Graphical convolution

Graphical convolution — or the flip-and-slide method — is the pictorial way of evaluating the convolution integral

$y(t)=(x\ast h)(t)={\int }_{-\infty }^{\infty }x(\tau )h(t-\tau )d\tau .$

The integral asks us to plot two things on the same $\tau$-axis: $x(\tau )$ (the input, unchanged) and $h(t-\tau )$ (the impulse response, flipped and shifted). Multiply pointwise. Integrate over $\tau$. The integral is the area under the product. Each value of $t$ gives a different shifted $h$, a different product, and a different area — that’s what makes $y(t)$ a function of $t$.

Constructing h(t − τ)

$h(t)\stackrel{t\to -\tau }{\to }h(-\tau )\stackrel{\text{shift by }t}{\to }h(-(\tau -t))=h(t-\tau ).$

First step replaces $t$ with $-\tau$, reflecting $h$ about the vertical axis. Second step is a shift: $h(t-\tau )$ as a function of $\tau$ has argument zero when $\tau =t$, so the signal’s origin is at $\tau =t$.

In total: we flipped $h$ about the vertical axis and slid it so its origin is at $\tau =t$. As $t$ increases, the flipped impulse response moves to the right.

The general procedure

1. Plot $x(\tau )$ on a $\tau$-axis.
2. Plot $h(-\tau )$ (flipped $h$) on the same axis.
3. For each value of $t$, shift the flipped $h$ to $h(t-\tau )$ and identify the overlap region with $x(\tau )$.
4. Compute the area of the product $x(\tau )h(t-\tau )$ over the overlap region. That’s $y(t)$.
5. As $t$ varies, the overlap region’s shape changes. Break $y(t)$ into piecewise regions of $t$ corresponding to those shape changes.

Worked example

Take $x(t)=2\mathrm{rect}(t/2)$ (rectangle of height 2 on $-1\le t\le 1$) and $h(t)=2-2t$ for $0\le t\le 1$, zero elsewhere (triangle from height 2 at $t=0$ down to 0 at $t=1$). Find $y(t)$.

Flip $h$: $h(-\tau )$ is a triangle on $\tau \in [-1,0]$, with height 2 at $\tau =0$, height 0 at $\tau =-1$. Slide by $t$: $h(t-\tau )$ is a triangle on $\tau \in [t-1,t]$, same shape.

Now slide this triangle across the rectangle $x(\tau )$ on $[-1,1]$:

• $t<-1$: triangle entirely left of rectangle. $y(t)=0$.
• $-1\le t<0$: partial overlap on right side of triangle. Overlap is $(-1,t)$. In overlap: $x=2$, $h(t-\tau )=2(1-t+\tau )$. Product: $4(1-t+\tau )$. Integrating: $y(t)=2-2t^2$. (Hand-check: $t=-1\to 0$ ✓; $t=0\to 2$ ✓.)
• $0\le t\le 1$: triangle entirely inside rectangle. Overlap is the whole triangle. Integral = area of triangle × height of rectangle = $(1/2\cdot 1\cdot 2)\cdot 2=2$. Flat top.
• $1\le t\le 2$: triangle half-in, half-out on right side. By symmetric algebra, $y(t)=2(2-t{)}^2$. (Hand-check: $t=1\to 2$ ✓; $t=2\to 0$ ✓.)
• $t>2$: triangle entirely right of rectangle. $y(t)=0$.

Final shape: zero outside $(-1,2)$, rising smoothly as $2-2t^2$ on $(-1,0)$, flat at 2 on $(0,1)$, falling smoothly as $2(2-t{)}^2$ on $(1,2)$. A rounded trapezoid.

Three things to internalize

• $h(t-\tau )$ slides to the right as $t$ increases.
• $y(t)$ is the area under the product as a function of $t$.
• You break the calculation into regions of $t$ corresponding to where the overlap region changes shape. Typically three to five regions per problem.

For simple convolutions (rect with rect, exponential with step), this method is fast and reliable. For complicated signals, switching to the Fourier or Laplace domain (where convolution becomes multiplication via the convolution theorem) is usually easier.