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Phase Spectrum

  1. Aug 12, 2010 #1
    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. jcsd
  3. Aug 13, 2010 #2
    [tex]Re[f(k)+g(k)]=Re[f(k)]+Re(g(k))[/tex]

    and the same with imaginary part. Does this answer the question, or I misunderstood something?
     
  4. Aug 13, 2010 #3
    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???
     
  5. Aug 13, 2010 #4
    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...
     
  6. Aug 13, 2010 #5
    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
  7. Aug 13, 2010 #6
    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

    [tex]e^{j\theta}=\cos\theta+j\sin\theta[/tex]

    2) The inverse of a complex number is

    [tex]\frac{1}{a+jb}=\frac{a-jb}{a^2+b^2}[/tex]

    3) Oh and finally recall that [tex]j^2=-1[/tex] !!!!!!!

    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.
     
  8. Aug 24, 2010 #7
    ok, but how do i do with the dirac function????
     
  9. Aug 25, 2010 #8
    Treat the delta function just like an ordinary real function.
     
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