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

1. Oct 3, 2006

### 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

2. Feb 10, 2010

### msd213

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.

3. Feb 10, 2010

### elibj123

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]