This source states that the unit step function in the Z-domain is \$\frac{z}{z-1}\$. However, in its derivation it states \$\sum_{k=0}^{\infty} z^{-k} = \frac{z}{z-1}\$.
But doesn't that last relation only hold true for \$z>1\$? I don't see how that condition is met. I know that the poles have to be in the unit circle for the system to be stable, but I don't see if and how that connects to the condition above.
What am I missing?
Here is one way to look at it.
If you consider your derived expression as a transfer function in z $$\frac{Y(z)}{X(z)}=\frac{z}{z-1}$$
you can rewrite in terms of the delay operator (1/z) as
$$\frac{Y(z)}{X(z)}=\frac{1}{1-z^{-1}}$$
then cross multiply to get the difference equation $$y(k)-y(k-1)=x(k)$$ or $$y(k)=x(k)+y(k-1)$$ To get the impulse response x(k)=1 for k=0 but zero for k >0
from which y(k)=1 for k>=0, zero elesewhere, aka the step function.
