(New page: = Practice Question on sampling and reconstruction (related to Nyquist rate) = The signal <math> x(t)= e^{j \pi t }\frac{\sin (3 \pi t)}{\pi t} </math> is sampled with a sampling peri...) |
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| − | = Practice Question on sampling and reconstruction (related to Nyquist rate) = | + | = [[:Category:Problem_solving|Practice Question]] on sampling and reconstruction (related to Nyquist rate) = |
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The signal | The signal | ||
<math> x(t)= e^{j \pi t }\frac{\sin (3 \pi t)}{\pi t} </math> | <math> x(t)= e^{j \pi t }\frac{\sin (3 \pi t)}{\pi 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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| + | x(w) = (1/2pi) F(e^jtpi)*F(sin(3tpi)/tpi) | ||
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| + | = (1/2pi) [2pi delta(w-pi)] * [u(w+3pi)-u(w-3pi)] | ||
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| + | = u(w-pi+3pi) - u(w-pi-3pi) | ||
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| + | = u(w+2pi) - u(w-4pi) | ||
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| + | w<sub>m</sub>=4pi | ||
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| + | Nyquist Rate = 2w<sub>m</sub> = 8pi | ||
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| + | Since we should sample w<sub>s</sub> > 8pi | ||
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| + | w<sub>s</sub> = 2pi/T > 8pi | ||
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| + | T < 1/4 in order to be able to reconstruct the signal using Nyquist.<br> --[[User:Ssanthak|Ssanthak]] 13:01, 21 April 2011 (UTC) | ||
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| + | :<span style="color:green">Instructor's comment: But would it be possible to sample below the Nyquist rate and still be able to reconstruct the signal from its samples? -pm </span> | ||
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=== Answer 2 === | === Answer 2 === | ||
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| − | == | + | The signal could still be reconstructed as long as T < 1/3, since the unshifted signal would have w<sub>m</sub> = 3pi, and therefore T < (1/2)(2pi/w<sub>m</sub>) = 1/3. As long as w<sub>s</sub> is slightly bigger than 3pi, there will not actually be overlap in the frequency response, so it can be filtered later.<br> |
| − | Write it here. | + | |
| + | --[[User:Kellsper|Kellsper]] 18:05, 21 April 2011 (UTC) | ||
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| + | Answer 3 | ||
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| + | Write it here. | ||
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Latest revision as of 10:29, 11 November 2011
Contents
The signal
$ x(t)= e^{j \pi t }\frac{\sin (3 \pi t)}{\pi 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
x(w) = (1/2pi) F(e^jtpi)*F(sin(3tpi)/tpi)
= (1/2pi) [2pi delta(w-pi)] * [u(w+3pi)-u(w-3pi)]
= u(w-pi+3pi) - u(w-pi-3pi)
= u(w+2pi) - u(w-4pi)
wm=4pi
Nyquist Rate = 2wm = 8pi
Since we should sample ws > 8pi
ws = 2pi/T > 8pi
T < 1/4 in order to be able to reconstruct the signal using Nyquist.
--Ssanthak 13:01, 21 April 2011 (UTC)
- Instructor's comment: But would it be possible to sample below the Nyquist rate and still be able to reconstruct the signal from its samples? -pm
Answer 2
The signal could still be reconstructed as long as T < 1/3, since the unshifted signal would have wm = 3pi, and therefore T < (1/2)(2pi/wm) = 1/3. As long as ws is slightly bigger than 3pi, there will not actually be overlap in the frequency response, so it can be filtered later.
--Kellsper 18:05, 21 April 2011 (UTC)
Answer 3
Write it here.
