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2 0 0 1 T 0 π 0 π 0 T0 o o T0 0 x 0 y 0 1 x 0 π 0 π/2 π 3π/2 x+y 0 π/2 3π/2 x−y 0 π/2 π/2 3π/2 x 0 π/4 0 Chapter 2 2.1-1 Both ϕ(t) and w0(t) are periodic. The average power of ϕ(t) is P = 1 ∫ T ϕ2(t) dt = 1 ∫ π e−t/2 2 dt = 1−e−π . The average power of w (t) is P = 1 ∫ T0 w2(t) dt = 1 ∫ T0 1 · dt = 1. 2.1-2 (a) Since x(t) is a real signal, Ex = ∫ 2 x2(t) dt. Solving for Fig. S2.1 -2(a), we have E = ∫ 2(1)2 dt = 2, E = ∫ 1(1)2 dt + ∫ 2(−1)2 dt = 2 Ex+y = ∫ 1(2)2 dt = 4, E x−y = ∫ 2(2)2 dt = 4 Therefore, Ex±y = Ex + Ey. Solving for Fig. S2.1 -2(b), we have E = ∫ π (1)2 dt + ∫ 2π (−1)2 dt = 2π, E = ∫ π/2(1)2 dt + ∫ π (−1)2 dt + ∫ 3π/2(1)2 dt + ∫ 2π (−1)2 dt = 2π E = ∫ π/2(2)2 dt + ∫ 3π/2(0)2 dt + ∫ 2π (−2)2 dt = 4π E = ∫ π/2(0)2 dt + ∫ π (2)2 dt + ∫ 3π/2(−2)2 dt + ∫ 2π (0)2 dt = 4π Therefore, Ex±y = Ex + Ey. 2 2 2 0 0 0 (a) -2 (b) -2 (c) x t y t
2 0 (a) 2 0 -2 (b) 2 0 -2 (c) x t y t
Fig. S2.1 -2 (b) E = ∫ π/4(1)2 dt + ∫ π (−1)2 dt = π, E = ∫ π (1)2 dt = π
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4 g g y y 3 1 1 3 2 ∫ ∫ x+y 0 π/4 0 π/4 x y 2T θ 2T 2 Pg = lim T0→∞ (C1 cos (ω1t + θ1) + C2 cos (ω1t + θ2)) 1 2 2 2 T →∞ C1C2 T0 ω1t θ1 ω1t θ2 1 2 2 2 C1C2 T0 2 ω1t θ1 θ2 θ1 − θ2 2 2 4 t3 2 dt / 4 −2 −t dt / 4 −2 t3 2 dt / / 4 ct3 2 dt c2/ = ∫ T 0 0 0 E = ∫ π/4(2)2 dt +∫ π (0)2 dt = π, E = ∫ π/4(0)2 dt +∫ π (−2)2 dt = 3π Therefore, E E + E , and Exˆ±yˆ = Exˆ ± Eyˆ are not true in general. 2.1-3 1 T0 Pg C2 cos2 (ω0t + θ) dt = C2 ∫ T0 [1 + cos (2 ω0t + 2θ)] dt T0 0 2T0 0 = C2 "∫ T0 + ∫ T0 cos (2 + 2 ) # = C2 [ + 0] = C2 2.1-4 If ω1 = ω2, then g2(t) = (C1 cos (ω1t + θ1) + C2 cos (ω1t + θ2))2 = C2 cos2(ω1t + θ1) + C2 cos2(ω1t + θ2) + 2C1C2 cos (ω1t + θ1) cos (ω1t + θ2) 1 2 1 ∫ T0 2 = C2 + C2 + lim 2 1 ∫ T0 cos ( + ) cos ( + ) = C2 + C2 + lim 2 1 ∫ T0 1 cos (2 + + ) + cos ( ) C2 C2 2C1C2 = 1 + 2 + 0 + cos (θ1 − θ2) = C 2 + C2 + 2C1C2 cos (θ1 − θ2) 2 2.1-5 = 1 ∫ 2 ( ) = 64 7 (a) = 1 ∫ 2 ( ) = 64 7 (b) = 1 ∫ 2 (2 ) = 4(64 7) = 256 7 (c) = 1 ∫ 2 ( ) = 64 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 Eg = 2π sin2 0 1 2π t dt = 0 1 2π dt − 2 cos 2t dt = π + 0 = π −2 −2 0 0 ∫ 2 0 x−y x±y 0 0 dt ω0t dt T0 dt 0 dt T →∞ dt 2 Pg P−g P2g Pcg 4 ∫ ∫ ∫ ∫ n n 2 → ∞ / 2 Pg = 5 2. g (c) Eg = 4π sin2 2π 1 4π t dt = 2π 1 4π dt 2π cos 2t dt = π + 0 = π (d) Eg = 2π (2 sin t)2 0 dt = 4 1 2π 2 0 1 2π dt − 2 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 1 ∫ T/2 g(t)g∗(t) dt T →∞ T −T/2 = lim 1 ∫ T/2 Σ Σ DkD∗rej(ωk−ωr)t dt T →∞ T −T/2 k=m r=m = lim 1 ∫ T/2 Σ Σ DkD∗rej(ωk −ωr )t dt + lim 1 ∫ T/2 Σ |Dk| dt T →∞ T −T/2 k=m r=m,r /=k T →∞ T −T/2 k=m The integrals of the cross -product terms (when k = 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 1 ∫ T/2 Σ |Dk|2 dt = Σ |Dk| T →∞ T −T/2 k=m k=m 2.1-8 2 (a) From Eq. (2.5a), the power of a signal of amplitude C is P = C , regardless of phase and frequency; therefore, √ √ 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, C2 + C2 , therefore, P = 100 /2 + 256 /2 = 50 + 128 = 178; the rms value is √Pg = √
178. 1 2 2 2 g (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 √Pg = √51. (d) g(t) = 10 cos (5t) cos (10t)= 10(cos (15t)+cos (5t)) = 5 cos (15t) + 5 cos (5t) Therefore, Pg = 25/2 + 25/2 = 25; the rms value is √Pg = 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 √Pg = 5. (f) |g(t)|2 = cos2(ω0t) Therefore, Pg = 1/2 = 0.5; the rms value is √Pg = √0.5 ∫ 2 — 2 ∫ 0 n n n n n 2 Pg = 100 /2 = 50; the rms value is
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2 0 0 1 T 0 π 0 π 0 T0 o o T0 0 x 0 y 0 1 x 0 π 0 π/2 π 3π/2 x+y 0 π/2 3π/2 x−y 0 π/2 π/2 3π/2 x 0 π/4 0 Chapter 2 2.1-1 Both ϕ(t) and w0(t) are periodic. The average power of ϕ(t) is P = 1 ∫ T ϕ2(t) dt = 1 ∫ π e−t/2 2 dt = 1−e−π . The average power of w (t) is P = 1 ∫ T0 w2(t) dt = 1 ∫ T0 1 · dt = 1. 2.1-2 (a) Since x(t) is a real signal, Ex = ∫ 2 x2(t) dt. Solving for Fig. S2.1 -2(a), we have E = ∫ 2(1)2 dt = 2, E = ∫ 1(1)2 dt + ∫ 2(−1)2 dt = 2 Ex+y = ∫ 1(2)2 dt = 4, E x−y = ∫ 2(2)2 dt = 4 Therefore, Ex±y = Ex + Ey. Solving for Fig. S2.1 -2(b), we have E = ∫ π (1)2 dt + ∫ 2π (−1)2 dt = 2π, E = ∫ π/2(1)2 dt + ∫ π (−1)2 dt + ∫ 3π/2(1)2 dt + ∫ 2π (−1)2 dt = 2π E = ∫ π/2(2)2 dt + ∫ 3π/2(0)2 dt + ∫ 2π (−2)2 dt = 4π E = ∫ π/2(0)2 dt + ∫ π (2)2 dt + ∫ 3π/2(−2)2 dt + ∫ 2π (0)2 dt = 4π Therefore, Ex±y = Ex + Ey. 2 2 2 0 0 0 (a) -2 (b) -2 (c) x t y t
2 0 (a) 2 0 -2 (b) 2 0 -2 (c) x t y t
Fig. S2.1 -2 (b) E = ∫ π/4(1)2 dt + ∫ π (−1)2 dt = π, E = ∫ π (1)2 dt = π
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4 g g y y 3 1 1 3 2 ∫ ∫ x+y 0 π/4 0 π/4 x y 2T θ 2T 2 Pg = lim T0→∞ (C1 cos (ω1t + θ1) + C2 cos (ω1t + θ2)) 1 2 2 2 T →∞ C1C2 T0 ω1t θ1 ω1t θ2 1 2 2 2 C1C2 T0 2 ω1t θ1 θ2 θ1 − θ2 2 2 4 t3 2 dt / 4 −2 −t dt / 4 −2 t3 2 dt / / 4 ct3 2 dt c2/ = ∫ T 0 0 0 E = ∫ π/4(2)2 dt +∫ π (0)2 dt = π, E = ∫ π/4(0)2 dt +∫ π (−2)2 dt = 3π Therefore, E E + E , and Exˆ±yˆ = Exˆ ± Eyˆ are not true in general. 2.1-3 1 T0 Pg C2 cos2 (ω0t + θ) dt = C2 ∫ T0 [1 + cos (2 ω0t + 2θ)] dt T0 0 2T0 0 = C2 "∫ T0 + ∫ T0 cos (2 + 2 ) # = C2 [ + 0] = C2 2.1-4 If ω1 = ω2, then g2(t) = (C1 cos (ω1t + θ1) + C2 cos (ω1t + θ2))2 = C2 cos2(ω1t + θ1) + C2 cos2(ω1t + θ2) + 2C1C2 cos (ω1t + θ1) cos (ω1t + θ2) 1 2 1 ∫ T0 2 = C2 + C2 + lim 2 1 ∫ T0 cos ( + ) cos ( + ) = C2 + C2 + lim 2 1 ∫ T0 1 cos (2 + + ) + cos ( ) C2 C2 2C1C2 = 1 + 2 + 0 + cos (θ1 − θ2) = C 2 + C2 + 2C1C2 cos (θ1 − θ2) 2 2.1-5 = 1 ∫ 2 ( ) = 64 7 (a) = 1 ∫ 2 ( ) = 64 7 (b) = 1 ∫ 2 (2 ) = 4(64 7) = 256 7 (c) = 1 ∫ 2 ( ) = 64 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 Eg = 2π sin2 0 1 2π t dt = 0 1 2π dt − 2 cos 2t dt = π + 0 = π −2 −2 0 0 ∫ 2 0 x−y x±y 0 0 dt ω0t dt T0 dt 0 dt T →∞ dt 2 Pg P−g P2g Pcg 4 ∫ ∫ ∫ ∫ n n 2 → ∞ / 2 Pg = 5 2. g (c) Eg = 4π sin2 2π 1 4π t dt = 2π 1 4π dt 2π cos 2t dt = π + 0 = π (d) Eg = 2π (2 sin t)2 0 dt = 4 1 2π 2 0 1 2π dt − 2 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 1 ∫ T/2 g(t)g∗(t) dt T →∞ T −T/2 = lim 1 ∫ T/2 Σ Σ DkD∗rej(ωk−ωr)t dt T →∞ T −T/2 k=m r=m = lim 1 ∫ T/2 Σ Σ DkD∗rej(ωk −ωr )t dt + lim 1 ∫ T/2 Σ |Dk| dt T →∞ T −T/2 k=m r=m,r /=k T →∞ T −T/2 k=m The integrals of the cross -product terms (when k = 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 1 ∫ T/2 Σ |Dk|2 dt = Σ |Dk| T →∞ T −T/2 k=m k=m 2.1-8 2 (a) From Eq. (2.5a), the power of a signal of amplitude C is P = C , regardless of phase and frequency; therefore, √ √ 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, C2 + C2 , therefore, P = 100 /2 + 256 /2 = 50 + 128 = 178; the rms value is √Pg = √
178. 1 2 2 2 g (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 √Pg = √51. (d) g(t) = 10 cos (5t) cos (10t)= 10(cos (15t)+cos (5t)) = 5 cos (15t) + 5 cos (5t) Therefore, Pg = 25/2 + 25/2 = 25; the rms value is √Pg = 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 √Pg = 5. (f) |g(t)|2 = cos2(ω0t) Therefore, Pg = 1/2 = 0.5; the rms value is √Pg = √0.5 ∫ 2 — 2 ∫ 0 n n n n n 2 Pg = 100 /2 = 50; the rms value is
