# fourrier transform proof

• Apr 16th 2008, 07:50 AM
allah's_slave
fourrier transform proof
is there a proof 4 fourrier's theory;that any signal is composed of infinite number of sinusoids?

another question : what is the reason that made fourrier think of signals in that way? i mean , of course he faced some problem or had a certain idea before he put his theory
• Apr 17th 2008, 02:23 AM
CaptainBlack
Quote:

Originally Posted by allah's_slave
is there a proof 4 fourrier's theory;that any signal is composed of infinite number of sinusoids?

For a continuous absolutely integrable function $f(x)$ with absolutly integrable Fourier transform $[\mathcal{F}f](\xi)=F(\xi)$ the result follows from:

$f(0)=\int_{-\infty}^{\infty} F(\xi)~d\xi \ \ \ \ \ \ \ ...(1)$

by the translation theorem.

To prove $(1)$ you need a sequence of well behaved functions $u_n(x)$ with known Fourier transforms $U_n(\xi)$, which as $n \to \infty$ approximates the behaviour of a $\delta$ functional.

A suitable sequence can be constructed from Gaussians with decreasing spread parameters (who's FTs form a sequence of Gaussians with increasing spread parameters).

Then to prove $(1)$ you consider:

$\lim_{n \to \infty} \int f(x)u_n(x) e^{i 2 \pi \xi x}dx$ $\ = \lim_{n \to \infty }(F*U_n)(\xi)$

By construction the limit on the left hand side is $f(0)$, and on the right hand side you can exchange the limit and integral and so find it goes to right hand side of $(1)$.

(Of course this is much more direct it you run amok with generalised functions or distributions, when you use $\delta(x)$ instead of $\{ u_n(x)\}$, and $1$ instead of $\{U_n(\xi)\}$)

(And also this is modulo the odd factor depending on how the FT has been defined)

RonL