Recall the definition of the z-transform for some discrete-time infinite length signal $x[n]$:$X(z)=\sum\limits_{n=-\infty}^{\infty}x[n]z^{-n}$Given that this is an infinite sum (unlike, say, the sum of $N$ terms in a DFT), it will not necessarily converge for all $x[n]$ and/or all $z$. For some $x[n]$, it will converge for all values of $z$. For example, if $x[n]$=\delta[n]$, then:
$X(z)=\sum\limits_{n=-\infty}^{\infty}\delta[n]z^{-n}=1$
For some, $x[n]$ and $z$, the sum may not converge. For example, if $x[n]=1$ and $z=1$, the sum is unbounded. As it happens, if indeed $x[n]=1$ then there is no nonzero $z$ such that the sum converges.
The region of convergence
So for some $x[n]$ (like $x[n]=\delta[n]$) the z-transform sum will converge for all $z$, and for some $x[n]$ (like $x[n]=1$) the z-transform will not converge for ANY $z$. As you might suspect, there are some $x[n]$ for which the sum converges for some $z$. These values of $z$, those for which the z-transform exists, are known as the z-transform's
region of convergence . Consider, for example the z-transform for the signal $x_1[n]=\alpha^n u[n]~,~\alpha=.8$:
$\begin{align*}
X_1(z)&=\sum_{n=-\infty}^{\infty} x_1[n] \, z^{-n}\\&= \sum_{n=0}^{\infty} \alpha^n \, z^{-n}\\&= \sum_{n=0}^{\infty} (\alpha \, z^{-1})^n\\&=\begin{cases}
\frac{1}{1-\alpha \, z^{-1}}&|\alpha z^{-1}|\lt 1\\
\textrm{undefined}&\textrm{else}
\end{cases}\end{align*}$
So, the z-transform $X_1(z)$ only exists for $|\alpha z^{-1}|\lt 1$, or put another way, for $|z|\gt \alpha$; it is not defined for other values of $z$. This should not be too terribly shocking, for we are familiar with other functions that only exist for certain values, e.g. $\log{x}$ only for $x\gt 0$, or $\frac{1}{x}$ only for $x\neq 0$.
However, there is a difference between the z-transform and its region of convergence and functions like $\log{x}$ or $\frac{1}{x}$. Where the defined regions of existence of the values of $\log{x}$ or $\frac{1}{x}$ are quite clear, there is nothing intrinsic about the function $\frac{z}{z-\alpha}$ to suggest it only exists for $|z|\gt\alpha$ (of course it is apparent that $z$ cannot equal $\alpha). Why in the world would it be a problem if $z=\alpha/2$, for then we would simply have $\frac{\alpha/2}{\alpha/2-\alpha}$, right? The reason is that the function was a simplification of the sum, a simplification that existed only for certain $z$.
To see why that is the case, consider another signal related to the one we just considered. Let $x_2[n]=-\alpha^{n} u[-n-1]~,~\alpha=.8$:
We will now consider its z-transform:
$\begin{align*}X_2(z)&= \sum_{n=-\infty}^{\infty} x_2[n] \, z^{-n}\\&= \sum_{n=-\infty}^{-1} -\alpha^{n} \, z^{-n}\\&= -\sum_{n=1}^{\infty}\alpha^{-n} \, z^{n}\\&= -\sum_{n=0}^{\infty} (\alpha^{-1}\, z)^{n} + 1 \\[2mm]&=\begin{cases} \frac{-1}{1-\alpha^{-1} \, z} + 1&|\alpha^{-1}z|\lt1\\
\textrm{undefined}&\textrm{else}\end{cases} \\&= \begin{cases} \frac{-1}{1-\alpha^{-1} \, z} + \frac{1 - \alpha^{-1} \, z}{1-\alpha^{-1} \, z}&|\alpha^{-1}z|\lt1\\
\textrm{undefined}&\textrm{else}\end{cases} \\&= \begin{cases} \frac{1}{1-\alpha \, z^{-1}}&|\alpha^{-1}z|\lt1\\
\textrm{undefined}&\textrm{else}\end{cases} \\
\end{align*}$
So for $x_2[n]$, its z-transform is defined only for $|\alpha^{-1}z|\lt 1$, or equivalently, $|z|\lt\alpha$, and for those values it is $\frac{1}{1-\alpha z^{-1}}$. Now we see another reason why the region of convergence is so important; it is not merely added information about the z-transform, it is intrinsic to its very definition. The z-transforms for $x_1[n]$ and $x_2[n]$ are completely different, which is to be expected as those are completely different signals. However, the only difference between the transforms is the region of convergence:$\begin{align*}
x_1[n]=\alpha^{n} u[n]&\leftrightarrow X_1(z)=\begin{cases} \frac{1}{1-\alpha \, z^{-1}}&|z|\gt\alpha\\
\textrm{undefined}&\textrm{else}\end{cases}\\
x_2[n]=-\alpha^{n} u[-n-1]&\leftrightarrow X_2(z)=\begin{cases} \frac{1}{1-\alpha \, z^{-1}}&|z|\lt\alpha\\
\textrm{undefined}&\textrm{else}\end{cases}
\end{align*}$
Types of rocs
We have seen a few different types of ROCs for the z-transforms of various signals. If the z-transform for a discrete-time signal exists, then the region of convergence will fall into one of these three categories:
--if the signal has a finite non-zero duration, then the region of convergence will be the entire z-plane, with the possible exception of $z=\infty$ (if the signal has nonzero values at time indices less than zero) and $z=0$ (if the signal has nonzero values at time indices greater than zero).--if the signal is of infinite duration as $n$ approaches $\infty (but not $-\infty$), then the region of convergence will extend outside some disk (e.g., $|z|$\gt 1)
--if the signal is of infinite duration as $n$ approaches $-\infty (but not $\infty$), then the region of convergence will be inside some disk (e.g., $|z|$\lt 1)--if the signal is of infinite duration as $n$ approaches both $\infty$ and $-\infty, then the region of convergence--if it exists--will be an annulus (e.g., $1\lt |z|\lt 2$)
Questions & Answers
if three forces F1.f2 .f3 act at a point on a Cartesian plane in the daigram .....so if the question says write down the x and y components ..... I really don't understand
a fixed gas of a mass is held at standard pressure temperature of 15 degrees Celsius .Calculate the temperature of the gas in Celsius if the pressure is changed to 2×10 to the power 4
