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= Practice Question on sampling and reconstruction (related to Nyquist rate) = | = Practice Question on sampling and reconstruction (related to Nyquist rate) = | ||
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The signal | The signal | ||
<math> x(t)= e^{j \frac{\pi}{2} t }\frac{\sin (6 \pi t)}{ t} </math> | <math> x(t)= e^{j \frac{\pi}{2} t }\frac{\sin (6 \pi t)}{ t} </math> | ||
| − | is sampled with a sampling period < | + | is sampled with a sampling period <span class="texhtml">''T''</span>. For what values of T is it possible to reconstruct the signal from its sampling? |
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== Share your answers below == | == Share your answers below == | ||
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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! | 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! | ||
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=== Answer 1 === | === Answer 1 === | ||
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| − | + | Using the table and a time shift | |
| − | | + | X(w) = pi[u(w+6pi-pi/2) - u(w-6pi-pi/2)] |
| − | + | = pi[u(w+11pi/2) - u(w-13pi/2)] | |
| − | + | <br> | |
| − | + | Thus the signal is band limited wm = 13pi/2, so ws > Nyquist Rate = 2wm = 13pi | |
| − | T | + | Since T = 2pi/ws |
| + | T < 2pi/13pi = 2/13 | ||
| + | <br> | ||
| − | T < 2/13 in order to recover the orginal signal | + | T < 2/13 in order to recover the orginal signal |
| − | --[[User:Ssanthak|Ssanthak]] 12:38, 21 April 2011 (UTC) | + | |
| + | --[[User:Ssanthak|Ssanthak]] 12:38, 21 April 2011 (UTC) | ||
=== Answer 2 === | === Answer 2 === | ||
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Write it here | Write it here | ||
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=== Answer 3 === | === Answer 3 === | ||
| − | Write it here. | + | |
| + | Write it here. | ||
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Revision as of 08:39, 21 April 2011
Contents
The signal
$ x(t)= e^{j \frac{\pi}{2} t }\frac{\sin (6 \pi t)}{ t} $
is sampled with a sampling period T. For what values of T is it possible to reconstruct the signal from its sampling?
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
Using the table and a time shift
X(w) = pi[u(w+6pi-pi/2) - u(w-6pi-pi/2)]
= pi[u(w+11pi/2) - u(w-13pi/2)]
Thus the signal is band limited wm = 13pi/2, so ws > Nyquist Rate = 2wm = 13pi
Since T = 2pi/ws
T < 2pi/13pi = 2/13
T < 2/13 in order to recover the orginal signal
--Ssanthak 12:38, 21 April 2011 (UTC)
Answer 2
Write it here
Answer 3
Write it here.
