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Mathematics LibreTexts

5.3: The Inverse Laplace Transform

( \newcommand{\kernel}{\mathrm{null}\,}\)

To Come

In The transfer Function we shall establish that the inverse Laplace transform of a function h is

L1(h)(t)=12πe(c+yi)th((c+yi)t)dy

where i21 and the real number c is chosen so that all of the singularities of h 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, as

x(t)=L1((sIB)1)(L(g+x(0))

As an example, let us take the first component of L(x), namely

Lx1(s)=0.19(s2+1.5s+0.27)(s+16)4(s3+1.655s2+0.4078s+0.0039)

We define:

Definition: Pole

Also called singularities, these are the points ss at which Lx1(s) blows up.

These are clearly the roots of its denominator, namely

1/100329/400±27316and1/6

All four being negative, it suffices to take c=0 and so the integration in Equation 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

x1(t)=211.35et100(0.0554t3+4.5464t2+1.085t+474.19)et6+e(329t)400(262.842cosh(273t16))+262.836sinh(273t16))

Screen Shot 2020-10-04 at 4.27.22 PM.png
Figure 1. The 3 potentials associated with the RC circuit model figure.

The other potentials, see the figure above, possess similar expressions. Please note that each of the poles of L(x1) appear as exponents in x1 and that the coefficients of the exponentials are polynomials whose degrees is determined by the order of the respective pole.


This page titled 5.3: The Inverse Laplace Transform is shared under a CC BY 1.0 license and was authored, remixed, and/or curated by Steve Cox via source content that was edited to the style and standards of the LibreTexts platform.

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