Fourier transform of an time integral

Discussion in 'Homework Help' started by epsilonjon, May 20, 2012.

  1. epsilonjon

    Thread Starter Member

    Feb 15, 2011
    65
    1
    Question:

    Derive the relationship

    \int^t_{- \infty} f(\tau) d \tau \Leftrightarrow \frac{F(\omega)}{j \omega} + \pi F(0) \delta (\omega)

    (where \Leftrightarrow means "Fourier transforms into").

    Attempt:

    I have already proved the relationship

    \frac{dg(t)}{dt} \Leftrightarrow j \omega G( \omega)

    so define h(t) = \frac{dg(t)}{dt}. Then we have

    \int^t_{- \infty} h(\tau) d \tau = \int^t_{- \infty} \frac{dg}{d \tau} d \tau = [g(\tau)]^t_{- \infty} = g(t) - g(-\infty)

    so

    g(t) = \int ^t _{- \infty} h(\tau) d \tau + g(- \infty)

    Using the Fourier differentiation relationship above we get

    \mathcal{F}[h(t)] = j \omega \mathcal{F}[g(t)] = j \omega \mathcal{F} \left [ \int ^t _{- \infty} h(\tau) d \tau + g(- \infty) \right ] = j \omega \mathcal{F} \left [ \int ^t _{- \infty} h(\tau) d \tau \right ]  + j \omega \mathcal{F}[g(- \infty)]
    = j \omega \mathcal{F} \left [ \int ^t _{- \infty} h(\tau) d \tau \right ]  + j \omega 2 \pi g(-\infty) \delta(\omega)

    so

    \mathcal{F} \left [ \int ^t _{- \infty} h(\tau) d \tau \right ] = \frac{\mathcal{F}[h(t)]}{j \omega} - 2 \pi g(-\infty) \delta(\omega)

    I'm not sure if i've done something wrong here or if somehow this is equivalent to the correct relationship? Could someone help please :confused:

    Thanks!
    Jon.
     
    Last edited: May 21, 2012
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