Quiz2 sol - z t we perform partial-fraction expansion Z s = 1 s 2 2 s 2 = A s B s 2 C s 1 =-1 s 1 s 2 1 s 1 Thus z t =-1 t e-t u t 2 We have H s =

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ECEN 314 Quiz # 2 Georghiades November 3, 2009 Note: Quiz is open-book, 15 minutes long. 1. Let x ( t ) = tu ( t ) and y ( t ) = e - t u ( t ) . Compute the Laplace transform of the following function: z ( t ) = x ( t ) * y ( t ) . 2. Compute the inverse Laplace transform, h ( t ), of H ( s ) = 1 s 2 + 2 s + 2 . Solution 1. Convolution in the time-domain is multiplication in the frequency-domain. Thus, Z ( s ) = X ( s ) Y ( s ) . We have X ( s ) = - d ds 1 s = 1 s 2 , and Y ( s ) = 1 s + 1 . Thus, Z ( s ) = 1 s 2 ( s + 1) . To compute
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Unformatted text preview: z ( t ), we perform partial-fraction expansion: Z ( s ) = 1 s 2 + 2 s + 2 = A s + B s 2 + C s + 1 =-1 s + 1 s 2 + 1 s + 1 . Thus, z ( t ) = (-1 + t + e-t ) u ( t ) . 2. We have H ( s ) = 1 s 2 + 2 s + 2 = 1 ( s + 1) 2 + 1 , and therefore from the frequency translation property h ( t ) = e-t sin( t ) u ( t ) . 1...
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This note was uploaded on 01/01/2012 for the course ECEN 314 taught by Professor Halverson during the Spring '08 term at Texas A&M.

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