# Practice Question on the Nyquist rate of a signal

$x(t)= e^{-j \frac{\pi}{2}t} \frac{\sin (3 \pi t)}{\pi t} \$>

If you answered "yes", what is the Nyquist rate for this signal?

$\mathcal X (\omega) = \mathfrak {F} \Big(e^{-j \frac{\pi}{2}t}\Big) * \mathfrak F \Big(\frac{\sin (3 \pi t)}{\pi t}\Big)=2\pi\delta(\omega-\frac{\pi}{2}) * \Big(u(\omega+3\pi)-u(\omega-3\pi)\Big) = 2\pi \Big(u(\frac{\pi}{2}+3\pi)-u(\frac{\pi}{2}-3\pi)\Big)$

$\mathcal X (\omega) = 2\pi$

So this signal is not band limited.

As such, there can be no Nyquist rate for this signal.

--Cmcmican 23:30, 30 March 2011 (UTC)

Instructor's comment: Reality check: the only signal whose Fourier transform is a constant is an impulse. So your answer cannot possibly be correct. -pm

I cannot figure out how to use equation editor, so sorry this answer is not as pretty as the one above...

X(w) = F(exp(-j(pi/2)t) * F(sin((3pi)t)/(tpi))

= 2pi (delta(w + pi/2) * (u(w + 3pi) - u(w - 3pi))

= 2pi (u((w + pi/2) + 3pi) - u((w + pi/2) - 3pi))

In other words, it is the FT of a sinc function with w_m = 3pi, but shifted to the left by pi/2.  Graphed, it would look like a box with

X(w) = {       1,     -7pi/2 < w < 5pi/2

0,     else

So it is band limited, and the Nyquist rate is still 2w_m = 2(3pi) = 6pi

TA's comment: That's correct. Good job!
Instructor's comment: Actually, this answer is not correct. The maximum frequency of the signal is 7 pi/2, and therefore the Nyquist rate is 7pi. -pm

X(w) = F(e^(-jtpi/2))*F(sin(3tpi/tpi))

= 2pi(delta(w-pi/2)*[u(w+3pi)-u(w-3pi)])

= 2pi[u(w+7pi/2)-u(w-5pi/2)]

This will appear to be a box starting at w = -7pi/2 and ending at w = 5pi/2

w= max(abs(w)) = 7pi/2

Nyquist = 2wm = 7pi

I see how you can shift the box to make the max w be 3pi and Nyquist 6pi, however I simply thought this was a case where the signal was not symetric and thus Nyquist could be violated a little bit.  How do we know when to shift it and when to just leave it

--Ssanthak 11:27, 19 April 2011 (UTC)

TA's comment: Well, in this problem the given signal is a complex-valued signal which basically what makes Nyquist theorem a little tricky to apply here. One way to look at this is by trying to avoid aliasing in frequency domain, which implies the condition that $\omega_s-7\pi/2>5\pi/2$ or equivalently $\omega_s>6\pi$.

I thought that the Nyquist rate would have to be 7pi in this case, even though aliasing wouldn't occur until 6pi? In homework 9 question 2, we were able to violate Nyquist and still recover a signal. -sr

Instructor's comment: You are correct sr: one can go below the Nyquist rate (7pi) when sampling this signal and still be able to reconstruct it from its sample. As you said, it is because the signal is "shifted" in the frequency domain. -pm

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