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My textbook has proved the convolution theorem as follows:

Theorem

If $f(t)$ and $g(t)$ are two functions of exponential order (so that their Laplace transforms exist), and writing $\mathcal{L} \{ f \} = F(s)$ and $\mathcal{L} \{ g \} = G(s)$ as the two Laplace transforms, then $\mathcal{L}^{-1} \{ FG \} = f * g$, where $*$ is the convolution operator.

Proof

Suppose that $f(t)$ and $g(t)$ are two functions of exponential order.

Then

$$ \mathcal{L} \{ f(t) * g(t) \} = \int_0^\infty e^{-st} \int_0^t f(\tau) g(t - \tau) \ d\tau \ dt ,$$

by the definition of Laplace transform and convolution.

Rewriting of this double integral to facilitate changing the order of integration gives

$$\mathcal{L} \{ f(t) * g(t) \} = \int_0^\infty \int_0^t e^{-st} f(\tau)g(t - \tau) \ d \tau \ dt$$

And then integrating with respect to $t$ first gives

$$\mathcal{L} \{ f(t) * g(t) \} = \int_0^\infty \int_{\tau}^\infty e^{-st} f(\tau) g(t - \tau) \ dt \ d\tau = \int_0^\infty f(\tau) \left\{ \int_{\tau}^\infty e^{-st} g(t - \tau) \ dt \right\} \ d\tau$$

[...]

I'm wondering how/why the limits of integration on the second integral changes from $0$ and $t$ to $\tau$ and $\infty$?

I would greatly appreciate it if someone could please take the time to clarify this.

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1 Answer 1

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Sketching a graph for the region of integration is a big help.

Note that the region of integration for $$ \int_0^\infty \int_0^t e^{-st} f(\tau)g(t - \tau) \ d \tau \ dt$$ is a triangular region below that line $\tau=t $ where t goes from $0$ to $\infty$ and $\tau $ goes from $0$ to $t$

When you change the order of integration on the same region, starting with $\tau$, the region is on the right side of the line $t=\tau$ so you have $\tau$ going from $0$ to $\infty$ and $t$ goes from $\tau$ to $\infty$

Thus you have $$ \int_0^\infty \int_{\tau}^\infty e^{-st} f(\tau) g(t - \tau) \ dt \ d\tau $$

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