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As we have in Laplace transform that the roots decide the stability of the system i.e. if the roots are complex and lie in the left side of the plane you get a sinusoidal response with decreasing amplitude

similarly is there any significance of the roots , zeros and ROC of the z-transform and the stability criteria . All i read in books is how to find the ROC and the properties of z transform like linearity ,time reversal,time shifting . It has nowhere mentioned why we are even using z-transform

my apologies if this question is too basic or if it shouldn't belong here.

user3132001
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    the Z-transform has the same significance to discrete-time signals and systems as does the Laplace transform has on continuous-time signals and systems. the Z-transform is exactly the same as the Laplace transform applied to the ideally sampled signal: $$x_\text{s}(t) = x(t) \ \sum\limits_{n=-\infty}^{\infty} \delta(t-nT) = \sum\limits_{n=-\infty}^{\infty} x(nT) \delta(t-nT)$$ and where $$x[n] \triangleq x(nT)$$. when you compute the Laplace transform of $x_\text{s}(t)$, you will get the Z transform of $x[n]$ with the substitution of $$ z \triangleq e^{sT} $$. that's all $z$ is. – robert bristow-johnson Apr 06 '15 at 19:05

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First of all, I think you're reading the wrong books. Almost any basic text on DSP has a chapter on the $\mathcal{Z}$-transform and its significance to describe linear time-invariant (LTI) discrete-time systems. If you're looking for good (and free) books, take a look at this answer.

I will not repeat all the details you can find in those books (and in many other places), but let me just point out a few very basic things to get you started. Each (single) pole $p$ of the transfer function $H(z)$ of a causal LTI discrete-time system contributes a term

$$c\cdot p^nu[n]\tag{1}$$

to the system's impulse response, where $c$ is some constant, $p$ is the (possibly complex) pole, and $u[n]$ is the discrete-time unit step function. From (1) it is clear that this contribution only decays with time if $|p|<1$. So for a causal system to be stable we require that all the poles of the transfer function are inside the unit circle of the complex plane, i.e. they have magnitudes smaller than $1$. So if you're looking for analogies with the Laplace transform, the inside of the unit circle corresponds to the left half plane of the complex variable $s$. Furthermore, the unit circle of the $z$-plane corresponds to the $j\omega$-axis. Knowing these two things, it becomes very easy to carry over everything you know about transfer functions of continuous-time systems (Laplace transform) to the discrete-time domain ($\mathcal{Z}$-transform).

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Matt L.
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