Re: [Leica] Film Demise

["SonC (Sonny Carter)" <cartersn@nsula.edu>]

Is that Metric pixels, Imperial pixels, or S.A.E. pixels?

Regards,

SonC


- ----- Original Message -----
From: <leica@davidmorton.org>
To: <leica-users@mejac.palo-alto.ca.us>
Sent: Sunday, April 22, 2001 8:58 AM
Subject: RE: [Leica] Film Demise


> 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.
>