ECE 160Solution09. Modern Digital and Analog Communication Systems

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2.1-1 Both ϕ(t) and w0(t) are periodic. The average power of ϕ(t) is Pg = T1 R0T ϕ2(t) dt = π1 R0π e-t/2
2 dt = 1-πe-π . The average power of w0(t) is Pg = T10 RoT0 wo2(t) dt = T10 R0T0 1 · ... dt = 1. 2.1-2 (a) Since x(t) is a real signal, Ex = R02 x2(t) dt. Solving for Fig. S2.1-2(a), we have Ex = R02(1)2 dt = 2, Ey = R01(1)2 dt + R12(-1)2 dt = 2 Ex+y = R01(2)2 dt = 4, Ex-y = R12(2)2 dt = 4 Therefore, Ex±y = Ex + Ey. Solving for Fig. S2.1-2(b), we have Ex = R0π(1)2 dt + Rπ2π(-1)2 dt = 2π, | Ey = R0π/2 | (1)2 dt + Rπ/ π 2(-1)2 dt + Rπ3π/2(1)2 dt + R32π/ π 2(-1)2 dt = 2π Ex+y = R0π/2(2)2 dt + Rπ/ 3π/ 2 2(0)2 dt + R32π/ π 2(-2)2 dt = 4π Ex-y = R0π/2(0)2 dt + Rπ/ π 2(2)2 dt + Rπ/ 3π/ 2 2(-2)2 dt + R32π/ π 2(0)2 dt = 4π Therefore, Ex±y = Ex + Ey. 0 2 2 2 0 (a) -2 (b) 2 0 -2 (c) 0 2 2 2 0 (a) -2 (b) 2 0 -2 (c) π 2 π34 π 4 x(t y ) + (t) π 2 π34 π 4 π π x(t y ) - (t) Fig. S2.1-2 (b) Ex = R0π/4(1)2 dt + Rπ/ π 4(-1)2 dt = π, Ey = R0π(1)2 dt = π 1 Ex+y = R0π/4(2)2 dt+Rπ/ π 4(0)2 dt = π, Ex-y = R0π/4(0)2 dt+Rπ/ π 4(-2)2 dt = 3π Therefore, Ex±y 6= Ex +Ey, and Exˆ±yˆ = Exˆ ± Eyˆ are not true in general. 2.1-3 Pg = | 0 ZC2 cos2 (ω0t + θ) dt = 2T0 Z[1 + cos (2ω0t + 2θ)] dt 0 | 0 T0 | C2T0 1 T= C2 2T0 "Z0T0 dt + Z0T0 cos (2ω0t + 2θ) dt# = 2CT20 [T0 + 0] = C22 2.1-4 If ω1 = ω2, then g2(t) = (C1 cos (ω1t + θ1) + C2 cos (ω1t + θ2))2 = C12 cos2(ω1t + θ1) + C22 cos2(ω1t + θ2) + 2C1C2 cos (ω1t + θ1) cos (ω1t + θ2) Pg = limT0→∞ 1 T0 Z0T0 (C1 cos (ω1t + θ1) + C2 cos (ω1t + θ2))2 dt = C12 2 + C22 2 + lim T →∞ 2C1C2 1 T0 Z0T0 cos (ω1t + θ1) cos (ω1t + θ2) dt = C12 2 + C22 2 + lim T →∞ 2C1C2 1 T0 Z0T0 1 2 cos (2ω1t + θ1 + θ2) + cos (θ1 - θ2) dt = C12 2 + C22 2 + 0 + 2C1C2 2 cos (θ1 - θ2) = C12 + C12 + 2C1C2 cos (θ1 - θ2) 2 2.1-5 Pg = 14 Z-22(t3)2 dt = 64/7 (a) P-g = 1 4 Z-22(-t3)2 dt = 64/7 (b) P2g = 1 4 Z-22(2t3)2 dt = 4(64/7) = 256/7 (c) Pcg = 1 4 Z-22(ct3)2 dt = 64c2/7 Changing the sign of a signal does not affect its power. Multiplication of a signal by a constant c increases the power by a factor of c2. 2.1-6 Let us denote the signal in question by g(t) and its energy by Eg. (a),(b) For parts (a) and (b), we write E g = Z02π sin2 t dt = | 0 | 2π | dt - | 0 | 2π | cos 2t dt = π + 0 = π1 2 Z12 Z2 (c) E g = Z2π4π sin2 t dt = 1 2 Z2π4π dt - 12 Z2π4π cos 2t dt = π + 0 = π (d) E g = Z02π (2 sin t)2 dt = 4 1 2 Z02π dt - 1 2 Z02π cos 2t dt = 4[π + 0] = 4π Sign change and time shift do not affect the signal energy. Doubling the signal quadruples its energy. In the same way, we can show that the energy of kg(t) is k2Eg. 2.1-7 Pg = lim T→∞ 1 T Z-T/ T/22 g(t)g∗(t) dt = lim T→∞ 1 T Z-T/ T/22 kX=nm rX=nm DkD∗rej(ωk-ωr)t dt = lim T→∞ 1 T Z-T/ T/22 kX=nm r=Xm,r n 6=k DkD∗rej(ωk-ωr)t dt + lim T→∞ T1 Z-T/ T/22 kX=nm |Dk|2 dt The integrals of the cross-product terms (when k 6= r) are finite because the integrands (functions to be integrated) are periodic signals (made up of sinusoids). These terms, when divided by T → ∞, yield zero. The remaining terms (k = r) yield Pg = lim | -T/ T/22 | Xn | |Dk|2 dt = | Xn|Dk| T→∞ k=m | k=m 1 T Z2 2.1-8 (a) From Eq. (2.5a), the power of a signal of amplitude C is Pg = C22 , regardless of phase and frequency; therefore, Pg = 100/2 = 50; the rms value is pPg = 5√2. (b) From Eq. (2.5b), the power of the sum of two sinusoids of different frequencies is the sum of the power of individual sinusoids, regardless of the phase, C212 + C222 , therefore, Pg = 100/2 + 256/2 = 50 + 128 = 178; the rms value is pPg = √178. (c) g(t) = (10 + 2 sin (3t)) cos (10t)=10 cos (10t) + 2 sin (3t) cos (10t) = 10 cos (10t) + sin (13t) - cos (7t) Therefore, Pg = 100/2 + 1/2 + 1/2 = 50 + 0.5 + 0.5 = 51; the rms value is pPg = √51. (d) g(t) = 10 cos (5t) cos (10t)=10(cos (15t2)+cos (5t)) = 5 cos (15t) + 5 cos (5t) Therefore, Pg = 25/2 + 25/2 = 25; the rms value is pPg = 5. (e) g(t) = 10 sin (5t) cos (10t)=5 (cos (15t) - cos (5t)) = 5 cos (15t) - 5 cos (5t) Therefore, Pg = 25/2 + 25/2 = 25; the rms value is pPg = 5. (f) |g(t)|2 = cos2(ω0t) Therefore, Pg = 1/2 = 0.5; the rms value is pPg = √0.5 3 2.1-9 (a) Power Pg = 14 R04 1 · dt = 1, and the rms value is pPg = √1 = 1 (b) Power Pg = 101π hR0π 1 · dt + Rπ9π 0 · dt + R910 π π 1 · dti = 101π [π + 0 + π] = 15; and the rms value is p1/5. (c) Power Pg = 1 T Z0T g2(t) dt = 1 6 Z01 g2(t) dt + Z12 g2(t) dt + Z24 g2(t) dt + Z45 g2(t) dt + Z56 g2(t) dt = 1 6 1 + Z12 (-t + 2)2 dt + 0 + Z45 (t - 4)2 dt + 1 = 1 6 1 + 1 3 + 0 + 1 3 + 1 = 4 9 and the rms value is p4/9 = 2/3 2.2-1 If a is complex with real part 0, a = iα; then, g(t) = e-iαt and |g(t)|2 = 1 Pg = limT →∞ R-T/ T/ 22 1 · dt = lim | T = 1. Hence it is a power signal. It is not an energy signal sinceT1 T →∞ T1 E g = R-∞ ∞ |g(t)|2 · dt = ∞. If a is real, then both Eg = R-∞ |e | -αt|2 · dt = ∞ and Pg = ∞.∞ 2.2-2 Let c = a + jb, where a, b are real valued. Therefore, |e-ct| = |e-(a+jb)t| = |e-at · e-jbt| = |e-at| · |e-jbt| = |e-at| · 1 = |e-at| E g = Z-∞ |e | -ct|2 · dt = Ze-2at · dt = ∞∞ -∞ ∞ Pg = lim T →∞ 1 T Z-T/ T/22 |e-ct|2 · dt = lim T →∞ T1 Z-T/ T/22 e-2at · dt = ∞ Therefore, e-ct is neither energy nor a power signal for a complex value of c with nonzero real part. 2.3-1 g2(t) = g(t - 1) + g1(t - 1), g3(t) = g(t - 1) + g1(t + 1), g4(t) = g(t - 0.5) + g1(t + 0.5) The signal g5(t) can be obtained by (i) delaying g(t) by 1 second (replace t with t - 1), (ii) then time-expanding by a factor 2 (replace t with t/2), (iii) then multiplying by 1.5. Thus g5(t) = 1.5g(2t - 1). [Show More]

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