# Inverse FT Question

1. Dec 9, 2011

### jp1390

Hi, quick question regarding my solution of this Inv. FT:

$$Y(\omega) = \frac{1}{j}[sinc(\frac{2\omega}{\pi} - \frac{1}{2}) - sinc(\frac{2\omega}{\pi} + \frac{1}{2})]$$

Recognizing that that this looks a lot like this property of the FT:

$$x(t)sin(\omega_{0}t) \leftrightarrow \frac{j}{2}[X(\omega + \omega _{0}) - X(\omega - \omega _{0})]$$

Rearranging to get in that form, factoring out a -1, which = j*j:

$$Y(\omega) = \frac{2j^{2}}{2j}[sinc(\frac{2\omega}{\pi} + \frac{1}{2}) - sinc(\frac{2\omega}{\pi} - \frac{1}{2})] = 2\frac{j}{2}[sinc(\frac{2\omega}{\pi} + \frac{1}{2}) - sinc(\frac{2\omega}{\pi} - \frac{1}{2})]$$

$$x(t) = p_{\tau}(t) \leftrightarrow X(\omega) = \tau sinc(\frac{\tau \omega}{2\pi})$$ τ = 4 in this case

Knowing this information, we can find ω0:

$$Y(\omega) = (\frac{1}{2})\frac{j}{2}[4sinc(\frac{4}{2\pi}(\omega + \frac{\pi}{4})) - 4sinc(\frac{4}{2\pi}(\omega - \frac{\pi}{4}))]$$

Therefore ω0 = ∏/4... and y(t) is found to be:

$$y(t) = \frac{1}{2}p_{4}(t)sin(\frac{\pi t}{4})$$ where p4(t) is a pulse with a duration of 4 seconds and is centered on the origin with a height of 1.

The solution manual says the answer is:

$$y(t) = \frac{1}{2}p_{4}(t)sin(\frac{\pi t}{2})$$

Can anyone see where I went wrong or if the solution manual has an error? Thanks!

Last edited: Dec 9, 2011
2. Dec 10, 2011

### jp1390

bump to the top!