Digital Signal Processing (4th Edition by Proakis & Manolakis) Solution

432 Pages·2011·2.38 MB·English

Chapter 1
1.1
(a) One dimensional, multichannel, discrete time, and digital.
(b) Multi dimensional, single channel, continuous-time, analog.
(c) One dimensional, single channel, continuous-time, analog.
(d) One dimensional, single channel, continuous-time, analog.
(e) One dimensional, multichannel, discrete-time, digital.
1.2
(a) f = 0.01π = 1 periodic with N = 200.
2π 200 ⇒ p
(b) f = 30π( 1 )= 1 periodic with N = 7.
105 2π 7 ⇒ p
(c) f = 3π = 3 periodic with N = 2.
2π 2 ⇒ p
(d) f = 3 non-periodic.
2π ⇒
(e) f = 62π( 1 )= 31 periodic with N = 10.
10 2π 10 ⇒ p
1.3
(a) Periodic with period T = 2π.
p 5
(b) f = 5 non-periodic.
2π ⇒
(c) f = 1 non-periodic.
12π ⇒
(d) cos(n) is non-periodic; cos(πn) is periodic; Their product is non-periodic.
8 8
(e) cos(πn) is periodic with period N =4
2 p
sin(πn) is periodic with period N =16
8 p
cos(πn + π) is periodic with period N =8
4 3 p
Therefore, x(n) is periodic with period N =16. (16 is the least common multiple of 4,8,16).
p
1.4
(a) w = 2πk implies that f = k. Let
N N
α= GCD of (k,N), i.e.,
k =k α,N =N α.
′ ′
Then,
k
′
f = , which implies that
N
′
N
N = .
′
α
3
(b)
N = 7
k = 01234567
GCD(k,N) = 71111117
N = 17777771
p
(c)
N = 16
k = 0123456789101112 ... 16
GCD(k,N) = 16121412181214 ... 16
N = 168164168162168164 ... 1
p
1.5
(a) Refer to fig 1.5-1
(b)
3
2
1
a(t)
> x 0
−
−
−
−1
−2
−3
0 5 10 15 20 25 30
−−−> t (ms)
Figure 1.5-1:
x(n) = x (nT)
a
= x (n/F )
a s
= 3sin(πn/3)
⇒
1 π
f = ( )
2π 3
1
= ,N =6
p
6
4
3
t (ms)
10
0 20
-3
Figure 1.5-2:
(c)Refer to fig 1.5-2
x(n)= 0, 3 , 3 ,0, 3 , 3 ,N =6.
√2 √2 −√2 −√2 p
(d) Yes.n o
100π
x(1)=3=3sin( ) F =200 samples/sec.
s
F ⇒
s
1.6
(a)
x(n) = Acos(2πF n/F +θ)
0 s
= Acos(2π(T/T )n+θ)
p
But T/T =f x(n) is periodic if f is rational.
p
⇒
(b) If x(n) is periodic, then f=k/N where N is the period. Then,
k T
p
T =( T)=k( )T =kT .
d p
f T
Thus, it takes k periods (kT ) of the analog signal to make 1 period (T ) of the discrete signal.
p d
(c) T =kT NT =kT f =k/N =T/T f is rational x(n) is periodic.
d p p p
⇒ ⇒ ⇒ ⇒
1.7
(a) Fmax =10kHz Fs 2Fmax =20kHz.
⇒ ≥
(b) For F =8kHz,F =F /2=4kHz 5kHz will alias to 3kHz.
s fold s
⇒
(c) F=9kHz will alias to 1kHz.
1.8
(a) Fmax =100kHz,Fs 2Fmax =200Hz.
≥
(b) F = Fs =125Hz.
fold 2
5
1.9
(a) Fmax =360Hz,FN =2Fmax =720Hz.
(b) F = Fs =300Hz.
fold 2
(c)
x(n) = x (nT)
a
= x (n/F )
a s
= sin(480πn/600)+3sin(720πn/600)
x(n) = sin(4πn/5) 3sin(4πn/5)
−
= 2sin(4πn/5).
−
Therefore, w =4π/5.
(d) y (t)=x(F t)= 2sin(480πt).
a s
−
1.10
(a)
Number of bits/sample = log 1024=10.
2
[10,000 bits/sec]
F =
s
[10 bits/sample]
= 1000 samples/sec.
F = 500Hz.
fold
(b)
1800π
Fmax =
2π
= 900Hz
FN = 2Fmax =1800Hz.
(c)
600π 1
f = ( )
1
2π F
s
= 0.3;
1800π 1
f = ( )
2
2π F
s
= 0.9;
But f = 0.9>0.5 f =0.1.
2 2
⇒
Hence, x(n) = 3cos[(2π)(0.3)n]+2cos[(2π)(0.1)n]
(d) = xmax−xmin = 5−(−5) = 10 .
△ m 1 1023 1023
−
1.11
x(n) = x (nT)
a
100πn 250πn
= 3cos +2sin
200 200
(cid:18) (cid:19) (cid:18) (cid:19)
6
πn 3πn
= 3cos 2sin
2 − 4
(cid:16) (cid:17) (cid:18) (cid:19)
1
T = y (t)=x(t/T )
′ a ′
1000 ⇒
π1000t 3π1000t
= 3cos 2sin
2 − 4
(cid:18) (cid:19) (cid:18) (cid:19)
y (t) = 3cos(500πt) 2sin(750πt)
a
−
1.12
(a) For F =300Hz,
s
πn πn
x(n) = 3cos +10sin(πn) cos
6 − 3
(cid:16)πn(cid:17) πn (cid:16) (cid:17)
= 3cos 3cos
6 − 3
(cid:16) (cid:17) (cid:16) (cid:17)
(b) x (t)=3cos(10000πt/6) cos(10000πt/3)
r
−
1.13
(a)
Range = xmax xmin =12.7.
−
range
m = 1+
△
= 127+1=128 log (128)
2
⇒
= 7 bits.
(b) m=1+ 127 =636 log (636) 10 bit A/D.
0.02 ⇒ 2 ⇒
1.14
samples bits
R = (20 ) (8 )
sec × sample
bits
= 160
sec
F
F = s =10Hz.
fold
2
1volt
Resolution =
28 1
−
= 0.004.
1.15
(a) Refer to fig 1.15-1. With a sampling frequency of 5kHz, the maximum frequency that can be
represented is 2.5kHz. Therefore, a frequency of 4.5kHz is aliased to 500Hz and the frequency of
3kHz is aliased to 2kHz.
7
Fs = 5KHz, F0=500Hz Fs = 5KHz, F0=2000Hz
1 1
0.5 0.5
0 0
−0.5 −0.5
−1 −1
0 50 100 0 50 100
Fs = 5KHz, F0=3000Hz Fs = 5KHz, F0=4500Hz
1 1
0.5 0.5
0 0
−0.5 −0.5
−1 −1
0 50 100 0 50 100
Figure 1.15-1:
(b) Refer to fig 1.15-2. y(n) is a sinusoidal signal. By taking the even numbered samples, the
samplingfrequencyisreducedtohalfi.e.,25kHzwhichisstillgreaterthanthenyquistrate. The
frequency of the downsampled signal is 2kHz.
1.16
(a) for levels = 64, using truncation refer to fig 1.16-1.
for levels = 128, using truncation refer to fig 1.16-2.
for levels = 256, using truncation refer to fig 1.16-3.
8
F0 = 2KHz, Fs=50kHz
1
0.5
0
−0.5
−1
0 10 20 30 40 50 60 70 80 90 100
F0 = 2KHz, Fs=25kHz
1
0.5
0
−0.5
−1
0 5 10 15 20 25 30 35 40 45 50
Figure 1.15-2:
levels = 64, using truncation, SQNR = 31.3341dB
1 1
0.5 0.5
−−> x(n) 0 −−> xq(n) 0
−0.5 −0.5
−1 −1
0 50 100 150 200 0 50 100 150 200
−−> n −−> n
0
−0.01
n)
> e(−0.02
−
−
−0.03
−0.04
0 50 100 150 200
−−> n
Figure 1.16-1:
9
levels = 128, using truncation, SQNR = 37.359dB
1 1
0.5 0.5
−−> x(n) 0 −−> xq(n) 0
−0.5 −0.5
−1 −1
0 50 100 150 200 0 50 100 150 200
−−> n −−> n
0
−0.005
n)
> e( −0.01
−
−
−0.015
−0.02
0 50 100 150 200
−−> n
Figure 1.16-2:
levels = 256, using truncation, SQNR=43.7739dB
1 1
0.5 0.5
−−> x(n) 0 −−> xq(n) 0
−0.5 −0.5
−1 −1
0 50 100 150 200 0 50 100 150 200
−−> n −−> n
−3
x 10
0
−2
n)
> e(−4
−
−
−6
−8
0 50 100 150 200
−−> n
Figure 1.16-3:
10
(b) for levels = 64, using rounding refer to fig 1.16-4.
for levels = 128, using rounding refer to fig 1.16-5.
for levels = 256, using rounding refer to fig 1.16-6.
levels = 64, using rounding, SQNR=32.754dB
1 1
0.5 0.5
−−> x(n) 0 −−> xq(n) 0
−0.5 −0.5
−1 −1
0 50 100 150 200 0 50 100 150 200
−−> n −−> n
0.04
0.02
n)
> e( 0
−
−
−0.02
−0.04
0 50 100 150 200
−−> n
Figure 1.16-4:
11