Hi, I might have some input on this for you. The Fourier transform of a Gaussian function is yet another Gaussian profile with an inverse sigma 1/s standard deviation. Nevertheless, it is still a Gaussian profile and it occupies the whole frequency range. Basically, it is not a band-limited signal. So, no matter with what frequency rate you perform sampling operation, it will alias the signal. The reason they took 3/s is because to mitigate the effect of aliasing in sampling and also preserve most of the information of the profile.  Thanks

Hi, I might have some input on this for you. The Fourier transform of a Gaussian function is yet another Gaussian profile with an inverse sigma 1/s standard deviation. Nevertheless, it is still a Gaussian profile and it occupies the whole frequency range. Basically, it is not a band-limited signal. So, no matter with what frequency rate you perform sampling operation, it will alias the signal. The reason they took 3/s is because to mitigate the effect of aliasing in sampling and also preserve most of the information of the profile.  Thanks

Thanks Mehdi, qualitatively I agree that a broad Gaussian peak will have a narrow Fourier transform and a narrow Gaussian peak will have a broad Fourier transform. If you look at the attached file, we still see "s = standard deviation" which appeared in the original Gaussian function. So the more specific question is what is the expression for width of Fourier transform of the Gaussian in terms of standard deviation? Regards. 

Hi Farooq,

The relationship between a fourier transform of a guassian function from its space to the reciprocal space (time domain to frequency domain, etc.) is given by

h(t) = (a / pi)1/2 exp (-at2)         H(f) = exp (-pi2f2 / a)

where f = 1/t and pi = 3.1416

I hope that is helpful.

Best,

John

Thanks Dr. John, Please see the attached file. What I am interested in is in the relationship between the standard deviation of the Gaussian peak (in the time domain) and the standard deviation of the Fourier transform of the Gaussian in the frequency domain. They are inversely related but what is the exact relation?

Hi Farooq,

Plotting the two functions above, for any constant "a" you would find that the FWHM for

  h(t) = (a / pi)1/2 exp (-at2)  is  2.355*(1/2a)1/2

and for

   H(f) = exp (-pi2f2  /  a)  is  2.355*(a / 2pi2)1/2

Here (1/2a)1/2 is the standard deviation for for the first Gaussian, and (a / 2pi2)1/2 is the standard deviation for the second Gaussian.

Dear Farooq

I remember I saw somewhere that for signals like Gaussian peak or Lorentzian peak (that aren’t bandlimited), we consider the part of peak which contains 99.9% of the peak’s area (Assume a Gaussian peak with zero mean, and assume a frequency w_m. The integration of the Gaussian peak from - w_m to +w_m should give 0.999. Since erf(2.32)=0.999, then w_m=3.29*sigma. Note that erf() is the error function in MATLAB). So if we sample a Gaussian peak with the sampling frequency   2*w_m  = 2*3.29*sigma_f  =  2*3.29/sigma_t, then only 0.1% of  Gaussian peak area will overlap.

I think the 3/s   =  3/sigma_t  =  3*sigma_f in your question might be this 3.29*sigma_f.

I should say all above is just a guess. Good luck

http://www.wescottdesign.com/articles/Sampling/sampling.pdf

Thanks Jack, It is a very nice article. I had come across it before and it clarifies many misconceptions. We already published this work.

https://www.researchgate.net/publication/288918598_Sampling_frequency_response_times_and_embedded_signal_filtration_in_fast_high_efficiency_liquid_chromatography_A_tutorial

Article Sampling frequency, response times and embedded signal filtr...

Hi M, Thanks; next time I will check the date. Keep up the good work. Jack