After studying z transform from different books and literature on internet I want to ask few which makes me confuse.

a) From the Discrete Time Fourier Transform we have drive equation for z transform. $$X(z)= \sum _ {n=-\infty}^{+\infty} x[n]z^{-n}$$ where $z$ is represented in polar form $z=re^{j\omega}$ I want to know that why we represent $z$ in polar coordinates? as in some books it is written than $z$ is complex $z=\sigma + j \omega$

b) ROC for $z$ transform is same as Laplace Transform? In laplace transform we check that direction of $t$ (i.e. if we have $u(t)$ than the $Re[s] > a$)?

I think it is common (in signal processing books) to write the z transform in polar form, to make clear its relationship with the fourier transform, that is z-transform equal to fourier transform on the unit circle, that is when r=1, then:

Ztransf-> $z=r*e^{jw}=e^{jw}$<- fourier transform or just

$z=e^{jw}$

• can you please check this question Dec 6 '15 at 6:19

The variable $z$ is complex, and it can be represented either in polar form $z=re^{j\phi}$ or in terms of its real and imaginary parts. The polar form is preferred because the region of convergence (ROC) is determined by $r$, the magnitude of $z$. This becomes clear if you write the term inside the sum of the definition of the $\mathcal{Z}$-transform as

$$x[n]z^{-n}=x[n]r^{-n}e^{-jn\phi}\tag{1}$$

From $(1)$ it is clear that the value of $r$ determines the convergence of the series. This is why the ROC of the $\mathcal{Z}$-transform is specified in terms of $r=|z|$. For the bilateral $\mathcal{Z}$-transform the ROC is a ring ($a<|z|<b$, with $b>a\ge 0$), whereas for the unilateral $\mathcal{Z}$-transform, the ROC is outside a circle centered in the origin of the complex plane ($|z|>a$, $a>0$).

Note that for the Laplace transform the complex variable $s$ occurs in the exponent, so it's the real part of $s$ that determines the ROC of the Laplace transform. These are the correspondences between the Laplace transform and the $\mathcal{Z}$-transform in terms of ROCs:

\begin{align}&s\text{-plane }&&z\text{-plane}\\\\ &\text{right half-plane} && \text{region outside a circle (centered at the origin)}\\ &\text{vertical strip} && \text{ring (centered at the origin)}\\ &\text{left half-plane} && \text{region inside a circle (centered at the origin)} \end{align}

The $j\omega$-axis in the $s$-plane corresponds to the unit circle in the $z$-plane, i.e. if the unit circle is part of the ROC, then the DTFT exists, and if the corresponding sequence is the impulse response of an LTI system, then the system is stable.

I think because of the use of the dtft and the way we explain it on z plane , by the time we prefer to use z transform instead of Fourier Transform.

there is a lot of discussion about the exact form of z , if you introduce z for analytical sequences , it's better to use the polar coordinates,because you work with ring with epsilon radius ,if you don't use this introduction ,you'll have more sequences like sinc to power of 2 which you can't find any z-t by the first definition.