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

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Discussion Overview

The discussion centers on calculating the Fourier transform of the function sin(a*t)*exp(-t/b), where 'a' and 'b' are constants. Participants explore methods for manual calculation, potential discrepancies in results from different sources, and the implications of varying definitions of the Fourier transform.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant expresses doubt about the result obtained from WolframAlpha and seeks a manual calculation method.
  • Another participant suggests using Euler's formula to convert sin(a*t) to its complex form and indicates that the Fourier transform can be computed using the integral from 0 to +∞.
  • A participant provides a proposed result for the Fourier transform, indicating specific peaks and widths related to the parameters a and b.
  • Some participants note that the result may vary if the function is defined differently, such as using e^{-|t|/b} and integrating over the entire real line.
  • There are concerns about the readability of the results from WolframAlpha, with one participant mentioning difficulty in interpreting the output and suggesting that the definitions used may differ.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the Fourier transform's result, as there are multiple approaches and interpretations presented. Disagreements arise regarding the definitions and methods used for the Fourier transform.

Contextual Notes

There are unresolved issues regarding the assumptions made about the function's domain and the specific forms of the Fourier transform being discussed. The differences in definitions and the conditions under which the transforms are calculated may affect the results.

Leonid92
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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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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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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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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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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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Please find attached the screenshot of WolphramAlpha's result:
242849
 
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.
 
Last edited:
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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
 

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