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Fourier expansion and Fourier transform

Fourier expansion
Let x_T(t) represent a continuous periodic function

$\begin{displaymath}x_T(t+mT)=x_T(t),\;\;\;\;\;(m=0,\;\pm 1,\;\pm 2, \cdots ) \end{displaymath}$

with its period equal to T. Its Fourier expansion is

$\begin{displaymath}x_T(t)=\sum_{n=-\infty}^{\infty} X(n)e^{j2\pi nf_0t} \end{displaymath}$

where f₀=1/T and X(n), $(n=0, \pm 1, \pm 2, \cdots)$ are the expansion coefficients defined as

$\begin{displaymath}X(n)=\frac{1}{T} \int_{-T/2}^{+T/2} x_T(t) e^{-j2\pi nf_0t} dt \end{displaymath}$

These coefficients can also be considered as the discrete, non-periodic spectrum of x_T(t)

$\begin{displaymath}X(f)=\sum_{n=-\infty}^{\infty} X(n)\delta(f-nf_0) \end{displaymath}$

where f₀=1/T is the interval between two neighboring frequency components.
Fourier transform of Non-periodic, continuous function
The Fourier of a non-periodic, continuous function x(t) is

$\begin{displaymath}x(t)=\int_{-\infty}^{\infty} X(f) e^{j2\pi ft} df \end{displaymath}$

where X(f) is the non-periodic continuous spectrum

$\begin{displaymath}X(f)=\int_{-\infty}^{\infty} x(t) e^{-j2\pi ft} dt \end{displaymath}$

These two equations are a Fourier transform pair (inverse and forward).
The convolution theorem
The convolution of two functions x(t) and y(t) is defined as

$\begin{displaymath}x(t)*y(t) \stackrel{\triangle}{=} \int_{-\infty}^{\infty} x(\... ...-\tau) d\tau = \int_{-\infty}^{\infty} y(\tau) x(t-\tau) d\tau \end{displaymath}$

It can be shown that the spectrum of x(t)*y(t) is the product of their spectra X(f) and Y(f):

$\begin{displaymath}\int_{-\infty}^{\infty} [x(t)*y(t)] e^{-j2\pi ft} dt = X(f) Y(f) \end{displaymath}$

Also, the spectrum of the product of the two functions is the convolution of their spectra:

$\begin{displaymath}\int_{-\infty}^{\infty} [x(t) \, y(t)] e^{-j2\pi ft} dt = X(f)*Y(f) \end{displaymath}$

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Ruye Wang
1999-09-28