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=== 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> | + | 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> | | + | <sub> </sub> <sub></sub>| | | <sub> </sub> | | <sub> </sub> | |
| − | ----------|----------|----------|----------|----------|----------|----------|----------|----------|----------|----------|----------|----------|----------|----- w | + | |----------|----------|----------|----------|----------|----------|----------|----------|----------|----------|----------|----------|----------|----------|----- w |
| − | -w<sub>m</sub> -w<sub>s</sub> 0 ws -w<sub>m</sub><sub></sub> <sub> </sub> | + | -w<sub>m</sub> -w<sub>s</sub> 0 ws -w<sub>m</sub><sub></sub> <sub> </sub> |
| − | <sub></sub> | + | <sub></sub> |
| − | (each |----------| represents pi distance) | + | (each |----------| represents pi distance) |
| − | + | Answer 3 | |
Write it here. | Write it here. | ||
Revision as of 14:03, 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.
