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The Inverse Laplace Transform

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

Summary: This module introduces the Inverse Laplace Transform. Building on the groundwork done in the Laplace Transform module, this module gives some background into the Inverse Laplace method, and uses MATLAB's ilaplace command to find the actual solutions to a problem.

To Come

In The Transfer Function we shall establish that the inverse Laplace transform of a function hh is

-1ht=12πe(c+yi)th(c+yi)tdy h t 1 2 y c y t h c y t
(1)
where i2-1 2 -1 and the real number cc is chosen so that all of the singularities of hh lie to the left of the line of integration.

Proceeding with the Inverse Laplace Transform

With the inverse Laplace transform one may express the solution of x=Bx+g x B x g , as

xt=-1sIB-1(g+x0) x t s I B g x 0
(2)
As an example, let us take the first component of x x , namely x 1 s=0.19(s2+1.5s+0.27)s+164(s3+1.655s2+0.4078s+0.0039) . x 1 s 0.19 s 2 1.5 s 0.27 s 1 6 4 s 3 1.655 s 2 0.4078 s 0.0039 . We define:
Definition 1: poles
Also called singularities, these are the points ss at which x 1 s x 1 s blows up.
These are clearly the roots of its denominator, namely
-1 /100 ,    -329 /400±27316 ,    and    -1/6 . -1 100,   ± -329 400 2 73 16 ,  and  -16.
(3)
All four being negative, it suffices to take c=0 c 0 and so the integration in Equation 1 proceeds up the imaginary axis. We don't suppose the reader to have already encountered integration in the complex plane but hope that this example might provide the motivation necessary for a brief overview of such. Before that however we note that MATLAB has digested the calculus we wish to develop. Referring again to fib3.m for details we note that the ilaplace command produces x 1 t=211.35et100(0.0554t3+4.5464t2+1.085t+474.19)et6+e(329t)400(262.842cosh273t16)+262.836sinh273t16 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 2 73 t 16 262.836 2 73 t 16

Figure 1: The 3 potentials associated with the RC circuit model figure.
Figure 1 (fib3_fig1.png)

The other potentials, see the figure above, possess similar expressions. Please note that each of the poles of x 1 x 1 appear as exponents in x 1 x 1 and that the coefficients of the exponentials are polynomials whose degrees is determined by the order of the respective pole.

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