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<p>This actually doesn't make sense to me using multiplication theorom (mimis version seen below)</p> | <p>This actually doesn't make sense to me using multiplication theorom (mimis version seen below)</p> | ||
<math>F(x_1(t)x_2(t)) = \frac {1} {2\pi} X_1(\omega)*X_2(\omega)</math> | <math>F(x_1(t)x_2(t)) = \frac {1} {2\pi} X_1(\omega)*X_2(\omega)</math> | ||
| − | <p>I took <math>x_1(t) = \cos(\frac{\pi}{2})</math> and <math> x_2(t) = rect(\frac{t}{2})</math><br/> | + | <p>I took <math>x_1(t) = \cos(\frac{\pi *t}{2})</math> and <math> x_2(t) = rect(\frac{t}{2})</math><br/> |
This resulted in <br/> | This resulted in <br/> | ||
<math>x_1(f) = \frac{1}{2}( \delta (f - \frac{1}{4}) + \delta (f + \frac{1}{4}))</math><br/> | <math>x_1(f) = \frac{1}{2}( \delta (f - \frac{1}{4}) + \delta (f + \frac{1}{4}))</math><br/> | ||
Revision as of 17:29, 10 February 2009
1 a)
$ x_(t) \,\!= \cos(\frac{\pi}{2})rect(\frac{t}{2}) $
Based on the Prof Alen's note page 179
$ x_(f) \,\!= \frac{1}{2}( \delta (f - \frac{1}{4}) + \delta (f + \frac{1}{4}))sinc(t/2) $
- Would you know how to compute this FT without a table if asked? --Mboutin 10:45, 9 February 2009 (UTC)
An answer to this 1a) question is stated in the discussion [1]
This actually doesn't make sense to me using multiplication theorom (mimis version seen below)
$ F(x_1(t)x_2(t)) = \frac {1} {2\pi} X_1(\omega)*X_2(\omega) $
I took $ x_1(t) = \cos(\frac{\pi *t}{2}) $ and $ x_2(t) = rect(\frac{t}{2}) $
This resulted in
$ x_1(f) = \frac{1}{2}( \delta (f - \frac{1}{4}) + \delta (f + \frac{1}{4})) $
$ x_2(f) = 2sinc(2f) $
By The multiplication theorem I get that these 2 functions need convolved but do not understand how that results in it being multiplied by sinc(t/2)
--Drestes 22:28, 10 February 2009 (UTC)
b)
$ x_(t) \,\!= repT[x0_(t)] = \frac {1}{T} \sum_{k} cos(\frac{\pi}{2})rect(\frac{t}{4}) $
Based on the Prof Alen's note page 184
$ x_(f) \,\!= \frac{1}{T}\sum_{k} ( \delta (f - \frac{1}{4}) + \delta (f + \frac{1}{4}))( \delta (f - \frac{k}{4})) $
