Is there an inverse Z transform for: 1/z-1 ?

  1. hay guys -really struggling to find an inverse Z transform for: 1/(z-1)

    doesn't seem to exist in the table of z transforms - so is this in fact possible to invert?? In case you're wondering - this forms part of a tut question.

    thanks

    John
     
  2. jcsd
  3. I'm wondering about something similar. I plotted some points in the transformation and it seems to be a circle but I can't manipulate it to get it in a form where I could find the radius or center.
     
  4. For a casual sequence it will be a Laurent Series: |z|>1


    [tex]\frac{1}{z-1}=\frac{1}{z}\frac{1}{1-1/z}=\frac{1}{z}\sum^{\infty}_{k=0}z^{-k}=\sum^{\infty}_{k=1}z^{-k}[/tex]

    Then by recalling the definition of the Z Transform:

    [tex] a[k \leq 0]=0 [/tex]
    [tex] a[k \geq 1]=1[/tex]

    Or using the step signal it's a[k]=u[k-1].

    For an anti-casual sequence it will be a simple Taylor Series: |z|<1

    [tex]\frac{1}{z-1}=-\sum^{\infty}_{k=0}z^{k}[/tex]

    So

    [tex]a[k \geq 1]=0 [/tex]
    [tex]a[k \leq 0]=-1[/tex]

    Or a[k]=-u[-k]
     
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