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Solução Lathi 2a ed - Sistemas Lineares - chapter 05, Resumos de Engenharia Elétrica

Solução dos exercicios do livro Sinais e Sistemas Lineares 2a ed, Lathi

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Baixe Solução Lathi 2a ed - Sistemas Lineares - chapter 05 e outras Resumos em PDF para Engenharia Elétrica, somente na Docsity! Chapter 5 Solutions 511. 5.1-2. 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 no-n mn (1/2 X=DE o tln! = Di = Dol 1/0º = "EEB Thus, . 1-28 X =——; C 0. (3) = ro 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 = e27k/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; IN No rea Figure $5.1-1: Pole-zero plot for «fn] = (= 1)"(ufn] — uln — 81) (a) xt o Dream pt 213 Nm (b) 4º sinanuln] = 0 for all. Hence x[]=0 (e) 4º cosmnu Hence x = = is a sequence 0,41,0,-42,0,4,0, —7 Hence x() = (Err) (Gs Da + ) fg pg ; , (e) "cos ZEufn] is a sequence 10,-2,0,48,0, 48,0, 98, AB 2 A 68 xt = 1-5 - a OI (1-5 =) = 5 bh (6) s 2 6fn — 2h] = ófn] + 46[n — 2) + 166[n — 4) + --- k=0 214 Therefore (a) Therefore (e) Therefore elnj= [o cos lu fn—1] = (2)-" cos ou [n) — 5 tn] H2-025) 025201) xtd= 2-052+025 — 22-0,52+0.25 elnj=nyuln-]=nyuln)-0=n9utn] YZ Xtf]= = (f) Because n(n — 1)(n—2)=0forn=0,1, and 2 an) = n(n = Dn = 92%u fr = m] = n(n— 1)(n— 22)" [n] n=0,1,or 2. Therefore and (g) Using the 2 fr) =(9"n(n — Dn — 2)2"u [n)) xtg=( [ES a] 6z E-pCE- ain] = (1) nufn] + ufo l E (2en+1) Jus i iba Pa IO dr I nim : ie 4 “| Raio) a entries in Sec. B.7-4, we obtain 22? 2 2 z (2-1)? 2-1 2-1 -23 40222 2 xt] PANA (h) ajn]= So tó -2k41] k=0 xg=D til" =5) (5 kófn — 2k + n) 2" n=0 n=0 Mk=0 Interchanging the order of summation and noting that =2k-1 on-m+=[0 oa We obtain xt] = Su (Sean) n=0 = 9 pt k=0 - Sel) o. 1/2? = “aja? 2 e) 51-5. (a) xt z-4002 1 = FT E-DG- 202 2-3 XE = 255073 el) = Out (b) xt Z-4 Ba 1/3 = = m-Delgo o tz-272-3 E 1 z XE = cstç02737-3 et) = =5ótl+ [2 ace) vtn (e) xt e2-2 a 1 | > o) es xt] = 2 sn) = [0”-2Jutnl (a) (2-2 22-24 1 2,1 Xhl= Z o 2 z ata Hence zln)=ôln-1)-25n-2]+ 6-3] (e) xt o 2+3 o 92 2 O E-DE-DE- z-3 52 2 94 Xe = 57010 "5=a faro 5 9 zm) = 5 =)" + ser uln (9 xt 42 03 z (G+DG-22 2+ Multiply both sides by 2 and let z — co. This yields 0=34+k+0 => k=3 1a z E Xb = 3a tico Zn) = BON -3()" + 2n(2"Ju fr) (8) xt 1424008 1 k 2 > “E-ode- 087 7-02"7-08' &-08) Multiply both sides by z and let z — oo. This yields O=14+k=> k=-1 z z EA 2 00 7-08"“(-082 [o2r - (087 + nos] un 4 xt) k ztn) (h) We use pair 12e with 4= 1,B=2,0= 0.5, |y| = 1. Therefore r=vVi=2 B= af] = 20)" cos( +oeln]= 2cos(TE +SJelnl 219 Therefore «[0] = 0, 2] =, el] =, 28]=37º,--:, and a [n] = nyºu [n) 5.1.8. (a) We can express xt), xp) xt) = xt) + HU AB z Z Let X,[2) and Xal2] be the numerator and the denominator polynomials of X[z] with powers M and N, respectively. IFM = N, then the long division of Xn with Xa in power series of 2! yields 2/0) = 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), 2(1), --:, fm — 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 =4. 521. (a) Xl]=1414& ++ 555. This sum can be found using the result in Sec. B.7. ,as (jm 1-2" *Xd=q/)D1 CI=a (b) an) = uln-uln-m) Po mz o =2" XxX = ao Zi 5.29. alnj=6[n=1)+26/n-2]+36[n 3) +46[n-4]+35n —5]+26[n 6) +6[n =] Therefore 12 3 4,3,/2,1 XxX = ctatatatatata 2 +22 +92! +40 432242241 Z7 Alternate Method: alo) = ntufnj-ufno5) +(cn+8)fuln-5]-uln—9]) = muln)-2mufn= 5] +nuln—9]4+8u[n 5] 8ufn— 9] nun) = Mn — 5Jufn — 5] + Sufn— 5] tn = 9)ufn 9] +9ufn —9] + 8u[n 5] - 8ufn — 9) nun) -2(n- 5Jufn—5]+(n—9ufn—9)-2ufn — 5] + ufn 9] | Therefore z 2z z 2z z x = coros toe Ze! FR-D SE [e 241 2(2-1)+(2—0)) 222 - 22 +41] Reader may verify that the two answers are identical. 5.23. (a) rêyu [n] Repeated application of Eq. (5.21) to yºu [n] > 5, yields = E Yz nyuln) >» —s In (2-9)? 2.m vaz +49) ryuln) ++» bl (2-9? Lety=1 nun] > E (b) Consider un) <> E z-7 Application of multiplication property to this pair yields nyufn) + e (+) - Es One more application of multiplication property to this pair yields a a YZ o vdzÃo) rêruies cega) = TS <) Application of Eq. (5.21) to n2yºu [n) <=> JEE5P (found in part a) yields G-) a [e + 2] AP A az +) da | (2-9)? (2-8 Now setting y = 1 in this result yields a(22 44241) rêuln] = Toy (a) = auln]-ufn—m)) = a"uln)- aa Du xH = to (e) atnj= ne "ufn-m) = (nom+me20mDufa -m = Ma - me Paufn -m) 4 me re ufn -m] il Xe == (8 en) = (n- 20.5)" un — 4) E sn 442705" tufn — 4] - sn — ay(o5)"-Sufn — 4] + (0.5)"-*ujn — 4] Application of shift property yields 1 0.5. z dn 5 Ie lr=055 o 0.25 1 2— 0.25 Xt=5k-05)! 20-05)" 26-05) 5.2-4. Pair2: ulnj=ó[n)+6fn-1]+6[n—2]+6[n = 3] +++ 1 1 1 1 unjes1+5+5 Z Repeated application of Eq. (5.21) to pair 2 yields pair 3, 4, and 5. Application of Ea. (5.20) to pair 2 yields pair 6, and application of time-delay (5.15) to pair 6 yields pair 7. Repeated application of Eq. (5.21) to pair 7 yields pair 8 and 9. 5.2:5. There is a typographic error in the book problem: cos(zn/A) should be cos(an/4). Given (Pair 11b with |y] = 1) sin Bnu[n] <> cos fiz FI cos Bnuln] = —sin (8n - 5) un and cos nun) = —sin (Br = 5) ul = —sin Go - 2] ufn = -sin Gt nun — 2] +sin (5) ótnl + sm 3) ófn=1) = -sm[Sn- 2) uln-2]+0n] + gi =] E From Pair 11b, and shift property CAR 1 fm sn [5-9 dn] ET VE VET) 2(z 224 with y[O)=-M,a is lim; nico eAGaay = 0. Thus, 0.75 Plot 1 corresponds to tos m the time-domain, $25/22 has the form of a sinusoid. Thus, either plot 3 22 -2/V2 or plot 4 is possible. Using the IVT, the initial value is lim; .oo 521 Thus, Pajvê 224 Plot 4 corresponds to The apparent repeated root at z = 1 of TER ai suggests a signal that grows linearly in the time-domain. Thus, plot 6 or plot 12 are possible. To dis- tinguish between the two, first determine whether or not a root at z = 1 really - - . 287) 8 28-52 : 28 bas exists. Pirst, lim; EE SgEÊ = lim, oa fo = lime apso — - E 0/0. “This is indeterminant so use L'HospitaPs rule: lim, E E : ftet sara : 245 i É lim;— qÊRaas = lim misêmo = 0/0. This too is . : ; a . . E sas 448 indeterminant so use L'Hospital's rule again: lim; Eus = A (485244) lim, qi = tim.ni OTTO a505) sos = 20/10 = 2. Since Ehebetpar 24 00, no root existsatz=1 and lim. srs sao Plot 6 corresponds to a In the time-domain, 5 is a growing exponential due to the root outside the unit circle. Thus, plot 16 or plot 17 is possible. Using the IVT, the initial value às lim, oo Tg = 1. Thus, Plot 16 corresponds to ——. z-11 In the time-domain, 1a is the convolution of a decaying sinusoid and a unit step. Thus, plot 7 or plot 11 are possible. Using the IVT, the initial value is lim, oo tros) = O. Thus, 0.257" Plot 7 corresponds to 557 755] vira) -qybnl=2[n+1] Pufn-1] x=-5 snes rt o gbe)esW+M: 227 The z-transform of the system equation is = (]+Mz-4Y [e] " Grid = and -Mz Pz ve = ot EoDe-n Ya Mo P MS Zz Do zoy GE-)E-D) 2-7 v-llzoy 2-1 z P [as z Yi = nistA vln] = [air + POD um r=9-1 The lomn balance is zero for n = N, that is, y [N] = 0. Setting n = N in the above equation we obtain Pa" — “| z yiN]= [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) = afn) “To determine the initial condition y[-1], we set n = O in this equation and substitute v(0] = 1 to obtain i+wl=afj=e =u-il=(e-0/2 The z-transform of the delay form of equation yields vtg+2 [Era + SD -—& Rearranging the terms yields stofu-ae 6 z 242 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 2e2 e le £€ rr Z42" Qe+DE+3) Qe+r(z-e)) Hence + 22; x 4+-* z+2 2erlz+2 2e+1z— 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 e 2e+1 zero-state comp. “The total response is 1 or pet util = 3 [t+ MD" +e Jin (a) 'The system equation in delay form is 2oto) = 3yln = 1) + vln = 2] = 4efn] = o(a 1] Also vt) es YE vh-ojes iva ln 2] + YE] + cn) es xbl= nojo 2 z— 0.25 0.25 The z-transform of the equation is , 3 Io 43 We-Vi+aridtl=co 72-02) or , 1 (2-*+5)rta and re a(82 2.75) 2 O (22-32+1)(2-0.25) a(8z — 2.75) 2 —0.5)(2— 1)(z— 0.25) 5/2 1/3 4/3 z-1/2 vt) = E +it05p - 360257] un] - [riem - St") ut -. z ie ros "dE DE+05) Hero Mem HO zero-input zero-state - Ma MB, 1/90 1/9 1/12 - 5" (+05) 2-1 2+05 (2+05)2 A zero-input zero-state Therefore (1/92 (82 (1/92 (0/92 | 0NDz yYjlzl= LD Sm pI d="505 "(+055 "= +05 * (+05) ol zero-input zero-state Ja qem MOS 1 (-0.5)" vin)=4 (5055) +35 5 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 3 op Entcosr] un) + Dufn) Nam steady-state transient 5.3-6. The system in delay form is yin) =3yfn=-]+2yln=2]=<2[n-1] Also sb) es YE) vino nes dyid+? vin-2 += Syt+2+3 cn) + XE aln= 1) + AXE] xt]= The z-transform of the system equation is rti-s brta+3) eafortasõ+s] 3 2 4 1 (1-Sna)rei=-S+s = 5-3) Ye -32+12 o -32+12 so 3/2 + É El&ADdE-) E DE-DE-3 = t7-3 232 9 z z 3 z Ya = 55-10 *g-atoz- uti = [cer + Ser) ut 5.3-7. The system equation in delay form is vt]=2yln-]+2y[n-2]=2[n-2] st) e YE un) + Avfg+a vn» Srb+o 1 z cln-2) > 5XEl and X= The z-transform of the difference equation is 1 1 1 1 vtg-2 [Brig +] +2 [erta+5] = 5 (22-22+2) — 22-42+43 2 YtH= re) 22? 4243 1 2 (-N(-2+2) 2-1 rg =: z(2-1) 2112-242 For the second fraction on the right-hand side, we use pair 12e with A=1, B= 1, a=-1, l2=2. This yieldsr = 1,8 = 5, and 0 =0. Therefore vtnj= f +(v7r cos(Gm)] ufn] 5.3-8. The equation in advance form is yln+2]+2y[n+1]+2y[n)=<2[n+1)+20(n) vt) > Y[ gh+)eszp yin+7]e>2r en) é X(] en+)eszX]-z and Xt]= “The z-transform of the difference equation is 2 2vtj-ssdy prot) = Ec dE Hed Zoe *tgrel are (2 +22 +42)Y [7] 2+ de+2) A2+2) z0e Therefore Y [2] o z+2 o 0.318 —0.318z — 0.502 z rm ze” 2+%+2 q 233 z(0.318z + 0.502) z Y = GE — ti +22+2 For the second fraction on the right-hand side, we use pair 12c with 4 = 0.318, B=0.502,0=1,|]?=2and r=0.367 f=cos” vin)= [osistor —o.367(V 3)" cos(SEm - 0.525 un) 5.3-9. In transform domain, H(z) = 2125; and Y(2) = 271 Ha =2: Since Y(2) = H(z)X(2), =+2" we know X(2) = Y(2)/H(2) = 857. Thus, X(2) = -3512. Using tables, a[n] = 3 (Pulo) — He 2p= tudo — 11) = 8 (at-ayrmtu 1)) or E = —36fn) + 71(-2)" un — 3). 5.3-10. Taking the 2-transform of bfm] = (1.01)b[m — 1) + p[m] and solving for B(z) yields B(z) = P(z)y-rd=- Thus, P(2) is required to solve for b[m]. One way to represent Sally's deposit schedule is pfm] = 100 (ufm] — Sx-o ô[m — (12k + 11)]). Defined this way, Sally deposits one hundred dollars on the first day of every month m except for Decembers, (m = 12k + 11 for k = (0,1,2,...)). Taking the z-transform yields P(z) = 100 (= — DW=n oo De-o ólm — (12k + mz”) = 100 (= - Di£o Decos 6lm — (126 + 1D]e”) - 100 (, - Dio 0), Substituting P(2) imto the expres sion for B(z) yields B(z2) = 100 (== — Dio 2 0) - E oo ra O 191 109 o arts) 100 (m= =) + Bio Eta) = 100 (= Qis=t + 126 r + DDk-o InTois The first two terms are easily inverted using a table of z-transform pairs, while the last sum is inverted using tables and the shifting property. blm] = 100 (ionconyrum — 100ufm] — > (1.017 02H Dufm — (12% + E) . k=o 53-11. (a) Note, hifn) = (=1 + (0.5)") uln] = —(1)"u[n] + (1/2)"u[n]. Thus, two real poles are evident at z = 1 and z = 1/2. Since h[n) is not absolutely summable, the system is not BIBO stable. Thought of another way, the pole on the unit-circle makes the system marginally stable, at best. Marginally stable systems are not BIBO stable. (b) Notice, h>[n] = (9)" (u[n] — uln — 10)) is a finite duration, causal signal. Thus, holn] has no poles (other than at zero). Since h>[n] is absolutely summable, the system is BIBO stable. 53-12. (a) Let unJ=55k Fed 234 Use of the result in Prob. 5.2-8a yields — 2 Hz+1) Dto tor RP EP Hence co n(n+1) 2 (ii) Let aln) = n2uln) Then fe41) az E = =p Use of the result in Prob. 5.2-8a yields 4z (E -1t - p= 3 Bm(n-1) 2n(n-1(n—2) 2º +83? +m m(n+I(n+) x =" 2 3 = G 6 5.3-16. Let . afn) = nºufn) Then 2 42 +41) xe Cami Use of the result in Prob. 5.2-8a yields n ao , +42+1) z 7z 122 6 pes EE E Leer Cuptep ep le- Hence n DDD nam EmA? 358 = nt Into son(a- 1-2) PDT =D itiri tor mini k=0 4 É 4 - 4 5.9-17. Let aln) = na"ufn) Then az xtj= —s = Use of the result in Prob. 5.2-8a yields r az? a Zz z z Etr S ota plo mat te Dg | 237 Hence no a A atart “9-1 Dtt=qónali=e +(a— 1)na?] stato al 53-18. (a) ctn]) = eu] XE] + ez? Ye = HH qoogroDE-os Therefore vtd ez o 0.186 1.13 = É E-JE+O(-os) z+02 z-08 z z z YE) = 13255 0186-1135 yin] = [132(e)" — 0.186(-0.2)7" — 1.13(0.8)"] u [nl] (b) From the given Hlz], we can write (22 — 0.62 — 0.16)Y [2] = zX[2) Hence, the corresponding difference equation of the system is vln + 2) — 0.6y[n + 1] — 0.16y[n] = z[n +) or al) = 0.6yln — 1) = 016y[n — 2] = af 1) 5.3-19. (a) o o. z(22+3) YH=XBt= o(s) Therefore re 9/2 2 o z-3 9 rel = *+37-3 vt) = [E-rer+Fer)ui (b) From the given H[2], we can write (2 —52+6)Y [2] = (224 3)X[2] Hence, the corresponding difference equation of the system is vln + 2) — 5yfn + 1) + Gy[n] = 2z[n + 1] + 3x[n] 238 or vln] — Syfn — 1) + Gy[n — 2] = 2e[n. — 1) + 3zfn — 2] 5.3-20. AM cases use the same transfer function. From the given H[2] (after dividing the numerator and the denominator by 6), we can write Hence, the corresponding difference equation of the system is vin 42] Bolo 1]+ gufnl = Soto +) ato] o bd Suln = 1] + E -“9=5efn-)-ofn-2] =4ul (3)ºu [n] so that X [7] = =p» ond 6z(5z — 1) =(52-1) Va=HMA = os -s:+ 07 E-De-DE-D Therefore 1 1 1 = yr as() + 36(5)” vt) = [12 agr 2065" | uti = a2[4" = 4(3)" + (2) un] (b) Here the input is 4--2ufn — 2) which is identical to the input in part (a) delayed by 2 units. Therefore the response will be the output in part (a) delayed by 2 units (time-invariance property). Therefore uln] = 12 [4-0 — a(gy (=D + 36978) un — 2] (c) Here the input can be expressed as =4"0-Dy cf = 16(4)“u [n] This input is 16 times the input in part (a). Therefore the response will be 16 times the output in part (a) (linearity property). Therefore vn] = 192 [4º — 4(3)" + 3(2)") un) (d) Here the input can be expressed as afj=4"ufln-2]= mta Dun 2] This input is 4; times the input in part (b). Therefore the response will be 5; 239 (e) — 22 — = “H(2-162+08) 1.25z — 1.62+0.8 For the second fraction on the right-hand side, A = 1.25, B = 0,0 = —0.8, ht =0.8, l= 3, and T=2795 B= cos (2ES )=0.464 0 =tan(-2) = —1.107 Afn) = 1.256fn] + 2195(59º cos(0.464n — 1.107)u [n] 5.3-25. (a) Noting that H(z) = 27*&, H(2) = no =] =78 72. Thus, t (b) Since h=:[n] is absolutely summable, the system inverse is stable. However, hn) 0 for n < 0 so the system is not causal. ho = 6[n +3] — ó[n + 2). (c) For systems that have time as the independent variable, it is only possible to realize causal systems. Shifting h-![n] by three makes it causal and therefore realizable. That is, implement Açao") = h7Hn 3] = ôfn]— ó[n—1], as shown in $5.3-25c. Within a delay factor, this implementation functions as the system inverse. x[n) tr] - -1 Figure $5.3-25c: Block realization of h causal f 5.41. For convenience, let y = “2 and rewrite hfn] (a) By inspection, the structure is a parallel implementation of two modes, which are easily identified in Afn). The transfer function of the structure is H(z) 1 Taking the transform of h[n), H(2) = 51 + = (1º + (19º) ul). 1 FAST É TRAZ Thus, and Ag = = * () tola] = hln) + (1) = Do blbeto = = (DES + (19 Jaln 3). Thus, Loans erre) 149 voln] = 4—— un 43), where y= 2 otrl ( 1-4 = mn+3) v8 qria Written another way, yoln] = 2R (15 Jufn + 3]. 242 5.42. (a) HH 322 — 1.87 32-06) ( 2.) (25) 2-z+016 (2-02(2-08) lz-02/1z-0. Parallel form: To realize parallel form, we could expand H (2) or H [2] /z into partial fractions. In our case: 1.22 — 0.48 Ht=+ 7 qa-08) 4 É Hl]=3+ — SA + id (2-02) (2-08) Alternatively we could expand H [2] /z into partial fractions as: 1 Hs 3(2-0.6) z (2-0.2)(2-0.8) 2 and Hs The realizations are shown in Figure S5.4-2. (b) Refer to Figure S5.4-2. dz+22 00 S2+22 (2+0.2)(2+0.8) 22+2+016 1 52+22) + z+02/1 2+08 /. AM the realizations are shown in Figure S5.4-3. Refer to Figure S5.4-3. HE] (b 5.4-4. (a) 3.82-1.1 HkJ= — [= 2008203770 005 For a cascade form, we express H [2] as: 1 3.82- 11 Hl= (— 52) (5 a) For a parallel form, we express H [2] as: —2 + 22+3 z-02 22-0.62+0.25 Ht= AM the realizations are shown in Figure S5.4-4. (b) Refer to Figure $5.4-4. 5.45. (a) Note: the complex conjugate poles must be realized together as a second order factor 243 | . XE% E 3 YE33 canonia “ vectr x£81 yo Ê YCsT = E Series Ga og -D6 x La] pa YU3] p— Parate) Po br GR 5s54-3b XI34 Id A Carne. Aee Yr33 a Parallel ê -08 est BB dy XI] Fig S54-4 -44 Pavalle Le fg. S S4-4b X0%3 Canonte. asireel— XE3T Series XT33 2 Yi31 Cualte) 22. XU33 Canonie. rec II- XI r E<Ê A Series ta -> xy AE YBS Taralte | Lt VA do A-6A 25 Fig. 55 4-6 B “ XE3] YA RARA! + seres pm XLg] Lyra? and 2u [e] = ct ( O4sin Q ) 1-0.4cos Q 322 — 1.8z Hlj= 22 "002. = 251016 3e%9 1.88% | (3cos 20 — 1.8cos N) + (3sin 2 — 1.8sin N) e%P ei? +0.16 (cos22-cos NL + 0.16) + j(sin 20 — sin N) 2 359 — 1.829 3e UP Em = [Eacsaçosl | É 2.0256 — = .8 cos 32 cos 8) + 0.32 cos 20 o 1/2 Therefore Ir tem=[ 12.24 — 10.8 cos Q | 2.0256 — 2.32 cos 9 + 0.32 cos 20 and , 3sin20 — 1 8sin 2 sin20 — sin O ZH (9) = tan (Bi “tai [em] = tam (qem Tr) tm A cos20— cos + 016 05,2 ++ 05,1 Hd=1+5+5 a + 5 z H[P] = 140508420 U8 4 20789 4 0,5e-188 4 ori eTiZsA [e?? SM +0.5e159 + 240. so +0.5e"7l 592 +eT32 58 o 32 so Ze 258 [pes + Jose +cos | Q 302 52 Aos + cos = 4 2cos Therefore |H [69]|= 3 3 and LH[e9] = -250 (b) Using the same procedure as in Prob5.45a, we obtain: .Q 3 50 sing +sin > + 2sin — rté= pasmo and (H[P]=-250 7/2. 5.5-3. The advance operator form of this equation is Etyn]= : [E!4+ ES 4 E2+ E +41] 252 and utgj=) drir2+z+] 5 A 1 [ei + ei30 + 620 pe 41 sy H[oº] = Potente = gerson [8 pe 414 e 4 ei] geri [1+2c0sN + 2c0529] 5.5-4. (a) The z-transform of the two equations yield: (1). ( + e) Ye =X] (ii. (1 - E) vYtg=X() Hence the transfer functions of these filters are (1).H[2] = = (ii).H[2] = . Consider the first system. () sa : mJ==E == 2709 140958 1+09cos = 70.9sm A 1 -09sinQ H[9]| = >=, H [&º] = — tan! | ta [e] TBI + LêcosO” te Dm |7409c0sn Eu 1 1 Hliy= EL 2 Als A [= 50-05" 109058 0.9c0s + 0.9sin A 1 0.9sin QL SM] — PM] = tan la [e] VÍBI= IAcon” 2H [28] = —ton [rega Filter(i) has a zero at the origin and a pole at —0.9. Because the only pole is near Q = m(z = —1), this is a highpass filter, as verified from the frequency response shown in Figure $5.5-4a. Filter(ii) has a zero at the origin and a pole at 0.9. Because the only pole is near Q = 0(z = 1), this is a lowpass filter, as verified from the frequency response shown in Figure $5.5-4b. (b) (i) For 2 =0.017 H (90) = = 0.5966 v1.8 0.017 253 For 9 = 0.997 H (e5989*] (ii) For 2 = 0.017 H [901=] = For 9 = 0.997 H [099%] = Filter(i) gain at No is tl = Filter(ii) gain at 7 — Ng is 1 1 === = 9.58 v1.81 +1.8cos 0.997 1 = 9.58 1.81 — 1.8cos0.01m 1 = 0.5966 v1.81 — 1.8cos 0.997 + 1.81 — 1.8cos No 1 |H| = ===E----===--+= = > Vi81=18cos(r- No) 1.81 + L8cos(do) 5.55 z+0.8 H = El--05 (a) Efe] 2 +08 (cosN+0.8) +jsinQ EP] = ST = CET St Ismi 05 (csN-05) +jsmQ Les fem) = 1 [89] Hi (e-59] LH [9] (b) 2=05 IH [605] |? IH [e] ) (2º +0.8)(e12 + 0.8) 1.64+1,6cosN (0º -0.5)(e-12 0,5) 1.25-cosQ sin Q sinQ = tan) — tan MS (naõ) aos) 1.64 + 1.6cos(0. 64 +1.6c05(0.5) 4, 1.25 — cos(0.5) 2.86 254 (b) fs > 2fn = 100 kHz, T<7j=10us. 5.5-10. Taking the ztransform of yjn] = S2(0.5)tzfn — k) yiclds Y(2) Dolo. SEX (e)a=* = X(z) Dyco(0.5/z)*. For |z] > 1/2, this becomes Y(z) X(2) 55" Thus, the system function is H(z) = 8) = br. (a) Using H(z) and letting z = e!º, the magnitude response is |H(e)] = —- em À = À . 1-0.5e=> - (1-0.5 cos(N))2+(-0.5 sin(N))? 1-cos(N)+0.25(cos?(N)+sinZ(9)) Thus, (b) Ee] = MATLAB is used to plot |H (e). >> Omega = linspace(-pi,pi,501); >> Hmag = 1./sqrt(5/4-cos(Omega)); >> plot(Omega,Hmag,ºk?); axis([-pi pi O 2.5]); grid; >> xlabel("'NOmega”); ylabel(” IH(eLjNOmega)) |); >> set(gca, 'xtick”, [-pi:pi/4:pil,'xticklabel?, [?-p E POOL OG OD O GD Gr aço apr po, *fontname”,º symbol); * N Figure 85.5-10a: |H(eº)] = FE: Using H(z) and letting z = e2º, the phase response is LH(eº) = Lila = =L(1- 0.5cos(9) — 0. Em Thus, -0.5sin(9) IM = (me LH(e) = acta (a) MATLAB is used to plot ZH(e?º). >> Omega = linspace(-pi,pi,501); >> Hang = -atan2(-0.5+sin(Omega),1-0.5+cos (Omega) ); >> plot(Omega,Hang,'k');axis([-pi pi -0.6 0.6]); grid; >> xlabel(ºNOmega'); ylabel('angle H(e"(jNOmega)) [rad]'); >> set(gca,xtick”, [-pi:pi/4:pi],'xticklabel?, [º-p *;... 257 Doo Op OP OP op op Sp po. *fontname”, symbol); (c) Since H(2) = HE) = s=9551» an equivalent difference equation description is vln] — 0.5y[n — 1] = a[n). From this equation, an efhicient block representationis found, as shown in Figure $5.5-10c xr] Ra Z1 12 Figure 85.5-10c: Block representation of y[n] — 0.5y[n — 1] = zfn). To O M—s vlLHLe'2a (Q> NL. Figure 55.6-1 5.6-1. Figure $5.6-1 shows a rough sketch of the amplitude and phase response of this filter. For the case (a), the poles are in the vicinity of 2 = Z. Therefore, the gain |H [ei] is high in the vicinity of Q = 7/4. In the case (b), the poles are in the vicinity of Q=7. therefore, the gain [H[ci2) is high in the vicinity of Q = 7. For case (a), the phases of the two poles are equal and opposite at 92 = 0. Hence ZH[e?) starts at O 258 5.6-2. 5.6-4, 5.6-5 5.6-6. (for Q = 0). As 9 increases, the angle due to both poles increase. Hence, ZH[e'º) increases in negative direction until it reaches the value —27 at Q = 7. For case (b), similar behavior is observed. Note that angle —27 is the same as 0. The two systems are very similar and have identical steady-state characteristics. There is an important difference, however, between the two systems. The system yfn] — ylr — 1] = z[n] — afn — 1) is first-order and can support an initial condition; the system y[n] = z[n) is zero-order and cannot support an initial condition. If the initial condition of the first system is non-zero, the output of the two systems can be quite different. (a) From the magnitude response plot, it is clear that this is a lowpass filter. Low frequencies near Q = O are attenuated, and high frequencies near 9 = 7 are passed with unity gain. (b) From the magnitude and phase response plots, H(e7"/2) = L.e/87/2, Thus, the output to ai[n] = 2sin(n + 3) is x 2" tn] vZsin(çn +71) = -v2sin( (c) Notice, H(e777/4) = H(e-3*/4). From the magnitude and phase response plots, H(e717/4) x: 0.071e72:43, Thus, the output to zo[n] = cos(ZEn) is x 7: ya[n] = 0.071 costtn +2.43) Refer to Figure $5.6-4 and the solution to 5.M-1. Refer to Figure 85.6-5 and the solution to 5.M-4. Hty= Kit z -—a Figure 85.6-6a shows the realization, Figure $5.6-6b shows the pole-zero configuration, and Figure $5.6-6c shows the amplitude response of the filter. Observe that the pole at a is close to = 0. Hence, there is the highest gain at de. There is a zero at —1, which represents 9 = 7. Hence, the gain is zero at 9.= 7. This is a lowpass filter. 58 o nfesj=n (é (a) a en cosfi-a+jsin 2 WH [é] |=K (L+cosf) For a = 0.2 2 cos) 39) = Ky) TOS va [e 1.04-0.4cosQ The de gain is LH] = 2.5 For 3 dB bandwidth |H[e9)p? = 1 H(e79]]? = 3.125K?. Hence 2(1+cos9) 1.04 —0.4cos A 3.125K2 = K? [ ] => N = 1.176 259 5.6-9. 5.6-10. (a) (b (a (b ) han] = (Diuln] = eim hn] Use of the property in Eq. (5.20) with y = e$i”, we obtain Hal Hifetra. Hence Ho(9] = Hi[e/(P+r] Figure $5.6-9 shows the frequency response of an ideal lowpass filter with cutoff frequency Nc. Figure $5.6-9 also shows the same frequency response shifted by a (with 2m-periodicity). It is clear that the shifted response corresponds to an ideal high-pass filter with cutoff frequency 7 — Pc. Let H(2) represent the original filter, either highpass or lowpass. The transformed filter has system function Hp(2) = H(2)| = H(-z). The basic character of the transformed filter can be assessed by its magnitude response, |Hp(e'?)| = tH (= = |H(e7 e) = |H(e!º+5)]. That is, the magnitude response of a transformed filter is just the magnitude response of the original filter shifted in frequency by 7. If the magnitude response of a digital LPF is shifted in frequency by 7, it becomes a highpass filter. Similarly, if the magnitude response of a digital HPPF is shifted in frequency by 7, it becomes a lowpass filter. Put another way, an original passband centered at 92 = O (LPF) is shifted by the transformation to a passband centered at 9 = 7 (HPF), and vice-versa. E H(7) = Dºcohlnzo” is the original filter, Hy(z) = H(-2) = DE hfn](=2)" = 52 o(=1)"h[n]z””. Thus, the transformed filter has 4 m=00 impulse response hrln] = (DAL Put another way, the same transformation is accomplished by simply negating the values of hfn) for every odd integer n. 5.6-11. The bilincar transformation states s = 24). Rearranging yields z = 1ÉTs/2, T+) sing Y) I-Ts/2 Thus, s = qu maps to z = XTR, (a) The magnitude of this transformation is [2] (b) Using the solution to 5.6-1a, we know that z = ItmwT/2) UtawT/2] 1 a] = mta = 1. Since only the unit circle has |2| = 1, the bilinear transform maps s = jw onto the unit ircle in the z-plane. ci LtawT/2 FETiS describes the unit circle in the complex plane. This allows us to write z = e? = JEMTB. Thus, the 262 bilinear transformation maps (—c0 < w < 00) to (-7/2 < Stx/2) in a monotonic, although non-linear, manner according to N = £z = «a = arctan(wT/2) — arctan(—wT'/2) = Zarctan(wT/2). 5.7-1. For the system in Eq. (3.15a), the transfer function H(s) is given by 1 H()= and h(t) = eu(t) Using impulse invariance criterion, we obtain the equivalent digital filter transfer func- tion from Table 5.3 corresponding to H(s) = «x; as Tz Tz (1-7) HI] assuming T>0 Dea The approximation used in Chapter 3, Eq. (3.15c), yields 5 z Mtl= É z+a Substituting 4 = Ez and a = jris, we obtain à Tz z Ter Rd=""— Tea AsST>0 1-cT. Hence o 2 I+eT a TA-cT)z Tz fgs "0H wu tool =D 0-0 which is same as that found by impulse-invariance method. 5.72. (a) 15+20 75+20 1 (5/2 AS +7s+10) As+2)(s+5) s+2 s+5 Ho(s) = Using Table 12.1, we get HE) z 5 z T [=> tazresr cen] First we select T: H0)=1 andior s>5=> Hi(s)= = 5 7 and Holjol=o > (b) We shall choose the filter bandwidth to be that frequency where 7 |Haliwo)] isl%of |Ha(0)]. Hence mo” 0.0! and w)=350, T= E 263 Substituting this value of T in H(z] yields He] = 0.008976 [A + 2 5 z- 097474 =-—— | = 0.031416: z-0.9822 "27— Sm] 00814162 [ | z? — 1.9383z + 0.9391 1) Canonical realization: 0.031416z2 — 0.03062z Hd = 2 1.95852 + 0.9591 2) parallel realization: Hj= 0.008976z 0.02244z 2 2709822 *2-0.9561 The canonical and the parallel realizations are shown in Figure $5.7-2. ——o: 3116 canon ica] 321 P L (O Y [o-52289] Parallel 0.456) Figure $5.7-2 5.7-3. Ha(s) = aa Using Table 12.1, we get Hf]= |>——— ABr 22 — 2ze-Tivicos (E) pe-vZ We now select T: Ha(0)=1 and for highs, Hu(s) = 5 5 1 and |Ho(jw)| = for high w 264 Sing Figure $5.7-5 The unit step response ga(t) is given by: gal) = 0 [BM] ga [oe o gafio Therefore ga()=(1-e“u(t) and gAnT) = (1- een] Also, g[n), the response of H[z] to ufn) is given by: gn) = 27 (ut) Since gfn] = ga(nT), ut = AA eua) z— erwT 1 = erur Therefore Hl]=120 z Teu Using the above argument, we can generalize: dm HE) (b) and 267 (b) Hence, Wit] = z nda Wi<ã [cosa = 17 + sn? um] V20 = cosuT)] 2sme7] The ideal integrator amplitude response is Italo) = À Observe that this amplitude response is identical to that found by the impulse invariance method in Prob. 5.7-4. Hence, this amplitude response and the ideal integrator amplitude response are the same as those in Figure $5.7-4. The only difference between the answers obtained by these methods is that the phase re- sponse of the step invariance response difers from that of the impulse invariance method by a constant Z. For a differentiator Ha(s) The unit ramp response r(t) is given by MO= LA P(Ha(o) = 045 (8) = u(t) + Now we must design H[z] such that its response to input nTu[n] is ufn), that is Zlufn]] = H(2]ZnTufn)) Tp o HE- z-1) For an integrator 1 H(s)=— (=. The unit ramp response r(t) is given by = E = Pulo s Now we design H[z) such that its response to nTufn] is n2T2u[n), that is z ( AnêTtul ) = HEZ(nTun)) or , Têdz+1) Ta Deo 2H 22-13 1? te Hence T/2+1 He) (5) 268 5.7-9. (a) The w-axis is given by s = jw. Rewriting the transformation s = E HA; ki Por H()=5 0 HE=TD — = a; + jk, then eNT = emTeikiT, When À; is in LHP, a; < O and [eMT| = esT < 1. Hence if À; is in LHP, the corresponding pole of He] is within the unit circle. Clearly if Ha(s) is stable, the corresponding H[] is also stable. (b) Z— as or Thus, we need to show that =oy describes a circle centered at (1/2,0) with a radius of 1/2. For a complex variable z, the equation |z — 1/2]? = (1/2)? describes a circle centered at (1/2,0) with a radius of 1/2. The transformation rule z = substituted into this expression. (2-1/2(2 1/2) 1 1 “1 1 2/1 rmT 7 and substituting s = ju yields z = 1 Ir 8 lz=1/2]º A : Tre + ago tam td = 1 + SUSWT)-AnwT) (1 = Trust 2 pa7) 0 =qoT) 4 = trsm + agroro ta = 1/4=(1/2º Since the equation is satisfied, the transformation rule z = to a circle centered at (1/2,0) with a radius of 1/2. Notice that different values of w can map to the same value of z (aliasing), which makes an inverse transformation very problematic. 1=20+ T 7 maps the w-axis asz = +. Next, notice First, rewrite the transformation s = iz Po = 1=sT— ss TE É RCE > ET rot For o < 0, the denominator 1 — 20T + (0? + w2)T? > 1 and |z|? < 1. That is, the left-half plane of s (o < 0) is guaranteed to map to the interior of the unit circle in the z-plane. = [n] = (0.8*u[n] + 2ºu[-(n + 1)) mn) zaln) an) «> as izl> 0.8 z go[n) > lzl<2 269 >> plot(real(zz),imag(zz),"ko?,real(zp), imag(zp),'kx'); >> xlabel('Re(z)?); ylabelC' Im(z)'); >> axis([-1.5 1.5 -1.5 1.5]); axis equal; grid; -os] “35 = “os o , 15 2 [e Figure $5.9-4a: Pole-zero plot for H(2) = There are two possible regions of convergence, both of which exclude the thrge system poles: || < 3/2 or |z| > 3/2. (b) The poles and zeros of H-(z) are just the zeros and poles, respectively, of H(2). Thus, the three zeros of H(z) satisfy 2º = 2, or z = 3/2e227/3 for k = (0,1,2). There are two finite poles at z = 0 and z = 1/2 as well as a pole at infinity. MATLAB is used to create the corresponding pole-zero plot. >> k = [0:2]; zz = 3/2xexp(j*2*pi*k/3); zp = [0,1/2]; >> plot(real(zz) imag(zz),'ko”,real(zp), imag(zp),'kx'); >> xlabel('Re(z)'); ylabel(ºIm(z)'); >> axis([-1.5 1.5 -1.5 1.5]); axis equal; grid; Za oo “SE AA 5 Figure $5.9-4b: Pole-zero plot for H1(2) = =) There are two possible regions of convergence, both of which exclude the three system poles: 0 < |z|<1/20r 1/2< |z] < oo. 272 5.95. 5.9-6. 5.9-8. 5.9-9. It is known that afn] has a mode (1/2)" and that z1[n] = (1/3)"z[n] is absolutely summable. For this to be true, the mode at (1/2)" must be right-sided. That is, (1/2)"(1/3)"u[n] is absolutely summable but (1/2)"(1/3)"u[-n) is not. It is also known that z2[n] = (1/4)"z[n] is not absolutely summable. For this to be true, there must be a pole somewhere in the annulus 3 < [2] < 4 that corresponds to a left-sided signal; such a mode when multiplied by (1/3)" is still absolutely summable but when multiplied by (1/4) is not absolutely summable. Thus, ef) is a two-sided signal. Im polar form, the known pole is at 2 = /18/16€7*/4, To be absolutely summable, the signal's region of convergence must include the unit circle, |2| = 1. (a) Yes, the signal can be left-sided. Since the known pole is outside the unit circle, a region of convergence that includes the unit circle (needed for absolute summa- bility) implies that the known pole corresponds to a left-sided component of the signal. . (d) No, the signal cannot be right-sided. If the known pole outside the unit circle is right-sided, the region of convergence cannot include the unit circle and the signal cannot be absolutely summable as required. S Yes, the signal can be two-sided. Let the known pole correspond to a left-sided component and let there be another pole within the unit circle that corresponds to a right-sided component. Such a signal is two-sided and has a region bf convergence that includes the unit circle, which ensures the signal is absolutely summable as required. (d) No, the signal cannot be finite duration. Pinite duration signals cannot have poles in the region 0 < [z] < oo. Such poles, such as the pole known to exist, correspond to time-domain components with infinite duration. (a) = a: Next, N(2) = H(2)Xi(2) = a Thus, Yi(2)/z = 5a — ia + a or Yi(2) = 6, + 385. Inverting yields n 1 nb (1/2)Pufn] + 5 (8/4) und (b) The idea of frequency response is used to determine the ontput in response to the everlasting exponential input z2[n] = (3/4). That is, H(z = 3/4) = 34=U3 = é = É Thus, 1 vali E (3/4)". The given signal is z[n) = (—1)"u[n — no] + a"u[-n]. 1f Jal = 2, the z-transform X(2) = (—1)tzno 27 4 =E has region of convergence 1 < |z| < Ja] = 2, as desired. Thus, the necessary constraint is lal= There is no constraint on the integer no, other than it be finite. (a) Xi(z) = 3 til” = Des (ND TPul-n)+óf-n)z" = Dino (5) + DX able] = DO (2) 41 Por fe] < 1, this be comes 0-3/z Kl)=1+7557=1+ ROC |2]<1. 273 6) Xo(z) = Dezinimo = ol)" cos(n + Duln) = Do3"0.5 (emu + eimoth(n+1)) a" = 4 0.5€7 (12) + Dito0.5e7a (22)", por |z] > 1, this becomes 0.5eº = X(2) = Ta 1; ROC |H> 1. (0) Xa(z) = tan” = Dito (10.5(€" -euf-n+ 12" = te-00 10.5 ((£)" — (E)"3. For [2] < e and |2] < €7!, this becomes Xa(2) — PURE = Ef). Ts, E - tr rocpI<e . (4) Xa(2) = Dh-oo Dn. oo (2)" ófm — 24] = 4, this becomes Dic (Etc ool29)"An 26) 2 = - For |22| < 2) = Disco (É 42 “dz Xa(z) = ; ROC |z]<2. 4 Note: |2%| = 2, so the region of convergence is |2]? < 4 or |z| < 2. 5.9-10. (a) The signal Xy (2) = == one must be real. Using the region of convergence, we know that a real root must be either 0.5, —0.5, 2, or —2. Trying these values, we find that z = —0.5 and 2 = —2 are both roots of the denominator. The remaining root can be found by noting that the denominator 1 + d8z-1.4 12-2.. 12-3 must be equal to (+0.52"D( +220D(14 4273) Equatin the Der of z73 on each side yields -1/3=A. Thus, Xi(2) = tros nm nnc=3 Expanding yields Xi(2) = Tr + Er + 1288, Using the region of convergence (0.5 < [2] < 2) and tables, the inverse is 3 has three finite poles, of which at least .a n Beomilno 2 n calo] == (1/2ufo) — (-2"ufon = 1) + So (1/3 un (b) Xa (2) = = em = “Pam = ep(s q + Tê) Using the region of convergence (0.5 < |2] < 2) and tables, die i inverse is zo[n] = BU ulm +3]- Han -4. s . SS (Me) = mms = medos - (=p + =p) Tr =D 1 (a + xi). Since hn is stable, the region of convergence for H(2) must be [2] > 1/2. Úsing this ROC and z-transform tables, the inverse transform is dalr] = (1/2""ufn =) = (=1/29)"ufn — 1). 274 X(2) (2? + 1). Thus, the system function is H(2) = SE) = ot) = 0.25 (125, x) 2 dz (a) MATLAB is used to create the pole-zero diagram. (b) H(eº) = qa >> zz = roots([1 O 1]); zp = roots([4 O -1]); >> theta = linspace(0,2+*pi,201); >> plot(real(zz), imag(zz), 'ko',real(zp), imag(zp), 'kx”, cos(theta),sin(theta),ºk”); >> xlabel(”Re(z)'); ylabel('Im(z)'); grid; >> axis([-1.1 1.1 -1.1 1.1]); axis equal; 08 04 oa £o -08 2H Figure $5.M-la: Pole-zero diagram for H(2) = Fil. rea MATLAB is used to plot the magnitude response z=e8 (e. >> Omega = linspace(-pi,pi,501); >> z = exp(j*Omega); H = (2.72+1)./(4+2.72-1); >> plot (Omega, abs(H),ºk'); axis([-pi pi O 1]); grid; >> xlabel("NOmega?); ylabel(” |H(e-(jN0mega)) |”); >> set(gca,'xtick”, [-pi:pi/4:pil,'xticklabel”, ['-p POR OP OP O jr apr o po, *fontname”,' symbol”); The pole-zero plot of Figure $5.M-la and the magnitude response plot of Figure S5.M-1b confirm that this is a band-stop system. Yes, the system is asymptotically stable. Referring to Figure S5.M-la, all the system poles are within the unit circle. Yes, the system is real. Since the system is expressed as a constant-coefficient linear difference equation with real coefficients, the impulse response h[n] and system are both real. For an input of the form zf[n) = cos(9m), the greatest possible amplitude of the output corresponds to the greatest gain shown in the magnitude response plot of Figure $5.M-Jb. Thus, 2/3 is the greatest output amplitude given an input of <[n) = cos(Qm). This output amplitude occurs when Q = kr, for any integer k. Ay an + 2] + 2! Figure $5.M-1b: Magnitude response plot for 4yfn + 2] — y[nl) x[n] Xin 1/4 2 + A z 1/4 À -14 E Figure $5.M-1g: TDFII implementation of y[n] — 0.25yfn — 2) = 0.25a[n] + 0.25cfn — 2]. (g) Inverting H(z) = SE) = 0.257Ht2>, provides y[n] — 0.25yfn — 2] = 0.25x[n) + 0.25x[n — 2], which is a convenient form for implementation. Figure 85.M-1g illustrates a TDFII implementation of the system. M.2. (a) H(2) = ofsta = COLE! Thys, there are zeros at z = esr/4 and —33n/4 z=e , and there are poles at z = 0.9e737/4 and z = 00. >> zz = roots([1 O -5]); zp = 0.9+exp(j+3+pi/4); >> theta = linspace(0,2+pi,201); >> plot(real(zz), imag(zz),'ko”,real(zp) imag(zp),'kxº,... cos(theta),sin(theta),ºk'); >> xlabel("Re(z)?); ylabel('Im(z)º); >> axis([-1.1 1.1 -1.1 1.1]); axis equal; grid; Since this is a complex system, poles and zeros need not occur in complex con- jugate pairs. Substituting z = 2 into H(z) = — bon, MATLAB is used to create the (b 967 magnitude response plot. >> Omega = linspace(-2+pi,24pi,1001); z = exp(j+Omega); >> H = (2.02-5)./(2-0.9+exp(j+3+*pi/4)); >> plot(Omega,abs(H),º'k?); axis([-2+*pi 2+pi O 21)); >> xlabel(*WOnega”); ylabel(? |H(e"(jNOmega)) |); grid; 278 ta) 2 Figure $5.M-2a: Pole-zero diagram for H(z) = —Est=a- 1/4:2+pil,?xticklabel”, ['-2p”; Po 0 09; 5 *5? 2p'], "fontname”, 'symbol?) >> set(gca, 'xtick”, [-2+pi Door o; po Figure $5.M-2b: Magnitude response plot for H(z) = —F=ta- From Figure $5.M-2b, the system appears to be a type of bandpass filter. This particular filter tends to pass only positive frequency inputs near Q = 3x/4; the corresponding negative frequencies near Q = —37/4 are significantly attenuated. Since the magnitude response is not an even function of O, the output of the filter will be complex-valued. 5.M-3. (a) H(z) = 555/55) has four finite zeros and two finite poles. The zeros are the four 2(27+0.819) roots of unity, and the poles are at z = 0.9e77/4 and 2 = 0.9e-337/4, MATLAB is used to compute the pole-zero plot. >> zz = roots([1 0 0 0 -1]); zp = roots([1 O j+*0.81)); >> theta = linspace(0,2+pi,201); >> plot(real(zz), imag(zz),'ko',real(zp), imag(zp),'kx,... cos(theta),sin(theta),ºk'); 279 >> plot(Omega,Hmag,'k'); axis([-pi pi O 2.5]); grid; >> xlabel(ºNOmega'); ylabel(” |H(e-LjNOmega)) |º); >> set(gca,'xtick”, [-pi:pi/4:pil,'xticklabel”, [?-p ?;.. POOR OD OD Gr o op po, *fontname”, symbol”); Figure 85.M-4b: |[H(eº)| = VE E Standard filter types do not provide a good description of this filter. The system appears most like a bandstop filter, but its stopband attenuation is quite poor. The system boosts the gain of low and high frequencies more than it attenuates the middle frequencies. ey E , 2 2 (c) Inverting H(z) = bs = ma + a yields fm) = 0.5 ((1/V2)" + (-1/V3) u 5.M-5. Label the first summation block output as v[n]. Thus, v[n] = zfn — 1] + iytn] or V(2) = 27! X(2) + 1Y (2). The output of the second summation block is yin +1) = alnJtuln=1) or zY(z) = X(2)+2"!V(2). Combining the two equations yields zY (2) = X(2) +20 (2 X(2) + 4Y(2)) or Y (2) (2- 121) = X(2) (1+272). Multiplying both sides by 2") yields Y(2) (1 — 4z-2) = X(2) (2-2 4.253). Thus, H(2) = HD = 2tpzo 22 - 5/2 52 -1.52/2 -1 52/2 Es is =mGas + ia+ ce =Prrtda+ ia: Taking the inverse transform yields 4 bn] = 40 = + 501/2"upn o 1)+ Hatha) Since hfn] = 0 for n < 0, the system is causal. The system has three poles located at 2 = 0, 2 = 1/2, and z = —1/2. Since all three poles are inside the unit circle, the system is stable. Rr 2 5.M-6. Since h[n] = 6[n—1]+ó[n+1], H(2) = 2!+z and |H (e) = = 2] cos(9). >> Omega = linspace(-pi,pi,501); 282 >> Hmag = 2*abs(cos(Omega)); >> plot(Omega,Hmag,'k'); axis([-pi pi O 2.5]); grid; >> xlabel(?NOmega'); ylabel(” IH(e(jNOmega)) |'); >> set(gca, 'xtick*, [-pi:pi/4:pil,?xticklabel”, ['-p *; Po op o OP jr Op DP po. *fontname”,*symbol?); Figure 85.M-6: |H(e'?)| = 2] cos(9)]. Using Figure $5.M-6, this system is Dest described by a bandstop filter with gain. That is, low and high frequencies have à boosted gain of two and middle frequencies near Q = 1/2 are attenuated. 5.M-7. (a) >> Omega = linspace(-pi,pi,501); z = exp(j+Omega); >> H = cos(1./z); >> plot(Omega,abs(H),'k'); axis([-pi pi O 2]); grid; >> xlabel (*NOmega'); ylabel(” |H(e-(jNOmega)) |'); >> set(gca,'xtick?, [-pi:pi/4:pil,ºxticklabel”, [?-p *; Pop op op OP dr op dp po. “fontname”,º symbol); Figure $5.M-Ta: Magnitude response for H(z) = cos(z-!) 283 d From Figure S5.M-Ta, it appears that this system behaves somewhat like a band- pass filter. Frequencies near Q = 7/2 are boosted while frequencies near Q = km are somewhat attenuated. (eat A Maclaurin series expansion of cos(z) is Jyco “aor- Substituting n = 2" vields H(z) = cos(2-3) = D$£o CE. Inverting vields e (1) 1 1 1 hnJ=5> Tr ln — 2] = ófn] — sôfn — 2] + pôln — 4] — gole +... & Em 1 i i >> n = [0:10]; h = zeros(size(n)); >> k = [0:5]; h(2*k+1) = (-1).-(k)./(gamma(2+k+1)); >> stem(n,h,ºk?); xlabel('n'); ylabel(?h[n]'); >> axis([-.5 10.5 -.6 1.1]); Pigure $5.M-7b: Impulse response hfn] for H(z) = cos(z71). (c) Using Figure S5.M-7b, it is clear that the impulse response quickly decays to zero. Thus, only the first few terms from h[n) are needed for a good approximation. Note that h[n] = ó[n] — ófn — 2]/2.+ ó[n — 4]/24. An FIR difference equation is found by letting h[n] = y[n) and ófn] = zfn). That is, yin] = efn] — zfn — 2]/2 + afn — aJ/24. This is a fourth-order FIR filter with only three non-zero coefficients. The mag- nitude response is casily computed using MATLAB. >> Omega = linspace(-pi,pi,501); z = exp(j*Omega); > H=1-2.0(-2)/2 + 2.0(-4)/24; >> plot(Omega,abs(H),'k); axis([-pi pi O 2]); grid; >> xlabel(ºNOmega'); ylabel(” |H(e(jNOmega)) |"); >> set(gca, 'xtick”, [-pi:pi/4:pil,*xticklabel?, ['-p *;. POP O O OP op apr as poT. *fontname”, 'symbol?); Visually, Figure S5.M-Tc is indistinguishable from Figure $5.M-7b. Thus, the FIR filter appears to closely approximate the original system. Roe) Figure S5.M-8c: Order-8 Butterworth BPF with passband between 57/24 and 117/24. (d) Order-8 Butterworth BSF with stopband between 57/24 and 117/24. Notice that the command butter requires the parameter N = 4 to be used to obtain a (2N = 8)-order bandstop filter. >> Omega c = [5*pi/24,11+pi/24]; Omega = linspace(-pi,pi,1001); >> [z,p,k] = butter(4,0mega c/pi,'stop'); * >> subplot(121),plot(real(p),imag(p),'kx”,... real(z) imag(z), ko” ,cos(Omega),sin(Omega),'kº); >> axis([-1.1 1.1 -1.1 1.1)); axis equal; >> xlabel(*Re(z)?); ylabel('Im(z)); >> [B,A) = butter(4,0mega c/pi,'stop”); >> HBS = polyval(B,exp(j+Omega))./polyval(A,exp(j+Omega)); >> subplot(122),plot (Omega, 20+10g10 (abs (HBS)),k'); >> axis([-pi pi -40 2]); grid; >> xlabel(ºNOmega'); ylabel(” |H (BSH(e"(jNOmega)) |); >> set(gca,?xtick”, [-pi:pi/3:pil,'xticklabel?,[?-p ?;... »oDpo op O pr pr po, *fontname”, symbol”); Figure S5.M-8d: Order-8 Butterworth BSF with stopband between 51/24 and 117/24. 287 5.M-9. Factored form is used to plot roots, and standard transfer function form is used to compute magnitude response plots. (a) Order-8 Chebyshev Type I LPF with Q = 7/3. >> Omega c = pi/3; Omega = linspace(-pi,pi,1001); >> [z,p,k] = chebyi(8,3,0mega c/pi); >> subplot(121),plot(real(p) ,imag(p),'kx',... real(z),imag(z),'ko”,cos(Omega),sin(Dmega),'k?); >> axis([-1.1 1.1 -1.1 1.1]); axis equal; >> xlabel('Re(z)'); ylabelC' Im(z)º); >> [B,A] = cheby1(8,3,0mega c/pi); >> HLP = polyval(B,exp(j*Omega))./polyval(A,exp(j*Omega)); >> subplot (122) ,plot (Omega, 20+10g10(abs(HLP)),?k?); >> axis([-pi pi -40 2]); grid; >> xlabel("NOmega”); ylabel( IH (LPJ(e-(jNOmega)) |?); >> set(gca,'xtick”, [-pi:pi/3:pil,'xticklabel”, [?-p ?;... » Op op Op dp dp pa. *fontname”, 'symbol”); reta Figure S5.M-9a: Order-8 Chebyshev Type I LPF with M = 7/3. (b) Order-8 Chebyshev Type I HPF with Q. = 1/3. >> Omega c = pi/3; Omega = linspace(-pi,pi,1001); >> [z,p,k] = cheby1(8,3,0mega c/pi,"high?); >> subplot(121),plot(real(p), imag(p),ºkx”,... real(z),imag(z),'ko”,cos(Omega),sin(Omega),ºk?); >> axis([-1.1 1.1 -1.1 1.1]); axis equal; >> xlabel(Re(z)'); ylabel('Im(z)º); >> [B,A] = cheby1(8,3,0mega c/pi,"high'); . >> HHP = polyval(B,exp(j+Omega))./polyval(A,exp(j+*Omega)); >> subplot(122) plot (Omega, 20+10g10(abs (HHP)),k); >> axis([-pi pi -40 2]); grid; >> xlabel('NOmega”); ylabel(? |H (HPJ(e"(jNOmega)) |"); >> set(gca, 'xtick”, [-pi:pi/3:pil,'xticklabel”, ['-p ?;... Pop op Op pr pr po), *fontname”,'symbol?); 288 as dE “ o . q Bete, 2 Figure $5.M-9b: Order-8 Chebyshev Type I HPF with Mc = 1/3. (c) Order-8 Chebyshev Type I BPF with passband between 57/24 and 117/24. No- (d tice that the command cheby1 requires the parameter N = 4 to be used to obtain a (2N = 8)-order bandpass filter. >> Omega c = [5+pi/24,11+pi/24]; Omega = linspace(-pi,pi,1001); >> [z,p,k] = cheby1(4,3,0mega c/pi); x >> subplot(121),plot(real(p), imag(p),'kx?,... real(z),imag(z),ºko”,cos(Omega) ,sin(Omega),ºk”); >> axis([-1.1 1.1 -1.1 1.1]); axis equal; >> xlabel('Re(z)?); ylabel(" Im(z)'); >> [B,A) = cheby1(4,3,0mega c/pi); >> HBP = polyval(B,exp(j+*0mega))./polyval(A,exp(jtOmega)); >> subplot(122) plot (Omega, 20*+10g10 (abs (HBP)),'k”); >> axis([-pi pi -40 2]); grid; >> xlabel(?NOmega”); ylabel(” IH (BPJ(e"(jNOmega)) |); >> set(gca,'xtick”, [-pi:pi/3:pil,'xticklabel?, ['-p º;. Doo op Op à; Pp... *fontname”, *symbol?); Order-8 Chebyshev Type I BSF with stopband between 57/24 and 117/24. No- tice that the command cheby1 requires the parameter N = 4 to be used to obtain a (2N = 8)-order bandstop filter. >> Omega c = [5+pi/24,11+*pi/24]; Omega = linspace(-pi,pi,1001); >> [z,p,k] = cheby1(4,3,0mega c/pi,'stop'); >> subplot(121),plot(real(p), imag(p), 'kx”,... real(z),imag(z),'ko”,cos(Omega),sin(Omega),ºkº); >> axis([-1.1 1.1 -1.1 1.1]); axis equal; >> xlabel('Re(z)'); ylabel(' Im(z)'); >> [B,A] = cheby1(4,3,0mega c/pi,'stop'); >> HBS = polyval(B,exp(j*Omega))./polyval(A,exp(j*0mega)); >> subplot(122) plot (Omega, 20+10g10(abs (HBS)),ºk'); >> axis((-pi pi -40 2]); grid; >> xlabel(? Omega”); ylabel(? IH (BSJ(e-LjN0mega)) |); >> set(gca, 'xtick”, [-pi:pi/3:pál,'xticklabel?, ['-p *;... Por OPG Gr PR pI... 289 (c may st E CER a = Reta, a Figure S5.M-10a: Order-8 Chebyshev Type II LPF with Lc = 7/3 >> HHP = polyval(B,exp(j+Omega))./polyval (A,exp(j+Omega)); >> subplot (122) plot (Omega, 20+10g10 (abs (HHP)),k”); >> axis([-pi pi -40 2]); grid; >> xlabel(º Omega”); ylabel(”? |H (HP)(e"(jNOmega)) |"); >> set(gca, 'xtick', [-pi:pi/3:pil,ºxticklabel?,['-p ;... POOR O GP Pr 52 pI. + *fontnamé*,* symbol”); Man Reta a Figure $5.M-10b: Order-B Chebyshev Type IL HPF with Q, = 7/3. Order-8 Chebyshev Type II BPF with passband between 57/24 and 117/24. Notice that the command cheby2 requires the parameter N = 4 to be used to obtain a (2N = 8)-order bandpass filter. >> Omega c = [5*pi/24,11+pi/24]; Omega = linspace(-pi,pi,1001); >> [z,p,k] = cheby2(4,20,0mega c/pi); >> subplot(121),plot(real(p),imag(p),'kx”,... real(z),imag(z),'ko”,cos(Omega) ,sin(Omega),*k'); >> axis([-1.1 1.1 -1.1 1.1]); axis equal; >> xlabel('Re(z)”); ylabel( Im(z)'); >> [B,A] = cheby2(4,20,0mega c/pi); 292 >> HBP = polyval(B,exp(j*Omega))./polyval(A,exp(j*Omega)); >> subplot(122) plot (Omega, 20+10g10 (abs (HBP)),?k?); >> axis([-pi pi -40 2]); grid; >> xlabel(?NOmega”); ylabel(? |H (BPJ(e(jNOmega)) |"); >> set(gca,'xtick?, [-pi:pi/3:pil,'xticklabel?, ['-p ?;... Doo Op O apr Op po), *fontname”,?symbol'); DT fim ab Ss d+ a * not Figure $5.M-10c: Order-8 Chebyshev Type II BPF with passband between 57/24 and 17/24. (d) Order-8 Chebyshev Type II BSF with stopband between 57/24 and 117/24. Notice that the command cheby2 requires the parameter N = 4 to be used to obtain a (2N = 8)-order bandstop filter. >> Omega c = [5+pi/24,11+pi/24]; Omega = linspace(-pi,pi,1001); >> [z,p,k] = cheby2(4,20,0mega c/pi,'stop'); >> subplot (121) ,plot(real(p),imag(p),'kxº,... real(z), imag(z), 'ko”,cos(Omega), sin(Omega),ºk'); >> axis([-1.1 1.1 -1.1 1.1]); axis equal; >> xlabel('Re(z)'); ylabel( Im(z)'); >> [B,A] = cheby2(4,20,0mega c/pi,'stop'); >> HBS = polyval(B,exp(j+Omega))./polyval(A,exp(j*Omega)); >> subplot (122) plot (Omega, 20+10g10(abs (HBS)),ºkº); >> axis([-pi pi -40 2]); grid; >> xlabel(*NOmega”); ylabel(” |H (BS)(e"(jNOmega)) |'); >> set(gca,xtick”, [-pi:pi/3:pi],'xticklabel?,['-p *;... POP OP O Op op po. *fontname”, symbol); To demonstrate the effect of increasing Rs, consider magnitude response plots for Chebyshev Type HI LPPs with Rs = (10, 20,30). >> Omega c = pi/3; Omega = linspace(-pi,pi,1001); >> [B,A] = cheby2(8,10,0mega c/pi); >> HLP1 = polyval(B,exp(j+Omega))./polyval(A,exp(j+Omega)); >> [B,A] = cheby2(8,20,0mega c/pi); >> HLP2 = polyval(B,exp(j+Omega))./polyval(A,exp(j*Omega)); >> (B,4) = cheby2(8,30,0mega c/pi); o 293 ima) at note, a Figure S5.M-10d: Order-8 Chebyshev Type IL BSF with stopband between 57/24 and 1x/24. >> HLP3 = polyval(B,exp(j+Omega))./polyval(A,exp(j*Omega)); >> plot (Omega, 20+10g10 (abs (HLP1)),*k-?,... Omega, 20+10g10 (abs (HLP2)), *k Omega, 20+10g10 (abs (HLP3)), *Xk >> axis([-pi pi -40 2]); gria; >> xlabel("NOmega'); ylabel(” IH (LPHeLjNOmega)) 1º); >> legend('R s = 10º,ºR.s = 20º,ºR.s = 30",0); >> set(gca,'xtick”, [-pi:pi/3:pil,'xticklabel?, ['-p 25... POP OP OD dr Sp poT,. *fontname”, symbol); Figure $5.M-10d: Changing R, for a digital Chebyshev Type II filter. Thus, increasing Rs tends to broaden the transition bands of the filter. 5.M-11. Factored form is used to plot roots, and standard transfer function form is used to compute magnitude response plots (a) Order-8 Elliptic LPF with Q. = 7/3. >> Omega c = pi/3; Omega = linspace(-pi,pi,1001); 294
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