Fourier Transform of a Product of functions

Intuition

The Fourier Transform, $\mathcal{F}\{f(t)\cdot g(t)\}\ne \mathcal{F}\{f(t)\}\cdot \mathcal{F}\{g(t)\}$ . Instead it is given by the convolution operation "$\ast$"

Mathematically

$\mathcal{F}\{f(t)\}=F(\omega )$ and $\mathcal{F}\{g(t)\}=G(\omega )$, then

$$\mathcal{F}\{f(t)\cdot g(t)\}=(F\ast G)(\omega )$$

The convolution of $F(\omega )$ and $G(\omega )$ is defined as:

$$(F\ast G)(\omega )={\int }_{-\infty }^{\infty }F(\nu )G(\omega -\nu )d\nu$$

Here, $\nu$ iterates over all possible frequency components:

• $F(\nu )$ is the value of the Fourier Transform of one function at the frequency $\nu$.
• $G(\omega -\nu )$ is the value of the Fourier Transform of the other function, shifted by $\nu$, evaluated at $\omega$.

The product $F(\nu )G(\omega -\nu )$ is then integrated over all frequencies. What this does is to sum up all possible interactions between different frequency components of the two functions, where $\nu$ allows each frequency component of $F$ to interact with a correspondingly shifted component of $G$. This “shifting” and summing up are essential to the convolution process, determining how the frequency content of one function modifies that of another.