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

Exam (elaborations) TEST BANK FOR Digital Signal Processing 4th Edition by J. Proakis and D. Manolakis (Instructor Solution Manual) 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 = 0.01 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 ) = 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. 7 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. 8 100 −1 −0.5 0 0.5 1 F0 = 2KHz, Fs=50kHz 50 −1 −0.5 0 0.5 1 F0 = 2KHz, Fs=25kHz Figure 1.15-2: −1 −0.5 0 0.5 1 levels = 64, using truncation, SQNR = 31.3341dB −− n −− x(n) −1 −0.5 0 0.5 1 −− n −− xq(n) −0.04 −0.03 −0.02 −0.01 0 −− n −− e(n) Figure 1.16-1: 9 −1 −0.5 0 0.5 1 levels = 128, using truncation, SQNR = 37.359dB −− n −− x(n) −1 −0.5 0 0.5 1 −− n −− xq(n) −0.02 −0.015 −0.01 −0.005 0 −− n −− e(n) Figure 1.16-2: −1 −0.5 0 0.5 1 levels = 256, using truncation, SQNR=43.7739dB −− n −− x(n) −1 −0.5 0 0.5 1 −− n −− xq(n) −8 −6 −4 −2 0 x 10−3 −− n −− e(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. −1 −0.5 0 0.5 1 levels = 64, using rounding, SQNR=32.754dB −− n −− x(n) −1 −0.5 0 0.5 1 −− n −− xq(n) −0.04 −0.02 0 0.02 0.04 −− n −− e(n) Figure 1.16-4: 11 −1 −0.5 0 0.5 1 levels = 128, using rounding, SQNR=39.2008dB −− n −− x(n) −1 −0.5 0 0.5 1 −− n −− xq(n) −0.02 −0.01 0 0.01 0.02 −− n −− e(n) Figure 1.16-5: −1 −0.5 0 0.5 1 levels = 256, using rounding, SQNR=44.0353dB −− n −− x(n) −1 −0.5 0 0.5 1 −− n −− xq(n) −0.01 −0.005 0 0.005 0.01 −− n −− e(n) Figure 1.16-6: 12 (c) The sqnr with rounding is greater than with truncation. But the sqnr improves as the number of quantization levels are increased. (d) levels 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. 13 14 Chapter 2 2.1 (a) x(n) =  . . . 0, 1 3 , 2 3 ,1↑ , 1, 1, 1, 0, . . .  . Refer to fig 2.1-1. (b) After folding s(n) we have -3 -2 -1 0 1 2 3 4 1 1 1 1 1/3 2/3 Figure 2.1-1: x(−n) =  . . . 0, 1, 1, 1,1↑ , 2 3 , 1 3 , 0, . . .  . After delaying the folded signal by 4 samples, we have x(−n + 4) =  . . . 0,0↑ , 1, 1, 1, 1, 2 3 , 1 3 , 0, . . .  . On the other hand, if we delay x(n) by 4 samples we have x(n − 4) =  . . .0↑ , 0, 1 3 , 2 3 , 1, 1, 1, 1, 0, . . .  . Now, if we fold x(n − 4) we have x(−n − 4) =  . . . 0, 1, 1, 1, 1, 2 3 , 1 3 , 0,0↑ , . . .  15 (c) x(−n + 4) =  . . .0↑ , 1, 1, 1, 1, 2 3 , 1 3 , 0, . . .  (d) To obtain x(−n + k), first we fold x(n). This yields x(−n). Then, we shift x(−n) by k samples to the right if k 0, or k samples to the left if k 0. (e) Yes. x(n) = 1 3 δ(n − 2) + 2 3 δ(n + 1) + u(n) − u(n − 4) 2.2 x(n) =  . . . 0, 1,1↑ , 1, 1, 1 2 , 1 2 , 0, . . .  (a) x(n − 2) =  . . . 0,0↑ , 1, 1, 1, 1, 1 2 , 1 2 , 0, . . .  (b) x(4 − n) =  . . . 0, 1 2↑ , 1 2 , 1, 1, 1, 1, 0, . . .   (see 2.1(d)) (c) x(n + 2) =  . . . 0, 1, 1, 1,1↑ , 1 2 , 1 2 , 0, . . .  (d) x(n)u(2 − n) =  . . . 0, 1,1↑ , 1, 1, 0, 0, . . .  (e) x(n − 1)δ(n − 3) =  . . .0↑ , 0, 1, 0, . . .  (f) x(n2) = {. . . 0, x(4), x(1), x(0), x(1), x(4), 0, . . .} =  . . . 0, 1 2 , 1,1↑ , 1, 1 2 , 0, . . .  (g) xe(n) = x(n) + x(−n) 2 , x(−n) =  . . . 0, 1 2 , 1 2 , 1, 1,1↑ , 1, 0, . . .  =  . . . 0, 1 4 , 1 4 , 1 2 , 1, 1, 1, 1 2 , 1 4 , 1 4 , 0, . . .  16 (h) xo(n) = x(n) − x(−n) 2 =  . . . 0,− 1 4 ,− 1 4 ,− 1 2 , 0, 0, 0, 1 2 , 1 4 , 1 4 , 0, . . .  2.3 (a) u(n) − u(n − 1) = δ(n) =   0, n 0 1, n = 0 0, n 0 (b) Xn k=−∞ δ(k) = u(n) =  0, n 0 1, n ≥ 0 ∞X k=0 δ(n − k) =  0, n 0 1, n ≥ 0 2.4 Let xe(n) = 1 2 [x(n) + x(−n)], xo(n) = 1 2 [x(n) − x(−n)]. Since xe(−n) = xe(n) and xo(−n) = −xo(n), it follows that x(n) = xe(n) + xo(n). The decomposition is unique. For x(n) =  2, 3,4↑ , 5, 6  , we have xe(n) =  4, 4,4↑ , 4, 4  and xo(n) =  −2,−1,0↑ , 1, 2  . 17 2.5 First, we prove that ∞X n=−∞ xe(n)xo(n) = 0 ∞X n=−∞ xe(n)xo(n) = ∞X m=−∞ xe(−m)xo(−m) = − ∞X m=−∞ xe(m)xo(m) = − ∞X n=−∞ xe(n)xo(n) = ∞X n=−∞ xe(n)xo(n) = 0 Then, ∞X n=−∞ x2(n) = ∞X n=−∞ [xe(n) + xo(n)]2 = ∞X n=−∞ x2e(n) + ∞X n=−∞ x2o(n) + ∞X n=−∞ 2xe(n)xo(n) = Ee + Eo 2.6 (a) No, the system is time variant. Proof: If x(n) → y(n) = x(n2) x(n − k) → y1(n) = x[(n − k)2] = x(n2 + k2 − 2nk) 6= y(n − k) (b) (1) x(n) =  0,1↑ , 1, 1, 1, 0, . . .  (2) y(n) = x(n2) =  . . . , 0, 1,1↑ , 1, 0, . . .  (3) y(n − 2) =  . . . , 0,0↑ , 1, 1, 1, 0, . . .  18 (4) x(n − 2) =  . . . ,0↑ , 0, 1, 1, 1, 1, 0, . . .  (5) y2(n) = T [x(n − 2)] =  . . . , 0, 1, 0,0↑ , 0, 1, 0, . . .  (6) y2(n) 6= y(n − 2) ⇒ system is time variant. (c) (1) x(n) =  1↑ , 1, 1, 1  (2) y(n) =  1↑ , 0, 0, 0, 0,−1  (3) y(n − 2) =  0↑ , 0, 1, 0, 0, 0, 0,−1  (4) x(n − 2) =  0↑ , 0, 1, 1, 1, 1, 1  (5) y2(n) =  0↑ , 0, 1, 0, 0, 0, 0,−1  (6) y2(n) = y(n − 2). The system is time invariant, but this example alone does not constitute a proof. (d) (1) y(n) = nx(n), x(n) =  . . . , 0,1↑ , 1, 1, 1, 0, . . .  (2) y(n) =  . . . ,0↑ , 1, 2, 3, . . .  (3) y(n − 2) =  . . . ,0↑ , 0, 0, 1, 2, 3, . . .  (4) x(n − 2) =  . . . , 0,0↑ , 0, 1, 1, 1, 1, . . .  19 (5) y2(n) = T [x(n − 2)] = {. . . , 0, 0, 2, 3, 4, 5, . . .} (6) y2(n) 6= y(n − 2) ⇒ the system is time variant. 2.7 (a) Static, nonlinear, time invariant, causal, stable. (b) Dynamic, linear, time invariant, noncausal and unstable. The latter is easily proved. For the bounded input x(k) = u(k), the output becomes y(n) = nX+1 k=−∞ u(k) =  0, n −1 n + 2, n ≥ −1 since y(n) → ∞ as n → ∞, the system is unstable. (c) Static, linear, timevariant, causal, stable. (d) Dynamic, linear, time invariant, noncausal, stable. (e) Static, nonlinear, time invariant, causal, stable. (f) Static, nonlinear, time invariant, causal, stable. (g) Static, nonlinear, time invariant, causal, stable. (h) Static, linear, time invariant, causal, stable. (i) Dynamic, linear, time variant, noncausal, unstable. Note that the bounded input x(n) = u(n) produces an unbounded output. (j) Dynamic, linear, time variant, noncausal, stable. (k) Static, nonlinear, time invariant, causal, stable. (l) Dynamic, linear, time invariant, noncausal, stable. (m) Static, nonlinear, time invariant, causal, stable. (n) Static, linear, time invariant, causal, stable. 2.8 (a) True. If v1(n) = T1[x1(n)] and v2(n) = T1[x2(n)], then α1x1(n) + α2x2(n) yields α1v1(n) + α2v2(n) by the linearity property of T1. Similarly, if y1(n) = T2[v1(n)] and y2(n) = T2[v2(n)], then β1v1(n) + β2v2(n) → y(n) = β1y1(n) + β2y2(n) by the linearity property of T2. Since v1(n) = T1[x1(n)] and 20 v2(n) = T2[x2(n)], it follows that A1x1(n) + A2x2(n) yields the output A1T [x1(n)] + A2T [x2(n)], where T = T1T2. Hence T is linear. (b) True. For T1, if x(n) → v(n) and x(n − k) → v(n − k), For T2, if v(n) → y(n) andv(n − k) → y(n − k). Hence, For T1T2, if x(n) → y(n) and x(n − k) → y(n − k) Therefore, T = T1T2 is time invariant. (c) True. T1 is causal⇒ v(n) depends only on x(k) for k ≤ n. T2 is causal ⇒ y(n) depends only on v(k) for k ≤ n. Therefore, y(n) depends only on x(k) for k ≤ n. Hence, T is causal. (d) True. Combine (a) and (b). (e) True. This follows from h1(n) ∗ h2(n) = h2(n) ∗ h1(n) (f) False. For example, consider T1 : y(n) = nx(n) and T2 : y(n) = nx(n + 1). Then, T2[T1[δ(n)]] = T2(0) = 0. T1[T2[δ(n)]] = T1[δ(n + 1)] = −δ(n + 1) 6= 0. (g) False. For example, consider T1 : y(n) = x(n) + b and T2 : y(n) = x(n) − b, where b 6= 0. Then, T [x(n)] = T2[T1[x(n)]] = T2[x(n) + b] = x(n). Hence T is linear. (h) True. T1 is stable ⇒ v(n) is bounded if x(n) is bounded. T2 is stable ⇒ y(n) is bounded if v(n) is bounded . 21 Hence, y(n) is bounded if x(n) is bounded ⇒ T = T1T2 is stable. (i) Inverse of (c). T1 and for T2 are noncausal ⇒ T is noncausal. Example: T1 : y(n) = x(n + 1) and T2 : y(n) = x(n − 2) ⇒ T : y(n) = x(n − 1), which is causal. Hence, the inverse of (c) is false. Inverse of (h): T1 and/or T2 is unstable, implies T is unstable. Example: T1 : y(n) = ex(n), stable and T2 : y(n) = ln[x(n)], which is unstable. But T : y(n) = x(n), which is stable. Hence, the inverse of (h) is false. 2.9 (a) y(n) = Xn k=−∞ h(k)x(n − k), x(n) = 0, n 0 y(n + N) = nX+N k=−∞ h(k)x(n + N − k) = nX+N k=−∞ h(k)x(n − k) = Xn k=−∞ h(k)x(n − k) + nX+N k=n+1 h(k)x(n − k) = y(n) + nX+N k=n+1 h(k)x(n − k) For a BIBO system, limn→∞|h(n)| = 0. Therefore, limn→∞ nX+N k=n+1 h(k)x(n − k) = 0 and limn→∞y(n + N) = y(N). (b) Let x(n) = xo(n) + au(n), where a is a constant and xo(n) is a bounded signal with lim n→∞ xo(n) = 0. Then, y(n) = a ∞X k=0 h(k)u(n − k) + ∞X k=0 h(k)xo(n − k) = a Xn k=0 h(k) + yo(n) clearly, P n x2o(n) ∞ ⇒ P n y2 o(n) ∞ (from (c) below) Hence, limn→∞|yo(n)| = 0. 22 and, thus, limn→∞y(n) = a Pn k=0 h(k) = constant. (c) y(n) = X k h(k)x(n − k) ∞X −∞ y2(n) = ∞X −∞ " X k h(k)x(n − k) #2 = X k X l h(k)h(l) X n x(n − k)x(n − l) But X n x(n − k)x(n − l) ≤ X n x2(n) = Ex. Therefore, X n y2(n) ≤ Ex X k |h(k)| X l |h(l)|. For a BIBO stable system, X k |h(k)| M. Hence, Ey ≤ M2Ex, so that Ey 0 if Ex 0. 2.10 The system is nonlinear. This is evident from observation of the pairs x3(n) ↔ y3(n) and x2(n) ↔ y2(n). If the system were linear, y2(n) would be of the form y2(n) = {3, 6, 3} because the system is time-invariant. However, this is not the case. 2.11 since x1(n) + x2(n) = δ(n) and the system is linear, the impulse response of the system is y1(n) + y2(n) =  0,3↑ ,−1, 2, 1  . If the system were time invariant, the response to x3(n) would be  3↑ , 2, 1, 3, 1  . But this is not the case. 23 2.12 (a) Any weighted linear combination of the signals xi(