University of California, San Diego Spring 2014 ECE 45 Homework 2 Solutions 2.1: Suppose we model a system by the differential equation y(t) + 2 d x(t) d2 y(t) = + x(t) dt dt2 1. What is the transfer function, H(ω) Represent x(t) and y(t) as phasors thus: Y + 2jω X = (jω)2 Y + X √ 1 − 2jω 1 + 4ω 2 −j tan−1 (2ω) Y = = e H(ω) = X 1 + ω2 1 + ω2 2. What is y0 (t) when x0 (t) = 4 y0 (t) = |H(0)|x0 (t) = 4 3. What is y1 (t) when x1 (t) = cos(t/2) y1 (t) = |H(1/2)| cos(t/2 + ∠H(1/2)) = 4√ 2 cos(t/2 − π/4) 5 4. What is y2 (t) when x2 (t) = 4 + cos(t/2) x2 (t) = x0 (t) + x1 (t) so y2 (t) = y1 (t) + y0 (t) 2.2: Are the following steady-state input-output pairs consistent with the properties of LTI systems? 1. cos(25t) → H(ω) → 99.5 sin(25t − p π) Yes. The output is an amplitude-scaled and phase-shifted version of the input. 2. 2 cos(4t) → H(ω) → 1 + 4 cos(4t) No. There is a new frequency component, ω = 0 in the output. 3. 4 → H(ω) → −8 Yes. The output is an amplitude-scaled and phase-shifted version of the input. 4. 4 → H(ω) → 8j No, we cannot have imaginary values as output signals. 5. 4 → H(ω) → cos(3t) No. There is a new frequency component, ω = 3 in the output. 6. sin(πt) → H(ω) → cos(πt) + 0.1 sin(πt) Yes. The output is an amplitude-scaled and phase-shifted version of the input. 7. sin(πt) → H(ω) → cos(πt) + 0.1 sin(2πt) No. There is a new frequency component, ω = 2π in the output. 8. sin(πt) → H(ω) → sin2 (πt) sin2 (πt) = 21 (1 − cos(2t)) No. There are two new frequency components, ω = 0 and ω = 2 2.3: Determine the Bode Plot (magnitude and phase) of the transfer function H(ω) = 225j (ω 2 − 2500jω − 106 ) (1 + 10−6 jω) (ω − 2000j) (ω − 500j) (1 + 10−6 jω) = 225j (ω 2 − 60000jω − 9 108 ) (25 + 5jω) (ω − 30000j)2 (25 + 5jω) H(ω) = 225j ω ω ) (1 + j 500 ) (1 + 10−6 jω) (−2000j)(−500j) (1 + j 2000 ω (25)(−30000j)2 (1 + j 30000 )2 (1 + j ω5 ) H(ω) = ω ω ) (1 + j 500 ) (1 + j 10ω6 ) j (1 + j 2000 ω )2 (1 + j ω5 ) 100 (1 + j 30000 Critical Points [ω, |H(ω)|] a = [5, −40dB] b = [500, −40 − 20 (log(500) − log(5))] = [500, −40 − 20 (2)] = [500, −80 dB] c = [30000, −80 + 20 (log(30000) − log(2000))] = [30000, −80 + 20 log(15)] ≈ [30000, −56 dB] d = [106 , −56 − 20 (log(106 ) − log(30000))] = [106 , −56 − 20 log(100/3)] ≈ [106 , −86 dB] [ω, ∠H(ω)] e = [0.5, π/2] f = [50, 0] g = [200, π/4 (log(200) − log(50))] = [200, π/4 log(4)] ≈ [200, 0.15π] h = [3000, 0.15π + π/2 (log(3000) − log(200))] = [3000, 0.15π + π/2 log(15)] ≈ [3000, 3π/4] i = [20000, 3π/4 − π/4 (log(20000) − log(5000))] = [20000, 3π/4 − π/4 log(4)] ≈ [20000, 0.6π] j = [105 , 0.6π − π/2 (log(105 ) − log(20000))] = [105 , 0.6π − π/2 log(5)] ≈ [105 , π/4] k = [2 105 , π/4 − π/4(log(2 105 ) − log(105 ))] = [2 105 , π/4 − π/4 log(2)] ≈ [2 105 , π/6] l = [107 , π/2] 2.4: Using the Bode Plot below, estimate the output for the given inputs 1. x(t) = 2 cos(4000t) |HdB (4000)| ≈ −50dB = 20 log |H(4000)| √ 10 1000 33.75o = 3π 16 |H(4000)| ≈ 10−50/20 = ∠H(4000) ≈ 45o √ Thus y(t) = 10 500 3 4 = cos(4000t + 3π ) 16 2. x(t) = sin(7 104 t + π/4) |HdB (7 104 )| ≈ −20dB = 20 log |H(7 104 )| |H(7 104 )| ≈ 10−20/20 = 0.1 ∠H(7 104 ) ≈ −45o 1 2 = −22.5o = −π 8 Thus y(t) = 0.1 sin(7 104 t + π/8) 3. x(t) = cos2 (12500t) = 1/2 + 1/2 cos(25000 106 t) |HdB (2.5 104 )| ≈ −30dB = 20 log |H(2.5 104 )| |H(2.5 104 )| ≈ 10−30/20 = ∠H(2.5 104 ) ≈ −45o = √ 10 100 −π 4 |HdB (0)| ≈ −40dB = 20 log |H(0)| |H(0)| ≈ 10−40/20 = Thus y(t) = 1 200 √ + 10 200 1 100 cos(25000 106 t − π/4) 2.5: Determine the Bode Plot of the transfer function √ ( 2 (50 + j50) ω 2 ) 100 ejπ/4 ω 2 H(ω) = = ω ω ω ω (1 + j 500 ) (1 + j 1000 ) (1 + j 500 ) (1 + j 1000 ) Critical Points: [ω, |H(ω)|] a = [10−1 , 0 dB] b = [500, 40 (log(500) − log(10−1 ))] = [500, 40 log(5000)] = [500, 40 log(103 ) + 40 log(5)] = [500, 120 + 40 log(5)] ≈ [500, 148 dB] c = [1000, 148 + 20 (log(1000) − log(500))] = [1000, 148 + 20 log(2)] ≈ [1000, 154 dB] [ω, ∠H(ω)] d = [50, π/4] e = [100, π/4 − π/4 (log(100) − log(50))] = [100, π/4 − π/4 log(2)] ≈ [100, π/6] f = [5000, 0.175π − π/2 (log(5000) − log(100))] = [5000, π/6 − π log(50)] ≈ [5000, −2π/3] g = [104 , −3π/4] 2.6: Determine the output of the system H(ω) = when the input is x(t) = P∞ n=−∞ 1 |ω| ≤ 4π 0 else cos(π n t)+sin(2π n t+nπ) 2 |H(ω)| = 1 |ω| ≤ 4π and ∠H(ω) = 0 0 else ∞ X |H(π n)| cos(π n t + ∠H(π n)) + |H(2π n)| sin(2π n t + nπ + ∠H(2π n)) y(t) = 2 n=−∞ −→ H(πn) = 1 for n = ±4, ±3 ± 2, ±1, 0 and H(2πn) = 1 for n = ±2, ±1, 0 both are 0 for all other n. cos(πnt) = cos(−πnt) so y(t) = 4 2 X X cos(π n t) sin(2π n t + nπ) + 2 2 n=−4 n=−2 cos(πnt) 2 + cos(−πnt) 2 = cos(πnt), and cos(0) = 1. Thus 4 X cos(π n t) 1 = cos(πt) + cos(2πt) + cos(3πt) + cos(4πt) + 2 2 n=−4 sin(πnt + π/n) = − sin(−(πnt + π/n)) so sin(πnt + π/n) + sin(−πnt − π/n) = 0, and sin(0) = 0 Thus 2 X sin(2π n t + nπ) =0 2 n=−2 so y(t) == cos(πt) + cos(2πt) + cos(3πt) + cos(4πt) + 1 2 2.7: Determine the period and fundamental frequency of the following functions: 1. f (t) = 4 + cos(3t) T0 = 2π/3 s and ω0 = 3rad/s 2. f (t) = 8 + 4 e−j4 t + 4ej4 t = 8 + 8 cos(4t) T0 = π/2 s and ω0 = 4rad/s 3. f (t) = 2 e−j2 t + 2 ej2 t + 2 cos(4t) = 4 cos(2t) + 2 cos(4t) T0 = π s and ω0 = 2rad/s 2.8: Determine the Fourier coefficients Fn and the average power of the function: 1 0<t<2 f (t) = 2 2<t<4 Where f (t) is periodic with a period of T = 4 T = 4 → ω0 = π/2 1 Fn = 4 = Z 4 jn π2 t f (t) e 0 1 dt = 4 2 Z jn π2 t e 0 1 dt + 4 Z 4 π 2 ejn 2 t dt 2 1 1 (e−jnπ − 1) + (e−jn2π − e−jnπ ) −jn2π −jnπ Note: e−jnπ = (−1)n and e−jn2π = 1 for all integers n Fn = 1 1 j ((−1)n − 1) + (1 − (−1)n ) = (1 − (−1)n ) −jn2π −jnπ 2πn 0 n is even and n 6= 0 Fn = j/nπ n is odd F0 is not well-defined, so we must calculate it separately: Z Z Z 1 4 1 2 1 4 F0 = f (t) dt = 1 dt + 2 dt = 3/2 4 0 4 0 4 2 Pavg 1 = 4 Z 0 4 1 |f (t)| dt = 4 2 Z 0 2 1 1 dt + 4 Z 4 4 dt = 5/2 2 2.9: Calculate the Fourier Series of f (t) T0 = 1 → ω0 = 2π, and in a period from −0.5 to 0.5: f (t) = 1 − |2 t| Z 1 2 Fn = −jn2πt f (t) e 0 Z −jn2πt dt = (2t + 1) e Z 0 −jn2πt = 2t e =e (1 − 2t) e−jn2πt dt 0 Z dt − − 12 −jn2πt dt + − 12 − 21 1 2 Z 1 2 −jn2πt 2t e Z 1 2 dt + 1 e−jn2πt dt − 12 0 1/2 0 1/2 jt jt j e−2jπnt 1 1 −jn2πt + −e + + 2π 2 n2 πn −1/2 2π 2 n2 πn 0 2πn −1/2 1 1 j 1 j 1 j jπn −jπn = 2 2 −e ( 2 2 − )− e ( 2 2+ )− 2 2 + (e−jπn − ejπn ) 2π n 2π n 2πn 2π n 2πn 2π n 2πn Note: ejnπ = e−jnπ = (−1)n 1 (−1)n Fn = 2 2 − 2 2 = π n π n 0 2 π 2 n2 n is even and n 6= 0 n is odd Not well defined for F0 so we must calculate it separately: Z 1 2 F0 = Z 0 f (t) dt = − 21 Z (2t + 1) dt + − 21 −→ f (t) = 1/2 + 1 2 (1 − 2t) dt = 1/2 0 X n odd 2 π 2 n2 ejn2πt 2.10: Using the results of 2.9, determine the Fourier Series of g(t). You can express g(t) as a shifted and scaled version of f (t) and use the properties of the Fourier Series. g(t) = 8 (f (t − 1/4) − 1/2)) = 8 f (t − 1/4) − 4 Thus by the scaling and time-shift properties of the Fourier Series: 8F0 − 4 n=0 0 n is even = Gn = 16 −jnπ/2 −jn2π 41 e n is odd n 6= 0 8Fn e π 2 n2 Note: e−jnπ/2 = (−j)n −→ g(t) = X 16 (−j)n ejn2πt 2 n2 π n odd 2.11: Determine the Fourier Series of f (t) = | sin(t)| T0 = π, ω0 = 2. Over one period, f (t) = sin(t) Note: By Euler’s Formula: ejθ − e−jθ sin(θ) = 2j Z Z π 1 1 π −jn2t sin(t) e dt = (ej(1−2n)t − e−j(1+2n)t ) dt Fn = π 0 j2π 0 j(1−2n)t π 1 e e−j(1+2n)t −1 ej(1−2n)π − 1 e−j(1+2n)π − 1 = − = − j2π j(1 − 2n) −j(1 + 2n) 0 2π (1 − 2n) (1 + 2n) Note: e(1±2nπ) = ejπ = −1 1 Fn = π 1 1 + 1 − 2n 1 + 2n −→ f (t) = = 2 1 π 1 − 4n2 ∞ X 2 1 ejn2t 2 π 1 − 4n n=−∞
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