Optimizing Exposure Times: Balancing Efficiency and Image Quality

[Andy Resnick]

Indeed, DDP does both a gamma stretch AND an unsharp mask, rendering any quantitative analysis basically impossible. At least for me.

As for your numbers from Fiji's analysis, I can't really make any sense out of them. All I did was choose a single pixel in my image for a quick analysis on. I also don't have a color camera to complicate things, just a monochrome camera with a filter wheel.

In APP, I retain the saturation adjustment but do not use any sharpening option- AFAIK, in APP the histogram stretch is independent of an unsharp mask.

I can't make sense of the numbers, either. I don't think the Bayer filter greatly complicates the issue (since a 16-bit/ch image is created by a 2 x 2 average of 14-bit data), but using a single pixel for analysis goes against my original query.

 

[Drakkith]

I used a single pixel just for simplicity. Everything I've said before still applies to your images overall. Based on my previous post where I looked at the SNR of different exposures, I'd say you benefit more from 10s subs vs 15s subs since you lose so many more images of the latter.

 

[Andy Resnick]

I think we are talking about two different types of noise.

Edit: I played around with the camera settings and kept parameterizing the frames in Fiji until I started to get consistent results...

As it happens, my original question actually concerns the effects of dark current: I found the (accurate) phrase "dark current [noise] and noise from skyglow are what is limiting faint object detection" here, see the replies by both sharkmelley and rnclark.

My camera sensor has a relatively high amount of dark current- about 1 e^-/s per pixel, so mitigating this could be beneficial.

So this is possibly good- at least, it's a potential path towards image improvement.

 

[Drakkith]

As it happens, my original question actually concerns the effects of dark current:

Oh? Did I miss that particular question? I don't see it in the thread.

My camera sensor has a relatively high amount of dark current- about 1 e-/s per pixel, so mitigating this could be beneficial.

Oh wow. That is very high. Are you sure about that number? That would mean your dark current is comparable to, or even higher than, your target signal except perhaps on brighter DSO's. My cooled Atik One 9.0 camera has a dark current of about 0.0002 e- /s per pixel at -10 degrees c (at least that's what the datasheet says), so dark noise is negligible in most of my images.

I found the (accurate) phrase "dark current [noise] and noise from skyglow are what is limiting faint object detection" here, see the replies by both sharkmelley and rnclark.

Yes, dark current can have a significant effect on the quality of your images. Consider a situation in which the dark current is the same as the signal. Our noise for a single image would then be:
$noise = \sqrt{sig + dc + ron^2}$

Using our previous numbers, with 10 seconds of exposure, that's: $noise = \sqrt{8.2 + 8.2 + 2.62 ^2} = \sqrt{16.4 + 6.86} = \sqrt{23.26} = 4.82$.

Compare this to our previous situation where we assumed negligible dark current:
$noise = \sqrt{8.2 + 2.62^2} = \sqrt{8.2 + 6.86} = \sqrt{15.06} = 3.88$

As you can see, dark current adds a significant amount of noise. Increasing it by nearly 25% in this case. Though we haven't included sky noise, so it's not quite so bad percentage-wise.

 

[Andy Resnick]

Ok- it took maximum effort, but I have some definite conclusions. The results are (for me) counterintuitive, so I needed to double-check and then generate some data (over the past 3 nights)

Note- I am using a DSLR and no narrow-band filters on a lens equivalent to a 100mm diameter refractive telescope. YMMV.

I first asked (what I thought) was a simple question: how many (visible band) photons emitted by a distant star are absorbed by the sensor per second? I used a combination of blackbody radiation (standardized to the sun: 43% of the emitted light is visible, the solar constant is 1361 W/m^2) and the relationship between flux and magnitude to generate a table like this:

And then convert to e-/pixel based on the (measured) FWHM size of the airy disc and assuming 50% quantum efficiency:

The next problem was to determine the amount of noise present. I gave up on this, because the specifications are not available and what I could find online was not really trustworthy (IMO). If I had to, I would estimate the total noise somewhere between 8 e- and 12 e-.

But- what I can see is that, for example imaging @ 400mm, that a magnitude 18 star should be well-separated from the noise even with 13s exposures. Not so much at 800mm, which will require longer exposures, but I can still comfortably image magnitude 16-17 objects at exposure times I can manage (typically 8-10 seconds)

So that brings me back to my efficiency calculation: given 'H' hours acquiring images, what fraction of those images are retained for stacking? That gives me a scond table:

So what I conclude is that, at 400mm imaging, I am better off exposing for 13s rather than 20s- the increased number of images results in a lower variance of the background.

So that's what I tried: imaging at 13s ISO 200 instead of 20s ISO 64 (also: lower read noise at ISO 200 compared to ISO 64, something I don't understand....)

The resulting stacks (comparing 20s subs and 13s subs), both with equal total integration time, showed that the two stacks appear nearly identical. This was confirmed with image analysis. The major distinction is that the 13s stack had a lower variance (which is what I wanted- easier to separate out light pollution).

Anyhow.. thanks to everyone for their comments, this has been educational!

 

[Gleb1964]

.. Consider a situation in which the dark current is the same as the signal. Our noise for a single image would then be:
$noise = \sqrt{sig^2 + dc^2 + ron^2}$

I think, the noise should be:
$noise = \sqrt{sig + dc + ron^2}$

 

[Drakkith]

I think, the noise should be:
$noise = \sqrt{sig + dc + ron^2}$

Ah, yes, you are correct. Noise is either:
$\sqrt{sig + dc + ron^2}$
in terms of signal

or

$\sqrt{shotnoise^2 + darknoise^2 + ron^2}$
in terms of noise

I've edited my previous post to correct my math. In the updated math, the dark current ends up adding 25% more noise instead of 40%.