Do you know if there is a relation between the phase of the Fourier Transform and the phase of the Analytic signal?

I mean, if i have a signal s(t) then...




phaseFourier= ?(phaseAnalytic)


  • 1
    $\begingroup$ As Peter has said - you're are comparing the phase in the frequency domain (from the FFT) with that of the Analytic signal in the time domain. There are several issues - one is your Hilbert transform is going to have a delay (possibly non-linear phase). The phase in time is instantaneous so it is very sensitive to noise. The phase in the frequency domain is only evaluated in specific frequency points. In general these will not be the same. $\endgroup$
    – David
    Dec 21, 2020 at 20:45
  • $\begingroup$ @David Thanks for your comment. Yes, I agree with what you have said. I am aware of the fact that the phase of the Fourier transform and the phase of the Analytic signal are different. Given that I understand this, I was asking if there is some sort of relation between them. I mean, if there is some operator that produce one from the other. Apparently there is no such a thing, as most comments below do not address the question about this relation. $\endgroup$ Dec 22, 2020 at 2:17

1 Answer 1


Discussing the relationship between the Fourier transform and the Hilbert transform, let us consider an example function. It can be any square integrable function; for readability of the text, we select a function with both transforms expressible via simple analytical functions.

Let a signal s be a sinc function of time, s(t) = sinc(t). The Fourier transform of the sinc function is a rect function $$ \mathscr{F}\{s(t)\}(ω)={{1\over{\sqrt{2π}}}·rect\left({ω\over{2π}}\right)} $$ which is a function of frequency: OP's FT_s is a function of frequency, or, for the discrete case, an array with indices representing frequency bins.

The Hilbert transform of the sinc function $$ \mathscr{H}\{s(t)\}(t)={{1-cos(t)}\over{t}} $$ is the function of time, because the Hilbert transform is not a transform of the kind that maps time domain functions to frequency domain functions -- as the Fourier transform does -- but "a specific linear operator that takes a function, u(t) of a real variable and produces another function of a real variable H(u)(t)" (citation from the Wikipedia article on the Hilbert transform).

Comparing these formulas, one sees that the question "if there is a relation between the phase of the Fourier Transform and the phase of the Analytic signal?" is rather pointless.

FT_s, as defined by OP, is an array of values against frequency bins, and atan(Imag(FT_s),Real(FT_s)) is also an array indexed by the same bin enumerables, corresponding to a continuous-time frequency. Therefore, atan(Imag(FT_s),Real(FT_s)) is an array of phases of the signal frequency components.

A_s is a function of time, and atan(Imag(A_s),Real(A_s)) is also the function of time. It is the instantaneous phase of an analytic signal, and it is not the phase of the Fourier transform frequency component, not in any way.

The OP-defined two continuous-variable "phase functions" and corresponding discrete "phase arrays", phaseFourier and phaseAnalytic, are defined on different domains: phaseFourier is defined on a frequency domain, phaseAnalyticis defined on a time domain. Still, you may want to see the graphs of these function for a sinc(t) signal.

For s(t) = sinc(t), the Fourier transform (OP's FT_s) values is a real constant up to a frequency of 1/(2π) and a zero for higher frequencies. Therefore, the array of phases of frequency components (OP's phaseFourier) is zero up to 1/(2π) and undefined in the region where the Fourier transform amplitude is zero. For sinc(t), one can safely assign all phases of FT components to zero in the entire frequency domain. The FT_s graph is trivial, and shown in every textbook and course notes on signal processing.

The instantaneous phase (OP's phaseAnalytic) graph may be shown less frequently, so here you are:

sinc instant phase

Summing up, one may agree that, talking about the relationship between the Fourier transform and the Hilbert transform, more appropriate is to indicate that the Hilbert transform is a multiplier operator. The multiplier of H is $$ σ_{\mathscr{H}}(ω)=-i·sgn(ω) $$ (see https://en.wikipedia.org/wiki/Hilbert_transform#Relationship_with_the_Fourier_transform).

  • $\begingroup$ Thank you so much for your answer and for the time you spent on this. It is so clear and complete. As I mentioned, I am aware that these two "phases" are different, and to hear from you that my question is pointless is really helpful. Thank you again. $\endgroup$ Dec 22, 2020 at 20:21
  • $\begingroup$ You're welcome. Stay tuned, gain and share your expertise through participation in dsp.SE! $\endgroup$
    – V.V.T
    Dec 23, 2020 at 4:16

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