| Line 22: | Line 22: | ||
<math>x(t) = rep_4(x_1(t))</math> | <math>x(t) = rep_4(x_1(t))</math> | ||
| − | We know that the CTFT of x1(t) is: | + | We know that the CTFT of x1(t) is:<br/> |
| − | <math>x_1(f) = sinc(2(f-\frac{1}{4} ) + sinc(2(f + \frac{1}{4})</math> | + | <math>x_1(f) = sinc(2(f-\frac{1}{4} ) + sinc(2(f + \frac{1}{4})</math><br/> |
We also know from alabechs notes section 1.4.1 that | We also know from alabechs notes section 1.4.1 that | ||
| − | <math> rep_T(x_1(t)) \Rightarrow \frac{1}{T} comb_{1/T}(X_1(f))</math> | + | <math> rep_T(x_1(t)) \Rightarrow \frac{1}{T} comb_{1/T}(X_1(f))</math><br/> |
Put all the pieces together and you get something that looks like<br/> | Put all the pieces together and you get something that looks like<br/> | ||
Revision as of 17:54, 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) $ $ x_1(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:
$ x_1(f) = sinc(2(f-\frac{1}{4} ) + sinc(2(f + \frac{1}{4}) $
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)
