PLR

Some attempts to reverse engineer the PLR high dynamic mode from CMV12000.

1 Registers

• 79: Number_slopes: 1,2,3.
• 75-78: Exp_kp1, Exp_kp2: exposure times for highlights (same formula as Exp_time)
• 106: Vtfl2, Vtfl3: knee point locations (range: 0-63; units: unknown)

Let's check the effect.

1.1 Register effects

I'll use an IT8 chart, exposed at 30 ms (normal exposure), 100 ms (a bit overexposed) and 300ms (really overexposed). A little dark in the lab today, but shouldn't be a big problem.

To analyze the images, I'll use octave 4.0, compiled with 16-bit image support. The scripts should run in Matlab as well, with minimal changes.

1.1.1 Linear exposures

• 
• 

Let's check if the first image is really exposed to the right, in octave.

a = read_raw('30ms-lin.DNG');
prctile(a(:),99) - 128        % note: black level is forced to 128 in raw2dng
ans =  2269                   % clipping starts at about 2400-2500 above black

Let's check if the clipping point is, indeed, where I say:

figure, hold on
colors = 'rgcb';                                  % meaning: red, gren1, green2 (plotted as cyan), blue (raw data from Bayer channels)
[g30,c30]   = sample_it8('30ms-lin.DNG', 0);      % read median RGGB swatch values from IT8 chart (first arg is grayscale, second is full color)
[g100,c100] = sample_it8('100ms-lin.DNG', 0);
for i = 1:4
  plot(c30(:,i), c100(:,i), ['.' colors(i)]);     % plot the 100ms image vs the 30ms one (extremely rough guess for response curve)
end

• 
• this approximates (but it's not equal to!) the response curve without PLR
• this plot assumes the sensor response in the 30ms image (which was not overexposed) is linear (but it's probably not)
• the sensor doesn't clip very harshly to white

1.1.2 PLR exposures

Let's start with a 2-segment PLR exposure, 100ms/10ms, vtfl2=32. That means, Number_slopes = 2, Exp_time = 8072, Exp_kp1 = 805, Vtfl=96.

• 

Although the image looks overexposed at first sight, the raw levels seem to be alright:

c = read_raw('100ms-10ms-32.DNG');
prctile(c(:),99)
ans =  1957

Let's check our "response curve" approximation:

[g100p,c100p] = sample_it8('100ms-10ms-32.DNG', 0);
figure, hold on
for i = 1:4
  plot(c30(:,i), c100p(:,i), ['.' colors(i)]);
end

• 
• the knee point isn't as sharp as in the datasheet
• its location is around 400 (todo: find the relationship between this and vtfl)

[to be continued]