HW4_solution - MAE107 Homework#4 Solution Prof M'Closkey...

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MAE107 Homework #4 Solution Prof. M’Closkey Problem 1 Solution I’m providing the details of the analytical solution since the “flip and drag” graphical approach is straightforward. Since the impulse response is causal and the input is zero prior to t = 0 the convolution reduces to (note the integral limits) y ( t ) = t 0 h ( t τ ) u ( τ ) dτ. We consider three time intervals in order to build the analytical solution. Case 1. “ t ” is confined to the interval I 1 = [0 , 2]. Note that u ( t ) = 1 on this interval and that the argument of h in the convolution also coincides with I 1 . Thus, we need to derive and expression for h on I 1 : h ( t ) = 1 t/ 2 , t I 1 , so y ( t ) = t 0 h ( t τ ) u ( τ ) = t 0 (1 ( t τ ) / 2) = τ 1 2 + 1 4 τ 2 t 0 = t 1 4 t 2 , t I 1 Case 2. “ t ” is confined to the interval I 2 = [2 , 4]. Note that h ( t ) = 2 t/ 2 for t I 2 . y ( t ) = t 0 h ( t τ ) u ( τ ) = 2 0 h ( t τ ) u ( τ ) 1 + t 2 h ( t τ ) u ( τ ) 0 = t t 2 h ( σ ) (set σ = t τ ) = 2 t 2 h ( σ ) + t 2 h ( σ ) = 2 t 2 (1 σ/ 2) h on I 1 + t 2 (2 σ/ 2) h on I 2 = σ 1 4 σ 2 2 t 2 + 2 σ 1 4 σ 2 t 2 = 1 t I 2 1
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Case 3. “ t ” is confined to the interval I 3 = [4 , 6]. y ( t ) = t 0 h ( t τ ) = t t 2 h ( σ ) (same steps as Case 2.) = 4 t 2 h ( σ ) + t 4 h ( σ ) 0 = 4 t 2 (2 σ/ 2) = 9 3 t + 1 4 t 2 t I 3 When t < 0, y ( t ) = 0 and when t > 6, y ( t ) = 0. Summarizing the cases, y ( t ) = 0 t < 0 t 1 4 t 4 t [0 , 2] 1 t [2 , 4] 9 3 t + 1 4 t 2 t [4 , 6] 0 t > 6 The plot of y is shown below: 2 1 0 1 2 3 4 5 6 7 8 0.5 0 0.5 1 1.5 seconds y 2
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Here is the Matlab code for generating this figure: T = 0.01; t1 = [-2:T:0]; y1 = zeros(1,length(t1)); t2 = [0:T:2]; y2 = t2- 0.25*t2.^2; t3 = [2:T:4]; y3 = ones(1,length(t3)); t4 = [4:T:6]; y4 = 9 - 3*t4 + 0.25*t4.^2; t5 = [6:T:8]; y5 = zeros(1,length(t5)); t = [t1 t2 t3 t4 t5]; y = [y1 y2 y3 y4 y5]; figure(1) plot(t,y,’b’,’LineWidth’,2); grid on axis([-2 8 -0.5 1.5]) xlabel(’seconds’) ylabel(’y’) Problem 2 Solution We need to compute χ ( ω ) = −∞ h ( t ) e jωt dt, where h ( t ) = 0 t < 0 . 1 5 t [ 0 . 1 , 0 . 1] 0 t > 0 . 1 . Thus, χ ( ω ) = −∞ h ( t ) e jωt dt = 0 . 1 0 . 1 5 e jωt dt = 5 ( e j 0 . 1 ω e j 0 . 1 ω ) . The code to graph this frequency response function is listed below: clear close all om = logspace(-1,3,500); H = 5*(exp(-j*0.1*om) - exp(j*0.1*om))./(-j*om); 3
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figure(1); loglog(om/2/pi,abs(H),’LineWidth’,2) grid on axis([0.01 100 1e-3 1e1]) xlabel(’Hz’) ylabel(’magnitude’) figure(2) semilogx(om/2/pi,unwrap(angle(H))*180/pi,’LineWidth’,2) grid on axis([0.01 100 -20 200]) The Bode plots are shown in Fig. 1. Note the deep notches start at 5 Hz and are thereafter repeated every 5 Hz. This is due to the fact that sinusoids with frequencies 5 k Hz, k = 1 , 2 , 3 , . . . , have a whole number of periods in the interval [ 0 . 1 , 0 . 1]. Since the impulse response is constant over this interval, the integral has zero net area. For example, using an equivalent representation of the
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