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Calculate Power Margin Required to Prevent Satellite Link Drop

  1. Sep 5, 2009 #1

    I've been looking at some example solution and still haven't found the physical/mathematical explanation for the given solution.


    Due to slow satellite stabilization rotation there are big differences between maximums and minimums of received signals. Calculate the power margin required to ensure the probability of link drop lower or equal to 1%.


    P/Pmax = sin^2(w*t)
    w*t = pi/2 * probability; probability = 1%

    a = 10*log(Pmax/P) = 20*log(sin(wt)) = 20*log(sin(pi/2*probability)) = 36.08 dB

    P = current power
    Pmax = maximum power
    w = angular frequency
    t = time

    Physical explanation (my interpretation):

    The power is proportional to square of the electric field strength, thus:

    P = k * (abs(E))^2, where E = Emax*sin(wt)

    so P = Pmax*sin^2(wt)

    But from here I don't understand why the solution is wt = pi/2*probability!

    There is certainly minimum power required in order to maintain the link at given parameters the function sin^2(wt) should be integrated over one period (2pi) and then the horizontal line (power margin) should be set at such level to divide upper and lower surface at ratio 99/1.

    I would really appreciate if anyone of you could theoretically explain the given solution (36.08 dB)

    Thank you!

  2. jcsd
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