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Suppose we have 2 identical gaussian pulse signals in time domain, offset by time delay $\tau$. After taking fourier transform of both into frequency domain, I want to phase shift one of them such that when I do an inverse FT back, the two pulses are now matching. I know I can do this using fourier shift theorem,

$f(t-\tau) = e^{-i \tau \omega} * \mathcal{F}(f)$.

If I multiply $e^{i \tau \omega}$ then IFT, I'll get the offset pulse to no longer be offset.

I've tried this on 2 offset sinusoids, it works. Question is, If I have two single gaussian pulses, what will $\omega$ be? For sinusoid example, it's straightforward but what would be the "frequency of a gaussian pulse"?

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  • $\begingroup$ Would you mind writing what $f(t)$ is for this gaussian pulse? is it just $f(t)=e^{-t^2}$ or not? $\endgroup$ – TSF Jun 28 at 7:49
  • $\begingroup$ It's not. It's just the pdf of a gaussian times some amplitude $\endgroup$ – MinYoung Kim Jun 28 at 8:29
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If you want to obtain $f(t - \tau)$ form the fourier transfor of $f$, say $\hat f$, you should multiply it by $\hat g(\omega) = e^{i\omega \tau}$ in the transformed domain. In this case, you should multiply the two functions $\hat g, \hat f$ pointwise. Obviously the product will be zero outside the support of $\hat f$. For the case of a sinusoid the support is only its frequency.

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  • $\begingroup$ I don't think you understand what I am asking. I am asking, for g($\omega$), what will $\omega$ be? $\endgroup$ – MinYoung Kim Jun 28 at 9:54
  • $\begingroup$ $\omega$ will not have a fixed value since it ranges in $\mathbb R$ which is the domain of $\hat f$. It is equal to the frequency $\omega_0$ of the sinusoid when $f$ is a sinusoid because its fourier transform is a delta function, which is zero for every $\omega \neq \omega_0$ $\endgroup$ – giovanni gajac Jun 28 at 10:53
  • $\begingroup$ dsprelated.com/freebooks/mdft/DFT_Definition.html I guess this is what I was looking for. $\omega_k = 2 \pi k / N$, with $N$ = number of samples, $k$ = 0,1,2 ... N-1, so k = np.arange(N) on python. $\endgroup$ – MinYoung Kim Jun 28 at 11:09

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