Continuous-Time Convolution

Introduction

Continuous-time convolution is the mathematical operation that computes the output of any LTI system given its impulse response and any input signal. The convolution integral takes two functions and produces a third function that describes how the shape of one is modified by the other. While the formula may appear intimidating at first glance, mastering convolution computation — both analytically and graphically — is essential for understanding how signals interact with systems.

The convolution of input $x(t)$ with impulse response $h(t)$ gives the output:

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

The Convolution Integral

The continuous-time convolution is defined as:

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

Both forms are equivalent (by commutativity). The variable $\tau$ is a dummy integration variable; the result is a function of $t$.

Interpretation

At each output time $t$:

1. Take the impulse response $h(\tau)$ and reverse it to get $h(-\tau)$
2. Shift the reversed function to position $t$: $h(t-\tau)$
3. Multiply this shifted, reversed function by the input $x(\tau)$
4. Integrate (sum) the product over all $\tau$

This "flip, shift, multiply, integrate" procedure is the essence of graphical convolution.

Graphical Convolution Method

The graphical method provides visual insight and is especially useful for signals defined in pieces (rectangles, triangles, ramps):

Step 1: Sketch $x(\tau)$ and $h(\tau)$ separately.

Step 2: Flip one function (say $h$) to get $h(-\tau)$.

Step 3: For each value of $t$, slide $h(t-\tau)$ across $x(\tau)$ and compute the integral (area of overlap, weighted by the product).

Step 4: Identify regions where the overlap changes character (begins, is fully engaged, ends) and compute the integral for each region separately.

Worked Example: Two Rectangular Pulses

Let $x(t) = u(t) - u(t-2)$ (rect pulse from 0 to 2, height 1) and $h(t) = u(t) - u(t-3)$ (rect pulse from 0 to 3, height 1).

Step 1: Both are rectangular pulses. Flip $h$: $h(-\tau)$ runs from $-3$ to $0$. Then $h(t-\tau)$ runs from $t-3$ to $t$.

Step 2: Identify overlap regions:

• $t < 0$: no overlap → $y(t) = 0$
• $0 \leq t < 2$: $x$ starts at 0, $h$ starts entering from left. Overlap = $[0, t]$. $y(t) = t$
• $2 \leq t < 3$: $x$ is fully inside $h$. Overlap = $[t-3, t] \cap [0, 2]$. Since $t < 3$, overlap = $[0, 2]$. $y(t) = 2$
• $3 \leq t < 5$: $h$ starts sliding past $x$. Overlap = $[t-3, 2]$. $y(t) = 2-(t-3) = 5-t$
• $t \geq 5$: no overlap → $y(t) = 0$

Result: A trapezoidal pulse that rises from 0 to 2, stays at 2, then falls from 2 back to 0.

Analytical Convolution

For signals with analytical expressions, substitute directly into the integral and evaluate:

Example: Exponential Convolved with Step

$x(t) = u(t)$, $h(t) = e^{-at}u(t)$ with $a > 0$.

$$y(t) = \int_{-\infty}^{\infty} u(\tau) \cdot e^{-a(t-\tau)}u(t-\tau)d\tau$$

For $t > 0$: both $u(\tau) = 1$ (need $\tau \geq 0$) and $u(t-\tau) = 1$ (need $\tau \leq t$):

$$y(t) = \int_0^t e^{-a(t-\tau)}d\tau = e^{-at}\int_0^t e^{a\tau}d\tau = e^{-at} \cdot \frac{e^{at}-1}{a} = \frac{1-e^{-at}}{a}$$

For $t < 0$: no overlap exists, so $y(t) = 0$.

Complete answer: $y(t) = \frac{1}{a}(1-e^{-at})u(t)$ — the familiar step response of a first-order system.

Duration of the Convolution Result

If $x(t)$ has duration $T_x$ (nonzero from $t_{x1}$ to $t_{x2}$) and $h(t)$ has duration $T_h$ (nonzero from $t_{h1}$ to $t_{h2}$):

• Result starts at: $t_{start} = t_{x1} + t_{h1}$
• Result ends at: $t_{end} = t_{x2} + t_{h2}$
• Result duration: $T_y = T_x + T_h$

The convolution of two finite-duration signals has duration equal to the sum of individual durations (minus one sample in discrete-time).

Properties Used in Computation

These properties simplify convolution calculations:

• Convolution with impulse: $x(t) * \delta(t) = x(t)$ (identity element)
• Convolution with shifted impulse: $x(t) * \delta(t-t_0) = x(t-t_0)$ (delay)
• Convolution with step: $x(t) * u(t) = \int_{-\infty}^{t} x(\tau)d\tau$ (running integral)
• Commutativity: $x * h = h * x$ (choose whichever is easier to flip)

Practical Tips

1. Choose which function to flip: Flip the simpler one to make the overlap analysis easier.
2. Draw carefully: Sketch both functions on the same axis and identify the limits of integration precisely.
3. Watch signs: In $h(t-\tau)$, increasing $t$ slides the function to the right.
4. Check endpoints: Verify $y(t)$ at the boundaries between regions — it should be continuous for convolution of finite signals.
5. Verify total area: $\int y(t)dt = (\int x dt)(\int h dt)$ — a useful sanity check.

Key Takeaways

• Continuous convolution: $y(t) = \int x(\tau)h(t-\tau)d\tau$ — flip, shift, multiply, integrate
• Graphical method: identify overlap regions as $t$ varies, compute integral in each region
• Duration of result = sum of input durations
• The step response is the running integral of the impulse response
• Convolution with $\delta(t-t_0)$ simply delays the signal by $t_0$
• Always check your result for continuity and correct total area