Difference between revisions of "HW 3 Question 1" - Rhea

Revision as of 05:41, 11 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)

• This answer is incorrect and does not match the answer we acquired..Please check the answer stated in the discussion below [1]

An answer to this 1a) question is stated in the discussion [2]--Drestes 10:41, 11 February 2009 (UTC)

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) $

• The answer from part a please check the referred link as the answer stated above is incorrect.

$ 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)

• Can you write your answer using a comb operator? --Mboutin 10:45, 9 February 2009 (UTC)
• How did you get to that answer? Please add some intermediate steps. --Mboutin 10:50, 9 February 2009 (UTC)