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 [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]