How to calculate the Fourier transform of sin(a*t)*exp(-t/b) ?

In summary, WolphramAlpha site gives the solution for the Fourier transform of the function sin(a*t)*exp(-t/b). However, the user has doubts about the result. They need to calculate the Fourier transform manually in order to be sure.
  • #1
Leonid92
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TL;DR Summary
How to calculate Fourier transform of sin(a*t)*exp(-t/b) ?
Hi all,

I need to calculate Fourier transform of the following function: sin(a*t)*exp(-t/b), where 'a' and 'b' are constants.
I used WolphramAlpha site to find the solution, it gave the result that you can see following the link: https://www.wolframalpha.com/input/...Fourier+transform"}+->+{"Calculator",+"dflt"}
But I have doubt about this result. Could you please tell, how to calculate the Fourier transform of mentioned function manually? And is there another reliable site/program where I can find Fourier transform of any function?
 
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  • #2
I presume your function starts at ## t=0 ##. First, convert ## \sin(at) ## to complex form with Euler's formula. Next ## \hat{F}(\omega)=\int\limits_{-\infty}^{+\infty}F(t)e^{-i \omega t} \, dt ##. I believe the integrals are readily workable, where you only need to integrate from ##0 ## to ## +\infty ##.
 
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  • #3
And I can't get to your "link" above, but I get ## \\ ##
##\hat{F}(\omega)=-\frac{1}{2}[\frac{(\omega-a)+ib'}{(\omega-a)^2+b'^2}-\frac{(\omega+a)+ib'}{(\omega+a)^2+b'^2} ] ## where ## b'=\frac{1}{b} ##.
 
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  • #4
The above function ## \hat{F}(\omega) ## basically has peaks at ## \omega=\pm a ##, with a width (a spread around ## \omega=\pm a ##) that is approximately ## b' ##.
 
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  • #5
Charles Link said:
The above function ## \hat{F}(\omega) ## basically has peaks at ## \omega=\pm a ##, with a width (a spread around ## \omega=\pm a ##) that is approximately ## b' ##.
Thank you very much!
 
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  • #6
Please find attached the screenshot of WolphramAlpha's result:
242849
 
  • #7
It's very hard to read. The result will be slightly different if you use ## e^{-|t|/b} ##, and let ## t ## run from ##-\infty ## to ##+\infty ## . It looks like they get their final ## \sqrt{\pi} ## in the numerator, but it's much too hard to see... They define their F.T. slightly differently...Even with a magnifying glass I can not make out their complete result.
 
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  • #8
Charles Link said:
It's very hard to read. The result will be slightly different if you use ## e^{-|t|/b} ##, and let ## t ## run from ##-\infty ## to ##+\infty ## . It looks like they get their final ## \sqrt{\pi} ## in the numerator, but it's much too hard to see... They define their F.T. slightly differently...Even with a magnifying glass I can not make out their complete result.
Sorry for bad quality of image. Please find attached the image with better quality.
WolframAlpha (1).png
 

1. How do I calculate the Fourier transform of sin(a*t)*exp(-t/b)?

The Fourier transform of a function f(t) is given by F(ω) = ∫f(t)e^(-iωt)dt, where ω is the frequency and i is the imaginary unit. In this case, we can use the Fourier transform of the function exp(-t/b) = b/(b^2 + ω^2) to calculate the Fourier transform of sin(a*t)*exp(-t/b). This gives us F(ω) = (a*b)/(b^2 + (a - iω)^2).

2. What is the purpose of calculating the Fourier transform of a function?

The Fourier transform is a mathematical operation that allows us to represent a function in terms of its frequency components. This is useful in many areas of science and engineering, such as signal processing, image processing, and quantum mechanics. It helps us understand the behavior of a function in the frequency domain, which can provide insights into its properties and behavior in the time domain.

3. Can the Fourier transform of sin(a*t)*exp(-t/b) be calculated analytically?

Yes, as shown in the answer to the first question, the Fourier transform of sin(a*t)*exp(-t/b) can be calculated analytically using the Fourier transform of the function exp(-t/b). This is possible because the Fourier transform has many useful properties, such as linearity and time-shifting, that allow us to simplify complex calculations.

4. What are the applications of the Fourier transform of sin(a*t)*exp(-t/b)?

The Fourier transform of sin(a*t)*exp(-t/b) has many applications in science and engineering. For example, it can be used in signal processing to analyze signals that contain both sinusoidal and decaying components. It can also be used in quantum mechanics to study the behavior of quantum systems that involve oscillations and decays.

5. Is there a specific method or formula for calculating the Fourier transform of sin(a*t)*exp(-t/b)?

Yes, as mentioned in the answer to the first question, we can use the Fourier transform of the function exp(-t/b) to calculate the Fourier transform of sin(a*t)*exp(-t/b). However, there are also other methods and formulas that can be used, such as the convolution theorem, which states that the Fourier transform of the product of two functions is equal to the convolution of their individual Fourier transforms. The choice of method or formula may depend on the specific problem and the desired level of accuracy.

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