repetition of z[-n] is represented
by
z[((-n))N) =z[((-n+lN))N], l: integer
where the right side is justified since z[n) (and z[-n]) is periodic with period N.
The above statement holds true for any choice oft. Therefore, for l = 1:

z{((-n)).] = :r:[((-n + N))N]
8.23. We have :r:[n] for 0::; n $P. We desire to compute X(z)I•=•-"'""N' nsing one N-pt DFT. (a) Suppose N > P (the DFT size is larger than the data segment). The technique nsed in this case is often referred to as zero-padding. By appending zeros to a small data block, a larger DFT may be nsed. Thus the frequency spectra may be more finely sampled. It is a common misconception to believe that zero-padding enhances apectral resolution. The addition of a larger block of data to a larger D FT would enhance this quality. So, we append N • = N - P zeros to the end of the sequence as follows: :'[n] = { :r:[n], 0::; n$ (P- 1)
0,

:r:[n]

P$n$N

--~~·IL-__N_~_t_o_FT -J~---.
__

X[k]

(O $k$ N)

303

(b) Suppose N > P, consider taking a DFT which is smaller than.the data block. Of course, some
aliasing is expected. Perhaps we could introduce time aliasing to offset the effects.
Consider the N-pt inverse DFT of X[k],
N-1

N1 L

z[n] =

X[k]W,V"",

0 ~ n ~ (N- 1)

lo=O

Suppose X[k] was obtained as the result of an infinite summation of complex exponents:

z(nj

=~ ~

(f:

k=O

m=O

z(mje-;(2d/N)m) W,V""

Rearrange to get:

L

z[n] =

oo

(

z[m]

L ,-;(2•/N)(m-n)i

~

N-1

)

k=O

m=-oc

Using the orthogonality relationship of Example 8.1:
00

00

L

=

z[n]

L

z[m]

o[m-n+rNJ

r=-oo

m=-oc
00

L

=

z[n]

z[n- rNJ

r=-oo

So, we should alias z[n] as above. Then we take the N-pt DFT to get X[k].

=

8.24. No. Recall that the DFT merely samples the frequency spectra. Therefore, the fact the /m{X[k]} 0
for 0 ~ k ~ (N - 1) does not guarantee that the imaginary part of the continuous frequency spectra is
also zero.

For example, consider a signal which consists of an impulse centered at n

= 1.

0~ n ~ 1

z[n] ;, o[n- 1],
The Fourier transform is:
X(eiw)
Re{X(eiw)}
/m{X(eiw)}

Note that neither is zero for ali 0 ~ w

~

=

= cos(w)
= -sin(w)

2. Now, suppose we take the 2-pt DFT:

= w;, o ~ k ~ 1

X[kJ

= { 1,
-1,

So, /m{X[k]} = 0,

,-;w

'lk. However, /m{X(eiw)}

k=O
k=1

# 0.

Note also that the size of the DFT plays a large role. For instance, consider taking the 3-pt DFT of

=

6[n- 1}, 0 ~ n
X[k] = Wf,
O~k~2
z[n]

1,

=

.-;(b/3)'

~

k =0
k=1

{ .-J(.. /l)' k

=2

2

304
Now, Jm{X[k]}

#- 0, for k = 1 or k = 2.

8.25. Both sequences z[n] and y(n] are of finite-length (N = 4).
Hence, no aliasing takes place. From Section 8.6.2, multiplication of the DIT of a sequence by a complex
exponential corresponds to a circular shift of the time-domain sequence.
Given Y[k] =

w1• X[k], we have
y[n] = z[((n- 3)) 4 ]

We use the technique suggested in problem 8.28. That is, we temporarily extend the sequence such that
a periodic sequence with period 4 is formed.

1

1

1

I I rt I I rr I I r r
3/4

3/4

-4 -3 -2 -1 0

1

2 3

3/4

4

5

1/2

6

%(n]
n

7

Now, we shift by three (to the right), and set all values outside 0 S n S 3 to zero.

I
1

--~·~·~~I~t~2-~L~4~~·~··------~~~J
3/4

-2 -1 0

1

2

3

4 5

8.26. (a) When multiplying the DIT of a sequence by a complex exponential, the time-domain signal
undergoes a circular shift.

For this case,
Y[k] = W:' X[k],

0SkS5

Therefore,

y[n] = z[((n- 4))o],
4
2

•

I

-1 0

1

t
1

•2 •3

OSnSS

3

II • •
4

5 6

y[n]
n

7

(b) There are two ways to approach this problem. First, we attempt a solution by brute force.

X[k] =
W(k]

=

4+3Wf+2Wt•+wzt,

Wf=e-;<><•t•>and0SkS5

JU{ X[k]}

= ~ (X[k] + X"[k])

= ~ (4+3Wf +2Wf + w:• +4+3w.-• +2w,-•• + w,- 30 )

305

Notice that

w~

=
w-•
N
=
W[k] =

e-·;(2·lll:fN}

eJ(2d/N)

= e-i(2r/N)(N-t) = w::-t

4+~
2'

[w.• + w!-•] + [w•• + w•-••] + !2[w.,.
' + w•-••]
6

6

6

6'

So,
w[n]

=

40[n) +

~ ( o[n- 1) + o[n- SJ) + o[n- 2] + o[n- 4)

+ ~ ( o[n - 3] + o[n - 3])
w[n]

=

3
3
40[n) + 2o[n- 1] + o[n- 2) + 6[n- 3) + 6[n - 4] + 26[n - 5),

Sketching w[n):
4

•
-1

I
0

3/2 1

1

TT

T T

1

3

2

w[n)

1 3/2

4

T• •
5 6 7

n

As an alternate approach, suppose we use the properties of the DIT as listed in Table 8.2.

= Xe{X[k)}
X[k] + X"[k)
=
2
= ~ IDFT{X[k)} + ~ IDIT{X"[k)}
= ~ ( :r[n) + x"[((-n))NJ)

W[k]

w[n)

For 0 :5 n :5 N - 1 and :r[n) real:
w[n)

= ~ (:r[n) + x[N- nJ)

4

•
-1

3

I• • ' I I • •
1

0

1

2

3

2

4

5 6

:r[N- n], for N

=6

n

7

So, we observe that w[n] results as al>ove.
(c) The DIT is decimated by two. By taking altemate points of the DFT output, we have half as
many points. Tbe influence of this action in the tin>e domain is, as expected, the appearance of
aliasing. For the case of decimation by two, we shall find that an additional replica of x[n) surfaces,
since the sequence is now periodic with period 3.

306

From part (b):
Let Q[k]

X[k]

= 4 + 3W." + 2Wi' + Wl•,

0$k$ 5

Q[k]

=4 + 3Wf + 2Wf' + w:•,

0 :S k :S 2

= X[2k],

Noting that

W:" = Jl1'•
q[n]

= 56(n] + 36[n- I]+ 26{n- 2],

.li i. ..

-1 0
8.27.

0 :S n :S 2

5

I

2

3

4

q[n]

n

5

(a) The linear convolution, z 1 [n] • z 2 [n] is a sequence of length 100 + 10- I = 109.

7

8

9

10

10

9

8

6

2

3

4

zl[n] • z,[n]

7

6

5

5

4
3

2

T;

; T
-1 0

I

2

3

4

5 6

7

8

n

99 100 101 102 103 104 105 106 107 108 109

9

(b) The circular convolution, z 1 (n] ~ 2 [n], can be obtained by aliasing the first 9 points of the linear
convolution above:

10 10 10 10 10 10 10 10 10 10

10

llllllllll L.
01234

56789

99

(c) Since N? 109, the circular convolution zl[nJ@,[n] will be equivalent to the linear convolution
of part (a).
8.28. We may approach this problem in two ways. First, the notion of modulo arithmetic may be simplified
if we utilize the implied periodic extension. That is, we redraw the original signal as if it were periodic
with period N 4. A few periods are sufficient:

=

i j i f i i i f ii i
-4 -3 -2 -1 0

1

2

3

4

5

6

i(n]

f
7

n

307

To obtain z 1 [n] = z[((n- 2)) 4 ], we shift by two (to the right) and only keep those points which lie in
the original domain of the signal (i.e. 0 $n$ 3):

6

ii li

• •

-2 -1 0

1

2

3

z 1 [n]

•4 •5

n

To obtain :r 2[n] = z[(( -n)) 4 ], we fold the pseudo-periodic version of :r[n] over the origin (time-reversal),
and again we set all points outside 0 $n$ 3 equal to zero. Hence,

6

.. !iii.
-2 -1 0

Note that :r[((0)) 4]

1

2

3

:r2[n]
n

I

4 5

=z[O], etc.

In the second approach, we work with the given signal. The signal is confined to
0 $n$ 3; therefore, the circular nature must be maintained by picturing the signal on the circumference of a cylinder.
8.29. Circular convolution equals linear convolution plus aliasing. First, we find y[n]

=:r,[n] • :r2[n]:

8
6

6

5

y[n]

4

4

3

3

2

T~

n

01234567891011
Note that y[n] is a ten point sequence (N

= 6 + 5- 1).

(a) For N = 6, the last four non-zero point (6 $n$ 9) will alias to the first four points, giving us
Yt [n] = z, [nj(§)r2 [n]

308
8
10

6

8
6

II• (n]

4

-1

0

1

2

3

4

5

6

7

"

(b) For N = 10, N ;::: 6 + 5- 1, so no aliasing occurs, and circular convolution is identical to linear
convolution.
8.30. We have a finite length sequence, whose 64-pt DFT contains only one nonzero point (for k = 32).
(a) Using the synthesis equation Eq. (8.68):
N-1

L

1

:r:[n] = N

X[k]W,V"".

0 :0: n :0: (N- 1)

1:=0

Substitution yields:
:r:[n]

=

! X(32JW6<

=

.!_,.;j<(32)n

=
:r:[n] =

32

"

64

_!_eJ .. n
64

0:0: n :0: (N -1)

!<-1)",

The answer is unique because we have taken the 64-pt DFT of a 64-pt sequence.
(b) The sequence length is now N = 192.
:r:(n]

:r:[n]

=

64 :0: " :0: 191

This solution is not unique. By taking only 64 spectral samples, :r(n] will be aliased in tin2e.
As an alternate sequence, consider
:r'(n] = ! G ) (-1)",

0:0: n :0: 191

8.31. We have a 1~point sequence, :r[n]. We want a modified sequence, :r1 (n], such that the
:ri[n] corresponds to
X;(k] = X(z)l.=!•""••t••>+<•t>•ll

Recall the definition of the Z-transform of :r[n]:
X(z)

=

00

L

n.=-oc

:r[n]z-•

1~pt.

DFT of

309
Since z[n] is of finite duration (N

= 10), we assume:
z[n]

= { nonzero,
0,

Therefore,

X(z)

0$n$9
otherwise

= L• z[n]z-•
n=O

Substituting in z

=!&1<•••/IO)+(•/IO)J:
X(z)],=!•W'"'"'>+<•I">l

9
( 1 . (2. . /10)+(•/10)) )-n
= L:;z[n]
2e'l

n=O

We seek the sigual z 1 {n], whose 1!)-pt. DFT is equivalent to the above expression. RecaJ.J the analysis
equation for the DFT:

X,{k)

•

= L;z.{n]Wt.;',

0$k$9

n=O

Since

wton = e-i( 2 " /lD)A:n' by comparison

= z{n] ( I .;<•/IO) ) -•

z 1{n]

2

=

8.32. We have a finite-length sequence, z{n] with N
8. Suppose we interpolate by a factor of two.· That
is, we wish to double the size of :r{n] by inserting zeros at all odd values of n for 0 $n$ 15.

Mathematically,
y{n]

= { z{n/2],

n even,

0,

n odd,

=

LY{nJWt&",

0$n$ 15

The 16-pt. DFT of y{n]:
15

Y{k]

0$k$ 15

n=O
7

=

L;z{n]Wff"
n=O

7

Y{k]

= L;z{n]Wf",

0$k$ 15

n=O

Therefore, the 16-pt. DFT of the interpolated sigual contains two copies of the 8-pt. DFT of z{n]. This
is expected since Y{k] is now periodic with period 8 (see problem 8.1). Therefore, the correct choice is

c.
As a quick check, Y[OJ

= X{O].

310
8.33. (a) Since
:t2[n]

=

{

z[n],
-z[n- N],

0 :S n :S N- 1
N :S n :S 2N- 1

0,

otherwise

If X[k] is known, z 2 [n] can be constructed by :

X[k)

•I

N-pt IDFT f--:~:[n')l-------,

1

Scale by -1

1

I

Shift by N

I

L--{~co~n~ca~t~e~na~t~e}-.......,[~2~N~-p~t~D~FT~}---

X2[k]

(b) To obtain X[k] from X,[k], we might try to take the inverse DFT (2N-pt) of X,[k), then take the
N-pt DFT of :r,[n] to get X[k].
However, the above approach is highly inefficient. A more reasonable approach may be achieved if
we examine the DFT analysis equations involved. First,
2N-1

X,[k)

=

L

:r,[n)W;'N,

0

:S k :S (2N - 1)

n=O
N-1

=

L :r[n)W;';)

n=O
N-1

X,[k)

=
=

L :r[n)w;:f2)n,

0 :S k :S (N- 1)

n=O

X[k/2),

0 :S k :S (N- 1)

Thus, an easier way to obtain X[k] from X1[k] is simply to decimate X1[k] by two.

X1[k]

---.t·IL..__• __,~-----·
-2_

X[k)

311
8.34. (a) The DFT of the even part of a real sequence:
H :z:[n] is of length N, then :z:.[n] is of length 2N- 1:

:z:.(n] -

x.(kJ =

{ :[n]+t[-nJ,

(-N+1)~n~(N-1)
otherwise

NL-1

(:z:[n] + :z:[-n]) w:>n
2
ZN-1,

( -N + 1) :S k :S (N- 1)

n=-N+l

~

=

:z:[-n]W:._
- 2 - 2N-1

L..J

+

n=-N+l

~ :z:[n]w:>n
L..J

2

2N-1

n=O

Letm= -n,
N-1

x.[kJ = "L..J

( )

N-1

n=O

x.(kJ

n.=O

~ :z:(n]cos u:~"1)

=

Reeall

( )

!!'_
+ "'""
2 w:-l:n
2N-1
L..J ~
2 w,kn
2N-1

n=O

N-1

X[k] =

L

:z:[n]W~",

0 :S k :S (N -1)

n=O

and

Re{X[k]}
So: DFT{:z:.(n]}

(2 k )
= N-1
~ :z:[n] cos ';.. n

# Re{X(k]}

(b)

Re{X[k]}

X[kJ + x·[kJ

=

2
N-1

2L
1

=

:z:(n]W~" +

1

2L

n=O

:z:(n]WN'"

n=O

N-1

~L

=

N-1

(:z:[n] + :z:[N-

n])W~"

n=O

So,

Re{X(k]}

=DFT { ~(:z:[n] + :z:[N- n])}

8.35. From condition 1, we can determine that the sequence is of finite length (N
X(~"')

=

= 5). Given:

1 +A, cosw + A2 cos2w

= 1 + A,(~"+ e-'"'l + A• (e'..., + e-i'"')
2

2

From the Fourier analysis equation, we can see by makhing terms tbat:

A1
A2
:z:[n] = 6[n] + T(o[n
- 1] + 6(n + l J) + 2(6[n2] + 6(n + 2]}

312
Condition 2 yields one of the values for the amplitude constants of condition 1. Since z[n] • cS[n - 3] =
:r[n- 3] 5 for n
2, we know z[-1] 5, and also that z[1] z[-1] 5. Knowing both these values
tells us that A1 = 10.

=

=

=

=

=

For condition 3, we perform a circular convolution between z[((n-3))s] and w[n], a three-point sequence.
For this case, linear convolution is the same as circular convolution since N = 8 ?: 6 + 3 - 1.
We know z[((n- 3))s]

= z[n- 3], and convolving this with w[n] from Fig P8.35-1 gives:
22

A2 + 15

lli
+ 11
2

1>- + 13

5+A>

lli

J

1>-

'

0
For n

2

1

= 2, w[n] • z[n- 3] = 11 so A2 = 6.

3

Thus, z[2]

5

7

n
8

= z[-2] = 3, and we have fully specified z[n]:

5

l lI

I

•

6

5

3

3

-3

4

z[n]

1

T

-2

-1

0

n

I

2

1

3

8.36. We have the finite-length sequence:
z[n]

= 20[n] + cS[n -

1] + cS[n - 3]

(i) Suppose we perform the 5-pt DFT:

X[k]
where

= 2 + Wt + Wi',

0 :0 k :0 5

w; = .-;('fl•.

(ii) Now, we square the DFT of z{n]:

Y[.I:J

=
=

X 2 [.1:J
2 + 2Wt +2Wff•
+2Wf + W;"+

wr

+ 2w1• + w:• + wt•.

Using the fact

w:• = w;
Y[.I:J = 3 + 5Wt + w;• + 4Wi' + wt•,

W!• = 141' =

o ::; .1: ::; 5

1 and

0 :0 k ::; 5

313
(a) By inspection,

y(n]

= 36(n] + 56(n- 1] + o(n- 2] + 46[n- 3] + o[n- 4],

0:::; n :::; 5

(b) This procedure performs the autocorrelation of a real sequence. Using tbe properties of tbe DFT,
an alternative method may be achieved with convolution:

y(n] =IDFT{X 2(k]} = z(n]• z[n]
The IDFT and DFT suggest that the convolution is circular. Hence, to ensure there is no aliasing, the
2: 2M- 1 where M is the length of z[n]. Since M = 3, N 2: 5.

size of the DFT must be N

8.37.

(a)
g,[n] = z[N- 1 - n],

G,[k] =

0:::; n:::; (N- 1)

I'l-l

L

z[N- 1- n]W~n,

0 :S k :S (N- 1)

n=O

Letm=N-1-n,
/'1-1

G,[k]

= L

z[m]W~(/'1-l-m)

m=O

=

/'1-1

w~d/1'1 X(ei'"')l..,z(b>/1'1)
G,[kJ = H1[k]
(b)

g,[n] = (-1)"z[n],

0:::; n:::; (N -1)

1'1-1

G2[k] =

L (-1)"z[n]W;',

n=O
1'1-1

=

L

z[n]w~fl·w~·

n=O
1'1-1

=

L

z[n]w:.,'+fl•

n=O

=

X(ei'"')l..,=b(Hf)/N

q,[k] = He[k]

0 :S k :S (N- 1)

314
(c)

O$n$ (N-1)
{ :(n],
g,(n] = :[n - N], N $n$ (2N -I)
0,
otherwise
2N-1

L :(n]W2~•
= ,..,..,

G3(k]

0$k$ (N -I)

N-1

2N-l

N-l

=N
N-l

L :fn]w;~ +
= ,..,..,
=

L

L

L

:fn]Wf.Y +

n=O
N-l

=

L

:(n -

NJWt.V

:fm]W;~m+Nl

m=O

:(n] (I+ w;p) W2~

n=O

G3[k]

N-1

L

=

(I+ w~tN/2))

=
=

{I+ (-l)t) X(eiw)lw=(d/N)
H3[k]

,..,..,

z[n]W~Im/2)

(d)
94 [nl

·

G4 [k]

z[n]+z[n+N/2], O$n$(N/2-I)

={

0,

otherwise

=

0$k :S (N- I) N/2-1 = L N/2-1 z[nJWt/2 + n=O N/2-1 = L L z[n+N/2]W,t/2 n=O N-1 z[nJWti 2 + L z[m]W~j';-N/ 2 ) m=N/2 A=O N-l = L z[n]W~tn n=O G4 [k] = = X(eiw)lw=(bt/N) Ha[k] (e) gs[n] = { :[n], 0 :S n$ (N -I)
0,
N $n$ (2N -I)
0,
otherwise

315
2/11-1

Go[k]

= L

Go(k]

L x[n]W,I'N
= .....,
= X(eiw)[w=(d/N)
= H2[k]

0 ~ k ~ (N -1)

:[nJW:N,
.....
N-1

(f)

go[n]

n even, 0 ~ n ~ (2N -1)

= { :[n/2],
0,

nodd

2N-l

= L

Go[k]

0

:r[n/2]WfN,

~

k

~

(N -1)

n:O

N-1

= L

:r[n]W~n

n=O

=
=

Go[k]

X (eiw)lw=(2d/N)
H1[k]

(g)

97[n]

= x[2n],

0 ~ n ~ (N/2- 1)

f-1

G1[k]

= L

0

..... x[2nJWM2,

= ~ :r[n]
=

N-1

L:x[n]

~

k

~

(N- 1)

c ~1)n) w~n
+

( 1 + W(N/2)n)

;

w~n

n=O

N-1

= ~ ~ :r[n] (w;:• + w~•+Nt•l)
G1[k]

= ~ [x+Nt•lJ]
= Ho[k]

8.38. From Table 8.2, the N-pt OFT of an N-pt sequence will be real-valued if

x(n]

For 0

~

n

~

=:((( -n))N]·

(N- 1), this may be stated as,

x[n]

= x(N- n],

0 ~ n ~ (N- 1)

316
For this case, N = 10, and
::r[1]
::r[2]

=
=

::r[9]
::r[S]

The Fourier transform of ::r[n] displays generalized linear phase (see Section 5.7.2). This implies that for
::r[n] .,< 0, 0 ::; n ::; (N- 1):
::r[n] = ::r[N - 1 - n]
For N

= 10,
::r[O]
::r[1]
::r[2]

=

=
=

::r[9]
::r[S]
::r[7]

To satify both conditions, ::r[n] must be a constant for 0::; n ::; 9.
8.39. We have two 100-pt sequences which are nonzero for the interval 0::; n ::; 99.
If x, [n] is nonzero for 10::; n S 39 only, the linear convolution
::r,[n] • :r2[n]
is a sequence of length 40 + 100- 1

= 139, which is nonzero for the range 10 ::; n S 139.

A 100-pt circular convolution is equivalent to the linear convolution with the first 40 points aliased by
the values in the range 100 ::; n $139. Therefore, the 100-pt circular convolution will be equivalent to the linear convolution only in the range 40::; n ::; 99. 8.40. (a) Since ::r[n] is 50 points long, and h[n] is 10 points long, the linear convolution y[n] = :r[n] • h[n] must be 50 + 10 - 1 59 pts long. = (b) Circular convolution= linear convolutin +aliasing. If we let y[n] = ::rjn] • h[n], a more mathematical statement of the above is given by 00 ::r[n] @.[n] = I: y[n + rN], 0::; n::; (N -1) r=-oo For N =50, ::r[nJ@.[n] We are given: :r[n] @>.In] =11[;.] + y[n +50], =10 Hence, y[n] + y[n + 50] = 10, = Also, y[n] 5, 0 ::; n ::; 4. Using the above information: 0 ::; n ::; 49 317 n=O n=4 n=5 n=8 y(O) + y[SO) = 10 y(SO) = 5 !1[4) + 11[54) = 10 11[54) = 5 11[5) + 11[55) = 10 11[55) = ? + 11[58) = 10 = ? = 10 = 10 y[8) 11[58) n=9 11[9) n=49 11[49) To conclude, we can determine 11[n) for 9 S n S 55 only. (Note that 11[n) for 0 S n S 4 is given.) 8.41. We have B A ----~------~-----------L------~--- n 0 39 9 30 n 10 19 (a) The linear convolution z[n]•11[n) is a 40 + 20- 1 =59 point sequence: ...-...---____ B•C A•C ----L-----------~--------~----------~--10 28 40 Thus, z[n)•y[n) = w[n) is nonzero for 10 S n S 28 and 40$ n $58. 58 n 318 (b) The 40-pt circluar convolution can be obtained by aliasing the linear convolution. Specifically, we alias the points in the range 40$ n $58 to the range 0$ n $18. Since w(n] = ~(n]•y(n] is zero for 0$ n $9, the circular convolution g(n] = ~[nJ@\t[n] consists of ooly the {aliased) wlues: w(n] = ~(n]• y(n], 40$ n :$49 Also, the points of g(n] for 18$ n $39 will be equiwlent to the poi.nts of w(n] in this range. To conclude, 8.42. w(n] = g(n], 18:$n:$39 w(n +40] - g(n], O$n:$9 (a) The two sequences are related by the circular shift: h2(n] = h,(((n + 4))a] Thus, H2 [kJ and = w.-•• n,[kJ = lw.-•• H,[k]t = IH,[k]l IH,[k]l So, yes the magnitudes of the S-pt OFTs are equal. (b) h,[n] is nearly like (sinx)fx. Since H 2 (k] H,[k]. h,[n] is a better lowpass filter. = .,.. 8.43. (a) Overlap add: If we divide the input into sections of length L, each section will have an output length: L + 100 - 1 = L Thus, the required length is L = 256 - 99 + 99 = 157 If we bad 63 sections, 63 x 157 = 9891, there will be a remainder of 109 points. Hence, we must pad the remaining data to 256 and use another OFT. Therefore, we require 64 OFTs and 64 IDFTs. Since h(n] also requires a OFT, the total: 65 OFTs and 64 IDFTs (b) Overlap save: We require 99 zeros to be padded in &om of the sequence. The first 99 points of the output of each section will be discarded. Thus the length after padding is 10099 points. The length of each section overlap is 256- 99 157 L. We require 65 x 157 10205 to get all10099 points. Because h(nj also requires OFT: = = = a 66 OFTs and 65 IDFTs (c) Ignoring the transients at tbe beginning and end of the direct convolution, each outpnt point requires 100 multiplies and 99 adds. overlap add: # mult = 129(1024) = 132096 # add = 129(2048) = 264192 319 overlap save: 131(1024} # mult = #add = = 131(2048) = 134144 268288 direct convolution: # mult #add = 100(10000} = 99(10000} = = 1000000 990000 8.44. First we need to compute the values Q[O] and Q[3]: Q[OJ Xl(1} = Xl(eJW)lw=O = co = 1 1-4 I:x>fn]= - -1 n=O = 4 3 x.(-1} = x.(eJW)lw=r = L r.[n]( -1}n = Q[3] co =<> 4 3 = One possibility for Q[k], the six-point DFT, is: Q[k] We then .find q[n], for 0 ~ 4 4 = 3"[k] + 3"fk - 3]. n <6 : q[n] = ~ L' Q[k]e'ftn >=<> = 6~(~3 + 3~(-1}n} = ~(1 + (-l}n) 9 otherwise it's 0. Here's a sketch of q[n]: 4/9 4/9 4/9 l.l.I. 012345 q[n] n 320 8.45. We have: DFIHz2 [n]} = X 2 [k] .....n. • = "\" L...z 2 [n]e-'' 0 Then: z,[O] = 1 • 7Lx,[kJ k=O 1 • = 7 })Re{Xa[k]} + jlm{X,[k]}) k=O = 1 • 7 L Re{Xo[k]} , since z 2 [0] is real. k=O = g[O]. To determine the relationship between z 2 [1] and g[1], we first note that since z 2 [n] is real: X(ei"') = X"(e-i"'). Therefore: X[k] = X"[N- kJ, k = 0, ... ,6. \\'e thus have: g[1] 1 • = 7 LRe{X,[k]}w,-• .... = !7 t, x,[kJ +2 x;[klw:-• 7 . k=O = !7 L 1:=0 = [ x, klw-• 7 2 . ! L x,[N- klw:-• 7 +7 k=O 2 1 1 • 2"''[1] + 14 L x,[kJW,' .=0 • X,[kJW,-u = ~z,[1] + 1~ L .=0 = = = 8.46. (i) This corresponds to :;[n] 1 2(z,[1] + za[6]) 1 (z,[1] + 0) 2 1 2"''[1]. = zj[(( ..:n))N], where N = 5. Note that this is only true for z 2[n]. (ii) X;(e'"') has linear phase corresponds to z;[n] having some internal symmetry, this is only true for x,[n]. 321 (ill) The DFT has linear phase corresponds to %;(n) (the periodic sequence obtained from z;(n]) being symmetric, this is true for z 1 [n] and x 2 [n] only. 8.47. (a) (b) V[k] = X(z)l "'=2~tj( 2•\+•) n=oc = •••• I: :r(n]z-nl z=2r' , ,--.--=• L x[n]z-nl , '·~·· ..... =• L x(n)(2eii )-n,-j1fin = =• L v(n]e-;1fin. = 1 n.=-oo = 1 ll=2e' ..... n=O We thus conclude that (c) 3 w(n] = = ~ L W[k]w;•n •=<> 3 ~ L(X(k] + X(k + 4J)e+''f•n 1:=0 3 = ~L 3 X(k]e+J'f•n + ~L k=O 3 = ~ L X[k]e+i'f>n + ~ k=O = ! X(k + 4Je+''f•n lo=O 7 L X(k]e+i'f•n 1:=4 7 LX(k]e+i'f»n 4>=0 = 2x[2n). We thus conclude that w[n) = 2x[2n). (d) Note that Y[k) can be written as: Y[k] = = X(k) + (-!)• X(k] X(k] + w:• X(k). 322 Using the DFT properties, we thus conclude that y[n] 8.48. = r[n] + z[((n - 4))s]. (a) No. x[n] only has N degrees of freedom and we have M ~ N constraints which can only be satisfied if r[n] = 0. Specifically, we want X(d'ii') = DFTM{r[n]} = 0. Since M ;:: N, there is no aliasing and x[n] can be expressed as: 1 z[n] Al-l ....2: =M X[k]Wt;, n Where X[k] is the M-point DFT of :r[n], since X[k] therefore the answer is NO. = O, ... ,M -1 . = 0, we thus conclude that r[n] = 0, and (b) Here, we only need to make sure that when time-aliased to M samples, :r[n] is all zeros. For example, let z[n] = o[n] - o[n - 2] then, Let M = 2, then we have X(ei'f 0 ) X(d'f') = = 1- 1 =0 1- I= 0 8.49. z 2 [n] is r,[n] time aliased to have only N samples. Since z,[n] = (31 )"u[n], We get: --<11:._ z,[n] 8.50. (a) Let n = { 1=1Tl" 0 n= O, ... ,N -1 otherwise = 0, ... , 7, we can write x[n] as: z[n] = I+ !(dtn + e-ifn)- !(dlf• + .-;lfn) 2 4 = 1 + !d'f"+ !d'f"7 - !d'fn3- !d'f"' 2 2 4 4 = !(s + 4d'f• + 4d'f•7 _ :W"fn3 _:lei¥"') 8 7 = ! 2: Xs[k]d'i"• 80=0 We thus get the following plot for Xs[k]: 323 8 Xa[k] 4 4 k 0 (b) Now let n 1 2 5 6 7 = 0, ... , 15, we can write v[n] as: 1 1 ...,_ .... v[n] = 1 + -(e', n + .-nn) _ -(e' , n + 0 - 1 , ") "L ·• 2 4 1_;hn2 +-r.-lll 1 _;"'-nH --C"'H 1 _;>"-'n!O 2 2 4 4 = = 1 = _!_ ~ 11,16 (k]e''fi-n 16 L.., +-~16 2..(16 + 8eitin2 16 + seitinH- (eitin6- 4e'ifn!O) •• &=0 We thus get the following plot for V16 (kj: 16 v,.[kJ 8 8 k 0 1 2 3 4 5 6 7 8 -4 9 10 11 12 13 14 15 -4 (c) 0:5k:515 where X(&w) is the Fourier transform of z[n]. Note that z[n] can be expressed as: z[n] = y[n]w[n] where: y[n] '"' 1 31m = 1 +cos(-)-cos(-) 4 2 4 324 and w(n] is an eight-point rectangular window. IX1o[k]l will therefore have as its even points the sequence JX8 (k]l .The odd points will correspond to the bandlimited interpolation between the even-point samples. The values that we can find exactly by inspection are thus: IX1o[k]l = IXs(k/2]1 k = 0,2,4, ... ,14. 8 IX1o[k]l I;> 4 4 2 2 "' 0 1 2 3 4 "'5 "' 6 7 "' 8 "' "'11 : •. 10 9 "' 12 13 14 15 8.51. We wish to verify the identity of Eq. (8.7): ~ L N-1 .,Ji;(&-r)n ={ n=O (a) For k- r k- r = mN, m: integer 1, 0, otherwise = mN, .,J~(&-r)n = .,J~(mN)n eibmn Since m and n are integers; .,J~(&-r)n = 1 , fork- r = mN So, 2_ N-1 L .,J~(·-r)n Nn=O 1 N-1 = -Nn=<> I;1 = 1 , for k - r = mN (b) ·~IN N-1 1 " ' ·~In- 1-e'Y N L.... e' - 1- eiif 1 n=O = This closed form solution is indeterminate for l mN only. For the case when l = mN, we use L 'Hopital 's Rnle to find: 1 _ ei2-' lim ·~, l-..mN 1- eJ-,r = = [=:::;:] N N l=mN k 325 (c) For the case when k- r # mN: JV-1 L ~ ~\f(l-r)n = n=O = 1- ~2r(l-r) 1 _ e>' \f(l-r) 0 Note that the denominator is nonzero, while the numerator will always be zero for k- r 8.52. (a) We know from Eq. (8.11) that if £ 1 [n] = f[n- m], we have: JV-1 X,[k] If we substitute r =n - = L f[n- m]W,t" ..... m into this equation, we get: N-1-m X1[k] L = f[r]W~(r+m) r=-m N-1-m = L w,tm f[r]W,t' r=-m (b) We can decompose the summation from part (a) into Using the fact that f[r] and W,t' are periodic with period N: -1 L f[r]W,t' r=-m Substituting l =r + N -1 L = f[r + N]W~(r+N) r=-m -1 L f[r]W~' = JV-1 L f[l]W,t' l=N-m r=-m (c) Using the result from part (b): X1[k] = Wf,m f I: LN-m 1 f[r]Wff + JVY:- f[r]W~'] r=O JV-1 = w,tm L f[r]Wff r=O = Wt/"X[k] Hence, if :ii[n] = x[n- m], then X 1[k] = w~m X[k]. 8.53. (a) 1. The DFS of i"[n] is given by: JV-1 :Ex"[n]W~" n.=O = JV 1 )" ( ~i[n]W,.V"" = x·[-kJ # mN. 326 2. The DFS of z'[-n]: Lt+l N-1 L z'[-n]W~" = n=O z[IJW;J). = X'[k] 3. The DFS of &{z[n]}: ~ z[n] + z'[n] w"" L- 2 N n=O = ~ ( X[kJ + X'[-kJ) = x,[k] 4. The DFS of j/m{f[n]}: ~ z[n] - x' [n] w•n L- 2 n=O ) = 21 (-X[k]- X'[-k] = Xo[k] N (b) Consider z[n] real: 1. Re{X[k]} = X[k] + X'[k] 2 From part (a), if f[n] is real, = DFS {f'[n]} DFS {z[n]} DFS {z[-n]} = DFS {f'[-n]} So, = = X[kJ X[-kJ &{X[k]} = = X'[-kJ X'[k] X'[k] + X[-k] 2 &{X[-k]} (i.e. the real part of X[k] is even.) 2. X[kJ- X'[kJ = /m{X[k]} 2 = x·[-k]- x[-kJ = /m{X[-k]} 2 (i.e., the imaginary part of X[k] is odd.) 3. IX[k]l = = VX[k]X'[k] Vx•[-A:]X[-k] = IX[-kJI 327 (i.e., the magnitude of X[k] is even) 4. = d[kJ arctan (Im{~[k]}) ~{X[k]} Im{X[-k]}) = arctan ( ~{X(-k]} = -d[-k] (i.e., the angle of X[k] is odd.) 8.54. 1. Let :r[n] (0 ~ n ~ N- 1) be one period of the periodic sequence f(n]. The Fourier transform of this periodic sequence can be expressed as: 00 X(eJ"') = L f[n]e-i"'" =-oo Recall the synthesis equation, Eq. (8.12): N-1 L f[n] = }_ X[k]W,Y•" Nt=<> Substitution yields: (~% X(k]W,Y•") .-'"'" X(elw) = j;oo Rearranging the summations and combining terms: The infinite summation is recognized as an impulse at w = (27rk/ N): X(ei"') =!.. ~ X(k]6 (wNt=<> 2"k) N 2. Since :r(n] corresponds to one period of f(n], we must apply a rectangular window (unit amplitude and length N) to the periodic sequence. Thus, to extract one period from f[n]: :r(n] = f(n]w[n] w[n] = { 0 ~ n ~ (N -1) ' 0, otherwise 1 The window has a Fourier transform: 00 W(e'"') = L w(n]e-i"'" .... -oo N-1 = L e-Jwn a=O = 1- e-;wN 1- .-,., = = .-j'f ei'f - .-j'f .-i.,(¥Jsin(w~) sin(~) 328 3. Since :r[n] = z[n]w[n], the Fourier transform of :r[n] can be represented by the periodic convolution (see Eq. (8.28)). X(eiw) = 2.1• d/1.!. ~ 2.. N ~ k.. -111' X[kJo ( 8 _ 2lrk) sin [~(w- 8)] ,-;(¥ ) Hence, N-1 z*[n] = ~ L X*[k]W~" •=<> substituting: N-1 L \:rln1\' n=•/N)••, 0 ~ k ~ (N -1) n=O N-l X.,[k] = L :r[n]e-i(2d/N+r/N)n n=O = N-l E z[nje-j(wn/N)e-J(2,.../N)I:n n=O So, x.,[n] = x[n]e-;(wn/N). (b) x.,[N- k] = N-l L x[n]e-;(rn/N).-;(.. /N)(N-•)•, 0 ~ k ~ (N- 1) n=O N-l = L :r[n]e-;(rn/N)ei(2r/N)• n=O x.,[N- (k + 1)] = N-l L :r[n]e-i(rn/N).-i(2r/N)(N-•-•)n n=O = L :r[n]e-;(rn/N) N-l 0 -;(2r/N)(N-l)nei(2r/N).,. n=O N-l = L :r[n]ei<•n/N)ei(2r/N).,. . n=O = X.it[k] So, X.,[k] = X.it[N- (k + 1)] , for 0 ~ k ~ (N- 1) and :r[n] real. 333 (c) (i) N -k-1 is odd when k is even. H R(k] G = (k] = XM(2k], we may obtain XM(k] from R[k] as follows: { R[l:/2], k even R•((N- (k + 1))/2], k odd where we note that R•((N- (k + 1))/2] = Xit(N- (k +I)] fork odd. (ii) R(kJ = x... [2kJ (N/2)-1 = L z[n]e-i(4d/N+•/N)n n=O N-1 = L z[n]e-,<•n/N)e-i('Mf) (N/2)-1 L = N-1 L z[n]e-11•n/Nie -i('-'ffl' )•+ x(n]e-i<•n/N)e -i('-'ffl') n=N/2 n.=O (N/2)-1 = L (z[n]e-il•n/N) +z[n+N/2Je-il•n/Nie-'l•/21)e-;('Mf) n=O So, r(n] = (x[n] - jx(n + N f2])e-il•n/N), 0<- n <- (N2 -1) (d) = = x,.,[kJ x, ... [n] x1 ... [kJX, ... [kJ N-1 Z1M[r]x, ... [((n- r))N] •=<> L From part (a): :z:1M[n] = :z:1[n]e-il•n/N) Z2M[n] = :z:,[n]e-i(•n/N) Z3M[n] = z 3[n]e-;( .. /NI So, :z: 3(n] = N-1 ei<•n/N) L z.(r]z2[((n- r))N]e-ii•/N)[((n-r))H+r] r=O N-1 = L z.[r]:z:,[((n- r))N]e-i<•/N}[((n-r))H-(n-•)] r=O Since, ((n-r))N ={ n-r, n>r - N+n-r, nM[n] = z,[n]e-;<•n/N) 0 = then the modilied circular convolution is equivalent to the modified linear convolution: ZtM[n] ~>M[n] = ZtM[n] • Z>M[n) (i.e. no aliasing occurs.) = Z3M(nj ZtM(n) * Z2M[n) - N-1 = Thus, :z:,[n) = L ZtM[r)z2M[n- r) N-1 ei(on/N) L :z:,[n]z,[n- r)e-j(•n/Nie-J(•n/N)(n-r) r=O N-1 = L z,[r]:z:,[n- r)e-;(r/N)(n-r) r=<> So, N-1 z,[n) 8.62. =L :r,[r]:z:,[n- r)e-><•fN)(n-r) = :z:1 [n) • :r2[n) (a) We wish to compute z[n)@j.[n): let :r1 [n] z,[n) h 1 [n) h 2 [n] = = = = :r[n), :r[n + 32], h[n], h[n + 32], 0 0 0 0 :5. n :5. 31 :5. n :5. 30 :5. n :5. 31 :5. n :5. 30 z[n). h[n) = z,[n]. h,[n] + :r,[n). h,[n). o[n- 32] + z,[n]. h,[n). o[n- 32] + z,[n]. h,[n]. 6[n- 32]. o[n- 32] Let !11 [n] !f2[n] !13[n] !lt[n] = = = = ·~1·11'"' = :r,[n] • h1 [n] :r,[n) • h,[n) = z 1 [n] 2[h) z,[n) • h,[n) :r,[n] ,[n] :r 2[n) • h2[n) :r,[n) 64 ,[n] = 335 We can compute each of the above circular convolutions with two 64-pt DFI's and one 64-pt inverse DFT. = = y(n) :r[n)• h(n) u,(n) + 112(n- 32) + y3(n- 32) + ll<[n- 64) So :r[nJ@).[n) = y(n) + y(n + 63), 0 :S n :S 62 The total computational cost is 12 DFTs of size N = 64. (b) Using two 128-pt DFTs and one 128-pt inverse DFT: = :r[n) @!n) = :r[n)• h(n) y(n) The 63-pt circular convolution: :r[nJ@.[n) = y[n) + y[n + 63), 0 :S n :S 62 (c) Using the 64-pt DFT method of part (a): #mult = 4(12)(64log2 (64)) = 18432 #mult = 4{3)(128log2{128)) = 10752 Using 128-pt DFI's: Direct convolution: 63 #mult =2 L =• n - 63 = 3969 8.63. From each circular convolution, the first 49 points will be incorrect. Therefore, we get 51 good points and the input must be overlapped by 100 - 51 = 49 points. {a) V = 49 (b) M =51 {c) The points extracted correspond to the range 49 Distorting filter: h[n) :S n :S 99. = o[n)- !6(n- "<>) 8.64. {a) The Z-transform of h[n) 00 H{z) = L h(nV" n=-oo H{z) The N-pt DFT of h(n]: {N = 1- !.-... 2 = 4no) •no-1 H(k) = L h(n]w;,:-e, n=O = H(k) = 1-!w;no 2 ... 1 - !e-i(•/>l• 2 o :S k :S {4no - I) 336 (b) = H;(z) h;[n] = 1 + 1,_...,, 112 t. f"" G (1)-no for causality lzl > 2 o[n - kno] The filter is IIR. (c) G[k] 1 1 = H[k] = 1 _.,-;<•l•l•, 0$ k $(4no- 1) The impulse response, g[n], is just h;[n] time-alia.sed by 4no points: g[n] g[n] = (1+116+2!6+···)o[n]+G+;2+5~2+··}[n-no] + G~ = 16 8 4 2 o[n] + o[n- no]+ o[n- 2no] + o[n- 3no] 15 15 15 15 + + 1:24 + .. -) o[n - 2no] "" G + 1!8 + ~8 + .. -) o[n - 3no] (d) Indeed, G[k]H[k] = 1, 0$ k $(4no- 1) However, this relationship is only true at 4no distinct frequencies. This fact does not imply that for all w: (e) y[n] = g[n] • h[n] M 8 4 2 8 = 156[n] + 156[n- no]+ 156[n- 2no] + 156[n- 3no]- 156[n- no] 4 2 1 o[n- 2no]- o[n- 3no]- o[n- 4no] 15 15 15 16 1 o[n]- o[n- 4no] 15 15 y[n] = 8.65. (a) We start by computing Xn[k + NJ: N-1 Xn[k+ NJ = L i[n]HN[n(k+ N)] n=O = I; i[n](cos(2"(nk; nN)) + sin(2.-(nk + nN)) n=O = N-1 L i[n](cos(21mk) + sin(2wnk )) ..,.. N N N-1 = L i[n]HN[nk] = Xn[k]. N 337 We thus conclude that the DHS coefficients form a sequence that is also periodic with period N. (b) We have: I N I N N-1 .,..L XH[k]HN[nk] = N-1 N-1 L(L z(m]HN(mk])HN(nk] 1:=0 m=O } N-1 N-l m=O .t=O L z[m) L HN(mk]HN(nk] N = = ~z[n]N = z[n]. Where we have used the fact that }:~,;;;,1 HN[mk]HN[nk] = N only if ((m))N it's 0. This completes the derivation of the DHS synthesis formula. = ((n))N, otherwise (c) We have: = cos(27f'{a;N))+s"m(27r(a;N)) = cos(N = = ( cosN +smN HN[a+NJ 2•a 271'a + 2") + s"m( N + 21r) 2.-a) . (2.-a) HN(a]. And: HN[a+b] = = = = = cos(2.-(';.,+b))+s"m(2.-(';.,+b)) (CN[a]CN[b]- SN[a]SN[b]) + (SN[a]CN[b] + CN[a]SN[b]) CN[b](CN[a] + SN[a]) + SN[b](-SN[a] + CN[a]) CN[b](CN[a] + SN[a]) + SN[b](SN[-a] + CN[-a]) CN[b]HN[a] + SN[b]HN[-a] = CN[a]HN[b] + SN[a]HN[-b] ( s"mce HN[a + b] HN[b +a] ) = Where we have used trigonometric properties. (d) We have: N-1 DHS(z[n- no]) = I: z[n- no]HN[nk] -=0 N-1-no = L n.=-ftcl z(n)HN((n +no)k) 338 N-1-no L = i(n](HN[nk]CN(nok] + HN(-nk]SN(nok]) n=-no N-1-no L = CN(nok] N-1-no N-1 = CN[nok] L i(n]HN[-nk] N-1 i(n]HN[nk] + SN[nok] =<> CN(nok]XH[k] = L i(n]HN(nk] + SN(nok] L i[n]HN[-nk] =<> + SN(nok]XH(-k] Where we have used the periodicity of HN(nk] and i(n]. (e) We have: N-1 DHT{i,[n]} = DHT{L :r.(m]:r,[((n- m))N]} m=O N-1 N-1 XHl[k] = L (L :t1[m]:r,[((n- m))N])HN[nk] N-1 = = N-1 L :t1[m] L :r,[((n- m))N]HN(nk] N-1 L :r.(m]DHT{:r•[((n- m))N]} m=O N-1 = L :r.(m](XH2[k]CN[mk] + xH.[((-k))N]SN[mk]) (using P8.65-7) N-1 = L = N-1 :r![m]XH>[k]CN(mk] + 1: + 1: L :r![m]XH2[(( -k))N]SN[mk] m=O :t1(m]XH>(k](HN(mk] =0 + HN(-mk]) 2 :r![m]XH>[((-k))N](HN[mk] - HN[-mk]) 2 m=O = 1 1 2xH.[k](XH![kJ +XH![((-k))N]) + 2xH,[((-k))N](XH!(kJ- xH1[((-k))NJl = 21 xH1[k](XH,[kJ + xH.[((-k))Nll + 21 xH.(((-k))N](XH•[kJ- xH.[((-k))NJl This is the desired ·convolution property. (f) Since the DFT of :r[n] is given by: N-1 X[k] = L ...... = N- 1 L =0 :r[n]e-1'iF 2..kn 2,-kn :r(n](cos(--~d + jsin(---:;.,--)) 339 1 ~ 2r kn . 2" kn = ...., .L.. o:[n](cos( """N) - j sm( """N)) N-1 -· L o:[nj(CN[kn)- jSN(kn]) =• then: N-l L = ~(X[kj +X((( -k))N]) o:[n)CN[knj =D N-1 L o:[n)SN[knj ...., = - 1 j (X[k)- X[(( -k))N)) 2 We thus get: N-1 XH(k) L = o:[n](CN[kn) + SN(kn)) This allows us to obtain XH[k) from X[k). (g) We have: Xg(kj = N-1 L o:[n](CN[kn) + SN(kn)) =0 Therefore: N-l L o:[n]CN[kn] = o:[n]SN[kn] = =D N-l L =D We thus get: N-1 X(k] = I; o:[n]e-i'iP =D N-1 = L =D o:[n](CN[knj- jSN(kn)) 340 = ~(Xs[k] + Xs[((-k))N]l- j~(Xs[k]- Xs[((-k))N]) = 1 . 1 . 2 2 2 2 (-- l)Xs[k] + (- + l)Xs[((-k))N] This allows us to obtain X[k] from Xs[k]. 8.66. (a) The DTFI' is given by: = . X(ei") + X(ei").-; .. N = X(ei"')(l + e-i"'N) X(ei") The DIT is just samples of the DTFI': X[kJ = = .i(ei"ll ..=w X(ei2•12N)(l + (-l)t) Therefore: .i[kJ = { 2X[!~ k even k odd (b) The original system computes the following: X[k]H[k] = { 2X[!JH[k~ ' keven k odd We thus want: X[k]G[k] G(k] = = 2X[k]H[2k] 2H[2k] 2 g[n] = = k = o, ... ,N -1 2N-1 L h[n]e-;9 ..... k =O, ... ,N -1 2(h[n] + h(n + N]) System A time aliases and multiplies by 2. For system B, we need: Y[k] = { W(~~ k even k odd 341 Thus: y(n] = 1 w(n] , o::;n::;N-1 N::;n::;2N-1 w[n-N] otherwise 0 System B regenerates the 2N -point sequence by repeating w(n]. 8.67. (a) We have: = { 1 , Jwl::; ~ 0 otherwise , Since h(n] is FIR, we assume it is non-zero over 0::; n ::; N. The phase of H(e 1 ~) should he set such that h(n] is symmetric about the center of its range, i.e. ~- Therefore, the phase of H(ei~) should he ei "f' . So one possible H(k] may be: 0::; k ::; i4N otherwise 4N-~::;k-s_4N that is: 0 < k < !f. &t• H(k] = 1 o~he~ 0 : eit• ' 4N- !f. < k < 4N 2 - - (b) System A needs to perform the following operations: X(k]H'(k] Y2(k] = 1 0 X(k- 3N]H'(k- 3N} 0::; k::; ~ otherwise 4N-!f.n/(2N-2) 1=0 N-l = L 1 ( Xn/(2N-2) 2N - 2 .=<1 + 2N-3 L •=N X where a[k] is given by: o[k]= U k=OandN-I I$.1: $N-2. This completes the derivation. 8.69. v[n] therefore, for k = z2[2n] =0, I, ... , N- 1: V(k] I = i(X2[k] + X2[k + N]). Using equation (8.I68), we have: V[k] = ~(X2(k] + X2[k + N]) = e'~ Re{Xfkje-;ifr} + .,i"'tlNl Re{X(k + N]e-; • . x,[k]e'...... ,<,N> 1 Ncx•2[o] + 2 1 u,;(X"[O] + N-1 2N-1 Cl k=N+l L x"[k]ei..t<•N>.,......,<,N> _ L N-1 L · 2N-l = 1 N(X"[O] + 2 N-1 X''[kj.,id(2n+1)/(2N) _ 1=-1 L N-1 + L X"[k]e-j•l(..,+1)f(2N)) i=l = 2N(X''[O] + N-1 L L X"[kj.,i•(2N--.,;• ....t<•"'>J x•'[k](ei••<••+1){(2N) i=l + .-jd(2n+l)/(2N))) i=l 1 = -2N (X'2[0] + N-1 ~ X' 2(k] cos( 1rk(2n + 1))) ~ 2N .f:=l . Furthermore: z[n] = z,[n] N-1 = ~ L .=0 for n = 0, 1, ... , N - 1 P[k]X''[k]cos(""k(~';,+ 1)) where P[k] is given by: P[k] = { ~ k=O 1~k~N-l. This completes the deri\'B.tion. 8. 71. First we derive Parsev..l's theorem for the DFT. Let x[n] be an N point sequence and define y[n] as follows: = :r:[nJ®z*[((-n))N]· y[n] Using the properties of the DFT, we have: Y[k] = X[k]X*[k] = fX[k]f 2 • ·Note that: y(O] = L fz[n]l 2 n and using the DFT synthesis equation, we get: 1 y[O] = N N-1 ....L Y[k] . O~n~N-1 345 Parseval's Theorem for the OFT is therefore: N-1 L jz[n]l' = ~ L n IX[k]l". A:=O (a) Note that: N-l N-1 L IX'1 [k]l 2 =L jX,[k]j 2 n=O and, using equation (8.164): 2N-3 L N-1 =2 L JX,[k]j 2 IX"[k]j 2 -IX"[OJJ 2 -IX" IN -1]j 2 • Using the OFT properties: 2N-3 L jz,[n]l' = 2Nl- 2 L JX,[k]J' '=0 n and, using equation (8.161): 2N-3 L N-1 Jz,[n]l 2 =2 L jz[n]l' -lz[O]I 2 - )z/N -1]1 2 • n=O n=O We thus conclude: u./- N-1 2 (2 L N-1 JX"fk]l' -IX''[OJI'- IX"[N- 1]1 2 ) = 2 n=O L jz[n]l'- lr[OJI'- lriN- !]I'. n=O (b) Using equation (8.171). N-1 L /'rl-1 IX'2 [k]l 2 =L jX,[k]j 2 . n=O n=O Note that, using equation (8.167): 2N-l L N-1 jX,fkJI' =2 L IX[k]l 2 -IX[O]j 2 , and, using equation (8.!66): 2N-l L N-1 jz,[n]l' =2 L Using the OFT properties: 2N-l L n=O lz[n]j 2 • n=O n=O l jz,[n]l' = 2N 2N-l L jX,[kJI'. .... We thus conclude: N-1 2~(2 L .... N-1 IX[k]J 2 -IX[O]j 2 ) =2 L jz[n]j 2 • 347 Solutions - Chapter 9 Computation of the Discrete-Time Fourier Transform 349 9.1. There are several possible approaches to this problem. Two are presented below. Solution #1: US.. the program to compute the DFT of X(k], yielding the sequence g[n]. N-1 g[n] =L X[k]e-j2dn/N 1=0 Then, compute 1 z[n] = Ng[((N- n))N] for n = 0, ... , N- I. We demonstrate that this solution produces the inverse DFT below. z[n] = 1 Ng[((N- n))N] = ~ N-1 L X[k]e-j2w1(N-n)/N 1=0 ~ = N-1 L X[kjd2dn/N 1=0 Solution #2: Take the complex conjugate of X[k], and then compute its DFT using the program, yielding the sequence f[n]. N-1 /[n] X"[kje-j2wln/N =L i:=O Then, compute z[n] = ~/"[n] We demonstrate that this solution produces the inverse DFT below. o:[n] = ~f"[n] = ~ N-1 L X[kj&••ln/N 1=0 9.2. (a) The "gain" along the emphasio:ed path is -W}. (b) In general, there is only one path between each input sample and each output sample. (c) o:[O] to X[2]: The gain is I. o:[l] to X[2]: The gain is W}. o:[2] to X[2]: The gain is -WX, =-I. x[3] to X[2]: The gain is -WX,W} -W}. x[4] to X[2]: The gain is WX, I. x[5] to X[2]: The gain is WX,W} = W}. x[6] to X[2]: The gain is -WX,WX, =-I. x[7] to X[2]: The gain is -WX,WX,W} -W}, as in Part (a). Now = = = 7 X[2] = L z[n]w;n n=O = = z[O] + z[1]W/ + z[2JW: + z[3JW: + z[4JW: + z[S]W/ 0 + z(6]W/ 2 + z[7JWi 4 z(O] + z(1]W/ + z[2](-1) +z(3](-W/) + z(4](I) + z[SJWi + z(6]( -1) +z(7)(- Wf) 350 Each input sample contributes the proper amount to the output DFI' sample. 9.3. (a) The input should be placed into A[r] in bit-reversed order. A[O] A[1] A[2] A[3] A[4] A[S] A[6] A[7] = = = = = = = = :r[O] :r[4] :r[2] :r[6] :r[l] :r[S] :r[3] :r[7] Tbe output should then be extracted from D[r] in sequential order. X[k] = D[k], k =0, ... , 7 (b) First, we find the DFT of (- WN )" for N = 8. ' = 2::<-Ws)"w,;• X[k] n=O = ' :L<-1J"w,;w,;• n=O ' = :L:•n/2- u[n] = 2m. Thus, the frequency of this oscillator is w 2" = 2"" 352 9.5. Multiplying out the terms, we find that (A- B)D+ (C -D)A =AD- BD +AC- AD= AC -BD =X (A- B)D+ (C+ D)B =AD- BD +BC +BD =AD+ BC =Y Thus, the algorithm is verified. 9.6. Answer 3 = 2. Decimation in Time: The figure is the basic hutter:fly with r Decimation in Frequency: The figure is the end of one butter:fly and the start of a second with r = 2. 9. 7. The figure corresponds to the flow graph of a Sec:ond-order recursive system implementing Goertzel's algorithm. This system finds X[k] for k = 7, which corresponds to a frequency of 14>< 7r "'• = 32 = 16 9.8. This is an application of the causal version of the chirp transform with N M Wo f:>w We therefore have for for Wn Wn = wo +nD.w or = = = = 20 The length of z[n] 10 The number of desired samples T 2• The starting frequency 2• 21 The frequency spacing between samples y[n + 19] = X(eiw• ), y(n] = X(eiw• ), n n = 0, ... , 9 = 19, ... , 28 = wo + (n- 19)J:>w. 9.9. In this problem, we are using butter:fly flow graphS to compute a DFT. These computations are done in place, in an array of registers. An example flow graph for aN = 8, (or v log2 8 3), decimation-intime DFT is provided below. = = 353 -1 -1 -1 (a) The difference between l 1 and lo can be foood by using the figure above. For example, in the first stage, the array elements A[4] and A[SJ comprise a butterfly. Thus, l 1 - lo = 5- 4 = 1. This difference of 1 holds for all the other butterflies in the first stage. Looking at the other stages, we find stage m = 1: l 1 -lo = 1 stage m = 2: stage m = 3: l1 - lo = 2 l 1 -lo = 4 From this we find that the difference, in general, is l, -lo =2m->, form= 1, ... ,v (b) Again looking at the figure, we notice that for stage 1, there are 4 butterflies with the same twiddle factor. The lo for these butterflies are 0, 2, 4, and 6, which we see differ by 2. For stage 2, there are two butterflies with the same twiddle factor. Consider the butterflies with the W,l' twiddle factor. The lo for these two butterflies are 0 and 4, which differ by 4. Note that in the last stage, there are no butterflies with the same twiddle factor, as the four twiddle factors are unique. Thus, we found stage m = 1: t:l.lo = 2 stage m = 2: stage m = 3: t:l.lo = 4 n/a From this, we can generalize the result t:J.lo =2m, ·form= l, ... ,v -1 9.10. This is an application of the causal version of the chirp transform with N M ""' Aw Letting W = = = = 12 The length of z[n] 5 The number of desired samples ~; The starting frequency h The distance in frequency between samples 10 = .-;&. we must have r[n] = .-;wonwn'/2 = .-;'ftn.-;tfn'/2 9.11. Reversing the bits (denoted by -+) gives 0 1 2 3 4 5 6 = = = = = = 0000 0001 0010 0011 0100 0101 = = 0110 12 = = = = = 1000 13 = 1101 14 = 1110 7 8 9 10 11 15 = 0111 1001 1010 1011 1100 1111 .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... 1001 = = = = = = = = = = 0101 = 5 1101 0111 = = = = 13 1111 = 15 0000 1000 0100 1100 0010 1010 0110 1110 0001 0011 1011 0 8 4 12 2 10 6 14 1 9 3 11 7 The new sample order is 0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15. 9.12. False. It is possible by rearranging the order in which the nodes appear in the signal !low graph. However, the computation cannot be carried out in-place. 9.13. Only them = 1 stage will have this form. No other stage of aN= 16 ra.dix-2 decimation-in-frequency FFT will have a W1& term raised to an odd power. · 9.14. The possible values of r for each of the four stages are ~=1, r=O m=2, m=3, m=4, r=0,4 T 0,2,4,6 r = 0,1,2,3,4,5,6, 7 = 355 9.15. Plugging in some values of N for the two programs, we find Program A Program B N 2 4 20 80 4 16 8 16 32 64 256 1024 64 4096 240 640 1600 3840 Thus, we see that a sequence with length N = 64 is the shortest sequence for which Program B runs faster than Program A. 9.16. The possible values for r for each oftbe four stages are m=2, r=O r =0,4 m=3, r =0,2,4,6 m=4, r m= 1, =0, 1,2,3,4, 5,6, 7 where WN is the twiddle factor for each stage. Since the particular butterBy shown bas r which have this butterfiy are m=3,4 = 2, the stages 9.17. The FFT is a decimation-in-time algorithm, since the decimation-in-frequency algorithm bas only terms in the last stage. !Vf2 9.18. lf the N 1 = 1021 point DFT was calculated using the convolution sum directly it would take Nf multiplications. lf the N 2 = 1024 point DFT was calculated using the FFT it would take N2log2 N2 multiplications. Assuming that the number of multiplications is proportional to the calculation time the ratio oftbe two times is N'f 1021 2 N2log2 N2 = 1024log2 1024 = l01. 8 "' 100 which would explain the results. 9.19. X(e'••l•) corresponds to the k = 3 index of a length N second-order recursive system for Goertzel's algorithm, a = 2cosC~k) = 2 cos C"i3)) = -¥2 b = -W~ = = -e-itnr/8 1 +j V2 = 8 DFT. Using the 6ow graph of the 356 9.20. First, we derive a relationship between the X 1 (ei") and X(ei") using the shift and time reversal properties of the DTFT. = o: 1 [n] z[32- n] X,(.,;w) = X(e-i")e-i3 ""' Looking at the figure we see that calculating y[32] is jnst an application of the Goertzel algorithm with k 7 and N 32. Therefore, = = y[32] = x,[7] = X, (ei" ) 1.,= 'H' = X(e-i")e-i"321 w=fi = X(e-ilt)e-i2 = X(e-ilt) Note that if we put z[n] through the system directly, we would be evaluating X(z) at the conjugate location on the unit circle, i.e., at w = +77r/l6. 9.21. = 0, for n < 0 and n > N- 1. From the figure, we see that Y>[n) = :t[n] + Wty.[n- 1] Starting with n =0, and iterating this recursive equation, we lind Y>[OJ = z[OJ 11•[1] = :[1] + wtz[O] Y>[2] = i[2] + W~z[1] + WYz[O] (a) Assume z[n] Y>fN] = :t[N] + W~z[N- 1) + · · · + W~(N-l):t[1) + Wf/' z[O) N-1 = o+ L w~fN). 9.29. (a) We offet two solutions to this problem. Solution #1: Looking at the DFT of the sequence, we find X[k] = N-1 L z[n]e-j2dn/N =0 = (N/2)-1 L N-1 z[n]e-j2dn/N + (N/2)-1 L = = z[n]e-j2dn/N (N/2)-1 z[n]e-;•••nJN n=O = L n.=N/2 n=O + L z[r + (N/2)]e-;2d[•+(N/2)J/N r=O (N/2)-1 L z[n][l- (-1)•je-j2•h/N =0 0, k even Solution #2: Alternatively, we can use the circular shift property of the DFT X[k] = -X[k]e-;<~>• = -(-t)•X[k] = (-1)>+ 1 X[k] When k is even, we have X(k] = -X[k] whlch can only be true if X(k] = 0. tO lind 363 (b) Evaluating the DIT at the odd-indexed samples gives us N-1 X[2k + 1] = L ..... :~:[n]e-;(2•/N)(2H1)n N/2-1 = L N-l L :<[n]e-i2•n/N .-;2•1n/(N/2) + n=O :<[n]e-i2•n/N .-i2•>n/(N/2) n=N/2 = DIT N/ 2 { :<[n]e-;(2•/N)n} + N/2-1 L '"" :<[1 + (N/2)fe-i2•{1+(N/2)}/N 0 -;2d{I+(H/2)]/(N/2) N/2-1 = DITN/2 { :t[n]e-;(2•/N)n} + ( -1)( -1) = DIT H/2 { 2:<[n]e-;(2r/N)n} L :t[l]e-;2•1/N 0 -;2rti/(H/2) = for k 0, ... , N /2- 1. Thus, we can compute the odd-indexed DIT values using one N /2 point DIT plus a small amount of extra computation. 9.30. (a) Note that we can write the even-indexed values of X[k] as X[2k] for k the definition of the D IT, we find X[2k] = 0, ... , (N/2) - 1. From N-1 L = = :t[n]e-;2r(2t)n/N ..... L ..... N/2-1 + :t[n]e-;IPi,>n N/2-1 L :t[n + (N/2)je-;,Ji•>>ne-;,Ji,• N/2-1 = L = ..... (:t[n] + :t[n + (Nf2)J)e-irPfu>n Y[k] Thus, the algorithm produces the desired results. (b) Taking theM-point DIT Y[k], we find Y[k] = = .M-1 00 L L oo :t[n + rM]e-i2dn/M M-1 L L :t[n +rM]e-j2d(n+rM)/Me>"2•(rJt)t/JI r=-oo n=O Let I= n + rM. This gives 00 Y[k] = L :t[l]e-;,..,/.11 l=-oo = xcei2••fMJ = Thus, the result from Part (a) is a special case of this result if we let M N /2. In Part (a), there are only two r terms for which y[n] is nonzero in the range n = 0, ... , (N /2) - 1. 364 (c) We can write the odd-indexed values of X[k] as X(2k definition of the DFT, we find X[2k + 1] = = = fork = 0, ... , (N/2)- 1. From the N-1 L z[n]e-i..(2H1)n/N ....., N-1 L z[n]e-i2rn/Ne-i2r(2>)n/N ....., (N/2)-1 (N/2)-1 L z[n]e-i2•n/N.-i,Jj,,>n + L z[n + (N/2)]e-j2r{n+(N/2)]/Ne-;;!r,>ln+(N/2)) n=O n=O (N/2)-1 = + 1] L [C:r[n]- :r[n + (N/2)])e-i~n] .-;rM.r>n n=O Let n _ { (z[n]- z[n + (N/2)])e-i< 2•/N)n, 0$ n $(N/2)- 1 y[J-0, " oth ennse Then Y[k] follows. = X[2k + 1]. Thus, The algorithm for computing the odd-indexed DIT values is as step 1: Form the sequence y[n] ={ (:r[n]- z[n + (N/2)])e-i(.. /Nln, 0$ n::; (N/2) -1
0,
otherwise

step 2: Compute the N/2 point DFT of y[n], yielding the sequence Y[k].
step 3: 'l'he odd-indexed values of X[k] are then X[k] = Y[(k- 1)/2), k = 1, 3, ... , N- 1.
9.31.

(a) Since x[n) is real, z[n) = x"[n], and X[k] is conjugate symmetric.
N-1

X[k]

= L x"[n]e-i~>n
..,..
=
=

(

)"
N-1
~ z{n]ei~>ne-i~Nn

X"[N-k]

Hence, XR(k) = XR{N- k) and Xr(k] = -Xr(N- k].
(b) In Part (a) it was shown that the DIT of a real sequence z[n] consists of a real part tbat has even
symmetry, and an imaginary part that has odd symmetry. We use this fact in the DIT of the
sequence g[nJ below.
G{k]

= X 1{k] + jX2[k]

=

(X1ER(k)

+ jX1or{k]) + j(X2ER[k) + jX2or(k])

= X1ER[k]- Xwr{k] +i (X1or[k] + x.ER[k])
real part

imaginary part

In these expressions, the subscripts "E" and "0" denote even and odd symmetry, respectively, and
the subscripts "R" and "1" denote real and imaginary parts, respectively.

365

Therefore, the even and real part of G[k] is

GER[k)

= X1ER[k)

the odd and real part of G[k] is

GoR[k] = -X,or[k]
the even and imaginuy part of G[k] is

GEr[k] = X2ER[k)
and the odd and imaginary part of G[k] is

Gor[k]

= X1or[k]

Having established these relationships, it is easy to come up with expressions for X1[k] and X2[k].

X 1[k]
x.[k]

=
=
=
=

X1ER[k] + jX,or[k)
GER[k] + jGor[k]
X2ER[k] + jX,or[k)
GEr[k]- jGoR[k]

(c) An N = 2" point FFT requires (N/2) log2 N complex multiplications and N log2 N complex additions. This is equivalent to 2N log2 N real multiplications and 3N log2 N real additions.
{i) The two N-point FFTs, X 1 [k] and X 2 [k], require a total of 4Nlog2 N real multiplications and
{ii) Computing theN -point FFT, G[k], requires 2N log2 N real multiplications and 3Nlog2 N real
additions. Then, the computation of GER[k], GEr[k], Gor[k], and GoR[k] from G[k] requires
approximately 4N real multiplications and 4N real additions. Then, the formation of X1 [.I:] and
X,[k] from GER[k], GEr[k], Gor[k], and GoR[k] requires no real additions or multiplications.
So this technique requires a total of approximately 2N log2 N + 4N real multiplications and
3N log2 N + 4N real additions.
(d) Starting with

N-1

X[k]

=L

z[n]e-j2dnfN

"'""

and separating :[n] into its even and odd numbered parts, we get

X[k]

=

L

z[n)e-j2wlm/N +

n even

L

z[n)e-;2wlm/N

nodd

Substituting n = 2! for n even, and n = 2l + 1 for n odd, gives

(N/2)-1

X[k]

= L

l=O

=

(N/2)-1
:[2l)e-i2•1l/(N/2)

+

L :[2! + l)e-;2d(2l+l)/N
l=O

(N/2)-1

(N/2)-1

l=O

l=O

L :[2l]e-j2dl/(N/2) + ,-;2rt/N L

={
(e) The algorithm is then

X1[k) + .-;•••IN X,[.l:),
X,[k- (N/2))- .-;•dfN X 2 [.1:- (N/2)],

:[2! + l)e-j2dl/(N/2)

366

step 1: Form the sequence g[n] = z[2n] + jz[2n + 1], which has length N /2.
step 2: Compute G[k], the N/2 point DFT of g[n].
step 3: Separate G[k] into the four parts, fork= 1, ... , (N/2)- 1
GoR(kj

=

1
2(GR(k]- GR((N/2)- k])

GsR(kJ

=

2(GR[k] + GR[(N/2)- k])

Gor[k]

=

Gsr[k]

=

1
1

2

(Gr[k]- Gr[(N/2)- k])

~(Gr[k] + G1 [(N/2)- k])

which each have length N /2.
step 4: Form
X,[k]
X,'(k]

=

GsR[k] + jGor[kJ

= e-;2>•/N (Gsr(k]- jGoR[k])

which each have length Nf2.
step 5: Then, form
X[k] = X.[k] + X,'[k],
step 6: Finally, form
X[k] = X"[N- k],
Adding up the computational requirements for each step of the algorithm gives (approximately)
step
step
step
step
step
step

1:
2:
3:
4:
5:
6:

0 real multiplications and N real additions.
0 real multiplications and 0 real additions.

In total, approximately N log2 ~ + 4N real multiplications aad !N log2 ~ + 4N real additions are
required by this technique.
The number of real multiplications aad real additions required if X[k] is computed using one Npoint FFT computation with the imaginary part set to zero is 2N log2 N real multiplications aad
9.32.

(a) The length of the sequence is L + P- 1.
(b) In evaluating y[n] using the convolution sum, each nonzero value of h[n] is multiplied once with
every nonzero value of z[n]. This can be seen graphica.lly using the flip and slide view of convolution.
The total number of real multiplies is therefore LP.
(c) To compute y[n] = h(n] • z(n] using the DFT, we use the procedure described below.
step 1: Compute N point DFTs of z[n] and h[n].
step 2: Multiply them together to get Y[k] = H[k]X[k].
step 3: Compute the inverse DFT to get 11[n].
Since y[n] has length L + P - 1, N must be greater thaa or equal to L + P - 1 so the circular
convolution implied by step 2 is equivalent to linear convolution.

367

(d) For these signals, N is large enough so that circular convolution of z[n] and h[n] and the linear convolution of :z;[n] and h[n] produce the same result. Counting the number of complex multiplications
for the procedure in part (b) we get
DFT of z[n]

(N/2)log2 N

DFT ol h[n]

(N/2)1og2 N

Y[k] = X[k]H[k]
Inverse DFT of Y[k]

N
(N/2) log., N

(3N/2)log.,N +N
Since there a.re 4 real multiplications for every complex multiplication we see that the procedure
takes 6Niog., N + 4N real multiplications. Using the answer from part (a.), we see that the direct
method requires (N /2)(N/2) N' /4 real multiplications.
The following table sbows that the sma.llest N
2• for which the FFT method requires fewer
multiplications than the direct method is 256.

=

9.33.

=

N

Direct Method

FFI' method

2

1

20

4

4

64

8

16

176

16

64

448

32

256

1088

64

1024

2560

128

4096

5888

256

16384

13312

(a) For each L point section, P- 1 samples a.re discarded, leaving L - P
complex multiplications a.re:

L point FFI' of input:

(L/2) log2 L

Multiplication of 61ter and section DFI':

L

samples. The

= v2• /2

= 2•

L point inverse FFT:

(L/2) log2 L

Total per section:

2•(v + 1)

Therefore,
Complex Multiplications
Output Sample

+ 1 output

2• ( v

= v2• /2

+ 1)

= 2• - P + 1

Note we assume here that H[k] ba.s been precalculated.

(b) The figure below plots the number of complex multipllcations per sample \'etS1I.i v. For v = 12,
the number of multiplies per sample reaches a. minimum of 14.8. In comparison, direct evaluation
of the convolution sum would require 500 complex multiplications per output sample.

368
.5125

f

"'20
'S
CL

~

15

8.

!! 10

S!

Is
'5
~

oL-~--~--L-~--~--L-~--~~~~

10

11

12

13

14

15
v

16

17

18

19

20

Although " : 9 is the first valid choice for overlap-save method, it is not plotted since the value is
so large (in the hundreds) it would obscure the graph.
(c)

lim 2"(v+1)
11-+CIO

2" - P

:

+1

lim

v+1

11-+IXI }

:

+ ;:)

"

Thus, for P : 500 the direct method will be more efficient for "
(d) We want

> 500.

.;,2:,.:"(,;_";.+..:.1~) < P.
2"-P+1-

Plugging in P: L/2: 2•- 1 gives
2"(v + 1)

<

2• -2•- 1 + 1 -

._ _
2 1

As seen in the table below, the FFI' will reqnire fewer complex multiplications than the direct
method when " : 5 or P : 2' : 16.

I Overlap/Save I Direct I
I 2"-1 I
I" I
,..~ .. +1~
2"' 2"'-1 1

1

2

1

2

4

2

4

3

6.4

4

8.9

8

5

11.3

16

9.34. This problem asks that" we find eight equally spaced inverse OFT coefticients nsing the chirp tranSform
. algorithm. The book derives the algorithm for the forward DFI'. However, with some minor tweaking,
it is easy to formulate an inverse DFI'. First, we start with the in?erSe DFI' relation
N-1

:r[n] = .!_

L

X(kjel'ralo/N

NbO

:r[nt)

:

N-1

.!. L

N>=O

X{kjd2wn,./N

369
Next, we define

where l

.O.n

=

1

nt

=

no + l.O.n

= 0, ... , 7. Substituting this into the equation above gives
z[nt]

=

Defining
we find
N-1

z[nt]

=~ L

X(k]ei'•no>!Nw-tt

4=0

Using the relation

lk

=

~[l' +k2 -

(k

-tl'l

we get
N-1

z[nt] =

2.
"X[k)ei2 •no. >tNw-t't'w-"'''w<•-tJ't>
NL.
•=<>

Let
Then,

z(nt] =

~w-t't>

CE

G(k]w<•-tJ'I')

From this equation, it is clear that the inverse DFI' ca.a be comptned using the chirp transform algorithm.
All we need to do is replace n by k, change the sign oC each oftl>e expoltelltial terms, and divide by a
factor of N. Therefore,

m1 (k)
m,[k)
h(k)
Using this system with no
desired sample:;. wilere

= ei2·d:no/Nw-~: /2
= "'-1:2/2
= N2.w•'t•

2

= 1020, and l = 0, ... , 7 will result in .a sequence y(n) which will contain the
y[O]
y[1]
y[2]

=
=
=
=
=
=

z[1020)
z[1021]
z{1022]
z(1023)

y(3)
y[4)
y(5]
y(6)

:t(l)
= :[2)

y(7)

== z[3)

z[O)

370
9.35. First note that

:r;[n]

= { :r[n],

iL:;; n :S iL + 127,

0,

otherwise

= { :r[n + iL],
0,

0 :S n :S 127,
otherwise

Using the above we can implement the system with the following block diagram.

x[n] ~

Shift
by -iL

Multiply

w[n ] = u[n]-u[n-128)

I

FFT-1

. J 256-pt

r-Multiply

h[n)

~

FFT-1
256-pt

---+

IFFf-2

~ 256-pt --+

y;[n]

The FFT size was chosen as the next power of 2 higher than the length of the linear convolution. This
insures the circular convolution implied by multiplying DFTs corresponds to linear convolution as well.

Neonv

=
=

=

N., + Nh - 1
128+ 64- I

191

NpPT = 256
9.36.

(a) The flow graph of a decimation-in-frequency radix-2 FFT algorithm for N

= 16 is shown below.

371
x(OJ

X(OJ

w;,

x(1J

x(2J

X(8J

X(4J

w;.

x(3J

x(4J

X(12]

X(2]

w;,

x(S]

X(10]

X(6]

x(6J

w;,

x[7J

x(8J

X(14]

X{1J

w;,

x(9J

X{9J

X{ 51

xf10J

w;,

x[11]

x[12]

X(13]

X{3J

w;.

x(13J

X(11]

X[7J

x(14]

w;,

x(15]

(b) The prw>ed Bow graph is shown below.

X[15)

372
X(O]

x!O]

w:.

>11]

X[B]

X[4]

w:.

X[12]

X[2]

w:.

X(10]

X[6]

w:.

X[~4]

X[1]

w:.

X[9]

X!5]

w:.
w:.

X[13]

X(3]

w:.

X[11]

X[7]

w:.

X[15]

(c) The pruned butterflies can be used in ( 11 - I') stages. For simplicity, assume that N /2 complex
multiplies are required in each unpruned stage. Counting all W~ terms gives
Number of multiplications

=

(Unpruned multiplications)+ (Pruned multiplications)
N .,_,

= 1'"2+ 2;2"'
1=1

373

9.37.

(a) Starting with the equation

f[n]

= :r[2n+ 1]- :r[2n -1] +Q, n = 0,1, ... , 2N

-1,

where
2

Q=N

f-1

L

:r[2n+ 1],

n=O

we note that :r[2n + 1]

= h[n], and :r[2n- 1] = h[n- 1] for n = 0, 1, ... , lf - 1.
f[n]

= h[n]- h[n- 1] + Q,

= 0, 1, · · ·, 2N -

n

Taking the N (2 point DFT of both sides gives

f-1
F[k] = H[k]- wt1,H[k] + Q

= H[k](1- Wf/) +

L

Wtj,

~ Qo[k]

So
F[O]

=

NQ
2

N2

= 2N
f-1

1

~
.t.... :r[2n + 1]
n=O

= L x[2n+ 1]
n=O
= H[O]
Therefore,
X[k]
X[O]
X[N/2]

=
=
=
=
=

G[k] + WtH[k]

= G[O] + F[O]

G[O] + H[O]

w;12 H[N/2]
G[O] + w; H[O]
G[N/2] +

1'

G(O]- H[O]

= G[O] -

F[O]

(b) The equation,

F[k] = H[k](1for k

Wf/) + ~ Q6(k]

# 0 becomes
Ffk}

= Hfk){l -

wJ,•)

= H[kJWt(wN•- Wt)

1.

We then get

374
So

Therefore,
X{k]

=

G(k] + WtH{k]

=

G(k]+

=

G(k]-

F(k]

w-•
N - w•
N
F{k]

j

2sin(21rk/N)

Clearly, we need to compute X[O] and X(N/2] with a separate formula since the sin(27rk/N) = 0
fork= 0 and k = Nf2.
(c) For each stage of the FFT, the equations
X(O]
X[N/2]

=
=

G(OJ + F(OJ
G[OJ - F(O]

require 2 real additions each, since the values G[O] and F[O] may be complex. We therefore require
a total of 4 real additions to implement these two equations per stage.
For a single stage, the equation
X[k] = G(k]-

!;

F[.l:]
2 sm(27rk/N)

k

"#- 0, N/2

requires (N - 2) /2 multiplications of the purely imaginary "twiddle factor" terms by the complex
coefficents of F[k] fork"#- 0, N/2. The number of multiplications were halved using the symmetry
sin(2,.(k + N/2)/N) = - sin(27rk/N) and the fact that F[k] is periodic with period Nf2. Since
multiplying a complex number by a purely imaginary number takes 2 real multiplies, we see that
the equation requires a total of (N- 2) real multiplies per stage.
We also need (N- 2) complex additions to add the G(k] and modified F[k] terms for k f. 0, N /2.
Since a complex addition requires two real additions, we see that the equation takes a total of
2(N - 2) real additions per stage.
Putting this all together with the fact that there are log2 N stages gives us the totals
Real Multiplications =

(N - 2) log2 N
2N log2 N .

Note that this is approximately half the computation of that of the standard FFT.
(d) The division by sin(27rk/N) fork near 0 and N/2 can cause X[k] to get quite large at these values
of k. Imagine a signal zl[n], and signal z2[n] formed from z,[n] by adding a small amount of white
noise. Using this FFT algorithm, the two FFTs X 1 [k] and X 2 [k] can vary greatly at such values
of k.
9.38.

(a)
N-1

X(2k] =

=

L z{n]W~
.....

(N/2)-1

L

(z!nJW~ +z(n + (N/2)]w~•n + z(n + (Nf4)]w;+w:,•<•+w;,•<•+w:,•<•+<•NI•»)
(N/4)-1

L

=

{(z[n]- z(n + (N/2)])- j(z[n + (N/4)]- z[n + (3Nf4)])}w;w~•·

n=O

In the derivation above, we used the fact that WJ:I< = -j, W;:'14 = j, Wf:/2 = -1, and W},N = 1.
Since wt,•• = Wf<'J4 , X[4k + 1] has been expressed as a N/4 point DFT. But we need to multiply
the sequence (z(n]- z[n + (N/2)])- j(z[n + (N/4)]- z[n + (3Nf4)]) by the twiddle factor W;,
0::; n ::; (N/4) -I before we compute the N /4 point DFT.
The other odd-indexed terms can be shown in the sa.me way to be
(N/4)-1

X[4k

+ 3] =

L

{(z[n]- z[n + (N/2)])

n=O

+ j(z[n + (N/4)]- z[n + (3N/4)])}W~·w:,••,

k = 0, 1, ... ,(N/4) -1.

Parts (a) and (b) show that we can replace the computation of anN point DFT with the computation of one N /2 point DFT, two N /4 point DFTs, and some extra complex arithmetic.
(c) Assume N = 16 and define
g(n]
f,(n]
/,[n]

=
=
=

z[n] + z[n + (N /2)], n = 0, 1, ... , (N /2)- 1
z[n]- z[n + (N /2)], n = 0, 1, ... , (N/4)- I
z[n+(N/4)]-z(n+(3N/4)], n=O,!, ... ,(N/4)-1

A diagram for computing the values of X[k] looks like:

)

376
g(O)
>!OJ

X[O]

x(1 J

X[S]

x(2)

X[4]

>13)

X[12)
8-point OFT

>141

X[2]

>IS]

X[10]

>161

X[6]

x(7J

X[14)

~.

x(B]

X[1)

,

w,.

x(9]

X[9]

x(10)

w..

X[S]

>111)

w:.

X[13)

~.

X[3)

w:.

X[11)

>113]

w,.

>114)

4;>oi'11 OFT

w,.

>115]
-j

If we carry on applying the split-radix principle, we get the next diagram:

X[7]

X[15]

377
X[OJ

~.

X[8)

lC[4)

~.

X[12)

X[2)

~.

X[10)

lC[6)

~.

X[14)

-1
lC[1)

~.

X[9)

~.

X[5)

~.

X[13)

X[3)

~.

X(11)

~.

lC[7J

~.

X[15)

(d) The fiow diagram for the regular radix-2 dec:im&tion-in-frequency algorithm is shown in the next
16. Not conming trivial multiplications by W~, we find that there are 17 complex
figure for N
multiplications total. Of these 17 complex multiplications, 7 are multiplications by
= -j.
Since a multiplication by - j can be done with zero real multiplications, and a complex multiplication requires 4 real multiplications, we find that the total number of real multiplicationS for the
decimation-in-frequency algorithm to be (10)(4) 40.

=

Wt,

=

Taking a look at the split-radix algorithm, we find again that there are 17 complex multiplications.

378

In this case, however, 9 of these are by wt.
multiplications to implement this ftow graph.

= -j.

Thus, it takes a total of (8)(4)

= 32 real
lqO)

w:.

lCI8J
X[ol)

w:.

lq12)

lCI2J

w:.

lq10)

X{6)

w:.

lq14)

lq1)

w:.

lq6)

X[5J

w:.

X(13)

X(3)

w:.

X(11)

X(7[

w:.
9.39.

(a) Noting that
arctan(:r)

'"'' < 1

X(15)

379

we notice that, to first order, the 9; and 9;+1 differ by a factor of 2. ·(Note tbat these formulae are
in radians). This approximate factor of 2 for sucessive 9; is confirmed by looking at some values
of 9;: 9o = 45°, 81 = 26.6°, 92 = 14.0°, 93 = 7.1°. So we have a set of angles whose values are
decreasing by about a factor of 2.
You can add and subtract these 8; angles to form any angle 0 < 8 < "/2. The error is bound by
= arctan(2-M), the angle tbat would be included next in the sum. H the error were greater
than 9M, then one of the a; terms must bave been incorrect. The inclusion of the Mth term must
bring the sum closer to 9.

9M

(b) An algorithm to compute 9, is described below.
Qo

= +1

-B = ao9o
fori=1toM-1
if (8> 9)
Oi

= -1

Qj

= +1

else

end for
Using this algorithm, the sequence a 1 is found to be

IQ: I~ 1-~ I~ I~ 1-~ 1-~ I~ I; 1-~ 1-~ I~~ I
(c) Note that (X+ jY)(1 + ja;2-') =(X- a,Y2-') + j(Y + a;X2-'). Hence, the recursion is simply
multiplying by M complex numbers of the form (1 + ja;2-'). These can be represented in polar
form:
(1

+ ja,2-')

= ,,IJ + 2-2<.,Ja . ..-....(2-'l
=

G;.tJail;

Multiplication of polar numbers produces a sum of the phases, a;9,.
M-1

B=

L

a,9,

i=O

(d) Multiplication of polar numbers produces a product of the magnitudes, G;.
M-1

GM

= II
i=O

9.40. (a) Starting with the definition of the DFT,
N-1

X[k] =

L

n=O

:z:(n]w;;•

v'1 +2-2i

380
N-1

L z[nJW1,..

=

X[3k]

N/3-1

2N/3-l

L

=

N-1

z[nJW1nk +

a=N/3

ft.=O

Substituting m

L

z[nJW1,.. +

L

z[nJW1nk

,_2N/3

= n- N /3 into the second summation, a.nd m = n- 2N/3 into the third summation

gives
N/3-1
X[3k]

L

=

N/3-1

L

z[nJW1,.. +

N/3-1
z[m +

Nf3JW1"'•wg• +

m=O

n.=O

L

z[m + 2Nf3JW1"'•w~•

m=O

N/3-1

L

=

(z[n]+z[n+N/3]+z[n+2N/3])W1nk

"""'

N/3-1

L

=

(z[n] + z[n + N/3] +z{n + 2N/3JlWM3

n=O

Define the sequence

zi[n]

=z[n] + :r[n + N/3] + :r[n + 2N/3]

The 3-point DFT of zi[n] is Xi[k] = X[3k].
(b) This part is similar to part (a). First, :r2 [n] is found. Starting again with the definition of the
DFT,
N-1

L

X[k] =

z[n]w~•

n={l

X[3k

+ 1]

N-1

L

=

.....

z[nJw;<3>+1l

N/3-1

L

=

z[n]w;< 3>+ 1l +

L

N-1

L

z[n]w;< 3>+1l +

n.=N /3

ft=O

Substituting m

2N/3-1

z[n]w;< 3>+1l

n.=2N/3

= n-N/3 into the second summation, a.nd m = n~2Nf3 into the third summation

gives

N/3-1
X[3k + 1]

=

L

z[nJw;<3t+l) +

N/3-1

L

z[m + N/3JW1m+N/SH 3H 1)

m=O

n=()

N/3-1

L

+

m=O

z[m + 2N/3JwJ.."'+2N/3)(3HI)
.

N/3-1

=

L

N/3-1
z[n]w;wg•w~' 3

m=O

N/3-1

+

=

L

z[m + 2N/3Jw;+•>

n=()

N/3-1

= L

2N/3-1

z[n]w,:;t•>+•>

+

n=O

Substituting m
gives

L

N-1

z[n]w,:;<>t+2l

+

n=N/3

L

z[n]w,:;t>t+•>

n=2N/3

= n- N/3 into the second summation, and m = n- 2N/3 into the third summation
N/3-1

X[3k + 2]

=

L

N/3-1

:r[n]w,:;t•>+•> +

L

z[m + Nf3JW~m+N/3)(3'+2)

m=-if-

n=O

N/3-1

L

+

=

:r(m + 2 N/ 3 ]W~m+2N/3)(3t+2)

N/3-1

L:

N/3-1

:r[n]w,:;< .. +•> +

n=O

L:

:r[m + Nf3]w;< .. +•>wff•w~Nt•

m=O

N/3-1

+

L:

:r[m + 2N/3Jw;<.. +•>w,V•w'~ 13

m=O

=

N/3-1

L

(z[n] + :r[n + Nj3]W~N/l + z[n + 2N/3]w'~' 3 )W,:;<3t+'l

n=O

N/3-l

=

L

13

(:r[n] + z[n + Nf3JW;:'

+ z[n + 2N/3]w'~ 13 )Wl-"WM 3

n=O

Define the sequence
z 3[n] = (:r[n] + z[n + Nf3]W~N/l + z[n + 2N/3]w'~ 13 )Wl-"

=

Tbe 3-point OFT of :r3[n] is X3[k] X(3k + 2].
(c) To draw tbe radix-3 butterfly, it helps to derive the output of the buttertly first. From the definition
of the OFT,
X[k]

=

L• z[n]W3"
n=()

=

X[O]
z[O] + z[1] + z[2]
X[1] = z[OJ + z[1]WJ + z[2]Wf
X[2]
z[O] + z[1]Wf + z[2JW:

=

The butterfly for the 3 point OFT is drawn below.

= z[O] + z[1]Wf + z[2JWJ

382

x, _,IPJ

(d) Using the results from parts (a) and (b), the flow graph is drawn below.

383

x1(0}

X(OJ

x1(1}

X(3J

x,(2)

vof.

xJOI

w'

•,!11

•

vi,

x,f2J

vof.

xjOJ

vi,

xpl

w:

xp!

(e) The system consisting entirely of N = 3 DFTs is drawn below .

•

X(6]

X(1]

N•3DFT

X(4]

X[7J

X(2]

N=30FT

XIS].

X(8]

384

x1(0(

lC(O)

~

N=3

><(1)

X) OJ

N=3

OFT

OFT

~

><(2)

X)6)

x1[1)

><(3)

w'

•

N=3

><(4)

OFT

X)1)

x,f1l

Na3

OFT

w:

><(5)

X)4)

X[7)

X[2]

><(6)

x[7J

><(8)

X)3)

N•3

N=3

OFT

OFT

----------ol

X[S]

X[8]

(f) A direct implementation of the 9 point DFI' equation requires 9" = 81 complex multiplications.
The system in part (e), in constrast, requires 4 complex multiplications for each 3 point DFI', and
an additional 4 from the twiddle factors, if we do not count the trival ~ multiplications. In total,
the system in part (e) requires 28 complex multiplications. In general, a r&dix-3 FFI' of a sequence
of length N 3" requires appro:r;imaUly

=

Number of complex multiplications

=

(4

complex multiplications)
3-pt DFI'

(N 3-pt DFI's) (v stages)+
3

stage

N twiddle factors
)
. h .ddl facto )
rs
( stage with twiddle factors (v - 1 stages WJt IW1 e

385

Replacing v with log, N, and simplifying this formula gives
Number of complex multiplications=

~Nlog, N- N

i,

Note that this formula for a radix-3 FFT is of the form N log, N. The constant multiplier, ·is
significantly larger than that of a radix-2 FFT. This is because a radix-2 butterfly bas no complex
multiplications, while in part (c) we found that a radix-3 butterfly has 4 complex multiplications.
Also note that this formula is an upper bound, since some of the N twiddle factors in the v - 1
stages will be trivial. However, the formula is a good estimate.
9.41.

(a)
N-1

X(k] = h"(k] ~)r[n]h"(n])h(k- n]
N-1

=

2

e-id fN

L

.....

2
2
z{n]e-jwn2/N ei•(t -2.Cn+n )/N

N-1

L x[n]e-;•.,.fN

=

"'""
(b)
N-1

X[k

+ NJ = h"[k + NJ L x[n]h"[n]h[k + N- n]
=0

h"(k + Nj =

=

2
.-jr(t+N) /N
e-jw(1: 2 +2J:N+N 2 }jN

= e-id 2 /N e-ittN
= h"[k]e-,•N
So
X[k

+ NJ

N-1

=

h"[k]e-i•N

=

X(k]

L

x[n]h"[n]h[k- n]ei•N

(c) From the figure

x{k)

---i(X}----!1

~]

xJkl
y{k]
1\•

h [k]

we define the signals z1[k] and x2[k) to be

x,(k] = x[k]h"[k]

386
00

= 2:

"'•lkJ

,.,[t]i.[k- ll

l=-oo
N-1

L

=

....

:z:[t]e-i•t'IN .,;•<•-tl' IN

k E [0, ... , 2N- 1]

Therefore,

y[k]

=

h"[k]:z:,[k]

= .-;d'IN

ke[N, ... ,2N-1]

N-1

L

:z:[t]e-;•t'IN.,i•<•-tl'IN

kE [N, ... ,2N- 1]

l..O
N-1

=

L :z:[t]e-;•dtiN k [N, ... , 2N- 1]
'""'
X[k+NJ
kE[O, ... ,N-1J

=

X[k]

=

E

{d)
2N-1

H(z)

= L

.=0

=

.,id'INz-•

M-12M-1

L L

r=O

l=O

J4-12M-1

==

L L
r=O

=

2
ei7t(r+lM) fNz-(r+lM)

ei'"2 jN ei2•rl/M ei•l2 z-r z-LM

l::=O

1
3:: .,;rr'IN .-•
=...

2 cos"\

e2
lm{X[k]}

sin (J)k
z-1

e1

e3

-1
(b)

Re{X(k]} = Re{y.(N]}
Im{X(k]} = lm{yo[N]}
Since the output of interest is the Nth sample, we need only consider the .ariance at time N.
The noise e,[n] is input to both hR(n] and hr(n]. Using the techniques from chapter 6, we find the
variance of the noise is
N

cr~[NJ =

~(n]
n=O

389
N

= .,;, + u!, L

crj[N]

h~(n].

n=O

Let 8=2dfN.
N
2

L:cos 8n
n=O

N

~ L(eifn + e-ifn)2

=

n=O
N

= !. L(ei12n + 2 + .-if2n)
4n=O
1 ( 1 _ ei21(N+I)

= 4 1- ei28 + 2(N + 1) +
= ~(1 + 2(N + 1) + 1)

1 _ e-i21(N+l))
1- e-;28

= N +1
2

Similarly, L~=<> h1{n]
Therefore,

= N f2.
uh(n]

2-28

=

!2(1 + (N/2) + 1)

=

2-28
l2(N +4)/2
8

2
= ~;

u~[n]

=

(1 + (N/2))

2-28
l2(N +2)/2

9.43.
N-1

X(k]

=L

N-1

:r[n]cos(2.. kn/N)- j

L

:r(n]sin{2.-kn/N).

n-o

n=O

For k # 0, there are N - 1 multiplies in the computation of the real part and the imaginary part:

[k)} + Re{Y3 [N- k)})
1
x,[A:J = 2[lm{Ys[A:]}
+ Im{Y.[N- A:)})
x.[A:J = ~[lm{Ys[A:]}- Im{Y3[N- A:]}]/ sin(2.. k/N),

k

x,[A:J = -~[R.e{Y3[k]}- Re{Y3[N- k)})/sin(2.-k/N),

# O,N/2
k

# O,N/2

9.48. First, we find an expression for samples of the system function H(z ).

H(z)

=

~M

b -r
-'-~ ,z
1 - I:t=1 a,z-l

~M b ,-;2•>T·~---=
1- L...t=l
~" a,e-;2d.l/N

Now assume N, M

~

511. Let b[n)

= bn and
a[n)

={

1
'
a,.,

Let B[k], A[k] be the 512 pt DFTs of b[n), and a[n]. Then

H(d'd./512) = B[k]

A[k]

9.49.

(a) It is interesting to note that (linear) convolution and polynomial multiplication are the same
operation. Many m&thematical software tools, like Matlab, perform polynomial multiplication
using convolution. Here, we replace
L-1

p(z) =

L

Jl-1

a;z',

=L

q(z)

;.:()

with

L-1

p[n] =

L

b;z'

;.:()

Jl-1

a;c5[n - i],

q[n]

=

;.:()

Then,

r[n)

L
..., b;6[n -

i)

= p[n] • q[n).

The coefficients in r[n) will be identically equal to those of r(z). We caD compute r[n] with circular
convolution, inste.ad of linear convolution, by zero padding p[n) and q[n] to a length N = L+ M -1.
This zero padding ensures that linear convolution and circular convolntion will give the same result.
(b) We can implement the circular convolution of p[n] and q[n] using the following procedure.
step 1: Take the DFTs of p[n] and q[n] using the FFT program. This gives P{A:] and Q[k).
step 2: Multiply to get R[A:J = P[A:JQ[A:J.
step 3: Take the inVerse DFT of R(A:] using the FFT program. This gives r[n].

395
Here, we assumed that the FFT program also computes inverse DFTs. If not, it is a relatively
simple matter to modify the input to the program so that its output is an inverse DFT. (See
problem 9.1 ).
While it may seem that this procedure is more work, for long sequences, it is actually more efficient.
The direct computation of r(n] requires approximately (L + M) 2 real multiplications, since a; and
b, are real. Assorning that a length L + M FFT computation takes [(L + M)/2] log2 (L + M)
complex multiplications, we count the complex multiplications required iD the procedure described
above to be
Operation
FFTs of p(n] and q(n]

=

Complex Multiplications
2[(L + M)/2] log2 (L + M)

R(k] P[k]Q[k]
Inverse FFT of R(k]

=(L + M) log,(L + M)

L+M
[(L

+ M)/2]1og2 (L + M)

= (3(L + M)/2] log,(L + M) + (L + M)
Since a complex multiplication is computed using 4 real multiplications, the number of real multiplications required by this technique is 6(L+ M) log2 (L+ M) +4(L+ M). Plugging in some values
for (L + M) = 2", we find
L+M Direct FFT
2

4

20

4

16

8

64
256
1024

64
176

16
32
64

4096

«8
1088
2560

Thus, for (L + M) 2: 64, the FFT approach is more efficient.
(c) The binary integers u and v have corresponding decimal values, which are
L-1

"decimal

=

vdecimal

=

L ,..2•

M-1

L

Vi2i

i=O

Note the resemblance to p(x) and q(x) of part (a). We form the signals
L-1

u(n]

=

v[n]

=

-....

L u;o[n- i]
J t-1

L v;o[n - i]

and use the procedure described in part (b). This computes the product u · v in binary. For
L = 8000 and M = 1000, this procedure requires approximately

#

real multiplications

= 6(8000 + 1000) log,(8000 + 1000) + 4(8000 + 1000)

=

7.45

X

lOS

In contrast, the direct computation requries 8.1 x 107 real multiplications.

396

(d) For the (forward and inverse) FFI's, the mean-square value of the output noise is (L + M)o-~.
While ~ will be sma.ll, as there are 16 bits, the noise can be significant, since L + M is a large
number.
9.50.

(a) Using the definition of the discrete Hartley transform we get

=

HN(a + N)

GN[a + N) + SN[a + N)

=

cos(2.-a/N+2,-)+sin(2.-a/N+2.-)
= cos(21ta/N) +sin(2ra/N)
= CN(a] + SN[a]
= HN[a]
HN[a +b)

=
=

CN(a + b] + SN[a+b]

=

cos(21tafN) cos(27rb/N) - sin(2,-a/N) sin(2lrb/N)

cos(2,-afN + 2lrb/N) + sin(2..ajN + 21rbjN)

+ sin(21ta/N) cos(2.-b/N) + cos(2ra/N) sin(2,-b/N)
Grouping the terms in the last equation one way gives us
HN[a + b]

=

[cos(2lra/N) +sin(2,-a/N)]cos(2,-bfN)
+[cos( -21ta/N) +sin( -27ra/N)] sin(2,-b/N)

=

HN[a)GN(bJ

+ HN(-a)SN(b]

while grouping the terms another way gives us

= [cos(21tb/N)+sin(2,-b/N))cos(2.-a/N)

HN[a+bJ

+[cos( -27rb/N) +sin( -21rb/N)]sin(2,-a/N)

=

HN(b}GN(aJ

+ HN[-b}SN(a)

(b) To obtain a fast algorithm for computation of the discrete Hartley transform, we can proceed as
in the decimation-in-time FFI' algorithm; i.e.,
(N/2)-1

XH(k]

= L

(N/2)-1
:t[2r]HN[2rk] +

L

:t[2r + l)HN((2r + l)k)

r=O

r=O

(N/2)-1

= L

(N/2)-1
:t[2r)HN[2rkj +

r=O

:L

:t(2r + l}HN[2rk]CN[k}

r=O

(N/2)-1

L

+

:t(2r + l)HN[((-2rk))N)SN(kj

r=O

Now since HN[2rk]

= HN;2 [rk], we have
(N/2)-1

XH(k]

= L

(N/2)-1
:t[2r}HN;2(rk] +

r=O.

L

z{2r + l]HN;z[rk)CN(k]

r=O

(N/2)-1
+

=

L

z{2r + l}HN;z[((-rk))N;2)SN(k)

F(k] + G[k]CN(k] + G[(( -k))N;2)SN[k)

397
where
(N/2)-1

F[A:J

= L

z[2r]HNt 2 [rkJ

r=O

is the N /2-pomt DHT of the e\'ell-mdexed pomts and
(N/2)-1

G[A:]

= .....,
L

z[2r

+ 1]HN/2[rk]

is the N /2-pomt DHT of the odd-mdexed pomts. As m the derivation of the decimation-in-time
FFT algorithm, we can continue to divide the sequences m half if N is a power of 2. Thus the
mdex'mg will be exactly the same except that we have to access G[(( -A:))N/2] as well as G[k] and
F[k]; i.e., the "butterlly" is slightly more complicated. The fast Hartley transform will require
N log2 N operations as in the case of the DFT, but the multiplies and adds will be real instead of

complex.
9.51. (a)

1

0

0

0

0

0

0

1 -1
0 0

0
I

0
1

0
0

0
0

0
0

0
0

0

F•

F3

=

=

0

1

0
I

0
0

0
0

0
0
0
0

1 -1

0

0

0

I

0 0

0

I

-1

0

0

0

0

0

0

0
0

0 0
0 0 0
1 0 I
0 I 0
1 0 -1

0

0

1

0

0

0 0
1 0

0 0 1 -1
0 0 0
0

0 0

0

0

0 0

0

141

0

0 0

0

0

Wi

0

0
0
0
0
0 0

0

0 0
1 0

0
0

1 0

I

0 0

0

0

0

0

0 0

0

0

0 0

1

0 0

0

-1 0 0 0
0 0 0
0 I 0 I
0 0 0 0 0 I 0
0 0 0
0 1 0 -1
0000010
1000100
01000
I
0
00100
01
0001000
I 0 0 0 -1 0
0
0 1 0 0 0 -1 0
0 0 1 0 0
0 -1
0 0 0 1 0
0
0
0 1

0

0 0
0 0

0

0

1 0
0 1

1

0
0

T•

=

0 0 0

0 0 1
0 0 w~
0
0 0 0 0
0
0
0
0
0
0
0
0
0
0 0
0
0
1 0
0
0 141 0
0 0
WJ 0

1

0 0

0

0
-1

0 0

0 0

0 0

0 0

0
0
0

1
0

0
0
-1

(b)

QH = Ff/TfF:TfF:
= F 1TiF•T2F3

0
0

0
0

0
0
0
0
0
0
0

Wf
0
0
0
0
0

0

Wf

0
0

0

0

Wi

398
where • denotes conjugation. Drawing the tlow graph,

we

get

xfO]

)iOJ

xf1]

-1

)i1]

)i2]

xf2]

xf3]

)i4]

xf4]

xf5]

y(6]

xf6]

x(7J

This structure is the decimation in frequency FIT with the twiddle factors conjugated and therefore
calculates
N · IDIT{z[n]}

1tQ

8 calculates the IDFT, we should realize that cas(c) Knowing that Q calculates the DIT and
cading the two should just return the original signal. More formally we have

F(lF, = 21 Fl1F2 = 2I FfF, = 21
T{'T, =I TfT2 =I
(1/N)Q 8 Q

=

(1/N) (F{IT{'FfTfFf) (F,T2F2T,F,)

= (1/N)(NI)

=
where N
9.52.

I

= 8 in this case.

(a) First, we derive the circular convolution property of the DFT. We start with the circular convolution of z[n] and h[n].
N-1

y[n] =

L z[m]h[((n- m))N]

m=O

Taking the DIT of both sides gives
N-l /11-l

Y[k]

= L L

z[m)h(((n- m))N)W"'

...0 m=O
N-1
N-1

= L :z:[m) L h{((n- m))N)W"'
m=O

.....

399

Using the circular shift property of the DFT,

N-1
Y[k] =

L

:r[m]H[k]WN'•

m=O

=

N-1
H[k]

L :r[m]WN'•

= H[k]X[k]
Next, the orthogonality of the basis vectors is shown to he a necessary requirement for the circular
convolution property. We start again with the circular convolution of :r[n] and h[n]

N-1
y[n] =

L :r[m]h[((n- m))N]

Substituting the inverse DFT for :r[m] and h[((n- m))N] gives

y[n]

=
=

Y: (~ Y:
Y: (~ Y:

m=O

1: 1=0

m=O

.k1=0

X[k1JW..;•·m)

(~ ~ H[k ]w;,;••«n-m))N)
2

I:2=0

X[k1Jw..;••m)

(~ ~ H[k ]w;,;••<•-m))
2

.t 2:0

= ~2 ~ ~ (x[k1]H[k2]W_N••• ~ w..;••mw~•m)
t,=O &2=0

With orthogonal basis vectors,

Y:

m=O

WN'

= { N,

n=O

k =0
;< 0

0,

k

so the right-most summation becomes

N-1

L

WN.tlmwt2m

m=O

N-1

2:

=

w;:2-i:J)m

m=O

= No[k1-

k,J

Therefore,

y[n] =

N-1 N-1

~

L L (X[k1]H[k Jw;••• N o[k1 - k
2

2 ])

il=Ol:2=0

=

~

N-1

L X[k,]H[k1]w_N•··

•·=<>

N-1

= L

:r[m]h[((n- m))N]

m=O

Therefore, the circular convolution property holds as long as the basis vectors are orthogonal.
(b)

400
k
k
k
k

=

=
=

0:
. I:

=

=
=
=

((I+ I+ I+ I))17
((I+ 4 + I6 + 64))17

2:

((I+ I6 + 256 + 4096)),7

The relation holds for P

((85))17
((4369))17

=

3: ((I + 64 + 4096 + 262I44)),7

=I7, N =4, and WJV =4.

=
=
=

((4)),7

((266305))17

=

(c)

(~ z[n]4n>))

= (

X[k]

17

=

=

((I· I+ 2 · 4• + 3 · I6• + 0 · 64•))17
((I +2·4• +3·16•))17

Using this formula for X(k),
X[O)
X[1]
X[2)
X[3)

=
=
=
=

=
=
=
=

((I+2+3)h7
((1 + 2. 4 + 3. I6))17
((I+ 2 · I6 + 3 · 256))17
((1 + 2. 64 + 3. 4096)),7

H[kJ

((6)h7
((57))17
((801))17
((124I7)),7

=
=
=
=

6
6
2
7

(t

= ( h[nW•))
17
= ((3·I + 1·4· +0·I6. +0·64·))17
= ((3+4.)h7

Using this formula for H[k),
H[O]
H[I]
H[2)
H[3]
Multiplying terms Y[k]

=
=
=
=

((3+I))17

=

((4))17

((3 + 4))17

=

((7))17

((3 + I6))17

=
=

((I9)h7
((67)),7

((3 + 64))17

=
=
=
=

4
7
2
I6

= X[k]H[k] gives
Y[O]
Y[1]
Y[2]

Y(3]

=
=
=
=

((24)),7

=

7

((42)),7

=

8

((4)h7

=
=

4

((112))17

IO

(d) By trying out different values,

((N- 1 N)h7

N- 1

=

w;;'

= 13

13

=((W;i WJVJl11 =((13 · 4))17 =((52)),7 = 1
1

4
0
0
0

401
(e)

y(n]

= ( ( 13

~ Y(k] 13,..))
17

=

((13(7 ·1 + 8 .13n + 4 · 169n + 10 · 2I9r])),7

Using this formula for y(n],

=

y(O] = ((13(7 + 8 + 4 + 10]))17
y(l]

=
=

((13(7 ·1 + 8. 13 + 4. 169 + 10. 2197])),7

= ((29584I)h7

y(2]
((13(7. 1 + 8 ·169 + 4. 28561 + 10. 4826809])),7
y(3] = ({13(7 . 1 + 8 . 2197 + 4 . 4826809 + 10. 10604499373]) h7
Performing manual convolution y(n]

({377)),7

=
=

=

= 3

9.53. (a) The tables below list the values for n and k obtained with the index maps.

n,

I

0 \I
0 I (• I

12

0

i2

k, 0 0 2 4

1 j3 j 4j5

1

1 2

1 3

5

As shown, the index maps only produce n = 0, ... , 5 and k = 0, ... , 5.
(b) Making the substitut>on we get
X(k]

= X(k 1 + 2k2 ]
= L• z[nJwi••+•••Jn
2

=

I

L L

:r:{3nl

n:z=Ont=O

(c) Expanding out the We terms we get

(d) Grouping the terms we get

The interpretation of this equation is as follows

+ n2]Wi.kt+2l::z)(3na+n:z)

7

((628988009))17

The results agree.

k,

3

({1378836141137)) 17

= z(n] • h(n] gives
y(OJ = 3
y(l] = 7
y(2] = 11
y(3] = 3

n,

=
=

11

402

(i) Let G[k,, n2] be the N

= 2 point DFTs of the inner parenthesis; i.e.,
1

G(k 1, n 2] =

L

z(3n 1 + n 2JW:' "',

n,=O

This calculates 3 DFTs, one for each column of the index map associated with n. Since the
DFT size is 2, we can perform these with simple butterflies and use no multiplications.
(ii) Let G[k,, n2] be the set of 3 column DFTs multiplied by tbe twiddle factors.

0 ~ k, ~ 1,
{ O~n 2 ~2.
(iii) The outer sum calculates two N

= 3 point DFTs, one for each of the two -values of k,.
2

X(k, +2k2]

=L

1
G(k1,n2]W3""',

n:a=D

(e) The signal flow graph looks like

The only complex multiplies are due to the twiddle factors. Therefore, there are 10 complex
multiplies. The direct implementation requires N 2 = 62 = 36 complex multiplies (a little less if
you do not count multiplies by 1 or -1).

403

(f) The alternate index map can be found be reversing the roles of n and It; i.e.,
n

= n 1 + 2n2

k=3k,+k2

for n 1

= 0, 1; n2 = 0,1,2

for .1: 1 = 0, 1;

lt2

= 0, 1, 2

405

Solutions - Chapter 10
Fourier Analysis of Signals
Using the Discrete Fourier Transform

407
10.1. (a) Using the relation
2..k
0 • =NT'
we find that the index k = 150 in X[k] corresponds to a continuons time frequency of

2.. (150)
(1000)(10 4 )

-

=

=

(b) For this part, it is important to realize that the k 800 index corresponds to a negative continuoustime frequency. Since the DFT is periodic in k with period N,
2.. (800 - 1000)
1000(10-4)

=
=
10.2. Using the relation

n. = 2.-k

NT

or

k

f• =NT
we find that the equivalent analog spacing between frequencies is
I

t:.j =NT

Thus, in addition to the constraint that N is a power of 2, there are two conditions which must be met:

j. > 10, 000 Hz (to avoid aliasing)

*

(given)

< 5Hz

These conditions can be expressed in the form
10,000 <

1

T < 5N

The minimal N = 2" that satisfies the relationship is
N

= 2048

for which
10, ooo Hz <
Thus, Fmin = 10,000 Hz, and Fmax
10.3.

TI < 10, 24o Hz

= 10,240 Hz.

(a) The length of a window is
L

= ( 16, ooo sa:les) (20 x 1o-• sec)
= 320 samples

(b) The frame rete is the number of frames of data processed per second, or equivalently, the number of DFT computations done per second. Since the window is advanced 40 samples between
computations of the DFT, the frame rate is
frame rate =

( 16 000 samples) ( 1 frame processed)
'

= 400~
sec

sec

40 samples

408
(c) The most straightforward solution to this problem is to say that since the window length Lis 320,
we need N 2: Lin order to do the DFT. Therefore, a value of N = 512 meets the criteria of N 2: L,
N = 2". However, since the windows overlap, we can find a smaller N.
Since the window advances 40 samples between computations, we really only need 40 114lid samples
for each DFT in order to reconstruct the original input signal. If we time alias the windowed data,
we can use a smaller DFT length than the window length. With N =256, 64 samples will be time
aliased, and remaining 192 samples will be valid. However, with N = 128, all the samples will be
aliased. Therefore, the minimum size of N is 256.
(d) Using the relation
1

A!= NT'
the frequency spacing for N

= 512 is
AJ = 16• OOO
512

and for N

= 256 is
A!

= 31.25 Hz

000
•
= 16256
= 62.5 Hz

10.4. (a) Since :z:[n] is real, X[k] must be conjugate symmetric.

X(kj

= X"[((-k))N)

We can use this conjugate symmetry property to find X[k] for k = 200.

=. X"[k]

X[((-k))N]
X[((-800))10oo]
X[200]

+ j)"

=

(1

=

1- j

(b) Since an N -point DFT is periodic in k with period N, we know that

X[SOO] = 1 +j
implies that
X[-200]

=1 + j

Using the relation

we find

n-200 =
=
nooo =
=

-211"(200)
(1000)(1/20, 000)
2.-(200)
(1000)(1/20, 000)

Consequently,
X<(j n)

ln=-••<4000)

x.u n) ln=••.n
This describes a line with slope >. and intercept Wo· Thus,

>. = t.y = (0.5.-- 0.257r) = 41 34
l!.:t

{19000
Wo

O)

.

X

10.16. Using
1

l!.f= NT
and assuming no aliasing occured when the continuous-time signal was sampled, we find that the frequency spacing between spectral samples is

l!.f

=
=

1

(1024)(1/10, 000)
9.771b

or

10.17. We should choose Method 2.

414
Method 1: This doubles the number of samples we take of the frequency wria.ble, but does not change
the frequency resolution. The size of the main lobe from the window remains the same.
Method 2: Thi3 improves the fre'luency resolution ,;...,. the ,...;, lobe from the window get. mudler.
Method 3: This increases the time resolution (the ability to distinguish events in time), but does not
affect the frequency resolution.
Method 4: This will decrease the frequency resolution since the main lobe from the window increases.
This is a strange thing to do since there are samples of x[n] that do not get used in the transform.
Method 5: This will only improve the resolution if we can ignore any problems due to sidelobe leakage.
For example, changing to a recta.ngu1ar window will improve our ability to resolve two equal
amplitude sinusoids. In most eases, however, "!' need to worry about sidelobe levels. A large
sidelobe might mask the presence of a low amplitude signal. Since we do not know ahead of time
the nature of the signal we are trying to analyze, changing to a rectangular window may actually
make things worse. Thus, in general, changing to a rectangular window will not necessarily increase
the frequency resolution.
10.18. No, the peaks will not have the same height. The peaks in V2(&"') will he larger than those in Vi (e;"').
First, note that the Fourier transform of the rectangular window bas a higher peak than that of the
Hamming window. H this is not obvious, consider Figure 7.21, and recall that the Fourier transform of
an L-point window tu[n], evaluated at DC (w = 0), is
L-l

W(&O)

= L w{n]

"""

Let the rectangular window he WR[n], and the Hamming window he WH(n]. It is clear from the figure
(where M
L+1) that

=

L-l

L-1

"""

.....

L WR{n] > L WH{n]

Therefore,

WR(&O)

> WH(.,.i 0 )

Thus, the Fourier transform of the rectangular window has a higher peak than that of the Hamming
window.
Now recall that the multiplication of two signals in the time domain corresponds to a periodic convolution
in the frequency domain. So in the frequency domain, V1 (&"') is the convolution of two scaled impulses
from the sinusoid, with the Fourier transform of the L-point Hamming window, WH(&"'). This results
in two scaled copies of WH(ei-'), centered at the frequencies of the sinusoid. Similarly, V2 (&"') consists
of two scaled copies of W R ( ei"'), also centered at the frequencies of the sinusoid. The scale factor is the
same in both cases, resulting from the Fourier transform of the sinusoid.
Since the peaks of the Fourier transform of the rectangular window are higher than those of the H•mming
window, the peaks in V2 (&"') will be larger than those in V1 (ei"').
10.19. Using the approximation given in the chapter

_ 24.-(A., + 12)
L155Am~ + 1
we lind for A.,

= 30 dB and A,., = ~ rad,
L ::

24.-(30 + 12)
155(.. /40)

=

261.1 .... 262

+1

415
10.20. (a) The best sidelobe attenuation expected nnder these constraints is
24.-(A,, + 12)
155.0.,..
24r(A., + 12)
512 :::
155(.-/100)
A,, ::: 21 dB

L :::

+1
+1

(b) The two sinusoidal components are separated by at least ..-{50 radians. Since the largest allowable
mainlobe width is .-/100 radians, we know that the peak of the DFT magnitude of the weaker
sinusoidal component will not be located in tbe mainlobe of the DFT magnitude of the stronger
sinusoidal component. Thus, we only need to consider the sidelobe height of the stronger component.
Converting 21 dB attenuation back &om dB gives
-21 dB = 20log10 m
m = 0.0891
Since the amplitude of the stronger sinusoidal component is 1, the amplitude of the weaker sinu·
soidal component must be greater than 0.0891 in order for the weaker sinusoidal component to be
seen over the sidelobe of the stronger sinusoidal component.
10.21. We have
v[n]

=

cos(2,.n/5)w[n]
ei2•n/• + .-;..n/5]
= [
w[n]
2

The rectangular window's transform is
W(eiw)

= sin{16w) .-;w3l/2
sin(w/2)

In order to label V(eiw) correctly, we must find the mainlobe height, strongest sidelobe height, and the
first nulls of W(e'w).

Mainlobe Height of W(eiw): The peak height is at w
W(ei 0 )

= 0 for which we can use l'hopital's rule to find

= 32cos(16w)

I

cos(w/2) w=O

= 32

Strongest Sidelobe height of W(eiw): The strongest sidelobe height for the rectangular window is
13 dB below the main peak height. Therefore, since 13 dB = 0.2239 we have
Strongest Sidelobe height = 0.2239(32) "' 7.2
First Nulls of W(eiw): The first nulls can be found be noting that W(eiw)
Thus, the first nulls occur at

2.w=±32

Therefore, JV(eiw)llooks like

= 0 when sin(16w) = 0

416

I V(J">} I

16

-2KIS + 2Jrl'32

2><15 + 2lri32

_,
"

Ill

Note that the numbers used above for the heights are not exact because we are adding two copies of
W(e'w) to get V(eiw) and the exact values for the heights will depend on relative phase and location of
the two copies. However, they are a very good approximation and the error is small.

10.22. The 'instantaneous frequency' of :r[n], denoted as >.[n], can be determined by taking the derivative
with respect to n of the argument of the cosine term. This gives
>.[n]

>.[n]

271"

! [";

=
+ 1000sin(~)]
" (>rn)
= 4" + scos
8000
= !+.!.cos(~)
8 16
8000

Once A[n]/2,. is known, it is simple to sketch the spectrogram, shown below.
0.5
0.45
0.4
0.35
0.3

..,"

~ 0.25

0.2
0.15
0.1
..

0.05

0
2000

4000

6000

·-8000

10000

Sample number (n)

12000

14000

16000

417
Here, we see a cosine plot shifted up the frequency (>./27r) axis by a coustant. A$is customary in a spectrogram, only the frequencies 0 ::; >.f2" :S: 0.5 are plotted. 10.23. In this problem, we relate the DFT X(k] of a discrete-time signal z[n] to the continuous-time Fourier transform X.(jO) of tbe continuous-time signal z.(t). Since z(n] is obtained by sampling z.(t), z(n] = X(ei"') = z.(nT) f f (;f + ;2;) Xc T=-CIC Over one period, as5llllrlng no aliasing, this is for which is equivalent to X(ei"') = { +x. (i'f), +x. (j~), -:If S. w :s; " for 0 S w < .for.- :s;w < 2" Since the DFT is a sampled version of X(ei"'), we find X[k] = { 'Xc: (·2d) 'f" 1 "FiT ' 1X ( ·2•<•-Nl) ~ c: J NT forO:S:k< I 'f for~:S:k:S:N-1 Breaking up the DFT into two terms .l.ilo< this is nnececessaJuy to relate the negative frequencies of X.(jO) to the proper indicies S k S N - I in X(k]. 'f Method 1: Using the above equation for X(k], and plugging in values of N = 4000, and T = 251JS, we find x,[k] = { 40,ooox. (;2" ·10 · k), foro::; k s 1999 40, ooox. (j21f · 10 · (k- 4000)), for 2000 :s; k S 3999 Therefore, we see this does not provide the desired samples. A sketch is provided below, for a triangular-shaped X.(jO). -1 ...... .t.t=10Hz 418 Method 2: This time we plug in values of N x.[l:J = 4000, and T = 50J.IS to find = { 20, ooox, U2.-. 5 ·A:), foro 5 A: 5 t999 20, OOOX, U2.- · 5 · (.I: - 4000)) , for 2000$ l: $3999 Therefore, we see this does provide the desired samples. A slretch is provided below. ---1 f- Af=5Hz Method 3: Noting that Z3[n] = z2[n] + z2 (n- ~].we get x.[kJ = x.[kJ + (-t)• x.[A:J X [l:] = { 2X2 [k], fork e~en 3 0, otherwiSe 40, OOOX, U2.. · 5 ·A:), for k even, and 0$ k $1999 X,[.l:] = 40,000X, U2.. · 5 · (l:- 4000)), fork even, and 2000$ k $3999 { 0, otherwise This system provides the desired samples only for k an even integer. A sketch is provided below. ---1 f- Af=5Hz (19 10.24. (a) In this problem, we relate the DFT X[k] of a discrete-time signal :t[n] to the continuous-time Fourier tranform X.(j!l) of the continuous-time signal :z:.(t). Since :z:[n] is obtained by sampling :r.(t), = X(ei~) = :r[n] :tc(nT) ff X. r=-oc (if+ i ;r) 2 Over one period, assuming no aliasing, this is for -1r ~ w :5 1r which is equivalent to Since the DFT is a sampled version of X(eiw), forO:Sk:SN-1 we find X[k) = { ,. 'x(·2••) c JW'f ' forO:Sk<~ 1 X ( ·2•~TN!) ,. c J ' for~:Sk:SN-1 Breaking up the DFT into two terms like this is necessary to relate the negative frequencies of X,(jn) to the proper indicies :S k :S N- 1 in X[k]. lf The effective frequency spacing is ~n 2>r = NT 2lf = (1000)(1/20,000) = 2"(20) radfs (b) Next, we determine if the designer's assertion that Y[k] = aX,(j2,. · 10 · k) is correct. To understand the effect of each step in the procedure, it helps to draw some frequency domain plots. Assume the spectrum of the original sit,nal :t,(t) looks like 420 1 Q -~~(,~~~~)---------+--------~~~,~~) Sampling this continuous-time signal will produce the discrete-time signal z[n], with a spectrum X(~ Next, we form X[k], O$k$250 W[k] = :s; k :s; 749 X[k], 750 :s; k :s; 999 0, { and find w[n] as the inverse DFT of W[k]. . X{k] = Samples of X( el "') o 251 421 W[k] 1/T 250 999k 750 500 Before going on, we should plot the Fourier transform, W(&~), of w[n]. It will look like W(r!") 1/T 1-----....:>.I.~.D.I.~.D.I.~.D.i.:__ _ _ _ ., 21t " W(ei~) goes through the DFT points and therefore is equal to samples of X,(jO) at these points for 0 ::; k ::; 250 and 750 ::; k ::; 999, but it is not equtJ to X,(jl1) between those frequencies. Furthermore, W(e'~) = 0 at the DFT frequencies for 251 :=; k ::; 749, but it is not zero between those frequencies; i.e. we can not do ideallowpass filtering using the DFT. Now we define ¥[n] = { w[2n], 0 ::; n ::; 499 0, 500 ::; n ::; 999 and let Y[k] be the DFT of y[n]. First note that Y(~) is Y(~) = !w(~l 2 ) + !w(ei1•) 2 which looks like 2 422 Y(ei"') ~~~~~~~~~~~~~~~~~w 2lt It 31t 4lt Y[k] is equal to samples of the Y(eiw) = Y(eiw)lw=..k/N = ~w (ei'lH) + ~w (e>~C¥l) Now putting all that we know together, we see that fork = 0, 1, ... , 500, Y[k] is related to X,(jO) Y[k] as follows. Y[k] = k even, k y! 500 i-tX,(j2" · 10 · k), j.X,(j2,. · 10 · k), 1 ~W(eid/Nj = + olJ.W(ej•(k-N)/N) k 500 k odd In other words, the even indexed DFT samples are not aliased, but the odd indexed values (and k = 500) are aliased. The designer's assertion is not correct. 10.25. (a) Starting with definition of the time-dependent Fourier transform, 00 L Y[n, ~) = y[n + m]w[m]e-;•m m=-oo we plug in M y[n + m] =L to get oo Y[n, ~) = = .... h(k]z[n + m - k] M L L h[k]z[n + m- k]w[m]e-;>m M oo le-o --oo L h(k] L z[n + m- k]w[m]e-;•m M = = L h[k]X[n ,... h[n] • X[n, ~) where the convolution is for the variable n. k, ~) (b) Starting with we find Y[n, >.) = .-i>•l~ h[k]X[n- k, >.)] = e-i>n l~ h[k]ei(•-•)> X[n- k,>.)] M L h[k]e-i» X[n - = k, >.) >=0 H the window is long compared toM, then a small time shift in X[n, >.) won't radically alter the spectrum, and X[n- k, >.) ::: X[n, >.) Consequently, M Y[n,>.) L h[k]e-i'" X[n, >.) ::: ::: H(ei')X[n, >.) 10.26. Plugging in the relation for c,.[m] into tbe equation for /(w) gives = /(w) L~ m=~- 1 ) [~ v[n]v[n + m]] e-1wm L-1 = L~ L L-1 n=O Let l L v[n] v[n + m]e-iwm m.=-(L-1) = n + m in the second summation. This gives I(w) l = LU = L~ L-1 n+(L-1) n=O l=n-(L-1) L v[n] L v[l]e-iw(l-n) L-1 n+(L-1) A=O t=n.-(L-1) L v[n]ei"'" L v[l]e-iwl Note that for all values of 0 ~ n ~ L- 1, the second summation will be over all non-zero values of v[l] in the range 0 s; l s; L - 1. M. a result, I(w) L-1 . L-1 _ = -U L v[n]e'wn L v[l]e-'"'' 1 = l=i -(M-1) Ii ;> i9 m M-1 0 which is the convolution of a rectangular signal, x[m] ...- 112 m -(M-1)/2 0 (M--1)/2 with itself. Tha.t is, wa[m] = :[m] • :[m]. Above, we found the Fourier tranform of a rectangular window, as where 2M - I was the length of the window. We can use this result to find the Fourier transform of :[m]. The signal :~:[m] is similar to the rectangular window, the difterence being 428 that it is scaled by 7i; and has a length 2M; 1 + 1 = M. X(ei"') = Therefore, 1 sin(wM/2) .,fM sin(w/2) = The time domain convolution, w8 (m] o:(m]•o:(m] corresponds to a multiplication, Ws(ei"') (X(ei"')] 2 in the frequency domain. As a result, Ws(ei"') = (X(ei"')) = 2 ..!_ [sin(wM/2)] _M 2 sin(w/2) A sketch of Ws(ei"') appears below. 0 It HannjngfHammjng: Starting with wH[m] = UIH(mj = (o + ,8 cos[nn/(M- 1)]) wR[m) (o + ~.,;•m/(M-1) + ~e-i"""/(M-1)) WR(m) We take the Fourier transform to find WH(ei"') = = oWR(ei"') 0 + ~ {WR(eilw-•/(M-1)]) + WR(eilw+•/(M-lllJ) sin[w (M- tlJ ~ [sin((w- -.f:r)(M- !JJ] sin(w/2) +2 sin((w- ...:,J/2) + ~ [sin((w + Jf.:r)(M- lll] 2 sin((w + Ji_ 1 )/2] (I) -It It = 429 (b) Rectangular: The approximate mainlobe width, and the approximate variance ratio, F, for the rectangular window are found below for large M. In part (a), we found the Fourier transform of the rectangular window as W (eiw) = sin[w(M- ~)] sin(w/2) R The numerator becomes zero when the argument of its sine term equals rn. (2M -1)w 2 = '"' 2'"' w = Plugging in n 2M-1 = 1 gives us half the mainlobe bandwidth. 21r ~ Mainlobe bandwidth = 2M-1 41r Mainlobe bandwidth = 2M-1 2.. Mainlobe bandwidth ::: (M-1) 1 L Q F = M 2 w (m] m:-(M-1) = ::: 1 -(2M -1) Q 2M Q Bartlett {triangular): The approximate mainlobe width, and the approximate variance ratio, F, for the Bartlett window are found below for large M. In part (a), we found the Fourier transform of the Bartlett window as W (eiw) B = _!_ [sin(wM/2)] M 2 sin(w/2) The numerator becomes zero when the argument of its sine term equals ..n. wM 2 = '"' w Plugging in n = 2m M = 1 gives us half the mainlobe bandwidth. 1 Mainlobe bandwidth 2 Mainlobe bandwidth 2r = M 4.. = M To compute F, we use the relations M-1 :[m M-1 = :[m• = m=O M(M -1) 2 M(M- 1)(2M- 1) 6 430 F 1 (M- ) 1 L = Q lml) 2 ( 1-M --CM-1) [2 'f: (1- !!:)' -1] 1 = .!. Q. ,...., 1 [ M-1 M M-1 M-1 ] 4 2 = -Q 2 m=O :E 1 -M...., - :Em+- :E m• -1 M',.... = .!_ [2M_ 4(M- 1}M 2(M- 1}M(2M- I) _ Q :: h [2M-2M+ :: + 2M 6M2 t] 2~] 2M 3Q HanningfHamming-. We can approximate the mainlobe bandwidth by analyzing tbe Fourier transform derived in Part (a). Looking at one of the terms from this expression, ~ 2 [sin[(w- 1f=;)(Msin[(w- u"_ 1 )/2] !>J] we note that tbe numerator is zero whenever the its argument equals wn, or "' = mr " M- {1/2) nw + M- 1 .- :: M+M :: O'(n +I) M So the mainlobe bandwidth for this term is 4Mainlobe bandwidth :: Mainlobe bandwidth :: ~ 2 "" ii Note tbat tbe peak value for this term occurs at a frequency w:: 1r/M. A similar analysis can be applied to tbe other terms in Fourier transform derived in Part {a). The mainlobe bandwidth for the term ~ [sin{{w+ Jl=r)(M -!)J] 2 sin{(w + u"_ 1 )/2] is also 2wfM. Note tbat the peak 'Value for this term occurs at a frequency w:: -rfM. Finally, the mainlobe bandwidth for the term a sin[w (M- ! )] . SlD(w/2) 431 is also 2" fM. Note that the peak value for this term occun at a frequency w A sample plot of these three terms, for {J = 2o and large M =0. is shown below. Thus, for large M, the mainlobe bandwidth is bounded by 2.. . 41< M < Mainlobe bandwtdtb < M Therefore, a reasonable approximation for the mainlobe bandwidth is . • 31< Mainlobe bandwtdtb o:: M F 1 = Q .,_, ( L o+{Jcos ( ""' )) M-1 2 m=-(M'-1) 1 = 1 .!.[ ~ a +2o{J ~ 2 Q m=-(M-1) cos(::\)+{J 'f cos•(;: 2 m=-(Al-1) m=-(Al-1) 1 )] Using the relation 1 1 cos2 8 =- +- cos28 2 2 we get F = -1 Q L [ Al-l =-+2 =-(Al-l) ~1 ) M- m=-(M-1) Al-l L: cos cos 21m1 )] M=I . --(Al-1) Noting that Jl-1 L ( ) cos M":1 = -1 • =-(Al-l) Jl-l L: m=-{AI-1} ( cos 2nn ) M-1 = 1 432 we conclude = F 2 2 2 2 .!_ [(2M- I)Q2 - 2Q/3 + /3 (2M- I)+ /3 Q ] ~ 2:(Q·+~·) 10.30. (a) Using the definition of the time-dependent Fourier trausform we find 13 L X(O,k] = :[m]e-j(2•/7)•m m=O • 13 L :(m]e-;(2•/T)•m + L:(l]e-j(2•/1)"' = L• :[m]e-j(O./T)•m + L• :r[m + 7]e-i(2•/7)>me-i2d • = L (:[m] + :[m + 7])e-j(2•/7)•m = l=1 m=O m=O m=O By plotting :[m] -'{mj ••• -1 0 2 1 we see that :[m] + :r[m + 7] 3 4 5 = I for 0 :S m :S 6. X[O,k] = 6 7 8 9 10 n Thus, L• (I)e-j(>•/T)•m m=O = l>J'T{l} = 7o(kJ (b) If we follow tbe same procedure we used in part (a) we find X(n, k] = 13 L :(n + m)e-;(2•/T)>m m=O • = L 13 :(n + m)e-;(2•/T)>m + I:=!n + ije-;(2•/7)>~ l=? m=O = L• (:(n + m) + :(n + m + 7J)e-iC••fT)>m m=O With n ;o: 0 we have·:r[n + m) + :[n + m + 7] = I for 0 :S m :S 6, and so X[n,k) = l>J'T{l} = 7o(kJ Therefore, for 0$ n $co we have • • = = I:x[n,k] .1:-=0 I:7o[kJ A:=O 7 10.31. (a) Sampling the continuous-time input signal = e"(h/l)tO't :z:(t) with a sampling period T = w-• yields a discrete-time signal :z:[n] = :z:(nT) = e"3•n/l In order for X,.[k] to be nonzero at exactly one value of k, it is necessary for the frequency of the complex exponential of :z:[n] to correspond to that of a OFT coefticent, '"• = 27rkfN. Thus, 3.- = 2.-k N = l6k 8 N 3 The smallest value of k for which N is an integer is k Xw[k] is nonzero at exactly one value of k is N = 3. Thus, the smallest value of N such that = 16 (b) The rectangular windows, w,[n] and w,[n], dil!er only in their lengths. w,[n] has length 32, and w,[n] has length 8. Recall that compared to that of a longer windov.·, the Fourier transform of a shorter window has a larger mainlobe width and higher sidelobes. Since the OFT is a sampled version of the Fourier transform, we might try to look for these features in the two plots. We notice that the second plot, Figure Pl0.31-3, appears to have a larger mainlobe width and higher sidelobes. As a result, we conclude that Figure Pl0.31-2 corresponds to w1 [n], and Pl0.31-3 corresponds to w,[n]. (c) A simple technique to estimate the value of Wo is to find the value of k at which the peak of jX,.[k]l occurs. Then, the estimate, is The Corresponding value of fio is - 2.-k l1o =NT This estimate is not exact, since the peak of the Fourier transform magnitude jX,.(e'w)l might occur between two values of the OFT magnitude IX,.[k]l, as shown below. o DTFT: IX (ei"')l OFT:IX}JI 434 The maximum possible error, Omax error, of the frequency estimate is one half of the frequency resolution of the DFI'. 1 2.2NT flmax error = = For the system parameters of N . NT = 32, and T = 10-4 , this is Omax error =982 rad/s (d) To develop a procedure to get an exact estimate Of flo, it helps to derive X,.[k]. First, let's find the Fourier transform of :t,.[n] = :t[n]w[n], where w[n] is anN-point rectangular window. N-l x.(~'"') L eJwone-'"'" = = Let w' =w- wo. n=O N-1 L ...... e-j(w-wo)n Then, X.,(eiw) N-1 L = ...... .-jw'n 1- e-iw'N = 1-e jw' {eiw' N/2 _ .-;w' N/2)e-jw' N/2 = (e = jw'/2- e-iw'/2)e '""'' sin.(w'N/2) -;w'(N-1)/2 sin(w' /2) • = sin[(w- .,.)N/2] -j(w-wo)(N-1)/2 sin[(w- .,.)/2] • Note that X,.(eiw) has generalized linear phase. Having established this equation for X,.(eiw), we now find X.,[k]. Recall that X,.[k] is sintply the Fourier transform X,.(eiw) evaluated at the frequencies w = 27fk / N, for k = 0, ... , N - 1. Thus, X,.[k] = sin((21rk/N- .,.)N/2] -;(0.4/N-wo)(N-1)/2 sin[(27rk/N- .,.)/2] e Note that the phase of X,.[k], using the above equation, is LX ,. [kl = (21rk/N- .,.)(N- 1) +m" 2 where the m,. term comes from the fact that the term lin{(2d:/N -.,.)N/2] lin{(2.,kfN ""')/2] can change sign (i.e. become negative or positive), and thereby offset the phase by .- radians. In addition, this term accounts for wrapping the phase, so that the phase stays in the range [-1r, .-]. 435 Re-expressing the equation for LX.,[k], we find wo= 2(LX,.[k] - m.r) N-1 2.-k +N Let X1,.[k] be the OIT of the 32-point sequence :r1.,[n] = :r[n]w.[n], and let X2.,[k] be the OFT of the S-point sequence :r,,.[n] = :r[n]w2 [n]. Note that the kth OFT ooeflicient of X,,.[k] corresponds to x ... [4k]. Thus, we can relate the 8 OIT coeffients of X 2,.[k] to 8 of the DIT ooeflicients in x ... [k]. Using the k = Oth OFT coefficient for simplicity, we find = 2(LX,..(OJ- m.r) _ 2(LX.,,[O]- p1r) 32-1 8-1 LX,.1[0]- m.LX,.,[OJ- p1r ""' = 15.5 - 3.5 A solution that satisfies these equations, with m and p integers, will yield a precise estimate of "'<>· We can accelerate solving these equations by determining which values of m and p to check. This is done by looking at the peak of IX.,(k]l in a procedure similar to Part (c). Suppose that the indices for two largest values of IX,.[k]l are k.n;n and k,... Then, we know that the peak of IX(e'w)l will occur in the range 2,- k,;n 21r k,.. --<·-~<--- N --u- N By re-expressing the equation for LX,.[k], we see that m,.;n = ;1 [ 2LX,.1 [k.nin J + ( ~2ll"k,.;n Wo • ) (N -1) ] ;(ux...[k.....J+C";=-Wt!)(N-1)] m,.... = In these equations, W. is the estimate found in Part (c). So we would look for values of m in the range [lm..,nJ, fm,....l]. Similar expressions bold for p. Once WQ is known, we can find flo using the relation flo =WQ /T. 10.32. For each part, we use the definition of the time-dependent Fourier transform, 00 X[n, >.) = L :r(n + m]w[m]e-i>m. m=-oc (a) Linearity: using :r[n] X[n, >.) =az1(n] + b:t2 [n], = L :r[n + m]w(m]e-i>m 00 = L (az.(n + m] + b:t,[n + m])w[m]e-i>m m=-oc 00 :r.[n + m]w[m]e-i'"' + b L :r,[n + m]w[m]e-i>m m=-oo = aX1 [n, >.) + bX,[n, >.) (b) Shifting: using y[n] = :r[n- no], Y[n, >.) = .. L m=-oo y[n + m]w[m]e-i-'m 436 00 = L = X[n - :[n - flm fl(), ).) (c) Modul4tion: using y[n) = .;... n:[n), y[n + m) = .;wom m=-oo L 00 = .;wo(n+mlz[n + m)w[m)e-;>m m.=-oo L .;... n:[n + m)w[m]e-;<>-wo)m 00 = m=-oo (d) Conjugate Symmetry: for z[n) a.nd w[n) real, 00 L X[n, ).) = z[n + m)w[m)e-;>m = L.~oo :[n + m]w[m]ei'"'] • = [X[n, -).)]' = X'[n, -).) 10.33. (a) We are given that -mean variables that are uucorrelated with each other, £ {e(n]e(n + m]} = cr~6(m], where cr~ = £{ e2 [nl} Putting this together, we get Since 2~ 1:• cos(2won + wom + 211)d8 = 0, we have ¢,,[m] = A• 2 cos(wom) (b) Since the Fourier transform of cos(wom) is ..O(w- wo) + cr~o(m] + .-o(w + wo) for !w! ::; "• 10.35. (a) Plugging in the equation !(.1:] = I(w•) = L1 JV[k)l 2 into the relation var(!(w)]::: .P;,(w) we lind that ftr [I JV(k]! var [!V(k)! 2 ] 2 ] ::: .P;.(w) ::: L 2 .P;.(w) 438 This equation can be used to find the approximate variance of JX(k]J 2 • We substitute the signal X(k] for V(k], the OFT length N for L, and use the power spectrum P•• (w) = -padding them to L 2048 samples will increase K by a factor of 10. Accordingly, the variance will decrease by a factor of 10. However, the frequency resolution will be reduced. (ii) U we increase the data record to 2,000,000 samples, we can keep the window length the same and increase K by a factor of 10. · = 10.37. (a) Taking the expected value of ~[m] = 21 ~· I(w)_,j"'mdw " -· = E{ 2~ L I(w).,;"'md N-1 :E :ra((m- n))N]Y,[n] = -=0 where :r'[n] = :r[-n]. Note that this is a circular convolution of :r,[-n] with y,[n]. Thus, we have expressed the circular correlation of z,[n] with y,[n] as a circular convolution of :r,[-n] with y,[n]. Now recall from chapter 8 that the circular convolution of two M point signals is equivalent to their linear convolution when N ;<: 2M - 1. Since we can express the circular correlation in terms of a circular convolution, this result applies to circular correlation as well. Therefore, we see that if N ;<:2M -1, e;[m] C.[m] forO~m~M-1 = Thus, the minimum value of N is 2M - 1. (c) A procedure for computing ~•• [m] is described below. step 1: Compute X,[k] and Y;(k], which are the N ;<: 2M- 1 point DFTs of :r,[n] and y,[n]. step 2: Multiply x,[k] and Y;"[k] point by point, yielding C,[k] G\[k] x,[k]Y,•[k]. step 3: Repeat the above two steps for all data (K times), then compute = . 1 K-1 ••• [k] = Q C,[k] L = forO~k~N-1 i=O step 4: Take theN point inverse DFI' of +•• [k] to get ~•• [m]. Assuming that a radix-2 FFI', requiring ~log, N complex multiplications is used to compute the forward and inverse DFTS, the number of complex multiplications is 2 · ~ log2 N · K = KN1og2 N, for step 1 KN, for step 2 N, for divide by Q operation in step 3 If log2 N for step 4 So the total number of complex multiplications is {K + f)Nlog 2 N + {K + l)N. (d) The procedure developed in part {c) would compute the cross-correlation estimate~•• without any major modifications. All we need to do is redefine 1/<[n] as y,[n] = y[n + iM], and :t;[n] is the same as it was before, namely ~~ Note that form (e) For N =2M, < 0, ~•• [m] = O$n$M-1 [n1= { z[n + iMJ, 0, M$n$N-1 ~.. [-m]. z[n + iMJ, for 0 S n S 2M - I = z[n + iM](u[n]- u[n- MJ) + z[n + iM](u[n- M]- u[n- 2M]) = z[n + iMJ(u[n]- u[n- MJ) + z[n- M + (i + 1)M](u[n- MJ- u[n- 2M]) = :t;[n] + Zi+t[n- M] yi[n] = Taking the DFT of this expression yields Y;[k] A procedure for computing~•• for 0 = X;[k] + (-1)•x,+l[k] S m S M- 1 is described below. step 1: Compute theN point DFT X;[k] fori= 0,1, ... ,K. step 2: Compute Y;[k] = X;[k] + (-1)• X;+l[k] fori= 0, 1, ... , K- 1. step 3: Let Ao[k] = 0 and compute A;[k] = A;- 1 [k] + X;[k]Y;"[k], i = 1, ... , K- I step 4: Deline V[k] = AK-l[k]. Compute 11[m], theN point inverse DFT of V[k]. step 5: Compute • 1 4J•• [m] = "Q"[m] Assuming that a rad.ix-2 FFT, requiring ~ log2 N complex multiplications is used to compute the forward and inverse DFTs, the number of complex multiplications is (K + 1)~1og2 N, 0, (K -l)N, for step I ~log2 N, N, for step 4 for step 2 for step 3 for the divide by Q in step 5 So the total number of complex multiplications is Kfl N log2 N + K N. Note that for large N and K, this procedure requires roughly half the number of complex multiplications as the procedure described in part (c). 10.39. (a) Using the relations, c[n,m] X[n, .\) = .. L -=-oo z[n + m]w[m)e-iM•d.\ 443 we find c(n,m] = 2.1• IX[n, A)j 2 ei'md>. 2• -· = ~ /_: X[n, >.)X[n, ->.)ei'md>. i: (~00 = ;.. = x[n + l]w[l]e-;>J) Ct- f. f. x[n + l]w[l]x[n + r]w[r] ( 2~ l=-oor=-oo = 00 L L x[n + l]w[l]x[n + r]w[r] ( -1 00 x[n + r]w[r]ei'r) ei'md>. 00 21r l=-ocr=-oo /_: .-i>J.,J>rei>md>.) 1" . ) .-J>(-I+rlei>md>. -'II' Using the Fourier transform relation, o[n- no];..--; .-;wn, we find 00 c[n, m] = 00 L L x[n + l]w[l]x[n + r]w[r]o[m -I+ r] l=-oor=-oo The 6[m - I + r] term is zero everwhere except when m - l + r = 0. Therefore, we can replace the two sums of land r with one sum over r, by substituting l == m + r. 00 c[n, m] = L :r[n + m + r]w[m + r]:r[n + r]w[r] r=-oo 00 = L x[n + r]w[r]:r[n + m + r]w[m + r] r=-oc (b) First, note that IX[n, >.)1 2 = X[n, ->.)X[n, .>.) = IX[n, ->.)J' Starting with the definition of c[n, m], c(n, m] = ..1"1'-· . -· 1 2 c[n, -m] = -1 2 we substitute .>.' = ->.to get c{n,-m] IX[n, .>.)1 2 .,J>m d>. IX[n, >.)J 2 .-J.,md>. = __!_ /.-•IX[n, ->.')12 .,J>'md>.' 2.- • 2.1• = 2.1• = = 211' 271'" IX[n,->.')1 2 ei>'md>.' -r -· c(n,m] IX[n, >.')12 .,i>'md>.' Thus, the time-dependent autocorrelation function is an even function o' Jll fur n fixed. Next, we use this fact to obtain the equivalent expression for c{n, m]. 00 c[n, m] = L = L :r[n + r]w[r]z[m + n + r]w[m + r] 00 :r[n + r]w[r]:r[-m + n + r]w[-m + r] r=-oo Substituting r' = n + r gives 00 = L = L :r[r']w[r' - n]:r[r'- m]w[(r'- m)- n] 00 :r[r']:r[r' - m]w[r' - n]w[-(m + n- r')] '1""'=-oo = L"" :r[r'J:r[r' - m]h.n[n - r'J r';-cc where h,.{r] = w[-r]w[-(m + r)] (c) To compute c[n,m] by causal operations, we see that h,.{r] =w[-r]w[-(m + r)] requires that w[r] must be zero for < 0 > 0 -r r and w[r] must be zero for < 0 m+r > 0 r > -m -(m+r) Thus, w{r] must he zero for r > min(O, -m). If m is positive, then w[r] must he zero for r This is equivalent to the requirement that w[-r] must he zero for r < 0. (d) Plugging in w[-r] into hm{r] = { 0,a' ' r;::o r 0. 445 > a2, Again we have assumed that m is positive. H [zl H,.(z) = h,.[r] = then a'" 1- a2 z 1 a'"6[r] + a 2 h,.[r- 1] Using this in the equation for c[n, m] gives 00 L c[n, m] = z[r]z[r - m]h,.[n - r] r=-oo 00 L = z[r]z[r- m] (c'"6[n- r] + c 2 h,[n- r - 11) r=-oo 00 = a'"z[n]z[n- m] + a 2 L z[r]z[r - m]h,[n- r - 1] r=-oo = c'"z[n]z[n- m] + c 2 c[n- 1, m] A block diagram of this system appears below. x{n] ----.,----~----\ir----.,-------+ c[n,m] z-m (e) Next, consider the system w-r[ ] -{ ra', r;o:O 0, r <0 h,.[r] = {ra' u[r]} {(r + m)a'+'"u[r + m]} r 2:: O;r 2:: -m To get the z-transform H,.(z), recall the z-transform property: rz[r] +-> -z•~•l. Using this property, we find ra2'u[r] ~ (1-a•z 1)2 a 2 z- 1(1 +a2z-1) (1-a•z Again we have assumed that m is positive. Thus, 1)3 446 = = a"'+2 z- 1 (1 + a 2 z- 1 + m - ma2 z'- 1 ) ~1- ,.zz-1)3 a"'+2 (1 + m)z- 1 + a"'+4(1- m)z- 2 1-3a2 z- 1 +3a•z-•-a6 z 3 Cross-multiplying and talring the inverse z-tra.nsform gives h,.[r]- 3a2 h,.[r- 1] + 3a4 h,.[r- 2]- a6 h,.[r- 3] = a"'+2(1 + m)6[r- 1] + a'"+4(1- m)6[r- 2] h,.[r] = 3a2 h,.[r- 1]- 3a4 h,.[r- 2] + a6 h,.[r- 3] + a"'+ 2 (1 + m)6[r- 1] + a'"+4(1- m)6[r- 2] Using this relation for h,[r] in c{n, m] = 00 L :r[r]:r[r- m]h,[n- r] r=-oo we get 00 c{n, m] = L :r[r]:r[r- m] (3a2 h,[n- r - 1]- 3a4 h,[n- r - 2] + a6 h,.[n- r - 3]) r=-oo 00 L + z[r]:r[r- m] (a"'+2 (1 + m)6[n- r - 1] + a'"+<(1- m)6[n- r - 2]) r=-oo = 3a2 c{n- 1, m] - 3a4 c[n- 2, m] + a6 c[n- 3, m] +a'"+ 2 (1 + m):r[n - 1]z[n- 1 - m] + a"'+4(1 - m)z[n- 2]z[n- 2-m] A block diagram of this system appears below_ y(n] ~ x[n] -1 z + c[n,m) + am+2(m+ 1)z-m 3a2 z-1 -3a4 z-1 z-1 X z-1 a6 am+4(1-m)z-m 10.40. (a) Looking at the figure, we see that X[n, >.) = {(:r[n]e-;"") • ho[nl} .,;>.n = [ f: --oo :r[n- m]e-;>.(n-m)ho[m]] .,;>.n 00 L = :r[n - m]ho(m].,;>.m m.-=-oc Let m' = -m. Then, -oo X[n, >.) = L m•=oo :[n + m1ho[-m'Je-;>.m' z-1 447 00 = L z[n + m']ho[-m1e-;>m' m'=-oo = if ho[-m] system. = w[m]. X[n,>.) Next, we show that for>. fixed, X[n,>.) behaves as a linear, time-invariant Linear: Inputting the signal oz 1 (n] + bz 2[n] into the system yields 00 L (oz,[n + m] + bz2[n + m]) ho[-m]e-;>m = --00 00 00 L oz,[n L + m]ho[-m]e-;>m + bz2[n + m]ho[-m]e-;>m = aX1 [n, .>.) + bX2[n, .>.) m=-oo m=-co The system is linearTime invariant: Shifting the input x[n] by an amount I yields 00 L x[n + m + l]ho[-m]e-;>m = X[n +I,.>.) m=-oo which is the output shifted by I samples. The system is time-invariant. Next, we find the impulse response and frequency response o£ the system. To find the impulse response, denoted as h[n], we Jet z[n] o[n]. = 00 h[nJ = 2: •In + mJwlmJ·-;•m m=-oo = w[-n]ei'" = ho[n]ei'" Taking the DTFT gives the frequency response, denoted as H(eJ~). H(eJw) = Ho(eJ(w-1)) (b) We find S(eiw) to be s[n] S(eJW) S(eJW) = = = (z[n]e-i'") • w[-n] X (ei)) W(e-jw) X (ei)) Ho(eJw) Note that most typical window sequences are lowpass in nature, and are centered around a frequency of w 0. Since H 0 (eiw) W(e-iw) is the Fourier transform o£ a window which is lowpass in nature, the signal S (e)w) is also lowpass. = = The signal s[n] = X[n, .>.) is mnltiplied by a complex exponential ei'". This modulation shifts the frequency response of S(eiw) so that it is centered at w = >.. = s(n]ei'" H(eJw) = S ( ei(w->)) h{n] Since S(eiw) is lowpass filter centered at w = 0, the overall system is a bandpass filter centered at w = >.. 448 (c) First, it is shown that the individual outputs Y•[n] are samples (in the,\ dimension) of the timedependent Fourier transform. 00 Y>[n] = L z[n + mjw[mje-j.\om m=-oc 00 L = z[n + m]w[mje-jbkm/N m=-oo = X[n, A)J.=2d/N Next, it is shown that the overall output is y[n] = Nw[O]z[n]. N-1 y[n] = L Y•[n] k=O N-1 = oc L L z[n + m]w[m]e-j2•>m/N i:=O m.=-oc CICI N-1 = L L x[n + m]w[m]e-j2dmfN N6JmJ = Nw[Ojx[n] (d) Consider a single cha.nnel, decimator expander x[n]---{ In the frequency domain, the input to the decimator is X ( ei<"'+>•l) Ho(ei"') so the output of the decimator is R-1 ~~X (ei<•l) Ho (eiwi)/R) The output of the expander is R-1 ~~X (.;•-2•1/R)) Ho (.;,)) X (ei•-2•1/R)) The overall system output is formed hy summing these terms over k. N-1 = L Y(eJw) .... Yk(eiw) = ~ R-lN-1 L L Go (ei•)) Ho (ei•->•1/R)) k=O l=l Aliasing Component Therefore, we require the following relations to he satisfied so that y(n] N-1 L Go (ei.•) = L --co :r[m]ho[n- m]e-;>,m 450 or, using>..= 21rkjN, co X[n, k] = L :~:(m]ho(n- m]e-i•••mfN m=-oo Therefore, the output of the decimator is co X(Rn, k] = L :~:[m]ho(Rn- m]e-i2•>m/N m=-oo Recall that in general, the output of an expander with expansion factor R is co L "'• (n] = :~:(l]o(n - lR] l=-oo This relation is given in chapter 3. Therefore, the output of the expander is co L X[Rl, k]o(n -lR] t•-oo This signal is then convolved with go(n], giving co co co L L X[Rl, k]o[m -lR]go[n- m] m=-ool=-oo =L X[Rl, k]g0 [n- lR] l=-oo Therefore, Y>[n] y[n] = ,t;.., go[n -lR] (J.., :~:[m]ho[RI- m]e-i••>mfN) ei 2•h/N = ~,f.;.., go(n -lR] C~.., :~:[m]ho[RI _ m]e-i2dm/N) eJ2•h/N N-l co co = L L go(n -lR] = :~:[m]ho[RI- m]e-jbk(m-n)/N m=-oo A:=O l=-oo co L co N-1 L L go[n- lR]ho[RI- m]:~:[m] l=-oo m=-CIO L eJ2•k(n-m)/N A:=O Now recall that I::~· ei2,.[m] = £ {y[n + m]y[n]} = £Ct. .. = h[k]z[n + m- k] ,t. h[l]z[n- 00 L L h[k]h[l]£{ z[n + m - k]z[n - I]} .b:-oo I= -oo .. = 00 L L h[k]h[/]9'>•• [1 + m- k] ~-ool=-oo Since z[n] is white noise, it has the autocorrelation function q,.. [l + m- k] =.,.;o[l + m- k] Substituting this into the expression for q,,.[m] gives 00 q,•• [m] = .,.; 00 L 2: h[k]h[l]o[l + m- k] 00 = .,.; L h[l + m]h[I] l=-oo Note that 00 4>•• [m] = .,.; L 1=-oo is also a correct answer, since 4>.. [m] 1]} = 4>.. [-m]. h[l - m]h[I] 452 (b) Taking the DTFT of 1/>n[m] will give the power density spectrwi. t,.(w). t .. (w) J;. {~ ..t.. = .. = oc oc L: ~ L: h[ij 1=-CIO Substituting k h[l+m]h[ij}·-h[t + mJ•_,..... m=-oo = I + m into the second summation gives oc t .. (w) oc L = u! L h[ij 1=-oo oc L: = ~ oc h[ij.;"'' l=-oo oc = ..~ L: h[k]e-jw(t-1) l=-oc L: h£"1·-;... 1:=-oo oc L: h[-n·-;..• l=-oo h[kJ.-;... l:=-oo = ..~w(.;"')H(.;"'l = ..~ IH(.;"'ll 2 (c) This problem can be approached either in the time domain or the z-transform domain. Time domain: Since all the ai: 's are zero for a MA process, Jl y[n] =L bto:[n - k] t=O so y[n] is nonzero for 0 :5 n :5 M. Note that the autocorrelation sequence, 1/>.,[m] = oc L y[n + m]y[n] n.=-oo can be re-written as a convolution 1/>,.[m] = where g[n] = y[-n]. oc L g[m- n]y[n] Therefore, 1/>,.[n] = 11[-n]•y[n] Since y[ -n] is nonzero for - M :5 n :5 0, and 11[n] is nonzero for 0 :5 n :5 M, we see that their convolution 1/>n[m] is nonzero only in the interva!Jml :5 M. Z-transform domain: Note that t,.(z) If all the at's = ~H(z)H.(z) = 0, then Jl Lb•z-• H(z) = .... t,.(z) = Lbtz-•L:b;z' Jl Jl l:=O tao 453 The relation for ~ .. (z) above is found by multiplying two polynomials in z. The highest power of z in ~ .. (z) is zM which arises from tbe multiplication of the k 0 and I M coefficients. The smallest power of z in ~,.(z) is z-M wbich arises from the multiplication of the k = M and I= 0 coeflicents. Thus, ,.,(m] is nonzero only in the interval 1m! SM. = = (d) For an AR process, = H(z) = IIi':, (1 Since ~ .. (z) ~ •• (z) a•z-') = ~H(z)H"(z) -- = rrl=1(1N -------~~·~----~ o.z- 1)(1- o;z) Thus, the poles for ~•• (z) come in conjugate reciprocal pairs. A sample pole-zero diagram appears below. X X lm X Re Nth order zero Nth order zero at z = ~ X X X By performing a. partial fraction expansion on ~ •• (z) we find that each pole pair contributes a. sequence of the form A.ol;' 1 ~c{ml ~·. M ~ and therefore N ,,,[m] = L A.al"'t l=l 454 (e) For an AR process, with bo = 1, H(z) = Y(z) = X(z) 1 1- ~ 1 ....-• which means that N y[n] = L 4tll[n - kJ + :r:[n] 0=1 The autocorrelation function is then ¢.,[m] = ¢.,[-m] = £ {y[n - m]y[n]} = £ { y[n- m] (t.aw[n- kJ + :z:[n])} N = L at£ {y[n - m]y[n- k]} + £ {y[n - m]:z:[n]} t=t N = L 4t9\yy[m- k] + 9'>,.[-m] t=l N = L ...;.. [m- k] + ¢,.[m] 1:=1 Form= 0, N q,.. [o] = L ...¢.,[-/.:] + ¢•• [o] bt The ¢ •• [OJ term is ¢.,[0] = = £ {:z:[n]y[n]} £ { :r:[n] (t. 4t11[n - k] + z[n]) } N = L "•£ {z[n]y[n- /.:]} + £ {z[n]z[n]} 1:=1 N = L "•£ {z[n]y[n - /.:]} +.; t=l Note that :r:[n] is uncorrelated with the y[n- k], for k = 1, ... , N. Therefore, ¢.,[0] =.; Thus, N q,..roJ = 1:...q,..r-~.:J +.; lost N = 1: ...q,..r~.:J + "" = n.=-oc I: z[nje'"(OJ = z[O]- 2 + 3 Plugging this into our equation for z[n] we find =SJ[n]- 26[n- 1] + 30[n- 4] z[n] 11.7. (a) Given the imaginary part of X(ei"'), we can take the inverse DTFT to find the odd part of z[n], denoted z 0 [n]. Im {X(ei")} = sinw + 2sin2w = .!..J"- .!..-;. + ~ .......,- ~.-;"'"' 2j = z.[n] Using the formula z[n] = = = 2] j yei"'"' + 2~ ....-- ;/-;.- j 7·-j""' 'DF/ 1 (;Im {X(tJ"l}] 'DFT-' [&""' + ~&· - ~.-;.- .-;""'] 1 1 o[n + 2] + 2o[n + 1]- 2o[n- 1]- o[n- 2] = 2z0 [n]u[n] + z[O]o[n], we find z[n] =-o[n- 1]- 26[n- 2] + z[0]6[n] Any z[O] will result in a correct solution to this problem. Setting z[O] z[n] = -"')(1 + 2ei"') = I - cosw + 4' •4- cosw l+4cosw+4 2 = = !z- !z) 1 (1)(1(I+ 2z-l)(I + 2z) H(z)H"(lfz•) Since h(n] is stable and causal and has a stable aod causal inverse, it must he a minimum phase system. It therefore has all its poles and zeros inside the unit circle which allows us to uniquely identify H(z) from JH(z)J 2 • · H(z) = 1+2z lzl > ~ h(n] I ( l)(n-1) = -26(nI]+ u[n- 1] 2 11.11. Note that z;[n] can be written as z;(n] = -46(n + 3] + 46(n - 3] Taking the DTFT of z;(n] gives X;(ei"') Since X(ei"') = 0 for -1r = --4ei.., + 4e_,.., = -4(2jsin3w) = -8jsin3w$ w < 0, we cao lind X(ei") using the relation
X(ei") = { 2jX;(e'"),
0,

0
-11'

< w < "
w < 0

:s

470

Thus,

X(.;•)

0 < w < ,.
< 0

= { 16sin3w,
0,

-1r\$W

Therefore, the real part of X (&"') is

X,(.;"') =

~[X(&"')+ X"(e-i"'J)
0 < "' < ..
-8sin3w, -.- :5 w < 0

= { 8sin3w,

11.12. (a) Factoring the magnitude squared response we get
_;"')J2
IH( e-

= -109 - -32 cosw = 1 = H(&"')H"(&"')

1 - '"' ) ( 1- -e'
1 ·.,)
-2 cosw + -1 = ( 1 - -e
3

9

3

3

Thus, one choice for H(ei"') and h[n] is
H(.;"')

= 1- ~.-;..
3

1

h[n] = o(n] - 3o[n- 1]
(b) No. We can fiud a new system by taking the zero from the original system and ftipping it to its
reciprocal location. This only changes the magnitude squared response by a scaling factor. If we
compensate for the scaling factor the two magnitude squared responses will be the same. Thus, we

find

=
h(n]

=

~(1- 3e-i"')

3

1

3

o[n] - 30[n - 1]

satisifies the given conditions.
11.13. Expressing XR(ei"') in terms of complex exponentials gives

XR(e'"')

=
=

1 + cosw + sinw- sin2w

1 + ~e'"' + ~.-;.. + ..!.,.;.. - ..!.,.-;.. - ..!_,.;,., + .!,.-j2w
2

2

2]

2] .

2]

2;

= _..!_,.;,., + ~e'"' + ..!_,.;., + 1 + ~.-;.. - ..!.,.-;.. + .!,.-j2w
2j

2

2j

2

2j

2j

Taking the inverse DTFT of XR(ei"') gives the conjugate-symmetric part of :r[n], denoted as :r,(n].

1
1
1
1
1
1
:r,[n] = - 2/(n + 2] + 2o[n + 1] + 2/[n + 1] + o[n] + 2o[n- 1]- 2/(n- 1] + 2/[n- 2]
Using the relation x(n] = 2:r,[n]u[n]- :r,[O]o[n],

:r[n]

= o[n] + o[n- 1] + jo[n- 1]- j6[n- 2]

471

We then find the conjugate-antisymmetric part,
:t0

[n]

:t0 [n]

as

= !2 (z[n]- :r•[-n])
= !2 (o[n] + o[n- 1] + jo[n- 1]- jo[n- 2]- o[n]- o[-n- 1] + jo[-n- 1]- jo[-n- 2])
1

= 2 (o[n - 1] + jo[n- 1]- jo[n- 2]- o[n + 1] + jo[n + 1]- jo[n + 2])
I

=

.

.

-2 (o[n + 1]- o[n- I])+~ (o[n + 1] + o[n- 1])- ~ (o[n + 2] + o[n- 2])

Taking the DTFT of z 0 [n] gives jXr(&"').

_!2 {&w- e-jw) + t2 (&w + e-jw) - t2 (&"" + .-i"")

=

= -jsinw+jcosw-jcos2w
So
11.14. First note that,
(a)

The inverse transform of XR(e'w) is z,[n], the even part of z[n]. This is true for any sequence
whether it is causal, anticausal, or neither.

(b) jX 1 (eiw) is the transform of z 0 [n], the odd part of z[n]. This is true for any sequence whether it
is causal, anticausa.l, or neither.

(c) For an anticausal sequence
z[n]

= 2z,[n]u[-n]- z,[O)o[n]

Using Euler's identity and (a),
XR(&w)

=

f: G)
1=0

= 1+ ~
.,,[n] = o[n] +

t

cos(kw)

t. or
~ t. G)·

(&•w

+ .-jkw)

(6[n + k] + o[n _ k])

Using (c) and then taking the odd part we get,
z[n]

=
=

2z,[n]u[-n]- z,[0]6[n]

cr
cr
00

o[n] +

'E

2

6[n + k]

1=1

z 0 [n]

=

z[n]- z[-n]
2

= 2I 'E 2
00

i=l

(o[n + k]- o[n- 1.:])

472

Now taking the DTFT and using (b),

~

=

jXJ(ei"')

t. 0r

~,

(ei .... _ .. ... )

"" (1).
2 sin(kw)
"" (1).
iL
2 sin(kw)

=

j {;

=

4=0

Thus,

. = L"" (1).
2 sin(kw)

....

XJ(e'"')

11.15. Given X;(ei"'), we can take the inverse DTFT of jX;(ei"') to find the odd part of z(n], denoted z.(n].

Im {X(ei"')}

:z: 0 (n]

Using the formula z[n]

=
=

sinw

..!..eJ"' _1_e-iw
2j

2j

= v:rr-' (jim {X(ei"')})
=

vn-' [~e>"'- ~·-'"']

=

2o(n + 1]- 2o(n- 1]

1

1

= 2z.[n]u[n] + z[O]o[n),
z[n]

= -o[n- 1) + z(O)o[n)

Since

L""

z[n] =3

n=-oo

-1 + z[O]
z[O]

Therefore,

z(n]

=

3

=

4

= 40(n]- o[n- 1]

11.16. Using Euler's identity and the fact that z.[n] is the inverse transform of XR(ei"') we have
XR(ei"')

z.[n]

=
=

2- 4cos(3w)
2- 2(ei"' + ·-'"')

=

-20(n + 3] + 20(n] - 20[n- 3]

Since z(n] is real and causal, it is fnlly determined by its e?eD part z.(n],

z[n]

= 2z.(n]u[n] - z.(0]6(n]
= 40(n]- 40(n- 3] - 20[n]

=

20[n] - 4o[n- 3]

473
Using this information in the second condition we find

"' :r[n]~•n
= L:
=-"'
"'
= L :r[n](-W
n.=-oc
= 2+4

X(~w)lw=•

'F

7

Thus, there is no real, causal sequence that satisfies both conditions.

11.17. There is more than one way to solve this problem. Two solutions are presented below.
Solution 1: Yes, it is possible to determine :r[n] uniquely. Note that X(k], the 2 point DFT of a real
signal :r(n], is also real, as demonstrated below.
1

L :r(n]e-i'•""'l2

=

X(k]

n=O

1

=

X(k]

L::rfn](-:W•
n=O

Thus,

X[O]
X(l]

:r(O] + z[l]
z[O]- z(l]

=
=

Clearly, if z[n] is real, then X[k] is real. Therefore, we can conclude that the imaginary part XI(k]
is zero.
Therefore, the inverse DFT of XR(k] is z[n], computed below.
1

z(n]

= ~ L XR[k]~2·""' 1 '
k=O

z(n]

=

z[O]

=

z[l]

=
=
=

Thus,

z(n]

1
2

1

L XR(k](-1)""'
k=O

1
2{XR(Oj + XR(l])
-1
1

2{XR(Oj - XR(1])
3

= -o(n] + 35[n- 1]

Solution 2: Start by making the assumption that X(k] is complex, i.e., X1[k] is nonzero and XR[k]
U[k] - 45(k- 1]. Then, because z.,[n] is the inverse DFT of XR[k] we find

z.,[n] =

! txR[k]~2 •""12
21

=

! L:xR[kJHl""'
2-

=

474
and
=

1
2(XR[Oj

=

-1

=

1
2(XR[Oj- XR[1])

=

3

+ XR[1])

z,.[n] = -o[n] + 36[n- 1]
Because z[n] is real and causal, we can determine it from z .. [n]

z[n]

=

z .. [n],

n=O

2:t.. [n],

0

< n 2, we cannot necessarily uniquely determine z[n] from
XR[k] unless we make additional assumptions about z[n] such as periodic causality. When N > 2
the two assumptions we used above leads t.o two different sequences with the same XR[k].
11.18. Sequence 1: Fork= 0, 1,2 we have

XR(k] = 90(kj + 66[k- 1] + 66[((k + 1)),)
and XR[k] = 0 for any other k. Using the DFT properties and taking the inverse DFT we find for
n =·1,2,3
z,.[n]

=

3 + 2 ( e><>•/3)n

+ .-j(2•/3)n)

= 3 + 4 cos(2,.n/3)
=

76[n]

+ o[n- 1] + o[n- 2]

If we let x[n] = z,.[n] we have the desired sequence.
Sequence 2: If we assume z[n] is periodically causal, we can nse the foUowing property t.o solve for
x[n] from z,.[n]:
z ..[o],
n=O
z[n] =
2:tq[n], 0 < n < ~
{
0,
otherwise
Note that this is only true for odd N. For even N, we would also need t.o handle the n = N /2
point as shown in the chapter. We have
z[n]

=
=

z ..[o],
n=0
2:tq[n], n = 1
{ 0,
otherwise
76[n] + 26[n- 1]

475

11.19. Given the real part of X[k], we can take the inverse DIT to find the even periodic part of z[n],
denoted Zcp[n].
Using the inverse DIT relation,
N-1

L

1

zq[n] = N

XR(k]W_,..

1=0

we find
1

zcp(Oj = 4(4+1+2+1)=2
!(Hj-2-j)=!
4
.
2
1
%cp[2] = -(4-1+2-1) = 1
4
Zcp(3] = -I (4 - ). - 2 +).) =-1
4
2
%cp[1] =

Thus,
1

1

2

2

zcp[n] = U[n] + o(n- 1] + o[n- 2] + o(n- 3]
Next, we can relate the odd periodic and even periodic parts of :(n] using

zcp(n],
0 < n < N/2
-:tcp{n], N/2 < n :S N- 1

:t.,.[n] =

{ 0,

otherwise

Performing this operation gives

z.,.[n]

1
1
= -o(n1]- -o(n- 3]
2
2

Taking the DIT of z.,.{n] yields jXJ(k]. Using the DIT relation,
N-1

iXI(k]

=L

:.,.[n]W"•

=<>

we find

D=

jXJ(O]

= (0 + ~ + 0-

jX1[1]

=

(o- ~ +0-

jX1 (2J

=

(o+~+o-D =o

jXJ(3]

=

(o+~+O+n =i

n

0

= -j

Thus,

iX1[kJ = -io(k- 11 + io(k- 3]
11.20. As the following shows, the second condition implies :(OJ= 1.

:t(OJ.

= !

t

X(k]ei(>•l•l""l

61=0

•

= ~ LX(k]
1=0

=

1

......

. 476

This condition eliminates all choices except :t 2 (n] and :t3 [n].
The odd periodic· parts of :t 2 (n] and :t 3 (n] for n = 0, ... , 5 are

:to(n]- :t2[((-n))•]

:t., [n]

=
2
= !3 (6(n-. 4]- 6(((n + 4))6]) - !3 (6(n- 5]- 6[((n + 5)).])

:t..,(n]

=

:t 3 [n]- :t3[(( -n)) 6 ]
2
1
(6(n- 1]- 6(((n + 1))6 ])

=

3

-

1

3 (6(n- 2]- 6(((n + 2)) 6 ])

For n < 0 or n > 5, these sequences are zero. Since the transform of :top(n] is jXI(k] we find for
k = 0, ... ,5

jX,,(kJ

= ~ ( ,-i(2r/6)<•

jX,,[k]

~ ( e-j(2r/6)5> _ ei!2r/6).. )

~j sin(4rk/3) + ~j sin(5rk/3)

= -

=

_ ei(2r/6)<•) _

j_!_ (-6[k- 2] + 6[k- 4])

../3

=

~

=

-~jsin(rk/3) + ~jsin(2.. k/3)

=

j_!_(-6[k-2]+6[k-4])

(e-j(2r/6)• -ei(2r/6l•) _

~

(e-j(2r/6)2> -ei(2r/6)2>)

../3

Thus, both x 2 [n] and :t 3 [n] are consistent with the information given.
11.21.

(a) Method 1:
We are given

=

XR(peiw)

U(p,w)

= 1+p- 1 acosw
av we have
7ii = !p&!
'

Since au

av
8w

=

v =

Since ~

-ap-• sinw + K(p)

= - ~ ~ we have,
ap- 2 sinw +

K'(p) = ap- 2 sinw

Thus,

K'(p)
K(p)

= 0
= C

477
Since z[n] is real V(p,w) is an odd function of w. Hence, V(p,O)
Therefore,
x~w)

=
=
=
=
=

X(z)

= 0, implying that C = 0.

U(p,w) + jV(p,w)
1 + p- 1ocos""- jp- 1asinw
1 + op- 1(cosw- j sinw)
1 +op- 1•-;..
1 + oz- 1

(b) Method 2: Since XR(ei"') is the transform of z,[n] we have

=
=

XR(ei"')

z,(n] =

1 + QCOSW
Q

.

1 + -t!w

2

Q

.

+ -e-Jw
2

Q

Q

o(n] + 2o(n + 1] + 2o(n- 1]

Because x(n] is real and causal, we can recover z 0 (n] from z,[n] as follows

=

x 0 (n]

=

n>O
n=O
-z,[n], nz;z,[m]

= tl>•••• [m].

= { j,-j,

0 J

1/T

' '-

H;(ei"')
Note that H;(e'"')

'lf Jnl

-

(d) From comparing the top and bottom figures in the answer to part (a), it is evident that the desired
complex system response is given by:

H(,.;w)={ 1, -1rz- 1 ) + ~)og(1- b•z)
k=l

k=l

II;

- :E Jog{1 - •••-

N.

1

l-

bl

:E log{1 - d•• l
1=1

(d) Using the power series expansion

L:
oo

log(1- z) = -

n

n=1

we find

log{1- Ctz- 1)

=

a" -n
- f: nz

1•1 > lal

1

n=1

log{1- .Bz)

""pn

= -L-;-z",

1•1 > 1.8- 1 1

n=1
-1

=

p-n
-n
n z'
n.=-oo

:E

1•1 > 1.8- 1 1

From the equations above we can identify the following z-transform pain;
a"
n

--u[n -1] +-+ log(1- az- 1 ),

p-n

-u[-n- 1] +-+ log{1- .Bz),
n

1•1 > lal
1•1 > 1.8- 1 1

We~ now take the inverse transform o:f X(z).

log{ A),

i[n] =

n=O
n>O
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