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We know that the CTFT of x1(t) is:<br/> | We know that the CTFT of x1(t) is:<br/> | ||
<math>x1(f) = sinc(2(f-\frac{1}{4} ) + sinc(2(f + \frac{1}{4})</math><br/> | <math>x1(f) = sinc(2(f-\frac{1}{4} ) + sinc(2(f + \frac{1}{4})</math><br/> | ||
+ | |||
+ | |||
*<span style="color:green"> -- i am a little confused with this step. </span> | *<span style="color:green"> -- i am a little confused with this step. </span> |
Revision as of 18:45, 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]
b)
This is how I came to my conclusion, I think it makes morse sense then the previous mentioned answer.
First take the x(t) from part a and call it $ x_1(t) $ $ x1(t) \,\!= \cos(\frac{\pi t}{2})rect(\frac{t}{2}) $
Now since this is a repeating function use the rep function to get $ x(t) = rep_4(x_1(t)) $
We know that the CTFT of x1(t) is:
$ x1(f) = sinc(2(f-\frac{1}{4} ) + sinc(2(f + \frac{1}{4}) $
- -- i am a little confused with this step.
- In part a we did the CTFT of x1(t) and we get
$ x(f) \,\!= \frac{1}{2}( \delta (f - \frac{1}{4}) + \delta (f + \frac{1}{4}))sinc(t/2) $
- so why here do you say the CTFT of x1(t) is
$ x1(f) = sinc(2(f-\frac{1}{4} ) + sinc(2(f + \frac{1}{4}) $
--Jwromine 23:45, 10 February 2009 (UTC)
We also know from alabechs notes section 1.4.1 that
$ rep_T(x_1(t)) \Rightarrow \frac{1}{T} comb_{1/T}(X_1(f)) $
Put all the pieces together and you get something that looks like
$ X(f) = \frac{1}{4} comb_{1/4}(sinc(2(f-\frac{1}{4} ) + sinc(2(f + \frac{1}{4})) $
--Drestes 22:53, 10 February 2009 (UTC)