Given are two cosines according to the following formula $x_i(t) = cos(2\pi f_i t)$ with $f_1 = 1Hz$ , $f_2 = 2Hz$ and $f_3 = 3Hz$ .

The two cosines are delayed by $\tau=0.1s$ to yield $y_i(t) = cos(2\pi f_i(t-0.1s))$. This corresponds to a phase shift and the delayed cosines can also be written as $y_i(t) = cos(2\pi f_it + \phi_i)$

Calculate the phase shifts $\phi_i$ for each cosine and verify that this corresponds to Time Shift theorem of Fourier Transform.

My work:

I found online some formula that supposed to calculate the $\phi_i s$. It is written like this $\phi_i=\tau *f *2\pi$ and calculated that $\phi_1 =\frac{2\pi}{10}$, $\phi_2 =\frac{4\pi}{10}$ and $\phi_3 =\frac{6\pi}{10}$.

Time Shift Theorem say If the original function g(t) is shifted in time by a constant amount, it should have the same magnitude of the spectrum, G(f). That is, a time delay doesn't cause the frequency content of G(f) to change at all. This should make sense. Since the complex exponential always has a magnitude of 1, we see the time delay alters the phase of G(f) but not its magnitude.

So the phase for these examples have changed but not the magnitude.

First of all, are the calculations and the formula correct? Does my arguments make sense for the Time shift theorem in regards to these three examples?

Could someone please explain what is the difference between the time delayed signal and the phase shifted signal?

Any help is much appreciated! Thanks!


1 Answer 1


Any signal $x(t)$ can be time-shifted: simply calculate $x(t + \Delta t)$.

A sinusoid can also be phase-shifted. Consider the cosine signal with phase $\phi$: $$x(t) = \cos(2\pi f_0 t + \phi).$$ Now, time shift it: $$x(t + \Delta t) = \cos(2\pi f_0 (t + \Delta t) + \phi) = \cos(2\pi f_0 t + 2\pi f_0 \Delta t + \phi).$$ The phase of this delayed cosine is $2\pi f_0 \Delta t + \phi$. The takeaway here is: for periodic sinusoids, a time-shift has a direct and straightforward relationship with a phase shift, and vice-versa. This is also true for complex sinusoids $x(t) = \exp(j2\pi f_0 t + \phi)$.

The definition of phase for non-sinusoidal signals is not as simple as that of sinusoids. For example, many signals can be written in the form $A(t)e^{j\phi(t)}$ where $A(t) > 0$ and their phase is defined as $\phi(t)$. Here, a time shift of $\Delta t$ results in a new phase $\phi(t + \Delta t)$. See a full discussion here and also here.

As an example of a slightly more complicated relationship between time shift and phase shift, consider the signal $$x(t) = \cos(2\pi f_0 t + \phi_0) + \cos(2\pi f_1 t + \phi_1).$$ The delayed signal is $$x(t - \Delta t) = \cos( 2\pi f_0 t + 2\pi f_0 \Delta t + \phi_0) + \cos(2\pi f_1 t + 2\pi f_1 \Delta t + \phi_1).$$ You can see that the time delay resulted in a different phase shift for each of the sinusoidal components of $x(t)$. Fourier tells us that all signals are made up of sums of sinusoids, and each one of them has a phase, so this approach can be generalized to all signals, even non-periodic ones, whose Fourier transform is a continuous sum of sinusoids.

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    $\begingroup$ I'll add that I agree with @Dan and we can (and do in SSQ) define phase for an even more general class of signals, with important distinctions from a simple time shift. $\endgroup$ Commented Apr 1, 2022 at 16:13

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