Exposure and histogram stretching

PS 6. Somewhat theoretical questions but I hope someone responds.
Using Photoshop, how do I emulate image (histogram) changes resulting from scanner exposure boost? This is not as simple as it may appear:
I ran some tests with my film scanner by increasing exposure by a fixed amount (0.1 ev increments). I then plotted the pixel value increase against ev increments and the result, instead of a straight line, was a slightly upward sloping curve. In other words the histogram did not just move to the right from scan to scan but was actually "stretched" (as expected). Pixels on the left "moved" more slowly than pixels on the right i.e. the distance between darkest and brightest pixels increased with exposure boost (more dynamic range).
Taking the baseline scan (before exposure increases) and applying brightness to it in Photoshop resulted in a different histogram. This time the whole histogram just shifted to the right, but the distance between the darkest and brightest pixels remained the same (same dynamic range).

So, how do I emulate this non-linear exposure histogram "stretching" in PS? I know, "use curves", but how exactly?

PS 6. Somewhat theoretical questions but I hope someone responds.
Using Photoshop, how do I emulate image (histogram) changes resulting from scanner exposure boost? This is not as simple as it may appear:

To expand the dynamic range try "levels": The darkest point in your image may be RGB (15,15,15), the brightest RGB (200,200,200) then the 186 intensities you have are spread over the full 256 value range (this might lead to banding if there are too few intensities used).

On Thu, 27 May 2004 07:35:13 GMT, (Xalinai)
wrote:

PS 6. Somewhat theoretical questions but I hope someone responds.
Using Photoshop, how do I emulate image (histogram) changes resulting from scanner exposure boost? This is not as simple as it may appear:

…

To expand the dynamic range try "levels": The darkest point in your image may be RGB (15,15,15), the brightest RGB (200,200,200) then the 186 intensities you have are spread over the full 256 value range (this might lead to banding if there are too few intensities used).

The problem is this will be done in a linear fashion i.e. the inserted "holes" will appear at regular intervals.

This is quite different to what happens when the scanner exposure is boosted. There are more "new values" in the highlights than in the shadows.

PS 6. Somewhat theoretical questions but I hope someone responds.
Using Photoshop, how do I emulate image (histogram) changes resulting from scanner exposure boost? This is not as simple as it may appear:
I ran some tests with my film scanner by increasing exposure by a fixed amount (0.1 ev increments). I then plotted the pixel value increase against ev increments and the result, instead of a straight line, was a slightly upward sloping curve. In other words the histogram did not just move to the right from scan to scan but was actually "stretched" (as expected). Pixels on the left "moved" more slowly than pixels on the right i.e. the distance between darkest and brightest pixels increased with exposure boost (more dynamic range).
Taking the baseline scan (before exposure increases) and applying brightness to it in Photoshop resulted in a different histogram. This time the whole histogram just shifted to the right, but the distance between the darkest and brightest pixels remained the same (same dynamic range).

So, how do I emulate this non-linear exposure histogram "stretching" in PS? I know, "use curves", but how exactly?

And while we’re at it, given an arbitrary starting pixel value and a specific EV exposure increase applied to it, does anyone happen to know the formula to calculate the resulting pixel value?
Don.

PS 6. Somewhat theoretical questions but I hope someone responds.
Using Photoshop, how do I emulate image (histogram) changes resulting from scanner exposure boost? This is not as simple as it may appear:

…

To expand the dynamic range try "levels": The darkest point in your image may be RGB (15,15,15), the brightest RGB (200,200,200) then the 186 intensities you have are spread over the full 256 value range (this might lead to banding if there are too few intensities used).

The problem is this will be done in a linear fashion i.e. the inserted "holes" will appear at regular intervals.

If all your original intensities fit in the range of 15 to 200 then your image only _has_ a maximum of 186 different intensities.
You see, if your image had exactly 186 pixels, each having one value in the given range, then using the full dynamic range would leave some values unused.

You can try and scan at twice the resolution needed and resample to 50% after using levels – that gives a fine homogenuous histogram as each new pixel value is calculated from four pixels in the original image. The more intelligent approach is to do it right.

This is quite different to what happens when the scanner exposure is boosted. There are more "new values" in the highlights than in the shadows.

Have you noticed that there is a midpoint setting in levels? This controls the distribution curve. Play with it for a while.
If you want to do it right, then correct the setting in the scanner software while you are in the deeper color range of the scanner. The function you need there is not exposure but setting of black and white points, even gamma correction might help.

If you start correcting in the 8bit area in PS you are already lost.

(Don) wrote in message news:…

And while we’re at it, given an arbitrary starting pixel value and a specific EV exposure increase applied to it, does anyone happen to know the formula to calculate the resulting pixel value?
Don.

Don,

I can only comment your last question, but not by "exposure":
x1 old left end of histogram
x2 old right end of histogram
u1 new left end
u2 new right end
x input
y old output
v new output

Old: y = x

New: v = x1 +(x-u1)*(x2-x1))/(u2-u1)

Test:
x=u1: v=x1
x=u2: v=x2

Two test points are sufficient, because it

Hi Gernot,

I’m afraid I don’t follow completely. Can you please explain what do you mean by input (x), old output (y) and new output (v)?
To explain it another way I was expecting the formula to look something like:

u1 = x1 + offset1
u2 = x2 + offset2

or in a general form:

new pixel value = old pixel value + offset

where "offset" is a function of both exposure and of old pixel value.
My tests show that "offset" is directly proportional to both the exposure and the starting (old) pixel value. So:
The larger the initial (old) pixel value, the larger the offset. The larger the exposure, the larger the offset.

Also, I just learned elsewhere that gamma plays an important part in all this and since I work in gamma 2.2 this may account for non-linear pixel value progression.

Don.

Don,

my previous suggestion was a stretching but not the expected one. A new attempt:

The darkest entry in a histogram is at x1, the lightest at x2. x1 should be mapped to u1, e.g. to u1=0.
x2 should be mapped to u2, e.g. to u2=1 (normalized instead of 255).
In a linear space this is a linear function:
v(x) = a + b*x

x is the input and v(x) the modified output.
We know two points:

a + b*x1 = u1
a + b*x2 = u2

The linear equations can be solved by Cramer

On 28 May 2004 00:17:12 -0700, (Gernot Hoffmann)
wrote:

Don,

my previous suggestion was a stretching but not the expected one. A new attempt:

The darkest entry in a histogram is at x1, the lightest at x2. x1 should be mapped to u1, e.g. to u1=0.
x2 should be mapped to u2, e.g. to u2=1 (normalized instead of 255).
In a linear space this is a linear function:
v(x) = a + b*x

x is the input and v(x) the modified output.
We know two points:

a + b*x1 = u1
a + b*x2 = u2

The linear equations can be solved by Cramer´s rule.

v(x) = a + x*b = (u1*x2-u2*x1)/(x2-x1) + x*(u2-u1)/(x2-x1)
Special case: u1=0, u2=1
v(x) = (x-x1)/(x2-x1)
v(x1)=0 and v(x2)=1 as expected.

Application for Gamma encoded values
Input x=0..255
Output v=0..255
Replace x by (x/255)^G
Replace v by Round(255*v^(1/G))

Best regards –Gernot Hoffmann

Hi Gernot,

OK, that makes sense now!

I still have to chase up "Cramer´s rule" on the Net, but at least I have to enough to get going.

Thanks very much, as always.

Don.

Don,

thanks for your friendly feedback.

I hope my idea is correct. Probably you will understand
that these contributions are always "under pressure".

Cramer