6. Z-Transforms: The Discrete-Time Cousin

The Z-transform is to discrete-time what Laplace is to continuous-time. Same job, different domain.

6.1 Definition

$X(z) = \sum_{n=-\infty}^{\infty} x[n]\, z^{-n}$

The variable $z$ is a complex number. You can think of $z$ as the discrete-time analog of the Laplace $s$, with the relationship $z = e^{sT_s}$ that comes from sampling a continuous-time signal: a Laplace exponential $e^{st}$ evaluated at $t = nT_s$ gives $e^{snT_s} = (e^{sT_s})^n = z^n$.

That mapping has a beautiful consequence: the imaginary axis in the s-plane (where $s = j\omega$, the home of all sinusoids) maps to the unit circle in the z-plane, $z = e^{j\omega T_s}$. The frequency response of a discrete-time system is what you get by evaluating $H(z)$ on the unit circle. The left half of the s-plane (stable region) maps to the inside of the unit circle (stable region for discrete-time).

6.2 Standard Z-transforms

Two different time signals (the third and fourth rows) have the same algebraic form but different regions of convergence. So the Z-transform is incomplete without specifying the ROC.

6.3 Properties

Just like Laplace and Fourier:

• Linearity (sums and scalings).
• Time shift: $x[n - k] \leftrightarrow z^{-k} X(z)$. Each one-sample delay multiplies the Z-transform by $z^{-1}$. So $z^{-1}$ is the algebraic embodiment of "delay by one sample."
• Convolution becomes multiplication: $x_1[n] * x_2[n] \leftrightarrow X_1(z)\, X_2(z)$.
• Differentiation in $z$: $n x[n] \leftrightarrow -z\, dX(z)/dz$.

6.4 Stability in the z-plane

For a discrete-time LTI system described by transfer function $H(z)$:

• All poles inside the unit circle ($|z| < 1$): stable.
• Poles on the unit circle: marginally stable.
• Any pole outside the unit circle: unstable.

This is the discrete-time analog of "left half plane equals stable" for continuous-time, derived directly from the $z = e^{sT_s}$ mapping.

6.5 Why Z-transform matters

When you implement a digital filter in software or in an FPGA, the difference equation looks like:

$y[n] = b_0 x[n] + b_1 x[n-1] + \cdots - a_1 y[n-1] - a_2 y[n-2] - \cdots$

Take the Z-transform of both sides (each $n-k$ shift becomes $z^{-k}$), get an algebraic equation, and solve for $H(z) = Y(z)/X(z)$:

$H(z) = \frac{b_0 + b_1 z^{-1} + b_2 z^{-2} + \cdots}{1 + a_1 z^{-1} + a_2 z^{-2} + \cdots}$

Now $H(z)$ tells you everything: stability (pole locations), frequency response (evaluate $H$ on the unit circle, $z = e^{j\omega}$), transient response. We will use the Z-transform extensively in Chapter 17 (DSP) when we design FIR and IIR digital filters.