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Probability density function of digital filter

  1. Mar 27, 2008 #1
    given that x has an exponential density function ie p(x) = exp (-x) and x(n) & x(m) are statistically independent.

    Now y(n) = x(n-1)+x(n)

    what is the pdf (probability density function) of y(n)
  2. jcsd
  3. Mar 27, 2008 #2
    The pdf of a sum of two independent variables can be obtained by the convolution of their respective pdf's.
  4. Mar 29, 2008 #3


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    Y(n) will be distributed as gamma(2,1) if X(n) has the pdf exp[-x(n)].
  5. Mar 31, 2008 #4
    I made a mistake while typing up the question :

    y(n) = [x(n-1) + x(n)]^2


    y(n) = x(n)^2 + x(n-1)^2

    So is the pdf of y(n) convolution of exp(-x(n)^2) and exp(-x(n-1)^2)

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