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,

$$\begin{align} \mathcal{Z}_u\{a x_1[n] + b x_2[n]\} &= \sum_{n=0}^{\infty} (a x_1[n] + b x_2[n]) z^{-n} \\ &= a \sum_{n=0}^{\infty} x_1[n] z^{-n} + b \sum_{n=0}^{\infty} x_2[n] z^{-n} \\ &= a X_1(z) + b X_2(z) \end{align}$$

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,

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

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

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

2. Time Shifting

Theorem 2.2 (Time Shifting)

For the unilateral Z-transform, a one-sample delay gives

$$\mathcal{Z}_u\{x[n-1]\} = z^{-1} X(z) + x[-1]$$

and more generally, for $k > 0$,

$$\mathcal{Z}_u\{x[n-k]\} = z^{-k} X(z) + \sum_{j=1}^{k} x[-j]\, z^{-(k-j)}$$

so that shifting brings in the initial values $x[-1], \ldots, x[-k]$. Only when all of these vanish does this reduce to

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

in which a delay of $k$ samples corresponds to multiplication by $z^{-k}$.

Proof

Take $k > 0$ and substitute $m = n - k$. The key point is that the lower limit drops from $n = 0$ to $m = -k$.

$$\begin{align} \mathcal{Z}_u\{x[n-k]\} &= \sum_{n=0}^{\infty} x[n-k] z^{-n} \\ &= \sum_{m=-k}^{\infty} x[m] z^{-(m+k)} \\ &= z^{-k} \left( \sum_{m=-k}^{-1} x[m] z^{-m} + \sum_{m=0}^{\infty} x[m] z^{-m} \right) \\ &= \sum_{j=1}^{k} x[-j]\, z^{-(k-j)} + z^{-k} X(z) \end{align}$$

In the last line the first sum was rewritten with $m = -j$. If the initial values $x[-1], \ldots, x[-k]$ all vanish, that term disappears and only $z^{-k} X(z)$ remains.

Note

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

$k < 0$: advance by $|k|$ samples. Advancing a sequence is mathematically unproblematic, but a real-time system that had to emit $x[n+|k|]$ at time $n$ would need future input, so in general it cannot be realised causally.

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,

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

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,

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

(b) By linearity,

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

Problem 2 (Time Shifting)

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

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

Solution

(a) Since $u[n-1]$ is zero for $n < 1$, we may compute directly from the definition:

$$\mathcal{Z}_u\{u[n-1]\} = \sum_{n=1}^{\infty} z^{-n} = \dfrac{1}{z-1}, \quad |z| > 1$$

Because $u[-1] = 0$, the zero-initial-value case of Theorem 2.2, $z^{-1} \cdot \dfrac{z}{z-1}$, gives the same answer.

(b) Similarly $u[-1] = u[-2] = u[-3] = 0$, so no initial-value term appears and

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

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,

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

Therefore,

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

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.

Frequently Asked Questions

What is the linearity property of the Z-transform?

For constants $a, b$ and sequences $x_1[n], x_2[n]$, the Z-transform satisfies $\mathcal{Z}\{ax_1[n]+bx_2[n]\}=aX_1(z)+bX_2(z)$. This allows you to compute transforms of individual terms separately and combine them, greatly simplifying the analysis of difference equations.

How does the time shifting property work in the Z-transform?

In general $\mathcal{Z}_u\{x[n-k]\}=z^{-k}X(z)+\sum_{j=1}^{k}x[-j]z^{-(k-j)}$, so initial values appear. Only when they all vanish does this become $z^{-k}X(z)$, in which $z^{-1}$ acts as a one-sample delay and each delay term in a difference equation becomes algebraic multiplication by $z^{-1}$.

Why does an initial-value term appear in the unilateral time-shift rule?

The unilateral Z-transform sums only over $n \geq 0$, so transforming $x[n-k]$ pushes the lower limit down to $m = -k$ and leaves $x[-1], \ldots, x[-k]$ behind. Concretely, $\mathcal{Z}_u\{x[n-1]\} = z^{-1}X(z) + x[-1]$. Only when all of these initial values vanish does the rule collapse to the simple form $z^{-k}X(z)$. This initial-value term is exactly what makes the unilateral transform convenient for difference equations with initial conditions.