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Integration time for poissonian signal

  1. Nov 13, 2011 #1

    RHK

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    1. The problem statement, all variables and given/known data

    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.

    2. Relevant equations

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

    3. The attempt at a solution

    I can not outline the problem... Any suggestion please?
     
  2. jcsd
  3. Nov 14, 2011 #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 :)
     
  4. Nov 16, 2011 #3

    RHK

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    Is there anybody that can help me?
     
  5. Nov 16, 2011 #4
    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
     
  6. Nov 17, 2011 #5

    RHK

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    Maybe you're right :) thanks
     
  7. Nov 17, 2011 #6

    Redbelly98

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    Yes, correct.
    (The "%" character does not show up in LaTeX, so I have edited this quoted version of your post: 5% --> 0.05)
    Yes, I agree.
    What equation is there for giving the magnitude?

    EDIT:
    I'm inclined not to move this thread; the statistics part of the problem has been answered.
     
    Last edited: Nov 17, 2011
  8. Nov 18, 2011 #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...
     
  9. Nov 18, 2011 #8

    Redbelly98

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    Yes, that is exactly what to do :smile:
     
  10. Nov 18, 2011 #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.
     
  11. Nov 18, 2011 #10

    Redbelly98

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    Looks good.
     
  12. Nov 19, 2011 #11

    RHK

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    Thank you!
     
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