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

Click For Summary
SUMMARY

The Fourier transform of the function sin(a*t)*exp(-t/b) can be calculated using Euler's formula to convert sin(a*t) into its complex form. The integral for the Fourier transform is defined as &hat;F(ω) = ∫[−∞, +∞] F(t)e^(−iωt) dt, but for this function, integration should be performed from 0 to +∞. The resulting transform is &hat;F(ω) = -1/2[(ω-a)+ib'] / [(ω-a)²+b'²] - [(ω+a)+ib'] / [(ω+a)²+b'²], where b' = 1/b. The peaks of this function occur at ω = ±a, with a spread around these peaks approximately equal to b'.

PREREQUISITES
  • Understanding of Fourier transforms and their properties
  • Familiarity with Euler's formula and complex numbers
  • Knowledge of integral calculus, specifically improper integrals
  • Experience using mathematical software or tools like WolframAlpha
NEXT STEPS
  • Learn how to compute Fourier transforms of other functions using WolframAlpha
  • Study the properties of Fourier transforms, including linearity and time-shifting
  • Explore the implications of using different limits of integration in Fourier transforms
  • Investigate the differences in Fourier transform definitions across various mathematical resources
USEFUL FOR

Mathematicians, engineers, and students studying signal processing or applied mathematics who need to compute or understand Fourier transforms of functions.

Leonid92
Messages
45
Reaction score
2
TL;DR
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?
 
Physics news on Phys.org
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 ##.
 
  • Like
Likes   Reactions: Leonid92
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} ##.
 
  • Like
Likes   Reactions: Leonid92
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' ##.
 
  • Like
Likes   Reactions: Leonid92
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!
 
  • Like
Likes   Reactions: Charles Link
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:
  • Like
Likes   Reactions: Leonid92
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
 

Similar threads

  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 3 ·
Replies
3
Views
11K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 6 ·
Replies
6
Views
25K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
7
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 17 ·
Replies
17
Views
7K