Convolution Integral

The convolution integral tells you how two functions combine when one is shifted across the other. In continuous time, it is defined by

$$(f * g)(t) = \int_{-\infty}^{\infty} f(\tau) g(t-\tau)\,d\tau$$

The fast intuition is simple: for each value of $t$, slide one function, find where the two functions overlap, multiply their values on that overlap, and add the result. If both functions are causal, meaning they are zero for negative time, this often becomes

$$(f * g)(t) = \int_0^t f(\tau) g(t-\tau)\,d\tau$$

for $t \ge 0$, as long as $[0,t]$ captures the full overlap. The main idea is practical: convolution turns sliding overlap into one number for each value of $t$.

Convolution Integral Definition And Intuition

Think of $f(\tau)$ as fixed and $g(t-\tau)$ as a flipped, shifted copy of $g$. As $t$ changes, the overlap changes, so the integral changes too.

That is the main difference from pointwise multiplication. You are not comparing the two functions at the same input. You are adding products across the whole region where the shifted copy overlaps the original.

Why The Overlap Determines The Limits

The limits in a convolution problem usually do not come from memorizing a template. They come from asking where both factors are nonzero.

That is why many convolution answers are piecewise. As $t$ moves, the overlap interval can grow, shrink, or disappear, so the integral has to change with it.

This is the part students often miss: the hard part is usually not the integration. It is finding the correct overlap interval first.

Convolution Integral Example: Two Unit Pulses

Let

$$f(t) = \begin{cases} 1, & 0 \le t \le 1 \\ 0, & \text{otherwise} \end{cases}$$

and let $g(t)$ be the same function. We want $(f * g)(t)$.

This example works well because the integrand is either $1$ or $0$, so the convolution is just the length of the overlap interval.

Using the definition,

$$(f * g)(t) = \int_{-\infty}^{\infty} f(\tau) g(t-\tau)\,d\tau$$

Because $f(\tau)=1$ only on $[0,1]$, and $g(t-\tau)=1$ only when $0 \le t-\tau \le 1$, the integrand is $1$ exactly where both conditions hold.

The second condition means

$$t-1 \le \tau \le t$$

So the overlap interval is

$$[0,1] \cap [t-1,t]$$

So $(f * g)(t)$ is the length of that overlap.

Case 1: $t < 0$

There is no overlap, so

$$(f * g)(t) = 0$$

Case 2: $0 \le t \le 1$

The overlap runs from $\tau=0$ to $\tau=t$, so

$$(f * g)(t) = \int_0^t 1\,d\tau = t$$

Case 3: $1 \le t \le 2$

The overlap runs from $\tau=t-1$ to $\tau=1$, so

$$(f * g)(t) = \int_{t-1}^1 1\,d\tau = 2-t$$

Case 4: $t > 2$

Again there is no overlap, so

$$(f * g)(t) = 0$$

Putting the pieces together,

$$(f * g)(t) = \begin{cases} 0, & t < 0 \\ t, & 0 \le t \le 1 \\ 2-t, & 1 \le t \le 2 \\ 0, & t > 2 \end{cases}$$

The result is a triangle. Its height grows while the overlap grows, then falls as the overlap shrinks.

Common Convolution Integral Mistakes

Forgetting The Shifted Input

The second factor is $g(t-\tau)$, not $g(\tau-t)$ and not just $g(\tau)$. The shift is the whole point of convolution.

Using The Wrong Limits

The safest method is to find where both factors are nonzero. If the overlap changes with $t$, the limits usually need a piecewise answer.

Treating Convolution As Pointwise Multiplication

Pointwise multiplication uses values at the same input. Convolution accumulates products across a whole interval.

Skipping The Condition Behind A Shortcut

The shortcut

$$(f * g)(t) = \int_0^t f(\tau)g(t-\tau)\,d\tau$$

works in common causal settings, but not for every pair of functions. Use it only when the support assumptions justify it.

Where The Convolution Integral Is Used

Use convolution when one quantity depends on how another is spread across nearby time or space.

In linear time-invariant systems, convolution gives the output from an input and an impulse response. In probability, if two independent random variables have densities, the density of their sum is a convolution of those densities. More broadly, convolution appears in smoothing, filtering, diffusion, and any setting where nearby values combine.

Try A Similar Convolution Problem

Try the same pulse example, but make the second pulse twice as tall:

$$g(t) = \begin{cases} 2, & 0 \le t \le 1 \\ 0, & \text{otherwise} \end{cases}$$

The overlap interval stays the same, but the integrand is now twice as large on that interval. If you can predict how that changes the triangular result before integrating, the core idea of convolution has clicked.