What does z⁻¹ mean in the Z-transform?

For the unilateral Z-transform the general rule is Zu{x[n-1]} = x[-1] + z⁻¹ X(z). When the initial value x[-1] is zero this reduces to z⁻¹ X(z), so under zero initial conditions z⁻¹ acts as the factor corresponding to a one-sample delay. Being able to treat each delay term algebraically as multiplication by z⁻¹ is the most practically useful property of the Z-transform.

Definition of the Z-Transform

Q: What is the unilateral Z-transform?

The unilateral Z-transform converts the n ≥ 0 part of a discrete-time signal x[n] into a function of the complex variable z, defined as X(z) = Σ x[n] z^{-n} (n = 0 to ∞). Because it converts difference equations into algebraic equations, it is widely used in digital filter design and analysis and in solving difference equations with initial conditions.

Q: What are the transfer function and poles in the Z-transform?

The transfer function H(z) is the ratio Y(z)/X(z) of the Z-transforms of output and input, expressing the input-output behaviour under zero initial conditions as a rational function in z. The values of z that make the denominator of H(z) zero are called poles. For a causal rational LTI system, all poles lying inside the unit circle (|z| = 1) is the condition for BIBO stability.

Introduction

Overview

The Z-transform converts a discrete-time signal into a function of the complex variable $z$. It plays the same role for discrete-time signals that the Laplace transform plays for continuous-time signals: it converts difference equations into algebraic equations, making it fundamental to a wide range of fields including digital filter design, time-series analysis, and control theory.

This chapter starts from the unilateral Z-transform, which is convenient both for causal signals and for difference equations with initial conditions, and aims to develop familiarity through concrete examples. The region of convergence and the bilateral Z-transform are treated in the Basic Level.

Definition of the Unilateral Z-Transform

Definition: Unilateral Z-Transform

The unilateral Z-transform of a discrete-time signal $x[n]$ defined for $n \geq 0$ is

$$X(z) = \mathcal{Z}\{x[n]\} = \sum_{n=0}^{\infty} x[n] z^{-n}$$

where $z$ is a complex variable.

The unilateral Z-transform transforms only the part of a signal with $n \geq 0$. It therefore applies to causal signals ($x[n] = 0$ for $n < 0$) and, just as usefully, to difference equations with non-zero initial conditions, where the values $x[-1], x[-2], \ldots$ appear in the shift formula. It is the form most often met in practice, including impulse responses of digital filters and the solution of difference equations.

Although $z$ is a complex number, at the introductory level it is sufficient to understand that "$z^{-1}$ represents a one-sample delay." The polar form of $z$ and its relationship to the Fourier transform are covered in Basic Level, Chapter 1.

Basic Z-Transform Pairs

We derive the Z-transforms of basic signals from the definition of the unilateral Z-transform.

Unit Impulse

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 $z$.

Unit Step Function

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 $$= \dfrac{1}{1-z^{-1}} = \dfrac{z}{z-1}$$ when $|z^{-1}| < 1$, i.e., $|z| > 1$.

Exponential Sequence

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

Sinusoidal Sequence

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 $|e^{\pm j\omega_0}| = 1$).

Z-Transform Table (Basic)

Signal $x[n]$ ($n \geq 0$)	Z-Transform $X(z)$	Convergence
$\delta[n]$	$1$	all $z$
$u[n]$	$\dfrac{z}{z-1}$	$\|z\| > 1$
$a^n u[n]$	$\dfrac{z}{z-a}$	$\|z\| > \|a\|$
$n\, u[n]$	$\dfrac{z}{(z-1)^2}$	$\|z\| > 1$
$n\, a^n u[n]$	$\dfrac{az}{(z-a)^2}$	$\|z\| > \|a\|$
$\cos(\omega_0 n)\, u[n]$	$\dfrac{z(z - \cos\omega_0)}{z^2 - 2z\cos\omega_0 + 1}$	$\|z\| > 1$
$\sin(\omega_0 n)\, u[n]$	$\dfrac{z \sin\omega_0}{z^2 - 2z\cos\omega_0 + 1}$	$\|z\| > 1$

Application Example: Digital Filter

The most important application of the Z-transform is the algebraic analysis of digital filters described by difference equations.

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. For the unilateral Z-transform the general rule is $$\mathcal{Z}_u\{y[n-1]\} = y[-1] + z^{-1} Y(z)$$ (the proof is given in Chapter 2). Assuming zero initial conditions ($y[-1] = 0$) this reduces to $z^{-1} Y(z)$, so under zero initial conditions a one-sample delay corresponds to multiplication by $z^{-1}$. Applying this to the difference equation above yields $$Y(z) = 0.5\, z^{-1} Y(z) + X(z)$$ Solving for $Y(z)$, $$H(z) = \dfrac{Y(z)}{X(z)} = \dfrac{1}{1 - 0.5z^{-1}} = \dfrac{z}{z - 0.5}$$ 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 characterizes that behaviour under zero initial conditions.

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 ($|z| = 1$). For a causal rational LTI system, having all poles inside the unit circle is the condition for BIBO stability: every bounded input produces a bounded output.

Figure 5: Block diagram of $y[n] = 0.5\,y[n-1] + x[n]$. The output $y[n]$ is fed back through the delay element $z^{-1}$ and the coefficient $0.5$ to the summer.

Pole-zero plot of H(z)=z/(z-0.5) — Figure 6: Pole-zero plot of $H(z) = \dfrac{z}{z-0.5}$. The red x marks the pole ($z = 0.5$) and the green o marks the zero ($z = 0$). The system is stable because the pole lies inside the unit circle (blue dashed line).

In this way, the Z-transform allows us to read off properties of a difference equation (stability, frequency response, etc.) from the arrangement of poles and zeros in the $z$-plane.

The Region of Convergence (ROC) and the bilateral Z-transform are covered in detail in Basic Level, Chapter 2. The relationship with the Laplace transform is treated in Intermediate Level.

References

Oppenheim, A. V., & Schafer, R. W. (2009). Discrete-Time Signal Processing (3rd ed.). Pearson.
Proakis, J. G., & Manolakis, D. G. (2006). Digital Signal Processing (4th ed.). Pearson.
Z-transform — Wikipedia

Frequently Asked Questions

Q1. What is the unilateral Z-transform?

The unilateral Z-transform converts the $n \geq 0$ part of a discrete-time signal $x[n]$ into a function of the complex variable $z$, defined as $X(z) = \sum_{n=0}^{\infty} x[n] z^{-n}$. Because it turns difference equations into algebraic equations, it is widely used in digital filter design and analysis, and in solving difference equations with initial conditions.

Q2. What are the transfer function and poles in the Z-transform?

The transfer function $H(z)$ is the ratio $Y(z)/X(z)$ of the Z-transforms of output and input, expressing the input-output behaviour under zero initial conditions as a rational function in $z$. The values of $z$ that make the denominator of $H(z)$ zero are called poles. For a causal rational LTI system, all poles lying inside the unit circle ($|z| = 1$) is the condition for BIBO stability.

Q3. What does $z^{-1}$ mean in the Z-transform?

For the unilateral Z-transform the general rule is $\mathcal{Z}_u\{x[n-1]\} = x[-1] + z^{-1} X(z)$. When the initial value $x[-1]$ is zero this reduces to $z^{-1} X(z)$, so under zero initial conditions $z^{-1}$ acts as the factor corresponding to a one-sample delay. Being able to treat each delay term algebraically as multiplication by $z^{-1}$ is the most practically useful property of the Z-transform.