Question about z transform

Question

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 $Re[s] > a$)?

Answer 1

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

$z=e^{jw}$

Answer 2

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.

Answer 3

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.