1
$\begingroup$

I need help on how to estimate an exposure time for a desirable signal to noise ratio (SNR). For example, given the magnitude and a desired SNR of a star we want to achieve, how long should the exposure time be? We may assume that the noise is only due to counting statistics, and ignore dark and flat noise. Any useful function to make this estimate?

$\endgroup$
4
  • $\begingroup$ This would depend a great deal on the hardware and viewing conditions and weather conditions. You really need to experiment with your equipment to develop a feel for it's behavior. It is truly amazing what a difference good image processing techniques make (a great example in this video about capturing Martian volcanoes on YouTube ). $\endgroup$ Commented Oct 5, 2020 at 19:19
  • 1
    $\begingroup$ I understand, that's why I asked for an equation $\endgroup$
    – user35427
    Commented Oct 5, 2020 at 21:07
  • 1
    $\begingroup$ It's a good question, but I think you need to add some more information about what kinds of information are given. Should it start with photon flux at the sensor, or a star's magnitude and a telescope's point spread function, or is it an extended object like a planet? $\endgroup$
    – uhoh
    Commented Oct 6, 2020 at 10:50
  • $\begingroup$ Still no information about what kind of observations are being done. Is this optical imaging of a point source? $\endgroup$
    – ProfRob
    Commented Oct 8, 2020 at 7:21

1 Answer 1

0
$\begingroup$

Here's a simple start, and I suggest you must include your sensor's noise -- not flat-fielding, but just the accumulated readout noise in a pixel.

You can use the Poisson SNR rule for the incoming light, so optical noise $N_o = \sqrt{S_o}$. Just add the electronics noise, again as a simple model, $N_e = C_(readout) + k\tau$ where $\tau$ is the integration time and $k$ represents leakage current into the pixel well. $S_o$ also equals some constant times $\tau$ , so just combine the two noise terms in an RSS manner, i.e. $ N_{tot} = \sqrt{N_o^2 + N_e^2}$ .

$\endgroup$
3
  • $\begingroup$ Why are you saying "electronic noise" depends on exposure time? Readout noise is independent of exposure time. $\endgroup$
    – ProfRob
    Commented Oct 6, 2020 at 22:48
  • $\begingroup$ @RobJeffries The noise associated with the readout switches is independent, but the leakage current filling the pixel well is time-dependent, I'll add that. $\endgroup$ Commented Oct 7, 2020 at 12:25
  • $\begingroup$ The conventional term is "dark current". It is usually negligible compared to (e.g.) sky noise - which is also proportional to the square root of time. $\endgroup$
    – ProfRob
    Commented Oct 7, 2020 at 13:13

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .