Dynamic range of an image

In message ,
Peter Rongsted wrote:

Further more each pixel is only in one channel in raw. Remember the
Bayer filter.

I have ways here of looking at the image as 4 quadrants, each with a
1.58MP greyscale image from each r, G, G2, or B element of each RGGB
tile. In other words, the red channel in one quadrant, the blue channel
in another quadrant, and two green channel quadrants. They are all
aliased, of course. If you combine them into one 1.58MP image, it looks
horrible if there is any high-contrast sharpness in the image, with
glaring color fringes around all the edges (this is how George Preddy
claims all Bayer images start out, before being scaled up to 4x as many
pixels).
--


John P Sheehy

wrote:

In message ,
"Roger N. Clark (change username to rnclark)"
wrote:


DN is "Data Number." That is the number in the file for each
pixel. I'm quoting the luminance level (although red, green
and blue are almost the same in the cases I cited).

16-bit signed integer: -32765 to +32765

16-bit unsigned integer: 0 to 67535

Photoshop uses signed integers, but the 16-bit tiff is
unsigned integer (correctly read by ImagesPlus).


How do you know that photoshop isn't using unsigned integers, but just
not using the most significant bit? I could see more use for the
headroom in intermediate calculations than negative values.

Because all DNs in photoshop ar 1/2 those from ImagesPlus.
Photoshop maxes out at 32765 and ImagesPLus 65535 (note
correction to above, not 67535). If you
just ignore the top bit, half the image would be saturated
in photoshop. Photoshop had to divide all the
DNs by 2.

On octal dump of the tif files show values up in the 65,000 range,
so the original file is 16-bit unsigned integers.

Roger

wrote:

In message ,
"Roger N. Clark (change username to rnclark)"
wrote:


DN is "Data Number." That is the number in the file for each
pixel. I'm quoting the luminance level (although red, green
and blue are almost the same in the cases I cited).

16-bit signed integer: -32765 to +32765

16-bit unsigned integer: 0 to 67535

Photoshop uses signed integers, but the 16-bit tiff is
unsigned integer (correctly read by ImagesPlus).


How do you know that photoshop isn't using unsigned integers, but just
not using the most significant bit? I could see more use for the
headroom in intermediate calculations than negative values.

Because all DNs in photoshop ar 1/2 those from ImagesPlus.
Photoshop maxes out at 32765 and ImagesPLus 65535 (note
correction to above, not 67535). If you
just ignore the top bit, half the image would be saturated
in photoshop. Photoshop had to divide all the
DNs by 2.

On octal dump of the tif files show values up in the 65,000 range,
so the original file is 16-bit unsigned integers.

Roger

In message ,
"Roger N. Clark (change username to rnclark)"
wrote:

wrote:

How do you know that photoshop isn't using unsigned integers, but just
not using the most significant bit? I could see more use for the
headroom in intermediate calculations than negative values.

Because all DNs in photoshop ar 1/2 those from ImagesPlus.

I know that. And half of what they load in as in Mathcad as well.
That was taken for granted in my response. My question was in contrast
to you "unsigned" theory, which also relies on the same assumption
(halved values).

Photoshop maxes out at 32765

32,767, to be exact.

and ImagesPLus 65535 (note
correction to above, not 67535). If you
just ignore the top bit, half the image would be saturated
in photoshop. Photoshop had to divide all the
DNs by 2.

Which is the same thing as shifting the bits to the right, and putting
the leftmost (MSB) as an unused range.

On octal dump of the tif files show values up in the 65,000 range,
so the original file is 16-bit unsigned integers.

The 15 least significant bits for 16-bit signed and unsigned are exactly
the same for #s 0 through 32767. The only difference is that the MSB
adds more positive numbers in unsigned (32768 to 65535), and signals
negative numbers (-32768 to -1) in signed.

My question was, why do you think that the numbers are unsigned?
--


John P Sheehy

In message ,
"Roger N. Clark (change username to rnclark)"
wrote:

wrote:

How do you know that photoshop isn't using unsigned integers, but just
not using the most significant bit? I could see more use for the
headroom in intermediate calculations than negative values.

Because all DNs in photoshop ar 1/2 those from ImagesPlus.

I know that. And half of what they load in as in Mathcad as well.
That was taken for granted in my response. My question was in contrast
to you "unsigned" theory, which also relies on the same assumption
(halved values).

Photoshop maxes out at 32765

32,767, to be exact.

and ImagesPLus 65535 (note
correction to above, not 67535). If you
just ignore the top bit, half the image would be saturated
in photoshop. Photoshop had to divide all the
DNs by 2.

Which is the same thing as shifting the bits to the right, and putting
the leftmost (MSB) as an unused range.

On octal dump of the tif files show values up in the 65,000 range,
so the original file is 16-bit unsigned integers.

The 15 least significant bits for 16-bit signed and unsigned are exactly
the same for #s 0 through 32767. The only difference is that the MSB
adds more positive numbers in unsigned (32768 to 65535), and signals
negative numbers (-32768 to -1) in signed.

My question was, why do you think that the numbers are unsigned?
--


John P Sheehy

wrote:

In message ,
"Roger N. Clark (change username to rnclark)"
wrote:


wrote:


How do you know that photoshop isn't using unsigned integers, but just
not using the most significant bit? I could see more use for the
headroom in intermediate calculations than negative values.


Because all DNs in photoshop ar 1/2 those from ImagesPlus.


I know that. And half of what they load in as in Mathcad as well.
That was taken for granted in my response. My question was in contrast
to you "unsigned" theory, which also relies on the same assumption
(halved values).


Photoshop maxes out at 32765


32,767, to be exact.
oops, yes. My fingers don't seem to be typing the right keys
this weekend!


and ImagesPLus 65535 (note
correction to above, not 67535). If you
just ignore the top bit, half the image would be saturated
in photoshop. Photoshop had to divide all the
DNs by 2.


Which is the same thing as shifting the bits to the right, and putting
the leftmost (MSB) as an unused range.

Yes. Aren't we saying the same thing?


On octal dump of the tif files show values up in the 65,000 range,
so the original file is 16-bit unsigned integers.


The 15 least significant bits for 16-bit signed and unsigned are exactly
the same for #s 0 through 32767. The only difference is that the MSB
adds more positive numbers in unsigned (32768 to 65535), and signals
negative numbers (-32768 to -1) in signed.

My question was, why do you think that the numbers are unsigned?

Octal dumps of the tif file shows they are unsigned integers.

Roger

Roger N. Clark (change username to rnclark) wrote:

In another thread in this newsgroup, the dynamic range
of an image and DSLRs are being discussed. In case you
are interested, I put together a page showsing some results,
and were not following that thread, see below.

Here is a page I put together regarding dynamic range.
I went out in the backyard and snapped a few pictures of
a pretty random scene and found it had a dynamic range of
11 bits, and than includes all diffusely reflecting objects
(no bright metal reflections, no bright clouds in the scene).

http://clarkvision.com/imagedetail/dynamicrange

The canon 10D recorded 11 bits of dynamic range
(max signal / RMS noise near the minimum intensity levels).

I've added another figu

Figure 5. Jpeg image values plotted against 16-bit linear
image values shows the jpeg transfer function (the default contrast on the 10D).
Note that the jpeg saturates at about half the level of the
16-bit data. The 16-bit file has a 1 stop increase in dynamic range before
data saturation compared to the jpeg image.

The jpeg image has less dynamic range recorded (the lows and highs are
clipped), and less signal to noise. While the low end of the
jpeg is close to linear, the Jpeg has only 50 integer levels while the
16-bit tif image has about 1880 integer levels over the same intensity
range, or about 37 times finer intensity detail. Toward the bright
end, the 16-bit tiff image has even more detail than the low end of
the jpeg image. For example, in the jpeg data from DN 337 to 247,
the 16-bit image goes from 22800 to 25200 or 2400 DN. Thus the 16-bit
tiff imge has 2400/10 = 240 times the intensity detail!

"Roger N. Clark (change username to rnclark)" wrote
in :

I've added another figu

Figure 5. Jpeg image values plotted against 16-bit linear
image values shows the jpeg transfer function (the default contrast on
the 10D).


Nice!

It shows that you lose a little more than one stop in the bright part.

The first image (converted from RAW) - is that taken at 1/4000 s?


/Roland

Roger N. Clark (change username to rnclark) wrote:
[]
I've added another figu

Figure 5. Jpeg image values plotted against 16-bit linear
image values shows the jpeg transfer function (the default contrast
on the 10D). Note that the jpeg saturates at about half the level of
the 16-bit data. The 16-bit file has a 1 stop increase in dynamic
range before data saturation compared to the jpeg image.

An excellent addition. As the transfer characteristic is suspected to be
a gamma correction, would a log-log plot be more appropriate?

Cheers,
David

Roger N. Clark (change username to rnclark) wrote:
[]
I've added another figu

Figure 5. Jpeg image values plotted against 16-bit linear
image values shows the jpeg transfer function (the default contrast
on the 10D). Note that the jpeg saturates at about half the level of
the 16-bit data. The 16-bit file has a 1 stop increase in dynamic
range before data saturation compared to the jpeg image.

An excellent addition. As the transfer characteristic is suspected to be
a gamma correction, would a log-log plot be more appropriate?

Cheers,
David