We previously demonstrated that we can express any linear, time-invariant system with convolution or as difference equations. Yet, these representations can be challenging to work with, both theoretically and computationally. To address the theoretical complexity, we can transform signals into new representations that are easier to manipulate and interpret. The first transform we consider is the Z-transform.
The Z-transform excels at improving our understanding of the transient properties of signals. We will start to explore the Z-transform from a fairly theoretical perspective. As we go through the first half of the course, we will expand our conceptual understanding.
The bilateral Z-Transform is defined as $$ X(z) = \sum_{n=-\infty}^{\infty} x[n] z^{-n} \; , $$ where $z$ is a complex number, $z = \sigma e^{j \omega}$. The inverse bilateral Z-Transform is defined as $$ x[n] = \frac{1}{2 \pi j} \oint_{\mathcal{C}} X(z) z^{n-1} dz \; , $$ where $\mathcal{C}$ is a closed contour encircling the origin and lying entirely within the region of convergence (see below). The unilateral Z-Transform, used for causal signals ($x[n] = 0$ when $n < 0$), is defined as $$ X(z) = \sum_{n=0}^{\infty} x[n] z^{-n} \; . $$
The unilateral Z-Transform is defined for all causal signals ($x[n] = 0$ when $n < 0$) $$ X(z) = \sum_{n=0}^{\infty} x[n] z^{-n} \; , $$ where $z$ is a complex number $z = \sigma e^{j \omega}$.
Note that $z^{-n} = \sigma^{-n} e^{-j \omega n}$. Thus, $z^{-n}$ represents a decaying complex exponential. The magnitude $\sigma$ controls the decay or growth rate, distinguishing the Z-transform from the DTFT (which uses $|z| = 1$).
A general linear, time-invariant system can be expressed as: $$ \begin{align*} a[n] * y[n] &= b[n] * x[n] \\ \sum_{m=-\infty}^{\infty} a[m] y[n-m] &= \sum_{m=-\infty}^{\infty} b[m] x[n-m] \; . \end{align*} $$ In the Z-domain: $$ \begin{align*} A(z) Y(z) &= B(z) X(z) \\ Y(z) \sum_{m=-\infty}^{\infty} a[m] z^{-m} &= X(z) \sum_{m=-\infty}^{\infty} b[m] z^{-m} \; . \end{align*} $$ This leads to the transfer function: $$ \begin{eqnarray} H(z) &=& \frac{Y(z)}{X(z)} = \frac{B(z)}{A(z)} \\ &=& \frac{\sum_{m=-\infty}^{\infty} b[m] z^{-m}}{\sum_{m=-\infty}^{\infty} a[m] z^{-m}} \; . \end{eqnarray} $$ For a causal system with $M+1$ feedforward terms and $N+1$ feedback terms: $$ \begin{eqnarray} H(z) &=& \frac{\sum_{m=0}^{M} b[m] z^{-m}}{\sum_{m=0}^{N} a[m] z^{-m}} \\ &=& z^{-M+N} \frac{b_0}{a_0} \frac{\prod_{m=1}^{M} \left( z - z_{m} \right) }{\prod_{m=1}^{N} \left( z - p_{m} \right)} \\ &=& \frac{b_0}{a_0} \frac{\prod_{m=1}^{M} \left( 1 - z_{m} z^{-1} \right) }{\prod_{m=1}^{N} \left( 1 - p_{m} z^{-1} \right)} \; , \end{eqnarray} $$ where $a_0 = a[0]$ and $b_0 = b[0]$.
The ratio of $Y(z)$ to $X(z)$ is known at the transfer function of the system.
From the form $$ H(z) = \frac{b_0}{a_0} \frac{\prod_{m=1}^{M} \left( 1 - z_{m} z^{-1} \right) }{\prod_{m=1}^{N} \left( 1 - p_{m} z^{-1} \right)} \; , $$ the values $z_{m}$ are the **zeros** of $H(z)$—the values of $z$ such that $H(z) = 0$. The values $p_{m}$ are the **poles**—values of $z$ such that $H(z) = \infty$. Here, $M$ is the order of the numerator polynomial, and $N$ is the order of the denominator polynomial. Poles and zeros are typically plotted in the complex plane (real vs. imaginary axes).
The $z_{m}$ values are the zeros of $H(z)$ (i.e., the values of $z$ such that $H(z) = 0$). The $p_{m}$ values are the poles of $H(z)$ (i.e., the values of $z$ such that $H(z) = \infty$).
The figures below illustrate how we plot poles and zeros in the complex (real-imaginary) plane.
$$ \begin{align} \hline & x[n] &\qquad & X(z) &\qquad & \textrm{Region of Convergence} \\ \hline & \displaystyle a^n u[n] &\qquad & \displaystyle \frac{1}{1 - a z^{-1}} &\qquad & |z| > |a| \\ & \displaystyle - a^n u[- n - 1] &\qquad & \displaystyle \frac{1}{1 - a z^{-1}} &\qquad & |z| < |a| \\ & \displaystyle n a^n u[n] &\qquad & \displaystyle \frac{a z^{-1}}{(1-a z^{-1})^2}&\qquad & |z| > |a| \\ & \displaystyle - n a^n u[-n - 1] &\qquad & \displaystyle \frac{a z^{-1}}{(1-a z^{-1})^2}&\qquad & |z| < |a| \\ & \displaystyle \delta[n] &\qquad & \displaystyle 1 &\qquad & \textrm{All } z \\ & \displaystyle \delta[n - n_0] &\qquad & \displaystyle z^{-n_0} &\qquad & \textrm{All } z \\ & \displaystyle u[n] &\qquad & \displaystyle \frac{1}{1 - z^{-1}}&\qquad & |z| > 1 \\ & \displaystyle \cos(\omega_0 n) u[n] &\qquad & \displaystyle \frac{ 1 - z^{-1} \cos(\omega_0) }{1 - 2 z^{-1}\cos(\omega_0) + z^{-2} } &\qquad & |z| > 1 \\ & \displaystyle \sin(\omega_0 n) u[n] &\qquad & \displaystyle \frac{ z^{-1} \sin(\omega_0) }{1 - 2 z^{-1}\cos(\omega_0) + z^{-2} } &\qquad & |z| > 1 \\ & \displaystyle a^n \cos(\omega_0 n) u[n] &\qquad & \displaystyle \frac{ 1 - a z^{-1} \cos(\omega_0) }{1 - 2 a z^{-1}\cos(\omega_0) + a^2 z^{-2} } &\qquad & |z| > |a| \\ & \displaystyle a^n \sin(\omega_0 n) u[n] &\qquad & \displaystyle \frac{ a z^{-1} \sin(\omega_0) }{1 - 2 a z^{-1}\cos(\omega_0) + a^2 z^{-2} } &\qquad & |z| > |a| \\ \hline \end{align} $$
For each property, assume: $$x[n] \stackrel{Z}{\longleftrightarrow} X(z) \quad \textrm{and} \quad y[n] \stackrel{Z}{\longleftrightarrow} Y(z) $$ $$ \begin{align} \hline & \textbf{Property} &\qquad & \textbf{Time domain} &\qquad & \textbf{Z-domain} &\qquad \\ \hline & \textrm{Linearity} &\qquad & A x[n] + B y[n] &\qquad & A X(z) + B Y(z) &\qquad \\ & \textrm{Time Shifting} &\qquad & x[n-n_0] &\qquad & X(z)z^{-n_0} &\qquad \\ & \textrm{Z-scaling} &\qquad & a^n x[n] &\qquad & X(a^{-1} z) &\qquad \\ & \textrm{Conjugation} &\qquad & x^*[n] &\qquad & X^*(z^*) &\qquad \\ & \textrm{Time Reversal} &\qquad & x[-n] &\qquad & X(z^{-1}) &\qquad \\ & \textrm{Convolution} &\qquad & x[n] * y[n] &\qquad & X(z) Y(z) &\qquad \\ & \textrm{Differentiation in z-domain} &\qquad & n x[n] &\qquad & -z \frac{d X(z)}{d z} &\qquad \\ & \textrm{Initial Value Theorem} &\qquad & x[n] \textrm{ is causal} &\qquad & x(0) = \lim_{z \rightarrow \infty} X(z) &\qquad \\ & \textrm{Final Value Theorem} &\qquad & x[n] \textrm{ is causal and stable} &\qquad & x(\infty) = \lim_{z \rightarrow 1} [z-1] X(z) &\qquad \\ \end{align} $$