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Homework Help: How to calculate Fourier Transform of e^-a*|t|?

  1. Mar 31, 2014 #1
    1. The problem statement, all variables and given/known data
    Calculate (from the definition, no tables allowed) the Fourier Transform of [itex]e^{-a*|t|}[/itex], where a > 0.

    2. Relevant equations

    Fourier Transform:

    [itex]G(f) = \int_{-\infty}^{\infty} g(t)e^{-j\omega t} dt[/itex]

    3. The attempt at a solution

    I thought I'd break up the problem into the two cases of t (where it's negative and positive). However, when I calculated the portion where t > 0, I got:

    G(f) = \int_{0}^{\infty} e^{-at} e^{-j\omega t}dt = G(f) = \int_{0}^{\infty} e^{-(j\omega + a)t}dt = \frac{e^{-(j\omega + a)t}}{-(j\omega + a)}\bigg|_0^\infty = 0 - \frac{1}{-(j\omega + a)} = \frac{1}{j\omega + a}

    Which is nowhere close to what WolframAlpha says the answer should be:


    So I guess I'm confused on how I should even approach the problem. Any suggestions?
    Last edited: Mar 31, 2014
  2. jcsd
  3. Mar 31, 2014 #2

    rude man

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    I got closer but not the Wolfram answer. So Wolfram is clearly wrong!

    JK. There are two different definitions of the Fourier transform.

    What is g(t) for t < 0? What should be the limits of integration?

    What is g(t) for t > 0? What should THOSE limits of intetgration be?

    You need to rethink your limits of integration.

    I got the same answer as Wolfram except for a sqrt(2pi) coefficient. This is probably due to the fact that they define the Fourier transform differently from you.
  4. Mar 31, 2014 #3
    Thanks for your help!

    Ah, I guess since there's an absolute value around t, that actually make g(t) an even function? So g(t) is going to be the same equation and fourier transform for t < 0 and t > 0?

    Shouldn't it be zero to infinity?
  5. Mar 31, 2014 #4

    rude man

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    No. If t < 0 but g(t) has to have a negative exponent, what must g(t) be? after all, -a|t| < 0 for a > 0 for ALL t, -∞ < t < ∞.

    same question for when t > 0. the exponent must also be negative for all t > 0.
    why? for t < 0 how can the limits of integration go from t=0 to t=∞?
  6. Apr 2, 2014 #5
    OK, looking at it again, I now realize that g(t) is a decaying exponential, starting at y = 1 and then going down to follow the x-axis. It's an even function because of the absolute value around t, so it's the same on both sides of the y-axis.

    Does that mean that I can just do the integration from 0 to infinity on one side and then multiply by 2?
  7. Apr 2, 2014 #6
    I plugged the integral into wolfram alpha and got the same answer as I posted in my original question:


    Unless I'm just forgetting basic integration rules, it seems like since the function is the same on both sides of the y-axis that I can simply multiply the above by 2 to get the integral from -inf to +inf, which gives:

    [itex]G(f) = \frac{2}{j\omega + \alpha}[/itex]

    Which is not the correct answer.

    What am I doing wrong? Can anyone help?
  8. Apr 2, 2014 #7
    Not quite. The function [itex]g(t)[/itex] is even but you are integrating [itex]g(t)e^{j\omega t}[/itex] which isn't even, so you need to do the other half of the integral, or notice that the negative time part is the complex conjugate of the positive time part.
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