Chapter 2: Basic Properties of the Z-Transform

Basic Properties of the Z-Transform

Introduction

Overview

The Z-transform possesses many useful properties. Among them, linearity and time shifting are the most fundamental and indispensable for algebraically solving difference equations via the Z-transform. Combined with the Z-transform of exponential sequences introduced in Chapter 1, these properties enable us to efficiently compute the Z-transforms of a wide variety of signals.

In this chapter, we present the theorems, proofs, and application examples for these two properties.

1. Linearity

Theorem 2.1 (Linearity)

For arbitrary constants $a, b$ and signals $x_1[n], x_2[n]$,

$$\mathcal{Z}\{a x_1[n] + b x_2[n]\} = a X_1(z) + b X_2(z)$$

where $X_1(z) = \mathcal{Z}\{x_1[n]\}$ and $X_2(z) = \mathcal{Z}\{x_2[n]\}$.

This holds provided the Z-transforms of both signals are defined for the same values of $z$ (i.e., their regions of convergence overlap).

Proof

By the definition of the Z-transform,

$$\begin{align} \mathcal{Z}\{a x_1[n] + b x_2[n]\} &= \sum_{n=-\infty}^{\infty} (a x_1[n] + b x_2[n]) z^{-n} \\ &= a \sum_{n=-\infty}^{\infty} x_1[n] z^{-n} + b \sum_{n=-\infty}^{\infty} x_2[n] z^{-n} \\ &= a X_1(z) + b X_2(z) \end{align}$$

This equality holds for values of $z$ where both series converge. The details of the region of convergence (ROC) are covered in Basic Level, Chapter 2.

Example 2.1

Find the Z-transform of $x[n] = 2 \cdot 3^n u[n] - 5 \cdot 2^n u[n]$.

Solution:

As we learned in Chapter 1, $\mathcal{Z}\{a^n u[n]\} = \dfrac{z}{z-a}$ ($|z| > |a|$), so by linearity,

$$\begin{align} X(z) &= 2 \cdot \mathcal{Z}\{3^n u[n]\} - 5 \cdot \mathcal{Z}\{2^n u[n]\} \\ &= 2 \cdot \frac{z}{z-3} - 5 \cdot \frac{z}{z-2} \\ &= \frac{2z(z-2) - 5z(z-3)}{(z-3)(z-2)} \\ &= \frac{-3z^2 + 11z}{(z-3)(z-2)} \end{align}$$

This expression is valid for $|z| > 3$ (the range where both series converge).

Example 2.2

Find the Z-transform of $x[n] = 3 \cdot 2^n u[n] - 5 u[n]$.

Solution:

Using $\mathcal{Z}\{u[n]\} = \dfrac{z}{z-1}$ ($|z| > 1$),

$$X(z) = 3 \cdot \frac{z}{z-2} - 5 \cdot \frac{z}{z-1} = \frac{-2z^2 + 7z}{(z-2)(z-1)}, \quad |z| > 2$$

2. Time Shifting

Theorem 2.2 (Time Shifting)

For an integer $k$,

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

In other words, delaying a signal by $k$ samples multiplies its Z-transform by $z^{-k}$.

Proof

Substituting $m = n - k$,

$$\begin{align} \mathcal{Z}\{x[n-k]\} &= \sum_{n=-\infty}^{\infty} x[n-k] z^{-n} \\ &= \sum_{m=-\infty}^{\infty} x[m] z^{-(m+k)} \\ &= z^{-k} \sum_{m=-\infty}^{\infty} x[m] z^{-m} \\ &= z^{-k} X(z) \end{align}$$

Note

$k > 0$: delay by $k$ samples ($z^{-k}$ factor). This is the most common case in digital filter design.

$k < 0$: advance by $|k|$ samples ($z^{|k|}$ factor). This requires future values and is non-causal.

x[n] (original signal) n 0 1 2 3 Delay k=2 × z⁻² x[n-2] (delayed by 2 samples) n 0 1 2 3 4 A delay of k samples in the time domain corresponds to multiplication by z^{-k} in the Z-domain. This is the basis for converting difference equations into algebraic equations via the Z-transform.

Figure 1: Time shifting — delaying x[n] by k=2 samples gives x[n−2]

Example 2.3 (Delay System)

Find the Z-transform of $x[n] = \delta[n] + 2\delta[n-1] + 3\delta[n-2]$.

Solution:

Since $\mathcal{Z}\{\delta[n]\} = 1$, by the time shifting property and linearity,

$$\begin{align} X(z) &= \mathcal{Z}\{\delta[n]\} + 2\mathcal{Z}\{\delta[n-1]\} + 3\mathcal{Z}\{\delta[n-2]\} \\ &= 1 + 2z^{-1} + 3z^{-2} \\ &= \frac{z^2 + 2z + 3}{z^2} \end{align}$$

This expression is valid for all $z \neq 0$ (finite-length signals converge everywhere except possibly at $z = 0$).

Practice Problems

Problem 1 (Linearity)

Find the Z-transform of each signal.

(a) $x[n] = 3^n u[n] + 2^n u[n]$

(b) $x[n] = 4 \cdot (0.5)^n u[n] - 3 \cdot (0.8)^n u[n]$

Solution

(a) By linearity,

$$X(z) = \frac{z}{z-3} + \frac{z}{z-2} = \frac{z(z-2) + z(z-3)}{(z-3)(z-2)} = \frac{2z^2 - 5z}{(z-3)(z-2)}, \quad |z| > 3$$

(b) By linearity,

$$X(z) = 4 \cdot \frac{z}{z-0.5} - 3 \cdot \frac{z}{z-0.8} = \frac{4z(z-0.8) - 3z(z-0.5)}{(z-0.5)(z-0.8)} = \frac{z^2 - 1.7z}{(z-0.5)(z-0.8)}, \quad |z| > 0.8$$

Problem 2 (Time Shifting)

(a) Find $\mathcal{Z}\{u[n-1]\}$.

(b) Find $\mathcal{Z}\{u[n-3]\}$.

Solution

(a) Using $\mathcal{Z}\{u[n]\} = \dfrac{z}{z-1}$ ($|z| > 1$) and the time shifting property,

$$\mathcal{Z}\{u[n-1]\} = z^{-1} \cdot \frac{z}{z-1} = \frac{1}{z-1}, \quad |z| > 1$$

(b) Similarly,

$$\mathcal{Z}\{u[n-3]\} = z^{-3} \cdot \frac{z}{z-1} = \frac{z^{-2}}{z-1} = \frac{1}{z^2(z-1)}, \quad |z| > 1$$

Problem 3 (Linearity + Time Shifting)

Apply the Z-transform to both sides of the difference equation $y[n] - 0.5y[n-1] = x[n]$ (initial condition: $y[-1] = 0$) and find the transfer function $H(z) = Y(z)/X(z)$.

Solution

By the time shifting property, $\mathcal{Z}\{y[n-1]\} = z^{-1}Y(z)$, so

$$Y(z) - 0.5z^{-1}Y(z) = X(z)$$

Rearranging,

$$Y(z) = \frac{X(z)}{1 - 0.5z^{-1}} = \frac{zX(z)}{z - 0.5}$$

Therefore,

$$H(z) = \frac{Y(z)}{X(z)} = \frac{z}{z - 0.5}$$

Summary

  • Linearity: The Z-transform of a linear combination of signals equals the same linear combination of their individual Z-transforms
  • Time shifting: A delay of $k$ samples corresponds to multiplication by $z^{-k}$
  • These two properties allow us to convert difference equations into algebraic equations and solve them
  • Combined with the Z-transform table from Chapter 1, we can efficiently compute the Z-transforms of a wide variety of signals

Other properties of the Z-transform (z-domain differentiation, scaling, time reversal, conjugation, convolution) are covered in Basic Level, Chapter 1.