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Fourier Transform Issue

  1. Aug 19, 2013 #1
    Hi there,
    I am trying to get some practice with Fourier Transforms, there is a long way to go.
    For example, let me consider the function $$ \gamma (t) = \int_{-\infty}^{t} C(t-\tau) \sigma(\tau) \mathrm{d}{\tau}$$
    Defining the Fourier Transform as
    $$ \gamma(\omega) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} \gamma(t) e^{i\omega t} \mathrm{d}t$$
    I am supposed to compute with ease that
    $$ \gamma(\omega) = \sigma(\omega) \int_0 ^{\infty} C(t) e^{i\omega t} \mathrm{d}t$$,
    but I am struggling, because I can not apply the convolution theorem (as the first equation is a convolution only to actual time).
    I tried to use the definition writing
    $$\gamma(\omega) = \frac{1}{2\pi} \int_{-\infty}^{\infty} \int_{-\infty}^{t} C(t-\tau) \sigma(\tau) \mathrm{d}{\tau} e^{i\omega t} \mathrm{d}t$$
    hoping to invertt integration order and express the inner integral as a Fourier Transform, but again I am not getting anywhere. I tried a variable change $$t-\tau = u$$ and that helps in changing the integration limits to 0 and infinity, but still does not bring me to the desired result, any advice or hint?
    Many thanks as usual
  2. jcsd
  3. Aug 19, 2013 #2


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    Science Advisor

    You are almost there. eiωt = eiω(t-u) eiωu.
    After the change of variables and the order of integration switch you have the result, noting that the integral (-∞,∞) is the Fourier transform of σ.

    Further suggestion don't use the same symbol for a function and its Fourier transform.
  4. Aug 20, 2013 #3
    thanks for your help. I understand your suggestion, I will continue only for this post with the previous notation just not to confuse things further.
    I am getting what I would like to, but in a little bit of a shaky way, so I post the calc for a check...
    Here is the calculation:

    $$\gamma(\omega) = \frac{1}{2\pi} \int_{-\infty}^{\infty} \int_{-\infty}^{t} C(t-\tau) \sigma(\tau) \mathrm{d}{\tau} e^{i\omega t} \mathrm{d}t$$
    Using the change of variable $$ t - \tau = u$$ one obtains
    $$\gamma(\omega) = \frac{1}{2\pi} \int_{-\infty}^{\infty} \int_{0}^{\infty} C(u) \sigma(t-u) \mathrm{d}{u} \, e^{i\omega (t-u)} e^{i \omega u} \mathrm{d}t$$
    inverting the order of integration
    $$\gamma(\omega) = \frac{1}{2\pi} \int_{0}^{\infty} \int_{-\infty}^{\infty} C(u) \sigma(t-u) \, e^{i\omega (t-u)} e^{i \omega u} \mathrm{d}t \mathrm{d}{u}$$
    and the integrals can be separated now, the one between infinite limits of integration leading the Fourier transform of $\sigma$ (upon a trivial change of variables $$ t - u = z$$ at constant u, is this corrrect?), the final result being as desired
    $$\gamma (\omega) = \sigma(\omega) \int_{0}^{\infty} C(u)e^{i\omega u} \mathrm{d}u$$ Many thanks
    Last edited: Aug 20, 2013
  5. Aug 20, 2013 #4


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    Science Advisor

    It is correct.

    (Note - your Latex didn't work in one line).
  6. Aug 21, 2013 #5
    Mathman, many thanks much appreciated.
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