Question on exposure calculation for CCD imaging

Dear all,

I'm a newbie in digital imaging and I have a question for you regarding
the total exposure time for imaging.
Let's say I want to image a deep sky object, I shoot a single frame of
30 seconds and i find that the pixel value of the brightest feature of
the object is 92 out of 255 (8 bits per pixel AD conversion).
Let's say I want to have as final pixel value something like 40000.
May I just roughly say that I need 40000/92 frames (around 434 frames)
to have the desired dynamic coverage for the object?

thanks a lot
Nicola

On 23 May 2006 05:16:15 -0700, "Nicola"
wrote:

I'm a newbie in digital imaging and I have a question for you regarding
the total exposure time for imaging.
Let's say I want to image a deep sky object, I shoot a single frame of
30 seconds and i find that the pixel value of the brightest feature of
the object is 92 out of 255 (8 bits per pixel AD conversion).
Let's say I want to have as final pixel value something like 40000.
May I just roughly say that I need 40000/92 frames (around 434 frames)
to have the desired dynamic coverage for the object?

That is basically correct. However, you need to be aware that a low
dynamic range camera like this (limited to exposures a few seconds long)
will also result in a significant readout noise contribution to your
final image. So while the signal increases linearly with time, the noise
also increases (but at closer to the square root of the time). So your
signal will be 434 times larger, and your noise will be 21 times higher.
Your actual S/N, which is what really matters, will improve by 21 times-
certainly an improvement, but maybe not as much as you'd like
considering the long total exposure required.

Ideally, you'd like your only noise source to be the statistical photon
noise from the object itself (equal to the square root of the signal).
Dark current noise can be reduced to little or no significance by
cooling the camera, and readout noise is reduced to insignificance by
combining fewer, longer exposures.

_________________________________________________

Chris L Peterson
Cloudbait Observatory
http://www.cloudbait.com

On 23 May 2006 09:16:58 -0700, "Nicola"
wrote:

Dark current noise can be reduced to little or no significance by
cooling the camera, and readout noise is reduced to insignificance by
combining fewer, longer exposures.

Does dark frame subtraction do the job for the first issue, to a
certain extent?
I'm using a modified webcam but no cooling, though.

No, dark frame subtraction can only increase the noise, since the noise
in the dark frame itself becomes part of the image. By definition, noise
can't be removed- it is just a statistical uncertainty in the value of a
pixel. Dark frame subtraction removes a type of characterizable, fixed
pattern signal. Since that signal is unwanted, it is often called noise,
but it really isn't in a mathematical sense.

Dark current noise is equal to the square root of the dark current,
which is linearly dependent on exposure time. This particular noise
source is the same whether you take a single long exposure or multiple
short exposures with the same total time. That is, there is no dark
current noise penalty when combining images. It is readout noise that is
problematic when combining short exposures, since each new image adds
noise.

_________________________________________________

Chris L Peterson
Cloudbait Observatory
http://www.cloudbait.com