# Phase Spectrum

1. Aug 12, 2010

Hi!

I'm trying to understand how do i get the phase spectrum from a Fourier Transform. From this site

http://sepwww.stanford.edu/public/docs/sep72/lin4/paper_html/node4.html#lin4_swhfactm

this statement

"The phase spectrum is usually calculated by taking the arctangent of the ratio of imaginary to real parts of the Fourier transform."

Yeah, right. So, when i have a sum of functions that are Fourier Transforms, how do i know which is the real part and the imaginary part of the entire sum??????

2. Aug 13, 2010

### Petr Mugver

$$Re[f(k)+g(k)]=Re[f(k)]+Re(g(k))$$

and the same with imaginary part. Does this answer the question, or I misunderstood something?

3. Aug 13, 2010

it answer half of the things i asked. The unanswered part is how do i get the imaginary and the real parts of any fourier transform???

4. Aug 13, 2010

### Petr Mugver

I think I'm missing something. Ther real and imaginary part of a complex function are taken the same way you do for numbers... try giving an example so we can see where's the problem...

5. Aug 13, 2010

i have this function

[PLAIN]http://j.imagehost.org/0556/fun_ao_pre_trans_fourier.png [Broken]

the fourier transform is

[PLAIN]http://j.imagehost.org/0286/fun_ao_trans_fourier.png [Broken]

now, how do i get the real and the imaginary parts??

Last edited by a moderator: May 4, 2017
6. Aug 13, 2010

### Petr Mugver

Ah ok, the problem is then more about complex numbers than with Fourier transforms. You have to put the Fourier transform in the form a + jb, with a and b real numbers. Then a will be the real part and b the imaginary part. You go by steps:

1) The exponential decomposes like

$$e^{j\theta}=\cos\theta+j\sin\theta$$

2) The inverse of a complex number is

$$\frac{1}{a+jb}=\frac{a-jb}{a^2+b^2}$$

3) Oh and finally recall that $$j^2=-1$$ !!!!!!!

Using these 3 rules, you can, with a bit of patience, write your expression like a + jb. Try it yourself, if you don't get it we'll see.

7. Aug 24, 2010