I have read in many books that whether the signal is 1D or multidimensional ,

  1. The magnitude spectrum tells you how strong are the harmonics in the signal


  2. The phase spectrum tells where this harmonic lies in time domain for 1 D signal (and in space domain in case of multidimensional)

But I didn't find any justification or explanation for the above sentences. I want to counter check (understand ) these sentences about phase spectrum and magnitude spectrum. So can anybody help for it ?

  • $\begingroup$ These are complex images with no peculiar characteristic such as dominant frequencies, and they are fairly isotropic. All you can observe is the decay rate of the amplitude, similar in both cases. And there's nothing you can interpret in the phase image. $\endgroup$ – Yves Daoust May 12 '15 at 15:49
  • $\begingroup$ @Yves Daoust sir,i have edited the question,can you give answer for the question with the help of mathematics or any other way? $\endgroup$ – pandu May 12 '15 at 16:14
  • $\begingroup$ This question is too broad and is better addressed by the numerous freely available texts on the matter. $\endgroup$ – AnonSubmitter85 May 12 '15 at 17:17

The phase tells you nothing about localization. Every sine and cosine is global. What phase tells us is a spatial offset to each wave. You will need a "short time" or "windowed" fourier transform to achieve temporal or spatial locality. Or you can use another transform like a Wavelet transform which gives a tradeoff between "frequency" and spatial/temporal information.

  • $\begingroup$ This is not correct. Phase can indeed tell you something about localization. For example, in $e^{j2\pi f \alpha + j\pi \beta f^2}$, $\alpha$ will tell you where and $\beta$ will tell you how wide (or how localized). $\endgroup$ – AnonSubmitter85 May 12 '15 at 17:46
  • $\begingroup$ 1. You probably mean to have - sign on the second term in the exponent. if so, then 2. Multiplication with a gaussian is a kind of "windowed transform" just as I mentioned in the answer. $\endgroup$ – mathreadler May 12 '15 at 18:15
  • $\begingroup$ What I meant was that looking at each dual basis function individually you can't say anything about localization if you take a global Fourier Transform. No localization information in the individual basis functions. $\endgroup$ – mathreadler May 12 '15 at 18:17
  • $\begingroup$ The sign of the phase doesn't matter and it doesn't have to be windowed (anymore than an image is already windowed given its finite extent). You can work out the IFT of the above example using Fresnel integrals and you'll see that the extent is directly proportional to $\beta$. Perhaps I am reading things differently, but the question is asking about information from the phase of the transform, which implies the phase of all frequency samples in totality and not just a single value. Either way, the question is a poor one to begin with. It's too general and better addressed by a text book. $\endgroup$ – AnonSubmitter85 May 12 '15 at 18:31
  • $\begingroup$ Phase does not say anything about spatial localization. Say you did FFT of a whole music song, you would not be able to say where the A tunes are by just looking at the spectrum. You would need a short-time fourier transform or other transforms with better temporal resolution for that. $\endgroup$ – mathreadler May 13 '15 at 11:31

Ok, another attempt at explaining the basics of FFT phase and translation / time-shift.

For any (additive) part of the signal $f(x)$ which has fourier transform $$F\{f(x)\}(\omega)$$ is shifted to $f(x+x_0)$ will have Fourier transform : $$e^{-i x_0\omega}F\{f(x)\}(\omega)$$ So every coefficient $\omega$ has it's phase altered by the linear factor inside that exponential function.


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