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The Inverse Laplace Transform: Complex Integration

Module by: Steven J. Cox. E-mail the author

Summary: (Blank Abstract)

The Inverse Laplace Transform

If qq is a rational function with poles λ j j=1h λ j j 1 h then the inverse Laplace transform of qq is

-1qt12πiqzeztd z q t 1 2 z C q z z t
(1)
where CC is a curve that encloses each of the poles of qq. As a result
-1qt= j =1hres λ j q t j 1 h res λ j
(2)
Let us put this lovely formula to the test. We take our examples from discussion of the Laplace Transform and the inverse Laplace Transform. Let us first compute the inverse Laplace Transform of qz=1z+12 q z 1 z 1 2 According to Equation 2 it is simply the residue of qzezt q z z t at z=-1 z -1 , i.e., res-1=limit   z -1 deztd z =tet res -1 z -1 z z t t t This closes the circle on the example begun in the discussion of the Laplace Transform and continued in exercise one for chapter 6.

For our next example we recall x 1 s=0.19(s2+1.5s+0.27)s+1/64(s3+1.655s2+0.4978s+0.0039) x 1 s 0.19 s 2 1.5 s 0.27 s 16 4 s 3 1.655 s 2 0.4978 s 0.0039 from the Inverse Laplace Transform. Using numde, sym2poly and residue, see fib4.m for details, returns r 1 =( 0.0029 262.8394 -474.1929 -1.0857 -9.0930 -0.3326 211.3507 ) r 1 0.0029 262.8394 -474.1929 -1.0857 -9.0930 -0.3326 211.3507 and p 1 =( -1.3565 -0.2885 -0.1667 -0.1667 -0.1667 -0.1667 -0.0100 ) p 1 -1.3565 -0.2885 -0.1667 -0.1667 -0.1667 -0.1667 -0.0100 You will be asked in the exercises to show that this indeed jibes with the x 1 t=211.35et100(0.0554t3+4.5464t2+1.085t+474.19)et6+e-329t400(262.842cosh73t16+262.836sinh73t16) x 1 t 211.35 t 100 0.0554 t 3 4.5464 t 2 1.085 t 474.19 t 6 -329 t 400 262.842 73 t 16 262.836 73 t 16 achieved in the Laplace Transform via ilaplace.

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