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

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Homework Statement


Calculate (from the definition, no tables allowed) the Fourier Transform of [itex]e^{-a*|t|}[/itex], where a > 0.

Homework Equations



Fourier Transform:

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

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:

[itex]
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}
[/itex]

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

http://www.wolframalpha.com/input/?i=fourier+transform+exp(-a*abs(t))

So I guess I'm confused on how I should even approach the problem. Any suggestions?
 
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Answers and Replies

  • #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.
 
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  • #3
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Thanks for your help!

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

What is g(t) for t > 0? What should THOSE limits of intetgration be?
Shouldn't it be zero to infinity?
 
  • #4
rude man
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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?
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.
Shouldn't it be zero to infinity?
why? for t < 0 how can the limits of integration go from t=0 to t=∞?
 
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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=∞?
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?
 
  • #6
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I plugged the integral into wolfram alpha and got the same answer as I posted in my original question:

http://www.wolframalpha.com/input/?i=integrate+exp(-alpha*t)*exp(-i*omega*t)+from+0+to+infinity

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?
 
  • #7
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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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