What happens with signal in frequency spectrum when it is time shifted in time spectrum?

Question

I have some trouble to understand what is going on with signal in frequency spectrum when it is time shifted in time spectrum.

I am hoping that somebody will help me to understand that.

Thanks you very much.

Answer 1

Each frequency in the FT of a time shifted waveform is rotated in phase by an amount proportional to the frequency and proportional to the amount of time shift.

If you delay a pure sinusoid by 25% of its period, it's phase referenced to any fixed point in time will change by pi/2 radians. Delay a slightly higher frequency sinusoid by the same amount of absolute time, and it's phase will change more. So the phase change will change with frequency.

If looking at a 3d plot of the FT of a time shifted signal, it looks like taking the FT before the time shift and twisting it. The twist will be linear, e.g. proportional to frequency. The more time shift, the greater the amount twist per unit of frequency (or twist revolutions per graph width).

Answer 2

Just to give you yet another way to look at it: try to see what happens to the Fourier transform. If $X(\omega)$ is the Fourier transform (i.e. the spectrum) of $x(t)$

$$X(\omega)=\int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt$$

then the transform of a time-shifted version of $x(t)$ is

$$x(t-\tau)\Longleftrightarrow \int_{-\infty}^{\infty}x(t-\tau)e^{-j\omega t}dt= \int_{-\infty}^{\infty}x(t)e^{-j\omega (t+\tau)}dt= e^{-j\omega\tau}\int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt=e^{-j\omega\tau}X(\omega)$$

So the magnitude of the spectrum remains unchanged, whereas a linear phase term $-\omega\tau$ is added to the original phase.

Answer 3

The magnitude of the FT is unaffected by a time shift of x seconds, i.e. for $f(t-x)$ the phase of the FT is multiplied by $\exp{(-j2 \pi f x)}$.