Z-Transform

2004. Dolby engineers are designing Dolby Digital Plus for HD DVD. The spec: an IIR filter with -80 dB/oct roll-off and at most 2 ms delay. Solved in a single day, on paper. Sketch a pole-zero diagram, nudge a few points around the Z-plane. Poles inside the unit circle - stable. Zeros parked at the target frequencies - suppression locked in. The Z-transform turns filter design from equations into geometry.

• **Audio:** equalizer = a set of IIR filters, each designed by placing poles/zeros in the Z-plane
• **Telecom:** channel equalizers in 5G/WiFi use Z-analysis to compensate for channel distortion
• **Control:** digital PID controllers are designed via the transfer function H(z) with guaranteed stability

Ragazzini, Zadeh, and the birth of digital control theory

Ragazzini and Zadeh introduced the Z-transform in 1952 to analyze digital control systems - the discrete sibling of the Laplace transform. Formally it is a Laurent series, 19th-century complex analysis repurposed for engineering. Zadeh would later invent fuzzy logic, but the Z-transform stands as his defining contribution to DSP.

Предварительные знания

• Sampling and the Nyquist-Shannon Theorem

Z-Transform Definition

**The Z-transform** generalizes the DTFT: $X(z) = \sum_{n=-\infty}^{\infty} x[n] \cdot z^{-n}$, where z is a complex variable. The Fourier transform lives on a single circle ($z = e^{j\omega}$). The Z-transform roams the whole complex plane.

Why bother? Growing signals ($a^n$ with $|a|>1$) blow up the Fourier transform - it simply does not exist. The Z-transform extends the domain so unstable systems and transient responses become tractable: the analytic continuation of Fourier off the unit circle.

**Delay property, the workhorse:** $\mathcal{Z}\{x[n-k]\} = z^{-k} \cdot X(z)$. Delay by k samples = multiply by $z^{-k}$. Hence the nickname for $z^{-1}$: the "delay operator". Every digital filter on the planet is wired from delays, multipliers, and adders.

Region of Convergence (ROC)

**ROC (Region of Convergence)** - the set of z values where the series $X(z) = \sum x[n] \cdot z^{-n}$ actually converges. ROC is non-negotiable: the same X(z) formula maps to DIFFERENT signals depending on which region is chosen.

Take $X(z) = z/(z-a)$. ROC $|z| > |a|$ produces a right-sided $a^n \cdot u[n]$ (causal). ROC $|z| < |a|$ produces a left-sided $-a^n \cdot u[-n-1]$ (anti-causal). Identical formula, opposite signals.

**Stability rule:** an LTI system is BIBO stable iff the ROC of H(z) contains the **unit circle** |z| = 1. On that circle the Z-transform collapses back to the DTFT - the frequency response.

ROC always takes the shape of an annulus (or a disk, or the exterior of a circle) centered at the origin. Its boundaries trace the poles of X(z) - points where X(z) blows up to infinity.

Inverse Z-Transform

**The inverse Z-transform** recovers x[n] from X(z). Method of choice: **partial fraction decomposition**. Break a rational X(z) into a sum of simple fractions, then read each piece off the lookup table.

Steps: 1) Write X(z)/z. 2) Decompose into fractions $A/(z-p_1) + B/(z-p_2) + ...$. 3) Multiply by z: $X(z) = Az/(z-p_1) + Bz/(z-p_2) + ...$. 4) Each $Az/(z-p) \to A \cdot p^n \cdot u[n]$ (if causal, $|z|>|p|$).

**Formal inverse:** $x[n] = \frac{1}{2\pi j} \oint X(z) \cdot z^{n-1} \, dz$. Nobody actually evaluates the contour integral in practice - partial fractions or residues do the job.

Pin down the ROC and the inverse is unique. Without it, causal and anti-causal interpretations both fit. Physically realizable (causal) systems pick the exterior of the outermost pole.

Transfer Function

**Transfer function** $H(z) = Y(z)/X(z)$ - the Z-transform of the impulse response h[n]. It captures an LTI system completely. Convolution $y[n] = x[n] * h[n]$ collapses into multiplication: $Y(z) = X(z) \cdot H(z)$. A whole digital equalizer is just an H(z) with poles and zeros placed in the right spots.

H(z) is a rational function: $H(z) = B(z)/A(z)$, with B(z) supplying the zeros and A(z) supplying the poles. **Poles** decide stability; **zeros** sculpt the shape of the frequency response.

**Poles inside the unit circle = stability.** Discrete-domain echo of "poles in the left half-plane" from continuous systems. The map $z = e^{sT}$ sends the left half-plane ($\text{Re}(s)<0$) straight inside the unit circle ($|z|<1$).

Back to lesson-one convolution: in Z-domain it becomes multiplication. Cascade filters: $H(z) = H_1(z) \cdot H_2(z)$. Parallel branches: $H(z) = H_1(z) + H_2(z)$. The full architecture of DSP systems is engineered through pole-zero placement.

Substituting $z = e^{j\omega}$ traces a walk along the unit circle and gives the frequency response $H(e^{j\omega})$. Get close to a pole and the frequency spikes (peak); brush a zero and it drops out (notch). Filter design boils down to placing poles and zeros with intent.

Key Ideas

• **X(z) = sum x[n]*z^(-n)** - generalization of DTFT to the entire complex plane
• **ROC** determines which signal corresponds to a given X(z): one formula - different signals
• **Partial fractions** - the primary method for inverse Z: decompose, look up in table
• **H(z) = Y(z)/X(z):** poles inside the unit circle = stability. Convolution = multiplication in Z-domain

Related Topics

The Z-transform is the analytical tool of DSP:

• Signals and Systems — The Z-transform formalizes convolution and LTI systems in algebraic form
• Sampling and the Nyquist-Shannon Theorem — Connection z = e^(sT): sampling in the Z-domain = periodic mapping of the s-plane

Вопросы для размышления

• The Z-transform is analogous to the Laplace transform (z = e^(sT)). Why is stability in the s-domain Re(s) < 0, but in the z-domain |z| < 1?
• An FIR filter (all poles at z=0) is always stable. So why use IIR (with non-zero poles)? What is the advantage?
• Can the magnitude response |H(e^jw)| be roughly sketched from the pole-zero diagram of H(z)? How does the distance to poles/zeros affect the amplitude?

Связанные уроки

• dsp-01 — LTI systems and convolution - what the Z-transform turns into multiplication
• dsp-02 — Connection z = e^(sT): sampling creates the Z-plane from the s-plane
• dsp-04 — Filter design - placing poles and zeros in the Z-plane
• dsp-05 — IIR vs FIR: poles inside unit circle = stable IIR
• nm-05 — Laplace transform - the continuous-time analog of the Z-transform
• it-03 — Complex analysis: Laurent series - the mathematical basis of ROC
• ca-13