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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
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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This document was uploaded on 03/15/2014 for the course BDA 3703 at Tun Hussein Onn University of Malaysia.
- Spring '14