determine L−1[F(s)G(s)] in the following two ways: (a) using the Convolution Theorem, (b) using partial fractions. F(s)=1s,G(s)=1s−2

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determine L−1[F(s)G(s)] in the following two ways:   (a) using the Convolution Theorem, (b) using partial fractions.  F(s)=1s,G(s)=1s−2

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Step 1   Given Data   The first function is F(s)=1s.   The second function is G(s)=1s−2.   (a) Using the convolution theorem, the expression for the inverse Laplace is,  L−1[F(s)G(s)]=L−1[F(s)]⋅L−1(G(s))  =L−1(1s)⋅L−1(1s−2)  =1⋅e2t  =f⋅g  Step 2   The convolution product of two continuous function f and g is,  f⋅g=∫0tf(t−τ)g(τ)dτ  1⋅e2t=∫0t1⋅e2τdτ  =12e2t−12  Hence the Laplace inverse using convolution theorem is 12e2t−12.   Step 3   (b) The expression for the sum of the partial fraction of F(s)G(s) is,  F(s)G(s)=1s(s−2)  =12(s−2)−12s  Step 4   The Laplace inverse of F(s)G(s) using partial fraction is,  L−1(F(s)G(s))=L−1[12(s−2)−12s]  L−1(12(s−2))−L−1(12s)  =12e2t−12  Hence the Laplace inverse using partial fraction is 12e2t−12.

determine L^{-1}[F(s)G(s)] in the following two ways: (a) using the C

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