Baixe Resolução Sinais e Sistemas Lineares - BP Lathi e outras Notas de estudo em PDF para Engenharia Elétrica, somente na Docsity! Solution Manual for B. P. Lathi Signal Processing and Linear Systems Oxford University Press Oxford University Press Oxford New York Athens Auckland Bangkok Bogotá Buenos Aires Calcutta Cape Town Chenna! Dar es Salaam Delhi Florence Hong Kong Istanbul Karachi Kuala Lumpar Madrid Melbourne Mexico City Mumbai Nairobi Paris São Paulo Shanghai Singapore Taipei Tokyo Toronto Warsaw and associaled companies in Berlin Thadan Copyright O 2001 by Oxford University Press, Enc. Published by Oxford University Press, Inc, 198 Madison Avenue, New York, New York, 10016 dettp:// verve op-usa org Oxford is a registered trademark of Oxford University Press AN rights reserved. No part of'this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, withont the prior permission of Oxford University Press. Library of Congress Cataloging-in-Pubilication Data TSBN 0-19- Printing number: 987654321 Primted in the United States of America on acid.fico paper á 4 O t+ 5 7 ol 4 (es o + 2t»> 1.4-1 Al the siguals are shown in Fig. S1L4-1. 1.4-2 HO = MA Du + 1) = (O + (2 + Salt) — u(t— 2) = (84 Du(t +41) — Gtu(t) 4 Bu(t) + (28 A)u(t— 2) p= E) -ut- D+ (-B)fu(t=-2)-u(t-4)]=u(t)- (1º > 28 + BJu(t — 2) (2t — Bju(t — 4) Es [ Eropa= [ Pigdi = Er. Eno [ red Plojde = E; Emen = f. “te-mPa- f. C Ptode = Er Eray= Ff. tao)f a rh, fila)da = Eslo Es(at-) -f. Lf(ar — Pt = :f. fledr=Es/a. Esto) -f. Lrteza? ao [* Plejdt=0Es Euse É last = [" Pat = Er 1.4.4 Using the fact thar f(z)ó(z) = f(0)é(x), we have (a) 0 (b) Zó(w) (e) 16) (d) Po 1) (e) z5zów+3) (f) kó(w) (use L' Hôpital's rule) 1,4-5 In these problems remember that impulse 5(z) is located at z = 0. Thus, au impulse d(t — 7) is located at 7 =, and so on. (a) The impulse is located at + = t and f(7) at 7 = t is f(t). Therefore / frott = 1) dr = 10) (b) The impulse ó(r) isat 7 = Gand f(t- 1) at 7 = 0 is (4). Therefore / single ar = 10 Using similar arguments, we obtain (1 (dO (eye (85 (gH-1) (h)-e TO 41 Ls s & a | LO 1 O 2. t-— (2 €3 tê > 1.4-6 1.4-7 1.4-8 1.4-9 Fig. SL.4.7 (a) Recall that the derivative ol à function at the jump discontinuity is equal to an impulse of strength equal to the amount of discontinnity. Hence, df/dt contains impulses 46(t + 4) and 26(t — 2). in addition, the derivative is —1 over the interval (—4, 0), and is 1 over the interval (0. 2). The derivative is zero for t < —4 and t > 2. “Fhe result is sketched in Fig. S1.4-6a. (b)Using the procedure in part (a), we find d? f/dt? for the signal in Fig. P1.4-2a as shown in Fig. 81.4-6b. (a) Recall that the area under an impulse of strength k is k. Over the interval 0 < t< 1, we have 1 n= frdome 0<t<l o Over the interval O < t < 3, we have 1 + ua = [ tar+ fara 1<t<3 o 1 Att=3, the impulse (of strength unity) yields an additional term of unity. Thus, 1 3 + vo = [ nar+ f (dos / dz-3)dr=1+(-D)+1=0 +>3 o À ame (b) + vt) fh-se-n-me-n-sa-a+ que tu) cut Due ut- sm) Changing the variable t to —z, we obtain fe somna-- f “ olcajsta) de o f É olca)ó(a) de = 6(0) oo This shows that f sosga= | AlBa(=t) dt = (0) Therefore s(t) = 6(-8) Letting at = x, we obtain (for a > 0) [ é(t)5(at) dt = f Eyó(x) de = Lo(0) o a a a Similarly for a < O, we show that this integral is — 16(0). Therefore f. elt)ó(at) dt = Feto) =d f a(t)ó(t) dt a Therefore ólat) = mito 31.4-10 (a) / S(BCL) dt = =0- / &(njótt) dt = (0) SEDES — [ “esta 1.4-11 (2) s12 = +93 (b) «TJ cos 3t = 0.5/eC+28t 4 o--i3], Therefore the frequencies are 5, Using the argument in (b), we find the frequencies s12 = 2+33 (d)s=-2 ()s=2(1) s=0. iz “a Fig. S1.4-11 1.5-1 (a) fit) = 0.5fu(t) + u(-1)] = 0.5 and folt) = 0.5[0(1) — (= 0]. tb) felt S[tu(t) — tu( 0.54] and fo(t) = O5[tu(2) + tu(>8)] = 0.54 (e) felt) = 0.5[sin wot u(t) + sin(-wot)u(—8)| = 0.5[sit wot u(t) — sin vot u(—t)] and fo(t) = 0.5[sinwot u(t) — sin(=wot)u(- 0) = 0.5[sin ot u(t) + sin ant u(—t)] = OS sinwot. (a) fole ájcos wnt w(t) + cos(—wot)Ju(—t)] = O.5jcos wut u(t) + coswat u(—t)] = 0.5 cosuot aud fo(t Sfcos vot u(t) — cost —wat)u(-t)] 5[cos moi u(t) — coswntu(-t)). (e) Ft) = O.5fsinwot + sin(>wot]] = 0 and fo(t) = O.5[sinwot — sin(—wot)] = sinwot (E) Jo(t) = O.5[coswot + cos(--wot)| = coswat and fu(t) = D.5[coswot — cos(—wot)) = O £ GA) £ tia Va E, ol t+» jo t> || -1 oltos Ya fo fe fo ne +, fe 4 x + Rr t+ | e O ta “ea NV (e 0 | o (£) Fig. S1.5-1 ts 1.6-1 IE f(t) and W(t) are te input and output, respectively. of an ideal integrator. thes dt) = fíti + q pt + no= sema f siopa+ f flndr= u(o) +[ Hdr —o —oe o Da o zerrinput Sam ama (a) zero-stato and 1.7-4 Only (b), and (hyare linear. All the remaining are nonlinear. This can be verified by using the procedure «liscussed in Example 1.10. 1.7-2 (a) The system is time-invariant because the input /(£) yields the output y(t) = f(t — 2). Hence. if the input is f(t— TP), the outputis [(t— T— 2) =y(t— F). which makes the system time-invariant. (b) The system is time-varying. The input f(t) vields the output y(t) = f(=t). Thus. the output is obtained by changing the sign of tin f(t). Therefore, when the input is f(t - T). the outputis f(-4-T) = HM-E+T) = gli + T). which represents the original output advanced by 7 (not delayed by 7). 5 1.8-3 Substituting of (2) in (1) yields or or But Diferentiation of (2) yields end or and substituting this in (1) vields or m(b=Dyott) or ut) = auto Di+2D+2 DAS q Dito) (DD? +2D +2)n(o) = D2 ft) la:(t) — qo(H)]At = ASA Ro = Aleu(t) = goto ao(t) = Rh(f) . R do(t) = RA(t) = qlai(t) — ao(b)] (0+ 8) mto Sato (D + aJao(t) = agilt) a= mim ao(t) = suo) dm) = ã (1 - 25) (0 = ab AD ro) (D+ ah) = Sao) (3) 1) (2) 63) Chapter 2 2.2-2 2.2-3 2.2-4 The characteristic polynomial is A2 + 5146. The characteristic equation is A2+5A 46 =D. Also A2+5A 46 = (A+ 2)(A + 3). Therefore the characteristic roots are A = —2 and Àz = —3. The characteristic modes are e? and eT3!, Therefore valt) = e + coe t and 2 ae to(t) = Zee”? — 3eze” Setting t = 0, and substituting initial conditions yo(0) = 2, to(0) n —1 inthis equation yields ea+r=2 q=5 = —2e; — 3) = —1 a=-3 Therefore volt) = 5e* 3e7't The characteristic polynomial is AZ + 4X + 4, The characteristic equation is A2+4A+4 =0. Also 2 +4A+4 = (A 42)2, so that the characteristic roots are —2 and —2 (repeated twice). The characteristic modes are "2 aud te-%, Therefore -at =2t volt) = ce + cate and 2é at tolt) = —2e;e”* — Degte PE + ae” Setting é = O and substituting initial conditions yíelds 3=m q=3 => -4 = —2c, +e2 e2=2 volt) = (3 + 29e Therefore “The characteristic polynomial is MA+1) = A2+A. The characteristic equation is M(A+1) = 0. The characteristie roots are O and —1. The characteristic modes are 1 and e”!. Therefore volt) = cx + exe t and io(t) = —coe”* Setting t = 0, and substituting initial conditions yieids l=e+ez =» 1=-c, volt)=2-€" “Therefore + The characteristic polynomial is A? + 9. The characteristic equation is A? +9=Dor(A+33MA— 73) = 0. The characteristic roots are +;3. The characteristic modes are «73! and e” 23, Therefore volt) = ecos(3t + 0) and do(t) = —Sesin(3t + 9) Setting t = 0, and substituting initial conditions yields 0=ccoso ecos9 =0 c=2 6=—3esing esinô = —2 B=-7/2 Therefore 2.2.5 2.2-6 2.27 2.3-1 volt) = 2eos(3t — z =2sin3t The characteristic polynomial is A? 4 4) +13. The characteristic equation is A2+4A +13 = 0o0r (A+2-j3XA+ 2+33) = 0. The characteristic roots are —-2 + ;3. The characteristic modes are ce! 29! and cge!2-i)t Therefore volt) = rc" * cos(3t + 8) and do(t) = —2ee "2 cos(at + 8) — 3ce E sin(3t + 9) Setting i = 0, and snbstituting initial conditions yields 5=ccosf ecosê =5 10 . => => 15.98 = —2e cosg — 3esing csing = —8.66 = —n/3 Therefore volt) = 10e"* cos(3t — z The characteristic polynomial is A2M(A + 1) or Xº 4+ A2. The characterístic equation is A(A + 1) = 0. The characteristic roots are 0, O and —1 (0 is repeated twice). Therefore volt) = +cot+ege”t and volt) = co — ese! djo(t) = case”! Setting t = 0, and substituting initial conditions yields =e+es = =eo-ei=> m=2 -1=e ea=-1 Therefore volt)=5+2t- e”! The characteristic polynomial is (A+ 1)(A2 + 5A +6). The characteristic equation is (A + 1)(12+5A+6) = 0 or (A+INA+2)A+3) =0. The characteristic roots are —1, —2 and —3. The characteristic modes are e and cT%, Therefore volt) = cre! + eae! + eae! and e a volt) = —creT! — 2epe 2 — 3ege” iolt) = cet + 4rge U 4 Bege Setting t = 0, and substituting initial conditions yields 3= -i=-o-2eo-Ss) =» = d=c)+4e2+9%s o= nn +e+es e Therefore volt) = 6"! — Te E + ge The characteristic equation is 22 + 4) +3 = (A+1IMA+3) = 0. The characteristic modes are e! and e, Therefore gu(t) = crer! + eze dt Pnlt) = —ereT! = 3ege Setting t = 0, and substituting y(0) = 0, y(0) = 1, we obtain 1 O=m+ez ) ass 1= =» - 2.4-6 2.4-7 2.4-8 + tu(t) x u(t) = f ru(r)u(r — t)dr o The range of integration is 0 < 7 <t. Therefore r >Dandr —-t>0sothatu(r)=u(r-t)=1 and + A tu(t) + u(t) = Tdr=— t>0 o 2 and 1 tult) + u(t)= atuo) ti) + sintu(t) + u(t) = (f sinru(r)u(t 7) 4) ult) o Because 7 and é — + are both nonnegative (when 0 < 7 <t) u(T)=u(t-7T)=1,and e sintu(t) cult) = (/ snrér) uít) = (1 cost)u(t) o (ii) Sintilarly costu(t) x u(t) = (/ cosrdr) u(t)=sintu(t) o In this problem, we use Table 2.1 to find the desired convolution. (a) u(t) = A($) + f(t) = entu(t) + u(t) = (1 — enNu(t) (b) xt (e ft) =etult) xe tu(t) = tetu(t) (e) u(t) = e-tu(t) we Put) = (e! — e E u(t) (d) y(t) = sin3tu(t) we *u(t) Here we use pair 12 (Table 2.1) witha = 0,8 = 3,6 = —90º and A=—1. This yields é = tan” [5] = —1084º and sinBtu(t) eectult) = (otite et 18) (o = 0.94860"* — sen + 184º) (5) te) u(t) = (20 = ut) x u(t) = 20 Vult) + u(t) — cult) + ut) - [ie 1er o = (E ess ea) uít) (by (2e tt — e u(t) » ectult) = 2e ut) ne" tu(t) — e Xu(t) + e tult) - pi e —] ue) = (0 Put) (e) v(t) = (203 — e u(t) se Cult) = 26" *ult) we ult) — e Pu(t) xe Pult) = pi - 7] e(t) =[((2- e - 2e*u(t) 13 24-9 2.4-10 2.4-11 tule) e u(t) = ePult) eu(t) — 2te Cult) x cult) E (Edo 2) um eu Ho =(1-20e (a) For ut) = 4"? cos3tu(t) + u(t), We use pair I2 witha =2,8=3,6=0,A=0. Therefore é=tan! 5] = —56.81º and = cos(3t + 56.31º =a ee ut) = = [0.555 — e—* cos(3t + 56.31º)] u(£) (b) For y(t) = de? cos3tu(t) » e tu(t), we use pair 12 witha=2,6=3,8=0,and À=-—1. Therefore =tanT!j OD oct [5 and u(t) u)=4 [eee = e ce + 11867) v10 = (0.816! — e cos(8t + 71.56º)] u(t) =4 [= - e cos(3t + 1569) u(t) (a) v(t) = etu(t) + etu(t) = (e! e2u(t) (b) e Butt) = e%e-*u(s), and y(t) = eº [e-tu(t) x e2u(6)] = e%(e7! — cut) (0) e Hu(t—3) = e Se Ht-Bu(t— 3). Now from the result in part (a) and the shift property of the convolution [Eq ( 3a) n(t) = 78 [e Du(t) — e20-D] u(t — 3) (d) ft) =u(t)-u(t— 1). Now qn(£), the system response to u(t) is given by alt) = etult) e ult) = (L- cult) The system response to u(! — 1) is yr(t — 1) because of time-invariance property. Therefore the response y(t) to HO = v(t) -a(t— 1) is given by nO) =()-nlt-D)= (1a) - (= ef Dag) The response is shown in Pig. $2.4-11. “ut t— Cont) Fig. S2.4-11 14 “ [na input eutpu -t EA CNE lo Es | +— Fig. S2.4.12 24-12 Rn = et) + 2 ato) setu(—t) (1) setaior) + au tu(t) + ctu(-t) “u(=1) + fe te(t) + elu(-8)] =eult) 24-13 1 sa merto= [qui -mar Because w(t-1)=1 forr <tandisO forr > t, we need integrate only up to 7 =t. 1 ta =.|t 2,45 maos / aiqer= um Tt = tan t+5 Figure 82.4-13 shows 2h and «(t) (the result of the convolution) ) 4 el »* 4 o &> 9 eu ET Em) o to To Fig. 82.413 2.4-14 For t < 2r (see Fig. 824.14) + “= 10 eat) = [ siurér= 1 cost Osis2m o Fort > 2r, the area of one cycle is zero and HO g()=0 t>2r and t<0 Fig. S2.4-14 15 (e) (to) (g) th) 146 : o= f maq ta e D+5 t<1 : ay= [erar=1-e o<t<3 o t d)= f eta=et Dect 1>3 t-3 «)=0 t<0 This problem is more conveniently solved by inverting fi(t) rather than fz(t) em : ao= (r-dr=5 t>0 , 2 ter 1 on= [ (t-y)dr=50-") 0>t>-1 o 2 d()=0 for t>0 A(g=e f()=e flm=e, flt-m)=e0D, o o dt) = f e Dgr= ef edr= Fte et] ogt<l —1++ —1+ : 2 dt) = engane ttf Srar=ifpt-e] Opt>- + 3 , . «= [ cet dr mera f dr -2 - ed) =0 tS-2 =D) cipeDo 2.4-17 Indicating the input and corresponding response graphically by an arrow, we have fe) — vt) fe) > yuit-T) (by Time-invariance) HMO-IU-D > O) -u(t-T) (by linearity) Therefore tm AO = HED] po Eleto cult 1] The left-hand side is /(t) and the right-hand side is j(t). Therefore Fo) — sto Next we recognize that + + (8) x u(t) f fru(i — már= [ fer) dr eb) This follows from the fact that integration is performed over the range -oc < 7 <t, wherer <t. Hence u(t-7)= 1. Now the response to f f(r)dr is Lr(e) e ultda (t) = [$(8) x h(E)) su(t) => lt) x lê) 18 2.4-18 2.4-19 2.4-20 2.4-21 2.51 But as shown in Eq. (1), y(t) su(t)is JL, ur) dr. Therefore the response to input J! f(r)dr à Using the hints, we obtain FO + a(0) = pm ELMO = = DI 9t0 = 100 + fim E io(O = alt= Ts fim 5 [ECO = té T) = 60 Sucoessive applications of the above procedure yields 100 (8) + (1) = foto) The system response to u(t) is 9(t) and the response to step u(t — 7) is g(t — 7). The input f(4) is made up of step components. The step component at 7 has à height A which can be expressed as at = GLar= for The step component at n47 has a height f(n4A7)AT and it can be expressed as [Á(rAT)Ar]u(t - nAT). lts response Ay(t) às Ay(t) = (nan Aral — não) The total response due to all components is vt = dm 3 fnsralt-nar)Ar = fo Strato mér= ft) ea) An element of length Ar at point nAr has a charge f(nôT)AT (Pig. $2.4-20). The electric field due to this charge at point z is fináAr)Ar = da nAvd The total field due to the charge along the entire length is e SFinanar o Am, » dre(z — nAT)? foto - Cr fa Ho) — (x — 1)2 = = AT o NAT a am a-naT Fig. S2.4-20 For an ideal delay of T secs., the impulse response is h(t) = 6(t- T). Hence, from Eq. (2.48) (using the sampling property [Eq. (1.24b)) H(s)= f. sr Te dr = e" We can also obtain the same result using Eq. (2.49). Let the input to an ideal delay of T seconds be an everlasting exponential e'*, The output is e“t-7), Hence, according to Eq. (2.49), H(5) = et D) jett = emsT, NA TALIZ=(A+3MA+4) 19 25-2 2.5.3 The natural response is tn(t) = Ke 4 oe tt (0) Forf=u=u(l), uMD)=H(O)- Sos v(t) = Ke + Roe t+I dt) = —3K1e"* — 4Kae Setting t = O and substituting initial conditions, we obtain ea 1= -3k, — 4K» and ug)=je*!-letri o t>0 6) SO-etut), vl)=H-D= SE = u(t) = Ke * + oe 4 det dt) = —3K1e"*! — 4koe * — Setting t = 0, and substituting initial conditions yields 0=M +Ka+3 K=— = : 1=—3K — 4H0 — À and Hi) = (0 HM)=e Put), volt)=H(-2)=0 u(t) = Ki 4 Ke lt) = —3K1e"* — 4Kge Setting t = 0, and substituting initial conditions yields 0= Ki +Ks 1= 3H — 4K, and vt) = NM +61425=(A43- j4A +34 34) characteristic roots are —3 + 74 yn(t) = Ke * cos(4t + 0) For f(t)=u(t), golt)= H(O)= & so that vt) = Ke"“cos(at + 0) + & bit) = -3He * cos(at + 9) — 4Ke"* cos(at + 6) Setting t = 0, and substituting initial conditions yíelds K = 0.427 8=-—106.3 —» 2= -3Kcos8 —-4Ksin6 KsinB= 0=Kcos0+ & ) K cos8 = DO and ult) = 042707“ cos(at — 1063) + tz0 Characterístic polynomial is A? + 4A + 4== (A + 2)2. The roots are —2 repeated twice. Un(t) = (Ki + Kat) 20 271 2.7-2 2.7-3 tita — 1) = hr and gl) = L lA(r)ldr = 00 q This violates the assumption. (a) The time-constant (rise-time) of the system is Th = 107º. The rate of pulse communication < 7h = 10º pulses/sec. The channel cannot transmit million pulses/second. (b) The bandwidth of the channel is 1 B=7=10 Hz The channel car transmit audio signal of bandwidth 15 kHz readily. The received pulse width «= (0.54 0.1) = 0.6 ms. Each pulse takes up 0.6 ms interval. The inaximum pulse rate (ta avoid interference between successive pulses) is 1 GE x 10c5 E 1667 pulses/sec Using Eas. (2.60) and (2.61) (a) 1 -. TeT= -5=10 a 104 10º pulses/sec. (b) “Fhe bandwidth 7. = 1/Ti = 1/7, (c) The pulse transmission rate is 7 23 Chapter 3 3.1-1 3.1-2 3.1-3 â,1-4 3.1-5 Trivial. (a) In this case E. = fo dt = 3, and 1 Í tdt=0.5 (b) Thus, f(t) = 0.5x(t), and the error e(t) = t — 0,5 over (0 St < 1), and zero outside this interval. Also Es and E, (the energy of the error) are 1 pn ef Holt) dt = 1 1 es [ Piga= [ tdt=1/3 and Ee= [(osfa-n o o o The error (t — 0.5) is orthogonal to z(t) because fe -05(1)dt=0 o Note that E; = ? E, + Ee. To explain these results in terms of vector concepts we observe from Fig. 3.1 that the error vector e is orthogonal to the component cx. Because of this orthogonality, the length-square of f [energy of f(t)] is equal to the sum of the square of the lengtks of cx and e [sum of the energies of cr(t) and e(t)] In this case Es = fo f2(t)dt = Jo tê dt = 1/3, and VA 1 E) e(t) f(t) dt = fra-s Es do Thus, z(t) = 1.5f(t), and the error e(t) = r(t) — 15f(t) = 1— t.5tover (0 <t < 1), and zero outside this interval. Also E» (the energy of the error) is Ee = fi(1— 1.582 dt = 1/4. (a) In this case Es = [) sin? 2wt dê = 0.5, and Lp 1f eee nose = 3 f tsin 2midt = —1/m (b) Thus, f(t) = —(L/m)e(t), and the estor e(t) = t+ (1/n)sin 2xt over (0 < t < 1), and zero outside this interval. Also Ey and E, (the energy of the error) are : . : Es= | Pljdt= | Earor/3 and E | fo (Umsin dm Lo 9 o o 3 2r The error t+ (1/7) sin 2rt] is orthogonal to a(t) because 1 f sin 2mt[t + (1/m)sin 2at]dt = 0 , . Note that Ep = c?E, + Ee. To explain these results in terms of vector concepts we observe from Fig. 3.1 that the error vector e is orthogonal to the component cx. Because of this orthogonality, the length of f [energy of f(b)] is equal to the sum of the square of the lengths of cx and e [sum of the energies of ez(t) and e(t)]. (a) 1f z(t) and y(t) are orthogonal, then we 24 3.2-1 (1) (3) 3.3-1 showed [see Eg. (3.22)) the energy of z(t) + y(t) is Ex + Ey. We now find the energy of x(t) — y(t): f. teto ufa = [O petopar+ [7 oa f. entar [º 2) de = f. In(o)P at + f. too) de (3.22) The last result follows from the fact that because of orthogonality. the two integrais of the cross products a(t)y"(t) and 2"(Dy(t) are zero [see Eq. (3-20)]. Thus the energy of x(t) + y(t) is equal to that of x(t) — y(t) if x(t) and y(£) are orthogonal. (b) Using similar argument, we can show that the energy of ciz(t) + coy(t) is equal to that of crr(t) — cay(t) if a(t) and v(t) are orthogonal. This energy is given by [ci [2 Ez + lca|2Ey. (c) If =(t) = a(t) + v(t), then f. eo eta = [O popa prgpara [” =onrtjar e“(tyt) de = Es + Ey £ (sy + Eye) We shall compute cn using Eq. (3.25) for each of the 4 cases. Let us first compute the energies of all the signals. 1 Ej= / sin?2mtdi = 0,5 o In the same way we find Ep, = E = E = E = 05. Using Eq. (3.25), the correlation coefficients for four cases are found as 1 1 1 i à - E i -si =— veis sin 2mtsin 4rtdt= 0 (2) vetos / (sin 2mt)(— sin 2mt) dt 1 1 0.5 1 i ; - 2 i - i = 4 ima 0.707 sin 2wédt = 0 (4) Vest [f 0.707 sin 2mt dt fotons 2xt al 1.414/7 Signals z(t) and f>(t) provide the maximum protection against noise. (2,1), Eo(-1,2), fs(O, —2), fa(1,2), f5(2,1), and f(3.0). From the figure, we see that pairs (fs.f6), (fi, 4) and (fz, fs) are orthogonal, We can verify this also analytically. %2 Pig. 83,3-1 &-=(0x3)+(-2x0)=0 fti=(2xD+4(-1x2)=0 fefs=(CIx29)+(2x1)=0 25 (b5 Co t N ta ez j 1 em A Ee Ds ; IN => JE do doa “odor d+ Cal) sat A 2m . o «E * (e) CP q E 48 Fig. 83.4-3 (e) To =3, ug = 27/3. 1 f 1 a=>| tdt=> 3h õ 2. mz 3 2m | mm 2m a = 5 tcos tdi = gatos + sing” -1] 1 Zn 3 o 2m 2m mm tsiu tt = gras [sin 40 es 31 Therefore Co = | and 3 Arênê Zen dm 2m nf 288 cos 2gs — sin Zé (n= 525 =: > -sinS>| a Bn=t — ul É — 28 (f) To =, wo = 1/3, ao = 0.5 (by inspection). Even symmetry; dba = 0. af nm m=5[ 168) cos TE ae 2ffPar 2 nz 3 / cos EE ae f E- tes !Zea] 6 1 1 6 ” 2 1 5m 1 Tm ft) =05+ 5 (cos Go - qcosmt + qucos Tt quo Te) Observe that even harmonices vanish. The reason is that if the de (0.5) is subtracted from f(t), the resulting function has half-wave symmetry. (See Prob. 3.4-7). Figure 83.4-3f shows the plot of Ca. 3.4-4 (a) 2x Here To =, and wo = odiei 2. Therefore flt)= ao + Da cos 2nt + ba sin 2nt n=1 To compute the coefficients, we shall use the interval —7 to O for integration. Thus 1º wo=5[ et? dt = 0.504 e/2 cos Qnt dt = 0.504 (iris) T+ Têni 2 fm o 8n ba = if sin 2nt dt = 0,504 (rig) Therefore a foba - Co=00=0504, Cu= 2546 = 0.504 (é) , Ba= tan (=) =tan!4n On fit) = 0.504 + 0.504 » DR cos (2nt + tan”! 4n) (b) This Fourier series is identical to that in Eq. (3.56a) with £ replaced by —t. (e) IE f(t) = Co + SÍ Cu cos(nuot + 8n), then 6-8) = Co + 5) Cncos(-nuot + 0,) = Co + 5) Cu cos(nunt — On) Thus, time inversion of a signal merely chenges the sign of the phase 8,. Everything else remains unchanged. Comparison of the above results in part (a) with those in Example 3.3 confirms this conclusion. 3.4-5 (a) Here To = 7/2, and wo = $E = 4. Therefore F(t)= ag + ST an cos ânt + ba sin ánt = where 2 prz a= if et dt = 0.504 7 o m/2 Qu = 4 ertcos Antdt = 0.504 (rãs) zh 1+16n' and a/2 4 = “ 8n bda= E) etsin Antdi = 0.504 (7 isa) Therefore 29 Co=00=0.504, On = Vai +bê=0.504 (é) . On=-—tanlán (b) This Fourier series is identical to that in Eq. (3.56a) with t replaced by 24 (e) 1 f(t) = Co + 3) Cn cos(nwot + 9n), then 4at) = Co + 35 Ca cos(n(awo)t + 8n) Thus, time scaling by a factor a merely scales the fundamental frequency by the same factor a. Everything else remains unchanged. If we time compress (or time expand) a periodic signal by a factor a, its fundamental frequency increases by the same factor a (or decreases by the same factor «). Comparison of the results in part (3) with those in Example 3.3 confirms this conclusion. This result applies equally well 3.4-6 (a) Here To = 2, and wo = SE = 1. Also f(t) is an even function of t. Therefore SH) =00+5 ancos nat where, by inspection ao = 0 and from Eq. (3.66b) 1a é 4 nie! n even Qn= :/ (—2t+ 1) cos nntdt= ——s (cosnnt — No = sa roda Therefore 1 1 5 Cós St + goes Tri +.) (b) This Fourier series is identical to that in Eq. (3.63) with t replaced by t + 0.5. (e) IF f(t) = Co + 3) Cn cos(nwot + 85), then 84 1 10= [cos mt+ qeos Bmt+ f4+T) = Co + 3 Cn cosjnua(t + T) + 8h] = Co + 3 7 On cosfuot + (Bm + muoT)] Thus, time shifting by T merely increases the phase of the nth harmonic by nuoT. 3.4-7 (a) For half wave symmetry so=-s(1*2) and a (ho a [Toiz Fo and An $tt) cos nwgt dt = EA F(t) cos nwat dt + ft) cos nwot dt To ê To o To/2 Let 2 =t-— To/2 in the second integral. This gives To/2 To/2 To st pars f + + e a) [ (8) cos nao À 1(2+ E) cosmo (2+ E) do To/2 To/2 [ / St) cos muot dt + f =(s)l- cos nwoz] ae) o o Ta/2 = % [f Fít) cos muwot al In a similar way we can show that an= aj aj » To/2 á F(8) sim muspt dt =.) (6) To =8,wo= 5,09 =0 (by inspection). Half wave symmetry. Hence ator nz ilfiam n=5I[ segs HEra] = 5 | [ os Eva 4 nm, nmo nm = (cos TE +EsnE o 1) (n odd) nt EnTo 1) (n odd) dr | aj” —s nº 30 a, 3.4-9 34-10 19=) br sin TE Na By inspection, ao = 8. Because of odd symmetry a, = O. Because of half wave symmetry (see Prob. 3.47), 1 a n=5[ tom Eloa f (-t+2)sinHErat = nada : [3 (£) Here, we need both sine and cosine terms with wo = 1 and odd harmonics only. Hence, we must construct a pulse such that it has hal-wave symmetry, but neither odd nor even symmetry, has à value é over the interval O<t<1, and repeats every 27 seconds as shown in Fig. 83.4-8f. Observe that the first half cycle from O to 7) and the second half cycle (from 7 to 27) are negatives of each other as required in half-wave symmetry. By inspection, ao = 0. This yields Ht)= 5) ancosnt+ basinnt n=1 noda Because of half-wave symmetry (see Prob. 3.4-7), ap 2 . afro 2 ne | tcosntdt= Do(cosnnsinn-1) br=o | tsinntdi= >(sinn —ncosn) — nodd EA E A xn 2m a bed etag hi periodic? yes yes no yes no yes yes yes yes wo 1d x period 27 27 2 1407 SE 27 q o f() =e0+ 3 ancosZrnt + bnsinZmnt (es = E) nt 1 1 1 ao=1 [ soa= [ tdt= 5 0 an = 2f tcosêmntdt=0 nz1 (ninteger) : = / tsin2mnidt = — o mm Hence Le) 1 2 Figure $3,4-10 33 nn=1+L LAI. = 217 Do qsindane aa If E.(N) is the energy of the error signal in the approximation using first N terms, then From Eq. (3.40) eoo-r Proa (HE (ao (25) (Note that En = 1/2forn=1.2,.. and E9=1) + (em 2nt + sem dmt + ssin Gmt+ ) nte 1 11 a 1 pq= [fa-f=i-i-d A 4 3 4712 7 = 0.014378 34-11 Ft) = coro(t) + ami(t) + + erer(t) 1 SinceE, = f En(t)dt = 1 o eb Heyrott) di = f Heat de HO) a joo(t) — Gan(t) = Gaste) = quarto) Hence Also : 1 frou-s and En=l A 3 H Ec(N) is the energy of the error signal in the approximation using first A terms, then From Eq. (3.40) 1 EM=5 . 1 1 Elu)=q-c6-ci= qg = 00204 1 2 2. 20 A BO = 5d rd o ds a cum 1 a EM) =3 = = 0.001302 The Walsh Fourier series gives small error than the trigonometric Fonrier series (in prob. 3.4-10) for the same number of terms in the approximation. 3.4-12 FB) = copolt) + exprlt) + e: + csns(t) 1 o=5[ fidt=0 Also a=u=-=-=0 a Hence 1 Also f fdt=2 and using Eq. (3.40) 1 05 1 3 EM) = frios - 3º -5 2 lz 1a - E)= | dt qi qe =0.28125 (b) This is a scaled version (time-expansion by factor 27) of the signal f(£) in pair a. a t sft T|oft 3ft foto) =s(5) = (5) *g E (5) - 163] + 3.5-1 (9): To = 4,wo = 2/2. Also Do = O (by inspection). 1 -jam tum, 2 eira | ciedgsLsnDo |>l : , ma Da (b) To = 107, wo = 27/107 = 1/5 s0= 5) Daeft, where Da=ã te) ft) =Do+ ” Dne'"*, where, by inspection Do =0.5 27 2 n>0 Da=df teima sothe |Dil=-L. and Du=d? Br h 2 Zm Zen =" 2<0 (d) To=7,wo=2and Da= s= 50 Dre, where (e) T= wo. tt) =) Dae were £Dn 7a ma 1 4 “3024 [a =M/2 123 o» Fig. 83.5-3 Da = Delito! sothat |Daj=|Dn), and £Dr=LDa- jnwoT (b) s0)= L Dyetriot ÍO=na)= 3.5-5 (a) From Exercise E3.6a À cosnat -1$t<l Ho= LS The power of f(t) is Moreover, from Parseval's theorem [Eq. (3.82)] Aco AA =C2 a -f— — Pr=C8+5] *= (5) (5 (b) É the N-term Fourier series is denoted by z(t), thén cosnmt —-Isi<i For N=1,P,=0Olll;forN=2, P,=ó019323, ForN=g3, P,= 019837, whichis greater “than 0.198. Thus, N = 3 3.5-6 (a) From Exercise E3.6b 24 1 = (9 S osinnrt -n<t< He) (1 SO o sinnm n4t<m z na 38 The power of f(t) is 2 1 Pr=5 | (uniao 4 Vs Moreover, from Parseval's theorem [Eq. (3.82)] Cê ço 44? 2 n P=+45 5 =55 Gs T T (b) If the N-term Fourier series is denoted by a(t), then x 224, qu a()= (1) L n For N = 1, P;= 0202642, for N=2, Po = 0253342, for N=5 P,=O026584, forN=6 P= 0,30222.4?, which is greater than 0.342, Thus, N = 6. 3.5-7 The power of a rectified sine wave is the same as that of a sine wave, that is, 1/2. Thus Pj = 0,5. Let the 2N +1 term truncated Fourier series be denoted by f(t). The power P; is required to be 0.9975P; = 0.49875. Using the Fourier series coeficients in Exercise E3.10, we have N 4 N 1 = ——s = Sim-55 (ap 04885 n=-N n=-N Direct calculations using the above equation gives P; = 4/12 = 0.4053 for N = O (only de), P; = 0.49535 for No 1 (3 terms), and P; = 0.49895 for N = 2 (5 terms). Thus, à 5-term Fourier series yields a signal whose power is 99.79% of the power of the rectified sine wave. The power of the error in the approximation of f(t) by f(t) is only 0.21% of the signal power Pp. 3.6-1 Period To = 7, and wo = 2, and . ju 0.504 HGu)= orar andfom Eq (874) De Toa There p= =, DHU mui 41.08n oj2nt eretore, (= 5) Dailinugerts SO o ID 39 Chapter 4 4.1-1 4.1-2 4.1-3 Flu Ff soetta = [O nO cosotar sf” flt)sinwt dt HE f(t) is an even function of t, f(t) sinwt is an odd function of t, and the second integral vanishes. Moreover, (t) coswt is an even function of t, and the first integral is twice the integral over the interval O to oc. Thus when f() is even Flo)=2 [” sttyconatar (1) o Similar argument shows that wben J(t) is odd f() sinwt dt (2) IÉ $(2) is also real (in addition to being even), the integral (1) is real. Moreover from (1) F(-w) fo coswtdt = F(w) Hence F(w) is real and even function of w. Similar arguments can be used to prove the rest of the properties. 1 [O 1 so= 5 | Pés do= =/. [Plujje <Pt çõos quo + IF Pl) cosfut + LP(io)| dio + 5 f. VE (a) sinfut + LP(co)] dos Since |P(cw)] is an even function and ZF(w) is an odd function of w, the integrand in the second integral is an odd function of w, and therefore vanishes. Moreover the integrand im the first integral is an even function of w. and therefore 10= ap [Ftw)lcosfut + £Plw)] duo ' Because f(t) = fo(t) + felt) and et = cos wi + jsin wi Fw) = f Ult) + Ste Te dt = f. [o(t) + fe()] coswt dt — sf [okt) + fe(t)] sincot dt Because fo(t) coswt and fo(t)sinwt are even functions and fo(t) coswt and fe(t) sinwt are odd functions of t, these integrals (properties in Egs. (B.43), p. 38] reduce to Flw) 2 [ isomerat os E sutismuras (1) o o Also, from the results of Prob. 4.1-1, we have FU) = ftjcoswtdt and Flo) = -2 [ fot) sinwt dt (2) o From Egs. (1) and (2), the desired result follows. 40 4.31 4.3-2 (a) 1 u(t) <=> mó(w) + — mam ju so Elo) Application of duality property vields nô8(t) + 1 + 2mu(—w) 3t O — FO) am f(—u) or | += u(—w) Application of Eq. (4.35) yields 1 1 5 [seo - à 6=+ vlw) But &(t) is an even function, that is A(—t) = ó(t), and 1 j FLU) + al <=> u(iw) €b) cos wot <=> m[ó(w + wo) + 5(w — wo)] se Fe) Application of duality property yields m[6(t + wo) + &t — wo)] <=> 2m cos (—waw) = 2m cos (wow) ma? O mm? Fo 2nH(-0) Setting wo = T' yields SE+T)+ 5 —-T) > 2cos Two (e) sin wot <=> jr[ó(u + wo) — ó(w — wo)) “(8 Flu) Application of duality property yields ir[ó(t + wo) — 5(t — 1u0)] + 2x sin(—uwgu) = —2m sin(wow) me? Ft) 2mf(-w) Setting wo = T yields Mt+T)- (t—T) > 2jsin Tu Fig. (b) fi(t) = f(-t) and Filo Fis. (e) fa()= f(t- 1) + fit 1). Therefore Fa(uo) = [E (wo) + Eito)Je 2! = [F(u) + Flu? Zerdu E (cos + usinw — 1) Fig. (d) Jul) = Mt > D+ Alta) Fa(w) = Flu) E + (ue 1 = Lito scosu] = Sp sinê gsm (5) 43 Fig. (e) SD) = f(t- 1)+ A(t+ 3), and Flw 2 4 Fi(ujede/? eiuta = fede 2 jude 1] + ed + jue de 1 E! jue l+—= j ] Dona ft = alêesim 5] =sine(5) Fig. (£) $5() can be obtained in three steps: (1) time-expanding f(t) by à factor 2 (ij) then delaying it by 2 seconds, (iii) and multiplying it by 1.5 [we may interchange the sequence for steps (i) and (i)]. The first step (time-expansion by a factor 2) yields f (5) ++ 2F(2u) Second step of time delay of 2 secs. yields (68) The third step of multiplying the resulting signal by 1.5 yields 1 Ps — Qu — Ne dd = aos (l = Bo — e) Pol) = 157 (5) es Ea pum esta) É a es a . o N rd oo Ns , , : x , d£ GN | E o | [o . , sntuiê Ar ;? Ie ntuiá . Jr TD SDnetTD Utt-mD cost e 5 cos (t-E ut 5 Fig. 84.33 438 (a) S48) = rec (re) — ret (+ 22) rs (1) ms oie (87) rect (sl “e) es sino (ST) otirta and (b) From Fig. S4.3-3h we verify that S(t) = sintu(t) + sin(t — mju(t — m) Note that sin(t — m)u(t — m) is sin tu(t) delayed by 7. Now, sintu(t) es» Elô(w — 1) — ófw + 1)j+ qi and sin(t — mJu(t — 1) =» tialóla — 1) — Gl + DJ] + jeriro 44 4.3-4 Therefore Flu)= tato -D-ó(u+ D+ Recall that f(x)ó(z — x0) = f(zo)6(x — 70). “Therefore &(u + 11 47274) =0, and (L + eTe (c) From Fig. S4.3-30 we verify that Ft) = cost [ue -—u (e — 5)| = costu(t) — costu (e - 3) But sin(t— £) = — cost, Therefore Ft) = costu(t) + sin (e - 3) “ (t - 5) Flo) = Slblo—1)+ (o +1)] + i 1l-w ao ecirelz 1-uw? ” + (Eeu-n-serm+ Also because f(z)ô(z — zo) = f(xo)6(z — zo), Sw E e TRE = gu de De EM = j6(u 1) Therefore —ymw/2 1 ju cre Ta Sto gal ee F(w) (a) Meu) -ult-T]=eCu(t)- e Cut T) ESTE ME Dy(t- T) E erist = “juro jura “jura F(w) (- entetiomy From time-shifting property HE+T) 6 F(wjeT Therefore HEATIAFHE-T) 6 F(ujeeT 4 Fw eT = 2F(u) coswT We can use this result to derive transforms of signals in Fig. P4.3-4. (a) Here f(t) is a gate pulse as shown in Fig. S4.3-4a. F(4) = rect (5) + 2sine(u) Also T=3. The signalin Fig. P43-4ais f(t+3) + f(t— 3), and S(t+3) + Ft 3) += 4sinc(w) cos3w (b) Here f(t) is a triangular pulse shown in Fig. $4.3-4b. From the Table 4.1 (pair 19) A(O es ue (E H)=A (5) + sine (5) Also T =3. The signal in Fig. P4.3-4b is (t+ 3) + f(t— 3), and 443) A F(t— 3) 65 2sine? (5) cos 45 1 (53) (+) nó(u) 1 jo=A* mê - 5) —Sbto) + [ ata TCA lju=A Taking the inverse transform of this equation yields HO = 50" = Du(o) tb) at Mat, eu(t) <= m=-» and e u(t) <=> m-% EE S(t) = eMtu(t) x eRtu(t), then º 1 - E O. Fo =to-)Go =) jo= Mood Therefore HO = (e = Duo) 1 (e) ruges Lo and tu) es -— ju à ju da TE /(8) = eMtu(t) + eMtu(-t), then — oc LH XX To =A)Qu = jo-k ju-k Therefore 1) = qto + at] KM Note that because Az > D, the inverse transform of =. is etu(—t) and not —e*2tu(t). The Fourier transform of the latter does not exist because Az > 0. (a) 1 Jo tu(-t) ss — and Mtu(-t) Es — o ju-M ME f(t) = eNtu(—t) + eMtu(—t), then Fi nem = — ( du — Aju — Àa) qu ju—Az Therefore 1 160) = 1 = a(o) The remarks at the end of part (c) apply here also. 4.3-9 From the frequency convolution property, we obtain 1 Pes 5 E lu) + Evo) Because of the width property of the convolution, the width of P(w) + F(w) is twice the width of F(w). Repeated application of this argument shows that the bandwidth of /º(t) is nB Hz (n times the bandwidth of f(1)). 4.3-10 (a) o 7 Fto= [ eta f estg=-2n- cos wT])= sin? (5) e , jo w 2 48 4.3-11 4.4-1 tb) and Flw) = sino (5) [eu — qr doT/2) = 2jTsine (5) sin = = Bane (2) aa (e) af rd St+T)— 280) + 6(t— T) The Fourier transform of this equation yieids juF(u) = e -24e 7 = 9 -coswT]= -Asinê (5) Therefore ru = Bsinê (22) / seita and SE = o da Flu É f seeiot dr Changing the order of differentiation and integration yields E- gumes = f ste de Therefore -sent0) e SE (b) e ult) > Fora asumat d LN. —jte “u(t) zo (+ = Torar and at 1 te u(t) => Gu ro Hu) = — E ur (a) 1 Fe) = usa Yuj= 1 Aa Qu +DOw+2) juri ju+2 ult) = (6! — e u(e) 49 (b) F(w)= je + 1 Cu + 12 ut) = te Cult) Yt)= (e) Flwj= je Lo do 2 = Go DS 41 om us) = Setute) + Getu(—t) (d) 1 Flo) = mtu) + 1 1 vo= a [rico + >| =nó(u)+ mem because f(e)ó(2) = f(0)6(2)] 4.42 (a) Flw)= = and Hlw)= and sue 3 MA a Qu lBgu+t) Iljw+rl ju-2 Therefore vb)= ; [e-tute) + eXu(-8)] 6) 1 1 Flw) = mo 7 md H(w)= = 3 and | RR Ve)=G>nGu=]T-1 7-2 Therefore 4.4-3 Fio) = sinc(;m) and Falo Figure $4.4:3 shows Fi(w). Fa(i), Hy(w) and Ho(w). Now Yo) = Flw) (w) Vo(ww) = Fo(w)Ho(w) The spectra Yi(w) and Ya(«) are also shown in Fig. 84.43. Because y(t) = ga (t)ya(t), the frequency convolution property yields Y(w) = Yi (ac) + You). From the width property of convolution, it follows that the bandwidth of 50 4.5.3 4.6-1 Hence the filter is noncausal and therefore unrealizable. Since A(t) is a Gaussian function delaved by to, it looks as shown in the adjacent figure. Choosing to = 32E, A(0) = e 43 = G.01L or 1.1% of its peak value. Hence to = 32 is a reasonable choice to make the filter approximately realizable. 2x 10º outo vi 1015* From pair 3, Table 4.1 and time-shifting property. we get H(w) = A(t) = e tOtctot The impulse response is noncausal, and the filter is unrealizable. E - Figure S4.5-2 The exponentia! delays to 1.8% at 4 times constants. Hence to = 4/a = 4 x 1075 = 40gs is a reasonable choice to make this filter approximately realizable. The unit impulse response is the inverse Fourier transform of H(w). Hence, we have hít) = (a) O.Srect ( (b) sine 2(10,000rt) (e) 1 t 2x Toa) AM the three systems are noncausal (and, therefore, unrealizable) because all the three impulse responses start before t = 0. For (a). the impulse response is & rectangular pulse starting at t = — 107º. Hence, delaying the A(t) by 1 second will make it realizable. This will not change anything in the system behavior except the time delay of 1 jisecond in the system response. For (b), the impulse response is a sinc square pulse, which extends all the way to —oo. Clearly, this system cannot be made realizable with a finite time delay. The delay has to be infinite. However, because the sinc square pulse decaya rapidly (see Fig. 4.24d), we may truncate it at £ = 1074, and then delay the resulting A(t) by 1072, This makes the filter approximately realizable by allowing a time delay of 100 pseconds in the system response. For (c), the impulse response is 1, which never decays. Consequently. this filter cannot be realized with any amount of delay. anne [E tejo e [ou fe dt Letting É = Ja and consequently dt = Sadr polo feng NE ÇA o A lo 2vVino 2ovT Also from pair 22 (Table 4.1) Flu) = eua 1 fe 2, 1 fas Bea frota do ("e da Letting ow = 5 and consequently du = Sadr 11 a2j2 v2x 1 E=——+= dr= = Covil CEaavi” vã 53 4.6-2 Consider a signal S(t) = sinc(kt) and Flu) = qrecitçe) o oo 2 2 Es PAR = o frest (3)] de np = [ du= ef, TE 2(8) +» A(o), then the output Y'(u) = A(w)H(uw), where H (wo) is the lowpass filter transfer function (Fig S4.4-6). Because this filter band 47 — O, we may express it as an impulse function of area 4r AF. Thus. =Ia 4.6-3 H(w) = [47 4F]ó(w) and Y(w) = [47 A(0) A F]ó(w) = [47 A(0)5F]ó(u) Here we used the property f(z)d(z) = f(0)6(x) [Eq. (1.238)]. This yislds u(t) = 2A(DJAF Next, because f2(t) + A(w), we have atoy= fr fHteitat sothat A(O) f. Pdt= Es Hence, y(t) = 2E/AF. sta SD O Fig. S4.6-3 4.6-4 Recall that fd) = Ef. Fa(wje'! dy and f. fe dt = Fi(-w) Therefore fenmonma= 5 runtfo estos aa -5 f raon f” are anão= Ef iitco)tatondo Interchanging the roles of fi(t) and fa(t) in the above development, we can show that ' Vo fnionas do [" io)fatoo) do qe Ve 4.6-5 Application of duality property [Eq. (4.31)] to pair 3 (Table 4.1) yields 2a aj t2+a? += 2xe 54 The sigrial energy is given by 1 [O amenas E dou m=5[ [re fita = am [ tai quo = EE 7h a The energy contained within the band (0 to W ) is mr 2m By = ar f cotas q = Ty — o 2AW] , a E Ew = 0.99Ey, then . 3098 566 erro —s W = 208 aj = 836 a a 47-14 (i) For m(t) = cos 1000 Prss.so(!) = m(t) cos 10,000t = cos 1000t cos 10,000: = 5 [cos 9000% + cos 11, 000%] LSB USE (ii) For m(t) = 2 cos 1000€ + cos 2000% Pose-sc(t) = m(t) cos 10,000t = [2 cos 1000t + cos 2000] cos 10, 000: = cos 9000t + cos 11,000 + altos 8000t + cos 12, 000%) = [cos 9000t + i cos 80008] + [cos 11,000t + 5 cos 12, 000%) 2 lr LSB USB (iii) For m(t) = cos 1000t cos 3000 osa.se (t) = m(t) cos 10,000: = Ecos 20004 + cos 40008] cos 10, 000 = Álcos 80004 + cos 12, 0008] + Alcos6000t + cos 14,000 = 5 (cos 8000t + cos 60004] + i cos 12, 0D0t + cos 14, 0008) LSB us This information is summarized in a table below. Figure 84.7-1 shows various spectra. Mew Modulated sigual speetvum Dil As dor 14 - CDE TOGO =|? -qros aQroo DO o Il 7 141 tt. tt -20e8 Ou prob. AD top -98-8K, o 0 gu AS TR PS “Il | Lilles 44 tt TAI “AR O) ak 4x “9h “IZ2 -8 “EM o WU 1 8k 12K tuts Fig. S4.7-1 55 mio) 4 CA os tt iai =0E "1e0 . e geo neo TE =p 2 te Í f Lt le Dal Jé tt “tre der . o gecso nijser aja vo vs Ps ê a A e ate Eco 0 HOC EO IEET IE CÊ LC Fig. S4.7.6 NM Pros únit) [o t> - 3008 t—+ Wim to QE 10 toa. o | PM Po) Fig. 84.8-1 4.8-1 In this case 7, = 10 MHz. mp = 1 and m) = 8000. For FM AF = kme/2m = 2m x 109/2m = 105 Ha. Also £. = 10”. Hence, (Fi)max = 10 + 10º = 10.1 MHz. and (Fidmin = 10º — 10 = 9.9 MHz. The carrier frequency increases linearly from 9.9 MHz to 10.1 MHz over à quarter (rising) cycle of duration a seconds. For the next a seconds, when m(t) = 1, the carrier frequency remains at 10.1 MHz. Over the next quarter (the falling) cycle of duration a, the carrier frequency decreases linearly from 10.1 MHz to 9.9 MHz., and over the last quarter cycle, when m(£) = —1, the carrier frequency remains at 9.9 MHz. This eycles repeats periodically with the period 4a seconds as shown in Fig. S48-la. For PM AF = komp/2x = 507 x 8000/27 = 2 x 10 Hz. Also F. = 107. Hence, (Fimax = 107 + 2 x 10º = 10.2 MHz. and (Fimin = 107 — 2x 10º = 9.8 MHz. Figure S$4.8-1b shows rir(t). We conclude that the frequency remains at 10.2 MHz over the (rising) quarter cycle, where ri(t) = 8000. For the next a seconds, rn(t) = 0, and the carrier Frequency remains at 10 MHz. Over the next a seconds, where sh(t) = —8000, the carrier frequency remains at 9.8 MHz. Over the last quarter cycle ra(t) = O again, and the carrier frequency remains at 10 MHz. This cycles repeats periodically with the period 4a seconds as shown in Fig. 84.8-1. 4.8-2 In this case Fe = | MHz, mp = 1 and en; = 2000. For FM AF = kemp/27 = 20,0007/27 = 10º Hz. Also F; = 1 MHz. Hence, (FiJmex = 10º +10? = 1.01 MHz. and (Fiein = 10º — 10º = 0.99 MHz. The carrier frequency rises linearly from 0.99 MHz to 1.01 MHz over the cycle (over the interval — ço <t< 193, Then instantaneously, the carrier frequency falls to 0.99 MHz and staris rising linearly to 10.01 MHz over the next cycle. The cycle repeats periodicaliy with period 107? as shown in Fig. S4.8-20. For PM Here, because m(t) has jump discontinuíties, we shall use a direct approach. For convenience, we select the origin for m(t) as shown in Fig 84.8-2. Over the interval to 1º, we can express the message signal as m(t) = 2000t. Hence, 58 em 0! SS MO mia Duma Len fio) | Fig. S4.8-2 êrult) = cos Porcos + Fe] = cos ferqo% + 52000:] = cos [2n(10)ºt + 10007t] = cos [2x (20º + 500) £] At the discontinuity, the amount of jump is ma = 2. Hence, the phase discontinuity is kama = 7. Therefore, the carrier frequency is constant throughout at 10º + 500 Hz. But at the points of discontinuities, there is a phase discontinuity of x radians as shown in Fig. S4.8-2b. In this case, we must maintain kp < 7 because there is a discontinuity of the amount 2. For kp > 7, the phase discontinuity will be higher than 27 giving rise to ambiguity in demodulation. 4.8-3 In this case ky = 10007 and kp = 1. For m(t) = 2 cos 100t + 18 cos 20007t and m(t) = —200 sin 100% — 36, 0007 sin 2000mt Therefore mp = 20 and mj = 36,000m + 200. Also the baseband signal bandwidth B = 20007/2m = 1 kHz. For FM: AZ = kymp/27 = 10,000, and Bem = 4AF + B) = 2(20,000 + 1000) = 42 kHz. For PM: AF = kym//2m = 18,000 + 100 Hz, and Bpm = 2(4F + B) = 2(18,031.83 + 1000) = 38.06366 kHz. 4.8-4 qru(t) = 10 cos(wct + 0.1 sin 200074). Here, the baseband signal bandwidth B = 20007/27 = 1000 Hz. Also, cor(t) = ue + 2007 cos 2000rt Therefore, Au = 2007 and AF = 100 Hz and Bam = HAF + B) = 2(100 + 1000) = 2.2 kHz. 4.8-5 qeu(t) = 5 cos(wçt + 20 sin 10007é + 10 sin 20007). Here, the baseband signal bandwidth B = 2000/27 = 1000 Hz. Also, wi(t) = we + 20, 0007 cos 1000mt + 20, 0007 cos 2000rt “Therefore, Aw = 20, 0007 + 20, 0007 = 40,0007 and AF = 20 kHz and Bem = 2(AF+B) = 2(20,000+1000) = 42 kHz. 59 Chapter 5 5.1-1 The bandwidths of fi(t) and fa(t) are 100 kHz and 150 kHz, respectively. Therefore the Nyquist sampling rates for fa(?) is 200 kHz and for fa(t) is 300 kHz. Also fu(t) es &Fi(iw) + Fi(w), and from the width property of convolution the bandwidth of f?(t) is twice the bandwidth of fi(t) and that of fa*(2) is three times the bandwidth of fa(4) (se also Prob. 4.3-10). Similarly the bandwidth of fi(t) fa(t) is the sum of the bandwidth of fi(t) and a(!). Therefore the Nyquist rate for f1"(t) is 400 kHa. for f2'(£) is 900 kHz. for fi(t) fa(é) is 500 kHz. 5.12 (a) sinc?(1007t) <=» 0.018(qãz) The bandwidth of this signal is 200 7 rad/s or 100 Hz. The Nyquist rate is 200 Hz (samples /sec) (b) The Nyquist rate is 200 Hz. the same as in (a), because multiplication of à signal by a constant does not change its bandwidth. (e) sine(100mt) + 3sinc?(60rt) <=» 0.01 rect(g) + dy A(zigx) The bandwidth of rect(zã5-) is 50 Hiz and that of A(zã=) is 60 Hz. The bandwidth of the sum is the higher of the two, that is, 60 Hz. The Nyquist sampling rate is 120 Hz. (td) sine(507t) <=> 0.02 rect(rés=) sine(1007t) <=> 0.01 rect(sã-) The two signals have baudywidths 25 Hz and 50 Hz respectively. The spectrum of the product of two signals is 1/27 times the convolution of their spectra. From width property of the convolution, the width of the convoluted signal is the sum of the widths of the signais convolved. Therefore, the bandwidth of sinc(50rt)sinc(1007t) is 25 +50 = 75 Hz. The Nyquist rate is 150 Hz. 5.1-3 The spectrum of f(t) = sine(2007t) is F(w) = 0.005 rect(zé=). The bandwidth of this signal is 100 Hz (2007 rad/s). Consequently, the Nyquist rate is 200 Hz, that is, we must sample the signal at a rate no less than 200 samples/second. Recall that the sampled signal spectrum consists of (1/7) F(w) = “SP rect(qê=) repeating periadicaly with period equai to the sampling frequency *, Hz. We present this information in the following Table for three sampling rates: 7, = 150 Hz (undersampling), 200 Hz (Nyquist rate), and 300 Hz (oversampling). sampling frequency 7, | sampling interval T Flw) | comments | 150 Hz 0.006667 0.75rect (z84z) | Undersampling | 200 Hz 0.005 rect (qr) | Nyquist Rate | 300 Hz 0.003334 Tsrect (66) | Oversampling | The spectra of F(w) for the three cases are shown in Fig. 85.1-3, In the first case, we cannot recover f(t) ftom the sampled signal because of overlapping cycles. wlich makes it impossible to identify F(w) from the corresponding F(w). in the second and the third case, the repeating spectra do not overlap, and it is possible to recover F(w) from F(uw) using à lowpass filter of bandwidth 100 Hz. In the last case the spectrum F(u) = O over the band between 100 and 200 Hz. Hence to recover F(w), we may use a practical lowpass filter with gradual hitD=pee) . - 41 T=te02s Ta) vo €> (as Da el aim HT ds 5.1-9 51-10 5.2-1 tb) Figure $5.1-7 Bocause the spectrum F(%w) has a zero value in the band from 100 to 150 Hz, we cau use au ideal lowpass filter of bandwidtl B Hz where 100 < B < 150. But if B > 150 Hz, the filter will pick up the unwanted spectral components from the next cycle, and the output will be distorted. The signal f(t), when sampled by au impulse train, results in the sampled signal /(t)ôr(£) (as shown in Fig. 5.Ld). If this signal is transmitted through a filter (Fig, $5.1-7a) whose impulse response is A(t) = p(t) = rect(yéz). then each impulse in the input will generate a pulse p(t), resulting in the desired sampled signal shown in Fig. PS.1.7. Moreover, the spectrum of the impulse train f(t)ér(t) is + Joe co Fw — mus). Hence. the output of the filter in Fig. 85.1-7a is Flw) = H(vo) Ê õ Hen where H(w) = P(w) = 0.025sinc ($), the Fourier transform of rect (53). Figure 85.1-7b shows this spectrum consisting of the repeating spectrum F(w) multiplied by H(w) = 0.025sinc ($). Thus, each cycle is somewhat distorted. To recover the signal f(t) from the flat top samples, we reverse the process in Fig. S5.1-7a. First, we pass the sampled signal through a filter with transfer function 1/H (w). This will yield the signal sampled by impulse train. Now we pass this signal through an ideal lowpass filter of bandwidth B Hz to obtain f(t). (a) The bandwidth is 15 kHz. The Nyquist rate is 30 kHz. (b) 65536 = 2/8, so that 16 binary digits are needed to encode each sample. (c) 30000 x 16 = 480000 bits/s. (d) 44100 x 16.= 705600 bits/s. (a) The Nyquist rate is 2 x 4.5 x 10º = 9 MHz. The actual sampling rate = 1.2 x 9 = 10.8 MHz. (b) 1024 = 2!º, so that 10 bits or binary pulses are needed to encode each sample. («) 10.8 x 108 x 10 = 108 x 109 or 108 Mbits/s. Assume a signal f(t) that is simultaneously timelimited and bandlimited. Let F(t) = O for [iu] > 27B. Therefore F(wjrect(qêgr) = F(to) for B' > B. Therefore from the time-convolution property (4.42) 18) = f(t)» BB'sine(2mB'9) =2B'f(t) + sine(27B't) Because f() is timelimited. f(£) = O for [tl > T. But f(t) is equal to convolution of f(£) with sinc(2r8't) which is not timelimited. It is impossible to obtain a time-limited signal from the convolution of a time-limited signal with a non-timelimited signal. o = 20ms B= 10000 Hence 7, >2B = 20000 1 = 20990 * 50º To 20x 1053 No= 7 = 59x 1075 — 400 5.2-2 5.2-3 5.2-4 Since No must be a power of 2, we choose Ng = 512. Also T = 50ps, and To = NoT = 512 x 50us = 25.6ms, Fo = 1/To = 39.0625 Hz. Since f(t) is of 10 ms duration, we need zero padding over 15.6 ms. Alternatively, we could also have used 20 x 10? T=DD0— =3 5 ps 313 39.0625 us This gives To = 20 ms, Fo = 50 Hz. And F= Lo 25600Hz =7=" There are also other possibilities of reducing T as well as increasing the frequency resolution. For the signal /(t), 1 > “025 Let us choose T = 1/8. Also To = 4. Therefore, No = To/T = 32. The signal (t) repeats every 4 seconds with samples every 1/8 second. The samples are Tf(&T) = (1/8)/(k/8). Thus, the first sample is (at & = 0) 1x (1/8) = 1/8. The 32 samples are (starting at 4 = 0) To =4, >3x2"8 1735134141 B'64' 32' 64º 16º 64º 32º a O 0,0,0,0,0,0,0, 1131537 0,0,0,0,0,0,0,0,0, Gr dos GR TES ET TD The samples of f(t) and g(t) are shown in Fig. 85.2-2. 1 18) = eTtu(t) = (a) We take the folding frequency *, to be the frequency where |F(w)] is 1% of its peak value, which happens tobe 1 (at w = 0). Hence, [Fu = À =0015w=21B= 100 This yields B = 50/r, and T < 1/28 = 7/100. Let us round T to 0.03125, resulting in 32 samples per second. The time constant of e-* is 1. For To, a reasonable choice is 5 to 6 time constants or more. Value of To = 5 or 6 results in No = 160 or 192, neither of which is a power of 2. Hence, we choose To = 8, resulting in No =32x8= 256, which is a power of 2. (b) We take the folding frequency F, to be the 99% energy frequency as explained in Example 4.16. From the results in Example 4.16, (with a = 1) we have 080m ga 2 a This yields B = E = 10.13 Ha. Also T < 1/2B = 0.04936. This results in the sampling rate * = 20.26 Hz. Also To = 8 as explained in part (a). This yields No 0.26 x 8 = 162.08, which is not a power of 2. Hence, we choose the next higher value, that is No = 256, which yields T = 0.03125 and To = 8, the same as in part (a). => W = 63.664 = 63.66 rad/sec. 2 0-5 Application of duality property to pair 3 (Table 4.1) yields 64 Following the approach of Prob. 5.2-2, we find that the peak value of |F(w)| = 2me-l“lis 2x (occurring at w = 0) Aiso, 2re” I*l becomes 0.01 x 27 (1% of the peak value ) at w = In 100 = 4.605. Hence, B = 4.605/2r = 0.733 Hz, and T < 1/2B = 0.682, Also, H(0)=2 and 0-5 t»1 Choose Th (the duration of f(t)) to be the instant where f(4) is 1% of /(0). 2 2 =p = = 10 This results in No = To/T = 10/0.682 = 14.68. We choose No = 16, which is a power of 2. This vields T = 0,625 and To = 10. (b) The energy of this sigual is 2 [ont DEE) (2m)2e e do = 2m The energy within the band from «w = 0 to W is given by em Ew = et dy = Iu(l — e) But Ew = 0.99E, = 0.99 x 2m. Hence, 09M2m)=27(1-0 2) >W=2303 Hence, B = W/2r = 0.366 Hz. Thus, T < 1/2B = 1.366. Also, To = 10 as found in part (a). Hence, No = To/T = 7.32. We select No = 8 (a power of 2), resulting in No = 8 and T = 1.25. % + $ “ a 3 + +> & Há zé 3 k> 3% dhooooo 3 I ' 2 3 4 > K> & RA as 31 Figure 85.2-5 5.2-5 The widths of f(t) and g(t) are 1 and 2 respectively. Hence the width of the convolved signal is 1+2= 3. This means we need to zero-pad f(t) for 2 secs. and g(t) for 1 sec. making To = 3 for both signals. Since T = 0.125 3 =ma o No must be a power of 2. Choose No = 32. This permits us to adjust To to 4. Hence the final values are T = 0.125 and To = 4. The samples of f(t) and g(t) are shown in Fig. 85.2-5. No 65 te) F(s) 6.1.3 (a) 1 2+5 4 1 52043) sH2f5+3 +80 = (67% + etuço) 6) 38 +5 CO = aqar r= e=tant()=634º ft) = 3.018e7* cos(3t + 6.34º )u(t) te) (+ (+ FO=20508" E+DG-5 This is an improper fraction with b, = bz = 1. Therefore a b 02,32 Ho=ltoatgoa lost 18) = 6(8) + (3,2%! — 0,20 *Su(t) (d) E + 25 + 1.25 sOgZ sea To find k set s = 1 on both sides to obtain and Eck+2sr do = 25, 125 FO=--Dotatçra St) = 1.25(-1 + 28+ Butt) = As+B te) 28+1 2 +28+2 FO tentei snsa sr Multiply both sides by s axid let 5 — 00. This yields ="144 =» 4= Setting s = O on both sides yields 1- B l=148 — B 1 s+3 Mod=-Grtarara Bet cos(t — 63.4º)u(t) (e) 42 2d Cs s+L (s+1)2 To compute k. multiply both sides by 5 and let 5 — os. This yields and 2 1 Ss. s+1 (+IZ 2-Q+9e Tutti) (e) 1 1 ki ka» Fes 1 Fo= trocar atm ap ap RIM Multiplying both sides by s and let s —» oo. This yields =i+k =» ky=-1 1 1 dA bm, boda GHDG+HM s+I s+2 (+ (+ T+ Setting s = 0 and -3 on both sides yields inter do =» dka+2hs= 6 dtltkg>ka-1 => ko-kg=0 Solving these two equations simultaneously yields ka = ka = —1. Therefore 1 1 1 1 1 pgy=dL DL dl dn Oo Rr Gr GH 2.8 = -qair + qe Ito Comment: This problem could be tackled in many ways. We conld have used Eq. (B.64b), or after determining first two coefficients by Heaviside method, we could have cleared fractions. Also instead of letting 5 = 0 and —3, we could have selected any other set of values. However, in this case these values appear most suitable for numerical work. Ch) s+1 = (1/20) | (3 As+B PO) = Grao Os +21 G+2p frases Multiplying both sídes by s and let 5 > oo yields O=G+R+rA = a Setting s = 1 and —1 yields =» Wh+6A+6B= =» 20k-104+10B = 9 Solving these three equations in k, 4 and B yield k = 1, 4=$ and B = 4. Therefore ele eq 89 6.2-1 120 4/4 (UM so Fo rat pap sra ) For the last fraction in parenthesis on the right-hand side A=1.B=-la=2,.c=5.b=v5-1[=1 r= SER = IO =tan MG) =71,56º e! cos(t + TL.56º)u(t) s2 k 1/4 As+B Fi D]WwWw—>— = DS DD = Grs O 1 rip! Rana Multiply both sides by s and let 5 — o to obtain =k+A Setting s = O and 1 yields 0=h-142 => 20h+4B=5 dg=b- b+ 4 =» 16h+44+4B=3 3/4 14 1, s—1W s+1i (s+12 4 EESrEr) For the last fraction in parenthesis, A=1, B=-1G,a=1l,c=5,5=v5=1=2 r= [Sho 559 g=tan MI) = 70º Theréfore S(0) = K3 — itje"! + Se! cos(2% + 70º)Ju(t) = [É(3 — t) + 1.3975 cos(2t + 70º)Je"*u(t) (a) 1C) = u(t) -utt— 1) and F(s) = Llu(b)] — Ljulé — 1)] =1 el 5 =10-0º) (6) (8) = e Duçé — 7) Los E) = (e) HO) = Duo) = e"etult) Therefore F()=e"- + E (d) )=etut-m)=e Te E Dutt-r) Observe that e-E-Du(t — 7) is e *u(t) delayed by 7. Therefore E Je =( 1 )erterir Fls)=e (= e = 70 6.2-4 (a) att) = HO + ft = To) + ft — 270) ++ and Gls) = F(3) + Else + E(sje TO po = P(A e Do pe ET pe BT 4, = [E je*B|< lor Res >0 €b) Hb=ut)-u(t-2) and F(s)= 2 = 2) FT Gts) = s ( — E) 6.2-5 Pair2 e : y ult) -[ S(r)dr > (1) == o- 3 5 Pair3 . - tu(t) -/ u(r)dr +=» o) =5 Pair 4: Use successive integration of tu(t) Pair 5: From frequency-shifting (6.33), we have ult) es À and Mult) + Pair 6: Because t + 1 tu(t) => Z and teu(t) += EP Pair 7: Apply the same argument to u(t), t'u(t), ..., aud so on. Pair 8a: cos btu(t) = aleit eat) o s+3 Pair 8b: Same way as the pair 8a. Pair 9a: Application of the frequency-shift property (6.33) to pair 8a cosbtu(t) +=+ ré yields sta ar e Ccosbtu(t) e Trairi Pair 9b: Similar to the pair 9a. Pairs 10a and 10b: Recognize thãt ai re“ cos(bt + 0) = re” “cos 8 cos bi — sin O sin bt) Now use results in pairs 9a and 9b to obtain pair 10a. Pair 10b is equivalent to pair 10a. 6.2-6 (a) (1) E = 60) -6t-2) sE()=1 e 1 -2a Fi)=q(1-e 2, Gi) Ea q %t-2)- 6-4) sF(s)=e 2 e Plo) = He et) 73 (b) & =u(t)-3u(t-2)+2u(t—3) spt) = 1 Cet ess (rop =0] Fls)= 50-36 4 208) 6.31 (a) (+35 +9Y() = sb) - 1 1 1 Vo=grario meio cr u(t) = (2! eat) tb) (sy (5) 25 1) +4(sY'(5) - 2) +4Y'(6) = (+ vv or º (92 ras +4)y(s)=25+10 and = 28410 284100 2 6 CO gaqasa Gr +31 Grp v(t) = (2+ 6t)e *u(t) (o) (r6)-s- D+66r6)-n+mra= (+92 + 2 or 2 (2 +65 +25)P (8) = +34 2 = Etetstio and +32 +50 2 —8 +20 Ve) = 5256425) Ts! ST LG4 E v(t) = [2 + 5.836e"* cos(4t — 99.86º)Ju(t) 6.3-2 (a) All initial conditions are zero. The zero-input response is zero. The entire response found in Prob. 6.3-2a is zero-state response, that is Veslt) = (e! = eBu(t) Yailt)=0 (b) The Laplace transform of the differential equation is (sr(s)-28-1)+4(9Y(5)-2) +4Y (5) = (5 + nL or s71 (ru +ar(s)-(28+9)=1 or 2 = (SC +45 +HAY()=28+9+1 Leterms input u+9 1 s2+45+4 52445+4 TH lo aero input aoroatato 2 5 1 tor ra it e r(s)= zero-input zeromtato —a -» D=(2+ nt)=(2+5t)e + te meroinput — 2erostate 7-4 (c) The Laplace transform of the equation is (y(s)-s- 1) +6(sy(s)- 1) + 25Y(5) = 25+ c|8 or (+68 +25)Y(s)= s+7 +25+ = E Lo. terms input s+7 4 50 2 +65+25 s(s2+65+25) Cm porem? mm ví) = aesoioput aorostute s+7 2 —2s + 13 (rare toras) v(t) = [V2e"* cos(4t — 23] + [2 + 5.154e7* cos(at — 112.839) AR 6 aoro-input zaro-stato 6.3-3 (a) Laplace transform of the two equations yields t (+ 3) (5) — 2h5(5) = —2Yi(5) + (25 + 4)Yn(5) = 0 Using Cramer's rule, we obtain v(gy= tl = tA2o 2 MS 1/6 Vo q r5s+4) a(s+IsH4) s s+l s+4 vas) 1 1 1/4 1/3, 1/12 ' = MA 1/3 Caros ssAlE4A) s s+1"S44 and —*u(t) 1 a + ge “ut 1f Hi(5) and Ha(s) are the transfer functions relating y:(£) and ya(t), respectively to the input f(t), thus s+2 1 (= ara Ud Mal) = rss ra (b) The Laplace transform of the equations are (s + B)Yi(s) — (s + 1)Yo(s —(s + 1) (5) + (2s + 1)Yo(s) = 0 Application of Cramer's rule yields vis) = EI = s+1 -10 OM | 026 Sra ri) o(s +0.385)(s+2618) s 240382 s+268 va(s) = —sE2 s+2 2 18 | 0.1056 dO (sr3s+T) ss +OJRIS 42618) s s+0382 5+2618 s+1 s+2 H(o= gra Md Hal) ara TT alt) = (1 07240 SB 02760 2 u(s) alt) = (2 — 1.894e 032 — 0 1056e 281% yu(t) 6.3-4 Att + the inductor current 71(0) = 4 and the capacitor voltage is 16 volts. After t = 0, the loop equations are 75 Fig. S6.4.4 6.4-1 Figure 86.4-1 shows the transformed network. The loop equations are + Eyrio) — dra(o) = EE Into tests bao “od [É] º [=] -1 és llyço] Lo 1 1 1 (IE rã + O (s+1)2 s+2s+2 vo(t) = ya(t) = (te! — le“*sini)u(t) or Cramer's rule yields 6.4-2 Before the switch is opened, the inductor current is 5A, that is y(0) = 5. Figure $6.4-2b shows the transformed circuit for t > O with initial condition generator. The current Y(s) is given by Y(s) (10/5)+5 5s+10 5/3 2 DO +2 Cs +973/s 550/) v(t) = (5 — Pe )u(t) 6.4-3 The impedance seen by the source f(t) is z()= E/CS) ( ds Lswo? — Le+(1/08) LOs2+1 s2+uwo? The current Y (3) is given by F(S) “O=76= ft) (a) Fossós vO= and ut) = 756) 52 wo?" Lup? €b) — — Ao A É Fís) Fra Yi)= Togo Sad ve) = qt) 78 Fig. S6.4-5 6.4-4 Att =0, the steady-state values of currents y, and yo is 71(0) = 2, y2(0) = 1. Figure 86.4-4 shows the transformed circuit for t > O with initial condition generators. The loop equations are (s+2)M(9) ras) = 248 -Yls)+(s+a(s)=1 Cramer's rule yields 282 + Hs +12 4 3/2 1/2 Nils) = SG +HI(ST3 s s+1 5+3 s2 +48 +6 2 3/2 1/2 PO) =IDGEDTS seItrSs n()=(4-5 val) = (2-5 — de “utt) + Le Puts) 6.4-5 The current in the 2H inductor at ! = Q is i0A. The transformed circuit with initial condition generators is shown in Figure 86.4-5 for t > 0. vis = RE ( Q0+io 20 atos WO Capisi Besos |54E+ = tan Hã) = 311º (e) = (1.168)e7!/ cos( Me — 314º)u(t) = 7,787e7 8 cos( 4 — 31.1º)u(t) The voltage vs(t) across the switch is ÉH 205 + 10 )= 0 + (+05) 5 382 48+1"" 5 (rr 32 (1 -8s+1 s 652 F1/39+1/3 vs(t) = 26(t) + (10 + 9.045e7*/º cos(2TLt — 152.2º)]u(t) Velo) = (8+ rt) = 20 =2 + 6.4-6 Figure $6.4-6 shows the transformed circuit with mutually coupled inductor replaced by their equivalents (see Fig. 6.14h). The loop equations are (8 + Di(S) — 28Ya(5) = To “25h (5) + (45 + Ya) =0 Cramer's rule yields 79 Fig. S6.4-.6 6 Fig. S6.4.7 40 Ya(s) = Cro) and volt) = valt) = 40e*u(t) 6.4-7 Figure 86.47 shows the transformed circuit with parallel form of initial condition generators. The admittance W(s) seen by the source is 2 , wo) = Brspa= Tttotlô The voltage across terminals ab is Vols) = Ko) d+3 0 3541 Wi Ty) Fam O 2 r4s +13 Also 1 3841 YW == >> DD — o(s) = 5 Vaso) = rp gs 115) and volt) = L76e * cos(3t + 29)u(t) Fig. 86.48 6.4.8 The capacitor voltage at t = O is 10 volts. The inductor current is zero. The transformed circuit with initial condition generators is shown for t > O in Fig. S6.4-8. To determine the current Y'(s), we determine Zas(), the impedance seen across terminals ab: Zao(o)= — E o dota 1 ( 1 ) Cas4i + sF2 2+ s+ so 6.5-2 6.6-1 camoucaL Form Y6s) Fig. S6.6-1 En this case H(s) is very close to 1/4. This is because the second ladder section causes a negligible load on the first. The Cascade rule applies only when the successive subsystems do not load the preceding subsystems. To determine the impulse response of either system, we apply f(t) = 6(t) at the input. The output y(t) is, by definition, the impulse response. . For series connection (Fig. 6.18b), if we apply /(t) = 6(t) at the input, the output of Ha (5) is Ai(t). This signal is applied to the input of Ha(s) with impulse response ha(t). Hence, the output of Ha(s) is A(t) = hr(t) + ha(t). For parallel connection (Fig. 6.18c), if we apply f(t) = 6(t) at the input, the outputs of Hi(s) and Ha(s) are ha(t) and ha(t), and the output of the summer is A(t) = hi(t) + ha(t). s E ( 1 )-aê- 3/2; , 88 s+1/15 43/1544) a+i o s+3 544 bas? + bas? + bis + bo st +azs! +ais + ao $+2s HO) radio rD” Also H(s) = withao=12,01=19,02=8, and to=0,b=2,ba=1. Figure 866-1 shows the canonical, series and parallel realizations. 6.6-2 (a) H() = 3s(s + 2) = 38 +65 . (E +D6242+2) sI+IFAs+2 o 3 ( s+2 J=- 3 65+6 s+T/Ast pas +2 s+1 s+25+2 83 bs tos ami wo) Tanivavg CANGNICAL. canonioal forem mm Fig. 86.6-3 and S6.6-4 85 and Choose C = Cy = 1074, R = 5000, R = 2000. This yields s+2 Eo)=-Cs ) This is followed by an op amp of gain —1 as shown in Fig. 6.6-8c. This yields s+2 s+5 H(s)= Fig. S6.6-8 6.6-9 One realization is given in Fig. 86.6-8c. For the other realization, we express H(s) as s+2 3 Ho)= Tio Tas We realize H(s) as à parallel combination of H1(s) = 1 and Ha(s) = —3/(s + 5) as shown in Fig. 6.6.9. The second stage serves as a summer for which the inputs are the input and output of the first stage. Because the summer has & gain —1, we need a third stage of gain —1 to obtain the desired transfer functions. 333. + Ab vá tok “Jor ru sed! > Fig. S6.6-9 as E 6) td) 100k0 100kA seoh o Jim AMA - AM Fm PS] do = + A “q te) FFig. 56.6-11 6.6-10 The transfer function here is identical to H(s) in Example 6.20 with a minor difference. Hence the op amp circuit in Fig. 6.32c can be used for our purpose with appropriate changes in the element values. The last suminer input resistors now are 19º kf? and 19º kf? instead of 50 kf) and 2092. 6.6-11 We follow the procedure in Example 6.20 with appropriate modifications. In this case ao = 13,a1 = 4, and bo =2, bi =5, and bo = 1 (in Example 6.20, we have ao = 10, a) = 4, and bo = 5, db) = 2, and bo =0). Because da is nonzero here, we have one more feedforward connection. Figure S6.6-11 shows the development of the suitable realization. 6.7-1 9 €) To)=araço > mos Mun=3 > (=05 Hence from Fig. 6.39, PO = 17% and wntr = 1.63 which yields t, = 0.526. Also 4 . . 1 . fo = Tom = 4/15 = 267 co= Bimil T(l=0 es=lim[t-T()/o= 7 ep=imil— T(s)l/S = 00 4 0) T)=maaa O m=2 Mun=3 = (=0% 89 - Henceé from Fig. 6.39, PO = 9% and wntr = 2.3 which yields t, = 1.15. Also 4 tez E W15=267 em -T()]=0 er = lim[l=T(8)]/s=0.75 cp = limit = T(5)]/5? = 00 95 no (O T)= mio O sl, Mun=10 => 05 Hence from Fig. 6.39, PO = 17% and cont, = 1.63 which yields tr = 0.1 Also n= É =4/5=08 es=liml=T(9)]=005 er= tim[l-T(s)j/e=00 cp = lim[l = T(s)]/52 = 00 6-2 Ka Sisto) Kiko Ti) = 1 | E] = RE +miy) Stas+ko PO=e ST VI-C 2 0.09 > € = 0.608. Moreover, 7 DenV1-Ç > wa = 5.04 for é = 0.608 Thus rMunstul=-sras+k> > a=6128, and K=254 The steady-state value of the output is given to be 2. But, the staedy-state value of the output is pas = lim ptska 1 as ras t ks Thus, the parameters are K; = 2, K> = 25.4 and a = 6.128. =K=2 Figure S6.7-3 6.7-3 The transfer function of the inner loop is 1/(s+2). Henee, the open loop transfer function of this unity feedback system is K K é K gy=—L. Tty=-882 = A (= +53 (5) Is SIU+K The characteristic roots are 1 + j/K7— 1. The root locus is shown in Fig. 86.7-3. Observe that for the characteristic polynomial s2 + 25 + K. Cun = land wi =X. But, ta = 4/Qua — 4. Hence, we cannot meet the settling time specification (ts < 1), regardiess of the value of K. We now find the steady-state errors. This being a unity feedback system, we could use parameters Kp, Ko and Ka. We have Kp= lim Gl) = 00 Ku = limsG(9)] = E a = Bmglo2G(5)] = 0 Hence, Also VRÇ=0504VK=10 + K=30367 Moreover. 20/K 0.4. For K = 393.67, we meet all the transient specifications. Also e, = O and we meet all the specifications for K = 393.67. Actually, any value 0f K in the range 333.94 S K < 393.67 will satisfy all the conditions 6.8-1 (a) Let A(t)= f(t)u(t) = etu(t) and folt) = (t)ul>t) = u(—t). Then Fi(s) has a region of convergence o > 1 And Fís) has a region o < 0. Hence there is no common region of convergence for F(s) = Fi(s) + Fats) (8) ht) —tu(t), and Fi(s) = — converges for o > —1. Also fa(t) = u(-t). and Fa(s) = —1 converges for herefore, the strip of convergence is -I<o<0 -— strip ef : - tanvera enc EA r Figure S6.8-1b 1 ) ast>occifRes>0 2 et : +41 ast— -oifRes<o Hence the convergence occurs at o = O (ju-axis) (a) | Mo= tra : ast>oojifRes> 1 "—o) ast> co ifRes<o Hence the region of convergenceis 1 < o < 0 (e) eee, 9) as t — oo for any value of s as t — —oo for any value of s Hence the region of convergence is the entire s-plane. 6.8-2 (a) fg=e = ectu(o) + ul) = fale) + fat) Fis) = + s> fa-t)=e-tu(t) and ij and o<1 Hence: F(5)= Fi() + Fa(s) = a (b) (e) td) (e) (9) 6.8-3 (a) 11) =e"Peost = e-tcostu(t) +e'costu(-t) = f(t) + fa(t) s+1 s+1 Hence J=—EE— aos) = SAL ence Fi(s) a ed fode OS Fls) = Fi(s) + Pas) s+1 so -“1<0<1 “PIT Goa H=eut) +etu(-t) Fi(s)= Fá) = s<2 Hence F(s)=Fi(s) + F(a) = Ts i<o<2 for t>0 Ht= for +<0 fdf)=e ut, fat)=u(-t). Hence Fi(s)= a g>-1 and Eos) = E DR) Allo s4+1 5 s(s+1) 1 ' A a A o andhencs: F(s)= flt)=1 for t>0 tu(=e) 19 Cio = for t<6 F(g=1 0>0 f(8) = coswotu(t) + etu(=8) = f(t) + fale) RO 0>0 1 and Fis) = po Fi(s)= e<1 - = (+08) F(s) = Fi(s) + Fo(s) = ET rod) 0<o<l 28 +45 F(s) CrNGID -3<o<-2 1 1 o “qto cêse<c2 The pole —2 lies to the right, and the pole —3 lies to the left of the region of convergence; hence the first term represents causal and the second term represents anticausal signal: But) — ePu(-t) fy)=e 9 6.8.4 tb) 28 - -D6-3) Fls) = 2<0o<3 ++ 2<0<3 =32" 5-3 T The pole at —2 lies to th left and that at 3 lies to the right of the region of convergence; hence Ho = cut) — eXtu(-t) (e) 25+3 P(s) = — 2888 - O=aanero 2 1 1 qto 20 Both poles lie to the left of the region of convergence, and 100) = (7! + Pao) (d) 2543 = st 2 fO=tanero CS 1 1 Sto SS Both poles lie to the right of the region of convergence, and hence: 1) = ("4a te) O 3-2 IT “(6 +D(s+3(s—5) E A Cs+l s4+3"s-5 F(s) -1<0<5 The poles --1 and —3 lie to the left of the region of convergence, whereas the pole 5 lies to the right: st) = (e! + e u(o) — eSu(-8) 22-26 1 1,2 C+DG-(8+2) s+41 s-10s+2 (3) Res > 1: AL poles to the left of the region of convergence. Therefore H8) = (et — et + 20 ut) (b) Res < —2: All poles to the right of the region of convergence. Therefore Tt pe! ge 2u(—t) ft)= (e) -1<Res< 1: Poles 1 and —2 to the left and pole t to the right of the region of convergence. Therefore H0) = (e7! + 2e Mult) + etu(-t) (d) -2< Res <—1: Poles —1 and 2 are to the right and pole —2 is to the left of the region of convergence. Therefore S(t) = 2e"Yu(t) + [et + elu(=t) 95