Z transform of finite signals I was trying to solve the Z-transform for u[n] - u[n-N], where u[n] means discrete unit step function, and N is some finite integer. I solved this using 2 methods.

Method 1 :

 

Taking z- transform using time-delay property and keeping in mind that delta[n] has z-transform = 1; I get:



which suggests that ROC is |Z| > 0

Method 2:

I know that z transform of $u[n] = \frac{z}{z-1}$ with ROC |Z| > 1

using this and the time-delay property on both u[n] and u[n-N] I say that:



which on making the denominators of the 2 fraction equal and simplifying becomes :


Which suggests ROC |z| > 1 .

The result in method-1 makes sense since the signal is a finite duration signal and taking z=0 would essentially mean a divide by zero situation while calculating the z transform. 

But Method-2 is something that results from simply applying the properties of z-transform on some pre-known result for a special signal. 

Why are the results different then?
Note that

$$1+z^{-1}+\ldots + z^{-(N-1)}=\sum_{n=0}^{N-1}z^{-n}=\frac{1-z^{-N}}{1-z^{-1}}\tag{1}$$

where I've used the formula for a finite geometric series.

So both your results are identical and correct.

The ROC is $|z|&gt;0$, which is the case for all causal sequences of finite length. Note that in the expression on the right-hand side of $(1)$ there is a pole-zero cancellation at $z=1$, so in fact there is no pole at $z=1$, hence the ROC $|z|&gt;0$.