Inverse Laplace transform

At a basic level, the inverse Laplace transform doesn’t need to be computed by hand. We can consult a table of transforms and work backwards. By definition it’s given by:

$$f(t)={\mathcal{L}}^{-1}\{F(s)\}=\frac{1}{2\pi j}{\int }_{\sigma -j\infty }^{\sigma +j\infty }F(s)e^{st}ds$$

As with the forward transform, the inverse transform is linear:

$${\mathcal{L}}^{-1}\{\alpha F(s)+\beta G(s)\}=\alpha {\mathcal{L}}^{-1}\{F(s)\}+\beta {\mathcal{L}}^{-1}\{G(s)\}$$

In MATLAB, we can use ilaplace(exp).¹

Computations

For simple $s$-domain expressions, we don’t need to compute the inverse Laplace transform by hand. In particular, for rational functions, we can perform partial fraction decomposition to get our function in a form that can be converted with $e^{-at}$.

Oftentimes, it’s easier and faster to use the definition to compute the transform, especially for more difficult functions. In other cases, PFD doesn’t reduce the function enough for the simple method. In this case, we use Cauchy’s residue theorem to get a variant of the definition, where we can instead compute the residues:

$$f(t)=\sum _{i=1}^n\text{Res}(F(s)e^{st};s=z_i)$$

where $z_i$ are the poles of the function.

Also: don’t forget to include the unit step function, since the Laplace transform is only defined for non-negative values of $t$. Note also that the residue method doesn’t properly account for time shifts of the unit step, so we must take extra care to apply the transform property prior to applying residue theorem.

Proof

By theorem, suppose $f(t)$ is continuous and $|f(t)|\le M{e}^{\alpha t}$ for some $M,\alpha$, i.e., is of exponential type $\alpha$. Then we can define $F(s)$ for $s\in \mathbb{C}$, more precisely:

$$F(z)={\int }_0^{\infty }{e}^{-zt}f(t)dt$$

is well defined and analytic for $\text{Re}(z)>\alpha$.

Then we say:

$$|{\int }_0^{\infty }{e}^{-zt}f(t)dt|\le M{\int }_0^{\infty }\left|e^{-zt}\right||f(t)|dt\le M{\int }_0^{\infty }{e}^{-\text{Re}(z)t}dt$$

This integral converges if $\text{Re}(z)>\alpha$. Moreover, one can show it is differentiable there:

$$F^{\prime }(z)={\int }_0^{\infty }(-t)e^{-zt}f(t)dt$$

toimprove wtf is this We should think about $s$ as a complex plane, which we’ll sometimes call $z$. In particular, it lives in the complex plane anywhere the real part is greater than $\text{Re}(\alpha )$. It’s analytic there.

By theorem, suppose $f(t)$ (as in the previous theorem) is exponential and of exponential type $\alpha$, then we can recover $f(t)$ by integrating on a vertical line $\text{Re}(z)=\sigma$ from $\text{Im}(z)=\pm \infty$. Then:

$$f(t)=\frac{1}{2\pi i}{\int }_{\sigma -i\infty }^{\sigma +i\infty }{e}^{zt}F(z)dz$$

where $\sigma >\alpha$.

The formula is obtained as followed. Let ${\gamma }_R$ be the closed contour of the vertical line and a semicircle towards the right. Apply Cauchy’s integral formula to $F(z)$, which we’ve seen is analytic inside ${\gamma }_R$ if $\sigma >\alpha$. Then:

$$F(s)=-\frac{1}{2\pi i}{\int }_{{\gamma }_R}\frac{F(z)}{s-z}dz$$

From the intuitive understanding of the integral formula as a sum of many functions that look like:

$$\frac{1}{z-z_0}$$

Something converges. Lost around minute 25 of this lecture by Prof Nachman.

Footnotes

1. ”MATLAB has a powerful inverse Laplace transform engine.” - Prof Najm ↩