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Solution Manual For Discrete-Time Signal Processing, 3rd Edition

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    |" aqxd 9/21/09 3:59 AM Page 1 & Nw © 2010 by Oppenheim. Published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2.1. (a) T(aln)) = glnlaln] « Stable: Let |2[n]| < M then [T{z[n]| < |g[n]|M. So, it is stable if |g[n]| is bounded. « Causal: yin] = g[njzi[n] and yo[n] = gln]za[n], so if zy[n] = za[n] for all n < no, then y1[n] = 2[n] for all n < no, and the system is causal. © Linear: Tani[n] +bzafn]) = gfnl(azan] + baafn] = ag[njai[n] + bg[njza[n] = aT(niln]) + bT(zafn)) So this is linear. o Not time-invariant: T(z[n —no]) = glnjzn — no) # yln = no] = gln — noje(n — no] which is not TL * Memoryless: y[n] = T(z[n]) depends only on the n‘* value of z, so it is memoryless. (b) T(zfn]) = Xion, 2lk] Not Stable: |z[n]| < M = |T(z[n])| < Tj, |z[k]| < |n — no|M. Asn = co, T — oo, so not stable. * Not Causal: T'(z[n]) depends on the future values of zn] when n < ny, so this is not causal. ® Linear: n T(azi[n) + baz) =) azi[k] + balk] k=no n n | = a} w]+b) mpm | > k=ng k=no Tan = aT(z:[n]) + bT(z2[n]) The system is linear. e Not TL: n T(zfn nol) = lk no k=no n-no = Yak k=0 n—no # yn—nol =) alk] k=no The system is not TIL. « Not Memoryless: Values of y[n] depend on past values for n > nq, so this is not memoryless. (©) T(zln)) RIN, zlK] o Stable: [T(zfnl)] < Tm, 2{k]] < Site, 2lklM < [2no + 1M for [ala] < M, so it is stable. « Not Causal: T'(z[r]) depends on future values of z[n], so it is not causal. eo Linear: ntno T(azy[n] + bz2[n]) = > az [k] + bza[k] k=n—no n+ng ning =a Y all+b Y z[k =aT(zi[n]) + bT(z2[n]) k=n—no k=n-no (+ 1 & Nw

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    Downloaded from StudyXY.com ® + StudyXY Sd Ye. o> \ | iF ’ pr E \ 3 S Stu dy Anything This ContentHas been Posted On StudyXY.com as supplementary learning material. StudyXY does not endrose any university, college or publisher. Allmaterials posted are under the liability of the contributors. wv 8) www.studyxy.com

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    ch02.qxd 9/21/09 3:59 AM Page 2 & NP © 2010 by Oppenheim. Published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. This is linear. o TL: nino T(eln—no] = 3 zlk—no k=n-no n = 3 =H k=n-ng = yln—no This is TL « Not memoryless: The values of y[n] depend on 2ng other values of z, not memoryless. (d) T(afn]) = z[n — no) « Stable: |T(z[n])| = [z[n — no]| < M if |z[n] < M, so stable. eo Causality: If ng > 0, this is causal, otherwise it is not causal. ® Linear: T(azi[n] +bzz[n]) = azi[n — no] + bzafn — ng] = aT(z[n]) + bT(z2[n]) This is linear. o TE: T(z[n — ng] = z[n — no — na) = y[n — ng). This is TL * Not memoryless: Unless ng = 0, this is not memoryless. (e) T(z[n)) = el" o Stable: |z[n]| < M, |T(z[n])| = |e*I")] < ell < eM this is stable. « Causal: It doesn't use future values of z{n], so it causal. * Not linear: T(azyn] + baofn]) = essiiniomaln] a — eomilnlgbaaln] a N% N% # aT(z1[n]) + bT (z2[n]) This is not linear. o TL T(zfn — no]) = e*"~"l = y[n — ng), so this is TL * Memoryless: y[n] depends on the n** value of z only, so it is memoryless. () T(z[n]) = azln] +b « Stable: |T'(z{n])| = |az{n] +b] < a|M| + |b], which is stable for finite a and b. « Causal: This doesn’t use future values of z[n], so it is causal. * Not linear: T(cz1[n] +dz2[n]) = aczi[n] + adza[n] +b # cI (zi[n]) + dT (z2[n]) This is not linear. 2 &

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    ch02. gqxd 9/21/09 3:59 AM Page 3 A IS5: © 2010 by Oppenheim. Published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. o TL T(z[n — no) = az[n — no] + b = y[n — nl. It is TL © Memoryless: y[n] depends on the n** value of z[n] only, so it is memoryless. (8) T(z[n]) = z[-n] « Stable: |T(z[n])] < |z[~n]| < M, so it is stable. « Not causal: For n < 0, it depends on the future value of z|n], so it is not causal. e Linear: T(azy[n] + bz2[n]) = azi[-n] + bzaz[-n] = aT(z:[n]) + bT(z2[n]) This is linear. e Not TI: T(a[n—no]) = az[-n—nd # yln — no) = z[-n+ ng] This is not TIL. « Not memoryless: For n # 0, it depends on a value of z other than the n** value, so it is not memoryless. (h) T(z[n]) = z[n] + un +1] Stable: |T'(z[n])| < M + 3 for n > —1 and |T(z[n])| < M for n < —1, so it is stable. o Causal: Since it doesn’t use future values of z[n], it is causal. * Not linear: T(azy[n] + basfn]) = aza[n] + baaln] + Buln + 1] # aT(z1[n)) + bT (z2[n]) EN This is not linear. A YY e Not TI: Av T(z[n —ng] = z[n—ng]+3uln+1] = yln—no = z[n—no]+3ufn—no +1] This is not TI. « Memoryless: y[n] depends on the n* value of z only, so this is memoryless. 3 & NP)

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    ch02.qxd 9/21/09 3:59 AM Page 4 & NP) © 2010 by Oppenheim. Published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2.2. For an LTI system, the output is obtained from the convolution of the input with the impulse response of the system: oo yln) = > h[k)z[n — Kk] k=—co (a) Since hlk] #0, for (No <n < Ny), Ny vin) = 3 hiklzln — K] k=No The input, z[n] # 0, for (N2 <n < N3), so z[n — kK] #0, for Na < (n—k) < Ng Note that the minimum value of (n — k) is N;. Thus, the lower bound on n, which occurs for k=Nois Ny=No+ Na. Using a similar argument, Ns = Ny + Ns. ‘Therefore, the output is nonzero for (No + Nz) <n < (Ny + Na). (b) If fn] # 0, for some n, <n < (no + N — 1), and hln] # 0, for some ny <n < (ny + M — 1), the results of part (a) imply that the output is nonzero for: (mo+m)<n<(no+m +M+N-2) A So the output sequence is M + N — 1 samples long. This is an important quality of the convolution A Av for finite length sequences as we shall see in Chapter 8. Av 4 & NP)

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    ch02.qxd 9/21/09 3:59 AM Page 5 an IS5: © 2010 by Oppenheim. Published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2.3. We desire the step response to a system whose impulse response is h[n] =a "u[-n], for0<a< 1 The convolution sum: - ln] = 3 hlklzln—&] k=—o0 The step response results when the input is the unit step: l= ={ b frn20 = “10, forn<0 Substitution into the convolution sum yields © yinl= Ya *u[-Kuln-k k=—co Forn <0: © yn) = Yat k=—oo £3 - Fa k=-n = ao" T 1-a For n > 0: [3] yin] =) ot Jan k=—co Jan N% ~ N% = Ya =o _ 1 CT 1-a 5 P 55! NP)

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    ch02.qxd 9/21/09 3:59 AM Page 6 & NP) © 2010 by Oppenheim. Published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2.4. The difference equation: . 1 yln] = vin — 11+ guln — 21 = 22[n — 1] ‘To solve, we take the Fourier transform of both sides. Y(e) — Sy (eye + FH e)e =2. X(e)emi The system function is given by: i Y (ev) wy = = 1 H(eY) X(e%) 2e~Iv T T-fev gfe The impulse response (for z[n] = 6[n]) is the inverse Fourier transform of H (ei). Hey=—3 +8 _ 1+ ge v 1-— Zev Thus, 1 1 Hin] = =8(3)"uln] + 8(3)" ul. Van Van NY NY 6 an &

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    ch02.qxd 9/21/09 3:59 AM Page 7 & NP) © 2010 by Oppenheim. Published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2.5. (a) The homogeneous difference equation: yln] — 5yln — 1] + 6y[n — 2] = 0 Taking the Z-transform, 1-5:71+6272=0 (1-211 -32"Y)=0. The homogeneous solution is of the form yaln] = A1(2)" + 42(3)". (b) We take the z-transform of both sides: Y(z)[1-52z71 +6272) = 2271 X (2) Thus, the system function is Y(z) 26) = x6) _ 227! TT 1-5z71+6z2 = 2 _,_2 To1-271 7 1-31 where the region of convergence is outside the outermost pole, because the system is causal. Hence the ROC is |z| > 3. Taking the inverse z-transform, the impulse response is hin] = —2(2)"u[n] + 2(3)"u[n]. (c) Let z[n] = u[n] (unit step), then 1 1 1 > X(@) == > and Y(z) = X(2) H(z) _ 2:71 To A-zH-22"0)(1-327) Partial fraction expansion yields 1 4 3 Ye = i= Tf The inverse transform yields: y[n] = uln] — 4(2)"u[n] + 3(3)"u[n]. 7 & NP)

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    ch02.qxd 9/21/09 3:59 AM Page 8 & NP © 2010 by Oppenheim. Published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2.6. (a) The difference equation: 1 yn] — Fin —1]=z[n] + 2z[n— 1] + zn -2] Taking the Fourier transform of both sides, Y(e*)1— 367) = X(&)[1 +2679 + e792], Hence, the frequency response is wv — Y(e) He?) = X (ef) _ 142e7iv 4 emi 1- Lew (b) A system with frequency response: 1— le—jw 4 g=i%w Hv) = —2- "72" 1+ ke 7v + Fei _ Ye) T X(ev) cross multiplying, . . 1, VE) + 3e + Je) = X (lL - ge + eI), and the inverse transform gives 1 3 1 yn] + vin —1]+ Fyn —2]=z[n] — 33m —1]+z[n-3]. a a N% N% 8 P IE

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    ch02.qxd 9/21/09 3:59 AM Page 9 & NP) © 2009 by Oppenheim. Published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2.7. z[n] is periodic with period N if z[n] = z[n + NV] for some integer N. (a) a[n] is periodic with period 12: (En) = GI (F)n+N) _ pil(Fn+2ak) = ork = 94 for integers k, N Making k = 1 and N = 12 shows that z[n] has period 12. (b) z[n] is periodic with period 8: EN) = GCE) M+N) _ i(3Ent2nk) 3 5 => 2k = “LN, for integers k, N =p N= Sh for integers k, N The smallest k for which both k and N are integers are is 3, resulting in the period N being 8. (¢) z[n] = [sin(wn/5)]/(wn) is not periodic because the denominator term is linear in n. (d) We will show that z[n] is not periodic. Suppose that z[n] is periodic for some period N: ECL = (FH) n+N) = J pntank) 3 => 2wk = —=N, for integers k, N 7 teg => N = 2v/2k, for some integers k, N' There is no integer k for which N is an integer. Hence z[n] is not periodic. Van Van NY NY 9 an &

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    |" qxd 9/21/09 3:59 AM Page 10 P IE © 2010 by Oppenheim. Published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2.8. We take the Fourier transform of both h[n] and zn], and then use the fact that convolution in the time domain is the same as multiplication in the frequency domain. i 5 HE®) = — 1+ ze=v Y(e®) = H(e™)X(e) _ 5 1 To ltleiw 1-le _ 3,2 oldie T1-lee 1 n, 1 n, yin] = 2(3)"uln] +3(~3)"uln] Pan Pan Av Av 10 A &

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    |" aqxd 9/21/09 3:59 AM Page 11 & NP) © 2009 by Oppenheim. Published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2.9. (a) Pirst the frequency response: Y(e) — Seven) + FEY) = FHKE) Y (ev) wy = HY) = X@%) _ Let To 1-fedv + Lette Now we take the inverse Fourier transform to find the impulse response: . -2 2 wy = —_— HE") =o leiv tio Teiw 1 1 hn] = =2(3)"uln] +2(3)"uln] For the step response s[n]: 0 sln] = > hkun — k] k=—co n = > h[k] k=—co _ _pl=a/t 1-1/2) = -2 1-1/3 ufn] +2 1-1/3 uln] = Lin Lin = +E" - 2p) (b) The homogeneous solution y(n] solves the difference equation when z[n] = 0. It is in the form Tat yn[n] = 3 A(c)", where the c's solve the quadratic equation Ia 5 I F 1 I - §+s=0 So for ¢ = 1/2 and c¢ = 1/3, the general form for the homogeneous solution is: 1 1, ln] = Ai (3)" + 42(3) (c) The total solution is the sum of the homogeneous and particular solutions, with the particular solution being the impulse response found in part (a): yin] = waln] + ypln] _ Lin Lin _olym, Li. = MG" + AE)" + =23) un] + 25) ula) Now we use the constraint y[0] = y[1] = 1 to solve for A; and Ap: yl) = A +42-2+2=1 yl] = A/2+A42/3-2/3+1=1 A+4 = 1 A/2+ 42/3 = 2/3 With A; = 2 and A; = —1 solving the simultaneous equations, we find that the impulse response i unl = 23)" ~ 3)" + ~2(3)"uln] + 2(3)"uln] 2 3 3 2 11 & NP)

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    ch02.qxd 9/21/09 3:59 AM Page 12 & NP © 2010 by Oppenheim. Published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any mens, without permission in writing from the publisher. 2.10. (a) y[n] = hin]xzn] = Y du[-k-1un-k k=—oc 3 a, n<-1 _ |) a 1 > a, n>-1 ko an 2 a<- _ J ime ns? - Ye n>-1 1-1/a’ (b) First, let us define v[n] = 2"u[-n — 1]. Then, from part (a), we know that 2" p< 1 win] = ufn] * fn] = { oh at Now, yn] = uln —4]xvln] = wn-4 273 n<3 = 1, n>3 (c) Given the same definitions for v[n] and w[n] from part(b), we use the fact that h[n] = 2" u[—(n— 1) — 1] = v[n — 1] to reduce our work: [Ss y[n] = z[n]*hfn] [Ss = zfn]sufn-1] = wn-1] _ [2° n<o0 - 1, n>0 (d) Again, we use v[n] and w[n] to help us. vin] = z[n] + hln] = (ufn] — un — 10) » v[n] = wn] —w[n—10] = (2u[—(n+1)] + u[r]) — (2"°u[-(n — 9)] + un — 10) 2(n+l) _9(n=9) p< 2 = 1-2(n-9), -1<n<8§ 0, n>9 12 & NP

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    ch02.qxd 9/21/09 3:59 AM Page 13 & Nw © 2009 by Oppenheim. Published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2.11. First we re-write z[n] as a sum of complex exponentials: m eIxn/4 _ g—inn/4 ln] = sin(F) = ———— Since complex exponentials are eigenfunctions of LTI systems, H(ei*/4)eim/4 _ f(e=in/4)e=imn/4 oy He De He Te J Evaluating the frequency response at w = +7 /4: . 1—eI%/2 I) = —/———=201-j)= —in/4 H(e'T) Yr 2(1- j) =2v2e 1-2 . if) = — = = x/4 H(e™i%) Tri 2(1 + j) = 2v/2e" We get: 2y/Be—in/Aeimn/4 _ 9, /Feix/4g=inn/4 ih = HE 7 = 2V2sin(rn/4 — 7/4). a a Nw) Nw) 13 & Nw

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    ch02.qxd 9/21/09 3:59 AM Page 14 & NP) © 2010 by Oppenheim. Published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2.12. The difference equation: yin] = nyfn — 1] + z[n] Since the system is causal and satisfies initial-rest conditions, we may recursively find the response to any input. (a) Suppose z[n] = dfn]: yin] = 0, forn <0 yf] =1 yi =1 v2] =2 yd =6 vid] =24 y[n] = h[n] = nlu[n] (b) To determine if the system is linear, consider the input: z{n] = ad[n] + bs[n] performing the recursion, yln] =0, forn <0 y[0]=a+b yll]=a+b v[2]=2(a +b) 4[3] = 6(a +b) Jan Jan NF; y[4] = 24(a +b) NF; Because the output of the superposition of two input signals is equivalent to the superposition of the individual outputs, the system is LINEAR. (c) To determine if the system is time-invariant, consider the input: z{n] = b[n— 1) the recursion yields y[n] =0, forn <0 [0] =0 yi] =1 v2 =2 y3]=6 vid =24 Using hln] from part (a), hn — 1] = (n — 1)'ufn — 1] # yln]lzpnj=sin—1) Conclude: NOT TIME INVARIANT. 14 & NP)

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    ch02.qxd 9/21/09 3:59 AM Page 15 an IS5: © 2009 by Oppenheim. Published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2.13. Eigenfunctions of LTI systems are of the form a”, so functions (a), (b), and (e) are eigenfunctions. Notice that part (d), cos(won) = .5(e7“°™ + e~7“0") is a sum of two a” functions, and is therefore not an eigenfunction itself. a Jay Av Av 15 P 55! NP)

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