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A homework problem of a free online course I am taking, asked to draw the magnitude and phase of $Y(\omega)$ where $y(t) = x(t) c(t)$ where $c(t)$ is $e^{j 3 \omega_{c} t}$ and where $X(\omega)$ is:

enter image description here

Simple phase modulation problem with a pretty simple exponential carrier.

Doing the Fourier Transform of the exponential carrier I got that it was an impulse of height $2 \pi$ at $\omega = 3 \omega_{c}$

Modulation (multiplication) equation in the frequency domain turns into $\frac{1}{2 \pi} X \left( \omega \right) * C \left( \omega \right)$ where $*$ is convolution.

(The $2 \pi$ cancels out the height of the impulse.) A frequency domain signal convolved with an impulse is just the signal moved over. So, my $\vert Y \left( \omega \right) \vert$ was just the $\vert X \left( \omega \right) \vert$ in the picture moved over where it centres at $3\omega_{c}$ like in the next picture.

For phase I broke up the equation into $Y \left( \omega \right) = \frac{1}{2 \pi} \left( 2 \pi e^{j0} \right) * \left( \vert X \left( \omega \right) \vert e^{j \angle X \left( \omega \right)} \right)$ because the carrier transform $C \left( \omega \right)$ has a zero phase.

The homework solutions showed the following as the right answer for $Y \left( \omega \right)$ magnitude and phase, and I'm not sure why the phase is shifted. My intuition wants to tell me that the phase will match that of the original input signal because the only other phase in the equation is zero, so I would have thought the phase of $Y \left( \omega \right)$ would be the same as of $X \left( \omega \right)$.

enter image description here

I understand the rule about convolving with an impulse shifting the signal, but I guess I don't understand how it would influence the phase. Does that rule apply to just the magnitude or the full signal including the phase? What about convolution does that for the phase?

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Does that rule apply to just the magnitude or the full signal including the phase?

The rules applies for the entire signal (magnitude, phase, real part, imaginary part, phase delay, group delay, level, etc.)

Does convolution do that for the phase?

Yes.

It's simply the shift property of the Fourier Transform. If $x(t) \leftrightarrow X(\omega)$ then

$$x(t) \cdot e^{j\omega_0 t} \leftrightarrow X(\omega-\omega_0)$$

It also follows directly from the convolution

$$f(t)*\delta(t-t_0) = f(t-t_0)$$

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  • $\begingroup$ Thank you for your response. From the sifting property it now seems plausible how the overall function, magnitude, phase, real part, and imaginary part that are a function are shifted. Someone edited my question which was "What about convolution does that to the phase?" I tried to go through a simple scenario in a notepad of the sifting property and I now understand why the phase is shifted. The phase is zero in those places where there is no overlap between the signal and where the impulse is non-zero 0*|X(w)|e^jphase(X(w)) = 0 Does that mean sifting doesn't quite exactly apply to the phase? $\endgroup$
    – Yulia
    Oct 27, 2023 at 20:15
  • $\begingroup$ To add to my question in the last comment, could it be that the first diagram was misleading to begin with and you can't have an existing phase outside of where the magnitude of the exponential, here |X(w)|, is non-zero? Or, are there some properties of zero magnitude and non-zero phase that I'm not aware of? $\endgroup$
    – Yulia
    Oct 27, 2023 at 20:22
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    $\begingroup$ @Yulia: for any complex number that has a magnitude of zero, the phase is "undefined". There isn't really a good way of drawing "undefined" so I don't think the diagram is "misleading". There just isn't a good way of drawing this correctly. $\endgroup$
    – Hilmar
    Oct 27, 2023 at 22:49

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