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Z-transform of a discrete convolution

  1. Jan 12, 2012 #1
    Hi,

    Suppose we have these two functions and their z-transforms are
    [tex]P(r,z)=\sum_{t=0}^{\infty}P(r,t)z^t[/tex]
    and
    [tex]F(r,z)=\sum_{t=0}^{\infty}F(r,t)z^t[/tex].
    Now we are going to transform the following convolution of P and F:
    [tex]\sum_{t'\le{t}}F(r,t')P(0,t-t')[/tex].
    The result is said to be
    [tex]F(r,z)P(0,z)[/tex].
    But I don't know how to obtain the result. There are two difficulties:(1)the upper limit of t' is t, which is finite. (2)the two summations are coupled.
    Anyone can help?
     
    Last edited: Jan 12, 2012
  2. jcsd
  3. Jan 12, 2012 #2
    It should be [itex] z^t[/itex], not [itex] t^z[/itex]; then it is just the standard power series multiplication.
     
  4. Jan 12, 2012 #3
    Sorry about the typo. But my problem remains. Could you show a bit more on the decoupling process?
     
  5. Jan 12, 2012 #4
    If you multiply 2 power series [itex] \sum_{k=0}^\infty a_k z^k [/itex] and [itex] \sum_{k=0}^\infty b_k z^k [/itex], the coefficients in the power series of of the product are given by the discrete convolution.

    To get the term with [itex]z^n [/itex] in the product you need to add up all products [itex] a_k z^k b_{n-k}z^{n-k} [/itex], [itex] 0\le k \le n [/itex]. Thus the coefficient at [itex]z^n[/itex] is given by [itex] \sum_{k=0}^n a_k b_{n-k} [/itex]

    In your case [itex] a_k = F(r,k) [/itex], [itex] b_k = P(0, k)[/itex].
     
  6. Jan 12, 2012 #5
    Thanks! Starting from the result is indeed easier!
     
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