Tidying up my final answer on z-transform

Discussion in 'Homework Help' started by u-will-neva-no, Jan 14, 2012.

  1. u-will-neva-no

    Thread Starter Member

    Mar 22, 2011
    230
    2
    Hey!

    The sequence of:  x[n] = 1, \frac{1}{2}^1,\frac{1}{2}^2,\frac{1}{2}^3,...

    was asked to be expressed using the z-transform:

    X(z) = 1 +\frac{1}{2}^1z^-^1 +\frac{1}{2}^2z^-^2+\frac{1}{2}^3z^-^3+...

    What I can't do is express the final result into the one on the sheet which is:
     \frac{z}{z-0.5}

    The notes say to use the linear difference equation:

    \frac{x[n]+x[n-1]}{2}

    I found x[n-1] just by subtracting one from each value for x[n] above so when I added the numerator I got 1 and the denominator as 2...clearly it is wrong.

    Thank you in advance for helping!
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    This is the sum of a geometric series.

    X(z)=1+\frac{1}{2z}+\frac{1}{4z^2}+\frac{1}{8z^3} + ....

    Since the series is infinite we may write ...

    \frac{1}{2z}X(z)=X(z)-1

    \frac{1}{2z}X(z)-X(z)=-1

    (\frac{1}{2z}-1)X(z)=-1

    (\frac{1-2z}{2z})X(z)=-1

    X(z)=\frac{-2z}{1-2z}

    X(z)=\frac{2z}{(2z-1)}

    X(z)=\frac{z}{(z-0.5)}
     
    u-will-neva-no likes this.
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