1

I'm trying to find the Fourier Transform of the following rectangular pulse:

$$ x(t) = rect(t - 1/2) $$

This is simply a rectangular pulse stretching from 0 to 1 with an amplitude of 1. It is 0 elsewhere. I tried using the definition of the Fourier Tranform:

$$ X(\omega) = \int_0^1 (1)*e^{-j\omega*t}dt $$

However carrying out the relatively simple integration and subbing in the bounds results for me in this:

$$ X(\omega) = \frac{1}{j\omega}[e^{-j\omega} - 1] $$

& unfortunately wolfram alpha has a different answer when I use it to compute this fourier transform. It's got the sinc function; I'd appreciate any help on this, if I've got some giant conceptual error. I have an exam on this stuff in a bit less than a week :/

Edit: also realized I used j; it's the same with i (the imaginary #)

bo60
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  • With the time shift the integral will be between 0.5 and 1.5. – Paul Dec 12 '15 at 18:43
  • I see... any idea why? The pulse goes from 0 to 1; it's just centered around 1/2 ... – bo60 Dec 12 '15 at 18:45
  • I was presuming that $rect(t)$ represented a square pulse with value 1 between 0 and 1. Then $rect(t-\frac{1}{2})$ represents the same pulse shifted right by 0.5. Could be wrong! – Paul Dec 12 '15 at 18:48
  • Oh ... Yeah no; Hadn't realized notation might be confusing. Just using what our prof taught us. Visually it's as below:

    http://www.wolframalpha.com/input/?i=rect+%28t+-+1%2F2%29

    – bo60 Dec 12 '15 at 18:51
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    My mistake. Your form is correct then. It could be re-written in terms of sinc and $\frac{w}{2}$ by taking out a factor of $exp(-j\frac{w}{2})$ – Paul Dec 12 '15 at 20:22

2 Answers2

0

They are equal to each other. Rewrite it using the following:

$$ ( e^{-j\omega} - 1 ) = e^{-j\omega/2}( e^{-j\omega/2} - e^{j\omega/2} ). $$

-3

You are doing a little mistake buddy, the Fourier transform of shifted rectangular function (given as ques) will be, $X(w)=\frac1{-jw}\left(e^{-jw} - 1\right)$. Then solve.