Graphical Convolution

Graphical convolution is a technique for performing the convolution of two signals. It requires knowing the piecewise representation of each signal. The final result will be a series of expressions for the convolution along with relational operators to indicate when each expression is relevant.

Introduction

The convolution operator $ * $ represents the follownig operation:

$ \begin{align*} y(t)&=x(t)*h(t)\\ ~&=\int_{-\infty}^{\infty}x(\tau)\,h(t-\tau)\,d\tau\mbox{~ or ~}\int_{-\infty}^{\infty}x(t-\tau)\,h(\tau)\,d\tau \end{align*} $

While convolution can be done analytically, there may be times when either $ x(t) $ or $ h(t) $ (or both) are sufficiently complicated or broken up into multiple piecewise functions that the math becomes unwieldy. In these cases, graphical convolution may provide a more straightforward way to find the final answer.

Graphical Convolution

At the heart of graphical convolution is the idea that you are finding the area under a curve, and that the area will be dependent o the product of an unshifted signal in $ \tau $ and a flipped and shifted signal in $ \tau $ - specifically, a signal with an independent variable transformation of $ -(\tau-t) $. Graphical convolution will allow us to see which parts of $ x $ and $ h $ overlap as $ t $ changes and thus what the integral limits and integrands should be at various times. The steps are as follows:

• Choose which function will have $ \tau $ as an argument and which will have $ t-\tau $. Typically, the more complicated function should get the $ \tau $. For the rest of this example, we will assume that we are using $ x(\tau) $ and $ h(t-\tau) $
• Determine the piecewise definition of each function and be sure to understand the transition points (that is, where the piecewise functions change from one thing to another).
  • For example, if $ x(t)=u(t)-r(t-2)+r(t-3) $, this can be written as:$ \begin{align*}x(t)&=\begin{cases} t\leq 0, & 0 \\ 0<t\leq 2, & 1 \\ 2< t\leq 3, & 3-t \\ t> 3, & 0\end{cases}\end{align*} $ Finding $ x(\tau) $ simply means replacing the $ t $ with $ \tau $. The signal $ x(\tau) $ thus has transitions at $ \tau=0, 2, 3 $
  • Now imagine that $ h(t)=r(t)-r(t-2) $. Writing this piecewise means:$ \begin{align*}h(t)&=\begin{cases} t\leq 0, & 0\\ 0<t\leq 2, & t \\ t> 2, & 2\end{cases}\end{align*} $ Finding $ h(t-\tau) $ requires a little more work and some simplification:$ \begin{align*} h(t-\tau)&=\begin{cases} t-\tau\leq 0, & 0\\ 0<t-\tau\leq 2, & t-\tau \\ t-\tau > 2, & 2\end{cases} & \text{Substition}\\ ~&=\begin{cases} \tau \geq t, & 0\\ t > \tau \leq t-2, & t-\tau \\ \tau < t-2, & 2\end{cases} & \text{Re-write relational operators in terms of }\tau\\ ~&=\begin{cases} \tau\leq t-2, & 2\\ t-2< \tau \leq t, & t-\tau \\ \tau\geq t, & 0\end{cases} & \text{order with increasing }\tau\end{align*} $

• Make a sketch of $ x(\tau) $