# 2.2. The z-transform – Digital Filters Design for Signal and Image Processing

## 2.2. The z-transform

### 2.2.1. Representations and summaries

With analog systems, the Laplace transform Xs(s) related to a continuous function x(t), is a function of a complex variable s and is represented by:

Chapter written by Mohamed NAJIM and Eric GRIVEL.

This variable exists when the real part of the complex variable s satisfies the relation:

with r = −∞ and R = +∞, r and R potentially characterizing the existence of limits of Xs, (s).

The Laplace transform helps resolve the linear differential equations to constant coefficients by transforming them into algebraic products.

Similarly, we introduce the z-transform when studying discrete-time signals.

Let {x(k)} be a real sequence. The bilaterial or two-sided z-transform Xz(z) of the sequence {x(k)} is represented as follows:

where z is a complex variable. The relation (2.3) is sometimes called the direct z-transform since this makes it possible to transform the time-domain signal {x(k)} into a representation in the complex-plane.

The z-transform only exists for the values of z that enable the series to converge; that is, for the value of z so that Xz(z) has a finite value. The set of all values of z satisfying these properties is then called the region of convergence (ROC).

DEMONSTRATION 2.1.– we know that the absolute convergence of a series brings about the basic convergence of the series. By applying the Cauchy criterion to the

series , the series absolutely converges if:

The series diverges if . If , we cannot be certain of the convergence.

From this, let us express Xz(z) as follows:

The series converges absolutely if:

or if:

As well, the series converges absolutely if:

or if:

If we write and , the z-transform Xz(z) converges if:

The quantities λmin and λmax now characterize the region of convergence (ROC) of the series Xz(z). The series diverges towards the strict exterior of the ROC.

We should remember that the region of convergence may be empty, as is sometimes the case where .

We can also represent, especially for causal sequences, the monolateral z-transform, Xz(z), from the sequence {x(k)} with:

DEMONSTRATION 2.2.– to establish the absolute convergence of the series, we can use another approach from the one previously shown with the bilateral transformation. It is based on d'Alembert's law. We use this law to understand the relation between two consecutive samples of the analyzed discrete-time signal.

We know that if the sequence converges towards a limit L that is strictly inferior to 1, the absolute convergence of is assured. If we apply this test to the z-transform, we get:

which gives us:

The ROC corresponds to all points in the complex-plane outside the central disk of radius λmin.

With discrete-time causal signals, such as:

x(k) = 0 for k < 0,

the one-sided (or unilateral) and the bilateral z-transforms are reduced to the same expression:

Now let us look at two examples of z-transforms.

EXAMPLE 2.1.– the unit step signal u(k) can be represented as:

u(k) =0 for k < 0 and u(k) =1 for k ≥ 0.

Its z-transform is written . The convergence is assured for |z| > 1, and we get the closed form expression of the z-transform .

EXAMPLE 2.2.– here we assume that the signal x(k) is represented by:

x(k) = α|k| with |α| < 1

We then get:

The absolute convergence of the series and is assured for . We then have:

When the signal is causal, we will obtain x(k) = αk for k ≥ 0 and x(k) = 0. Its z-transform then equals:

Figure 2.1. Representation of x(k)=α|k| and of the ROC of its z-transform Xz(z)

Figure 2.2. Representation of the causal signal x(k) = α|k| u(k) and of the ROC of its z-transform Xz(z)

### 2.2.2. Properties of the z-transform

#### 2.2.2.1. Linearity

The z-transform is linear. Actually, with the two sequences {x1(k)} and {x2(k)}. a1, a2, we have:

where Z[.] represents the operator “z-transform”, This result is valid, provided the intersection of the ROC is not empty.

DEMONSTRATION 2.3.-

The ROC of a sum of transforms then corresponds to the intersection of the ROCs.

EXAMPLE 2.3.– the linearity property can be exploited in the calculation of the z- transform of the discrete hyperbolic sinus x(k)=sh(k) u(k):

The ROC is represented by .

for |z| > exp(1).

#### 2.2.2.2. Advanced and delayed operators

Let Xz(z) be the z-transform of the discrete-time signal {x(k)}. The z-transform of {x(km)} is:

Delaying the signal by m steps thus brings about a multiplication by z-m in the z domain. The operator z-1 is called the basic delay operator, then simply the delay operator. With filters, we often see the following representation:

Figure 2.3. Delayed unitary operator

Usually, the ROC is not modified, except potentially at origin and at infinity.

DEMONSTRATION 2.4.– by definition . By changing the variables n = km, we get:

Advancing the m signal leads to a multiplication by zm of the transform in the domain of z. The operator z is called the advanced unitary operator or, more simply, the advance operator. The following representation shows this.

EXAMPLE 2.4.– now we look at the z-transform of discrete-time exponential signals x(k) = e−αk for k ≥ 0 and x(k) = 0 for k < 0 and y(k) = x(k-m) where m is a natural integer.

and

#### 2.2.2.3. Convolution

We know that the convolution between two discrete causal sequences {x1(k)} and {x2(k)} verifies the following relation:

The z-transform of the convolution product of the two sequences is then the simple product of the z-transforms of the two sequences:

The ROC of the convolution product is the intersection of the ROC of the z-transforms of {x1(k)} and {x2(k)}.

We see that this result is very often used in studying invariant linear systems, since the response of a system corresponds, as we saw in equation (1.34), to the convolution product of its impulse response by the input signal.

DEMONSTRATION 2.5.– since the product X1(z)X2(z) can be written as:

on the condition that the intersection of the ROC of the two series must not be empty.

#### 2.2.2.4. Changing the z-scale

Let us assume that Xz(z) is the z-transform of the discrete-time signal {x(k)}. With a given constant a, real or complex, the z-transform of the {akx(k)} is:

DEMONSTRATION 2.6.-

The ROC is then:

#### 2.2.2.5. Contrasted signal development

Let Xz (z) be the z-transform of the discrete-time signal {x(k)} with λmin <|z|< λmax. We then represent the sequence as {y(k)} = {x(− k)}. The z-transform of {y(k)} then equals:

DEMONSTRATION 2.7.-

The region of convergence is then written as:

#### 2.2.2.6. Derivation of the z-transform

By deriving the z-transform in relation to z−1 and then multiplying it by z−1, we return to the following characteristic result:

EXAMPLE 2.5.– now we will at the z-transform of the following discrete-time causal signal:

We can easily demonstrate that the z-transform of δ(k) for all values of z equals 1. By using advanced and delayed linearity properties, we find that:

#### 2.2.2.7. The sum theorem

If 1 is inside the ROC, we easily find that:

#### 2.2.2.8. The final-value theorem

Here we look at two sequences {x(k)} and {y(k)} such as y(k) = x(k + 1)−x(k),

by supposing the absolute convergence of the series .

From this we get the sum theorem . Now, we know that Yz(z) = (z − 1) Xz(z), and, by construction, .

From there, if , we have .

#### 2.2.2.9. Complex conjugation

Here we consider the two sequences {x(k)} and {y(k)} such as y(k) = x(k)*

#### 2.2.2.10. Parseval's theorem

provided that Xz(z) converges on an open ring containing the circle unity. The energy does not depend on the representation mode, whether it is temporal or in the z-domain.

### 2.2.3. Table of standard transform

 {x(k)} Xz(z) δ(k) 1 x(k) = ku(k) x(k) = k2u(k) x(k) = k3u(k) x(k) = k4u(k) x(k) = α|k| with |α| < 1. x(k) = k αku(k) x(k) = k2 αku(k) x(k) = k3 αku(k) x(k) = k4 αku(k) x(k) = sin(ω0, kTs).u(k) x(k) = cos(ω0, kTs).u(k) x(k) = αk, sin(ω0. kTs).u(k) x(k) = αk, cos(ω0. kTs).u(k) x(k) = kαk, sin(ω0. kTs).u(k) x(k) = kαk, cos(ω0. kTs).u(k) x(k) = [1- − cos(ω0. kTs)].u(k) x(k) = [1- (1 + a.k)e−a.k]u(k) x(k) = e−a.k. sin(ω0. kTs).u(k) x(k) = e−a.k. cos(ω0. kTs).u(k)

Table 2.1. z-transforms of specific signals