# Effect of sign change on ROC & z transform?

When we take ztransform of unit step it is z/[z-1] And our ROC is |z|>1 But if some how the minus sign between z and 1 changes to +, will our ROC still be same as old ( |z| >1 ) or will it be opposite?

Secondly the ztransform and ROC of u(-n) and u(n) are same or not?

You need to understand that a function such as $$F(z)=\frac{z}{z+1}$$ doesn't have a region of convergence (ROC). Only an infinite series has a ROC. E.g., the $$\mathcal{Z}$$-transform of a sequence $$x[n]$$ can be expressed by the series

$$\sum_{n=-\infty}^{\infty}x[n]z^{-n}\tag{1}$$

and this expression only makes sense if the series converges for certain values of $$z$$. The ROC is the region of the $$z$$-plane for which the series $$(1)$$ converges.

If you turn the problem around and if you're given a function of $$z$$ like $$F(z)$$ defined above, you can ask yourself if $$F(z)$$ is a valid $$\mathcal{Z}$$-transform of some sequence $$f[n]$$. It turns out that $$F(z)$$ is the $$\mathcal{Z}$$-transform of two different sequences, but with different ROCs. So a $$\mathcal{Z}$$-transform is only completely defined by the expression for the function $$and$$ the corresponding ROC.

Coming back to the function $$F(z)$$, it's straightforward to show that it is the $$\mathcal{Z}$$-transform of the sequence $$f_1[n]=(-1)^nu[n]$$ with the ROC $$|z|>1$$, and of the sequence $$f_2[n]=-(-1)^nu[-n-1]$$ with the ROC $$|z|<1$$.

Concerning your last question, as mentioned above, the expression for the $$\mathcal{Z}$$-transform including the ROC is unique, so the two sequences $$u[n]$$ and $$u[-n]$$ cannot have the same $$\mathcal{Z}$$-transform and ROC. It's actually pretty easy to come up with the $$\mathcal{Z}$$-transform of $$u[-n]$$:

\begin{align}\mathcal{Z}\big\{u[-n]\big\}&=\sum_{n=-\infty}^0z^{-n}\\&=\sum_{n=0}^{\infty}z^n\\&=\frac{1}{1-z},\qquad |z|<1\end{align}

which is of course different from the $$\mathcal{Z}$$-transform of $$u[n]$$. Note that the algebraic expression as well as the ROC are different.