Shot Noise - A question for the wise ...

Started Jul 21, 2012 | Discussions
Detail Man
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Shot Noise - A question for the wise ...
Jul 21, 2012

If you don't know the answer to my question below, then there would be no point in responding .

A fundamental limit to the optical intensity noise as observed in many situations (e.g. in measurements with a photodiode or a CCD camera) is given by shot noise. This is a quantum noise effect, related to the discreteness of photons and electrons. Originally, it was interpreted as arising from the random occurrence of photon absorption events in a photodetector, i.e. not as noise in the light field itself. Intensity noise at the shot noise level is obtained when the probability for an absorption event per unit time is constant and not correlated with former events .

However, the existence of amplitude-squeezed light, which exhibits intensity noise below the shot noise level (sub-Poissonian intensity noise), proves that shot noise must be interpreted as a property of the light field itself, rather than as an issue of photodetection only ...

http://www.rp-photonics.com/shot_noise.html
.

Squeezed light was first demonstrated in 1985 .

By: R.E. Slusher, L.W. Hollberg, B. Yurke, J.C. Mertz, and J. F. Valley, Phys. Rev. Lett. 55, 2409 (1985), Observation of Squeezed States Generated by Four-Wave Mixing in an Optical Cavity

http://www.squeezed-light.de/
.

Question : If what is often referred to as "photon shot noise" in photo-detectors is, in fact, only a property of the light that is allowed to illuminate the surface of the photo-detector - and not to some extent semiconductor shot noise generated within the photo-detectors themselves , then how and why is it said that a higher "quantum effieciency" results in lower levels of "photon shot noise" for a given level of light being transduced to electron charges ? Something does not add up

In the case of photon detection, the relevant process is the random conversion of photons into photo-electrons for instance, thus leading to a larger effective shot noise level when using a detector with a quantum efficiency below unity .

http://en.wikipedia.org/wiki/Shot_noise#Optics

Detail Man
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Re: Shot Noise - A question for the wise ...
In reply to Detail Man, Jul 21, 2012

Perhaps it is simply a matter of the fact that allowing more light to illuminate the photo-detectors on an image-sensor results in a higher ratio of detected light divided by photon-shot-noise (SNR) ?

That makes sense to me. However, it seems that the rather modest incremental improvements in quantum-efficiency seen in image-sensor designs results in only very marginal improvements in SNR.

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3DrJ
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Re: Shot Noise - A question for the wise ...
In reply to Detail Man, Jul 21, 2012

Detail Man wrote:

Question : If what is often referred to as "photon shot noise" in photo-detectors is, in fact, only a property of the light that is allowed to illuminate the surface of the photo-detector - and not to some extent semiconductor shot noise generated within the photo-detectors themselves , then how and why is it said that a higher "quantum effieciency" results in lower levels of "photon shot noise" for a given level of light being transduced to electron charges ? Something does not add up

In the case of photon detection, the relevant process is the random conversion of photons into photo-electrons for instance, thus leading to a larger effective shot noise level when using a detector with a quantum efficiency below unity .

http://en.wikipedia.org/wiki/Shot_noise#Optics

Don't understand the question. The second paragraph is contradictory to the statements in the question. You say ' "photon shot noise" in photo-detectors is, in fact, only a property of the light ', yet you quote "the relevant process is the random conversion of photons into photo-electrons", which obviously has to occur within the photo-diode to have photons in, electrons out. If it were "only a property of the light", then the photo-diode has nothing to do with the process?

Quantum efficiency is going to be less than unity, hence there will be some noise, if vanishingly small with input of lots of photons.

I'll shut up since you are right, I can't answer your question, but I'm wondering if anyone really can.

JRA

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Detail Man
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Re: Shot Noise - A question for the wise ...
In reply to 3DrJ, Jul 21, 2012

3DrJ wrote:

In the case of photon detection, the relevant process is the random conversion of photons into photo-electrons for instance, thus leading to a larger effective shot noise level when using a detector with a quantum efficiency below unity .

http://en.wikipedia.org/wiki/Shot_noise#Optics

... The second paragraph is contradictory to the statements in the question. You say ' "photon shot noise" in photo-detectors is, in fact, only a property of the light ', yet you quote "the relevant process is the random conversion of photons into photo-electrons", which obviously has to occur within the photo-diode to have photons in, electrons out. If it were "only a property of the light", then the photo-diode has nothing to do with the process?

Note (as I believe that you have) that the italicized sections in my posts represent quotations (with source references below). The fact that a process occurs within the photo-diodes does not (in itself) establish that a particular random noise arises within the photo-diode. In fact, it is said that it was established in 1985 (references provided) that this particular random noise does not arise within the photo-diode.

I believe that I have answered my own question posed. However, the very small improvements in SNR arising out of small incremental improvements in quantum efficiency seem to me quite marginal

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bobn2
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Re: Shot Noise - A question for the wise ...
In reply to Detail Man, Jul 21, 2012

Detail Man wrote:

Question : If what is often referred to as "photon shot noise" in photo-detectors is, in fact, only a property of the light that is allowed to illuminate the surface of the photo-detector - and not to some extent semiconductor shot noise generated within the photo-detectors themselves , then how and why is it said that a higher "quantum effieciency" results in lower levels of "photon shot noise" for a given level of light being transduced to electron charges ? Something does not add up

In the case of photon detection, the relevant process is the random conversion of photons into photo-electrons for instance, thus leading to a larger effective shot noise level when using a detector with a quantum efficiency below unity .

http://en.wikipedia.org/wiki/Shot_noise#Optics

It's not only a property of the light. The photon shot noise depends on the number of photons counted, so it depends on the pattern in which photons present themselves to be counted (a property of the light) and the efficiency of the camera in counting them (a property of the camera).
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Detail Man
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Re: Shot Noise - A question for the wise ...
In reply to bobn2, Jul 21, 2012

bobn2 wrote:

Detail Man wrote:

Question : If what is often referred to as "photon shot noise" in photo-detectors is, in fact, only a property of the light that is allowed to illuminate the surface of the photo-detector - and not to some extent semiconductor shot noise generated within the photo-detectors themselves , then how and why is it said that a higher "quantum effieciency" results in lower levels of "photon shot noise" for a given level of light being transduced to electron charges ? Something does not add up

In the case of photon detection, the relevant process is the random conversion of photons into photo-electrons for instance, thus leading to a larger effective shot noise level when using a detector with a quantum efficiency below unity .

http://en.wikipedia.org/wiki/Shot_noise#Optics

It's not only a property of the light. The photon shot noise depends on the number of photons counted, so it depends on the pattern in which photons present themselves to be counted (a property of the light) and the efficiency of the camera in counting them (a property of the camera).

Yet, it seems that a full 14.8% improvement in an image-sensor's quantum efficiency will result in what seems like an essentially insignificant improvement in photon-shot-noise related SNR (0.1 EV). A 31.9% improvement in an image-sensor's quantum efficiency will result in what (also) seems like a nearly insignificant improvement in photon-shot-noise related SNR (0.2 EV). "Where's the beef" ?

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bobn2
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Re: Shot Noise - A question for the wise ...
In reply to Detail Man, Jul 21, 2012

Detail Man wrote:

bobn2 wrote:

Detail Man wrote:

Question : If what is often referred to as "photon shot noise" in photo-detectors is, in fact, only a property of the light that is allowed to illuminate the surface of the photo-detector - and not to some extent semiconductor shot noise generated within the photo-detectors themselves , then how and why is it said that a higher "quantum effieciency" results in lower levels of "photon shot noise" for a given level of light being transduced to electron charges ? Something does not add up

In the case of photon detection, the relevant process is the random conversion of photons into photo-electrons for instance, thus leading to a larger effective shot noise level when using a detector with a quantum efficiency below unity .

http://en.wikipedia.org/wiki/Shot_noise#Optics

It's not only a property of the light. The photon shot noise depends on the number of photons counted, so it depends on the pattern in which photons present themselves to be counted (a property of the light) and the efficiency of the camera in counting them (a property of the camera).

Yet, it seems that a full 14.8% improvement in an image-sensor's quantum efficiency will result in what seems like an essentially insignificant improvement in photon-shot-noise related SNR (0.1 EV). A 31.9% improvement in an image-sensor's quantum efficiency will result in what (also) seems like a nearly insignificant improvement in photon-shot-noise related SNR (0.2 EV). "Where's the beef" ?

What a 14.8% improvement in QE gives depends on what it was before and how many photons were being captured before.

So, if we have a pixel with a QE of 50% capturing 1000 photons, then the shot noise SNR will be 5 stops. Increase that by 14.8% to 57.4% and the photon count goes up to 1148, for a SNR of 5.08 stops.

On the other hand, had we had a QE of 20% (400 photons) giving a SNR of 4.3 stops, had we increased it 14.8% to 23% to give 460 photons, we'd have had an SNR of 4.42 stops.

As the QE rises, we get more to the point of diminishing returns, and a 14.8% increase in QE isn't enough to make a large difference to the shot noise SNR.

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Steen Bay
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Re: Shot Noise - A question for the wise ...
In reply to Detail Man, Jul 21, 2012

Detail Man wrote:

bobn2 wrote:

Detail Man wrote:

Question : If what is often referred to as "photon shot noise" in photo-detectors is, in fact, only a property of the light that is allowed to illuminate the surface of the photo-detector - and not to some extent semiconductor shot noise generated within the photo-detectors themselves , then how and why is it said that a higher "quantum effieciency" results in lower levels of "photon shot noise" for a given level of light being transduced to electron charges ? Something does not add up

In the case of photon detection, the relevant process is the random conversion of photons into photo-electrons for instance, thus leading to a larger effective shot noise level when using a detector with a quantum efficiency below unity .

http://en.wikipedia.org/wiki/Shot_noise#Optics

It's not only a property of the light. The photon shot noise depends on the number of photons counted, so it depends on the pattern in which photons present themselves to be counted (a property of the light) and the efficiency of the camera in counting them (a property of the camera).

Yet, it seems that a full 14.8% improvement in an image-sensor's quantum efficiency will result in what seems like an essentially insignificant improvement in photon-shot-noise related SNR (0.1 EV).

Doubling the QE will 'only' increase the SNR by 41%, so guess you could say that a small improvement in QE only will result in an "essentially insignificant" improvement in SNR. I'd prefer to say that a 41% increase in SNR is a 'one stop improvement', since it'll mean that the camera now will have the same noise at iso200 as it had at iso100 before, but guess that's open for discussion.

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Detail Man
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Re: Shot Noise - A question for the wise ...
In reply to bobn2, Jul 21, 2012

bobn2 wrote:

It's not only a property of the light. The photon shot noise depends on the number of photons counted, so it depends on the pattern in which photons present themselves to be counted (a property of the light) and the efficiency of the camera in counting them (a property of the camera).

Are you saying that additional (semiconductor-generated) shot-noise components arise out of the currents which flow as a result of the transduction of photons to electron charges in the photo-detectors (and the MOSFETs existing within the interface itself) as a result of the process ? That might not be a surprise. I am interested in what is understood regarding the particulars themselves

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bobn2
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Re: Shot Noise - A question for the wise ...
In reply to Detail Man, Jul 21, 2012

Detail Man wrote:

bobn2 wrote:

It's not only a property of the light. The photon shot noise depends on the number of photons counted, so it depends on the pattern in which photons present themselves to be counted (a property of the light) and the efficiency of the camera in counting them (a property of the camera).

Are you saying that additional (semiconductor-generated) shot-noise components arise out of the currents which flow as a result of the transduction of photons to electron charges in the photo-detectors (and the MOSFETs existing within the interface itself) as a result of the process ? That might not be a surprise. I am interested in what is understood regarding the particulars themselves

No, I'm saying that the shot noise is also sensor dependent because the sensor can fail to count all of the available photons.

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Detail Man
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Re: Shot Noise - A question for the wise ...
In reply to bobn2, Jul 21, 2012

bobn2 wrote:

Detail Man wrote:

bobn2 wrote:

Detail Man wrote:

Question : If what is often referred to as "photon shot noise" in photo-detectors is, in fact, only a property of the light that is allowed to illuminate the surface of the photo-detector - and not to some extent semiconductor shot noise generated within the photo-detectors themselves , then how and why is it said that a higher "quantum effieciency" results in lower levels of "photon shot noise" for a given level of light being transduced to electron charges ? Something does not add up

In the case of photon detection, the relevant process is the random conversion of photons into photo-electrons for instance, thus leading to a larger effective shot noise level when using a detector with a quantum efficiency below unity .

http://en.wikipedia.org/wiki/Shot_noise#Optics

It's not only a property of the light. The photon shot noise depends on the number of photons counted, so it depends on the pattern in which photons present themselves to be counted (a property of the light) and the efficiency of the camera in counting them (a property of the camera).

Yet, it seems that a full 14.8% improvement in an image-sensor's quantum efficiency will result in what seems like an essentially insignificant improvement in photon-shot-noise related SNR (0.1 EV). A 31.9% improvement in an image-sensor's quantum efficiency will result in what (also) seems like a nearly insignificant improvement in photon-shot-noise related SNR (0.2 EV). "Where's the beef" ?

What a 14.8% improvement in QE gives depends on what it was before and how many photons were being captured before.

So, if we have a pixel with a QE of 50% capturing 1000 photons, then the shot noise SNR will be 5 stops. Increase that by 14.8% to 57.4% and the photon count goes up to 1148, for a SNR of 5.08 stops.

On the other hand, had we had a QE of 20% (400 photons) giving a SNR of 4.3 stops, had we increased it 14.8% to 23% to give 460 photons, we'd have had an SNR of 4.42 stops.

As the QE rises, we get more to the point of diminishing returns, and a 14.8% increase in QE isn't enough to make a large difference to the shot noise SNR.

Where P is photons detected (due to illumination levels and quantum efficiency), (linear) SNR is:

P / SQRT (P)

or

SQRT (P)

the derivative of which equals:

1 / ( (2) x ( SQRT (P) ) )

Only at low illumination levels (and/or low quantum efficiences) are significant differences observed.

That does explain why more significant improvements (for a given level of improvement of quantum efficiency) are observed at very low illumination levels, but not at higher illumination levels

Thus, a sensor's higher photo-site "full well capacity" will improve SNR by only marginal amounts ... and if the levels of Read/Dark Noise between two image-sensors are comparable (as appears to be the case where it comes to the E-M5 and the GH2 image-sensors), then it is only by virtue of interpolations of SNR measurements at low illumination levels by DxOMark that would appear to result in a higher RAW-level Dynamic Range specification for the E-M5 relative to the GH2.

DM ...

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3DrJ
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Re: Shot Noise - A question for the wise ...
In reply to Detail Man, Jul 21, 2012

Detail Man wrote:

bobn2 wrote:

It's not only a property of the light. The photon shot noise depends on the number of photons counted, so it depends on the pattern in which photons present themselves to be counted (a property of the light) and the efficiency of the camera in counting them (a property of the camera).

Are you saying that additional (semiconductor-generated) shot-noise components arise out of the currents which flow as a result of the transduction of photons to electron charges in the photo-detectors (and the MOSFETs existing within the interface itself) as a result of the process ? That might not be a surprise. I am interested in what is understood regarding the particulars themselves

Every electronic component, amplifier, gate, etc. produces noise, doesn't it? Noise associated with the photo-diode interface per se may not be the only source of noise that could be apparently attributable to photon-> electron conversion.

I, for one, don't really know what the details of the sensor photo-diode construction are, and certainly not well enough to make any predictions about how one influence or another is going to affect the output, or final S/N ratio.

I've used large photo-diodes in the past, but it seems almost certain that the microscopic diodes in the camera are going to have different properties due to size alone to consider.

These discussions are always interesting. I think it is good to be curious about how it works, but it reaches a point that it becomes impossible for anyone outside the loop (anyone not an Olympus R&D engineer) to really know how they do it.

If it helps a person make better photographs, gaining some idea of how the camera processes the image isn't a bad thing at all.

JRA

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gollywop
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Re: Shot Noise - A question for the wise ...
In reply to Detail Man, Jul 21, 2012

The fact that shot noise has been shown to be a property of light, it doesn't mean that the total shot noise in-camera cannot be from two independent sources: the light itself and the behavior characteristics of the sensor. Each could be Poisson distributed, as well as the total, since the sum of two independent Poisson distributions is also Poisson distributed.
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bobn2
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Re: Shot Noise - A question for the wise ...
In reply to 3DrJ, Jul 21, 2012

3DrJ wrote:

Detail Man wrote:

bobn2 wrote:

It's not only a property of the light. The photon shot noise depends on the number of photons counted, so it depends on the pattern in which photons present themselves to be counted (a property of the light) and the efficiency of the camera in counting them (a property of the camera).

Are you saying that additional (semiconductor-generated) shot-noise components arise out of the currents which flow as a result of the transduction of photons to electron charges in the photo-detectors (and the MOSFETs existing within the interface itself) as a result of the process ? That might not be a surprise. I am interested in what is understood regarding the particulars themselves

Every electronic component, amplifier, gate, etc. produces noise, doesn't it? Noise associated with the photo-diode interface per se may not be the only source of noise that could be apparently attributable to photon-> electron conversion.

Yes, you'd call those noises 'read noise', and they aren't signal dependent in the way that shot noise is.
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Detail Man
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Re: Shot Noise - A question for the wise ...
In reply to gollywop, Jul 22, 2012

gollywop wrote:

The fact that shot noise has been shown to be a property of light, it doesn't mean that the total shot noise in-camera cannot be from two independent sources: the light itself and the behavior characteristics of the sensor. Each could be Poisson distributed, as well as the total, since the sum of two independent Poisson distributions is also Poisson distributed.

Have had a look for some information regarding the various random noise sources in the P-N type photodiodes that bobn2 has indicated exist in our image-sensors.

Section 2.4 of the below-linked reference identifies thermal noise from the shunt resistance (Equation 2-7) as well as reverse-biased leakage (dark) current shot-noise (Equation 2-8). Photon shot-noise present when such photo-diodes are at some level illuminated (Equation 2-9) dominate:

http://sales.hamamatsu.com/assets/html/ssd/si-photodiode/index.htm

There also exist additional noise sources in the MOSFET transistors that comprise the photo-diode interface circuitry. It is unclear to me whether those noise-sources may become a factor in situations of very low levels of illumination of the photo-sensors. Perhaps bobn2 has some idea ?

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Eric Nepean
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Re: Shot Noise - A question for the wise ...
In reply to Detail Man, Jul 22, 2012

If shot noise were only a property of the light field illuminating the device, then shot noise should disappear when there is no light. Furthermore, devices which do not photodetect, and those which are not illuminated, should then not exhibit shot noise.

It seems odd that such a direct relationship should have gone unnoticed by those working in the area of noise in electronic circuits.
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Detail Man
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Shot Noise - Implications relating to DxOMark Dynamic Range Spec.
In reply to Detail Man, Jul 22, 2012

Detail Man wrote:

... a sensor's higher photo-site "full well capacity" will improve SNR by only marginal amounts ... and if the levels of Read/Dark Noise between two image-sensors are comparable (as appears to be the case where it comes to the E-M5 and the GH2 image-sensors), then it is only by virtue of interpolations of SNR measurements at low illumination levels by DxOMark occurring that would appear to result in a higher RAW-level Dynamic Range specification for the E-M5 relative to the GH2 .

DxOMark's method of deriving Dynamic Range specifications :

Each uniform zone on the chart (a “patch”) is measured for luminance (cd/m^2) with a certified luminance-meter; then all the values are input into DxO Analyzer software .

Once we measure the target and calibrate the DxO Analyzer software, the selected camera shoots an image of the noise target at different ISO settings, and we measure the noise for each color channel of the target image (R, Gr, Gb, B) .

We compute the mean gray level and noise values for each patch and for all images shot at different ISO settings. We then interpolate these numerical values for all gray levels to calculate and plot signal-to-noise ratio (SNR) curves, from which DxO Analyzer extracts the SNR 18%, the dynamic range, and the tonal range .

Source: http://www.dxomark.com/index.php/About/In-depth-measurements/Measurements/Noise
.

At Base ISO Gain :

In the absence of any illumination of the GH2 and EM5 image-sensors (linear SNR=0), RawDigger has reported an essentially identical Standard Deviation of Read/Dark Noise in the GH2 and the EM5

When inspecting nearly identical (scene, lighting level, lens, F-Number, and Shutter Speed) GH2 and EM5 images (where the RAW-scaling has been normalized for the -0.345 EV difference in the arithmetic Average recorded RAW-levels in the case of the EM5 shot, and which is a valid normalization in cases where Read/Dark Noise sources dominate over Photon Shot Noise sources),

... the Standard Deviation of the composite (of Read/Dark Noise and Photon Shot Noise) noise of the GH2 and the EM5 are equal somewhere in between a SNR of around 3 dB and a SNR of around 9 dB (perhaps at a SNR somewhere in the region of 6 dB), ...

... with the noise of the GH2 being lower than the EM5 below that (above-mentioned) SNR, and the noise of the EM5 being lower than the GH2 above that (above-mentioned) SNR (SNR as defined by DxOMark in their characterization of RAW-level performance). My observations below:
.

SNR ~ 3 dB (where Read/Dark Noise appears to dominate over Photon Shot Noise) :
.

The location of the Mean Value is shown as a vertical gray line in the histograms below:

.

(3 pixel-width) arithmetically averaged view; corresponding histograms allowing easier assessment:

.
.

SNR ~ 9 dB (where Photon Shot Noise appears to dominate over Read/Dark Noise) :

The location of the Mean Value is shown as a vertical gray line in the histograms below:

.

(3 pixel-width) arithmetically averaged view; corresponding histograms allowing easier assessment:

Notes : In such as case as the ~ 9 dB SNR case displayed directly above, where Photon Shot Noise sources appear to be dominating over Read/Dark Noise sources, the SNR of the EM5 has been disadvantaged relative to the GH2 by (potentially as much as) the square-root of 0.345 EV (0.173 EV). Therefore, it makes sense (in this particular case) to consider the displayed Standard Deviation of the EM5 noise as being (up to) 0.173 EV (12.7%) larger in magnitude than is actually the case .

The differences in arithmetic Average RAW-scaling were determined from the unprocessed RAW image-data prior to any non-linear tone-curve transfer-functions being applied in mapping into the RGB color-space of the 16-bit TIFFs .

While the non-linear transfer-function of sRGB color-space mapping has attenuated the mean value (as well as the noise-distributions) displayed by around -1.568 EV in the case of the ~ 3 DB SNR case with a sRGB mean value of 37, and attenuated the mean value (as well as the noise-distributions) displayed by around -2.730 EV in the case of the ~ 9 DB SNR case with a sRGB mean value of 79, since the mean values of the histograms are matched in value, the (estimated) SNRs, as well as the (differential) comparisons of the widths of the noise-distributions, appear to be valid.

DM ...

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intruder61
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Re: Shot Noise - A question for the wise ...
In reply to Detail Man, Jul 22, 2012

this is your answer.....work it out and then go take photos.

a_0y_t + a_1y_{t-1} + a_2y_{t-2} = f(t) or a_0y_{t+2} + a_1y_{t+1} + a_2y_t = f(t)

The solving of a second order difference equation is very similar to the method of solving a second order differential equation, which is discussed in this previous post here.

The General Solution of a second order difference equation has a Complementary Function and a Particular Solution.

The Complementary Function (CF) is found by writing the auxiliary equation

a_0m^2 + a_1m + a_2 = 0 and solving it to find the two roots of such a quadratic equation.

When there are two different roots, m1 and m2, the CF is written as

y_t = A\left( m^t_1 \right) + B \left( m^t_2 \right) where A and B are arbitrary constants.
When there are two equal roots, such that m1 = m2 = m, then the CF is written as
y_t = (A + Bt)m^t.
When there are two complex roots, u + iv and u - iv, then the CF is

y_t = r^t (A \cos t\theta + B \sin t\theta) where r = \sqrt{u^2 + v^2} and \tan \theta = \frac{v}{u}.

Based on the standard form of the second order difference equation, the Particular Solution depends on the form f(t), the function in t on the RHS of the difference equation.

Again, similar to the case of differential equations, the Particular Solution (PS) is the same form as f(t) but contains undetermined coefficients which are determined by comparing them with the function in the RHS of the given difference equation.

If the function is a constant, say c, then the general form of the PS is k, an unknown constant.

If the function is in the form c \left( a^t \right), then the general form of the PS is k\left( a^t \right).

If the function is in the form at+ b, then a general linear function in t, A_1t + A_2, is the general form of the PS.

A homogeneous second order difference equation has zero on the RHS. Its standard form is a_1y_{t+2} + a_2y_{t+1} + a_3y_t = 0.

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Detail Man
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Re: Shot Noise - A question for the wise ...
In reply to Eric Nepean, Jul 22, 2012

Eric Nepean wrote:

If shot noise were only a property of the light field illuminating the device, then shot noise should disappear when there is no light. Furthermore, devices which do not photodetect, and those which are not illuminated, should then not exhibit shot noise.

Shot noise exists in all semiconductor devices within which currents are flowing (whether they are photo-sensitive, or otherwise). In particular, a reverse-biased photo-diode will have a reverse-bias leakage current in the absence of any illumination (i.e., the shot-noise component of "dark-noise").

See Section 2.4 at: http://sales.hamamatsu.com/assets/html/ssd/si-photodiode/index.htm

... linked to at: http://forums.dpreview.com/forums/read.asp?forum=1041&message=42082133

... in which the case that you are making (in the case of externally-induced currents are flowing) is affirmed. Small "reverse-saturation currents" do exist in PN junction devices [on the order of 10^(-13) to 10^(-14) Amperes], and appear as a variable in the "diode equation". I read that rather low levels of current are integrated by photo-diodes in image-sensors. Perhaps you may be able to tell us whether such 10-100 femtoAmpere "reverse-saturation currents" are a factor ?

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Detail Man
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Re: Shot Noise - A question for the wise ...
In reply to intruder61, Jul 22, 2012

intruder61 wrote:

this is your answer.....work it out and then go take photos.

a_0y_t + a_1y_{t-1} + a_2y_{t-2} = f(t) or a_0y_{t+2} + a_1y_{t+1} + a_2y_t = f(t)

The solving of a second order difference equation is very similar to the method of solving a second order differential equation, which is discussed in this previous post here.

The General Solution of a second order difference equation has a Complementary Function and a Particular Solution.

The Complementary Function (CF) is found by writing the auxiliary equation

a_0m^2 + a_1m + a_2 = 0 and solving it to find the two roots of such a quadratic equation.

When there are two different roots, m1 and m2, the CF is written as

y_t = A\left( m^t_1 \right) + B \left( m^t_2 \right) where A and B are arbitrary constants.
When there are two equal roots, such that m1 = m2 = m, then the CF is written as
y_t = (A + Bt)m^t.
When there are two complex roots, u + iv and u - iv, then the CF is

y_t = r^t (A \cos t\theta + B \sin t\theta) where r = \sqrt{u^2 + v^2} and \tan \theta = \frac{v}{u}.

Based on the standard form of the second order difference equation, the Particular Solution depends on the form f(t), the function in t on the RHS of the difference equation.

Again, similar to the case of differential equations, the Particular Solution (PS) is the same form as f(t) but contains undetermined coefficients which are determined by comparing them with the function in the RHS of the given difference equation.

If the function is a constant, say c, then the general form of the PS is k, an unknown constant.

If the function is in the form c \left( a^t \right), then the general form of the PS is k\left( a^t \right).

If the function is in the form at+ b, then a general linear function in t, A_1t + A_2, is the general form of the PS.

A homogeneous second order difference equation has zero on the RHS. Its standard form is a_1y_{t+2} + a_2y_{t+1} + a_3y_t = 0.

Intruder ,

I have already "gone out and taken some photos", thanks. Your notes regarding second-order difference-equations are impressive in their sheer magnitude, but I have no idea what it is that you are attempting to say, and no idea what you are perhaps endeavoring to even relate them to.

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