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=== Answer 1 === | === Answer 1 === | ||
| − | x(w) = (1/2pi) F(e^jtpi)*F(sin(3tpi)/tpi) | + | x(w) = (1/2pi) F(e^jtpi)*F(sin(3tpi)/tpi) |
| − | = (1/2pi) [2pi delta(w-pi)] * [u(w+3pi)-u(w-3pi)] | + | = (1/2pi) [2pi delta(w-pi)] * [u(w+3pi)-u(w-3pi)] |
| − | = u(w-pi+3pi) - u(w-pi-3pi) | + | = u(w-pi+3pi) - u(w-pi-3pi) |
| − | = u(w+2pi) - u(w-4pi) | + | = u(w+2pi) - u(w-4pi) |
| − | w<sub>m</sub>=4pi | + | w<sub>m</sub>=4pi |
| − | Nyquist Rate = 2w<sub>m</sub> = 8pi | + | Nyquist Rate = 2w<sub>m</sub> = 8pi |
| − | Since we should sample w<sub>s</sub> > 8pi | + | Since we should sample w<sub>s</sub> > 8pi |
| − | w<sub>s</sub> = 2pi/T > 8pi | + | w<sub>s</sub> = 2pi/T > 8pi |
| − | T < 1/4 in order to be able to reconstruct the signal using Nyquist.<br> | + | T < 1/4 in order to be able to reconstruct the signal using Nyquist.<br> --[[User:Ssanthak|Ssanthak]] 13:01, 21 April 2011 (UTC) |
| − | --[[User:Ssanthak|Ssanthak]] 13:01, 21 April 2011 (UTC) | + | |
:<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> | :<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> | ||
| + | |||
=== Answer 2 === | === Answer 2 === | ||
| − | + | 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> | |
| + | |||
| + | | | ||
| + | |||
| + | ^^^^^^^^^^^^^| |^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^| |^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^| | ||
| + | |||
| + | <sub> </sub> <sub></sub>| | | <sub> </sub> | | <sub> </sub> | | ||
| + | |||
| + | ----------|----------|----------|----------|----------|----------|----------|----------|----------|----------|----------|----------|----------|----------|----- w | ||
| + | |||
| + | -w<sub>m</sub> -w<sub>s</sub> 0 ws -w<sub>m</sub><sub></sub> <sub> </sub> | ||
| + | |||
| + | <sub></sub> | ||
| + | |||
| + | (each |----------| represents pi distance) | ||
| − | == | + | <span class="Apple-style-span" style="font-family: sans-serif; line-height: 19px; white-space: normal; font-size: 18px; font-weight: bold; ">Answer 3</span> |
Write it here. | Write it here. | ||
Revision as of 14:00, 21 April 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.
|
^^^^^^^^^^^^^| |^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^| |^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^|
| | | | | |
|----------|----------|----------|----------|----------|----------|----------|----------|----------|----------|----------|----------|----------|----- w
-wm -ws 0 ws -wm
(each |----------| represents pi distance)
Answer 3
Write it here.
