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
2π = 1
200 ⇒ periodic with Np = 200.
(b) f = 30π
105 ( 1
2π ) = 1
7 ⇒ periodic with Np = 7.
(c) f = 3π
2π = 3
2 ⇒ periodic with Np = 2.
(d) f = 3
2π ⇒ non-periodic.
(e) f = 62π
10 ( 1
2π ) = 31
10 ⇒ periodic with Np = 10.
1.3
(a) Periodic with period Tp = 2π
5 .
(b) f = 5
2π ⇒ non-periodic.
(c) f = 1
12π ⇒ non-periodic.
(d) cos( n
8 ) is non-periodic; cos( πn
8 ) is periodic; Their product is non-periodic.
(e) cos( πn
2 ) is periodic with period Np=4
sin( πn
8 ) is periodic with period Np=16
cos( πn
4 + π
3 ) is periodic with period Np=8
Therefore, x(n) is periodic with period Np=16. (16 is the least common multiple of 4,8,16).
1.4
(a) w = 2πk
N implies that f = k
N . Let
α = GCD of (k, N ), i.e.,
k = k′α, N = N ′α.
Then,
f = k′
N ′ , which implies that
N ′ = N
α .
3

(b)
N = 7
k = 0 1 2 3 4 5 6 7
GCD(k, N ) = 7 1 1 1 1 1 1 7
Np = 1 7 7 7 7 7 7 1
(c)
N = 16
k = 0 1 2 3 4 5 6 7 8 9 10 11 12 . . . 16
GCD(k, N ) = 16 1 2 1 4 1 2 1 8 1 2 1 4 . . . 16
Np = 1 6 8 16 4 16 8 16 2 16 8 16 4 . . . 1
1.5
(a) Refer to fig 1.5-1
(b)
0 5 10 15 20 25 30
−3
−2
−1
0
1
2
3
−−−> t (ms)
−−−> xa(t)
Figure 1.5-1:
x(n) = xa(nT )
= xa(n/Fs)
= 3sin(πn/3) ⇒
f = 1
2π ( π
3 )
= 1
6 , Np = 6
4

0 10 20 t (ms)
3
−3
Figure 1.5-2:
(c)Refer to fig 1.5-2
x(n) =
{
0, 3√2 , 3√2 , 0, − 3√2 , − 3√2
}
, Np = 6.
(d) Yes.
x(1) = 3 = 3sin( 100π
Fs
) ⇒ Fs = 200 samples/sec.
1.6
(a)
x(n) = Acos(2πF0n/Fs + θ)
= Acos(2π(T /Tp)n + θ)
But T /Tp = f ⇒ x(n) is periodic if f is rational.
(b) If x(n) is periodic, then f=k/N where N is the period. Then,
Td = ( k
f T ) = k( Tp
T )T = kTp.
Thus, it takes k periods (kTp) of the analog signal to make 1 period (Td) of the discrete signal.
(c) Td = kTp ⇒ N T = kTp ⇒ f = k/N = T /Tp ⇒ f is rational ⇒ x(n) is periodic.
1.7
(a) Fmax = 10kHz ⇒ Fs ≥ 2Fmax = 20kHz.
(b) For Fs = 8kHz, Ffold = Fs/2 = 4kHz ⇒ 5kHz will alias to 3kHz.
(c) F=9kHz will alias to 1kHz.
1.8
(a) Fmax = 100kHz, Fs ≥ 2Fmax = 200Hz.
(b) Ffold = Fs
2 = 125Hz.
5

1.9
(a) Fmax = 360Hz, FN = 2Fmax = 720Hz.
(b) Ffold = Fs
2 = 300Hz.
(c)
x(n) = xa(nT )
= xa(n/Fs)
= sin(480πn/600) + 3sin(720πn/600)
x(n) = sin(4πn/5) − 3sin(4πn/5)
= −2sin(4πn/5).
Therefore, w = 4π/5.
(d) ya(t) = x(Fst) = −2sin(480πt).
1.10
(a)
Number of bits/sample = log21024 = 10.
Fs = [10, 000 bits/sec]
[10 bits/sample]
= 1000 samples/sec.
Ffold = 500Hz.
(b)
Fmax = 1800π
2π
= 900Hz
FN = 2Fmax = 1800Hz.
(c)
f1 = 600π
2π ( 1
Fs
)
= 0.3;
f2 = 1800π
2π ( 1
Fs
)
= 0.9;
But f2 = 0.9 > 0.5 ⇒ f2 = 0.1.
Hence, x(n) = 3cos[(2π)(0.3)n] + 2cos[(2π)(0.1)n]
(d) △ = xmax−xmin
m−1 = 5−(−5)
1023 = 10
1023 .
1.11
x(n) = xa(nT )
= 3cos
( 100πn
200
)
+ 2sin
( 250πn
200
)
6

= 3cos
( πn
2
)
− 2sin
( 3πn
4
)
T ′ = 1
1000 ⇒ ya(t) = x(t/T ′)
= 3cos
( π1000t
2
)
− 2sin
( 3π1000t
4
)
ya(t) = 3cos(500πt) − 2sin(750πt)
1.12
(a) For Fs = 300Hz,
x(n) = 3cos
( πn
6
)
+ 10sin(πn) − cos
( πn
3
)
= 3cos
( πn
6
)
− 3cos
( πn
3
)
(b) xr (t) = 3cos(10000πt/6) − cos(10000πt/3)
1.13
(a)
Range = xmax − xmin = 12.7.
m = 1 + range
△
= 127 + 1 = 128 ⇒ log2(128)
= 7 bits.
(b) m = 1 + 127
0.02 = 636 ⇒ log2(636) ⇒ 10 bit A/D.
1.14
R = (20 samples
sec ) × (8 bits
sample )
= 160 bits
sec
Ffold = Fs
2 = 10Hz.
Resolution = 1volt
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.
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0 50 100
−1
−0.5
0
0.5
1
Fs = 5KHz, F0=500Hz
0 50 100
−1
−0.5
0
0.5
1
Fs = 5KHz, F0=2000Hz
0 50 100
−1
−0.5
0
0.5
1
Fs = 5KHz, F0=3000Hz
0 50 100
−1
−0.5
0
0.5
1
Fs = 5KHz, F0=4500Hz
Figure 1.15-1:
(b) Refer to fig 1.15-2. y(n) is a sinusoidal signal. By taking the even numbered samples, the
sampling frequency is reduced to half i.e., 25kHz which is still greater than the nyquist rate. 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.
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0 50 100 150 200
−1
−0.5
0
0.5
1
levels = 128, using truncation, SQNR = 37.359dB
−−> n
−−> x(n)
0 50 100 150 200
−1
−0.5
0
0.5
1
−−> n
−−> xq(n)
0 50 100 150 200
−0.02
−0.015
−0.01
−0.005
0
−−> n
−−> e(n)
Figure 1.16-2:
0 50 100 150 200
−1
−0.5
0
0.5
1
levels = 256, using truncation, SQNR=43.7739dB
−−> n
−−> x(n)
0 50 100 150 200
−1
−0.5
0
0.5
1
−−> n
−−> xq(n)
0 50 100 150 200
−8
−6
−4
−2
0 x 10−3
−−> n
−−> e(n)
Figure 1.16-3:
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0 50 100 150 200
−1
−0.5
0
0.5
1
levels = 128, using rounding, SQNR=39.2008dB
−−> n
−−> x(n)
0 50 100 150 200
−1
−0.5
0
0.5
1
−−> n
−−> xq(n)
0 50 100 150 200
−0.02
−0.01
0
0.01
0.02
−−> n
−−> e(n)
Figure 1.16-5:
0 50 100 150 200
−1
−0.5
0
0.5
1
levels = 256, using rounding, SQNR=44.0353dB
−−> n
−−> x(n)
0 50 100 150 200
−1
−0.5
0
0.5
1
−−> n
−−> xq(n)
0 50 100 150 200
−0.01
−0.005
0
0.005
0.01
−−> n
−−> e(n)
Figure 1.16-6:
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(c) The sqnr with rounding is greater than with truncation. But the sqnr improves as the number
of quantization levels are increased.
(d)
levels 64 128 256
theoretical sqnr 43.9000 49.9200 55.9400
sqnr with truncation 31.3341 37.359 43.7739
sqnr with rounding 32.754 39.2008 44.0353
The theoretical sqnr is given in the table above. It can be seen that theoretical sqnr is much
higher than those obtained by simulations. The decrease in the sqnr is because of the truncation
and rounding.
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Chapter 10
10.1
(a) To obtain the desired length of 25, a delay of 25−1
2 = 12 is incorporated into Hd(w). Hence,
Hd(w) = 1e−j12w, 0 ≤ |w| ≤ π
6
= 0, otherwise
hd(n) = 1
2π
∫ π
6
− π
6
Hd(w)e−jwndw
= sin π
6 (n − 12)
π(n − 12)
Then, h(n) = hd(n)w(n)
where w(n) is a rectangular window of length N = 25.
(b)H(w) = ∑24
n=0 h(n)e−jwn ⇒ plot |H(w)| and 6 H(w). Refer to fig 10.1-1.
(c) Hamming window:
w(n) = (0.54 − 0.46cos nπ
12 )
h(n) = hd(n)w(n) for 0 ≤ n ≤ 24
Refer to fig 10.1-2.
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0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5
−150
−100
−50
0
50
−−−> Freq(Hz)
−−−> mag(dB)
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5
−4
−2
0
2
4
−−−> Freq(Hz)
−−−> phase
Figure 10.1-1:
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5
−150
−100
−50
0
50
−−−> Freq(Hz)
−−−> mag(dB)
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5
−4
−2
0
2
4
−−−> Freq(Hz)
−−−> phase
Figure 10.1-2:
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(d) Bartlett window:
w(n) = 1 − 2(n − 12)
24 0 ≤ n ≤ 24
Refer to fig 10.1-3.
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5
−50
−40
−30
−20
−10
0
−−−> Freq(Hz)
−−−> mag(dB)
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5
−4
−2
0
2
4
−−−> Freq(Hz)
−−−> phase
Figure 10.1-3:
10.2
(a)
Hd(w) = 1e−j12w, |w| ≤ π
6 , π
3 ≤ |w| ≤ π
= 0, π
6 ≤ |w| ≤ π
3
hd(n) = 1
2π
∫ π
−π
Hd(w)e−jwndw
= δ(n) − sin π
3 (n − 12)
π(n − 12) + sin π
6 (n − 12)
π(n − 12)
(b) Rectangular window:
w(n) = 1, 0 ≤ n ≤ 24
= 0, otherwise
305

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Solution Manual For Digital Signal Processing, 4th Edition

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