| Line 17: | Line 17: | ||
=== Answer 1 === | === Answer 1 === | ||
| − | from the Table | + | from the Table |
| − | x(w) = u(w+3pi)-u(w-3pi) | + | x(w) = u(w+3pi)-u(w-3pi) |
| + | <br> | ||
| + | Thus the signal is bandlimited with a w<sub>m</sub> = 3pi | ||
| − | + | We must sample above the Nyquist Rate which is equal to 2w<sub>m</sub> or 6pi | |
| − | + | w<sub>s</sub> > 6pi | |
| − | + | T = 2pi/w<sub>s</sub> < 2/6 = 1/3 | |
| − | T | + | |
| + | |||
| + | The signal can be reconstructed for all T < 1/3. | ||
=== Answer 2 === | === Answer 2 === | ||
Revision as of 08:08, 20 April 2011
Contents
The signal
$ x(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?
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Answer 1
from the Table
x(w) = u(w+3pi)-u(w-3pi)
Thus the signal is bandlimited with a wm = 3pi
We must sample above the Nyquist Rate which is equal to 2wm or 6pi
ws > 6pi
T = 2pi/ws < 2/6 = 1/3
The signal can be reconstructed for all T < 1/3.
Answer 2
Write it here
Answer 3
Write it here.
