Now, there is a special reason the transfer function is given the notation $H(z)$: it happens to be the z-transform of $h[n]$, the impulse response of the system. Like all z-transforms, it has a region of convergence. We have seen previously that the ROC for a z-transform will be one of the following:--no z (the z-transform does not converge for any z)
--the entire z plane (with the possible exception of z=0 or z=\infty)--all z outside a certain disc (e.g., $|z|\gt 1$)
--all z within a certain disc (e.g., $|z|\lt 1$)--z within an annulus (e.g., $1\lt|z|\lt 2$)
Poles play a very significant role in the ROC for a transfer function; in fact, they completely define the ROC. The ROC of a z-transform cannot contain a pole (for the function would not exist at that point), which means that once we know the poles of a system, we know the potential ROCs. We say POTENTIAL because there are a variety of possible ROCs for a transfer function with one ore more poles; to be precice, if there are $M$ disctinct poles, then there will be $M$+1 possible ROCs: the disc within the pole of smallest magnitude, the region outside the disc or radius equal to the largest pole, and then the $M$-1 annuli of regions between adjacent poles. If, suppose, the system in question has 2 poles, then there will be 3 possible regions of convergence:
CAPTION.CAPTION.CAPTION.CAPTION.CAPTION. As we have also seen previously, each type of ROC corresponds to a particular kind of discrete-time signal. ROCs which are a disc correspond to signals with a non-zero duration extending to $n=-\infty$, those that are all $z$ outside of a certain disc correspond to signals with a non-zero duration extending to $n=-\infty$, and those that are an annulus correspond to signals whose non-zero duration extends both to $n=-\infty$ and $n=\infty$. For the class of discrete-time LTI system that we have been just considering, the impulse response is causal, and thus if it is infinite, will have a duration that extends towards $n=\infty$. Thus the transfer function of such a system has a ROC that extends outward from the outermost pole.
Finding the transfer funcion, zeros and poles, and roc
Consider a discrete-time system represented by this block diagram:
CAPTION. We would like to find its transfer function, along with its poles and zeros. The first step will be to express the input/output relationship in the time domain. From there we can take the z-transform of each side and find $\frac{Y(z)}{X(z)}$, which is by definition the transfer function:
$\begin{align*}y[n]&=x[n]+3x[n-1]+\frac{11}{4}x[n-2]+\frac{3}{4}x[n-3]+y[n-1]-\frac{1}{2}y[n-2]\\
y[n]-y[n-1]+\frac{1}{2}y[n-2]&=x[n]+3x[n-1]+\frac{11}{4}x[n-2]+\frac{3}{4}x[n-3]\\
Y(z)-z^{-1}Y(z)+\frac{1}{2}z^{-2}Y(z)&=X(z)+3z^{-1}X(z)+\frac{11}{4}z^{-2}X(z)+\frac{3}{4}z^{-3}X(z)\\
Y(z)(1-z^{-1}+\frac{1}{2}z^{-2})&=X(z)(1+3z^{-1}+\frac{11}{4}z^{-2}+\frac{3}{4}z^{-3})\\
H(z)=\frac{Y(z)}{X(z)}&=\frac{1+3z^{-1}+\frac{11}{4}z^{-2}+\frac{3}{4}z^{-3}}{1-z^{-1}+\frac{1}{2}z^{-2}}\\&=\frac{z^{-3}(z^3+3z^{2}+\frac{11}{4}z+\frac{3}{4})}{z^{-2}(z^2-z+\frac{1}{2})}\\&=\frac{(z+1)(z+\frac{1}{2})(z+\frac{3}{2})}{z(z-\frac{1}{2}-\frac{j}{2})(z-\frac{1}{2}+\frac{j}{2})}
\end{align*}$So, having found the transfer function and factored the numerator and denominator, we see readily that the zeros of the system are $z=-1/2$, $-1$, $-3/2$, and the poles for the system are $z=0$, $1/2+j/2$, $1/2-j/2$. As the system is causal (from the block diagram we see the output is dependent only on present and/or past values of the input and output), the ROC will extend outward from the outermost pole. The two nonzero poles are equidistant from the origin, at a distance of $\frac{\sqrt{2}}{2}$, so the ROC is $|z|\gt \frac{\sqrt{2}}{2}$. A pole/zero plot, with the ROC shaded, is below:
CAPTION.
