Laplace transform via convolution

This is a draft page to describe how to get the Laplace transform from Convolution.

System Requirements

The Laplace transform applies to signals that are absolutely infinitely integrable when multiplied by a real exponential of some kind. The following passage and equation, which say that much better, are from Signals and Systems by Haykin and Van Veen (2nd ed, Wiley and Sons, 2005, p. 485):

"Hence, a necessary condition for convergence of the Laplace transform is the absolute integrability of \(x(t)e^{-\sigma t}\). That is, we must have

\( \int_{-\infty}^{\infty}|x(t)e^{-\sigma t}|dt<\infty \)

The range of \(\sigma\) for which the Laplace transform converges is termed the region of convergence (ROC)."

Note that \(\sigma\) here is real - more on that later. Overall, the Laplace transform may be used to calculate responses from linear, time invariant systems so long as the system and the inputs have some area of overlap of their regions of convergence.

Exposition

Our protagonist is you, the person who wants to find some way to determine the output of an LTI system to some mostly arbitrary input (where "mostly" here is due to the fact that the input needs to have a Laplace transform).

Rise

Standing in our hero's way is the convolution integral,

\( \begin{align} y(t)&=x(t)\ast h(t)=\int_{-\infty}^{\infty}x(\tau)~h(t-\tau)~d\tau\\ y(t)&=h(t)\ast x(t)=\int_{-\infty}^{\infty}h(\tau)~x(t-\tau)~d\tau\\ \end{align} \)

Except in specific instances (see Convolution_Shortcuts) this integral is not for the faint of heart. To begin, it will be useful to determine the response of an LTI system to a generalized exponential. Specifically, to determine how the system responds to:

\( x(t)=e^{st} \)

where \(s\) has a real part \(\sigma\) and an imaginary part \(\omega\); that is, \(s=\sigma+j\omega\). For an LTI system then, we will use the bottom of the two definitions above for convolution to get:

\( \begin{align} y(t)&=h(t)\ast x(t)\\ y(t)&=h(t)\ast e^{st}\\ y(t)&=\int_{-\infty}^{\infty}h(\tau)~e^{s(t-\tau)}~d\tau=\int_{-\infty}^{\infty}h(\tau)~e^{st}~e^{-s\tau}~d\tau \end{align} \)

Since one of the terms is not a function of \(\tau\), it can come out of the integral:

\( \begin{align} y(t)&=e^{st}\int_{-\infty}^{\infty}h(\tau)~e^{-s\tau}~d\tau \end{align} \)

The remaining integral is now solely a function of the impulse response and \(s\) - it is no longer dependent on time. What this means is that the response to a general exponential is that exact same general exponential multiplied by some scaling factor determined by the impulse response of the system and the exponent itself. Since that integral is a function of \(s\), we can re-write it as:

\( \begin{align} y(t)&=e^{st}H(s) \end{align} \)

where

\( \begin{align} H(s)&=\int_{-\infty}^{\infty}h(\tau)~e^{-s\tau}~d\tau \end{align} \)

This is the Laplace transform. Note: to make life slightly more confusing, often the dummy variable of integration used for the Laplace transform is \(t\) instead of \(\tau\), so... This is the Laplace transform:

\( \begin{align} H(s)&=\int_{-\infty}^{\infty}h(t)~e^{-st}~dt=\mathcal{L}\left\{h(t)\right\} \end{align} \)

Climax

That's all well and good, but \(e^{st}\) shares a feature with \(h(t)\) - mathematically fantastic and physically unrealizable. The real power comes from looking at how the Laplace transform of the response of an LTI system relates to the Laplace transform of the input to the system. Ready, Player One?

\( \begin{align} Y(s)&=\mathcal{L}\left\{y(t)\right\} = \mathcal{L}\left\{ \int_{-\infty}^{\infty}x(\tau)~h(t-\tau)~d\tau \right\}\\ &=\int_{-\infty}^{\infty}\left(\int_{-\infty}^{\infty}x(\tau)~h(t-\tau)~d\tau\right)~e^{-st}~dt \end{align} \)

You can reverse the order of integration, which will allow a term to come out of the now-innermost integral:

\( \begin{align} Y(s)&=\int_{-\infty}^{\infty}\left(\int_{-\infty}^{\infty}x(\tau)~h(t-\tau)~e^{-st}~dt\right)~d\tau\\ &=\int_{-\infty}^{\infty}x(\tau)~\left(\int_{-\infty}^{\infty}h(t-\tau)~e^{-st}~dt\right)~d\tau\\ \end{align} \)

With a transformation of variables on the inner integral (borrowed from the University of Mississippi):

\( \begin{align} \gamma &= t-\tau\\ t &= \gamma + \tau\\ dt &= d\gamma \end{align} \)

the inner integral changes and allows for some rearrangement of the terms:

\( \begin{align} Y(s)&=\int_{-\infty}^{\infty}x(\tau)~\left(\int_{-\infty}^{\infty}h(\gamma)~e^{-s(\gamma+\tau)}~d\gamma\right)~d\tau\\ &=\int_{-\infty}^{\infty}x(\tau)~\left(\int_{-\infty}^{\infty}h(\gamma)~e^{-s\gamma}~e^{-s\tau}~d\gamma\right)~d\tau\\ &=\int_{-\infty}^{\infty}x(\tau)~e^{-s\tau}~\left(\int_{-\infty}^{\infty}h(\gamma)~e^{-s\gamma}~d\gamma\right)~d\tau\\ \end{align} \)

Fall

This still seems awfully complicated - but wait! The inner integral is now exactly the Laplace transform of \(h(t)\), which allows the following substitution and rearrangement:

\( \begin{align} Y(s)&=\int_{-\infty}^{\infty}x(\tau)~e^{-s\tau}~H(s)~d\tau\\ Y(s)&=H(s)\int_{-\infty}^{\infty}x(\tau)~e^{-s\tau}~d\tau\\ \end{align} \)

and behold - all that remains of the integral is exactly the Laplace transform of \(x(t)\)... meaning...

Dénouement

\( \begin{align} Y(s)&=X(s)\cdot H(s) \end{align} \)

This shows that convolution (eww) in the time domain is merely multiplication (hurrah!) in the Laplace domain. We have just turned differential calculus into linear algebra!