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Inverse Fourier transform of ## \frac{1}{a+jw} ##

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  1. Nov 15, 2015 #1

    etf

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    Fourier transform is defined as
    $$F(jw)=\int_{-\infty}^{\infty}f(t)e^{-jwt}dt.$$
    Inverse Fourier transform is defined as
    $$f(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}F(jw)e^{jwt}dw.$$

    Let ##f(t)=e^{-at}h(t),a>0##, where ##h(t)## is heaviside function and ##a## is real constant.

    Fourier transform of this function is
    $$F(jw)=\int_{0}^{\infty}f(t)e^{-jwt}dt=\int_{0}^{\infty}e^{-at}e^{-jwt}dt=\frac{1}{a+jw}$$
    How can I calculate inverse Fourier transform of ##\frac{1}{a+jw}##, ##f(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{1}{a+jw}e^{jwt}dw##?
    Although ##\frac{1}{a+jw}## doesn't look complicated, there is no way I can calculate this integral. Generaly, I have problems calculating inverse FT. Any suggestion?
    Thanks in advance.
     
  2. jcsd
  3. Nov 15, 2015 #2

    Ray Vickson

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    What is the reason you want to calculate the inverse FT of your ##F(\omega)##? [BTW: it should be called ##F(\omega)##, not ##F(i \omega)## or ##F(j \omega)##.] You already know how you obtained ##F## from your original function ##f(t)=e^{-at}h(t)##. Are you just trying to practice, to see if you can do the integral and get back your ##f(t)##, or are you fundamentally misunderstanding something about about the subject?
     
    Last edited: Nov 16, 2015
  4. Nov 16, 2015 #3

    vela

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    It's traditional to use the Greek letter omega, ##\omega##, not ##w##. (I hope you don't say "double-u" when you're talking to others, especially your instructor.)

    It's straightforward to calculate that integral if you know how to do contour integration in the complex plane. I'm guessing you probably haven't seen that yet.
     
  5. Nov 16, 2015 #4

    etf

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    I'm just trying to practice...

    I know it's omega, just didn't write it :)
    I'm not familiar with contour integration yet, but I would like to see solution...
     
  6. Nov 20, 2015 #5

    vela

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    The idea is to consider the integral along the real axis as part of a closed contour in the complex plane. The rest of the contour doesn't contribute to the integral, so you have
    $$\frac{1}{2\pi}\int_{-\infty}^\infty \frac{1}{a+i\omega} e^{i\omega t}\,d\omega = \frac{1}{2\pi}\int_C \frac{1}{a+iz} e^{izt}\,dz.$$ The latter integral can be evaluated using the residue theorem.
     
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