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

432 Pages·2011·2.38 MB·English
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Preview Digital Signal Processing (4th Edition by Proakis & Manolakis) Solution

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

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