Convolution

This is just a draft at this point intended to capture the progression of showing how the convolution integral works. This is only a draft. If this were a real page, this message would be followed by absolutely perfect math and simply delightful commentary to completely explain the math.

Goal

The goal of this page is to describe how the convolution integral can be used to determine the response of some linear, time-invariant system to an arbitrary input.

System Requirements

Overall, the convolution integral will apply to linear, time invariant systems. These are also called LTI systems. No one knows why.

• The system must be linear, or we must use a linearized model for a non-linear system (generally by looking at small perturbations about some fixed point). Linear systems have the properties of homogeneity and superposition. A short (and incomplete - see draft notice above) way of describing a linear system is if two inputs to a system \(x_1(t)\) and \(x_2(t)\) lead to two outputs \(y_1(t)\) and \(y_2(t)\), then a linear combination of those inputs will lead to the same linear combination of those outputs. That is, if:

\( \begin{align} x_1(t)&\rightarrow y_1(t)\\ x_2(t)&\rightarrow y_2(t) \end{align} \)

then, for some constants \(a\) and \(b\),

\( \begin{align} ax_1(t)+bx_2(t)&\rightarrow ay_1(t)+by_2(t) \end{align} \)

Integration, differentiation, and multiplication by a constant are all linear.

• The system must be time-invariant. That is, if:

\( \begin{align} x(t)&\rightarrow y(t) \end{align} \)

then

\( \begin{align} x(t-t_0)&\rightarrow y(t-t_0) \end{align} \)

Hallmarks of time-varying systems include \(t\) outside of the argument of \(x\) or \(y\), time-scaling or time-reversal of \(t\), or specific times in the model (integrating from 0, etc.).

Integration, differentiation, and multiplication by a constant are all time invariant.

Definitions

Singularity Functions

Unit Step Function

The unit step function \(u(t)\) is defined as:

\( u(t)= \begin{cases} 0 & t<0 \\ 1 & t> 0 \end{cases} \)

It is undefined at exactly 0. The unit step function is basically off for negative arguments and on for positive arguments.

Impulse Function

The impulse function \(\delta(t)\) is defined as:

\( \delta(t)=\frac{d}{dt}u(t) \)

which is a little hard to understand. It is easier to quantify with the inverse relationship,

\( u(t)=\int_{-\infty}^{t}\delta(\tau)~d\tau \)

Considering the integral relates to the area under a curve, the fact that the step function is 0 before time 0 means the impulse function has no value before time 0. The unit step function jumps to 1 instantaneously at time 0, meaning the impulse function must have an area of 1 at time 0. Since the value of the unit step function does not change after time 0, the impulse must then be 0 for all positive times. This means:

\( \delta(t)= \begin{cases} 0 & t\neq 0 \\ \mbox{Area of 1} & t=0 \end{cases} \)

Physically Realizable Approximations

Eventually, there will be some nifty graphics showing approximations to a unit step and an impulse function. There will be an \(\epsilon\) and a \(\frac{1}{\epsilon}\) involved and it will be approximately glorious.

Impulse Response

For an LTI system , we will define the impulse response as the...response...to an impulse. And we will call it \(h(t)\), so basically, if there is an LTI system such that:

\( \begin{align} x(t)&\rightarrow y(t) \end{align} \)

then we will define \(h(t)\) such that

\( \begin{align} \delta(t)&\rightarrow h(t) \end{align} \)

Step Response

The impulse response is going to be very useful mathematically, but it is impossible to find directly experimentally since the impulse function is impossible to replicate). Fortunate, we can define something called the step response (which is the response to a step (which we can replicate physically)). The step response \(s_r(t)\) is defined as:

\( \begin{align} u(t)&\rightarrow s_r(t) \end{align} \)

and just as the relationships between the step and the impulse are through an integral and a derivative, the relationships between the step response and the impulse response are through an integral and a derivative:

\( \begin{align} s_r(t)&= \int_{-\infty}^{t} h(\tau)~d\tau \\ h(t)&=\frac{d}{dt}s_r(t) \end{align} \)

Singularity Integrals

With the definitions above, there are several different integrals involving singularity functions that may prove useful. On the Singularity Functions page there is a section on Impulse Functions as Part of Integrand that shows how the impulse impacts integrals. Specifically, when integrating with in impulse, first determine when the impulse is non-zero, check to see if the dummy variable does or might obtain the value needed to have the impulse be non-zero, and then substitute that value in to the rest of the integrand and pull the integrand out of the integral. Fortunately, for purposes of this page, there is only one integral with an impulse to worry about:

\( \int_{-\infty}^{\infty}x(\tau)~\delta(t-\tau)~d\tau \)

The impulse function is only nonzero at \(t=\tau\) and the dummy variable \(\tau\) will definitely hit that value. As a result, you can pull out the integrand when \(\tau=t\) and thus get:

\( \int_{-\infty}^{\infty}x(\tau)~\delta(t-\tau)~d\tau=x(t) \)

LTI Redux

If we define the impulse response such that:

\( \begin{align} \delta(t)&\rightarrow h(t) \end{align} \)

then for an LTI system, a scaled shifted impulse would result in a scaled shifted impulse response:

\( \begin{align} a\delta(t-t_0)&\rightarrow ah(t-t_0) \end{align} \)

Furthermore, if we use the fact that:

\( \int_{-\infty}^{\infty}x(\tau)~\delta(t-\tau)~d\tau=x(t) \)

then we can solve for \(y(t)\) in:

\( \begin{align} x(t)&\rightarrow y(t) \end{align} \)

by first substituting in the singularity integral representation of \(x(t)\):

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

and then recognizing that the response of an LTI system to a sum (or integral) of scaled, shifted impulses is similarly scaled, shifted impulse responses - that is:

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

The Convolution Integral

In other words ("words") for an LTI system with a known impulse response \(h(t)\) and an arbitrary input \(x(t)\), the output \(y(t)\) can be calculated with the convolution integral:

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

This integral is often represented by the convolution operator \(\ast\), so:

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

Note that convolution is commutative such that:

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

Documenting proof of this is left as an exercise for someone else.