Integration time for poissonian signal

Click For Summary

Homework Help Overview

The problem involves determining the integration time necessary to measure the flux of photons detected from a star using a photomultiplier, based on Poissonian statistics. The average detection rate is given as 5 photons per second, and the goal is to achieve a standard deviation of 5% of the average flux.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between mean and variance in a Poisson distribution, with one suggesting a calculation to determine the required number of photons and integration time. Others raise questions about the validity of the approach and seek further guidance on the second part of the problem regarding magnitude error.

Discussion Status

Several participants have offered insights and calculations related to the integration time and magnitude error. There is an ongoing exploration of the statistical aspects of the problem, with some participants expressing uncertainty and seeking additional confirmation or suggestions.

Contextual Notes

Participants note potential confusion regarding the problem setup and the application of statistical principles. There is mention of the need for further clarification on the error propagation related to magnitude calculations.

RHK
Messages
64
Reaction score
0

Homework Statement



Observing a star with a telescope where a photomultiplier is applied, 5 photons per second are detected on the average.
Assuming a Poissonian statistics and the the background is negligible, determine the integration time useful to measure the flux with a standard deviation of 5% of the average flux, and the determine the magnitude error.

Homework Equations



Poissonian statistic: [itex]P(n)=e^{-\lambda}{\lambda^n}/{n!}[/itex]

The Attempt at a Solution



I can not outline the problem... Any suggestion please?
 
Physics news on Phys.org
Maybe i found a solution.
For a poissonian distribution, the mean value and the variance are equal.
As the standard deviation is the 5% of the mean, this can be translated in:

[itex]√X = 5% X[/itex] that's X = mean = 400.

So, to collect 400 photons with a mean rate of 5 photons per second, then 80 seconds are needed.

What about this? And for the second point?
Please I need suggestions :)
 
Is there anybody that can help me?
 
you might have more help in the statistical section of the forum. I've done this stuff before, but this problem seems a bit screwy to me
 
Maybe you're right :) thanks
 
RHK said:
Maybe i found a solution.
For a poissonian distribution, the mean value and the variance are equal.
As the standard deviation is the 5% of the mean, this can be translated in:

√X = 0.05 X that's X = mean = 400.
Yes, correct.
(The "%" character does not show up in LaTeX, so I have edited this quoted version of your post: 5% --> 0.05)
So, to collect 400 photons with a mean rate of 5 photons per second, then 80 seconds are needed.
Yes, I agree.
What about this? And for the second point?
Please I need suggestions :)
What equation is there for giving the magnitude?

EDIT:
Liquidxlax said:
you might have more help in the statistical section of the forum. I've done this stuff before, but this problem seems a bit screwy to me
I'm inclined not to move this thread; the statistics part of the problem has been answered.
 
Last edited:
For the magnitude, can I use the relation: m = -2.5 log F ?
And then propagate the error?
I'm not sure at all...
 
RHK said:
For the magnitude, can I use the relation: m = -2.5 log F ?
And then propagate the error?
I'm not sure at all...
Yes, that is exactly what to do :smile:
 
Ok, let's try :)

The error for the magnitue can be found by using: [itex]\Delta m = |\frac{∂m}{∂F}| \Delta F[/itex]

where [itex]\Delta F = σ(F) = 0.05\times X = 0.05\times 400 = 20[/itex]

and the derivative [itex]|\frac{∂m}{∂F}| = 2.5 |\frac{∂}{∂F} log F| = \frac{2.5}{ln 10} \times 1/F[/itex]

So: [itex]\Delta m = \frac{1.086\times 20}{400} = 0.054[/itex]

That is about the same of the flux error.
It seems reasonable.
 
  • #10
Looks good.
 
  • #11
Thank you!
 

Similar threads

Replies
1
Views
3K
  • · Replies 4 ·
Replies
4
Views
5K
  • · Replies 8 ·
Replies
8
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 12 ·
Replies
12
Views
5K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 17 ·
Replies
17
Views
4K
Replies
2
Views
25K
  • · Replies 3 ·
Replies
3
Views
3K