# Practice Question on the Nyquist rate of a signal

$x(t)= 7 \frac{\sin (3 \pi t)}{\pi t} \$>

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

Yes, this signal is band limited. It is a sinc function, and its Fourier transform can be found using the table of formulas in the textbook on page 329.

$\mathcal X (\omega) = \begin{cases} 1 & \Big|\omega\Big| < 3\pi \\ 0 & \mbox{otherwise} \end{cases}$

This is band limited.

In addition, the $\omega_m$ is $3\pi$.

Therefore the Nyquist rate for this signal is $6\pi$.

--Cmcmican 23:11, 30 March 2011 (UTC)

Instructor's comments. You got the correct Nyquist rate, but there is a small mistake in the Fourier transform. Since the mistake is a non-zero constant factor, it does not change the bandwidth of the signal, and therefore you were able to obtain the correct max frequency of the signal. It would be ok to say that the Fourier transform is a non-zero constant multiple of a low-pass filter with gain 1 and cutoff $3 \pi$ and conclude from there; you would get full credit because finding the constant is not necessary to answer the question. -pm

From observation we can see $x(t)= 7 \frac{\sin (3 \pi t)}{\pi t}$ is a sinc function with $W = 3 \pi$ and multiplied by a constant $C = 7$ Therefore the Fourier Transform can be taken from the table of transforms as

$\mathcal X (\omega) =\left\{\begin{array}{ll}7, & \text{ if }|\omega| <3 \pi,\\ 0, & \text{else.}\end{array} \right. \$

From here it is fairly evident this function is band limited, with $\omega_m = 3\pi$.

Since $\omega_s \ge 2\omega_m$ must hold true for Nyquist to be satisfied, the Nyquist rate must be $6\pi$

--Darichar 21:12, 12 April 2011 (UTC)

TA's comment: Your approach is correct. Your only mistake is that you transfered the constant unintentionally wrong. The constant should be 7 and not $\frac{2}{7}$.