Inverse Fourier Transform Question

Discussion in 'Homework Help' started by jp1390, Dec 9, 2011.

  1. jp1390

    Thread Starter Member

    Aug 22, 2011
    45
    2
    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. jp1390

    Thread Starter Member

    Aug 22, 2011
    45
    2
    bump to the top!
     
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