How does high pass filter determine frequency

Question

I Am having a hard time understanding how the high pass filter determines high vs low frequency. What value is it measuring. ? Change in RGB values or Contrast ? I see the words sharpness or blur , but what values determine amount of blurriness or sharpness? Is there and actual frequency being measures or is that just a metaphor To compare it to the audio high pass filter ?
Any insight into the inner workings of this filter is appreciated.

Answer 1

Some further explanations:
Audio signals:
frequency = cycles per second
This is the frequency in the time domain.
An arbitrary signal can be represented by a bundle of harmonic signals (sine, cosine)
plus a DC-part (direct current, non periodical component).
A typical high-pass filter for audio signals:
The cut-off frequency fc is found at the transition between increasing gain
(low frequency) and fixed gain 1.0 (high frequency).
frequency   gain (attenuation)
0.01 fc   0.01
0.10 fc   0.1
1.00 fc   0.707
10.0 fc   1.0
100  fc   1.0
A typical high-pass filter for digital images:
frequency = cycles per unit length or (better) cycles per pixel
This is the frequency in the spatial domain.
The highest frequency is 0.5 cycles per pixel (alternating black and white pixels)
An arbitray row in a digital image can be represented by a bundle of harmonic signals
plus a DC-part.
Each channel R,G,B is filtered individually, one after the other.
Frequency and gain are as above.
A high-pass filter applies an attenuation <<1 to low frequency signal components
and removes the DC-component entirely. This would result in a black background.
Therefore the background is shifted to R=G=B=128 (8bit per channel) or
R=G=B=0.5 (normalized).
Sharp edges contain stronger high frequency components. These are retained.
http://en.wikipedia.org/wiki/High-pass_filter
In this doc we find a simple implementation:
y[0] := x[0]
   for i from 1 to n
     y[i] := a * y[i-1] + a * (x[i] - x[i-1])
   return y
x[i] are values R (or G or B) in a row at column i in the original image
y[i] are values R (or G or B) in the same row at column i after filtering.
Factor (a) contains indirectly the cut-off frequency.
This example does not yet apply the shift to R=G=B=128.
The implementation can be different (non-recursive filter instead of recursive, as above,
higher order instead of first order, using a filter kernel 3*3 or 5*5 pixels instead of working
in each row independently).
I'm not trying to explain how Photoshop works!  There are so many alternatives.
Best regards --Gernot Hoffmann