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A Step Backward ?

 
 
Roger N. Clark (change username to rnclark)
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      08-27-2006
Confused wrote:

> On Wed, 23 Aug 2006 08:23:24 -0600
> In message <(E-Mail Removed)>
> "Roger N. Clark" <(E-Mail Removed)> wrote:
>
>>>...
>>>This seems to establish a fact:
>>>the two cameras are very different.

>>
>>They are different only in the fact that one is larger than
>>the other in all aspects.
>>...

>
>>>>Are the pictures from the two cameras the same?
>>>
>>>No.

>>
>>Correct! The images from both cameras are absolutely
>>identical in every respect except that the image from
>>camera B, the larger camera, has higher a signal-to-noise
>>ratio (2x higher).

>
> I've been thinking about this and it doesn't make sense. (This duck
> is in the wrong pond <g> so I'll drop out of the exercise unless you
> want go through everything one step at a time. I tacitly assumed up
> front that whis was closed book test.)
>
> Cameras A and B are identical in all aspects except size.
>
> The software collecting the data will collect and calculate a
> relatively identical amount of data and signal to noise ratio.
> 4 x's the sensor area; 4 x's the data; 4 x's the noise.
>
> To me this is no different than collecting rain at the same location
> with 2 pans; one twice the size of the other. There will be virtually
> identical percentages of pollution in both pans. The larger pan will
> collect 4 x's the polluted water.


Actually, the idea of collecting rain drops is an excellent
analogy, assuming the drops are falling randomly, which is
probably the way it is.

The noise in the counted raindrops collected in any one
container is the square root of the number of rain drops.
This is called Poisson statistics (check wikipedia).
So if you double the count (photons in a pixel or rain drops
in a bucket), the noise in the count will go up by
square root 2. For example, put out 10 buckets and you
would find the level in the buckets that on average
collected 10,000 rain drops, would vary in the measured
amount of water by 1% from bucket to bucket (the standard
deviation):
square root (10,000)/10,00 = 0.01 = 1%

(10,000 rain drops is about 0.5 liter of water by the way.)

So with Poisson statistics, which is the best that can be
done measuring a signal based on random arrival times
(e.g.of photons), the
signal / noise = signal / square root(signal) = square root(signal).

So in our camera test, collecting 4x the photons increases
signal to noise ratio square root (4) = 2.

Fortunately, most digital cameras have such noise characteristics
except at near zero signal. This means that improving
noise performance can only come through increasing the photon
count. That can be done 3 ways: increasing quantum efficiency
(currently dcams are around 30%), fill factor (most are probably
already above 80%), or increase the pixel size (e.g. the larges
bucket collects more rain drops).

>>Next question:

>
> Just for grins...
>
>>Assuming there is no change in aberrations if you change f/stop,
>>what could be done to the above test images to make camera B
>>produce an image that is completely identical to that from camera A?
>>A: Assume the subject is static; no movement, so there
>> are two answers; extra credit for giving both.

>
> Increase the ISO.


Increasing the ISO does not change how many photons
a digital camera collects. ISO is only a post photon
collection gain to digitize the photon signal.

> Stop it down and decrease the shutter speed.


Yes! We have 4x the photons, so the two answers are:
1) stop down 2 stops to decrease the light level 4x.
2) decrease the shutter speed 4x.

>>B: Assume the subject is not static, then there is one answer.
>>What is it?

>
> Increase the ISO.
>
> Changing the shutter speed would significantly alter the image.
> (Not sure how that might change the amount of noise but changing the
> ISO is the obvious answer.)


While changing ISO changes the perceived image, it does not
change the number of photons collected. The one and only
answer is stop the lens down two stops. This reduces the
photon count and also happens to make the depth of field the
same as the smaller sensor camera, finally making the
results from the two cameras identical (total photons
per pixel as well as depth of field). Changing the
iso higher 2 stops would bring the digitized signal to
the same relative level as the small camera, but that could
also be done in post processing (again the photon count
and signal-to-noise ratio would be the same). The iso change
would also make the metering the same as the small camera,
then the metered shutter speeds would be identical too.

In real cameras, boosting the iso a good step as it reduces
A/D quantization and reduces the read noise contribution
to the signal.

Overall, you did quite well, only stumbling a little at the
end. Congratulations, A overall and A+++ for effort.

I'll assemble the questions and answers into one post in
case other people want to go through the exercise.

Roger
 
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David J. Littleboy
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      08-27-2006
"Roger N. Clark (change username to rnclark)" <(E-Mail Removed)> wrote:
>
> In real cameras, boosting the iso a good step as it reduces
> A/D quantization and reduces the read noise contribution
> to the signal.


It's an interesting exercise to find the correct exposure for a scene for
the a camera's highest ISO and then shoot that scene at each ISO setting on
the camera with the same exposure. Then correct the exposure in
postprocessing and compare the results.

It says that there's quite a bit more dynamic range to be had from our
sensors at low ISOs if the electronics were a bit less noisy and the A/D
converter had a few more bits. (Of course, we may not be willing to pay for
or lift the electronics required to do that...)

David J. Littleboy
Tokyo, Japan


 
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Roger N. Clark (change username to rnclark)
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      08-29-2006
Roger N. Clark:
> I'll assemble the questions and answers into one post in
> case other people want to go through the exercise.


Here it is:

Let's start from the beginning and work through things one step
at a time. First rule: don't jump ahead to conclusions.

The first question (this is not a trick question):

-----------
Given two cameras A and B: Camera B is a perfectly
scaled up version of camera A. Thus, the sensor in
camera B is twice as large as the sensor in camera A.
Both cameras have the same number of pixels, such that
the pixels in cameras B are twice the linear size
(4 times the area) of camera A.
Each camera has an f/2 lens and exposes a
scene with constant lighting at f/2.
The lens on camera B is twice the focal length as
camera A so that both cameras record the same
field of view.
Let's say that camera A collects 10,000 photons
in each pixel in a 1/100 second exposure at f/2.

How many photons are collected in each pixel in
camera B in the same exposure time on that same
scene with the same lighting at f/2?

Answer: 4 * 10,000 = 40,000 photos per pixel.
The number of photons collected scales as the area of
the pixel. The area of a pixel in camera B is 4 times
that of camera A, so the answer is 4 * 10,000.

---------
Now that we have established that the two cameras have
identical fields of view, just that one camera is twice size
of the other. Their spatial resolution on the subject is
identical. Now let's say we are taking a picture
of a flat wall so there are no depth of field issues.
The images have the same pixel count, the same
field of view, and the same spatial resolution
taken with the same exposure time and f/stop.

Are the pictures from the two cameras the same?

Answer: No. They are the same except camera B
has collected 4 times the photons.
The images from both cameras are absolutely
identical in every respect except that the image from
camera B, the larger camera, has higher a signal-to-noise
ratio (2x higher).

An analogy to collecting photons in a camera with pixels is like
collecting rain drops in buckets, assuming the drops are
falling randomly, which is probably the way it is. Larger
buckets collect more rain drops. If you measure the number
of drops collected from a bunch of buckets, you will find
that the amount in each bucket is slightly different.

The noise in the counted raindrops collected in any one
container is the square root of the number of rain drops.
This is called Poisson statistics (e.g. check wikipedia).
So if you double the count (photons in a pixel or rain drops
in a bucket), the noise in the count will go up by
square root 2. For example, put out 10 buckets and you
would find the level in the buckets that on average
collected 10,000 rain drops, would vary in the measured
amount of water by 1% from bucket to bucket (the standard
deviation):
square root (10,000)/10,000 = 0.01 = 1%

(10,000 rain drops is about 0.5 liter of water by the way.)

So with Poisson statistics, which is the best that can be
done measuring a signal based on random arrival times
(e.g.of photons), the
signal / noise = signal / square root(signal) = square root(signal).

So in our camera test, collecting 4x the photons increases
signal-to-noise ratio by square root (4) = 2.

Fortunately, most digital cameras have such noise characteristics
except at near zero signal. This means that improving
noise performance can only come through increasing the photon
count. That can be done 3 ways: increasing quantum efficiency
(currently dcams are around 30%), fill factor (most are probably
already above 80%), or increase the pixel size (e.g. the larger
bucket collects more rain drops).

----------------
Next question:

Assuming there is no change in aberrations if you change f/stop,
what could be done to the above test images to make camera B
produce an image that is completely identical to that from camera A?

A: Assume the subject is static; no movement, so there
are two answers; extra credit for giving both.

B: Assume the subject is not static, then there is one answer. What
is it?

Answer A:
We have 4x the photons, so the two answers are:
1) stop down 2 stops to decrease the light level 4x.
2) decrease the shutter speed 4x.

(OPTIONAL: Increase the ISO):
While changing ISO changes the perceived image, it does not
change the number of photons collected.

Answer B: The one and only answer is
stop the lens down two stops. This reduces the
photon count and also happens to make the depth of field the
same as the smaller sensor camera, finally making the
results from the two cameras identical (total photons
per pixel as well as depth of field). Changing the
ISO higher 2 stops would bring the digitized signal to
the same relative level as the small camera, but that could
also be done in post processing (again the photon count
and signal-to-noise ratio would be the same). The ISO change
would also make the metering the same as the small camera,
then the metered shutter speeds would be identical too.

In real cameras, boosting the ISO increase is a good step
as it reduces A/D quantization and reduces the read noise
contribution to the signal.

So, what was the result of the exercise? In making the images
from two different sized cameras identical in terms of resolution,
angular coverage, exposure time, and signal-to-noise ratio,
we find the final property: the depth of field is also identical.

I have added this discussion to:
The Depth-of-Field Myth and Digital Cameras
http://www.clarkvision.com/photoinfo/dof_myth

Roger
 
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Confused
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      08-30-2006
On Mon, 28 Aug 2006 19:57:52 -0700
In message <(E-Mail Removed)>
Roger N. Clark wrote:

Oh-key-doh-key

All right then.

LIFO

> So, what was the result of the exercise?
> ...
> we find the final property: the depth of field is also identical.


I really didn't know what the thread was about; didn't peek back or
ahead in the threads and didn't know what the purpose of the exercise
was.

> Overall, you did quite well, only stumbling a little at the
> end. Congratulations, A overall and A+++ for effort.


Oh no... say it isn't so! Putting in an effort and then failing an
exam (or a real world project like a space mission) is failing. <G>

> ...
> the image from camera B, the larger camera, has a higher
> signal-to-noise ratio (2x higher).


Shouldn't that read "lower signal-to-noise ratio" ?

(relatively less noise with larger pixels)
(4x data, 2x noise)

> 10,000 rain drops is about 0.5 liter of water by the way.


Wait one moment there!!!! Rain falls under diverse conditions and
I've seen rain drops of many sizes with my own eyes! How on earth (as
opposed to in a lab) can 10,000 rain drops always be about 1/2
liter????? I've seen those suckers bigger'n gum balls one day and
smaller than pin pricks the next! ;o)

Jeff
 
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