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Unformatted text preview: (Problem 4.26 a) (a) i. Given signals x ( t ) = te − 2 t u ( t ) and h ( t ) = e − 4 t u ( t ), find the output y ( t ) using the convolution theorem and the inverse Fourier transform. Using the table, we know that X ( jω ) = 1 (2 + jω ) 2 (11) and also that H ( jω ) = 1 4 + jω (12) Use the convolution theorem. Y ( jω ) = H ( jω ) X ( jω ) = 1 (2 + jω ) 2 (4 + jω ) (13) To find the inverse Fourier transform of Y ( jω ), use a partial fraction expansion. 1 (2 + jω ) 2 (4 + jω ) = A (2 + jω ) 2 + B 2 + jω + C 4 + jω (14) Then, we get that 1 = A (4 + jω ) + B (2 + jω )(4 + jω ) + C (2 + jω ) 2 (15) Substitute ω = j 4 which then leaves the following 1 = C (2 + j ( j 4)) 2 = C ( − 2) 2 = 4 C (16) So C = 1 4 . Substitute ω = j 2 which then leaves the following 1 = A (4 + j ( j 2)) = A (2) = 2 A (17) So A = 1 2 . Then, substitute ω = 0. So, we get 1 = 4 A + 8 B + 4 C (18) Solving this, B = − 1 4 ....
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This note was uploaded on 06/08/2009 for the course EE 301 taught by Professor Enright during the Fall '08 term at USC.
 Fall '08
 Enright

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