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7: Complex Analysis II
7.3: The Inverse Laplace Transform- Complex Integration

7.3: The Inverse Laplace Transform- Complex Integration

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The Inverse Laplace Transform

If \(q\) is a rational function with poles \(\{\lambda_{j} | j = \{1, \cdots, h\}\}\) then the inverse Laplace transform of \(q\) is

\[\mathscr{L}^{-1}(q)(t) \equiv \frac{1}{2 \pi i} \int q(z) e^{zt} dz \nonumber\]

where \(C\) is a curve that encloses each of the poles of \(q\)

\[\mathscr{L}^{-1}(q)(t) = \sum_{j = 1}^{h} res(\lambda_{j}) \nonumber\]

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

\[q(z) = \frac{1}{(z+1)^2} \nonumber\]

According to Equation it is simply the residue of \(q(z)e^{zt}\) at \(z = -1\) i.e.,

\[res(-1) = \lim_{z \rightarrow -1} de^{zt} dz = te^{-t} \nonumber\]

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

\[\mathscr{L} (x_{1}(s)) = \frac{0.19(s^2+1.5s+0.27)}{(s+1/6)^{4}(s^3+1.655s^2+0.4978s+0.0039)} \nonumber\]

from the Inverse Laplace Transform. Using numde, sym2poly and residue, see fib4.m for details, returns

\[r_{1} = \begin{pmatrix} {0.0029}\\ {262.8394}\\ {-474.1929}\\ {-1.0857}\\ {-9.0930}\\ {-0.3326}\\ {211.3507} \end{pmatrix} \nonumber\]

and

\[p_{1} = \begin{pmatrix} {-1.3565}\\ {-0.2885}\\ {-0.1667}\\ {-0.1667}\\ {-0.1667}\\ {-0.1667}\\ {-0.0100} \end{pmatrix} \nonumber\]

You will be asked in the exercises to show that this indeed jibes with the

\[x_{1}(t) = 211.35e^{\frac{-t}{100}}-(0.0554t^3+4.5464t^2+1.085t+474.19)e^{\frac{-t}{6}}+e^{\frac{-329t}{400}}(262.842 \cosh (\frac{\sqrt{73}t}{16})+262.836 \sinh (\frac{\sqrt{73}t}{16})) \nonumber\]

achieved in the Laplace Transform via ilaplace.