## is there an inverse Z transform for: 1/z-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

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 Quote by LM741 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
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.

 For a casual sequence it will be a Laurent Series: |z|>1 $$\frac{1}{z-1}=\frac{1}{z}\frac{1}{1-1/z}=\frac{1}{z}\sum^{\infty}_{k=0}z^{-k}=\sum^{\infty}_{k=1}z^{-k}$$ Then by recalling the definition of the Z Transform: $$a[k \leq 0]=0$$ $$a[k \geq 1]=1$$ 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 $$\frac{1}{z-1}=-\sum^{\infty}_{k=0}z^{k}$$ So $$a[k \geq 1]=0$$ $$a[k \leq 0]=-1$$ Or a[k]=-u[-k]