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Solution Manual For Digital Signal Processing, 4th Edition

Solution Manual For Digital Signal Processing, 4th Edition makes solving textbook exercises easier with step-by-step solutions and helpful tips.

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    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 π 2 π = 1 200 ⇒ periodic with N p = 200. (b) f = 30 π 105 ( 1 2 π ) = 1 7 ⇒ periodic with N p = 7. (c) f = 3 π 2 π = 3 2 ⇒ periodic with N p = 2. (d) f = 3 2 π ⇒ non-periodic. (e) f = 62 π 10 ( 1 2 π ) = 31 10 ⇒ periodic with N p = 10. 1.3 (a) Periodic with period T p = 2 π 5 . (b) f = 5 2 π ⇒ non-periodic. (c) f = 1 12 π ⇒ 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 N p =4 sin ( πn 8 ) is periodic with period N p =16 cos ( πn 4 + π 3 ) is periodic with period N p =8 Therefore, x(n) is periodic with period N p =16. (16 is the least common multiple of 4,8,16). 1.4 (a) w = 2 πk 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

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    (b) N = 7 k = 0 1 2 3 4 5 6 7 GCD( k, N ) = 7 1 1 1 1 1 1 7 N p = 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 N p = 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 ) = x a ( nT ) = x a ( n/F s ) = 3 sin ( πn/ 3) ⇒ f = 1 2 π ( π 3 ) = 1 6 , N p = 6 4

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    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 } , N p = 6. (d) Yes. x (1) = 3 = 3 sin ( 100 π F s ) ⇒ F s = 200 samples/sec. 1.6 (a) x ( n ) = Acos (2 πF 0 n/F s + θ ) = Acos (2 π ( T /T p ) n + θ ) But T /T p = 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, T d = ( k f T ) = k ( T p T ) T = kT p . Thus, it takes k periods ( kT p ) of the analog signal to make 1 period ( T d ) of the discrete signal. (c) T d = kT p ⇒ N T = kT p ⇒ f = k/N = T /T p ⇒ f is rational ⇒ x(n) is periodic. 1.7 (a) F max = 10 kHz ⇒ F s ≥ 2 F max = 20 kHz . (b) For F s = 8 kHz, F fold = F s / 2 = 4 kHz ⇒ 5 kHz will alias to 3kHz. (c) F=9kHz will alias to 1kHz. 1.8 (a) F max = 100 kHz, F s ≥ 2 F max = 200 Hz . (b) F fold = F s 2 = 125 Hz. 5

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    1.9 (a) F max = 360 Hz, F N = 2 F max = 720 Hz. (b) F fold = F s 2 = 300 Hz. (c) x ( n ) = x a ( nT ) = x a ( n/F s ) = sin (480 πn/ 600) + 3 sin (720 πn/ 600) x ( n ) = sin (4 πn/ 5) − 3 sin (4 πn/ 5) = − 2 sin (4 πn/ 5) . Therefore, w = 4 π/ 5. (d) y a ( t ) = x ( F s t ) = − 2 sin (480 πt ) . 1.10 (a) Number of bits/sample = log 2 1024 = 10 . F s = [10 , 000 bits/sec] [10 bits/sample] = 1000 samples/sec. F fold = 500 Hz. (b) F max = 1800 π 2 π = 900 Hz F N = 2 F max = 1800 Hz. (c) f 1 = 600 π 2 π ( 1 F s ) = 0 . 3; f 2 = 1800 π 2 π ( 1 F s ) = 0 . 9; But f 2 = 0 . 9 > 0 . 5 ⇒ f 2 = 0 . 1 . Hence, x ( n ) = 3 cos [(2 π )(0 . 3) n ] + 2 cos [(2 π )(0 . 1) n ] (d) △ = x max − x min m − 1 = 5 − ( − 5) 1023 = 10 1023 . 1.11 x ( n ) = x a ( nT ) = 3 cos ( 100 πn 200 ) + 2 sin ( 250 πn 200 ) 6

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    = 3 cos ( πn 2 ) − 2 sin ( 3 πn 4 ) T ′ = 1 1000 ⇒ y a ( t ) = x ( t/T ′ ) = 3 cos ( π 1000 t 2 ) − 2 sin ( 3 π 1000 t 4 ) y a ( t ) = 3 cos (500 πt ) − 2 sin (750 πt ) 1.12 (a) For F s = 300 Hz , x ( n ) = 3 cos ( πn 6 ) + 10 sin ( πn ) − cos ( πn 3 ) = 3 cos ( πn 6 ) − 3 cos ( πn 3 ) (b) x r ( t ) = 3 cos (10000 πt/ 6) − cos (10000 πt/ 3) 1.13 (a) Range = x max − x min = 12 . 7 . m = 1 + range △ = 127 + 1 = 128 ⇒ log 2 (128) = 7 bits. (b) m = 1 + 127 0 . 02 = 636 ⇒ log 2 (636) ⇒ 10 bit A/D. 1.14 R = (20 samples sec ) × (8 bits sample ) = 160 bits sec F fold = F s 2 = 10 Hz. Resolution = 1volt 2 8 − 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

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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. 8

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    0 10 20 30 40 50 60 70 80 90 100 −1 −0.5 0 0.5 1 F0 = 2KHz, Fs=50kHz 0 5 10 15 20 25 30 35 40 45 50 −1 −0.5 0 0.5 1 F0 = 2KHz, Fs=25kHz Figure 1.15-2: 0 50 100 150 200 −1 −0.5 0 0.5 1 levels = 64, using truncation, SQNR = 31.3341dB −−> n −−> x(n) 0 50 100 150 200 −1 −0.5 0 0.5 1 −−> n −−> xq(n) 0 50 100 150 200 −0.04 −0.03 −0.02 −0.01 0 −−> n −−> e(n) Figure 1.16-1: 9

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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: 10

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    (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. 0 50 100 150 200 −1 −0.5 0 0.5 1 levels = 64, using rounding, SQNR=32.754dB −−> n −−> x(n) 0 50 100 150 200 −1 −0.5 0 0.5 1 −−> n −−> xq(n) 0 50 100 150 200 −0.04 −0.02 0 0.02 0.04 −−> n −−> e(n) Figure 1.16-4: 11

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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: 12

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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. 13

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    14

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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 H d ( w ). Hence, H d ( w ) = 1 e − j 12 w , 0 ≤ | w | ≤ π 6 = 0 , otherwise h d ( n ) = 1 2 π ∫ π 6 − π 6 H d ( w ) e − jwn dw = sin π 6 ( n − 12) π ( n − 12) Then, h ( n ) = h d ( 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 . 46 cos nπ 12 ) h ( n ) = h d ( n ) w ( n ) for 0 ≤ n ≤ 24 Refer to fig 10.1-2. 303

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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: 304

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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) H d ( w ) = 1 e − j 12 w , | w | ≤ π 6 , π 3 ≤ | w | ≤ π = 0 , π 6 ≤ | w | ≤ π 3 h d ( n ) = 1 2 π ∫ π − π H d ( w ) e − jwn dw = 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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