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Answers to selected exercises for chapter 1

1.1

Apply cos(α + β) = cos α cos β − sin α sin β, then f1 (t) + f2 (t) = A1 cos ωt cos φ1 − A1 sin ωt sin φ1 + A2 cos ωt cos φ2 − A2 sin ωt sin φ2 = (A1 cos φ1 + A2 cos φ2 ) cos ωt − (A1 sin φ1 + A2 sin φ2 ) sin ωt = C1 cos ωt − C2 sin ωt, where C1 = A1 cos φ1 + A2 cos φ2 and C2 = A1 sin φ1 + A2 sin φ2 . Put A = p 2 2 C1 + C2 and take φ such thatcos φ = C1 /A and sin φ = C2 /A (this is possible since (C1 /A)2 +(C2 /A)2 = 1). Now f1 (t)+f2 (t) = A(cos ωt cos φ− sin ωt sin φ) = A cos(ωt + φ).

1.2

Put c1 = A1 eiφ1 and c2 = A2 eiφ2 , then f1 (t) + f2 (t) = (c1 + c2 )eiωt . Let c = c1 + c2 , then f1 (t) + f2 (t) = ceiωt . The signal f1 (t) + f2 (t) is again a time-harmonic signal with amplitude | c | and initial phase arg c. The power Pis given by Z π/ω Z ω A2 ω π/ω P = A2 cos2 (ωt + φ0 ) dt = (1 + cos(2ωt + 2φ0 )) dt 2π −π/ω 4π −π/ω A2 . = 2 The energy-content is E = The power P is given by P =
3 1X | cos(nπ/2) |2 = 1 . 2 4 n=0

1.5

1.6 1.7

R∞
0

e−2t dt = 1 . 2

1.8 1.9

The energy-content is E = sum 1/(1 − e−2 ).

P∞

n=0

e−2n , which is a geometric series with

a If u(t) is real, then the integral,and so y(t), is also real. b Since ˛Z ˛ Z ˛ ˛ ˛ u(τ ) dτ ˛ ≤ | u(τ ) | dτ, ˛ ˛ it follows from the boundedness of u(t), so | u(τ ) | ≤ K for some constant K, that y(t) is also bounded. c The linearity follows immediately from the linearity of integration. The time-invariance follows from the substitution ξ = τ − t0 in the integral Rt u(τ − t0 ) dτ representing the response to u(t − t0 ). t−1 Rt dCalculating t−1 cos(ωτ ) dτ gives the following response: (sin(ωt) − sin(ωt − ω))/ω = 2 sin(ω/2) cos(ωt − ω/2)/ω. Rt e Calculating t−1 sin(ωτ ) dτ gives the following response: (− cos(ωt) + cos(ωt − ω))/ω = 2 sin(ω/2) sin(ωt − ω/2)/ω. f From the response to cos(ωt) in d it follows that the amplitude response is | 2 sin(ω/2)/ω |. g From the response to cos(ωt) in d it follows that the phase responseis −ω/2 if 2 sin(ω/2)/ω ≥ 0 and −ω/2 + π if 2 sin(ω/2)/ω < 0. From

1

2

Answers to selected exercises for chapter 1

phase and amplitude response the frequency response follows: H(ω) = 2 sin(ω/2)e−iω/2 /ω. 1.11 a The frequency response of the cascade system is H1 (ω)H2 (ω), since the reponse to eiωt is ﬁrst H1 (ω)eiωt and then H1 (ω)H2 (ω)eiωt . b The amplitude response is | H1 (ω)H2(ω) | = A1 (ω)A2 (ω). c The phase response is arg(H1 (ω)H2 (ω)) = Φ1 (ω) + Φ2 (ω). ˛ ˛ √ a The amplitude response is | 1 + i | ˛ e−2iω ˛ = 2. b The input u[n] = 1 has frequency ω = 0, initial phase 0 and amplitude 1. Since eiωn → H(eiω )eiωn , the response is H(e0 )1 = 1 + i for all n. c Since u[n] = (eiωn + e−iωn )/2 we can use eiωn → H(eiω )eiωn to obtain that y[n] = (H(eiω )eiωn + H(e−iω )e−iωn)/2, so y[n] = (1 + i) cos(ω(n − 2)). d Since u[n] = (1 + cos 4ωn)/2, we can use the same method as in b and c to obtain y[n] = (1 + i)(1 + cos(4ω(n − 2)))/2. a The power is the integral of f 2 (t) over [−π/ | ω | , π/ | ω |], times | ω | /2π. Now cos2 (ωt + φ0 ) integrated over [−π/ | ω | , π/ | ω |] equals π/ | ω | and cos(ωt) cos(ωt + φ0 ) integrated over [−π/ | ω | , π/ | ω |] is (π/ | ω |) cosφ0 . Hence, the power equals (A2 + 2AB cos(φ0 ) + B 2 )/2. R1 b The energy-content is 0 sin2 (πt) dt = 1/2. The power is the integral of | f (t) |2 over [−π/ | ω | , π/ | ω |], times | ω | /2π, which in this case equals | c |2 . a The amplitude response is | H(ω) | = 1/(1 + ω 2 ). The phase response is arg H(ω) = ω. b The input has frequency ω = 1, so it follows from eiωt → H(ω)eiωt that theresponse is H(1)ieit = iei(t+1) /2. a The signal is not periodic since sin(2N ) = 0 for all integer N . iω b The frequency response H(eiω ) equals A(eiω )eiΦe , hence, we obtain iω iω 2 that H(e ) = e /(1 + ω ). The response to u[n] = (e2in − e−2in )/2i is then y[n] = (e2i(n+1) − e−2i(n+1) )/(10i), so y[n] = (sin(2n + 2))/5. The amplitude is thus 1/5 and the initial phase 2 − π/2. a If u(t) = 0 for...