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

1. Mar 31, 2014

### Nat3

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

2. Relevant equations

Fourier Transform:

$G(f) = \int_{-\infty}^{\infty} g(t)e^{-j\omega t} dt$

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:

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?

Last edited: Mar 31, 2014
2. Mar 31, 2014

### rude man

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.

3. Mar 31, 2014

### Nat3

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?

4. Mar 31, 2014

### rude man

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=∞?

5. Apr 2, 2014

### Nat3

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. Apr 2, 2014

### Nat3

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:

$G(f) = \frac{2}{j\omega + \alpha}$

Which is not the correct answer.

What am I doing wrong? Can anyone help?

7. Apr 2, 2014

### cpt_carrot

Not quite. The function $g(t)$ is even but you are integrating $g(t)e^{j\omega t}$ 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.