What is convolution; convolution integral

Convolution

Convolution is a mathematical operation that combines two signals to produce a third. In signals and systems it gives the output $y(t)$ of an LTI system as the input $x(t)$ "convolved" with the impulse response $h(t)$:

$$y(t) = x(t) * h(t)$$

It works as a weighted, shifted sum: each past value of the input is weighted by the system's impulse response and all contributions are added.

Expression for the convolution integral

Step 1 – Input as a sum of impulses. Approximate $x(t)$ by narrow pulses of width $\Delta$:

$$x(t) \approx \sum_{k=-\infty}^{\infty} x(k\Delta)\,\delta_\Delta(t-k\Delta)\,\Delta$$

where $\delta_\Delta(t)$ is a pulse of width $\Delta$ and height $1/\Delta$ (unit area).

Step 2 – Response to one pulse. Let $h_\Delta(t)$ be the response to $\delta_\Delta(t)$.

• Time invariance: $\delta_\Delta(t-k\Delta) \rightarrow h_\Delta(t-k\Delta)$.
• Homogeneity: $x(k\Delta)\Delta\,\delta_\Delta(t-k\Delta) \rightarrow x(k\Delta)\Delta\,h_\Delta(t-k\Delta)$.
• Additivity:

$$y(t) \approx \sum_{k=-\infty}^{\infty} x(k\Delta)\,h_\Delta(t-k\Delta)\,\Delta$$

Step 3 – Limit $\Delta \to 0$. Then $\delta_\Delta \to \delta(t)$, $h_\Delta \to h(t)$, $k\Delta \to \tau$, $\Delta \to d\tau$ and the sum becomes an integral:

$$y(t) = \int_{-\infty}^{\infty} x(\tau)\,h(t-\tau)\,d\tau$$

This is the convolution integral. Substituting $\lambda = t-\tau$ gives the equivalent form $y(t) = \int_{-\infty}^{\infty} h(\tau)\,x(t-\tau)\,d\tau$.

For causal system and causal input ($h(t) = 0$ and $x(t) = 0$ for $t < 0$):

$$y(t) = \int_{0}^{t} x(\tau)\,h(t-\tau)\,d\tau$$

Procedure (graphical): fold $h(\tau)$ to $h(-\tau)$, shift it by $t$, multiply by $x(\tau)$, and find the area of the product; repeat for each $t$.

Properties: commutative ($x*h = h*x$), associative ($(x*h_1)*h_2 = x*(h_1*h_2)$, used for cascade), distributive ($x*(h_1+h_2) = x*h_1 + x*h_2$, used for parallel connection), and $x(t)*\delta(t-t_0) = x(t-t_0)$.

Example: $e^{-at}u(t) * u(t) = \int_0^t e^{-a\tau}d\tau = \frac{1}{a}(1-e^{-at})u(t)$.