# Fourier Transform question, is it correct to remove the correcting factor?

1. Feb 19, 2010

### thecake

My math professor doesn't include the correcting (normalizing factor) neither in the Fourier transform nor in the inverse Fourier! He says that it's optional.
It is weird because I have seen it used in all the text books! Is it a big deal?

He defines the Fourier transform as:-

http://img685.imageshack.us/img685/3822/56167015.jpg [Broken]

And inverse as:-

http://img218.imageshack.us/img218/8430/25649157.jpg [Broken]

Last edited by a moderator: May 4, 2017
2. Feb 19, 2010

### xepma

You might want to take another look at the Fourier transforms in those pictures... They don't make sense.

As for your answer: yes, it does matter. The constant in front is a normalization factor. If you have a function f(x) and you Fourier transform it, followed by the inverse transform, then you should expect to get the same function back, f(x). But without the normalization factor you will always get back $2\pi f(x)$ -- which is why you need the normalization in front. Just try it out with some function like $\sin(\pi x)$

You can also use the following conventions

$$F(\omega) = \int f(x) e^{-2\pi i \omega x} dx$$
and for the inverse
$$f(x) = \int F(\omega) e^{2\pi i \omega x} dx$$

No normalization is needed, but the $$2\pi$$ still pops up somewhere.

3. Feb 19, 2010

### thecake

Well, that's what I have written in my notes, what's wrong?
Omega is the angular frequency
Though he never substitutes it.

So, should I go have a talk with him?

But the questions are multiple choice, so if I use a different form I'd get a different answer :(

Last edited by a moderator: Apr 24, 2017
4. Feb 19, 2010

### xepma

I'm afraid they can't be correct -- really. In the first you integrate over t, while the function f is a function of x. That means you can just pull this function out of the integral! The same goes for the inverse: you integrate over t, while the function F is a function of x... it simply doesnt make sense.

So:
$$\int f(x) e^{i\omega t} dt = f(x) \int e^{i\omega t} dt$$

Do you see why this is true?

If you're notes are incorrect, then you're teacher might still be right, depending on his conventions for the transform. It's best to ask him what conventions he uses precisely. I can assure you that what you've written here is either incorrect, or some information is missing.

:/

Last edited by a moderator: Apr 24, 2017
5. Feb 19, 2010

### thecake

You're right it is a function of (t) xD! Silly me. I've just checked.
So is he right after all?