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What is probability of two signals occuring at once?

  1. Feb 20, 2008 #1
    1. The problem statement, all variables and given/known data

    Two particle detectors are set up in such a way that their signals are compared. Each detector is only capable of detecting whether or not it has been hit by a particle, and will emit an electrical pulse when it detects something. The signals from the detectors are fed into a computer that records the number of pulses. Each pulse is 150 ns long. How often will the computer record pulses coming from both detectors simultaneously?

    The following parameters are known:
    the experiment is being run for a time [tex]T[/tex]
    the number of pulses from detector 1 is [tex]N_{1}[/tex]
    the number of pulses from detector 2 is [tex]N_{2}[/tex]
    each pulse lasts the same amount of time [tex]\delta t (= 150 ns)[/tex]

    Want to predict the number of simultaneous pulses [tex]N_{12}[/tex] that occur over the length of time [tex]T[/tex].

    To clarify, the distribution of pulses over time is random for each detector. Also, there is no relation between the number or frequency of pulses between the two detectors (the detectors are measuring random noise).

    2. Relevant equations

    multiplicity function: [tex] g(T,N) = \frac {T!}{(T-N)! N!} [/tex]

    3. The attempt at a solution

    I wasn't sure how to approach this, so I thought about the problem like this: I divided the time of the experiment [tex]T[/tex] into blocks of time [tex]\delta t = 150 ns[/tex] long. This way, each block of time either has a pulse or it doesn't. Since [tex]N_{1},N_{2}[/tex] are known, we know how many pulses exist - but know nothing of their distribution over time.

    The number of ways that the pulses can be distributed over the time axis for detector 1 is:
    [tex] g_{1}(\frac {T}{150 ns}),N_{1}) = \frac {\frac {T}{150 ns}!}{(\frac {T}{150 ns}-N_{1})! N_{1}!} [/tex]

    The number of ways that the pulses can be distributed over the time axis for detector 2 is:
    [tex]g_{2}(\frac {T}{150 ns}),N_{2}) = \frac {\frac {T}{150 ns}!}{(\frac {T}{150 ns}-N_{2})! N_{2}!} [/tex]

    Let [tex]P(N_{12})[/tex] be the probability of getting [tex]N_{12}[/tex] simultaneous pulses. I reasoned that:
    [tex]P(N_{12}) = \frac{R}{W}[/tex]

    where [tex]R[/tex] is the number of ways you can get [tex]N_{12}[/tex] simultaneous pulses, and [tex]W[/tex] is the total number of arrangements possible.

    Then, I said that the simplest case would when [tex]N_{1}=N_{2}[/tex]. In this case, the total number of arrangement's possible would be [tex]N_{1}^{N_{2}}[/tex].

    I still need to figure out how many different ways you can get [tex]N_{12}[/tex] simultaneous pulses. I'm not sure how to proceed. I'm not sure if I'm even going in the right direction with this, so any help would be appreciated. Thanks in advance.
     
  2. jcsd
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