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Integration time for poissonian signal

  • Thread starter RHK
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  • #1
RHK
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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?
 

Answers and Replies

  • #2
RHK
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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 :)
 
  • #3
RHK
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Is there anybody that can help me?
 
  • #4
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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
 
  • #5
RHK
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Maybe you're right :) thanks
 
  • #6
Redbelly98
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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:
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:
  • #7
RHK
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For the magnitude, can I use the relation: m = -2.5 log F ?
And then propagate the error?
I'm not sure at all...
 
  • #8
Redbelly98
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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:
 
  • #9
RHK
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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
Redbelly98
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Looks good.
 
  • #11
RHK
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Thank you!
 

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