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Suppose the signal is shifted by dt (signal 'starts' later, say after 1s instead of 0s), does that correspond to a positive or a negative phase shift df in the frequency domain?

There are certainly very detailed answers to this but I am still having trouble, especially with the sign conventions.

Thanks

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    $\begingroup$ Haven't you seen this? $\endgroup$
    – Matt L.
    Commented Feb 12, 2016 at 8:07

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According to the Fourier's transform properties, the Fourier's transform of $f(t+dt)$ would be $F(f)e(j2\pi fdt)$ So, basically, the spectrum (Fourier's transform magnitude, representing the frequency content of your signal) does not change. However, f Fourier's now has a phase-shift proportional to the frequency.

To answer your question : the phase-shift's sign is the time delay's sign.

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  • $\begingroup$ Thanks, just to be sure: so if I delay the start of my signal, that corresponds to $\ f(t-dt) $ and the Fourier transform is then $\ F(f) e(-j2πfdt) $. This means that the phase change induced by the time shift $\ dt $, as compared to the non-delayed signal, is $\ 2πfdt $, right? And $\ -2πfdt $ if the signal actually starts earlier ($\ dt $ is then negative) as compared to the non-shifted case. Thanks! $\endgroup$
    – armis
    Commented Feb 12, 2016 at 18:22
  • $\begingroup$ Almost : the phase change induced is $-2\pi fdt$ (the sign is the same as dt). Try to apply this to a simple function (e.g. cos(x)) $\endgroup$
    – MaximGi
    Commented Feb 12, 2016 at 19:26

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