FourierTransform function in mathematica returns Fourier transform of input function in omega. enter image description here

but I need to get Fourier transform of input function in hertz(frequency) that output should be something like this enter image description here However with some changes in FourierParameters and a command like this

FourierTransform[1, x, f,   FourierParameters -> {1, -1}]

I can't find my desire response.

As I know we can't divide our Fourier output by 2pi because it only amplitude of function not input range as you can see in picture has shown below (as I remember we had to Duality theory to correct our output) enter image description here

the maximum should be occur on (their position )/2Pi not in above position


closed as off-topic by Jens, chuy, Feyre, happy fish, J. M. is away Oct 15 '16 at 12:59

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question arises due to a simple mistake such as a trivial syntax error, incorrect capitalization, spelling mistake, or other typographical error and is unlikely to help any future visitors, or else it is easily found in the documentation." – Jens, chuy, Feyre, happy fish, J. M. is away
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ FourierTransform[1, x, f, FourierParameters -> {1, -1}]/(2 Pi) $\endgroup$ – Quantum_Oli Oct 14 '16 at 16:09
  • $\begingroup$ FourierTransform[1,x,2*Pi*f,FourierParameters->{1,-1}]? $\endgroup$ – N.J.Evans Oct 14 '16 at 16:19

What you want is probably this:

FourierTransform[((Cos[2 10*Pi*x] Sin[
       2*Pi*x])/(Pi x))^2, x, f, FourierParameters -> {0, -2 Pi}]

==> 1/8 (-22 Sign[-22 + f] + f Sign[-22 + f] + 40 Sign[-20 + f] - 
   2 f Sign[-20 + f] - 18 Sign[-18 + f] + f Sign[-18 + f] - 
   4 Sign[-2 + f] + 2 f Sign[-2 + f] - 4 f Sign[f] + 4 Sign[2 + f] + 
   2 f Sign[2 + f] + 18 Sign[18 + f] + f Sign[18 + f] - 
   40 Sign[20 + f] - 2 f Sign[20 + f] + 22 Sign[22 + f] + 
   f Sign[22 + f])

This is in the documentation for FourierParameters.

  • $\begingroup$ Thanks Jens , yes it correct the function argument very well but it doesn't return the proper amplitude for example for Cos[x] it return 2 delta functions with amplitude of Pi , but it should be 2 delta functions with amplitude of 1 $\endgroup$ – Ehsan Zakeri Oct 14 '16 at 16:51
  • $\begingroup$ You can adjust the amplitude to anything you want using the first entry in FourierParameters. See the docs for FourierTransform. $\endgroup$ – Jens Oct 14 '16 at 17:24

Not the answer you're looking for? Browse other questions tagged or ask your own question.