Chapter 1 Solution

Chapter 1 Solution - Chapter 1 Answers 1.1 Converting from...

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Unformatted text preview: Chapter 1 Answers 1.1. Converting from polar to Cartesian coordinates: is?” = ems" =_~_%e . #73" = ewe-1r); —% . ail: cosx(%) +Jsm = J, (’51: cos - Jsm = -1 e155 _=eJe =1, \/—2—e37_=\/§(cos.(§)+jsin(%))=1+j fie¥=¢§e¥=1+L fieflF=fie=F=1—j x/ieI‘F = 1 —j 1.2. Converting from Cartesian to polar coordinates: _ 5 = 5e10, —2 = 2e“, -33' = 36-17 in“? =e'ji, 1+j = x/ie’i, (1 52')? =2e'ji j(1_j)=er, alien, {2T~;;7~5’2=e—m m 1.3. (a) Eco =/ e'4‘dt = i, P0° = 0, because Eco < oo o co m |12(t)|2dt = / dt = 00, P00 = -m -w (b) z2(t) = ejm‘li'), |z2(t)| = 1. Therefore, Eco = / Tam Tam Tax '1‘ T lim ill—,Llefitszt = lim TIT/Jot = lim 1 =1 00 00 (c) z3(t) = cos(t). Therefore, E0° = / |z3(t)|2dt = / c052(t)dt = 00, —m -w 1 T 2 1 T l+cos(2t) 1 = I — = I -— _.————— = — Pee T1320 2T 4°08 W“ T1220” 2 2 (d) min] = (%)"uln],lz1[n]|2 = G)" ulnl. Therefore, Em = 55 min“? e 2 it)” = %« nz—oo n '0 P00 = 0, because Em < oo. 00 (e) 2:2[n] = ej(%+%), |:1:2[n]|2 = 1. Therefore, E0° = Z |x2[n]|2 = 00, ‘ nz—oo 1 N 1 N = ' 2= ' 1:], R” 13520 2N + 1 2 Mn“ 131?; 2N +1 2 n=—N n=—N W 00 7r (f) 13M] 2 cos(§n). Therefore, E},o = Z: l2:;;[1n]|2 = Z 0032(Zn) = 00, n=—oo n=-oo N N n . 1 2 1r _ . 1 l+cos(§n)) _} Pw‘AlT‘w2N+1";NC°S(4")’nl§“m2N+1n;N( 2 ‘2 1.4. (a) The signal :r[n] is shifted by 3 to the right. The shifted signal will be zero for n < 1 and n > 7. (b) The signal :c[n] is shifted by 4 to the left. The shifted signal will be zero for n < -6 and n > 0. 1.5., 1.6. 1.7. (c) The signal :r[n] is flipped. The flipped signal will be zero for n < —4 and n > 2. (d) The signal $[n] is flipped and the flipped signal is shifted by 2 to the right. This new signal will be zero for n < -2 and n > 4. (e) The signal :r[n] is flipped and the flipped signal is shifted by 2 to the left. This new signal will be zero for n <_ —6 and n > 0. (a) 1(1 — t) is obtained by flipping :z:(t) and shifting the flipped signal by 1 to the right. Therefore, :r(l — t) will be zero for t > —2. (b) From (a), we know that :c(1—t) is zero fort > —2. Similarly, 2(2—1‘.) is zero for t > —1. Therefore, a:(1 — t) + a:(2 -— t) will be zero for t > —2. (c) $(3t) is obtained by linearly compressing 1(t) by a factor of 3. Therefore, :r(3t) will be zero for t < 1. (d) 1(t / 3) is obtained by linearly stretching a:(t) by a factor of 3. Therefore, 9305/ 3) will be zero for t < 9. (a) z1(t) is not periodic because it is zero for t < 0. (b) 2:2[n] = 1 for all n. Therefore, it is periodic with a fundamental period of 1. (c) z3[n] is as shown in the Figure 31.6. Figure 81.6 Therefore, it is periodic with a fundamental period of 4. (a) 5mm} = émn] + $1[—n]) = %(u[n] — u[n — 41+ u[—n] — u[—n — 4]) Therefore, £v{:r:1[n]} is zero for In] > 3. (b) Since 12(t) is an odd signal, £U{$2(t)} is zero for all values of t. (c) 1 1 1 ,. l -7. €v{za[nl} = 5(z1lnl + Ill-n1) = 5K5) uln — 31—(5) ul-n — 311 Therefore, £u{:r3[n]} is zero when |n| < 3 and when |n| —+ 00. ((1) 5mm} = $0124“) + x4(—t)) = —;—[e-5tu(t + 2) — e5‘u(—t + 2) Therefore, £v{:r4(t)} is zero only when |t| —) oo. 1.8. 1.9. 1.10. 1.11. 1.12. (a) Re{:r1(t)} = -2 = 260‘ cos(0t + 7r) _ (b) Re{:c2(t)} = ficosfi) cos(3t + 2n) = cos(3t) = eo‘ cos(3t + ) (c) ’Re{:ra(t)} = e“ sin(3t + 1r) = e“ cos(3t + 3%) (d) 'Re{$4 (t)} = -—e'2' sin(100t) = e‘” sin(100t + 1r) = 6‘” cos(100f. + %) (a) :rl(t) is a periodic complex exponential. $1“) =J-ej101 ___ ej(10H-%) The fundamental period of 2:1(t) is % = (b) 12(t) is a complex exponential multiplied by a decaying exponential. Therefore, 2:2(t) is not periodic. (c) “[12] is a periodic signal. . . I3[n] = 6171": = 63an 2:3[n] is a complex exponential with a fundamental period of a} = 2. (d) z4[n] is a periodic signal. The fundamental period is given by N = m(3:75) By choosing m = 3, we obtain the fundamental period to be 10. = meg). (e) 15[n] is not periodic. 2:5[n] is a complex exponential with we = 3/5. We cannot find any integer m such that mtg—7;) is also an integer. Therefore, 2:5[n] is not periodic. x(t) = 2cos(10t + 1) —— sin(4t - 1) Period of first term in RHS = 2—3 = g Period of second term in RHS = 2T" = % Therefore, 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 1r. $[n] = l +9171" —ejzsl" Period of the first term in the RHS = 1 Period of the second term in the HHS = m(;%7) = 7 (when m = 2) Period of the third term in the RHS = m(%§§) = 5 (when m = 1) Therefore, the overall signal flu] is periodic with a period which is the least common multiple of the periods of the three terms in a:[n]. This is equal to 35. The signal a:[n] is as shown in Figure $1.12. :1:[n] can be obtained by flipping u[n] and then shifting the flipped signal by 3 to the right. Therefore, :r[n] = u[—n + 3]. This implies that M = —1 and no = —3. 10'] —L-I o I 2. 3 11 Figure $1.12 1.13. t t 0, t < —2 W) =/ 3(T)dt =/ (6(1 + 2) — 6(1 — 2))dt = 1, —2 g t g 2 "°° '°° 0, t, > 2 Therefore, 2 EOO=/ dt=4 —2 1.14. The signal z(t) and its derivative g(t) are shown in Figure $1.14. 1m 1 Figure 31.14 Therefore, 00 a) g(t)=3 Z 6(t—2k)—3 Z 6(t—2k—1) k:—oo [CZ-00 This implies that A1 = 3, t1 = 0, A2 = —3, and t2 = l. 1.15. (a) The signal zfln], which is the input to $2, is the same as y1[n]. Therefore, y2[n] = 1:2]71 — 2] + éxfln - 3] ll y1[n - 2] + éydn - 3] = 22:1[11 -— 2] +4xl[n — 3] + %(221[n — 3] + 4z1[n — 4]) = 21:1[71 — 2] + 51:1[71 — 3] + 2x1]n — 4] The input-output relationship for S is y[n] = 2:1:[n — 2] + 5:1:[n — 3] + 2:1:[n - 4] 1.20. (a) Given _ :1:(t) = cm —> y(t) = e7“ x(t) = e‘jzt —-——> y(t) = e”j3‘ Since the system is linear, (ej3t + e—j3t) . . 1 mm = gem + arm) —» y1(t) = 5 Therefore, z1(t) = cos(2t) —+ y1(t) = cos(3t) (b) We know that ——j ‘2: ej -j2t 12(t) = cos (2(t - %)) = _._.__—_e e] :- 6 Using the linearity property, we may once again write z1(t) = $4519” + eje-fl‘) ——» 3/1 (t) = é-(e‘jejat + eve-i3!) = cos(3t -- 1) Therefore, :n(t) = cos(2(t — 1/2)) ——+ y1(t) = cos(3t-1) 1.21. The signals are sketched in Figure $1.21. 16: -I) “(1‘3 1. 2. 1..(Zt+l) 2. z (rt/2) 3 rzmuwjum 2 04’ 0.6 1. a j— JL_JL 4' 6 8 to + o I .1. fly?- 3/2. t Figure $1.21 1.22. The signals are sketched in Figure 31.22. 1.23. The even and odd parts are sketched in Figure 81.23. 10 1.24. 1.25. 1.26. 1.27. Figure $1.24 The even and odd parts are sketched in Figure 81.24. (a) Periodic, period = 21r/(4) = 1r/2. (b) Periodic, period = 21r/(1r) = 2. (c) x(t) = [l + cos(4t — 21r/3)]/2. Periodic, period = 21r/(4) = 1r/2. (d) :1:(t) = cos(47rt)/2. Periodic, period = 21r/(41r) = 1/2. (e) z(t) = [sin(47rt)u(t) — sin(47rt)u(——t)]/2. Not periodic. (f) Not periodic. (3) Periodic, period = 7. (b) Not periodic. (c) Periodic, period = 8. (d) z[n] = (l/2)[cos(31rn/4) + cos(1m/4)]. Periodic, period = 8. (e) Periodic, period = 16. (3) Linear, stable. (b) Memoryless, linear, causal, stable. (c) Linear (d) Linear, causal, stable. (e) Time invariant, linear, causal, stable. (f) Linear, stable. (g) Time invariant, linear, causal. l2 1.35. 1.36. 1.37. 1.38. 1.39. We want to find the smallest No such that m(21r/N)N0 = Zn]: or N0 = kN / m, where k is an integer. If No has to be an integer, then N must be a multiple of m/k and m/k must be an integer. This implies that m/k is a divisor of both m and N. Also, if we want the smallest possible No, then m/k should be the GCD of m and N. Therefore, N0 = N/gcd(m, N). (a) If :r[n] is periodic ej“°("+N)T = ej“’°"T, where wo = 27r/To. This implies that 2—7:NT = 21rk => T — f- = a rational number. To E‘N (b) If T/To = p/q then z[n] = e72"(”/4). The fundamental period is q/gcd(p, q) and the fundamental frequency is 271' 21r w w T —q-gcd(p,q) = —Egcd(p,q) = flamed = —:Tgcd(p,q)- (c) p/gcd(p, q) periods of :c(t) are needed. (a) From the definition of ¢zy(t), we have foo :c(t + 'r)y(-r)dr —oo ¢1y(t) = = [00 y(—t + T)x(T)dT = ¢y1("t)- (b) Note from part (a) that ¢n(t) = ¢fl(-—t). This implies that ¢n(t) is even. Therefore, the odd part of ¢n(t) is zero. (‘3) Here: 451g“) = (1)11“ - T) and ¢yy(t) = ¢Iz(t)' (a) We know that 26A(2t) = 6642“). Therefore, . ' 1 Llano 6A(2t) -— 1131310 5613/20)- This implies that l (b) The plots are as shown in Figure $1.38. We have AiEOUA(t)5(t) = Ai-TOuA(O)6(t) = 0. Also, mundane) = gm). l7 We have —00 Therefore, 0, t<0 '.‘6(t—'r)=0 y(t) = 1, t > 0 u(T)6(t — r) = 6(t — T) . undefined for t = 0 1.40. (a) If a system is additive, then 0 = x(t) — a:(t) -—> y(t) - y(t) = 0. Also, if a. system is homogeneous, then 0 = ow) ——» y(t).0 = 0. (b) y(t) = z2(t) is such a. system. t (c) No. For example, consider y(t) = / z(r)d1 with 2(t) = u(t) —u(t— 1). Then :r(t) = 0 —00 for t > 1, but y(t)=1fort>1. 18 ...
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This note was uploaded on 02/06/2012 for the course EE 3610 taught by Professor Liang during the Fall '10 term at LSU.

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Chapter 1 Solution - Chapter 1 Answers 1.1 Converting from...

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