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Homework Help: Rate of Noise in Three Sensors

  1. Feb 7, 2012 #1
    1. The problem statement, all variables and given/known data

    Suppose you have two sensors with rates of noise [itex]R_{A}^{noise}[/itex] and [itex]R_{B}^{noise}[/itex]. Suppose an erroneous signal occurs if the two sensors have a noise trigger within time [itex]\tau[/itex] of each other. Show that the rate of false signals is [itex]R_{AB}^{noise}=2R_{A}^{noise}R_{B}^{noise}\tau[/itex]. How does this change if a third sensor is introduced?

    2. Relevant equations


    3. The attempt at a solution

    I will now drop the noise superscripts for convenience. Also note that the rates above are constants, and do not vary in time.

    The probability that each sensor has a false signal in a time dt is obviously R*dt. So the probability that they both fire ought to be [itex]R_{A}R_{B}dtds[/itex]. We are interested in number of counts per time, so we will need to take an integral to bring probability to counts and a time derivative to get a rate.

    [tex]R_{AB}=\frac{d}{dt}\int^{\tau}_{-\tau}R_{A}R_{B}dsdt = R_{A}R_{B}\int^{\tau}_{-\tau}ds = R_{A}R_{B}2\tau[/tex]

    So for three sensors I do the analogous thing, presuming I got the above right. Here's the catch - I'm not sure what do with the integral limits.

    [tex]R_{ABC}=\frac{d}{dt}\int \int R_{A}R_{B}R_{C}dudsdt[/tex]

    My only guess is that the first integral is integrated to the next integral's variable.

    [tex]R_{ABC}=\frac{d}{dt}\int^{\tau}_{-\tau} \int^{s}_{-s} R_{A}R_{B}R_{C}dudsdt = R_{A}R_{B}R_{C}\frac{d}{dt}\int^{\tau}_{-\tau}2sdsdt = R_{A}R_{B}R_{C}[s^{2}]^{\tau}_{-\tau} = 0[/tex]

    This is obviously not the correct answer. Any help would be appreciated.
    Last edited: Feb 7, 2012
  2. jcsd
  3. Feb 7, 2012 #2
    Maybe I didn't really do the analogous thing. I still have 3 pure numbers with [itex]R_{A}dtR_{B}dsR_{C}du[/itex], so by dimensional analysis I still only need one integral and one derivative, but what happens to the remaining differential time?
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