sihnal and systms solution, Schemes and Mind Maps of Law

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Chapter 1 Answers Mk is La. Converting from Cartesian to polar coordi 7, —2=2, plac 1+ 4 junj)eet, Bisel, ag (a) zn= [ ‘e-Mat = 1, Poo = 0, because Boo < 08 lo [r2(e)Pat £ dt = 00, Po = (b) z(t) = 2 Jza(t)| = 1. Therefore, Boo £ + a sina [ona ina [iam jim. 1 (c) zat) = cos(t). Therefore, Bao = [iestortae 3 £ cos%{t)dt = 00, rem ae ae ran | (APE eo “ |. cas?(¢)dt = Jim 4)" ula}, tealn]!? = (3)" ul}. Therefore, Boe = Y ele? (a) 2/9] = Pra Syst because Bee < 09 (e) zaln) = fF 48, [agfn)]? = 1. Therefore, Zoo = EX lanl? =o, im ——— P= lim = tuavter So beavis 3, wie0 (0 zal) = con fn). Therefore, Em = 5 fsa = So oa) =o . be cue 1_¢ Lente) 3 Pes jin ovat 2m Ge= Beatct &, (~ ie) 2 (a) The signal xn] is shifted by 3 to the sight. The shifted signal willbe zero for n <1 and n> 7 (b) The signal ln is sifted by 4 tothe I, ‘The sifted signal will be zero for n < —6 and n> 0. 2 = 2e™ cos(Ot + *) Vicos( {) cos(3t + 2x) = cas(3t) = e* cos(3t + 9) -t sin(3t + x) = et eos(3t + §) (1004) ‘sin(100t + ) = 7% cos(100# + $) (a) Retzi()} = (b) Re{z2(0)) (©) Retzat®)) (d) Refza(t)} = (a) 21(t) is a periodic complex exponential (ty = 50 = e109) ‘The fundamental period of 21(t) is %§ = 3 (b) z(t) is a complex exponential multiplied by a decaying exponential. Therefore zat) is not periodic. (c) za{n] is a periodic signal, ss alm] = 7" = 7” +2s[n] is a complex exponential with a fundamental period of 7 = 2 (a) z4{n] is a periodic signal, The fundamental period is given by N an( 524) = mB). Ry choosing m = 3, we obtain the fundamental period to be 10. (e) x(n) ie not periodic. zsin] is a complex exponential with wo = 3/5. We canebt find aay integer m such that m(2£) is also an integer. Therefore, =s(n] is not periodic. a(t) = 2eas(10t + 1) ~ sin(at — 1) Period of first term in RHS = 3 = Period of second term in RHS = 3% . Freee ot. the overall signal is periodic with a period which is the least common multiple of the periods of the first and second terms. This is equal to 7. 140%" oF" Period of the first term in the RHS = 1 Period of the second term in the RHS = m(;2%q) = 7 (when m = 2) Period of the third term in the RHS = m(2%z) = 5 (when m = 1) ‘Therefore, the overall signal z{n]} is periodic with a period which is the least common multiple of the periods of the three terms in z{n]. This is equal to 35. ‘The signal {ni as shown in Figure $1.12. 2[n] can be obtained by Bipping ufn] and then shuftng the Bippe signal by 3 t0 the right. Therefore, 2{n} = ul-m +S], This implies ¢hat M = Land no = -3 Lad. as. (c) The signal z{n] is flipped. The flipped signal will be zero for n < 4 and n > 2. (4) The signal z[n] is flipped and the Ripped signal is shifted by 2 to the right. This new signal will be zero for n < ~2 and n > 4 = _ (e) The signal ={n] is @ipped and the fipped signal is shifted by 2 to the left. This new signal will be zero for n < -6 and n > 0. ‘i " (a) 2(1 ~ ¢) is obtained by ipping 2(t) and shifting the Sipped signal b he tose eee ee (b) From (a), we know that 2(1~t) is zero for t > ~2. Similarly, 2(2—t) is zero for ¢ > 1. ‘Therefore, 2(1 — t) + 2(2— 1) will be zero for t > ~2. (6) {0 is obeaine by Uoesty compressing =(¢) by a factor of 3. Therefore, 2(3t) will be zero for ¢ <1. (d) 2(¢/3) is obtained by linearly stretching 2(¢) by a factor of 3. Therefore, =(t/3) will be zero for t <9. (a) 2,(t) is not periodic because it is zero for t <0. (b) za[n] = 1 for all n. Therefore, itis periodic with a fundamental period of 1. (c) 23(n] is as shown in the Figure S16. + Figure $1.6 Therefore, it is periodic with a fundamental period of 4 (a) Ev{ziln) $(euln) + 2f-n)) = Huh = uln - 4] + fn] - uf-n- 4) ‘Therefore, Ev{zi(n]} is zero for [n| > 3. ey Since z2(t) is an odd signal, £v{72(t)} is zero for all values of t. Ev{esinh) = extn] + aik-al) = Ld)win—3)- Hmufan ~ al] a ‘Therefore, £v{za[n]} is zero when [n| <3 and when |n| + oo. Evlzae)} = Healt) + 24(-t) = BlerSule +2) = u(—t+2) ‘Therefore, Ev{z4(t)} is 2er0 only when |e} -¥ 00. TE @ “eto rad na Figure $1.12 ‘ ten? (6(r +2) (7-2), Bets? ° 0, 52 2 a= [ a=s Bs ‘The signal x(t) and its derivative g(t) are shown in Figure $1.14 3 2 xt “ “7ore =befe rt ee Figure $1.14 Therefore, alt) =3 D> s(t—2k)-3 }> 4(t- 2-1) = Feats ‘This implies that Ay = 3, 0, Az = ~3, and tg = 1. (a) The signal za{n, which is the input to Sz, is the same as yi[n]. Therefore, vale) = aan -2) + 520in—3) = vile 2) + 5uln—3] = 2ay[p— 2} + dzrln - 3] + 3(2ey fn — 3] + Aziln ~ 4)) = 2Qey[n- 2] + $2y[n—3} + 22y[n 4) ‘The input-output relationship for S is ln] = 2z[n — 2] + 5e[n — 3} + 2z[n — 4]