Prerequisite:

The Discrete-Time Analog of the Laplace Transform

The Laplace transform converts a continuous-time signal $x(t)$ into a function of a complex variable $s$, turning differential equations into algebraic ones. The Z-transform does the same for discrete-time sequences $x[n]$, turning difference equations into algebraic ones in the complex variable $z$.

It is the natural tool for analyzing digital filters, discrete control systems, and recurrent computations. The connection to the Fourier transform and the Laplace transform is exact and illuminates when and why each tool applies.

Definition and Region of Convergence

Definition. The (bilateral) Z-transform of a sequence $x[n]$ is

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

where $z \in \mathbb{C}$. The region of convergence (ROC) is the set of $z$ for which this sum converges absolutely:

$$\text{ROC} = \left{z \in \mathbb{C} : \sum_{n=-\infty}^\infty |x[n]|,|z|^{-n} < \infty\right}$$

The ROC is always an annular region $r_1 < |z| < r_2$ in the complex plane (possibly degenerate: a ring, the exterior of a disk, the interior of a disk, or the entire plane).

ROC in the complex plane:

    Im
     |
  r2 |....  (outer boundary)
     |    ..        ..
     | ..      ROC      ..
     |.                   .
-----+---------------------+--- Re
     |.                   .
     | ..              ..
     |    ..        ..
  r1 |....  (inner boundary)
     |
     
ROC: r1 < |z| < r2

The specification of $X(z)$ is incomplete without the ROC. Two different sequences can have the same algebraic expression for $X(z)$ but different ROCs, leading to different inverse transforms.

Common Transform Pairs

Unit impulse: $x[n] = \delta[n]$ (equals 1 at $n=0$, zero elsewhere).

$$X(z) = 1, \quad \text{ROC: all } z$$

Unit step: $x[n] = u[n]$ (equals 1 for $n \geq 0$, zero for $n < 0$).

$$X(z) = \frac{1}{1-z^{-1}} = \frac{z}{z-1}, \quad \text{ROC: } |z| > 1$$

Causal geometric sequence: $x[n] = a^n u[n]$.

$$X(z) = \frac{1}{1-az^{-1}} = \frac{z}{z-a}, \quad \text{ROC: } |z| > |a|$$

Anti-causal geometric sequence: $x[n] = -a^n u[-n-1]$.

$$X(z) = \frac{1}{1-az^{-1}}, \quad \text{ROC: } |z| < |a|$$

Note that the causal and anti-causal sequences share the same algebraic form but differ in their ROC. This is why the ROC is not optional information.

Properties of the Z-Transform

Linearity

$$\mathcal{Z}{\alpha x[n] + \beta y[n]} = \alpha X(z) + \beta Y(z)$$

with ROC containing $\text{ROC}_X \cap \text{ROC}_Y$.

Time Shift

$$\mathcal{Z}{x[n-k]} = z^{-k} X(z)$$

This is the discrete analog of the Laplace shift theorem. The factor $z^{-1}$ represents a unit delay: multiplying by $z^{-k}$ in the $z$-domain corresponds to delaying the sequence by $k$ samples.

Convolution

$$\mathcal{Z}{(x * y)[n]} = X(z)\cdot Y(z)$$

where $(x*y)[n] = \sum_{k=-\infty}^\infty x[k],y[n-k]$. Convolution in time becomes multiplication in the $z$-domain, exactly as for the Laplace and Fourier transforms.

Scaling in the $z$-Domain

$$\mathcal{Z}{a^n x[n]} = X(z/a), \quad \text{ROC: } |a|,r_1 < |z| < |a|,r_2$$

Initial Value Theorem

If $x[n] = 0$ for $n < 0$ (causal sequence), then

$$x[0] = \lim_{z \to \infty} X(z)$$

Final Value Theorem

If $x[n]$ is causal and has a finite limit as $n \to \infty$, then

$$\lim_{n \to \infty} x[n] = \lim_{z \to 1}(z-1)X(z)$$

Inverse Z-Transform

There are three standard methods.

Partial Fraction Decomposition

When $X(z)$ is a rational function of $z$, express it as a sum of terms of the form $\frac{z}{z-a_k}$ (or $\frac{1}{1-a_k z^{-1}}$), then read off each inverse transform from the table of pairs, being careful to choose the causal or anti-causal form based on the ROC.

Long Division

Dividing the numerator by the denominator as a power series in $z^{-1}$ directly yields the coefficients $x[0], x[1], x[2], \ldots$. Useful when only the first few terms are needed.

Contour Integration

The formal inversion formula is

$$x[n] = \frac{1}{2\pi i}\oint_C X(z),z^{n-1},dz$$

where $C$ is a counterclockwise contour inside the ROC. By the residue theorem, $x[n] = \sum_k \text{Res}[X(z)z^{n-1},, z=z_k]$.

System Transfer Function and Filter Design

For a linear time-invariant (LTI) system with input $x[n]$, output $y[n]$, and impulse response $h[n]$, we have $y[n] = (h * x)[n]$, so in the $z$-domain:

$$Y(z) = H(z),X(z)$$

The transfer function $H(z) = Y(z)/X(z)$ completely characterizes the system.

FIR filters (Finite Impulse Response) have $H(z) = \sum_{k=0}^{M} b_k z^{-k}$, a polynomial in $z^{-1}$. The impulse response has finite duration. These are always stable and straightforward to design.

IIR filters (Infinite Impulse Response) have $H(z) = \frac{B(z)}{A(z)}$, a rational function. The denominator introduces poles, giving the system memory beyond any finite window. IIR filters can achieve steeper roll-off with fewer coefficients than FIR, but stability must be checked.

Poles, Zeros, and Stability

Write $H(z) = \frac{\prod_k (z - z_k)}{\prod_j (z - p_j)}$ where $z_k$ are the zeros and $p_j$ are the poles.

Pole-zero plot (example: Butterworth LP filter):

    Im
  1 |    x (pole)
    |         o (zero)
  0 +----+----+--- Re
    |   -1    1
 -1 |    x (pole)

Unit circle shown as dashed.
Poles inside unit circle => stable.

Theorem (BIBO Stability for Causal Systems). A causal LTI system is bounded-input bounded-output (BIBO) stable if and only if the ROC of $H(z)$ includes the unit circle $|z| = 1$, which for a causal system requires all poles to lie strictly inside the unit circle: $|p_j| < 1$ for all $j$.

Relationship to Laplace Transform and DTFT

Laplace connection. If a continuous-time signal $x_c(t)$ is sampled at rate $1/T$ to produce $x[n] = x_c(nT)$, then the Z-transform evaluated at $z = e^{sT}$ equals the Laplace transform of the sampled signal. The mapping $z = e^{sT}$ sends the imaginary axis of the $s$-plane (stability boundary for continuous systems) to the unit circle in the $z$-plane (stability boundary for discrete systems).

The bilinear transformation $s = \frac{2}{T}\frac{z-1}{z+1}$ is a widely-used method to convert an analog (Laplace-domain) filter design to a digital (Z-domain) one, mapping the left half-plane to the interior of the unit circle.

DTFT connection. The Discrete-Time Fourier Transform (DTFT) of $x[n]$ is

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

This is exactly the Z-transform evaluated on the unit circle: $X(z)\big|_{z=e^{i\omega}}$. The DTFT exists (as a function) if and only if the unit circle lies inside the ROC of the Z-transform.

Solving Linear Recurrences

Consider the second-order linear recurrence (difference equation):

$$y[n] - a_1 y[n-1] - a_2 y[n-2] = x[n]$$

Taking the Z-transform with the time-shift property:

$$Y(z) - a_1 z^{-1}Y(z) - a_2 z^{-2}Y(z) = X(z)$$

$$H(z) = \frac{Y(z)}{X(z)} = \frac{1}{1 - a_1 z^{-1} - a_2 z^{-2}} = \frac{z^2}{z^2 - a_1 z - a_2}$$

The poles are the roots of $z^2 - a_1 z - a_2 = 0$. Stability requires both roots inside the unit circle. Applying partial fractions and inverting gives the closed-form solution $y[n]$ as a function of the initial conditions and input.

Examples

Audio Effects. Reverb, equalization, and dynamic range compression are implemented as IIR or FIR filters. The biquad filter (a second-order IIR section) is the workhorse of audio DSP: stable, efficient, and composable.

IIR Filter Design. Standard designs (Butterworth, Chebyshev, elliptic) begin in the analog domain (Laplace), then are converted to digital via the bilinear transformation. The pole locations are engineered to achieve a target frequency response shape.

Recurrent Neural Networks. An RNN hidden state update $h_t = \sigma(W_h h_{t-1} + W_x x_t)$ is a nonlinear difference equation. For a linear RNN, the Z-transform gives the transfer function immediately; the eigenvalues of $W_h$ play the role of poles. Vanishing and exploding gradients correspond directly to poles inside or outside the unit circle - a stability analysis in disguise. This is why LSTM forget gates and gated state-space models (like S4, Mamba) are engineered to keep effective poles near the unit circle.

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