CramX Logo
Solution Manual For Digital Signal Processing, 4th Edition - Document preview page 1

Solution Manual For Digital Signal Processing, 4th Edition - Page 1

Document preview content for Solution Manual For Digital Signal Processing, 4th Edition

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.

Sarah Anderson
Contributor
4.2
0
12 months ago
Preview (16 of 427 Pages)
100%
Log in to unlock
Page 1 of 16
Solution Manual For Digital Signal Processing, 4th Edition - Page 1 preview imageChapter 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=1200periodic withNp= 200.(b)f=30π105(12π) =17periodic withNp= 7.(c)f=3π2π=32periodic withNp= 2.(d)f=32πnon-periodic.(e)f=62π10(12π) =3110periodic withNp= 10.1.3(a) Periodic with periodTp=2π5.(b)f=52πnon-periodic.(c)f=112πnon-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=kα, N=Nα.Then,f=kN,which implies thatN=Nα .3
Page 2 of 16
Solution Manual For Digital Signal Processing, 4th Edition - Page 2 preview image
Page 3 of 16
Solution Manual For Digital Signal Processing, 4th Edition - Page 3 preview image(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
Page 4 of 16
Solution Manual For Digital Signal Processing, 4th Edition - Page 4 preview image01020t (ms)3−3Figure 1.5-2:(c)Refer to fig 1.5-2x(n) ={0,32,32,0,32,32}, 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=fx(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=kTpN T=kTpf=k/N=T /Tpf is rationalx(n) is periodic.1.7(a)Fmax = 10kHzFs2Fmax = 20kHz.(b) ForFs= 8kHz, Ffold =Fs/2 = 4kHz5kHzwill alias to 3kHz.(c) F=9kHz will alias to 1kHz.1.8(a)Fmax = 100kHz, Fs2Fmax = 200Hz.(b)Ffold =Fs2= 125Hz.5
Page 5 of 16
Solution Manual For Digital Signal Processing, 4th Edition - Page 5 preview image1.9(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>0.5f2= 0.1.Hence,x(n)=3cos[(2π)(0.3)n] + 2cos[(2π)(0.1)n](d)=xmaxxminm1=5(5)1023=101023.1.11x(n)=xa(nT)=3cos(100πn200)+ 2sin(250πn200)6
Page 6 of 16
Solution Manual For Digital Signal Processing, 4th Edition - Page 6 preview image=3cos(πn2)2sin(3πn4)T=11000ya(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=xmaxxmin = 12.7.m=1 + range=127 + 1 = 128log2(128)=7 bits.(b)m= 1 +1270.02= 636log2(636)10 bit A/D.1.14R=(20 samplessec)×(8bitssample )=160 bitssecFfold=Fs2= 10Hz.Resolution=1volt281=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
Page 7 of 16
Solution Manual For Digital Signal Processing, 4th Edition - Page 7 preview image050100−1−0.500.51Fs = 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
Page 8 of 16
Solution Manual For Digital Signal Processing, 4th Edition - Page 8 preview image0102030405060708090100−1−0.500.51F0 = 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
Page 9 of 16
Solution Manual For Digital Signal Processing, 4th Edition - Page 9 preview image050100150200−1−0.500.51levels = 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
Page 10 of 16
Solution Manual For Digital Signal Processing, 4th Edition - Page 10 preview image(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
Page 11 of 16
Solution Manual For Digital Signal Processing, 4th Edition - Page 11 preview image050100150200−1−0.500.51levels = 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
Page 12 of 16
Solution Manual For Digital Signal Processing, 4th Edition - Page 12 preview image(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
Page 13 of 16
Solution Manual For Digital Signal Processing, 4th Edition - Page 13 preview image14
Page 14 of 16
Solution Manual For Digital Signal Processing, 4th Edition - Page 14 preview imageChapter 1010.1(a) To obtain the desired length of 25, a delay of2512= 12 is incorporated intoHd(w). Hence,Hd(w)=1ej12w,0≤ |w| ≤π6=0,otherwisehd(n)=12ππ6π6Hd(w)ejwndw=sinπ6(n12)π(n12)Then,h(n)=hd(n)w(n)wherew(n) is a rectangular window of lengthN= 25.(b)H(w) =24n=0h(n)ejwnplot|H(w)|and6H(w). Refer to fig 10.1-1.(c) Hamming window:w(n)=(0.540.46cos 12 )h(n)=hd(n)w(n)for 0n24Refer to fig 10.1-2.303
Page 15 of 16
Solution Manual For Digital Signal Processing, 4th Edition - Page 15 preview image00.050.10.150.20.250.30.350.40.450.5−150−100−50050−−−> 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
Page 16 of 16
Solution Manual For Digital Signal Processing, 4th Edition - Page 16 preview image(d) Bartlett window:w(n)=12(n12)240n24Refer 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)=1ej12w,|w| ≤π6,π3≤ |w| ≤π=0, π6≤ |w| ≤π3hd(n)=12πππHd(w)ejwndw=δ(n)sinπ3(n12)π(n12)+sinπ6(n12)π(n12)(b) Rectangular window:w(n)=1,0n24=0,otherwise305
Preview Mode

This document has 427 pages. Sign in to access the full document!