Solution Manual For Digital Signal Processing, 4th Edition

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Chapter 11.1(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=0.01π2π=1200⇒periodic withNp= 200.(b)f=30π105(12π) =17⇒periodic withNp= 7.(c)f=3π2π=32⇒periodic withNp= 2.(d)f=32π⇒non-periodic.(e)f=62π10(12π) =3110⇒periodic withNp= 10.1.3(a) Periodic with periodTp=2π5.(b)f=52π⇒non-periodic.(c)f=112π⇒non-periodic.(d)cos(n8) is non-periodic;cos(πn8) is periodic; Their product is non-periodic.(e)cos(πn2) is periodic with periodNp=4sin(πn8) is periodic with periodNp=16cos(πn4+π3) is periodic with periodNp=8Therefore, x(n) is periodic with periodNp=16. (16 is the least common multiple of 4,8,16).1.4(a)w=2πkNimplies thatf=kN. Letα=GCD of (k, N),i.e.,k=k′α, N=N′α.Then,f=k′N′,which implies thatN′=Nα .3

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(b)N=7k=0 1 2 3 4 5 6 7GCD(k, N)=7 1 1 1 1 1 1 7Np=1 7 7 7 7 7 7 1(c)N=16k=0 1 2 3 4 5 6 7 8 9 10 11 12. . .16GCD(k, N)=16 1 2 1 4 1 2 1 8 1 2 1 4. . .16Np=1 6 8 16 4 16 8 16 2 16 8 16 4. . .11.5(a) Refer to fig 1.5-1(b)051015202530−3−2−10123−−−> t (ms)−−−> xa(t)Figure 1.5-1:x(n)=xa(nT)=xa(n/Fs)=3sin(πn/3)⇒f=12π(π3 )=16, Np= 64

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01020t (ms)3−3Figure 1.5-2:(c)Refer to fig 1.5-2x(n) ={0,3√2,3√2,0,−3√2,−3√2}, Np= 6.(d) Yes.x(1) = 3 = 3sin( 100πFs)⇒Fs= 200 samples/sec.1.6(a)x(n)=Acos(2πF0n/Fs+θ)=Acos(2π(T /Tp)n+θ)ButT /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= (kf T) =k(TpT)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) ForFs= 8kHz, Ffold =Fs/2 = 4kHz⇒5kHzwill alias to 3kHz.(c) F=9kHz will alias to 1kHz.1.8(a)Fmax = 100kHz, Fs≥2Fmax = 200Hz.(b)Ffold =Fs2= 125Hz.5

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1.9(a)Fmax = 360Hz, FN= 2Fmax = 720Hz.(b)Ffold =Fs2= 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=1800π2π=900HzFN=2Fmax = 1800Hz.(c)f1=600π2π( 1Fs)=0.3;f2=1800π2π( 1Fs)=0.9;Butf2=0.9>0.5⇒f2= 0.1.Hence,x(n)=3cos[(2π)(0.3)n] + 2cos[(2π)(0.1)n](d)△=xmax−xminm−1=5−(−5)1023=101023.1.11x(n)=xa(nT)=3cos(100πn200)+ 2sin(250πn200)6

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=3cos(πn2)−2sin(3πn4)T′=11000⇒ya(t) =x(t/T′)=3cos(π1000t2)−2sin(3π1000t4)ya(t)=3cos(500πt)−2sin(750πt)1.12(a) ForFs= 300Hz,x(n)=3cos(πn6)+ 10sin(πn)−cos(πn3)=3cos(πn6)−3cos(πn3)(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 +1270.02= 636⇒log2(636)⇒10 bit A/D.1.14R=(20 samplessec)×(8bitssample )=160 bitssecFfold=Fs2= 10Hz.Resolution=1volt28−1=0.004.1.15(a) Refer to fig 1.15-1. With a sampling frequency of 5kHz, the maximum frequency that can berepresented is 2.5kHz. Therefore, a frequency of 4.5kHz is aliased to 500Hz and the frequency of3kHz is aliased to 2kHz.7

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050100−1−0.500.51Fs = 5KHz, F0=500Hz050100−1−0.500.51Fs = 5KHz, F0=2000Hz050100−1−0.500.51Fs = 5KHz, F0=3000Hz050100−1−0.500.51Fs = 5KHz, F0=4500HzFigure 1.15-1:(b) Refer to fig 1.15-2.y(n) is a sinusoidal signal.By taking the even numbered samples, thesampling frequency is reduced to half i.e., 25kHz which is still greater than the nyquist rate. Thefrequency 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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0102030405060708090100−1−0.500.51F0 = 2KHz, Fs=50kHz05101520253035404550−1−0.500.51F0 = 2KHz, Fs=25kHzFigure 1.15-2:050100150200−1−0.500.51levels = 64, using truncation, SQNR = 31.3341dB−−> n−−> x(n)050100150200−1−0.500.51−−> n−−> xq(n)050100150200−0.04−0.03−0.02−0.010−−> n−−> e(n)Figure 1.16-1:9

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050100150200−1−0.500.51levels = 128, using truncation, SQNR = 37.359dB−−> n−−> x(n)050100150200−1−0.500.51−−> n−−> xq(n)050100150200−0.02−0.015−0.01−0.0050−−> n−−> e(n)Figure 1.16-2:050100150200−1−0.500.51levels = 256, using truncation, SQNR=43.7739dB−−> n−−> x(n)050100150200−1−0.500.51−−> n−−> xq(n)050100150200−8−6−4−20 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 1.16-6.050100150200−1−0.500.51levels = 64, using rounding, SQNR=32.754dB−−> n−−> x(n)050100150200−1−0.500.51−−> n−−> xq(n)050100150200−0.04−0.0200.020.04−−> n−−> e(n)Figure 1.16-4:11

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050100150200−1−0.500.51levels = 128, using rounding, SQNR=39.2008dB−−> n−−> x(n)050100150200−1−0.500.51−−> n−−> xq(n)050100150200−0.02−0.0100.010.02−−> n−−> e(n)Figure 1.16-5:050100150200−1−0.500.51levels = 256, using rounding, SQNR=44.0353dB−−> n−−> x(n)050100150200−1−0.500.51−−> n−−> xq(n)050100150200−0.01−0.00500.0050.01−−> 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 numberof quantization levels are increased.(d)levels64128256theoretical sqnr43.900049.920055.9400sqnr with truncation31.334137.35943.7739sqnr with rounding32.75439.200844.0353The theoretical sqnr is given in the table above. It can be seen that theoretical sqnr is muchhigher than those obtained by simulations. The decrease in the sqnr is because of the truncationand rounding.13

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Chapter 1010.1(a) To obtain the desired length of 25, a delay of25−12= 12 is incorporated intoHd(w). Hence,Hd(w)=1e−j12w,0≤ |w| ≤π6=0,otherwisehd(n)=12π∫π6−π6Hd(w)e−jwndw=sinπ6(n−12)π(n−12)Then,h(n)=hd(n)w(n)wherew(n) is a rectangular window of lengthN= 25.(b)H(w) =∑24n=0h(n)e−jwn⇒plot|H(w)|and6H(w). Refer to fig 10.1-1.(c) Hamming window:w(n)=(0.54−0.46cos nπ12 )h(n)=hd(n)w(n)for 0≤n≤24Refer to fig 10.1-2.303

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00.050.10.150.20.250.30.350.40.450.5−150−100−50050−−−> Freq(Hz)−−−> mag(dB)00.050.10.150.20.250.30.350.40.450.5−4−2024−−−> Freq(Hz)−−−> phaseFigure 10.1-1:00.050.10.150.20.250.30.350.40.450.5−150−100−50050−−−> Freq(Hz)−−−> mag(dB)00.050.10.150.20.250.30.350.40.450.5−4−2024−−−> Freq(Hz)−−−> phaseFigure 10.1-2:304

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(d) Bartlett window:w(n)=1−2(n−12)240≤n≤24Refer to fig 10.1-3.00.050.10.150.20.250.30.350.40.450.5−50−40−30−20−100−−−> Freq(Hz)−−−> mag(dB)00.050.10.150.20.250.30.350.40.450.5−4−2024−−−> Freq(Hz)−−−> phaseFigure 10.1-3:10.2(a)Hd(w)=1e−j12w,|w| ≤π6,π3≤ |w| ≤π=0, π6≤ |w| ≤π3hd(n)=12π∫π−πHd(w)e−jwndw=δ(n)−sinπ3(n−12)π(n−12)+sinπ6(n−12)π(n−12)(b) Rectangular window:w(n)=1,0≤n≤24=0,otherwise305

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