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

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:

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