What is the Transform?
Why use the Transform?
Explanation
The -Transform plays the same role as the Laplace Transform in the discrete domain .
The Bilateral -Transform is given by:
\text{for }z\in \mathbb{C} $$ The Unilateral $\mathcal Z$-Transform is given by: $$ \color{green}\mathcal{X}(z)=\sum_{n=0}^{\infty}x[n]\cdot z^{-n}\\ \text{for }z\in \mathbb{C} $$ The $\mathcal Z$-transform of a signal $x[n]$ gives a difference equation as follows: $$ \mathcal{X}(z)=x[0]z^{-0}+x[1]z^{-1}+...+x[n]z^{-n} $$ All of the values of $z$ make the summation exist in a Region of Convergence (RoC). ## The Complex Variable $z$ $$ z=re^{j\omega}=r\cos(\omega)+j\sin(\omega) $$ where: - $r$ is the rate of decay or growth. - $\omega$ is the frequency of oscillation. ## Properties For a system impulse response $h(n)$ its Z-transform is given by: $$\mathcal H(z)=\sum_{n=-\infty}^{\infty}h(n)\cdot z^{-n}$$ Similarly, the Z-transform of a scaled delta function $a\delta(n)$ is given by: $$\mathcal H(z)=\sum_{n=-\infty}^{\infty}a\cdot\delta(n)\cdot z^{-n}=a$$ $$ x(n)\overbrace{\iff}^{\mathcal Z}\mathcal X(z) $$ - Linearity: $ax(n)+by(n) \underbrace{\iff}_{\mathcal Z} a\mathcal X(z)+b\mathcal Y(z)$ - Delay: $x(n-n_0) \underbrace{\iff}_{\mathcal Z} z^{-n_0}\mathcal X(z)$ - Convolution: $y(n)=h(n)*x(n) \underbrace{\iff}_{\mathcal Z}\mathcal Y(z)=\mathcal H(z)\cdot \mathcal X(z)$ ### Simple Examples of $\mathcal Z$-Transforms 1. Finite-length Sequence $$ \begin{bmatrix} 1&2&-1&3 \end{bmatrix} \overbrace{\iff}^{\mathcal Z} 1+2z^{-1}-1z^{-2}+3z^{-3} $$ 2. Impulse $$ \delta(n)\overbrace{\iff}^{\mathcal Z} \sum_{n=-\infty}^{\infty}\delta(n)z^{-n}=z^{-0}=1 $$ ## Stability of a System in the $\mathcal Z$ Domain  # Poles and Zeros $$ G(z)={p_0\prod_{\ell=1}^M(1-\xi_{\ell}z^{-1}) \over d_0\prod_{\ell=1}^N(1-\lambda_{\ell}z^{-1})} \\ \therefore \\ G(z)=z^{N-M}{ p_0\prod_{\ell=1}^M(z-\xi_{\ell}) \over d_0\prod_{\ell=1}^N(z-\lambda_{\ell}) } $$ ### We define a _pole_ as: $$z=\lambda_\ell \ ; \ G(z)\to \infty$$ ### We define a _zero_ as: $$z=\xi_\ell \ ; \ G(z)=0$$ Note that $G(z)$ has $M$ amount of finite _zeros_ and $N$ amount of finite _poles_. - $N>M \implies$additional $N-M$ _zeros_ at $z=0$ (origin) - $N<M \implies$additional $M-N$_poles_ at $z=0$ (origin) ## Example Questions Given the sequence $x(n)=u(n)$ find the $\mathcal Z$-transform of $x(n)$: We start with the definition of the $\mathcal Z$-transform $$ \mathcal X(z)=\sum_{n=0}^{\infty}u(n)z^{-n}=1+z^{-1}+z^{-2}+\dots $$ Using the MacLaurin series expansion we know that $$ 1+r+r^2+\dots={1\over 1-r} \ where \ |r|\lt1 $$ As such, we implement the same logic to our signal $u(n)$ and say that $$ \mathcal X(z)={1\over 1-z^{-1}}={z\over z-1} $$ As such we say that when $|z^{-1}|<1\implies|z|>1$ # Regions of Convergence (RoC) The region of convergence (RoC) is defined as the set of points in the complex plane $\mathbb{C}$ for which the $\mathcal Z$-Transform converges. Consider the impulse response $$ \mathcal{X}(z)={1\over 1+0.6z^{-1}} $$ For $\mathcal X(z)$ the region of convergence is given by $|z|>|0.6|$ # BIBO Stability # Estimating Frequency Response of a Digital System