# Necessary and sufficient conditions: Fourier Transform

1. Sep 17, 2008

### klondike

I don’t know if the question belongs to engineering or math but here it goes.
I was taught that a sufficient (not necessary) condition for existence of Fourier transform of f(t) is f(t) is absolutely integratble. I was wondering what are the “necessary and sufficient conditions” for FT of f(t) to exist. Some textbook said “it can be very involved” but what is that?

2. Sep 18, 2008

### morphism

This depends on what you mean by "Fourier transform." Usually (but there are many exceptions), the Fourier transform F is defined on the space L^1 of absolutely integrable functions. Are you trying to find the 'largest' domain for F?

3. Sep 18, 2008

### klondike

The Fourier Transform which I was taught is:

$$F(\omega)=\int _{-\infty}^{+\infty}f(t)e^{-j\omega t}dt$$
the one that I use to convert signals from timedomain to analyze and process in freq domain. All physically realizable signals that I have to deal with are "absolutely integrable".

Other forms of FT that I'm familiar with are those 2D or 3D FT that are used to solve linear multi-dimensional PDEs.

Can you please show an example of exceptions or point me to some URLs for further reading?

No. I was trying to find a signal (function of time) that doesn't satisfy "absolutely integrable" condition, but some sort of FT exists for such signal, for the sake of curiosity. As an analytical electrical engineer, I love engineering math in general but certainly can't afford much time digging into hardcore math.

thanks again.

Last edited: Sep 18, 2008
4. Sep 19, 2008

### morphism

Hmm. I don't know if this is what you're looking for, but try to look into Fourier transforms of distributions.

5. Sep 21, 2008

### klondike

Hmm, let's see. My problem is to find a function that is NOT "absolutely integrable" but its fourier transform exists. If I can find such function, that means "absolutely integrable" is not necessary condition for FT to exist.

Given the problem, let's say

$$F(\omega)=\int_{-\infty}^{+\infty}f(t)e^{-j\omega t}dt$$
$$f(t)=\dfrac{1}{2\pi}\int_{-\infty}^{+\infty}F(\omega)e^{j\omega t}d\omega$$

let
$$f(t)=\cos{\Omega t}$$
And clearly, f(t) is not absolutely integrable, but its FT does exist:

$$F(\omega)=\pi \delta (\omega -\Omega)+ \pi \delta (\omega+\Omega)$$

Make sense?

6. Sep 21, 2008

### Hurkyl

Staff Emeritus
That depends very much on what you mean by 'exist'! At the most elementary level (i.e. the math you learned in your basic calculus sequence in college), that Fourier transform doesn't exist.

To be able to use 'generalized functions' (such as the Dirac delta), you have to set up some additional mathematical scaffolding. And then to use the Fourier transform on them (or even things like derivatives), that's a bit more scaffolding.

If Wikipedia is to be trusted, if the Fourier transform is of primary interest, then the right scaffolding is the space of tempered distributions. The natural definition for the Fourier transform of a tempered distribution is implicitly through the following integral equation:

$$\int_{-\infty}^{+\infty} \mathcal{F} \{ f \} \phi = \int_{-\infty}^{+\infty} f \mathcal{F} \{ \phi \}$$

where f denotes any tempered distribution and $\phi$ denotes any 'well-behaved' function. It turns out that the Fourier transform of a tempered distribution is a tempered distribution -- so in this context, Fourier transforms always exist.

Every 'somewhat well-behaved' function can be viewed as a tempered distribution -- the basic requirement is that it doesn't grow 'too fast' at infinity. And every tempered distribution can be written as a limit of distributions that came from somewhat well-behaved functions. (e.g. you do this when you write the Dirac delta as a 'limit' of ordinary functions)