# Fourier shift of a Gaussian pulse

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"?

• Would you mind writing what $f(t)$ is for this gaussian pulse? is it just $f(t)=e^{-t^2}$ or not? – TSF Jun 28 at 7:49
• It's not. It's just the pdf of a gaussian times some amplitude – MinYoung Kim Jun 28 at 8:29

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.
• I don't think you understand what I am asking. I am asking, for g($\omega$), what will $\omega$ be? – MinYoung Kim Jun 28 at 9:54
• $\omega$ 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 $\omega \neq \omega_0$ – giovanni gajac Jun 28 at 10:53
• 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. – MinYoung Kim Jun 28 at 11:09