# What is the concept of Region of Convergence of Z-Transform

I have been trying to learn Z transform. But I haven't found any good source that will clear my concept about the region of convergence. I have found some keywords like unit circle, but I don't have a clear concept about Region of Convergence. I have knowledge about complex number, representation of complex numbers on a graph. I want to develop a clear concept about Region of Convergence and the other stuff. I want somebody to give me a clear concept from the scratch. I am considering myself a very beginner in this subject. Any URL to resource will also be helpful.

Mathematical Definition:

The Region of Convergence, or $$ROC$$, of a Z-Transform comprises all the values on the Z-plane for which the transformation converges. So, remember that:

$$\mathcal{Z}\{x[n]\}=\sum_{n=-\infty}^{+\infty}x[n]z^{-n}$$

You should also keep in mind that $$z\in\mathbb{C}$$, which means it can be written as:

$$z=|z|e^{j\theta}$$

Let us say that $$|z|$$ represents a determined radius $$r$$ in the complex plane, so $$z=re^{j\theta}$$.

Mathematically, the $$ROC$$ is defined as the set of values of $$z$$ for which the Z-transform sum is absolutely summable, or:

$$ROC=\left\{z:\left|\sum_{n=-\infty}^{+\infty}x[n]z^{-n}\right|<\infty\right\}$$

Time for some algebra:

\begin{align} \left|\sum_{n=-\infty}^{+\infty}x[n]z^{-n}\right|&<\infty\\ \left|\sum_{n=-\infty}^{+\infty}x[n]\left(re^{j\theta}\right)^{-n}\right|&<\infty\\ \left|\sum_{n=-\infty}^{+\infty}x[n]\left(\frac{1}{r}\right)^{n}e^{j\theta n}\right|&<\infty\\ \end{align}

The triangle inequality for absolute values state that:

$$\left|\sum_{n=-\infty}^{+\infty}x[n]\left(\frac{1}{r}\right)^{n}e^{j\theta n}\right|\leq\sum_{n=-\infty}^{+\infty}\left|x[n]\left(\frac{1}{r}\right)^{n}e^{j\theta n}\right|<\infty$$

Going on:

\begin{align} \sum_{n=-\infty}^{+\infty}\left|x[n]\left(\frac{1}{r}\right)^{n}e^{j\theta n}\right|&<\infty\\ \sum_{n=-\infty}^{+\infty}\left|x[n]\right|\left|\left(\frac{1}{r}\right)^{n}\right|\underbrace{\left|e^{j\theta n}\right|}_{1}&<\infty\\ \sum_{n=-\infty}^{+\infty}\left|x[n]\right|\left|\left(\frac{1}{r}\right)^{n}\right|&<\infty\\ \end{align}

As $$r$$ is already defined as $$|z|$$, it will always be non-negative. And we can also split the sum into two parts:

$$\sum_{n=-\infty}^{+\infty}\left|x[n]\right|\left|\left(\frac{1}{r}\right)^{n}\right|=\underbrace{\sum_{n=-\infty}^{-1}\left|x[n]\right|\left(\frac{1}{r}\right)^{n}}_{\text{anticausal}}+\underbrace{\sum_{n=0}^{+\infty}\left|x[n]\right|\left(\frac{1}{r}\right)^{n}}_{\text{causal}}$$

In order for the sum to converge, both parts must be finite, and although we may not know exactly what $$x[n]$$ is, it must be asymptotically exponentially bounded in order to assure a finite $$X(z)$$, which means that $$\exists K_{1,2},r_{1,2}\in\mathbb{R}_{\geq0}$$ such that:

$$|x[n]|\leq\begin{cases} K_1r_1^{n}, & \forall n<0\\ K_2r_2^{n}, & \forall n\geq0\\ \end{cases}$$

so starting by the anticausal part:

\begin{align} \sum_{n=-\infty}^{-1}\left|x[n]\right|\left(\frac{1}{r}\right)^{n}&<\infty\\ \sum_{n=1}^{+\infty}\left|x[-n]\right|\left(\frac{1}{r}\right)^{-n}&<\infty\\ \sum_{n=1}^{+\infty}\left|x[-n]\right|r^{n}&<\infty\\ \sum_{n=1}^{+\infty}\left(\frac{1}{r_1}\right)^{n}r^{n}&<\infty\\ \sum_{n=1}^{+\infty}\left(\frac{r}{r_1}\right)^{n}&<\infty\\ \end{align}

Which automatically implies, that for this sum to converge:

$$r

Now, for the causal part:

\begin{align} \sum_{n=0}^{+\infty}\left|x[n]\right|\left(\frac{1}{r}\right)^{n}&<\infty\\ \sum_{n=1}^{+\infty}\left(\frac{r_2}{r}\right)^{n}&<\infty \end{align}

And therefore, we can only that the sum converges for:

$$r>r_2$$

Conclusions so far:

Anticausal signals have "inwards" $$ROC$$, which contains the origin.

Causal signals have "outwards" $$ROC$$, which contains infinity.

If the signal is noncausal (has both causal and anticausal parts), the $$ROC$$ is an annulus corresponding to the intersection of the $$ROC$$s ($$r_2), provided that $$r_2, otherwise there is no $$ROC$$, and as consequence, no convergence.

Why is the $$ROC$$ important?

Some inverse Z-transforms may result in ambiguous discrete time domain functions. Therefore, the region of convergence of the Z-Transform brings important information about causality and stability of the original signal. It is always important that a Z-Transform comes along with its $$ROC$$, so that you don't get mistaken when trying to find its inverse.

Poles, Stability, Causality, Unit Circle and $$ROC$$:

The Unit Circle at the Z-plane is the set of points $$z$$ to which the Z-Transform equals the Discrete Time Fourier Transform ($$DTFT$$) and also, if you map it to the s-Plane, it corresponds to the Imaginary axis.

A Causal system is stable if all poles are inside the unit circle.

An Anticausal system is stable if all poles are outside the unit circle.

As consequences of the previous statements, the $$ROC$$ contains no poles and contains the Unit Circle.

A system is stable if the Unit Circle is inside its $$ROC$$. Which means that for stable anticausal signals $$r_1>1$$, for stable causal signals $$r_2<1$$ and for stable acausal (non causal) signals $$r_2<1.

The values of $$r_1$$ and $$r_2$$, represent the absolute values of smallest (innermost), and largest (outermost) poles, respectively.

• I know it's not related to the OP's question, but what happens when the coefficient of poles and zeros is the same?
– Rima
Feb 11 at 6:47

To extend @bertozzijr answer. I've a simple example. Which illustrates what's region of convergence.

Consider the signal:

$$x[n] = a^n u[n]$$

So, it's z-transform can be written as:

$$X(z) = \sum_{n=-\infty}^\infty x[n] z^{-n}$$ $$X(z) = \sum_{n=-\infty}^\infty a^nu[n] z^{-n}$$

It's actually the right sided sequence. As: $$u[n]$$ only exists for $$n\geqslant0$$. And it's called unit-step signal.

So, summation goes like:

$$X(z) = \sum_{n=0}^\infty a^n z^{-n} = \sum_{n=0}^\infty (az^{-1})^{n}$$

As the term on the right side of the above equation is simply the sum of infinite geometric sequence. So, the ratio must be less than 1,otherwise it won't converge. i.e inside the summation we have r. If we see from the point of sum of geometric sequence.So:

$$|az^{-1}|<1 \implies |z|>|a|$$

So that's, how the region of convergence can be found.