s p a s s p inverse laplace transform

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Unformatted text preview: Laplace Transform Case 2: Real and repeated roots. Consider F(s) written as B( s) B( s) F (s) = = A( s ) ( s + p ) n F(s) can be written as B(s) C1 C2 Cn F(s) = = + ++ n 2 n (s + p) (s + p) (s + p) (s + p) n B( s) =? ( s + p ) A( s ) s = − p Inverse Laplace Transform Case 2: Real and repeated roots. In general, the residue bn, bn­1, …, bi, …b1 are found from the following equation: C n −i di 1 =i i! ds n B(s) (s + p) A (s) s = − p i = 0,1,2, , n − 1 Inverse Laplace Transform Case 2: Real and repeated roots. Example. Find the inverse Laplace transform of F (s) = s 2 + 2s + 3 ( s + 3) 3 Inverse Laplace Transform Case 3: Complex roots. Partial fraction method of F(s) with real roots in the denominator can be used for complex and imaginary roots. However the residues are complex conjugates. After taking the inverse Laplace, the result can simplified by using the following Euler equations: e +e 2 jθ − jθ e −e 2j jθ = cos θ − jθ = sin θ Inverse Laplace Transform Case 3: Complex roots. Example. Find the inverse Laplace transform of 3 F(s) = 2 s( s + 2s + 5) 3 3 −t 1 f (t ) = − e cos 2t + sin 2t 55 2 Solving Differential Equations Obtain the differential equations Obtain the Laplace transform of the differential equations Use the Laplace...
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This document was uploaded on 03/15/2014 for the course BDA 3703 at Tun Hussein Onn University of Malaysia.

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