# How does shifting a signal in the time domain affect its frequency domain?

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

• Haven't you seen this? – Matt L. Feb 12 '16 at 8:07

## 1 Answer

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.

• 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! – Armantas Feb 12 '16 at 18:22
• 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)) – MaximGi Feb 12 '16 at 19:26