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

In summary, the problem is to calculate the Fourier Transform of e^{-a*|t|}, where a > 0, using the definition of the Fourier Transform. After breaking up the problem into two cases of t (positive and negative), the correct answer is G(f) = \frac{2}{j\omega + \alpha}. However, this is not the same as the answer given by WolframAlpha due to a difference in the definition of the Fourier Transform. The limits of integration for t < 0 should be from -∞ to 0, and for t > 0 should be from 0 to ∞.
  • #1
Nat3
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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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  • #2
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
Thanks for your help!

rude man said:
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?

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

Shouldn't it be zero to infinity?
 
  • #4
Nat3 said:
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=∞?
 
  • #5
rude man said:
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
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
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.
 

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

1. How do I calculate the Fourier Transform of e^-a*|t|?

To calculate the Fourier Transform of e^-a*|t|, you will need to use the definition of the Fourier Transform and apply it to the given function. This involves integrating the function with respect to time and then multiplying it by the appropriate constant. The final result will be a function of frequency.

2. What is the purpose of taking the Fourier Transform of e^-a*|t|?

The Fourier Transform allows us to convert a function from the time domain to the frequency domain. This can be useful in analyzing signals or systems that have different characteristics in the time and frequency domains.

3. Can I use any value of "a" when calculating the Fourier Transform of e^-a*|t|?

Yes, you can use any value of "a" when calculating the Fourier Transform of e^-a*|t|. However, the value of "a" will affect the shape and magnitude of the resulting function in the frequency domain.

4. Is there a specific formula for calculating the Fourier Transform of e^-a*|t|?

Yes, the formula for calculating the Fourier Transform of e^-a*|t| is: F(ω) = 2a/(a^2 + ω^2), where ω is the frequency variable. This formula can be derived using the definition of the Fourier Transform and properties of the exponential function.

5. How can I interpret the Fourier Transform of e^-a*|t|?

The Fourier Transform of e^-a*|t| represents the frequency content of the original function e^-a*|t|. It shows the amplitude and phase of the different frequency components present in the signal. The magnitude of the Fourier Transform at a particular frequency represents the strength of that frequency component in the original signal.

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