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[[Category:ECE301Spring2011Boutin]] [[Category:Problem_solving]] | [[Category:ECE301Spring2011Boutin]] [[Category:Problem_solving]] | ||
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| − | = Practice Question on the Nyquist rate of a signal = | + | = [[:Category:Problem_solving|Practice Question]] on the Nyquist rate of a signal = |
Is the following signal band-limited? (Answer yes/no and justify your answer.) | Is the following signal band-limited? (Answer yes/no and justify your answer.) | ||
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=== Answer 2 === | === Answer 2 === | ||
From observation we can see <math class="inline"> x(t)= 7 \frac{\sin (3 \pi t)}{\pi t} </math> is a sinc function | From observation we can see <math class="inline"> x(t)= 7 \frac{\sin (3 \pi t)}{\pi t} </math> is a sinc function | ||
| − | with <math class="inline"> W = 3 \pi </math> and multiplied by a constant <math class="inline"> C = | + | with <math class="inline"> W = 3 \pi </math> and multiplied by a constant <math class="inline"> C = 7 </math> |
Therefore the Fourier Transform can be taken from the table of transforms as | Therefore the Fourier Transform can be taken from the table of transforms as | ||
| − | <math>\mathcal X (\omega) =\left\{\begin{array}{ll} | + | |
| + | <math>\mathcal X (\omega) =\left\{\begin{array}{ll}7, & \text{ if }|\omega| <3 \pi,\\ 0, & \text{else.}\end{array} \right. \ </math> | ||
From here it is fairly evident this function is band limited, with <math class="inline">\omega_m = 3\pi</math>. | From here it is fairly evident this function is band limited, with <math class="inline">\omega_m = 3\pi</math>. | ||
Latest revision as of 09:28, 11 November 2011
Contents
Practice Question on the Nyquist rate of a signal
Is the following signal band-limited? (Answer yes/no and justify your answer.)
$ x(t)= 7 \frac{\sin (3 \pi t)}{\pi t} \ $>
If you answered "yes", what is the Nyquist rate for this signal?
You will receive feedback from your instructor and TA directly on this page. Other students are welcome to comment/discuss/point out mistakes/ask questions too!
Answer 1
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
Answer 2
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} $.
Answer 3
Write it here.
