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

by LM741
Tags: 1 or z1, inverse, transform
 P: 240 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]