(New page: = Practice Question on sampling and reconstruction (related to Nyquist rate) = The signal <math> x(t)= e^{j \frac{\pi}{2} t }\frac{\sin (6 \pi t)}{ t} </math> is sampled with a sampl...) |
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=== Answer 1 === | === Answer 1 === | ||
− | + | Using the table and a time shift | |
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+ | X(w) = pi[u(w+6pi-pi/2) - u(w-6pi-pi/2)] | ||
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+ | = pi[u(w+11pi/2) - u(w-13pi/2)] | ||
+ | |||
+ | <br> | ||
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+ | Thus the signal is band limited wm = 13pi/2, so ws > Nyquist Rate = 2wm = 13pi | ||
+ | |||
+ | Since T = 2pi/ws | ||
+ | |||
+ | T < 2pi/13pi = 2/13 | ||
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+ | |||
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+ | T < 2/13 in order to recover the orginal signal | ||
+ | --[[User:Ssanthak|Ssanthak]] 12:38, 21 April 2011 (UTC) | ||
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=== Answer 2 === | === Answer 2 === | ||
Write it here | Write it here |
Revision as of 08:38, 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?
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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.