Search the CONFOCAL archive at
http://listserv.acsu.buffalo.edu/cgi-bin/wa?S1=confocal

Another very simple explanation:

You need samples to be closer together to properly sample a narrower
Gaussian. Samples closer together are able to represent higher
frequencies. The narrower Gaussian contains higher frequencies.
It's cut-off frequency is higher.

Cris.


Andrew Resnick wrote:
> Search the CONFOCAL archive at
> http://listserv.acsu.buffalo.edu/cgi-bin/wa?S1=confocal
> 
> Here's how I explain it:
> 
> Without the math- because frequency is inversely related to distance, 
> the pulsewidth and bandwidth are inversely related.
> 
> With the math: Define a Gaussian function to be Exp(-pi*[x/b]^2).  This 
> is a Gaussian with area 'b', and FWHM of about 2*b.
> The Fourier transform integral is of the form Integral 
> (Exp[-p^2*x^2+q*x])dx = sqrt[pi]/p *Exp[q^2/4p^2]
> 
> Substituting in p = Sqrt[pi]/b, q = 2*pi*i*u (u is the conjugate 
> variable to x) gives b*Exp[-4*pi*b^2*u^2], which is a Gaussian with FWHM 
> of about 1/2*b, inversely proportional to the initial Gaussian.
> 
> Hope this helps,
> 
> Andy
> 
> At 09:23 AM 8/29/2007, you wrote:
>> Search the CONFOCAL archive at 
>> http://listserv.acsu.buffalo.edu/cgi-bin/wa?S1=confocal
>> Dear All,
>>
>> I'm plotting the Fourier transform of the Gaussian psf graphs,by the 
>> math and equations I know that cutoff frequency is equal to invers of  
>> psf width,but I'm trying to understand the concept behind it,I 
>> appreciate if you could help me with it!
>>
>> regards
>> Sarah
> 
> Andrew Resnick, Ph. D.
> Instructor
> Department of Physiology and Biophysics
> Case Western Reserve University
> 216-368-6899 (V)
> 216-368-4223 (F)


-- 
Cris L. Luengo Hendriks, PhD

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