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

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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 ) = 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 ] = 0 - e - 3 t e - st dt = 0 - e - st - 3 t dt = 0 - e ( - s - 3) t dt = 0 - e - ( s +3) t dt = - 1 s + 3 e - ( s +3) t 0 - = - 1 s + 3 e - ( s +3) - e - ( s +3)0 = - 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 We can apply the Inverse Laplace transform to each of the partial fraction of G ( s ): L - 1 [ G ( s )] = L - 1 3 2 s + 4 + L - 1 - 3 2 s + 6 g ( t ) = 3 2 e - 4 t u ( t ) - 3 2 e - 6 t u ( t ) 3
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3. Given the transfer function of the system: V ( s ) = 4 s + 16 s ( s 2 + 4 s + 8)
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