Hello Robin,

with high interest I watched the video of your recent speech at PAS (https://www.youtube.com/watch?v=3RH93UvP358) and read your tutorial on "Picking the correct exposure for Deep Sky" here in the forum (viewtopic.php?f=35&t=456).

There is one thing though that I do not understand, maybe you can help me along.

The physics of dark noise, read noise and shot noise is clear. I do not understand though why you only consider shot noise coming from the photons of the image, shouldn't there also be shot noise introduced by the dark noise (aka thermal noise)? I would assume that the thermal noise also follows a poisson distribution, i.e. the same math should apply, shouldn't it?

Or can one neglect this considering that the dark current is much lower than the signal coming from light pollution?

Thanks for your advice in advance,

Stefan

## Understanding Dark Noise and Shot Noise

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### Re: Understanding Dark Noise and Shot Noise

Hi,

You are correct that the electrons released into the pixels due to thermal noise will also have a poisson distribution, although in general this will be a significantly smaller amount of noise than from the light signal (we would normally cool a camera to reduce thermal noise until we get to the point that there is no point in cooling any further because the background noise from light pollution exceeds the thermal noise).

The real reason that we can ignore it from the calculations though is that the sum of two Poisson distributions is itself also a Poisson distribution, so the noise characteristics of the total number of electrons collected calculated as a Poisson distribution is identical to the noise characteristics that you would calculate if you separated out the thermal electrons and the light electrons and calculated their noise contribution separately.

Cheers, Robin

You are correct that the electrons released into the pixels due to thermal noise will also have a poisson distribution, although in general this will be a significantly smaller amount of noise than from the light signal (we would normally cool a camera to reduce thermal noise until we get to the point that there is no point in cooling any further because the background noise from light pollution exceeds the thermal noise).

The real reason that we can ignore it from the calculations though is that the sum of two Poisson distributions is itself also a Poisson distribution, so the noise characteristics of the total number of electrons collected calculated as a Poisson distribution is identical to the noise characteristics that you would calculate if you separated out the thermal electrons and the light electrons and calculated their noise contribution separately.

Cheers, Robin

### Re: Understanding Dark Noise and Shot Noise

Hi Robin,

thanks for your reply.

You are right, generally the light pollution dominates the dark current from the camera electronics.

But when using an uncooled camera such as a DSLR under very dark skies (e.g. Bortle 1 or 2), then the impact from both light pollution and dark current should be in the same range. Correct?

Assuming both would be about the same, then the influence would be double compared to only light pollution, i.e. the impact of noise would be scaled by a factor of sqrt(2). Right?

This would then feed directly into the equation, resulting in

viewtopic.php?f=35&t=456#p2053

in

[math]

Correct?

When using a DSLR in central Europe, or when using a cooled camera in the Namib desert, then this would not apply, of course...

Best

Stefan

thanks for your reply.

You are right, generally the light pollution dominates the dark current from the camera electronics.

But when using an uncooled camera such as a DSLR under very dark skies (e.g. Bortle 1 or 2), then the impact from both light pollution and dark current should be in the same range. Correct?

Assuming both would be about the same, then the influence would be double compared to only light pollution, i.e. the impact of noise would be scaled by a factor of sqrt(2). Right?

This would then feed directly into the equation, resulting in

viewtopic.php?f=35&t=456#p2053

in

[math]

Correct?

When using a DSLR in central Europe, or when using a cooled camera in the Namib desert, then this would not apply, of course...

Best

Stefan

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### Re: Understanding Dark Noise and Shot Noise

Hi Stefan,

Yes, that makes sense – although you'd fully be best off with separate terms for the sky rate and the thermal noise rate of electrons. The interesting point is that if you are adding in a significant amount of thermal noise then the shot noise will overcome the read noise even sooner (i.e. at shorter exposures) than it would just for sky noise.

One other thing to consider is that if you are using SharpCap brain function to measure the background electron rate, the thermal noise will automatically be included in the rate measured so there is no need to account for it separately.

Cheers, Robin

Yes, that makes sense – although you'd fully be best off with separate terms for the sky rate and the thermal noise rate of electrons. The interesting point is that if you are adding in a significant amount of thermal noise then the shot noise will overcome the read noise even sooner (i.e. at shorter exposures) than it would just for sky noise.

One other thing to consider is that if you are using SharpCap brain function to measure the background electron rate, the thermal noise will automatically be included in the rate measured so there is no need to account for it separately.

Cheers, Robin

### Re: Understanding Dark Noise and Shot Noise

Hi Robin,

I took the liberty to use the extensive description of yours in viewtopic.php?f=35&t=456, and to modify the math, assuming this time that one cannot neglect thermal noise in comparison to sky noise (Part 6):

...

We have assumed that we have measured the read noise of the sensor to be [math], so we can use the rule of adding in quadrature to get the total noise for the frame

[math]

Substituting in our calculation for the shot noise, we have.

[math]

Now, obviously we are taking the square root of the electron count from sky noise and then immediately squaring it again, getting us back where we started, which means we can simplify the frame noise value to

[math]

Phew - we now have a value for the noise in a single frame, let's see how we can go from there to the noise in our final stack of

We will make our final stacked image by adding all the sub frames. This gives us a clue as to how to calculate the noise - we already know how to combine noise figures together in this sort of case - we add them in quadrature. That gives us a noise figure for our final stacked frame of

[math]

The total number of frame noises that we are adding inside the square root will be equal to

[math]

Remembering that the number of sub-exposures times the sub-exposure length gives the total time [math], we finally have

[math]

And continuing in Part 8:

...

Following your calculations...

...

[math]

If we neglect the thermal noise over the sky noise we come up with the "old" equation:

[math]

If we assume that both are of the same magnitude, then we can simplify this to:

[math]

This is a factor of two smaller!

So, I interpret this that with an uncooled camera under very dark skies, i.e. where one cannot neglect the dark noise over the sky noise, the ideal single exposure time is half as long as with a cooled camera.

Is my understanding correct?

Thanks

Stefan

I took the liberty to use the extensive description of yours in viewtopic.php?f=35&t=456, and to modify the math, assuming this time that one cannot neglect thermal noise in comparison to sky noise (Part 6):

...

We have assumed that we have measured the read noise of the sensor to be [math], so we can use the rule of adding in quadrature to get the total noise for the frame

[math]

Substituting in our calculation for the shot noise, we have.

[math]

Now, obviously we are taking the square root of the electron count from sky noise and then immediately squaring it again, getting us back where we started, which means we can simplify the frame noise value to

[math]

Phew - we now have a value for the noise in a single frame, let's see how we can go from there to the noise in our final stack of

*n*frames.We will make our final stacked image by adding all the sub frames. This gives us a clue as to how to calculate the noise - we already know how to combine noise figures together in this sort of case - we add them in quadrature. That gives us a noise figure for our final stacked frame of

[math]

The total number of frame noises that we are adding inside the square root will be equal to

*n*, the number of sub-frames, so[math]

Remembering that the number of sub-exposures times the sub-exposure length gives the total time [math], we finally have

[math]

And continuing in Part 8:

...

Following your calculations...

...

[math]

If we neglect the thermal noise over the sky noise we come up with the "old" equation:

[math]

If we assume that both are of the same magnitude, then we can simplify this to:

[math]

This is a factor of two smaller!

So, I interpret this that with an uncooled camera under very dark skies, i.e. where one cannot neglect the dark noise over the sky noise, the ideal single exposure time is half as long as with a cooled camera.

Is my understanding correct?

Thanks

Stefan

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### Re: Understanding Dark Noise and Shot Noise

Hi,

Yes, the effect of dark noise if you can't get rid of it will be to reduce the recommended exposure time. Since you assumed the dark noise was the same as the sky noise you end up with a factor of two at the end of the calculation. If you assume that the dark noise is a different article of the sky noise then the exact answer will change but it will still be a reduction in recommendation of sub exposure length.

Cheers, Robin

Yes, the effect of dark noise if you can't get rid of it will be to reduce the recommended exposure time. Since you assumed the dark noise was the same as the sky noise you end up with a factor of two at the end of the calculation. If you assume that the dark noise is a different article of the sky noise then the exact answer will change but it will still be a reduction in recommendation of sub exposure length.

Cheers, Robin

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