Products in $s$ appear constantly: $Y(s) = G(s)\cdot\frac{1}{s^{2} + 4}$ when a spring is driven by $g(t)$. Which function of $t$ has a product as its transform? Not the product of the two functions, but their convolution.
Key idea
The convolution of $f$ and $g$ is $$(f * g)(t) = \int_0^{t} f(\tau)\,g(t - \tau)\,d\tau,$$ and $\mathcal{L}\{f * g\} = F(s)G(s)$. Convolution is commutative: $f * g = g * f$.
Why it works
Write the product as a double integral: $F(s)G(s)$ $= \int_0^{\infty}\int_0^{\infty} e^{-s(\tau + u)}f(\tau)g(u)\,du\,d\tau$. Substitute $t = \tau + u$ in the inner integral and switch the order of integration over the region $0 \le \tau \le t$: $$F(s)G(s) = \int_0^{\infty} e^{-st}\left(\int_0^{t} f(\tau)g(t - \tau)\,d\tau\right)dt.$$ In words: the response at time $t$ adds up the input at each earlier time $\tau$, weighted by how the system responds after a delay $t - \tau$.
Worked example: Compute a convolution
Find $1 * e^{2t}$ and check with transforms.
- $\int_0^{t} 1\cdot e^{2(t - \tau)}\,d\tau$ $= \left[-\frac{1}{2}e^{2(t - \tau)}\right]_0^{t} = \frac{e^{2t} - 1}{2}$.
- Check: $\mathcal{L}\{1\}\mathcal{L}\{e^{2t}\} = \frac{1}{s(s - 2)}$ $= \frac{1}{2}\left(\frac{1}{s - 2} - \frac{1}{s}\right)$, which inverts to $\frac{e^{2t} - 1}{2}$. ✓
Answer: $\frac{e^{2t} - 1}{2}$
Worked example: An inverse transform by convolution
Find $\mathcal{L}^{-1}\left\{\frac{1}{s(s^{2} + 9)}\right\}$.
- $\frac{1}{s} = \mathcal{L}\{1\}$ and $\frac{1}{s^{2} + 9} = \mathcal{L}\left\{\frac{\sin 3t}{3}\right\}$.
- So the answer is $\int_0^{t} \frac{\sin 3\tau}{3}\,d\tau = \frac{1 - \cos 3t}{9}$.
Answer: $\frac{1 - \cos 3t}{9}$
Worked example: An integral equation
Solve $y(t)$ $= 1 + \int_0^{t} \sin(t - \tau)\,y(\tau)\,d\tau$.
- The integral is $(\sin t) * y$, so transforming gives $Y = \frac{1}{s} + \frac{Y}{s^{2} + 1}$.
- $Y\cdot\frac{s^{2}}{s^{2} + 1} = \frac{1}{s}$, so $Y = \frac{s^{2} + 1}{s^{3}} = \frac{1}{s} + \frac{1}{s^{3}}$.
Answer: $y = 1 + \frac{t^{2}}{2}$
Common mistake
Assuming $\mathcal{L}\{f g\} = FG$. Try $f = g = t$: $\mathcal{L}\{t^{2}\} = \frac{2}{s^{3}}$, but $\mathcal{L}\{t\}\mathcal{L}\{t\} = \frac{1}{s^{4}}$, which is the transform of $t * t = \frac{t^{3}}{6}$. Products in $s$ correspond to convolutions in $t$, never to plain products.
Convolution describes every linear system at rest. For $ay'' + by' + cy = g(t)$ with zero initial values, $Y = H(s)G(s)$, where $H = \frac{1}{as^{2} + bs + c}$ is the transfer function. Its inverse $h(t)$ is the response to a unit impulse, since $\mathcal{L}\{\delta\} = 1$. Then $y = h * g$ (Duhamel's principle): the output is the input, chopped into many small impulses, each followed by its own copy of $h$. Engineers measure $h$ once and predict the response to any input.