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Chapter 6 Solutions 151 E / EL z [ol | K 2 did am Figure S6.1-1 298 CA 6.1-1. (a) To=4,wo=2 = E. Because of even symmetry, all sine terms are zero. n>3 Y; (b) (a) do) = 004 5)ancos(S76) n=1 ao = O (by inspection) on = [[ cos(S70) ai [Pcos (Ei) au] = A sntz Therefore, the Fourier series for x(t) is (= 8 (cos! Logs Pp Los BEE Lea RtS MOI 38 "5a 7 Here ba = 0, and we allow C, to take negative values. Figure S6.1-1a shows the plot of Ca. To = 107, wo = 2E = 1. Because of even symmetry, all the sine terms arc zero. eli) = + » an cos (59) + ba sin (5 1 . . do = 5 (by inspection) « 2 mom 1,5 nal 2 na a = mr) (cs (5º) dt= s(o)sin (58) Sesin(T) 2 [om . . . bd = mm / sm 2) dt=0 (integrand is an odd function of t) Here by = 0, and we allow Cy to take negative values. Note that Cn = an for n=0,1,2,3,--- Figure $6.1-1b shows the plot of Cn To=2m, up =1. o(t)=a0 + ancosnt+basinnt with ao =0.5 (by inspection) n=1 14 21 4 => — cosntdt = 0, [ —sinntdt = ho 2m o 27 and 1/. 1. 1. 1. a(t) = 05-+ (sint+>sin2t+ > sin3t+ >sin4t+ Ea 2 3 4 n 0541 [eos (t+ 5) + joos (245) + Scos (2845) 4.) The reason for vanishing of the cosines terms is that when 0.5 (the dc component) is subtracted from z(t), the remaining function has odd symmetry. Hence, the Fourier series would contain dc and sine terms only. Figure $6.1-1c shows the plot of Cy and 0n. To =, wo = 2and a(t) = ao = 0 (by inspection). a, =0 (n > To compute the coefficients, we shall use the interval 7 to O for integration. Thus 17º aq = / et/2 dt = 0.504 TJa 2(º. 2 2 /2 = 0,504 (— 2 a = af. cos 2nt dt 0504 (,rigas) 27º. 8n = À 1 gi =-0. —— ba fe sin Qnédt = 0.504 (q Therefore Co = ag=0.504 2 CG = Val+b=0.504 (=) º v1+ 16n? bn E On = tan (=) =tan!4n An — 2 t) = 0.50440.5045) —===>> cos (2nt + tan” 4 e(t) 5 L Ta ( an tan) (b) This Fourier series is identical to that in Eq. (6.154) with t replaced by —t. (e) Ef a(t) = Co + DD Cu cos(nuwpt + 0h), then )=Co + 3) Cu cos(-nuot + On) = Co + 3) Cn cos(nuwot — On) Thus, time inversion of a signal merely changes the sign of the phase 9,. Every- thing else remains unchanged. Comparison of the above results in part (a) with those in Example 6.1 confirms this conclusion. 613. (a Here To = 7/2, and wo = Ea = 4. Therefore a(t) = 00 + ancos Ant + by sin Ant na where 2 m/2 aq = E) etdt=0.504 Tlo 4a 2 an = / e cos antdt = 0508 (5) and 4 qu? 8 da= E) e-tsin 4nt dt = 0.504 (— SD ERA 1+ 16n2 Therefore Cy = ao = 0.504, Cu = AR +IZ = 0.504 ( —tan-l4n (b) This Fourier series is identical to that in Eq. (6.15a) with é replaced by 2t 301 6.1-4. 6.1.5. o (a (e) If z(t) = Co + E Ca cos(nuot + On), then e(at) = Co + 3 Cn cos(n(awo)t + 04) 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 a). Comparison of the results in part (a) with those in Example 6.1 confirms this conclusion. This result applies equally well Here To = 2, and wo = 2º =. Also a(t) is an even function of t. Therefore z(t)=00 + ancos nat na where, by inspection ag = O and from Eq. (6.18b) 1 , 4 1 o) n even m=5 |, A(-2t+1) cos natdt = aê (cosnat — Dlo = da, nodd Therefore * 84 Ls a; 1 et) = Ec feos t+ qcos 3at+ = cos mt + ag “os Tnt+ e | (b) This Fourier series is identical to that in Eq. (6.16) with t replaced by t + 0.5. (c) If a(t) = Co + 3) Cu cos(nwot + 0h), then T(t+T) = Co+5 Cu coslnwo(t + T) +0n] = Co +) Co cosfnwot + (On +nwpT)] Thus, time shifting by T merely increases the phase of the nth harmonic by nwoT. ay= e (148) For half wave symmetry and 2 To D) To/2 To an= z/ e(t) cos nwpt dt — z/ =(t) cos nwgt dt +[ =(t) cos ruvpt dt To Jo To Jo To/2 Let 7 =t — To/2 in the second integral. This gives 2 To/2 To/2 a = z|f, ati cosmoede+ [ ara Do cosni (r+ dr To [Jo o 2 2 2 To/2 To/2 - 2 TÁ (8) cos nupt dt + [ a(r)l- cosnupr] dr To [Jo o 4 To/2 -4 [ e(t) cosmugt dt lh 302 6.16. (a) (b) (a) Here, we need only cosine terms and wo = 3. Hence, we must construct a pulse such that it is an even function of t, has a value t over the interval O
ancos (5) t n=1 By inspection, we find ag = 1/4. Because of symmetry by = 0 and an= [ toostEtat= ch, [os (8) + sin (27) 1] Here, we need only sine terms and wo = 2. Hence, we must construct a pulse with odd symmetry, which has a value t over the interval 0 ) ancosnt = nodd By inspection, ao = 0. Because of even symmetry ba = O. Because of halftwave symmetry (see Prob. 6.1-5), “2 2 2 a/2 a 2 ” an= É f teosntat— (t-1)cosntdt| = A (cosna-1)+2 sin 2E o n/2 m n Here, we need only sine terms with wo = 7 and odd harmonics only. Hence, we must construct a pulse such that it is an odd function of t, has a value t over the interval O < t < 1, repeats every 4 seconds and has half-wave symmetry as shown in Fig. $6.1-6e. Observe that the first half cycle (from O to 2) and the second half cycle (from 2 to 4) are negatives of each other as required in half-wave symmetry. This will cause even harmonics to vonish. The pulse has an odd and half-wave symmetry. This yields o na 9= > basn a(t) » sino nodd By inspection, ao = 0. Because of odd symmetry a; = O. Because of half-wave symmetry (see Prob. 6.1-5), 8 nm sin — nodd 2 ap 7 2 y n=5), tsin rata [ (-t+3)sin Trat = n2m 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 half-wave symmetry, but neither odd nor even symmetry, has a value t over the interval O ) ancosnt + ba sinnt n=1 nodd Because of half-wave symmetry (see Prob. 6.1-5), ap 2 . an=— | teosntdt= (cosn+nsinn—1) 27 bh an ap 2 da = =| tsinntdt= ->>(sinn — ncosn) — nodd 2a do an? 305 modd (d) To=m,wo=2and Di =0 a)= > Dyeert, 1 [744 -i/2 where Dr= f Cem (> nO com 7 Ja ana 2 2 () T=3uwo =. a) =) Dre, 1/1) sam, 3 2za ( j2mn where D.=5), ted ted as [TS (Em + —1 Therefore 3 4r2n? 2mn dn. Zan out= sê [2 g tom fe ode) 2mn cos mr 2mn “52 cos “2 — sin “2 =tantt 3 3 3 and ZD, = tan! (SS cos SE + sing — (1) To =6, wo = 7/3 Do = 0.5 a()=054+ 5) Dre |) 3 ] 1[[m ent Do sem 2 ant al/ (+70 ar [ e a [ (ae a] 6 LJ» - EN : 3 LE cos E nên? 3 3 6.3-2. Note that the signal «(t) is defined as at O :/, Morcover, from Parseval's theorem Eq. (6.40) pa (; ja (tar )- (b) If the N-term Fourier series is denoted by 1o(t), then N= 1,4 =)" v)=5+) E cosnrt -1 (> DO — 2mn) (some c(l+4rên3) Hence the output y(t) is given by > DaH(j2an)eierrt vt) S (e- DL P2mn)G2an) sor » li drtad)Gam + 1) ' n=-c0 35 êle — 1 +) porn 2mn(e -1M2mn +), e(I + 472022 nõãoo 6.5-1. Equating the derivative (with respect to c) yields 2cly|? = 2x.y which yields the desired result 6.5-2. (a) e(t) = 2(t) — co(t) 316 Also [Pete - co dt= IA a to a(t)x(t) dt — ef (tu) dt t But , Ji e(De(t) dt c= tt SÊ 0) dt Substitution of c in the earlier equation yields to [ =(Ole(t) — celt)]dt = 0 ts Therefore x(t) and [z(t) — cx(t)) are orthogonal. (b) We can readily see result from Figure 6.17. The error vector e is orthogonal to vector x. (e) 2x x 4 2 27 4 2 * [ e(tdt = / (1- Leme) a- [ (1+ Zone) dt o o ” ” ” x 27 2m - If sintdt+ [ sintai] = [ sintdt=0 ” o m TJ 6.5-3. (a) If a(t) and y(t) are orthogonal, then we showed [see Eq. (6.67)] the energy of e(t) + y(t) is Es + Ey. We now find the energy of a(t) — y(t): Pora [P popa -f e(oy'(t) dt — Ff 2º (Oy(t) dt Pteoras [O quo a The last result follows from the fact that because of orthogonality, the two in- tegrals of the cross products z(t)Jy"(t) and x*(t)y(t) are zero [see Eq. (6.80)]. Thus the energy of z(t) + y(t) is equal to that of z(t) — y(t) if z(t) and y(t) are orthogonal. k feto vota (b) Using similar argument, we can show that the energy of c;x(t) + cay(t) is equal to that of crz(t) — coy(t) if a(t) and y(t) are orthogonal. This energy is given by laEs + eoPE,. (c) TE z(t) = 2(t) + y(t), then oo , co oo f to) +y()Pa = f I(O)P dt + f Iy(oP dt + [O a(Oy(B) dt + Ff "(tule dt 317 
