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I would like to shift in frequency domain the following function: $f(t)=\frac{1}{\sqrt{2\pi}\sigma}\exp(-\frac{(t-t_0)^2}{2\sigma^2})$.

As usual, frequency shift will introduce a new term $\exp(-j\omega_0t)$ and the new function will become (in time domain) $f(t)=\frac{1}{\sqrt{2\pi}\sigma}\exp(-j\omega_0t)\exp(-\frac{(t-t_0)^2}{2\sigma^2})$. Why this will not become $f(t)=\frac{1}{\sqrt{2\pi}\sigma}\exp(-j\omega_0(t-t_0))\exp(-\frac{(t-t_0)^2}{2\sigma^2})$ as expected if we first shift in frequency and then shift in time. What's wrong in the procedure? I assume that when shifting in time after shifting in frequency I have to change only the exponential argument of the gaussian $-t/2\sigma^2$

dimpep
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1 Answers1

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What you call frequency shift is actually partly undetermined because the phase is neglected. Exchanging the order of the two operations merely makes this more evident: the arbitrary phase factor chosen in one procedure differs from the arbitrary phase factor in the other procedure by a phase $\omega t_0.$ Note that this applies to any function $f(t),$ not just Gaussians.

The indeterminacy of the frequency shift becomes clearer if we consider a pure wave signal $f(t)=\exp(i\omega t).$ Multiplying by $\exp(i\omega_0t)$ does not merely change the frequency to $\omega+\omega_0,$ it also causes a change of phase, depending on the point of reference $t,$ with magnitude $\omega_0t.$ For example, if you wanted your signal to change its frequency from a specific time $t_0$ onwards then in order for the signal to remain continuous you would need to multiply by $\exp(i\omega_0(t-t_0))$ to compensate for the phase shift at $t_0.$

Justpassingby
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  • Thank you for the answer. I can't understand why this frequency shift is partially undetermined. I can understand that the phase is neglected in time domain but the frequency shift is also there – dimpep Jan 29 '16 at 08:46
  • One thing I cannot understand is how the shifts are working. For example, for a time shift in $f(t)$ (which essentially changes t to $t-t_0$ in frequency domain this shift reads as $\exp(-j\omega t)$. If then make another shift in frequency by $\omega_0$, in time domain we have to multiply by $\exp(j\omega_o t)$ or by $\exp(j \omega_o (t-t_0))$? Because after the first time shift, the variable is $t-t_0$. This point here confuses me. – dimpep Jan 29 '16 at 09:44
  • A frequency shift is not a uniquely determined operation, it is only completely determined if it is accompanied by a precise phase shift. One way to make the choice unambiguous is by requiring that the phase is not changed at time $t=t_0.$ Another way, equally valid but incompatible with the former, is to require that the phase is not changed at time $t=0.$ Depending on the choice you would multiply by $\exp(j\omega_0(t-t_0))$ or by $\exp(j\omega_0t).$ – Justpassingby Jan 29 '16 at 10:01
  • I assume that changing the phase equals to multiplication with $\exp(j\omega_0 t)$ and not changing the phase with $\exp(j\omega_0 (t-t_0))$ – dimpep Jan 29 '16 at 10:20
  • Changing the frequency equals multiplication with either of these, depending on your choice of a reference time where the phase should not change. – Justpassingby Jan 29 '16 at 10:48
  • So, If I decide to swift the time by $t_0$, this also has to be applied on the $\exp(j\omega t)$ term too or only on $\exp(\frac{-(t-t_0)^2}{2\sigma^2})$? I know that the time shift is just a phase shift but I am confused where to apply this – dimpep Jan 29 '16 at 12:50
  • If you apply the time shift to the $\exp(j\omega t)$ factor, as well, then you indicate that the frequency shift should respect the phase at the origin of the new time reference frame. If you don't, you indicate that the frequency shift should respect the phase at the origin of the old time reference frame. – Justpassingby Jan 29 '16 at 14:39
  • But what is the most correct? Or both are equal? – dimpep Jan 30 '16 at 16:50