# Convolution

In the article on the transfer function we have seen, that the relation between the input $x$ and the output $y$ of a system can be characterized by the transfer function $G$. We denote the Laplace transform of some quantity with capital letters. For zero initial conditions we can write \begin{align} G(s)=\frac{Y(s)}{X(s)} \end{align} in this case the relation between the input and the ouput is given by the equation. \begin{align} Y(s)=G(s)X(s) \end{align}

The convolution of $x$ and $g$ is denoted $x*g$ and is defined as the integral \begin{align} y(t)=\int_0^t x(\tau) \cdot g(t-\tau) d\tau \end{align}

The most important properties of the convolution are:

• Commutativity

$x * g = g * x \, \tag{1}$

• Associativity

$x * (g * h) = (x * g) * h \tag{2}$

• Distributivity

$x * (g + h) = (x * g) + (f * h) \,\tag{3}$

• Multiplicative identity

$x * \delta = x \,\tag{4}$

Furhter more associativity with scalar multiplication holds $a (x * g) = (a x) * g \, \tag{5}$ for any real (or complex) number $${a}\,$$.

## Convolution theorem

Let us now study the Laplace transfrom of the convolution \begin{align} Y(s)&=\int_0^{\infty} e^{-st} dt \left\{ \int_0^{\infty} x(\tau) \cdot g(t-\tau) d\tau \right\}\\ &=\int_0^{\infty} g(t-\tau) \, e^{-st} dt \left\{ \int_0^{\infty} x(\tau) d\tau \right\} \end{align} now we make the variable transformation $u=t-\tau$ to get \begin{align} &=\int_0^{\infty} g(u) \, e^{-s(u+\tau)} du \left\{ \int_0^{\infty} x(\tau) d\tau \right\}\\ &=\int_0^{\infty} g(u) \, e^{-su} du \, \int_0^{\infty} x(\tau) \, e^{-s\tau} d\tau \\ \end{align}

This fundamental result is called the convolution theorem and also holds for the Fourier transform of the convolution.

The convolution of two functions in the time domain is equivalent with their multiplication in the frequency domain. \begin{align} G(s)X(s)=\int_0^{\infty} \left\{ x*g \right\} \, e^{-st} dt \\ \end{align} Let $\mathcal{F}$ denote the (Fourier or Laplace) transfrom, then we can write the convolution theorem as $\mathcal{F}\{f*g\} = \mathcal{F}\{f\} \cdot \mathcal{F}\{g\}$

It also works the other way around:

$\mathcal{F}\{f \cdot g\}= \mathcal{F}\{f\}*\mathcal{F}\{g\}$

By applying the inverse Fourier transform $$\mathcal{F}^{-1}$$, we can write:

$f*g= \mathcal{F}^{-1}\big\{\mathcal{F}\{f\}\cdot\mathcal{F}\{g\}\big\}$