Archived posting to the Leica Users Group, 2001/04/22
[Author Prev] [Author Next] [Thread Prev] [Thread Next] [Author Index] [Topic Index] [Home] [Search]Austin Franklin wrote: "Sorry, but my statement is perfectly correct. Having 25 years of digital signal processing and both digital imaging and digital audio design and development experience, I believe I understand this topic quite well." I'm sorry, but you are quite wrong. In your original calculation you said: "Since you have to sample (scan) at 'slightly more than' 2x the frequency you want to get, that would be ((1 x (.5 x 5080)) x (1.5 x (.5 x 5080))) or ~9M pixels." What you have done here is apply the Nyquist criterion inappropriately. If we consider the simplified case where we are sampling a vertical grating at 5080 DPI, then Nyquist tells us that the absolute maximum frequency that we can reproduce is fs/2, where fs is our sampling frequency. Since our sampling frequency is 5080 DPI, the maximum frequency in your example is then 2540 line pairs per inch (lppi). But notice that this maximum frequency is in line *pairs* per inch, we have *two* pixels per cycle, as sampling theory tells us we must have. However in order to calculate the maximum number of pixels in the scanned image, you have used this *latter* figure, which is a count of the number of reproducible *cycles* not the number of pixels. You have divided the sampling frequency by two (in both dimensions) and used this to derive the total pixel count. If you are sampling a slide 1 inch by 1.5 inches at 5080 DPI, then the total pixel count is approximately 39Mpixels ((5080*1)*(5080*1.5) and the maximum reproducible spatial frequency is 2540 lppi.