Exam (elaborations) TEST BANK FOR Digital Signal Processing 4th Edition by J. Proakis and D. Manolakis (Instructor Solution Manual)

(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 = 0.01 2 = 1 200 ⇒ periodic with Np = 200. (b) f = 30 105 ( 1 2 ) = 1 7 ⇒ periodic with Np = 7. (c) f = 3 2 = 3 2 ⇒ periodic with Np = 2. (d) f = 3 2 ⇒ non-periodic. (e) f = 62 10 ( 1 2 ) = 31 10 ⇒ periodic with Np = 10. 1.3 (a) Periodic with period Tp = 2 5 . (b) f = 5 2 ⇒ non-periodic. (c) f = 1 12 ⇒ non-periodic. (d) cos(n 8 ) is non-periodic; cos( n 8 ) is periodic; Their product is non-periodic. (e) cos( n 2 ) is periodic with period Np=4 sin( n 8 ) is periodic with period Np=16 cos( n 4 +  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 = 2k 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 8 16 4 . . . 1 1.5 (a) Refer to fig 1.5-1 (b) 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) = n 0, 3 √2 , 3 √2 , 0,− 3 √2 ,− 3 √2 o ,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 ⇒ NT = 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 ) =