HW6sol - ECEN 314 Signals and Systems Solutions to HW 6...

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ECEN 314: Signals and Systems Solutions to HW 6 Problem 9.2 (a) X ( s ) = Z -∞ e - 5 t u ( t - 1) e - st dt = Z 1 e - (5+ s ) t dt = e - (5+ s ) s + 5 The ROC is Re { s } > - 5. (b) The laplace transform of g ( t ) works out as follows: G ( s ) = Z -∞ Ae - 5 t u ( - t - t o ) e - st dt = Z - t o -∞ Ae - (5+ s ) t dt In order for the above to be integrable, we need Re { s } < - 5. With this condition, the above integral then becomes: G ( s ) = - A ( s + 5) e ( s +5) t o Comparing G ( s ) with X ( s ) (from part (a)), we see that A = - 1 and t o = - 1. Problem 9.21 (a) e - 2 t u ( t ) L ←→ 1 s + 2 Re { s } > - 2 e - 3 t u ( t ) L ←→ 1 s + 3 Re { s } > - 3 X ( s ) = 2 s + 5 s 2 + 5 s + 6 The ROC is then Re { s } > - 2. We have a zero at s = - 5 / 2 and poles at s = - 2 , - 3. 1

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(b) e - 4 t u ( t ) L ←→ 1 s + 4 Re { s } > - 4 e - 5 t ( e j 5 t ) u ( t ) L ←→ 1 ( s + 5 - 5 j ) Re { s } > - 5 e - 5 t ( e - j 5 t ) u ( t ) L ←→ 1 ( s + 5 + 5 j ) Re { s } > - 5 Hence, X ( s ) = s 2 + 15 s + 70
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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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HW6sol - ECEN 314 Signals and Systems Solutions to HW 6...

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