Convolution and Laplace Transforms

What convolution is

The convolution of two functions $f$ and $g$ (on $[0, \infty)$) is $(f * g)(t) = \int_{0}^{t} f(\tau) \, g(t - \tau) \, d\tau.$

Picture it as a sliding integral: for each output time $t$, you flip $g$ about the vertical axis, shift it right by $t$, multiply pointwise with $f$, and integrate. As $t$ changes, the flipped-shifted $g$ slides along and the overlap area updates — that area is the value of $(f * g)(t)$.

Why it matters

For any causal linear time-invariant (LTI) system with impulse response $h(t)$ (see #191), the response to an arbitrary input $x(t)$ is $y(t) = (h * x)(t) = \int_{0}^{t} h(\tau) \, x(t - \tau) \, d\tau.$

In other words, once you know $h$, you know everything. This is the central formula of linear systems theory — signal processing, control, electrical engineering, optics, and image processing all live here.

The Laplace convolution theorem

$\mathcal{L}\{(f * g)(t)\} = F(s) \cdot G(s).$

Convolution in the time domain ↔ multiplication in the Laplace domain. This is remarkable: the complicated sliding-integral formula becomes a simple product after transforming. It's also one of the main reasons engineers think in the frequency / $s$-domain — convolutions become arithmetic.

The given problem

Compute $(f * g)(t)$ for $f(t) = e^{-t}$, $g(t) = t$, and verify the convolution theorem.

Step-by-step computation (direct)

$(f * g)(t) = \int_{0}^{t} e^{-\tau} \cdot (t - \tau) \, d\tau.$

Split: $= t \int_{0}^{t} e^{-\tau} \, d\tau \;-\; \int_{0}^{t} \tau \, e^{-\tau} \, d\tau.$

First piece. $t \int_{0}^{t} e^{-\tau} \, d\tau = t \bigl[-e^{-\tau}\bigr]_{0}^{t} = t \bigl(1 - e^{-t}\bigr).$

Second piece (integration by parts with $u = \tau$, $dv = e^{-\tau} d\tau$). $\int_{0}^{t} \tau \, e^{-\tau} \, d\tau = \bigl[-\tau e^{-\tau}\bigr]_{0}^{t} + \int_{0}^{t} e^{-\tau} d\tau = -t e^{-t} + (1 - e^{-t}).$

Combine. $(f * g)(t) = t \bigl(1 - e^{-t}\bigr) - \bigl[-t e^{-t} + 1 - e^{-t}\bigr]$ $= t - t e^{-t} + t e^{-t} - 1 + e^{-t}$ $= \boxed{\, t - 1 + e^{-t} \,}$

Verification via the Laplace convolution theorem

Transform each piece: $F(s) = \mathcal{L}\{e^{-t}\} = \frac{1}{s + 1}, \qquad G(s) = \mathcal{L}\{t\} = \frac{1}{s^{2}}.$

Product: $F(s) \cdot G(s) = \frac{1}{s^{2}(s + 1)}.$

Partial-fraction decomposition: $\frac{1}{s^{2}(s + 1)} = \frac{A}{s} + \frac{B}{s^{2}} + \frac{C}{s + 1}.$

Multiply through and solve:

• Set $s = 0$: $1 = B \cdot 1 \implies B = 1$ (after multiplying by $s^{2}$).
• Set $s = -1$: $1 = C \cdot 1 \implies C = 1$ (after multiplying by $s + 1$).
• Match $s^{2}$ coefficient in $1 = A s (s+1) + B(s+1) + C s^{2}$: $0 = A + C \implies A = -1$.

So $\frac{1}{s^{2}(s + 1)} = -\frac{1}{s} + \frac{1}{s^{2}} + \frac{1}{s + 1}.$

Inverse-transform: $\mathcal{L}^{-1}\!\left\{-\frac{1}{s} + \frac{1}{s^{2}} + \frac{1}{s + 1}\right\} = -1 + t + e^{-t} = t - 1 + e^{-t}. \; \checkmark$

Same answer. Two routes, one truth — the convolution theorem works.

Verification of properties at boundary points

• $t = 0$: $(f * g)(0) = \int_{0}^{0} = 0$. Our formula: $0 - 1 + 1 = 0$ ✓
• $t \to \infty$: $t - 1 + e^{-t} \to \infty$ (linearly). Makes sense — $g(t) = t$ is already unbounded, and convolving with a decaying $f$ doesn't tame that.

Properties of convolution

Very much like ordinary multiplication, with a few wrinkles:

• Commutative: $(f * g)(t) = (g * f)(t)$. Proof: substitute $\sigma = t - \tau$ in the integral.
• Associative: $(f * g) * h = f * (g * h)$.
• Distributive: $f * (g + h) = f * g + f * h$.
• Scalar-linear: $(c f) * g = c (f * g)$.
• Identity: $\delta * f = f$. The Dirac delta is the multiplicative identity for convolution.
• NOT idempotent / NOT pointwise: $f * f \ne f^{2}$ in general.

Connection to the impulse response recipe

For an LTI system with impulse response $h(t)$, the step response is $s(t) = (h * u)(t) = \int_{0}^{t} h(\tau) \, d\tau,$ i.e. the step response is the integral of the impulse response. That's why impulse responses are often shown alongside step responses — they're derivatives/integrals of each other.

For the simple RC low-pass filter with $h(t) = \dfrac{1}{RC} e^{-t/(RC)}$ (a decaying exponential), convolving with any input gives a smoothed version — the classic "low-pass" behaviour. Convolution is literally the math behind averaging filters.

A slick reformulation

Via the convolution theorem, any ODE-with-zero-initial-conditions problem $L y = g(t), \quad y(0) = y'(0) = \ldots = 0$ has solution $y(t) = (h * g)(t),$ where $h$ is the impulse response of $L$ (inverse Laplace of $1/L(s)$). No need to redo the partial-fraction algebra for each new $g$ — just convolve.

Common mistakes

• Wrong integration variable. It's $d\tau$, not $dt$. $t$ is the output time (held fixed) while $\tau$ sweeps across $[0, t]$.
• Upper limit $\infty$ instead of $t$. For causal functions on $[0, \infty)$, the upper limit is $t$ because $g(t - \tau) = 0$ for $\tau > t$.
• Forgetting the reflection. Convolution uses $g(t - \tau)$ (a reflection-plus-shift), not $g(\tau - t)$.
• Confusing convolution with pointwise product. $(f \cdot g)(t) = f(t) g(t)$ — completely different. $(f * g)(t)$ is an integral.

Try it in the visualization

Slide $t$ along the time axis and watch the flipped-shifted $g(t - \tau)$ move through $f(\tau)$. The area of their pointwise product is the value of $(f * g)(t)$ — animate this area growing, peaking, and evolving as $t$ advances.