Continuous-time convolution

Drag Position t slowly from left to right and watch the shaded overlap between the two pulses.

From a sum to an integral

On the last page you built a system’s output by flipping one sequence, sliding it past the other, and at every slide position multiplying the overlapping samples and adding them up: $y[n]=\sum_k x[k]\,h[n-k]$. Most of the signals I actually care about, a voltage, a sound pressure, aren’t lists of samples at all: they’re defined at every instant. I want the exact same flip-and-slide picture to work for them too.

It does, with one change. Where the discrete version added up finitely many overlapping products, the continuous version has infinitely many instants to add over, so “add them up” becomes an integral, a running area under a curve (you met this reading on the impulse and step page). Flip $h$, slide it across $x$ as before, and at each slide position the overlap area, not an added-up sum, is the output at that instant:

Here $x(t)$ and $h(t)$ are the two signals being combined, $t$ is the instant you’re reading the output at, and $\tau$ is a sliding time variable: a dummy variable the integral runs over, not a fixed instant, standing in for the sample index $k$ from the discrete sum. At each value of $t$, $h(t-\tau)$ is $h$ flipped (the minus sign in front of $\tau$) and shifted to sit at $t$; multiply it by $x(\tau)$ at every $\tau$ and total up the area under that product, and you have $y(t)$.

Go back to the two pulses at the top of the page. The shaded area grows as the sliding pulse starts to cover the fixed one, peaks once they line up exactly, then shrinks back to nothing, and the Overlap area, y(t) readout traces out exactly that rise and fall as the lower plot builds up.

Two pulses, one trapezoid

Two rectangular pulses are the simplest pair of signals to convolve by eye, and the shape you get tells you almost everything about how the general picture behaves. As the sliding pulse starts to overlap the fixed one, the overlap area climbs; if the two pulses are the same width, the overlap climbs all the way to full overlap and then falls straight back down, a plain triangle. If the sliding pulse is wider, once the narrower pulse is completely inside it the overlap area stops growing and holds flat for a while, a plateau, before falling as the pulses separate.

Set Second pulse width equal to the fixed pulse’s width and look at the output, then drag it wider.

At equal widths the output is a plain triangle, peak in the middle. As the second pulse widens, watch a flat Plateau width open up on top of the triangle while the Peak height stops growing, capped at the width of the narrower pulse.

Charging and discharging a capacitor

A capacitor stores electric charge the way a bucket stores water: pour current in, and the charge (and the voltage across it) rises. A resistor placed in series with it limits how fast current can flow in, the way a narrow tap limits how fast the bucket fills. Together, an RC circuit’s voltage doesn’t jump instantly to match its input; it eases toward it, and how quickly it gets there is set by the time constant $RC$, the resistance times the capacitance: roughly how long the circuit takes to get most of the way to a new value.

An RC circuit’s impulse response, its reply to a single instantaneous spike of current, is a decaying exponential, $h(t)=\tfrac{1}{RC}e^{-t/RC}u(t)$ for $t\ge0$: charge dumped in all at once leaks away smoothly afterward. Convolve that with a brief rectangular pulse of current and you get exactly the curve you’d expect from a real capacitor: the voltage climbs while the pulse is on (charging), then eases back down once it’s off (discharging), each leg governed by the same $RC$.

Drag Time constant down small, then large, and watch the output follow the input pulse.

With a small time constant the output nearly traces the input pulse’s rectangle, fast charge, fast discharge. With a large one, notice the output turns into a slow, rounded rise that’s barely started falling by the time the pulse has long since ended, with Voltage when the pulse ends dropping further below the pulse’s full height the larger the time constant gets.

Convolving with a shifted spike

Back on the impulse and step page, the sifting property told you that multiplying a signal by $\delta(t-t_0)$ and totalling the running area picks out exactly one value, $x(t_0)$. Convolution does that same sifting at every instant $t$ at once: $x(t)*\delta(t-t_0)=x(t-t_0)$. Convolving any signal with a perfectly narrow, unit-area spike sitting at $t_0$ just delays that signal by $t_0$, with no other change at all, because the spike is too narrow to blur together any two nearby values the way a wider pulse (like the RC circuit’s decaying exponential) does.

Drag Delay across its whole range and watch the bright curve.

The bright curve is the faint original curve, picked up and set back down, never stretched, flattened or blurred. Notice the Shifted peak value readout always reads out the original signal’s own peak value, however far you’ve slid it.

Worked example

Take two equal rectangular pulses, $x(t)=h(t)=1$ for $0\le t\le1$ (else $0$). Their convolution is a triangle: $y(t)=t$ for $0\le t\le1$, then $y(t)=2-t$ for $1\le t\le2$, peaking at $y(1)=1$. As a check, the total area under a convolution always equals the product of the two signals’ areas: a triangle of base $2$ and height $1$ has area $\tfrac12(2)(1)=1$, matching (area of $x$) $\times$ (area of $h$) $=1\times1=1$.

Now an RC circuit with $RC=1$ s, driven by a 1 s rectangular pulse starting at $t=0$ (dropping the $1/RC$ prefactor since $RC=1$). While the pulse is on, $y(t)=1-e^{-t}$; at $t=1$ s, the pulse’s end, $y(1)=1-e^{-1}=0.63212$. After the pulse ends, $y(t)=e^{-(t-1)}-e^{-t}$; at $t=2$ s, one further time constant later, $y(2)=e^{-2}(e-1)=0.23254$.

Finally, take $x(t)=e^{-t}u(t)$ convolved with $\delta(t-3)$: the delay property says this is just $x(t-3)=e^{-(t-3)}u(t-3)$. At $t=5$, that’s $e^{-2}=0.13534$.

Where you’ll meet this

That RC charge-and-discharge curve comes back in Chapter 6, where you’ll get the same answer a second way, straight from how the circuit is built. The delay property you just confirmed is what later chapters use to treat a time delay as a simple multiplication instead of redoing an integral every time. And the same sliding, integrating picture shows up outside signal processing too: it’s how you work out the chances for the total of two dice-like random quantities.

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