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Solution5

# Solution5 - EE 200 – Fall 2008(Weber Homework 5 Solutions...

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Unformatted text preview: EE 200 – Fall 2008 (Weber) Homework 5 Solutions 1. a. h(n) ... ... n b. y(n) = ∞ k=−∞ h(k)x(n − k) 7 = ∞ k=−∞ m=0 7 δ(k − m) cos δ(k − m) cos π (n − k) 4 π (n − k) 4 = 7 ∞ m=0 k=−∞ = m=0 cos 0 π (n − m) 4 = c. H(ω) = ∞ m=−∞ h(m)e−iωm 7 = 7 ∞ m=−∞ k=0 δ(m − k)e−iωm δ(m − k)e−iωm = 7 ∞ k=0 m=−∞ = k=0 e−iωk 1 − e−iω8 1 − e−iω = d. 1−1 1 − e−i 4 8 √ √ =0 = H(π/4) = π 1 − e−i 4 1 − ( 22 − i 22 ) π 1 2. a. h(n) 2 ... 1 ... n b. Use convolution to ﬁnd the relationship between the input and output. y(n) = = ∞ k=−∞ h(k)x(n − k) x(n) + 2x(n − 1) h(n) Setting x(n) = u(n) gives 0 1 y(n) = u(n) + 2u(n − 1) = 3 n<0 n=0 n≥1 3 ... 1 ... n c. Using the method in part b, the ramp generates the output h(n) 7 4 y(n) = r(n) + 2r(n − 1) = d. The frequency response. 0 3n − 2 n<0 n≥1 ... 1 ... n H(ω) = ∞ k=−∞ −iω0 h(k)e−iωk =e + 2e−iω1 = 1 + 2e−iω e. The response is periodic. H(ω + 2π) = = = = f. The response is conjugate symmetric. H(−ω) = 1 + 2eiω 2 1 + 2e−i(ω+2π) 1 + 2e−iω e−i2pi 1 + 2e−iω 1 H(ω) = (1 + 2eiω )∗ = H(ω)∗ g. The magnitude of the frequency response. |H(ω)| = = 1 + 2e−iω (1 + 2 cos(ω))2 + (2 sin(ω))2 1 + 4 cos(ω) + 4 cos2 (ω) + 4 sin2 (ω) 5 + 4 cos(ω) = |1 + 2 cos(ω) − 2i sin(ω)| = = h. The phase of the frequency response H(ω) = = = i. (1 + 2e−iω ) (1 + 2 cos(ω) − 2i sin(ω)) −2 sin(ω) tan−1 1 + 2 cos(ω) y(n) = |H(π/2)| cos(πn/2 + π/6 + H(π/2)) + |H(π)| sin(πn + π/3 + H(π)) |H(π/2)| = H(π2) = |H(π)| = H(π) = √ √ 5 + 4 cos(π/2) = 5 + 4 · 0 = 5 −2 sin(π/2) −2 tan−1 = tan−1 ≈ −1.107 1 + 2 cos(π2) 1 √ √ 5 + 4 cos(π) = 5 + 4 · −1 = 1 = 1 −2 sin(π) 0 tan−1 = tan−1 = −π or π 1 + 2 cos(π) −1 y(n) = = √ 5 cos(πn/2 + π/6 − 1.107) + sin(πn + π/3 + π) √ 5 cos(πn/2 + π/6 − 1.107) − sin(πn + π/3) 3. a. False. The output frequency may not be the same as the input frequency. b. False. This only tells about what happens for that one frequency. c. True. Y(n) is impulse response and the DFT of this is the frequency response. d. True. If x(n) = u(n) then x(n) − x(n − 1) is an impulse input so the output y(n) − y(n − 1) is the impulse response. The impulse response can be determined from this. e. True. The output produced by the x′ (n) is not simply a time-shifted version of the output produced by x(n). f. False. Nothing in the example of x(n) and y(n) says the system in not LTI. An impulse input is produced by x(n) − x(n − 2) + x(n − 4) − x(n − 6) + · · ·. This would results in an impulse response of h(n) = y(n)− y(n− 2)+ y(n− 4)− y(n− 6)+ · · · = δ(n)+ δ(n− 1)+ δ(n− 2)− δ(n− 4)+ δ(n− 6)− · · ·. 3 4. a. h(n) ... ... n b. x(n) = 1 ∞ k=−∞ y(n) = = = c. H(ω) = = = h(k)x(n − k) x(n) − x(n − 1) 0 ∞ m=−∞ −iω0 h(m)e−iωm − e−iω1 e 1 − e−iω d. See the accompanying plot of the frequency response. The response at a frequency of zero has zero magnitude which is consistent with the result in part b. 2 1.8 1.6 Magnitude of frequency response 1.4 1.2 1 0.8 0.6 0.4 0.2 0 -4 -3 -2 -1 0 1 2 3 4 Frequency 4 5. h(t) = H(ω) = = 0 0 ∞ A Tt A −T t 0≤t≤T + 2A T ≤ t ≤ 2T otherwise h(t)e−iωt dt A −iωt te dt + T T 2T T −∞ T A (− t + 2A)e−iωt dt T 2T 2T = = = = = = = A T A T A T A T A T 0 A te−iωt dt − T te−iωt dt + 2A T t=T t=0 T e−iωt dt t=2T i −iωt i t− e ω ω − i i −iωt t− e ω ω + 2T t=T i −iωt e ω t=2T t=T i i i i i −iω2T i i i i −iωT − − − e−iω2T 2T − + e−iωT T − + 2T T− e − e−iωT e ω ω ω ω ω ω ω ω ω iT −iωT 1 i2T −iω2T iT −iωT i2T −iω2T 1 1 1 i2T −iωT e + 2 e−iωT − 2 − e − 2 e−iω2T + e + 2 e−iωT + e − e ω ω ω ω ω ω ω ω ω 1 −iωT 1 1 1 e − 2 − 2 e−iω2T + 2 e−iωT ω2 ω ω ω A1 1 1 1 e−iωT − 2 e−iωT − 2 e−iωT + 2 T ω2 ω ω ω 2A (1 − cos ωT ) e−iωT T ω2 ﬁrst deﬁning h′ (t) = h(t + T ). This centers the response on the t axis between −T ≤ t ≤ 0 0≤t≤T otherwise This can also be done by −T and T . A T (T + t) A h′ (t) = (T − t) T 0 H ′ (ω) = = −T ∞ −∞ 0 h′ (t)e−iωt dt T 0 0 −T A (T + t)e−iωt dt + T e−iωt dt + A T A T A (T − t)e−iωt dt T T T = = = = = = A −T te−iωt dt − te−iωt dt 0 t=0 t=−T i A e−iωt ω t=T + t=−T i −iωt i t− e ω w − i i −iωt t− e ω w t=T t=0 i −iωT i i i i i i Ai A − − eiωT −T − − e−iωT T − + e − eiωT + ω Tω w ω w ω w ω A1 iT iωT iT −iωT 1 1 iωT 1 −iωT i −iωT e − eiωT + + − +2 e − 2e e − 2e A ω T ω2 ω ω ω ω ω A1 1 iωT 1 −iωT 1 − 2e − 2e +2 T ω2 ω ω ω 2A (1 − cos ωT ) T ω2 − i w 5 Since h(t) is h′ (t) delayed by T , the frequency response of h(t) is H ′ (ω) times the response of the delay by an amount T . 2A H(ω) = e−iωT (1 − cos ωT ) T ω2 6 ...
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