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Z transform of a discrete filter

  1. May 21, 2009 #1
    The question is to find the difference equation relating u(k) and y(k), which are input and output respectively.

    H(Z) is given as

    H(Z) = [tex]\frac{1+(1/2)z^{-1}}{(1-(1/2)z^{-1})(1+(1/3)z^{-1})}[/tex]


    Solution that is given:

    y(k)-[tex]\frac{1}{6}[/tex]y(k-1)-[tex]\frac{1}{6}[/tex]y(k-2) = u(k)+[tex]\frac{1}{2}[/tex]y(k-1)


    I have had many attempts and haven't be correct. The thing is i am applying the same method as questions of a similar nature which i got correct. Any help would be appreciated.
    Cheers
     
  2. jcsd
  3. May 21, 2009 #2

    Defennder

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    Homework Helper

    Ok, but what have you tried so far? All you need do is to write H(z)=Y(z)/U(z) and then express Y(z) (...) = U(z) (...) where ... refers to the appropriate expression terms of the transfer function. Then simply apply the Z-inverse transform to get the answer.
     
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