Chapter 2 examples - EEL4657 - Dr. Haniph Latchman Chapter...

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EEL4657 - Dr. Haniph Latchman Chapter 2: Example Problems 1. Given x ( t ) = e - 3 t use the definition of the Laplace Transform to find X ( s ). Solution The Laplace Transform for a signal x ( t ), t 0 - is defined by: L [ x ( t )] = X ( s ) = Z 0 - x ( t ) e - st dt Thus, substituting for x ( t ) and solving the integral will yield X ( s ) 1
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X ( s ) = L [ e - 3 t ] = Z 0 - e - 3 t e - st dt = Z 0 - e - st - 3 t dt = Z 0 - e ( - s - 3) t dt = Z 0 - e - ( s +3) t dt = - 1 s + 3 ± e - ( s +3) t ² ² ² ² 0 - ³ = - 1 s + 3 h e - ( s +3) - e - ( s +3)0 i = - 1 s + 3 [0 - 1] = - 1 s + 3 [ - 1] X ( s ) = 1 s + 3 2
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2. Given the transfer function: G ( s ) = 3 ( s + 4)( s + 6) Find g ( t ) using the Inverse Laplace Transform properties. Solution Use partial fraction decompositon to convert G ( s ), and use the ”cover-up” method to solve for a and b G ( s ) = 3 ( s + 4)( s + 6) = a s + 4 + b s + 6 For a : a = 3 s + 6 ± ± ± s = - 4 = 3 - 4 + 6 = 3 2 For b : b = 3 s + 4 ± ± ± s = - 6 = 3 - 6 + 4 = - 3 2 Substituting for a and b , we get: G ( s ) = 3 2 s + 4 + - 3 2 s + 6
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This note was uploaded on 01/11/2012 for the course EEL 4657l taught by Professor Staff during the Spring '08 term at University of Florida.

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Chapter 2 examples - EEL4657 - Dr. Haniph Latchman Chapter...

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