springps07_soln

springps07_soln - 18.03 PS7 Solutions-Spring 09 a0 an int...

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18.03 PS7 Solutions—Spring 09 1. a 0 2 + X n =1 a n cos nt + X n =1 b n sin nt = a 0 2 e 0 t + X n =1 a n 2 ( e int + e - int ) + X n =1 b n 2 i ( e int - e - int ) = a 0 2 e 0 t + X n =1 ± a n - ib n 2 e int + a n + ib n 2 e - int ² = X n = -∞ c n e int where c n = ( a n - ib n ) / 2 ,n > 0 a 0 / 2 ,n = 0 ( a n + ib n ) / 2 ,n < 0 . Hence c - n = ¯ c n . We can reverse all the steps above. Working backwards and solving for a n and b n , a 0 = 2 c 0 ; a n = 2 Re( c n ) ( n 1); b n = - 2 Im( c n ) ( n 1) . 2a. See Figures 1-3. 2b. f 0 1 ( t ) = - u ( t ) - ( t )+ δ ( t - 1) - 2 δ ( t +1). In fact, ( t ) is identically zero: for any test function g ( t ), R -∞ g ( t )( ( t )) dt = g (0) · 0 = 0 = R -∞ g ( t )(0) dt . Two generalized functions are equal if and only if their integrals with respect to any test function are equal; thus ( t ) = 0. We obtain 1 f 0 1 ( t ) = - u ( t ) + δ ( t - 1) - 2 δ ( t + 1) . f 2 ( t ) = (2 t - 4)[ u ( t - 2) - u ( t - 3)] + ( t - 1)[ u ( t - 3) - u ( t - 4)] = (2 t - 4) u ( t - 2) + (3 - t ) u ( t - 3) + (1 - t ) u ( t - 4), meaning that f 0 2 ( t ) = 2 u ( t - 2) + (2 t - 4) δ ( t - 2) - u ( t - 3) + (3 - t ) δ ( t - 3) - u ( t - 4) + (1 - t ) δ ( t - 4) = 2 u ( t - 2) - u ( t - 3) - u ( t - 4) + 0 δ ( t - 2) + 0 δ ( t - 3) - 3 δ ( t - 4) = 2 u ( t - 2) - u ( t - 3) - u ( t - 4) - 3 δ ( t - 4) where in the next-to-last line we used the formula h ( t ) δ ( t - a ) = h ( a ) δ ( t - a ). 2 Finally, f 0 3 ( t ) = X n = -∞ δ ( t - n ) . See Figures 4-6 for the graphs. 3a. The weight function for the equation x 0 - (1 / 10) x = q ( t ) is w ( t ) = e t/ 10 u ( t ). Because x (0 - ) = 0 and f 0 3 = X n = -∞ δ ( t - n ), the input is the portion that occurs for times t 0: X n =0 δ ( t - n ) and the output is x ( t ) = X n =0 w ( t - n ) = X n =0 e ( t - n ) / 10 u ( t - n ) = b t c X n =0 e ( t - n ) / 10 1 It’s also acceptable to write
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springps07_soln - 18.03 PS7 Solutions-Spring 09 a0 an int...

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