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Chapter 5 Solutions 5.1-1. Given the fact that the time-domain signal is finite in duration, the region of con- vergence should include the entire 2-plane, except possibly z = 0 or z = 00: oo -n T non mn (1/2) X=DE os tn! = Di = Dial 1/0)º = "EEB Thus, . 1-2 X =——; Cc 0. (3) = RO tel> In this form, X(z) appears to have eight finite zeros and one finite pole. The eight zeros are the eight roots of unity, or z = e274/8 for k = (0,1,...,7). The apparent pole is at 2 = —1. However, there is also a zero 2 = —1 (k = 4) that cancels this pole. Thus, there are actually no finite poles and only seven finite zeros, z = elmk/8 for k = (0,1,2,3,5,6,7). MATLAB is used to plot the zeros in the complex plane; the unit circle is also plotted for reference. >> k = [0:3,5:7); zz = exp(j+24pi+k/8); >> ang = linspace(0,2+4pi,201); >> plot(real(zz), imag(zz), 'ko”,cos(ang) ,sin(ang),*k”); >> xlabel('Re(z)?); ylabel( Im(z)º); >> axis([-1.1 1.1 -1.1 1.1]); axis equal; grid; ima) a E - O 85 + [o Figure $5.1-1: Pole-zero plot for «fn] = (=1)"(ufn] — uln — 8) 512. (a) xo Dream 213 jm (b) 4º sinanuln] = 0 for all n. Hence x[]=0 (e) 4º cos nu Hence xt) = = is a sequence 0,41,0,-42,0,4,0, 7 Hence x() = (Err) (Ge Dao ) POA E . (e) "cos ZEufn] is a sequence 1,0,-2,0,48,0, 48,0, 98, AB xt) = 1-5+45 = be (O “o Of +( (5 =) = 5a bh 2 A 8 8 2 (6) > 2 6fn — 2] = ófn] + 46[n — 2) + 166[n — 4) + --- k=0 214 Therefore xtg=0"* (x) 0) xtH=5 Anata - Da EE) po 1 From the result in part (1) it follows that x [g] = rat = (dr = al/z 5.13. Note that the signal zfn) = nufn) has z-transform X(2) = Da Thus, DS on(-3/2)7" is easily found by evaluating the x(2) Do n(=3/2)"" = Plos2 7 o = Rc] =-B=-S. ia 6 3on(-3/7)7" = —5 = 04 n= 5.14. (a) ufn) — ufn — 2] = ófn] + 6[n —1) Hence | : uti)-ujn=2] es 14 1= EL za (b) 1 vPun- = (yr ufn] — ófn] = ô[n — 1)) Hence un — 9] 6 É e. Y 2-1) (e) elnj= (tuto) (o (ufa 1]+ eu tn) 216 Therefore 4 1 z Xd="D5+cz- (a) el = [6995 cos Ft) ufn — 1) = (2)"" cos Fun) — 6] Therefore A 0.25) 0.25(z — 1) ztz— E X==D>052405 !=77-0574025 (9) efu=nyufn= 1])=ny'u[n]-0=n9"un] Therefore Ez *ei= op (f) Because n(n — 1)(n—2)=0forn =0,1, and 2 «n]= n=0,1,or 2. Therefore n(n = Dn = 92%ufn = m] = n(n— 1)(n — 22)" fr] e fr) = (9 Un(n — Dn — 2)2"u [n)) and xtg= (0 [ES a] 6z E-)e- (g) ain] = (1) nun] + NES ] 1: (2n+1) sus i iba 4 q» dr I nim : ia 4 “el Raio É Using the entries in Sec. B.7-4, we obtain 22? 2 2 z (2-1)? 22-12 2-1 -23 40222 - 2 xt] 27 sn) = [0”-2Jutnl (a) (4-2 22-24 1 2 1 Xl = z o. 2 “z ata Hence zln)= ó[n— 1] -26fn— 2] + 6[n.— 3] (e) xt 2+3 o 9/2 72 O E-DE-DE- z-3 52 2 9% Xl = 57010 ';-at2203 5 9 en] = E -“2" + ser un) (9) Xtg -e+2 3 Z (G+DG-22 2+1 Multiply both sides by z and let z — oo. This yields 0=3+k+0 => k=—3 A z zo xd = 3 rat» cl) = [8(-1)"-3(2)" + 20(2)"]u fm) (8) XE) 142+008 1 k 2 > “E-o)G- 087 7-02"7-08' E-08) Multiply both sides by z and let z — oo. This yields O=14k=> k=-1 z z EA PEA z-02 z-08"“(-082 [ozr -(087 + nos] un 4 xt) k ztn) (h) We use pair 12e with 4= 1,B=2,0= 0.5, |y| = 1. Therefore r=Vi=2 p= a tn] = 20)" cos(E +oJeln]= 2cos(T +SJelnl 29 () 6) (9) 27º — 0.32 + 0.25 1 Az+B — CH r06:4055)) 2! 274062425 Multiply both sides by x and let z —» 00. This yields 2=1+4 => 4=1 Setting z = 1 on both sides yields 1.95 1+B iso ltTa > B=009 2(2-0.9) X =1+>D[D[—— — d=1+ 240071405 For the second fraction on right side, we use pair 12c with 4 =1, B = —0.9, a=0.3, and |] = 0.5. This yiclds xt 2(32 — 23) o 2 “(E-1(2-62+25) Multiply both sides by z and let z — oo. This yields 0=-244 => 4=2 Set z=0 on both sides to obtain 46 B =-2+; — B=4 (2204) 22-62 +25 For the second fraction on the right-hand side, we use pair 12c with 4 = 2, B=-4a=-3 andh|=5 r=>— B= cos (5) =0.927 0=tan” 2+ VE (sy cos(o soma - 0.25) | un] xt] 3.832 411.34 1 Az+B E + 2 (e-9(-52+25) 2-2 22052425 Multiply both sides by z and let z — oo. This yields 0=1+4 => A=— 220 Therefore [0] = 0, cl] =, el] = 2, 2 8]=37º,--:, and af) =nyºu[n) 5.1-8. (a) We can express xt), xp) xto) = xo) + HU AB z E: Let X,[2) and Xa[2] be the numerator and the denominator polynomials of X [2] with powers M and N, respectively. IFM = N, then the long division of Xn with Xa in power series of 2! yields [0] = a nonzero constant. IF N — M = 1, the term «[0) = 0, but 2(1), «/2], --- are generally nonzero. In general, EN -M = m, then the long division show that all z[0), z(1), --:, s[m — 1] are zero. Only the terms from «[m] on are generally nonzero. The difference N — M indicates that the first N — M samples of zfn] are zero. (b) In this case the first four samples of xfn] are zero. Hence N — M =a4. 521. (a) Xl]=1414& ++ 555. This sum can be found using the result in Sec. B.7. ,as (/gmi=-1 1-2" *Xd=qD1 CI=21 (b) cn) = uln-uln-m) Po mz o =2" XxX = ao Zac 5.29, alnj=6[n=1)+26/n-2]+36[n 3) +46[n-4]+35n —5]+26[n 6) +6[n =] Therefore 123 4,3,/2,1 Xe = ctatatatatata 28 +22 +92! +40 4322 42241 Z7 Alternate Method: alo) = ntufnj-ufn=5) +(-n+8)fuln-5]-uln— 9) = mufn)-2nufn- 5] +nuln-9]+8u[n 5] 8ufn—9] nun) = Mn — 5Jufn — 5] + Sufn— 5] Hm 9)ufn 9] +9ufn — 9] + 8uln = 5]- 8ufn —9) nufn)-2(n— 5Jufn—5]+(n—9ufn—9)-2ufn— 5] + ufn 9] | Therefore z 2z z 2z z x = copos toe zen! 2R-D SER [9 241 2(2-1)+(2—1)) 222 - 22 +41] Reader may verify that the two answers are identical. 5.23. (a) rêy"u [n] Repeated application of Eq. (5.21) to qu [n] > 5, vields = E YZ nyuln) > —s tl (2-9)? 2.m va(2 +49) ryuln) ++ bl (2-n? Lety=1 nu (n] > Er (b) Consider yufn) <> E z-7 Application of multiplication property to this pair yields nyufn) ++ e (+=) - TE One more application of multiplication property to this pair yields n a YZ o vdzto) veres cf [Ea] = ES «) Application of Eq. (5.21) to n2y"u [n) <=> JEE5P (found in part a) yields E-) d [e + 2] AP A dz +) da | (2-9)? (2-3 Now setting y = 1 in this result yields a(22 +42 +41) rêuln] = Toy (a) = auln)-ufn—m)) = a"uln)- aa xa = ár tão (e) afnj= ne Pufn-m) = (nom+me mu fam = Ma - me Paufn -m) 4 me reu -m] il Xe Therefore 1 He-y) vdz (2-v22+1) cos ru fn] + -1 <> +1 v2a22 — 2241) 5.2-6. Application of time reversal to pair 6 yiels guns E = bI + z+7 225 5.28. 5.2-9. (a) (b) (a) (d) o (a) (e (£ So afkl = 5 alkluln — m) k=0 k=0 Application of time-convolution property to this result yields Sel e extd Let z|n] = ófn], which yields X 2] = 1. Application of the result in part (a) vields >= In the time-domain, =Erayz 1s à convolution of two causal, decaying exponcn- tials. Thus, either plot 1 or plot 13 is possible. Using the IVT, the initial value is lim, n.os ovas = 1 Thus, 2 Plot 13 corresponds to . P (-0752 : ; 2-097/V3. ano sinusoid. 1 o In the time-domain, ESSE is a decaying sinusoid. Thus, either plot 8 or plot 9 is possible. Using the IVT, the initial value is lim: .oo 255, Dea = Thus, 2092/02 2 0.922+0.81 Plot 9 corresponds to Note that Di.g2 28 = 14272 42744278 4.278. Thus, the time-domain signal is ófn) + óin — 2] + 6[n — 4] + 6fn — 6] + ófn — 8) and 4 Plot 18 corresponds to >) k=0 By inspection, 455 corresponds to a unit step that is shifted to the right by five. Thus, Plot 10 corresponds to 7 —z Using synthetic division on = yields (272 4208 42719 +...) Im the time- domain, the first non-zero term therefore occurs at n = 2. Thus, E Plot 15 corresponds to — - In the time-domain, TES is a convolution of two causal, decaying exponen- tials. Thus, either plot 1 or plot 13 is possible. Using the IVT, the initial value 226 The 2-transform of the system equation is = (]+Mz-4Y [e] ! G-Yt = and -Mz Pz ve = ot EoDe-n rt Mo, P MS Z» Cozoy (G-WG-D) 2-7 v-llzoy 2-1 z P [a z Yi = nistA ul] = [ap + POD um r=9-1 The loan balance is zero for n = N, that is, y [N] = 0. Setting n = N in the above equation we obtain P(y" — “| 7 vin) = [my + “This yields rM . Because part (b) requires us to separate the response into zero-input and zero-state components, we shall start with the delay operator form of the equations, as vln] + 2ytn — 1) = «(n) “To determine the initial condition y[-1], we set n = O in this equation and substitute v(0) = 1 to obtain i+w=alj=e = u-il=(e-0/2 The z-transform of the delay form of equation yields vtg+2 [Era + SD -—&, Rearranging the terms yields sfe-ae cs z 2+2 z-e “The term (1 — e) on the right-hand side is due to the initial condition, and hence rep- resents the zero-input component. the second term on the right-hand side represents the zero-state component of the response. Thus 228 5.3-3, 1-e 2e? e + + Z42 Qe+DE+3) Qe+ilz-e)) Hence 2 26 z e z z + z+2'2e+12z+2 2e4+1z-e “The first term on the right-hand side is the zero-input component and the remaining two terms represent the zero-state component. Thus ylg=(1-0) 2e? 2e+1 ul (-2)"uln) + vin) = (1-2 ufn) + zero-input comp < e 2e+1 zero-state comp. The total response is 1 9 pet) util = sm [t+ MD" +e Jin (a) 'The system equation in delay form is 2yto) = 3yln = 1) + vln = 2] = 4efn] = Se (n 1] Also st) es YE vh-oje do yln-2] + YE] + en) es Xbl= ope 2 z— 0.25 0.25 The z-transform of the equation is , 3 Io 4 3 Wt-Vi+aridtl=co 2-0257 or , 1 (2-*+5)rta and re a(82- 2.75) 2 — (22-32+1)(2-0.25) e(82 — 2.75) = —0.5)(z— 1)(z— 0.25) 5/2 1/3 4/3 z-1/2 uti) = [5 +i(05p - 360257] un) - [riem - Sta") ubo and 5.35. (a) (b) o 322+12252-3.75 46/3 4/3 25/2 2(2-0.25)(z-1)(2-0.5) 2-1 2-0.25 2-05 4 2 4 2 25 sz Ye = 7501735005 27005 vtd = [ System equation in delay form is 4yln+4y[n=]+yn-J=2[n-1] Also vb) + YE) ubomes iv] ub-ges Syd (etal=0) 1 g[n—1] += —— Z- gn) > 1 The z-transform of the system equation is artiririe Srt+= (1) trirtly ço 2rs (2) and re o 2-2) REC zZ 0 4z-1(24+2+0.25) 4(z-1)(2+0 1[4/9 13/9 5/6 i&-: 505 * (4 Ss] 4 1/4 2 13 2 5 z YH = ali oras totraspl vn) = [5 - (cos = Sgnt-03)"| ut To find the zero-input and the zero-state components, we observe that the only term arising because of the initial conditions is 1 on the left-hand side of Eq. (1). Hence, we can rewrite Eq. (2) with explicit zero-input and zero-state components as 422 +47+1 2 Yld="1+5 1 z-1 Here the term —1 on the right-hand side represents the zero-input terms and the second term on the right-hand side represents the zero-state component. Rearranging the equation, we obtain a = FERN [ + + -. z de ros "dE DE+05 Hero Memo zero-input zero-state 1/4 N 1/8 4 1/9 9 1/12 = 2405 (+05)? 2-1 2405 (2+0.5) zero-input zero-state Therefore (1/92 (2 | (1/92 (/9z | 0/Dz yYt]= — -os tros! 2-1 2405" (+05 zero-input zero-state Jon MOS 1 (05 n(-0.5)" vin] =4 (0.5) 1 +9 E) ç uln) zero-input zero-state (c) The terms which vanish as n — oo correspond to the transient component and the terms which do not vanish correspond to the steady-state component. Hence Sos êntcos] ufnj+ Sufn) Nam steady-state transient 5.3-6. The system in delay form is vin) =3yln= 1] +2yln=2]=2[n-1] Also vt) ++ YE] voe irtd+? vin-2 += Syt+2+3 cn) + XE aln= 1) + AXE] xt]= The z-transform of the system equation is rtd-s brta+3 eafartisê+s] 3 2 4 1 (1-Sa)rei=S+5 = 2-3) Ye -32+12 o -32+12 3 3/2 = É El&A+DdE- E DE-DE-3 = t7-3 232 
