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problem 6

# problem 6 - Problem Set 6 The following problems will...

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Problem Set 6 The following problems will familiarize you with the basics of the Laplace transform and its application to LTI systems. 1. Problem 9.21 OW2, parts (a), (d) and (j) Solution: -2 Re Im -3 -5/2 2 Re Im -2 2 -4 4 -3 Re Im 0 Part a 9.21 pole-zero plots Part d Part j 9.21 (a) x ( t ) = e - 2 t u ( t ) + e - 3 t u ( t ). Thus, X ( s ) = Z -∞ x ( t ) e st dt = Z 0 ( e - 2 t + e - 3 t ) e st dt = 1 s + 2 + 1 s + 3 = 2 s + 5 s 2 + 5 s + 6 The ROC is Re { s } > - 2. 1

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9.21 (d) x ( t ) = te - 2 | t | . Thus, X ( s ) = Z 0 -∞ te 2 t e - st dt + Z 0 te - 2 t e - st dt = - d ds Z 0 -∞ e 2 t e - st dt - d ds Z 0 e - 2 t e - st dt We know e - 2 t u ( t ) L ←→ 1 s + 2 , Re { s } > - 2 e 2 t u ( - t ) L ←→ - 1 s - 2 , Re { s } < 2 Thus, X ( s ) = - d ds 1 s + 2 + d ds 1 s - 2 = d ds 4 s 2 - 4 = - 8 s ( s 2 - 4) 2 where the ROC is - 2 < Re { s } < 2. 9.21 (j) x ( t ) = δ (3 t ) + u (3 t ) = δ ( t ) / 3 + u ( t ). Thus, X ( s ) = 1 3 + 1 s = s + 3 3 s and where the ROC is Re { s } > 0. 2
2. Problem 9.22 OW2, parts (d) and (g) Solution: 9.22 (d) We are given X ( s ) = s + 2 s 2 + 7 s + 12 with ROC - 4 < Re { s } < - 3. Using s 2 + 7 s + 12 = ( s + 3)( s + 4) partial fraction expansion we get X ( s ) = 2 s + 4 - 1 s + 3 From the given ROC we know that x ( t ) must be a two-sided signal.

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