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Unformatted text preview: MAE107 Homework #4 Solution Prof. M’Closkey Problem 1 Solution 1. The Fourier series coefficients are simply, c k = Z T u ( t ) e jkω t dt = Z 1 u ( t ) e jkω t dt = Z . 5 1 · e jkω t dt ( ω = 2 π ) = 1 jk 2 π 1 e jkπ , k 6 = 0 , This expression is valid for k 6 = 0. When k = 0, c = Z T u ( t ) dt = 0 . 5 . 2. The approximation of u given by the truncated Fourier series: u ( t ) ≈ 1 T 10 X k = 10 c k e jkω t , is shown in Fig. 1 The Matlab code is: t = [0.2:.001:1.2]; N = length(t); u = 0.5*ones(1,N); %% constant term for k = 1:10 c = 1/(j*k*2*pi)*(1exp(pi*j*k)); u = u + c*exp(j*k*2*pi*t) + conj(c*exp(j*k*2*pi*t)); end plot(t,u,’LineWidth’,2) grid on axis([0.2 1.2 0.2 1.2]) 1 ï 0.2 0.2 0.4 0.6 0.8 1 1.2 ï 0.2 0.2 0.4 0.6 0.8 1 Figure 1: Approximation of u using the first ten frequencies in the Fourier series. Problem 2 Solution 1. Timedomain approach. The impulse response of ˙ x + 3 x = 3 u. (1) is h ( t ) = 3 e 3 t μ ( t ) , so the solution to an initial value problem is x ( t ) = e 3 t x (0) + Z t 3 e 3( t τ ) u ( τ ) dτ, t ≥ . It is cumbersome to compute the “limiting” periodic solution of x from an arbitrary initial condition so we instead employ a “trick”: let’s find the initial condition that coincides with the limiting periodic solution. In other words, given the periodic input u , find x (0) such that x ( T ) = x (0), where T is the period of u . For the present problem T = 1 so we need to 2 compute α = x (0) = x (1) from α = e 3 α + Z 1 h (1 τ ) u ( τ ) dτ = e 3 α + Z . 5 3 e 3( t τ ) dτ = e 3 α + e 3 ( e 3 / 2 1) = ⇒ α = e 3 e 3 / 2 1 1 e 3 ! Now we can compute the periodic solution using α . For t ∈ [0 , . 5], x ( t ) = e 3 t α + Z t 3 e 3( t τ ) dτ = e 3 t α + 1 e 3 t , t ∈ [0 , . 5] . For t ∈ [0 . 5 , 1], x ( t ) = e 3 t α + Z t 3 e 3( t τ ) u ( τ ) dτ = e 3 t α + Z . 5 3 e 3( t τ ) u ( τ )  {z } 1 dτ + Z t . 5 3 e 3( t τ ) u ( τ )  {z } dτ = e 3 t α + e 3 t e 3 / 2 1 , t ∈ [0 . 5 , 1] Summarizing, one period of the periodic response is x ( t ) = ( e 3 t ( α 1) + 1 t ∈ [0 , . 5] e 3 t ( α + e 3 / 2 1 ) t ∈ [0 . 5 , 1] Fig. 2 shows one period of the analytical solution. 2. Fourier series approach. The periodic output computed using Fourier series is simply x ( t ) = 1 T ∞ X k =∞ H ( jkω ) c k e jkω t = 1 T ∞ X k =∞ 3 jk 2 π + 3 c k e jk 2 πt , where the c k were computed in the previous problem. The approximation using the truncated series, x ( t ) = 1 T 10 X k = 10 3 jk 2 π + 3 c k e jk 2 πt , is compared to the analytical solution in Fig. 2. As more terms are included, the truncated Fourier series converges to the analytical solution....
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This note was uploaded on 02/02/2011 for the course MAE 107 taught by Professor Tsao during the Spring '06 term at UCLA.
 Spring '06
 TSAO

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