Derive the convolution integral

The convolution integral gives the output $y(t)$ of a continuous-time LTI system for any input $x(t)$ in terms of its impulse response $h(t)$:

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

Step 1: Represent the input with narrow pulses

Define a unit-area pulse of width $\Delta$:

$$\delta_\Delta(t) = \begin{cases} 1/\Delta, & 0 \le t < \Delta \\ 0, & \text{otherwise}\end{cases}$$

The staircase approximation of $x(t)$ is a sum of shifted, scaled pulses:

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

x(t)     __
      __|  |__
   __|        |__     staircase of width-Delta
  |              |    pulses, height x(k.Delta)
--+--+--+--+--+--+--> t
     kD (k+1)D

Step 2: Response to one pulse

Let $\hat{h}_k(t)$ be the response to $\delta_\Delta(t - k\Delta)$. By linearity (superposition and scaling),

$$\hat{y}(t) = \sum_{k=-\infty}^{\infty} x(k\Delta)\,\hat{h}_k(t)\,\Delta$$

By time invariance, the response to a shifted pulse is the shifted response: $\hat{h}_k(t) = \hat{h}_0(t - k\Delta)$.

Step 3: Take the limit $\Delta \to 0$

As $\Delta \to 0$:

• $\hat{x}(t) \to x(t)$,
• $\delta_\Delta(t) \to \delta(t)$, so $\hat{h}_0(t) \to h(t)$, the impulse response,
• $k\Delta \to \tau$ (continuous variable), $\Delta \to d\tau$, and the sum becomes an integral.

$$\begin{aligned} y(t) &= \lim_{\Delta\to0}\sum_{k=-\infty}^{\infty} x(k\Delta)\,\hat{h}_0(t-k\Delta)\,\Delta \\ &= \int_{-\infty}^{\infty} x(\tau)\,h(t-\tau)\,d\tau \end{aligned}$$

The same result follows directly from the sifting property: $x(t) = \int x(\tau)\delta(t-\tau)d\tau$; the system maps $\delta(t-\tau) \to h(t-\tau)$ (time invariance) and the integral passes through (linearity).

Meaning and evaluation

• The output is a weighted sum of shifted impulse responses; $x(\tau)$ is the weight of the impulse at $\tau$.
• Steps to evaluate: (1) change variable to $\tau$; (2) fold $h(\tau)$ to get $h(-\tau)$; (3) shift by $t$ to get $h(t-\tau)$; (4) multiply by $x(\tau)$; (5) integrate over the overlap; repeat for each range of $t$.
• Using $\tau \to t - \tau$, $y(t) = \int h(\tau)x(t-\tau)d\tau$ also holds (commutative property).

Example: $x(t) = u(t)$, $h(t) = e^{-at}u(t)$, $a>0$: $y(t) = \int_0^t e^{-a(t-\tau)}d\tau = \frac{1}{a}(1 - e^{-at})u(t)$.