Fourier transform ##f(t) = te^{-at}##

In summary, the function you are trying to Fourier transform is singular at the point ##\omega = \frac{-\alpha}{i}##.
  • #1
DragonBlight
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Homework Statement
Finding the Fourier transform of ##f(t) = te^{-at}## if t > 0 and ##f(t) = 0## otherwise.
Relevant Equations
##F(w) = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} te^{-at}e^{-iwt} dt##
Doing the Fourier transform for the function above I'm getting a result, but since I can't get the function f(t) with the inverse Fourier transform, I'm wondering where I made a mistake.

##F(w) = \frac{1}{\sqrt{2 \pi}} \int_{0}^{\infty} te^{-t(a + iw)} dt##
By integrating by part, where G = -a - iw
##F(w) = \frac{te^{Gt}}{G}|_{0}^{\infty} - \int_{0}^{\infty} \frac{e^{Gt}}{G} dt##

##= \frac{1}{G^2}##

Thus,
##F(w) = \frac{1}{\sqrt{2 \pi}} \frac{1}{(-a -iw)^2}##
 
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  • #2
Is your function ##t^{-at}## as you write in the statement and title or ##t e^{-at}## as you have used in your computation?
 
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  • #3
I made I mistake. It's ##te^{-at}##
 
  • #4
So, here is an alternative way: Fourier's trick (differentiating with respect to a parameter inside the integral)

Let
$$
G(a,\omega) = \frac{1}{\sqrt{2\pi}} \int_0^\infty e^{-at} e^{-i\omega t} dt = -\frac{1}{\sqrt{2\pi}} \left[\frac{e^{-at-i\omega t}}{a+i\omega}\right]_0^\infty = \frac{1}{\sqrt{2\pi}}\frac{1}{a+i\omega}.
$$
Now,
$$
\frac{\partial G}{\partial a} = - \frac{1}{\sqrt{2\pi}}\int_0^\infty t e^{-at} e^{-i\omega t} dt = -F(\omega).
$$
It follows that
$$
F(\omega) = -\frac{1}{\sqrt{2\pi}} \frac{d}{da}\frac{1}{a+i\omega} =\frac{1}{\sqrt{2\pi}} \frac{1}{(a+i\omega)^2}.
$$

Why do you think there is a mistake?
 
  • #5
DragonBlight said:
since I can't get the function f(t) with the inverse Fourier transform, I'm wondering where I made a mistake.
As @Orodruin has noted, you calculated the Fourier transform correctly, so it sounds like the problem is in your calculation of the reverse Fourier transform. Can you show us your attempt at that?
 
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  • #6
DragonBlight said:
I made I mistake. It's ##te^{-at}##
Edited title and Homework Statement to correct this typo.
 
  • #7
vela said:
As @Orodruin has noted, you calculated the Fourier transform correctly, so it sounds like the problem is in your calculation of the reverse Fourier transform. Can you show us your attempt at that?

##f(t) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \frac{1}{\sqrt{2 \pi}} \frac{1}{(-\alpha -i\omega)^2} e^{i\omega t} d\omega##

I don't see how to find the residue since the imaginary part is ##\omega##. There's a singularity ##\omega = i \alpha##
I mean, if I had ##(-i\alpha - \omega)## I could find it.

I think I found. if I factor i at the denominator then I have a pole of order 2 at ##\omega = \frac{- \alpha}{i}##
 
Last edited:

FAQ: Fourier transform ##f(t) = te^{-at}##

1. What is the Fourier transform of f(t)?

The Fourier transform of f(t) is F(ω) = 1/(a + jω)², where j is the imaginary unit and ω is the frequency domain variable.

2. What does the parameter "a" represent in the function?

The parameter "a" represents the decay rate of the function. It determines how quickly the function decays as time increases.

3. How does the Fourier transform relate to the original function?

The Fourier transform is a mathematical operation that decomposes a function into its constituent frequencies. In this case, the Fourier transform of f(t) gives us the amplitude and phase of each frequency component in the function.

4. What is the significance of the exponential term in the function?

The exponential term, e^(-at), is a common factor in many physical systems and represents the decay or growth of a quantity over time. In this function, it causes the function to decay as time increases, making it useful for modeling systems with decay.

5. How is the Fourier transform used in real-world applications?

The Fourier transform has numerous applications in science and engineering, including signal processing, image processing, and data analysis. It allows us to analyze the frequency components of a signal or function, making it useful in fields such as telecommunications, audio and video processing, and medical imaging.

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