How many darks do I need? Some measurements

[han59]

Probably an old discussion but lets review it with some measurements:

 

The dark noise should only have a small influence on the total noise of the final image. Most noise is generated by the sky background. Under good conditions SQM = 20.4, I measure using my ASI1600MM-Cool the following noise (standard deviation) in a dark and in a light for an area where no stars are visible (local measurement using ASTAP):

Dark 1 x 200sec, σ = 15 (range 0..65535)
Light 1 x 200sec, σ = 130

The noise in the dark is roughly 12% of the light, which seems acceptable to me. That would argue for about the same amount of darks as lights. With a worse SQM, you can probably do 2.5 times less darks for each (magnitude) step. So under light polluted sky you can do with much less darks than lights.

If you are going to photograph with the H-alpha filter, it will be super dark. In a single H-alpha (7nm) light I measure a σ = 25r. Of these, 15 are self-noise and 10 of the incoming light. In good conditions and using an H-alpha filter, this is an argument to make much more darks than lights

Above for a monochrome camera. To measure with an OSC (color) sensor I think it is better to first split the 4 Bayer pixels into 4 files and then measure them separately.

Some measurements with my ASI1600MM-Cool, monochrome:

DARKS noise:
1 x 200 seconds, σ = 16
1 x 200 seconds - master dark, σ = 15
4 x 200 seconds combined - master dark, σ = 6.8    This is approximately 15 / square root (4)
41 x 200 seconds combined, σ = 5
90 x 200 seconds combined, σ = 3.8   This is a limit value that arises mainly from unevenness of the pixels. The noise will be smaller, approximately 15 / square root (90) is 1.6

 

STACKED LIGHTS noise (lights corrected with darks and flats):
11x200 seconds, σ = 70   (measured at a star free area, standard deviation in 0..65535 range, sky conditions could have been different)
18x200 seconds, σ = 36
18x200 seconds, σ = 40
40x200 seconds, σ = 26
42x200 seconds, σ = 30
44x200 seconds, σ = 25
58x200 seconds, σ = 20
95x200 seconds, σ = 16

Apparently the light noise decreases considerably while stacking more lights and I reach σ values up to 16 a 20. You do not want to stack these images with a single dark having a σ = 15. If you want to keep the dark noise added below 10% of σ = 16 then you need 100 darks because they give: 15 / square root (100) = 1.5 noise.

So this confirms for a good suburban site (SQM=20.4) you will need about the same amount (or more) darks then lights. For a more light polluted area you can take less darks since the noise from the skybackground will be abundant. For H-alpha work you better take more darks then lights.

Han

[vlaiv]

Can I make a few observations?

You need to be more careful when doing measurements - what is it that you are measuring in the first place.

For example - using single dark frame and doing standard deviation measurement on it is rather meaningless.

Single dark frame contains:

1. Bias signal

2. Bias noise

3. Dark signal

4. Dark noise

In order to really measure dark noise, you need to remove 3 other components. Using just ordinary numbers is fine, but you really want to convert your measurements in electrons - that way you'll get comparable results that can be tested against other things that you know.

For example, you say that sigma of single dark of 200s is ~15.

That does not tell us much, right? We know that read noise of that camera is around 1.7e or better (if you used unity gain or higher) and we also know that dark current at let's say -20C is ~0.0062e/px/s. In 200s exposure one can expect 1.24e of dark current signal and consequently square root of that as dark noise - ~1.1136e, which is interestingly enough - less than read noise of that camera.

Neither of the two numbers are close to number 15 that you had.

Simplest way to get read actual dark current noise measured is to set camera at unity gain - so you don't have to do conversion between ADUs and electrons (or rather just divide your subs with 16 because of 12bit format), take two dark subs, subtract the two dark subs (subtract second from first). Measure standard deviation, divide with square root of two (you stacked two subs when you subtracted them) and that will give you dark noise and read noise combined.

Do the same for two bias subs to get read noise.

Remove read noise from above dark+read noise and you'll get pure dark current noise that you can use to estimate dark current if you want.

That way, you'll get two numbers that you can compare to published figures of 1.7e of read noise and 0.0062e/px/s for dark current.

Just to show you the difference, here is what standard deviation of single dark sub for ASI1600 gives and what you get as combined dark + read noise from two dark subs.

First measurement is of single 60s / gain 139 / offset 64 sub. StdDev is 2.275 and mean value is ~62.8. Mean value is clearly offset from 0 so it's not pure noise it contains some signal as well (bias + dark current).

Second measurement is difference between two subs. Mean is now ~0 which is what you would expect when there is no signal but pure noise. We now have to divide stddev value with sqrt(2) to get actual value, and it is:

~2.81 / sqrt(2) = ~1.987e

We see that bias and dark signal were not uniform and had some variance to them as they increased stddev from 1.987e to 2.275e - we need to be careful to measure proper thing.

Now, I don't have any bias subs on me, so I'm just going to assume 1.7e of read noise and try to calculate dark current from that and 1.987e combined noise to see if we get comparable results to published ones.

dark_noise = sqrt(1.987^2 - 1.7^2) = ~ 1.0287e

So we get that our dark current noise is 1.0287e and associated dark current will be that number squared so ~1.058169e

This is for 60s sub at -20C, let's see what value do we get for single second as 1.058169 / 60 = ~0.01763615e/px/s

That is about x2.5 higher than published value, but I suspect that difference is due to me taking 1.7e as read noise, when in fact it is higher than that - around 1.8 or 1.9 for unity gain - usually ZWO published read noise values are rather optimistic. If we repeat calculation with 1.9e as read noise, we will get:  0.00563615e/px/s which is much closer to published 0.0062 and even a bit lower, so real value of read noise is probably a bit less than 1.9e - but like I said - best to be measured rather than assumed from graphs. Graph/published values are just good comparison reference value and should be taken with a grain of salt.

 

 

[han59]

That is a interesting method you describe. I will look into tomorrow. In principle it should be possible to calculate the required ratio  nrdarks /nnrlights   from  the local SQM value, focal ratio and camera noise figures and exposure time. This minimum ratio will indicate how many dark are required to keep the dark noise below 10% of total noise.

[vlaiv]

6 minutes ago, han59 said:

That is a interesting method you describe. I will look into tomorrow. In principle it should be possible to calculate the required ratio  nrdarks /nnrlights   from  the local SQM value, focal ratio and camera noise figures and exposure time. This minimum ratio will indicate how many dark are required to keep the dark noise below 10% of total noise.

I'm not sure I'm following your reasoning.

Every single light sub you take will have some dark current and hence dark current noise. Let that be X. Let's also leave other noise sources aside as we don't need them yet for what I'm about to point out here.

Let's say that you take 25 of darks for your master dark.

This means that master dark has X/5 of dark noise - as we stacked 25 of darks to get master dark.

This also means that once you calibrate your light sub with such master dark - total dark noise will be sqrt(X^2+(X/5)^2) = sqrt(X^2 + X^2/25) = sqrt(26 * X^2 / 25) = X * sqrt(26) / sqrt(25) = ~1.0198.

We increased our dark current noise by only 2% by using 25 of darks.

This is regardless of any shot, lp or read noise.

Want to reduce it to even more insignificant levels? Take more darks for your master dark. You can calculate how much you increase dark current noise with simple formula sqrt(N+1)/sqrt(N) where N is number of dark subs you take.

Want to make real difference - dither.

This is going to be a bit more tricky to explain, but it is really important.

What I've written above works only if you dither. If you don't dither then we can't add master dark like that. Take for example perfect guiding - you don't need to register any of the light subs - they are perfectly aligned. Dark value that you are subtracting from each pixel will be exactly the same for every single light pixel across your light subs.

You can pull that dark in front of sum and what you end up doing is polluting final stack with considerable amount of dark (and bias) noise.

Let's say you have X of dark noise and you take 25 darks for your stack. You also take 100 light subs (just because it's easy to calculate).

Perfect dither case, resulting stack will have 1.0198 * X / 10 = 0.10198. Why? Because we have seen that that 25 darks increases every light sub's dark noise by factor of 1.0198 and then we stack 100 of those and we reduce total amount of noise by x10.

We can do that because of perfect dither and fact that it makes every calibration independent (because it is shifted with respect to every other).

Now look what happens with perfectly aligned case:

we will have ( (L1 - MD) + (L2 - MD)  + (L3 - MD) + .... + (L100 - MD) )  / 100

(L1,2,3, .... being light subs and MD being master dark)

Now we can rearrange that as (L1 + L2+ L3 + ..... L100) / 100 + (MD+MD+MD+....+MD)/100

We can do this because in every calibration MD aligns with every other MD and they are the same - no shifts so each pixel gets onto itself - same value, no randomness.

Which is then stack_of_lums - 100 * MD / 100 = stack_of_lums - MD

Now let's see what sort of noise we have now?  sqrt( (X/10)^2 + (X/5)^2 ) = sqrt(  (X^2 + 4*X^2) / 100 ) = X * sqrt(5)/10 = X * ~0.2236

That is twice as much noise from dark current than in perfectly dithered case!

 

 

 

[han59]

The perfect guiding doesn't exist as far I'm aware.

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I tried to answer the question how many darks you will need to take to be sure the noise addition by the dark is minimal, lets say 10%.  Since you have to add noise by take the power you have to use this formula:  Noise^2 :=LightNoise^2+DarkNoise^2,  so 1.1^2:=1^2+darknoise^2. Then dark noise is should be sqrt(1.1^2-1^2)= 45% of the light noise maximum.  I did that wrong in the first post

In the lights the most noise comes from the incoming fotons, the so called sky glow background.  If you analyse one dark and one light you can estimate how many darks you need to reduce the dark noise influence to 10%.  So lets do the math correctly:

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DARK calculation:

The total noise in a dark is:

• Noise^2 :=ReadNoise^2+DarkNoise^2

I measure for the ASI1600, -15C  in unity gain these dark noise values in ADU's:

• 0 seconds binned 2x2,  σ = 13 ADU, this is 13/16= 0.82 electrons since the 12 bit is converted to 16 bit equals x16. That is 0.82*sqrt(4)=1.64 electrons for a single unbinned pixel
• 200 seconds binned 2x2,  σ = 15 ADU
• 400 seconds binned 2x2,  σ = 18 ADU, using above formula 18^2=13^2+ DarkNoise^2. Then Darknoise is 12.4 ADU or  (12.4/16)/200= 0.0039 electrons/seconds. For unbinned this is then 0.0078 electrons/seconds

This matches nicely with reported values.

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LIGHT calculation and number of darks required:

Now for the light:

• Noise^2 :=SkyNoise^2+ ReadNoise^2+DarkNoise^2

The skyglow in a ASI1600 behind a F/5.8 telescope, unity gain for a single light 200 seconds binned 2x2, unfiltered taken at SQM 20 is 6800 ADU and the measured 

σ =168 ADU. 

So 168^2 :=SkyNoise^2+ 13^2+12.4^2. The SkyNoise  is then 167 ADU or  (167/16)/200 elektron/seconde.  The shotnoise should be about sqrt(6800)=82 ADU and the remainder PRNL noise

If 100 lights are stacked the resulting noise is about 168/sqrt(100) is 16.8 ADU. If I want the darks adding only 10 % noise to the result, then (16.8*1.1)^2=16.8^2+ DarkNoise^2.  Then total dark noise should be then 7.7 ADU maximum. That would require only 4 darks since 15/sqrt(4)=7.5. This is a different result then before. So the number of darks required are much less.

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LIGHT H-alpha calculation and number of darks required:

If I repeat the above calculation for a light using a H-alpha filter 7nm, then the  sky noise is about 34 ADU. A stack of 100  lights will result in 3.4 ADU noise. This requires darks with a total 1.5 ADU noise to keep the influence below 10%, so  for H-alpha it requires 100 darks since 15/sqrt(100)=1.5.

 

Note this 7 nm H-alpha reduces the light with a factor 7nm/280nm= 0.025. The noise is then sqrt(0.025 ) is 0.158 time lower. This matches with 34 ADU/168 ADU is 0.2

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I better put this in a spreadsheet. Please correct me for any error.

 

Han

 

 

 

 

Edited August 6, 2020 by han59

[vlaiv]

7 minutes ago, han59 said:

The perfect guiding doesn't exist as far I'm aware.

No it does not, but that is really not the point. Neither does perfect dither - or rather you can never dither to integer shift value.

Those are two extreme cases that we use to asses what will happen. In reality something in between will happen and results will be in between - closer you get to either - closer the results will be to academic ones.

9 minutes ago, han59 said:

I tried to answer the question how many darks you will need to take to be sure the noise addition by the dark is minimal, lets say 10%.  Since you have to add noise by take the power you have to use this formula:  Noise^2 :=LightNoise^2+DarkNoise^2,  so 1.1^2:=1^2+darknoise^2. Then dark noise is should be sqrt(1.1^2-1^2)= 45% of the light noise maximum.  I did that wrong in the first post

In the lights the most noise comes from the incoming fotons, the so called sky glow background.  If you analyse one dark and one light you can estimate how many darks you need to reduce the dark noise influence to 10%.  So lets do the math correctly:

Your reasoning is somewhat flawed here.

By analyzing one light and one dark - you can't change contribution of dark current to the light frame. You can never do anything about their ratio.

Both LP noise and dark current noise raise at the same rate as their sources depend on time in the same way. Dark current accumulates with time at a certain rate and LP levels accumulate at a certain rate.

Take 60s long sub and take another that is 120s long - both LP level and dark current levels in second will be twice as high as in the first one and they will keep their ratio. If dark current noise is 10% of LP noise - it will be the same regardless of what is exposure length.

Adding calibration frames can only make this number worse - it can't improve it.

Next thing to understand is that for calculation of how many darks you need - you don't need LP levels at all. You already have "competing" noise - dark itself. Stack of 25 darks will have 5 times less noise than single dark - and that is already plenty of reduction so that their sum will increase by less than 2% (look at above calculation).

17 minutes ago, han59 said:

DARK calculation:

The total noise in a dark is:

• Noise^2 :=ReadNoise^2+DarkNoise^2

   

I measure for the ASI1600, -15C  in unity gain these dark noise values in ADU's:

• 0 seconds binned 2x2,  σ = 13 ADU, this is 13/16= 0.82 electrons since the 12 bit is converted to 16 bit equals x16. That is 0.82*sqrt(2)=1.64 electrons for a single unbinned pixel
• 200 seconds binned 2x2,  σ = 15 ADU
• 400 seconds binned 2x2,  σ = 18 ADU, using above formula 18^2=13^2+ DarkNoise^2. Then Darknoise is 12.4 ADU or  (12.4/16)/200= 0.0039 electrons/seconds. For unbinned this is then 0.0078 electrons/seconds

This matches nicely with reported values.

Again, how did you measure?

If you have 2x2 binned pixel, since this is CMOS sensor - it will have double the read noise. Since you are adding 4 pixels it will also have twice the dark current. If X is for a single pixel then adding 4 pixels together will be sqrt(X^2 + X^2 + X^2 + X^2) = sqrt(4*X^2) = X*sqrt(4) = 2*X

I would expect 2x2 binned from 0 seconds exposure if bias signal is removed to have about 3.4 as a result at unity gain (if we take 1.7e to be read noise per pixel).

Let's forget the binning and just check if above values match properly.

You say that you have read noise of 13 and you have total noise of 200s be 15 and total noise of 400s be 18.

Let's calculate dark current per second for both cases and compare them, shall we?

dark_current_200 = (15^2 - 13^2)/200 = 56/200 = 0.28ADU/s/px

dark current_400 = (18^2 - 13^2)/400 = 155/400 = 0.3875ADU/s/px

Dark current at 400s is higher by about 50% than that at 200s - that can't be right, can it? Here we did not do any comparison to other values - just checked your measurements one against another.

 

[han59]

Quote

Your reasoning is somewhat flawed here.

I don't see your point. What you write is the same as what I write.

 

Quote

I would expect 2x2 binned from 0 seconds exposure if bias signal is removed to have about 3.4 as a result at unity gain (if we take 1.7e to be read noise per pixel).

As you bin the noise reduces.  So 4 pixels give combined 1.7e/sqrt(4)=0.85e noise. In the binning the 4 pixel values are added together and then divided by 4. I just did as test with the ASI1600 at 0 Celsius:

bias unity gain at 0 Celsius, 1x1 binned - master bias=>  σ = 28 ADU measured or 28/16= 1.75 electron

bias unity gain at 0 Celsius, 2x2 binned - master bias=>  σ = 14 ADU measured or 14/16= 0.88 electron

or your method:

bias 1x1 - bias 1x1 =>  σ = 40 measured, so 40/sqrt(2)=28 ADU

bias 2x2 - bias 2x2 =>  σ = 20 measured, so 20/sqrt(2)=14 ADU

Quote

 

dark_current_200 = (15^2 - 13^2)/200 = 56/200 = 0.28ADU/s/px

dark current_400 = (18^2 - 13^2)/400 = 155/400 = 0.3875ADU/s/px

Dark current at 400s is higher by about 50% than that at 200s - that can't be right, can it? Here we did not do any comparison to other values - just checked your measurements one against another.

 

?? These where quick inaccurate local spot measurements. The 15 could 15.5 or so. In your calculation the sqrt is missing.

Dark current noise for 200 second bin2x2 is (15^2-13^2)^0.5 /(200sec*16)= 0.002339 electrons/sec. The dark current noise for single pixel is sqrt(4) larger equals 0.004677 electrons/sec..

Dark current noise for 400 second bin2x2 is (18^2-13^2)^0.5 /(400sec*16) = 0.001945 electrons/sec. The dark current noise for single pixel is sqrt(4) larger equals 0.003891 electrons/sec.

The dark current should be the same for 200 seconds and 400 seconds. This offset is just because of the inaccurate measurement. Next time I will measure noise in (dark - master dark)  or your method  noise in (dark -dark)/1.414.

Han

 

 

[vlaiv]

6 minutes ago, han59 said:

As you bin the noise reduces.  So 4 pixels give combined 1.7e/sqrt(4)=0.85e noise. In the binning the 4 pixel values are added together and then divided by 4. I just did as test with the ASI1600 at 0 Celsius:

bias unity gain at 0 Celsius, 1x1 binned - master bias=>  σ = 28 ADU measured or 28/16= 1.75 electron

bias unity gain at 0 Celsius, 2x2 binned - master bias=>  σ = 14 ADU measured or 14/16= 0.88 electron

or your method:

bias 1x1 - bias 1x1 =>  σ = 40 measured, so 40/sqrt(2)=28 ADU

bias 2x2 - bias 2x2 =>  σ = 20 measured, so 20/sqrt(2)=14 ADU

How do you do your binning? Bin x2 implies 4 pixels added together - not averaged. You can average them out but it is usually via software, and you are right if they are averaged - your math is correct with respect to expected stddev value.

I assumed you binned in drivers not in software.

9 minutes ago, han59 said:

In your calculation the sqrt is missing.

Dark current noise for 200 second bin2x2 is (15^2-13^2)^0.5 /(200sec*16)= 0.002339 electrons/sec. The dark current noise for single pixel is sqrt(4) larger equals 0.004677 electrons/sec..

Dark current noise for 400 second bin2x2 is (18^2-13^2)^0.5 /(400sec*16) = 0.001945 electrons/sec. The dark current noise for single pixel is sqrt(4) larger equals 0.003891 electrons/sec.

No, you have to differentiate dark current noise and dark current. Dark current is build up of signal and is modeled as Poisson process. Therefore noise associated with dark current signal is square root of that signal.

It is irrelevant if you observe single pixel as 1/4 of binned pixel or binned pixel as they behave the same in terms of noise - read noise of binned pixel will be combined noise of individual pixels and dark current noise will be combined dark current of individual pixels - you can calculate dark current of binned pixel the same - no need to divide with anything and if you want to divide - you'll divide with 4 rather than square root of that - as we are dealing with signal here - electron build up in 4 pixels combined is just sum of of individual electron buildups.

In above example, if all you are measuring is dark current rate per second - you should get the same value (roughly) regardless if you measure it for 100s, 200s or 400s - and that is not the case with your example.

I'm just pointing out that you should be careful of how you actually measure something to get the proper value.

15 minutes ago, han59 said:

I don't see your point. What you write is the same as what I write.

We mostly agree on how noise behaves and everything - except the fact that you for some reason want to relate LP noise levels to number of dark subs in master dark. I see no practical link between the two.

LP noise level has practical link to read noise and exposure duration for two reasons - LP is dominant noise source with respect to read noise in most cases and read noise value is per exposure - it does not depend on exposure length while LP noise depends on exposure length - accumulates with time.

You are suggesting that somehow number of darks in master dark can be related to LP noise levels. I'm just saying that there is no practical reason to do so, because:

a) both depend on time in the same way and ratio of dark noise and LP noise in single sub remains constant regardless of the exposure length.

b) stack of darks and related dark noise already has dominant noise source that will overpower it - single sub dark noise. No need to compare it with LP noise since single sub dark noise - that of light sub that we will calibrate - it will be much higher than master dark resulting noise.

I have shown you that you'll increase single sub dark noise by less than 2% if you use only 25 dark subs in your master dark. This is true if you image in SQM 18 or SQM 22 skies. Same thing will happen with dark noise alone for every sub you calibrate.

If same thing happens in SQM 18 and SQM 22 - why would we want to base number of darks in master dark on LP noise?

[han59]

Software binning for me is averaging both in drivers and software. You do the same for master darks and flats otherwise the levels are wrong.

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Yes there should distinction between dark current and dark noise. My measurement where noise measurements. Lets do it again:

Data:

2x2 bin, bias -15 Celsius, mean value 304 ADU, σ = 17.5 ADU

2x2 dark 200 sec, -15 Celsius,  mean value is 320 ADU,  σ = 21.9 ADU

 

Using the mean values:

The dark signal increases with 320-304 is 16 ADU or 16/16 is 1 electron. This also valid for a single pixel, so the dark current is then  1 electrons /200sec is 0.005 electrons/sec.

The noise is then sqrt(1 e)= 1 electrons or 16 ADU. For bin 2x2 the noise should be half or 8 ADU

 

Using the measured standard deviation values:

The measured noise should be 21.9^2=17.5^2+darknoise^2 =>  the darknoise measured is 16 ADU. 

 

 

The measured noise 16 ADU seems higher then predicted 8 ADU.  Random telegraph noise? It could also be the sudden increase of the mean in the first seconds as reported here. That would indicate that the delta mean is 1 electron too low.

I will do some checks at bin 1x1 when it is colder. At the moment I can't reach -15 Celsius.

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Quote

it will be much higher than master dark resulting noise.

The idea is to calculate it. It seems to matter for H-alpha images.

[han59]

Using new dark images taken last night, the dark noise values measured and calculated from the dark current match very accurately. See below. So I assume the math is now good and I can make the spreadsheet to calculate the influence of the darks noise on the final image.

Han

 

New data ASI1600MM-Cool:

dark, bin 1x1, 2 sec -15 Celsius, mean value 272 ADU, σ = 26.9 ADU

dark, bin 1x1,  600 sec, -15 Celsius,  mean value is 357 ADU,  σ = 45.3 ADU

 

The σ was measured by subtracting two darks of the same exposure divided by sqrt(2)

A exposure of 2 seconds was taken as bias since the mean values can drift  in the first second(s)

 

Using the mean values:

The dark signal increases with 357-272 is 85 ADU or 85/16 is 5.31 electron. The dark current is then 5.31 electrons/600sec is 0.00885 electrons/sec.

The noise calculated is then sqrt(5.31 e)= 2.3 electrons or 2.3*16 is 36.7 ADU

(dark noise follows a Poisson relationship to dark current, and is equivalent to the square-root of the number of thermal electrons generated within the image exposure time.)

 

Using the measured standard deviation values:

The measured noise should be

totalnoise^2 := readnoise^2 + darknoise^2

45.3^2 : = 26.9^2 + darknoise^2 =>  the darknoise measured is then 36.4 ADU. 

[han59]

The simulation is almost ready. But I noticed an interesting phenomena. Even using a single dark and many lights the final noise level is much lower then a single dark. This must be caused by the dithering. The noise in the single dark is averaged out while stacking. It is like a Gaussian blur is applied.

[han59]

The simulation is ready.

The influence of the darks seems almost steady. The noise from the master dark goes further down due to the dithering effect. So the master dark is each time shifted while stacking resulting in a lower noise addition then would be expected from the master dark itself.  The dithering seems to reduce the master dark noise with a factor (2/nr_lights)^0.5

• For imaging with no filtering only one darks seems sufficient.
• For imaging with a H-alpha filter only a few darks seem required.

 

It is possible to enter the SQM value and object  brightness to calculate the SNR value

Han

Attachment updated. See next post.

Edited August 11, 2020 by han59

[han59]

I have updated the spreadsheet calculation with flats and bias frames.

The input parameters are the camera noise parameters, binning, number of darks, flats, bias frames, filter bandwidth and sky SQM value

If the model is correct, only a few darks are required but the number of flats should be near the number of lights.

Han

Attachment is updated. See later posting.

Edited August 12, 2020 by han59

[vlaiv]

11 minutes ago, han59 said:

If the model is correct, only a few darks are required but the number of flats should be near the number of lights.

Out of interest - how did you model flats contribution as it is modulatory rather than additive (it impacts other noise sources not only with noise but signal value as well - both flat signal and target signal - there are cross terms that will be "noise").

[han59]

First step in modeling is for lights and dark frames:

signal_lights := (1/n)*Σ (Lights- masterdark)

noise_lights^2 := (Σ(light_noise^2 + dark_noise^2))^2

noise_lights := (lights_noise^2 + darks_noise^2)^0.5

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same for master flat and bias frames

signal_flats := (1/n)*Σ (flats - masterbias)

noise_flats := (flats_noise^2 + biases_noise^2)^0.5

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The flat is normalized to 1 and applied as divisor to the signal_lights

total_signal:=signal_lights /  signal_flats

 

For the noise without normalisation to 1:

total_noise^2 := (noise_lights^2 + (noise_flats * signal_lights/signal_flats)^2)/signal_flats^2

or since the flat is normalized to 1

total_noise^2 := noise_lights^2 + (noise_flats* signal_lights)^2

 

No further binning of blur of the flats was assumed.

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Edited August 12, 2020 by han59

[vlaiv]

6 minutes ago, han59 said:

same for master flat and bias frames

signal_flats := (1/n)*Σ (flats - masterbias)

noise_flats^2 := (Σ(flat_noise^2 + dark_noise^2))^2

What does this mean?

I don't really follow your equations. How do you represent following:

Let's assume we have a perfect flat that corrects vignetting of 70%. It will divide image in that area with 0.7.

All components get divided with 0.7 (what ever remains in single sub at point of flat calibration) - signal gets divided and all remaining noise sources get divided with 0.7.

I don't see that example in your above formula.

[han59]

It was late and I made a typing error, I will correct. The summation sign could be confusing since you averaging the flats to the same level as one flat. It should be:

noise_flats := (flats_noise^2 + biases_noise^2)^0.5

 

The vignetting was not take into account. The noise level is for the center of the image.

Edited August 12, 2020 by han59

[vlaiv]

25 minutes ago, han59 said:

It was late and I made a typing error, I will correct. The summation sign could be confusing since you averaging the flats to the same level as one flat. It should be:

noise_flats := (flats_noise^2 + biases_noise^2)^0.5

 

The vignetting was not take into account. The noise level is for the center of the image.

But how do you take into account duration of the flat and mean flat signal level?

If you for example have camera with 7k pixel well and you have another camera with 57k full well capacity.

Both of these have their flats taken at 75% histogram peak.

Single flat of first camera will have SNR of ~72.45, while single flat of second camera will have SNR of ~206.76.

Scale both flats so that signal is 1, noise will be very small.  In case of first camera, although bigger, noise will be ~0.0138.

This value is very small, but how does it impact SNR of image?

Let's say that signal in the image is 1e and it is perfect signal - no noise. We need to correct it with above flat. We will have:

1 / 1 +- 0.0138 => 1/0.9862 to 1/1.0138 => 1.014 to 0.9864

This is 0.014 above 1 and 0.0136 below 1. I'm showing you this because this is no longer symmetrical distribution - it is no longer Gaussian and you can't add it like Gaussian (or Poisson) distributions if you are not sure it will work like that.

 

 

[han59]

>>But how do you take into account duration of the flat and mean flat signal level?

The flat exposure was missing. But it has no real influence. I have added it.  See attached spreadsheet.

>If you for example have camera with 7k pixel well and you have another camera with 57k full well capacity.

>Both of these have their flats taken at 75% histogram peak.

>Single flat of first camera will have SNR of ~72.45, while single flat of second camera will have SNR of ~206.76.

>Scale both flats so that signal is 1, noise will be very small.  In case of first camera, although bigger, noise will be ~0.0138.

That done with the input B8, enter here 12, 14 bit. With this value the number of electrons is calculated.   75% * 65535/2^(16-nrbits)

>This value is very small, but how does it impact SNR of image?

The spreadsheet should give the answer.

>Let's say that signal in the image is 1e and it is perfect signal - no noise. We need to correct it with above flat. We will have:

>1 / 1 +- 0.0138 => 1/0.9862 to 1/1.0138 => 1.014 to 0.9864

>This is 0.014 above 1 and 0.0136 below 1. I'm showing you this because this is no longer symmetrical distribution - it is no longer Gaussian and you can't add it like Gaussian (or Poisson) >distributions if you are not sure it will work like that.

?? If you add an AC (alternating) signal to a DC signal the signal shape will not change.

------------------------------

There is an other problem with the simulation is that combining flats doesnt help after about 50 flats. The noise doesn't go done anymore.

Standard deviation as function of the number of combined flats:

 

 

SNR-Calculator 2020-8-12.zip