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# Bs f s a s f s f1 s f2 s fn

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Unformatted text preview: se Laplace transform of a complicated function, we can convert the function to a sum of simpler terms for which we know the inverse Laplace transform. B(s) F (s) = A( s ) F ( s ) = F1 ( s ) + F2 ( s ) + + Fn ( s ) f (t ) = −1 [ F1 ( s )] + −1 [ F2 ( s )] + + −1 [ Fn ( s )] = f1 (t ) + f 2 (t ) + + f n (t ) Inverse Laplace Transform We will consider three cases and show an F(s) can be expanded into partial fractions: • Case 1: Distinct and real roots • Case 2: Real and Repeated Roots • Case 3: complex conjugates Roots Inverse Laplace Transform Case 1: Real and distinct roots. Consider F(s) is written as. K ( s + z1 )( s + z 2 ) ( s + z m ) F (s) = ( s + p1 )( s + p2 ) ( s + p n ) for m<n If F(s) involves distinct poles only then can be written as B(s) C1 C2 Ck Cn F(s) = = + ++ ++ A(s) (s + p1 ) (s + p 2 ) (s + p k ) (s + p n ) B( s) =? ( s + p k ) A( s ) s = − p k Inverse Laplace Transform Case 1: Real and distinct roots. The residue Cn is found from B(s) C n = (s + p k ) A(s) s = − p k Example. Find the inverse Laplace transform of F (s) = s+3 s 2 + 3s + 2 Inverse...
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