Convolution Theorem

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Let $\mathbb F \in \set {\R, \C}$.

Let $f: \R \to \F$ and $g: \R \to \F$ be functions.

Let their Laplace transforms $\laptrans {\map f t} = \map F s$ and $\laptrans {\map g t} = \map G s$ exist.


$\map F s \map G s = \displaystyle \laptrans {\int_0^t \map f u \map g {t - u} \rd u}$

Proof 1

\(\displaystyle \laptrans {\int_0^t \map f u \map g {t - u} \rd u}\) \(=\) \(\displaystyle \int_{t \mathop = 0}^\infty e^{-s t} \paren {\int_{u \mathop = 0}^t \map f u \map g {t - u} \rd u} \rd t\) Definition of Laplace Transform
\(\displaystyle \) \(=\) \(\displaystyle \int_{t \mathop = 0}^\infty \int_{u \mathop = 0}^t e^{-s t} \map f u \map g {t - u} \rd u \rd t\)
\(\displaystyle \) \(=\) \(\displaystyle \lim_{M \mathop \to \infty} \int_{t \mathop = 0}^M \int_{u \mathop = 0}^t e^{-s t} \map f u \map g {t - u} \rd u \rd t\)
\(\text {(1)}: \quad\) \(\displaystyle \) \(=\) \(\displaystyle \lim_{M \mathop \to \infty} s_M\)

where $s_M$ is defined to be:

$\displaystyle \int_{t \mathop = 0}^M \int_{u \mathop = 0}^t e^{-s t} \map f u \map g {t - u} \rd u \rd t$

The region in the plane over which $(1)$ is to be integrated is $\mathscr R_{t u}$ below:


Setting $t - u = v$, that is $t = u + v$, the shaded region above is transformed into the region $\mathscr R_{u v}$ the $u v$ plane:



\(\displaystyle s_M\) \(=\) \(\displaystyle \iint_{\mathscr R_{t u} } e^{-s t} \map f u \map g {t - u} \rd u \rd t\) from $(1)$ above
\(\text {(2)}: \quad\) \(\displaystyle \) \(=\) \(\displaystyle \iint_{\mathscr R_{u v} } e^{-s \paren {u + v} } \map f u \map g v \size {\dfrac {\map \partial {u, t} } {\map \partial {u, v} } } \rd u \rd v\)

where $\dfrac {\map \partial {u, t} } {\map \partial {u, v} }$ is the Jacobian of the transformation:

\(\displaystyle J\) \(=\) \(\displaystyle \dfrac {\map \partial {u, t} } {\map \partial {u, v} }\)
\(\displaystyle \) \(=\) \(\displaystyle \begin{vmatrix} \dfrac {\partial u} {\partial u} & \dfrac {\partial u} {\partial v} \\ \dfrac {\partial t} {\partial u} & \dfrac {\partial t} {\partial v} \end{vmatrix}\)
\(\displaystyle \) \(=\) \(\displaystyle \begin{vmatrix} 1 & 0 \\ 1 & 1 \end{vmatrix}\)

Thus the right hand side of $(2)$ is:

$(3): \quad \displaystyle s_M = \int_{v \mathop = 0}^M \int_{u \mathop = 0}^{M - v} e^{-s \paren {u + v} } \map f u \map g v \rd u \rd v$

Let $\map K {u, v}$ be the function defined as:

$\map K {u, v} = \begin{cases} e^{-s \paren {u + v} } \map f u \map g v & : u + v \le M \\ 0 & : u + v > M \end{cases}$

This function is defined over the square region in the diagram below:


but is zero over the lighter shaded portion.

Now we can write $(3)$ as:

\(\displaystyle s_M\) \(=\) \(\displaystyle \int_{v \mathop = 0}^M \int_{u \mathop = 0}^M \map K {u, v} \rd u \rd v\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \lim_{M \mathop \to \infty} s_M\) \(=\) \(\displaystyle \int_{v \mathop = 0}^\infty \int_{u \mathop = 0}^\infty \map K {u, v} \rd u \rd v\)
\(\displaystyle \) \(=\) \(\displaystyle \int_{v \mathop = 0}^\infty \int_{u \mathop = 0}^\infty e^{-s \paren {u + v} } \map f u \map g v \rd u \rd v\)
\(\displaystyle \) \(=\) \(\displaystyle \paren {\int_{v \mathop = 0}^\infty e^{-s u} \map f u \rd u} \paren {\int_{v \mathop = 0}^\infty e^{-s v} \map g u \rd v}\)
\(\displaystyle \) \(=\) \(\displaystyle \map F s \map G s\)

Hence the result.


Proof 2

Convolution Theorem/Proof 2

Also presented as

Some sources give this as:

$\invlaptrans {\map F s \map G s} = \displaystyle \int_0^t \map f u \map g {t - u} \rd u$