2

I'm currently trying to do some Fourier transformations, or at least trying to understand them. The only thing I'm worried about is the complex part of the function. All I have is some basic, self thought, understanding about complex numbers.

As far as I'm concerned, a complex number exists of a real part and a complex part. In the following DFT it seems it has no real part. Is that correct?

alt text

I think we can simplify it to the following, since the others are just modifiers.

$e^{-i}$

But for some reason, im not getting the good value's. Perhaps i'm doing something wrong, but im not quite sure.

Edit:

I just realised this may be a bit vague, but my questions are

  1. Does the function F(k,l) return a complex number, with no real part.
  2. Is there a special rule about $e^{-i}$

Edit2:

Okay, I'm sorry for asking these basic questions, this is all way beyond the math I've learned at school xD. But I understand most of the DFT now. Only there's one thing I don't understand and that:

  1. What does the $F(k,l)$ part of the function define.
Timo Willemsen
  • 1,637
  • You should edit that to read a real part and an imaginary part. Complex is the number as a whole. – Raskolnikov Dec 01 '10 at 18:17
  • If you give your DFT/FFT routine an array whose elements are $f(i,j)$, the routine will return an array where each $f(i,j)$ is replaced by $F(i,j)$ (i.e., your data's Fourier transform). – J. M. ain't a mathematician Dec 03 '10 at 08:51
  • Okay, since I don't see how that practically works. Lets say I have a matrix with 4 entries in it. So I have F(0,0), F(1,0), F(0,1) and F(1,1). For every pixel in the matrix, I'd have to calculate the DFT. Doesn't it give it a complexity of $O(N^4)$ instead of $O(N^2)$ – Timo Willemsen Dec 03 '10 at 08:54

1 Answers1

4

Your sum has both real and imaginary parts. Because

$$e^{ix}= \cos(x) + i \sin(x) \; .$$

Is $f(a,b)$ in your formula a real number? Then the transform has real and imaginary components in general. (For certain choices of $f(a,b)$, you can have purely real numbers.)

If you mean $e^{-i}$, then use the formula I provided you and you should be able to compute it.

Raskolnikov
  • 16,108
  • Thanks a lot, I wasn't aware that $e^{ix} = cos(x) + i sin(x)$ – Timo Willemsen Dec 01 '10 at 18:54
  • 1
    @Timo: You might be interested in this other question: http://math.stackexchange.com/questions/3510/how-to-prove-eulers-formula-expi-t-costi-sint. – Hans Lundmark Dec 01 '10 at 19:22
  • 3
    @Timo: Additionally, if $f$ is real, $F$ will have a number of redundant components; you can in fact exploit the symmetry of FFT with real inputs so that only the non-redundant parts are computed (and the computation time is at least halved). – J. M. ain't a mathematician Dec 02 '10 at 11:04
  • @J.M. Thank you, I'm really interested in that. I've rewritten the function as following:

    $\cos(\pi({ka/n}+{lb/n})) - i \sin(\pi({ka/n}+{lb/n})$

    But however the part inside the $\cos$ and $\sin$ will have to be calculated every iteration right?

     – Timo Willemsen Dec 03 '10 at 06:56
  • 2
    @Timo: Nope. Try constructing a table of sines and cosines and you'll see for yourself that there's an awful lot of repeated data (only differing up to sign). That's because of the relationships satisfied by the sine and cosine. – J. M. ain't a mathematician Dec 03 '10 at 07:11
  • Very cool ^^ I'll go and check it out. Thanks a lot, this is really fun ^^ – Timo Willemsen Dec 03 '10 at 07:29
  • Okay I've created a table, and I've noticed that both the $cos(x)$ and the $sin(x)$ part are just repeating. So I could just create a set of constants (depending on the $N$) – Timo Willemsen Dec 03 '10 at 07:49
  • 1
    In fact, quite a number of well-tuned FFT programs internally cache a bunch of frequently-used trig function values in an array, since array accesses are usually cheaper than computing trig functions ab initio. – J. M. ain't a mathematician Dec 03 '10 at 08:46