Velocity Reviews > Photon noise: is it white?

# Photon noise: is it white?

Philip Homburg
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Posts: n/a

 10-21-2006
In article <(E-Mail Removed). com>,
acl <(E-Mail Removed)> wrote:
>Philip Homburg wrote:
>> In article <ehbpki\$2duu\$(E-Mail Removed)>,
>> Ilya Zakharevich <(E-Mail Removed)> wrote:
>> >The ultimate solution is to consider it as an honest
>> >quantum-mechanical system, and just calculate the answer. I'm too
>> >lazy to do it right now; maybe somebody already KNOWS the answer?

>>
>> I think it is in the end very simple: the number of photons that are
>> recorded by a sensor element has an uncertainty of the square root of
>> that number.

>
>What he is arguing is that there is a correlation in the noise detected
>by neighbouring sensor elements. He is wrong.

In practice there will be a correlation. Most images have relatively
little energy around Nyquist. So, neighboring sensor elements will have
'similar' values, which in turn results in similar photon noise.

Of course, in a Bayer pattern sensor, you have to consider neighboring
elements of the same color.

>> The thing is that traditional white noise is to a large extent independent of
>> the signal. However, photon noise is part of the signal. Wherever your
>> signal goes (diffraction, colored filters, etc) photon noise comes with it.
>>

>

I don't know where you see sources of white noise. I see them mostly in
electronics, and not much in other areas.

>What you are saying (that the noise goes with the signal) could be
>modelled phenomenologically (ie in a hand-waving way) by writing, for
>the signal, s(x)+f(x), where x is a 2-d vector, s is the signal and f
>is the noise, and specifying a zero-mean normal distribution for f (ie
>it is white). This will work because, at this level of approximation,
>nothing nonlinear ever happens to the signal (notwithstanding
>complicated signal processing). If there was nonlinear processing, we'd
>need all the moments of f.

Fortunately, optics tends to be linear.

>Anyway, I think I discussed this with you some time in the past (or
>maybe it wasn't you). And this is going seriously off topic, so I'll
>stop.

So, to get back on track, is 'f(x)' white or not?

--
That was it. Done. The faulty Monk was turned out into the desert where it
could believe what it liked, including the idea that it had been hard done
by. It was allowed to keep its horse, since horses were so cheap to make.
-- Douglas Adams in Dirk Gently's Holistic Detective Agency

Roger N. Clark (change username to rnclark)
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 10-21-2006
Ilya Zakharevich wrote:
> For a long time I lived under impression that photon noise IS white.
> Now I thought about it more, and I'm not absolutely sure. I can make
> two "English language" arguments, and by one of them the noise is
> white, by another its spectrum is (related to) the MTF of the lens.
>
> The problem is that a photon emitted by the source hits several
> sensels SIMULTANEOUSLY (distributed by PSP). At most one of the
> sensels will register the photon, so THERE IS a correlation between
> readings of different sensels. The question is whether this
> correlation survives when one considers not one photon, but a Poisson
> distribution of photons (possibly emitted by different sources).
>
> The ultimate solution is to consider it as an honest
> quantum-mechanical system, and just calculate the answer. I'm too
> lazy to do it right now; maybe somebody already KNOWS the answer?
>
> Thanks,
> Ilya

Start here
http://www.imatest.com/docs/noise.html

then study Poission statistics.

Roger

acl
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Posts: n/a

 10-21-2006
Philip Homburg wrote:
> In article <(E-Mail Removed). com>,
> acl <(E-Mail Removed)> wrote:
>>What he is arguing is that there is a correlation in the noise detected
>>by neighbouring sensor elements. He is wrong.

>
> In practice there will be a correlation. Most images have relatively
> little energy around Nyquist. So, neighboring sensor elements will have
> 'similar' values, which in turn results in similar photon noise.

Well yes, but that is because of the characteristics of the image
(neighbouring areas have similar intensities). This is not what he is
arguing, as far as I can tell. If I take what he says literally, he is
arguing that there is correlation because of the PSP, ie, because of the
wave-like nature of light (ie mostly because of diffraction, in this
case). Well yes there is. This is why there's an image formed. So yes,
as you say, this will result in in neighbouring areas to be correlated,
but this is trivial. That's why in my other post I specified that I
include things like diffraction etc in the definition of "image".

>
> Of course, in a Bayer pattern sensor, you have to consider neighboring
> elements of the same color.
>
>>>The thing is that traditional white noise is to a large extent independent of
>>>the signal. However, photon noise is part of the signal. Wherever your
>>>signal goes (diffraction, colored filters, etc) photon noise comes with it.
>>>

>
> I don't know where you see sources of white noise. I see them mostly in
> electronics, and not much in other areas.

Maybe you mostly look at electronics. In, for instance, physics, almost
nothing else is used except white noise. And when coloured noise
appears, it may be eliminated in favour of twice the number of equations
and white noise only; this is what people usually do to study such
systems. Also in theoretical biology, etc.

Think, for example, of the statistics of particles arriving on a surface
randomly (eg deposited from above) and the morphology of the resulting
configuration. There is a subsubfield of physics dealing with this,
motivated by molecular beam epitaxy, and which usually aims to derive
and study stochastic differential equations describing the growth of the
resulting pile of particles. There, there is intrinsic noise (same
reasons as for photons), and the necessary stochastic equations involve
white noise. This is derived, as opposed to postulated (but this was
correctly done only very recently).

Anyway, there is a very large body of literature in physics dealing with
processes with white noise. There is no question about that.

>
>>What you are saying (that the noise goes with the signal) could be
>>modelled phenomenologically (ie in a hand-waving way) by writing, for
>>the signal, s(x)+f(x), where x is a 2-d vector, s is the signal and f
>>is the noise, and specifying a zero-mean normal distribution for f (ie
>>it is white). This will work because, at this level of approximation,
>>nothing nonlinear ever happens to the signal (notwithstanding
>>complicated signal processing). If there was nonlinear processing, we'd
>>need all the moments of f.

>
> Fortunately, optics tends to be linear.
>
>>Anyway, I think I discussed this with you some time in the past (or
>>maybe it wasn't you). And this is going seriously off topic, so I'll
>>stop.

>
> So, to get back on track, is 'f(x)' white or not?

I had said it's normally distributed. White noise=noise with power in
all frequencies, so yes, it's white, uncorrelated noise. On second
thought, though, maybe it's better to use s(x)+sqrt(s) f(x), so that the
average of the square of the noise is proportional to the signal itself.
f is still white and uncorrelated here.

But this is hand waving. I'm sure it will give the right answers,
though, and that it can be derived as follows: We know the photon
arrival times are governed by Poisson stats. Given the exposure time and
the intensity of the signal, we can easily find the magnitude of the
standard deviation of the noise for each site. It will be something like
what I wrote above, I suppose. But I am drowning in work at the moment
and cannot actually check details.

By the way, to avoid confusion, what people normally mean with
Poisson-distributed "shot noise" for photons is a Poisson process in
time. It's different than the s(x) I have above, which models the fact
that noise is present and uncorrelated across pixels. Well, except
through the dependence on s.

Marvin
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Posts: n/a

 10-21-2006
Ilya Zakharevich wrote:
> For a long time I lived under impression that photon noise IS white.
> Now I thought about it more, and I'm not absolutely sure. I can make
> two "English language" arguments, and by one of them the noise is
> white, by another its spectrum is (related to) the MTF of the lens.
>

The noise arises mainly in the sensor, so it has nothing to
do with the lens. Technically, it is described as "white",
meaning that the noise pulses are randomly distributed in
time. The confusion comes from having two meanings for
"white", depending on the context.

Marvin
Guest
Posts: n/a

 10-21-2006
Ilya Zakharevich wrote:
> For a long time I lived under impression that photon noise IS white.
> Now I thought about it more, and I'm not absolutely sure. I can make
> two "English language" arguments, and by one of them the noise is
> white, by another its spectrum is (related to) the MTF of the lens.
>

A correction to my posting a few minutes ago. I said that
white noise is random with time. That is true for photon
counting detectors. For the kind of detectors in digicams,
it is more significant that the noise pulses are randomly
distributed in amplitude. In some cases, the pulses become
systematically less frequent as the amplitude increases, and
it may be called 1/f noise, meaning that the number of
pulses per unit time is inversely proportional to the amplitude.

Ilya Zakharevich
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Posts: n/a

 10-22-2006
[A complimentary Cc of this posting was NOT [per weedlist] sent to
Ilya Zakharevich
<(E-Mail Removed)>], who wrote in article <ehbpki\$2duu\$(E-Mail Removed)>:
> For a long time I lived under impression that photon noise IS white.
> Now I thought about it more, and I'm not absolutely sure. I can make
> two "English language" arguments, and by one of them the noise is
> white, by another its spectrum is (related to) the MTF of the lens.

Thanks to everybody how answered; however, even the people who
actually understand what is a "white noise" and "correlation" could
only make hand-waving arguments; AFAIU, nobody could claim a precise
answer... Hand-wave can even I; the problem was that I could wave in
two different ways.

Anyway, I made the calculation, and I can sleep well again: the noise
is indeed white, as I was supposing all the time. The calculation
attached.

Thanks again,
Ilya

================================================== =====

Consider an image of an evenly lighted while surface created by a lens.
Is the photon noise in the image white? In other words, is there a
correlation between photon count in the (neighboring) pixels?

It turns out that photon noise is indeed white, notwithstanding the
correlation between sensels when individual photons hit the sensor.

To simplify the situation so that analysis is handy, consider a sensor
with two sensels, A and B; model the evenly lighted white source by two
point sources, sA and sB. Assume sA is focused on sensel A, sB is
focused on sensel B. Assume that photons from sA hit sensel A with
probability (1+p)/2 and hit sensel B with probability (1-p)/2; likewise
for sB - with A and B exchanged.

What we need to do is to calculate the distribution of the sum S of
readings of sensels A and B, and the difference D between these readings.
If the photon noise is white, these distributions should be the similar.
S is obvious; I do not discuss it here.

It turns out that it makes sense to break D into two parts, D = D0 + D';
here each photon from sA increases D0 by p, each photon from sB decreases
D0 by p. Then expectation of D' is 0, the sigma^2 contributed by
each incoming photon is Var(D') = 1 - p^2.

Assume that sA emits N + Na photons, sB emits N + Nb photons, with
N being the expectation, and Na, Nb variations (of order of magnitude
sqrt(N)). Then D0 is p*(Na - Nb), and, for given values of Na and Nb,
Var(D') = (2N + Na + Nb)(1 - p^2).

The key step is that although D0 and D' are not independent, this
interdependence does not contribute into Var(D). Indeed,

Var(D) = Var(D0) + Var(D') + 2 Expectation(D0 * D').

To show that the last term vanishes, it is enough to show that
the conditional distribution of D' for the given value of D0 (i.e.,
of Na - Nb) has expectation 0; however, it has conditional expectation
0 even for fixed values of both Na and Nb.

Finally,
Var(D0) = p^2 * Var(Na - Nb) = p^2 * 2N,
Var(D') = (1 - p^2)(2N + Expectation(Na + Nb)) = (1 - p^2) * 2N.

Thus Var(D) is 2N, as expected for the white noise.

minnesotti
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 10-22-2006

Ilya Zakharevich wrote:
> For a long time I lived under impression that photon noise IS white.
> Now I thought about it more, and I'm not absolutely sure.

I also have a question: what is the diffrerence between "white noise",
"pink noise" and "grey noise" ?

David J Taylor
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Posts: n/a

 10-22-2006
minnesotti wrote:
> Ilya Zakharevich wrote:
>> For a long time I lived under impression that photon noise IS white.
>> Now I thought about it more, and I'm not absolutely sure.

>
> I also have a question: what is the diffrerence between "white noise",
> "pink noise" and "grey noise" ?

See, for example:

http://en.wikipedia.org/wiki/Pink_noise

http://en.wikipedia.org/wiki/White_noise

http://en.wikipedia.org/wiki/Grey_noise

David