Definition of the Z-Transform

The Z-transform of the unit impulse $\delta[n] = \begin{cases} 1 & n = 0 \\ 0 & n \neq 0 \end{cases}$ is $$\mathcal{Z}\{\delta[n]\} = \sum_{n=0}^{\infty} \delta[n] z^{-n} = z^0 = 1$$ which converges for all $z$.

The Z-transform of the unit step $u[n] = \begin{cases} 1 & n \geq 0 \\ 0 & n < 0 \end{cases}$ is $$\mathcal{Z}\{u[n]\} = \sum_{n=0}^{\infty} 1 \cdot z^{-n} = 1 + z^{-1} + z^{-2} + \cdots$$ This is a geometric series with first term $1$ and common ratio $z^{-1}$, converging to $$= \frac{1}{1-z^{-1}} = \frac{z}{z-1}$$ when $|z^{-1}| < 1$, i.e., $|z| > 1$.

The Z-transform of the exponential sequence $x[n] = a^n u[n]$ (where $a$ is a constant) is $$\mathcal{Z}\{a^n u[n]\} = \sum_{n=0}^{\infty} a^n z^{-n} = 1 + \frac{a}{z} + \left(\frac{a}{z}\right)^2 + \cdots$$ This is a geometric series with first term $1$ and common ratio $a/z$, converging to $$= \frac{1}{1-az^{-1}} = \frac{z}{z-a}$$ when $|a/z| < 1$, i.e., $|z| > |a|$. The unit step is the special case where $a = 1$.

We derive the Z-transform of $x[n] = \sin(\omega_0 n)\, u[n]$. By Euler's formula, $$\sin(\omega_0 n) = \dfrac{e^{j\omega_0 n} - e^{-j\omega_0 n}}{2j}$$ Applying the linearity of the Z-transform and the exponential sequence result $\mathcal{Z}\{a^n u[n]\} = \dfrac{z}{z-a}$ with $a = e^{j\omega_0}$ and $a = e^{-j\omega_0}$, $$\mathcal{Z}\{\sin(\omega_0 n)\, u[n]\} = \dfrac{1}{2j}\left(\dfrac{z}{z - e^{j\omega_0}} - \dfrac{z}{z - e^{-j\omega_0}}\right)$$ Combining the fractions, the denominator is $$(z - e^{j\omega_0})(z - e^{-j\omega_0}) = z^2 - (e^{j\omega_0} + e^{-j\omega_0})z + 1 = z^2 - 2z\cos\omega_0 + 1$$ and the numerator is $$z(z - e^{-j\omega_0}) - z(z - e^{j\omega_0}) = z(e^{j\omega_0} - e^{-j\omega_0}) = 2jz\sin\omega_0$$ Therefore, $$\mathcal{Z}\{\sin(\omega_0 n)\, u[n]\} = \dfrac{1}{2j} \cdot \dfrac{2jz\sin\omega_0}{z^2 - 2z\cos\omega_0 + 1} = \dfrac{z\sin\omega_0}{z^2 - 2z\cos\omega_0 + 1}$$ converging for $|z| > 1$ (since $|e^{\pm j\omega_0}| = 1$).

Example: First-Order IIR Filter

Suppose the relationship between input $x[n]$ and output $y[n]$ is given by $$y[n] = 0.5\, y[n-1] + x[n]$$ To apply the Z-transform to this difference equation, we use the time-shift property. Assuming zero initial conditions ($y[-1] = 0$), the unilateral Z-transform gives $$\mathcal{Z}\{x[n-1]\} = z^{-1} X(z)$$ In other words, a one-sample delay corresponds to multiplication by $z^{-1}$ (the proof is given in Chapter 2). Applying this to the difference equation above yields $$Y(z) = 0.5\, z^{-1} Y(z) + X(z)$$ Solving for $Y(z)$, $$H(z) = \frac{Y(z)}{X(z)} = \frac{1}{1 - 0.5z^{-1}} = \frac{z}{z - 0.5}$$ This $H(z)$ (the ratio of the Z-transforms of output and input) is called the transfer function. The transfer function expresses the input-output relationship of a system as a rational function in $z$ and completely characterizes the system.

The values of $z$ that make the denominator of $H(z)$ zero are called poles. In this example, the pole is at $z = 0.5$, which lies inside the unit circle ($|z| = 1$). A discrete-time system is stable (its output does not diverge) when all poles lie inside the unit circle.