# Time shifts and change in phase

My question here is in regards to implicit information/terminology.

Basically, whenever we say that "a shift in time means a change of phase in fourier space", is it implicit that the shift is always circular? Are there any exceptions to this? Caveats? etc?

• Are you asking about an infinite length transform (FT or DTFT) into Fourier space, or a finite length window (DFT/FFT)? – hotpaw2 Nov 15 '13 at 22:50
• @hotpaw2 Finite length DFT/FFT. Signal is a digital signal of length $N$. – TheGrapeBeyond Nov 15 '13 at 23:00

For a continuous-time signal $x(t)$ with Fourier Transform $X(f)$, the Fourier transform of $x(t-t_0)$ is $e^{-j2\pi ft_0}X(f)$, that is, the value of $X(3)$, say, gets changed to $e^{-j2\pi 3t_0}X(3)$ which has the same magnitude as $X(3)$ but a different phase.
For a discrete-time continuous-amplitude signal $x[n]$ with DFT $X[k]$ of length $N$, you have essentially discarded every piece of information conveyed to you by $x[-1], x[-2], \ldots$ and by $x[N], x[N+1], x[N+2], \ldots$ and have retained only $x[0], x[1], \ldots, x[N-1]$. If you phase-shift the DFT appropriately, then yes, you will get a circular shift of the $x[n]$ values. The result will be the linear shift of the $x[n]$ values only in the fortuitous circumstance that the sequence $x[n]$ is itself periodic with period $N$, and so the untrained observer such as myself cannot tell if that symbol that just shifted into my DFT wndow is $x[-1]$ coming from the left (linear shift) or $x[N-1]$ from the right (circular shift) because by golly they look exactly the same to me. Savvier denizens of this forum will give you more elaborate answers explaining how you can tell the difference between the two cases.