Properties of the Z-transform

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This section first describes the main properties of the ZT. Finally, the so called one-sides ZT will be introduce, which is useful for solving Difference Equations with initial conditions.

Assume the ZT of the sequences $x [n]$ and $y [n]$ are given by $X (z)$ and $Y (z)$ and their ROC by $R_{x}$ and $R_{y}$ respectively.

$\begin{array}{rcl} x [n] \circ - \circ X (z) with ROC R_{x} & and & y [n] \circ - \circ Y (z) with ROC R_{y} \end{array}$

Property: Linearity

The ZT, as the FTD, is a linear operation:

$w [n] = α x [n] + β y [n] \circ - \circ W (z) = α X (z) + β Y (z) with ROC R_{w} = R_{x} \cap R_{y}$

The resulting ROC $R_{w}$ of the sequence $w [n]$ will include the intersection of $R_{x}$ and $R_{y}$ . However the ROC $R_{w}$ may be larger.

Example

Compute the ROC of

w [n] = x [n] - y [n]

, with

x [n] = u [n]

and

y [n] = u [n - 1]

Both ROC's

X (z)

and

Y (z)

are all values of

z

outside the circle with radius 1. However, the ZT of

w [n] = x [n] - y [n] = δ [n]

is the entire

z

-plane.

Property: Shifting

Shifting a sequence, which can be delaying or advancing, multiplies the ZT by a power of $z$ :

$x [n - n_{0}] \circ - \circ z^{- n_{0}} X (z) with ROC R_{x}$

Because shifting a sequence does not affect its absolute sumability, shifting does not change the ROC. Therefor, the ZT of $x [n]$ and $x [n - n_{0}]$ have the same ROC, with the possible exception of adding or deleting the points $z = 0$ and $z = \infty$ .

Property: Time reversal

The ZT of the time reversed sequence $x [- n]$ is given by $X (z^{- 1}$ .

$x [- n] \circ - \circ X (z^{- 1}) with ROC 1 / R_{x}$

In case the ROC $R_{x}$ is given by the annulus in between $r_{-}$ and $r_{+}$ , the ROC of the time reversed sequence is given by the annulus in between $\frac{1}{r_{+}}$ and $\frac{1}{r_{-}}$ , which is denoted by $1 / R_{x}$ .

Property: Multiply by $a^{n}$

$a^{n} x [n] \circ - \circ X (a^{- 1} z) with ROC R_{y} = | a | R_{x}$

The proof of this property is as follows: If a sequence $x [n]$ is multiplied by a complex exponential $a^{n}$ the resulting ZT $Y (z)$ can be obtained by first combining the parameter $a^{- 1}$ and the variable $z$ resulting in a scaled version of $X (z)$ :

$\begin{array}{rcl} y [n] = a^{n} x [n] \circ - \circ & Y (z) & = \sum_{n = - \infty}^{\infty} a^{n} x [n] z^{- n} = \sum_{n = - \infty}^{\infty} x [n] (a^{- 1} \cdot z)^{- n} \\ = X (a^{- 1} z) with ROC R_{y} = | a | R_{x} \end{array}$

The resulting ROC is also scaled by the parameter $a$ .

Example

The ZT and ROC of the unit step function

x [n] = u [n]

is as follows:

\begin{array}{rcl} x [n] = u [n] & \circ - \circ & X (z) = \sum_{n = 0}^{\infty} z^{- n} \overset{| z | > 1}{=} \frac{1}{1 - z^{- 1}} R_{x} : | z | > 1 \end{array}

Calculate the ZT and its ROC of the scaled version

y [n] = a^{n} u [n]

of the unit step function.

Evaluating the ZT of the scaled version

y [n]

via the property leads to the following complex function

Y (z)

from which the ROC is outside a circle with radius

a

\begin{array}{rcl} y [n] = a^{n} x [n] & \circ - \circ & Y (z) = X (a^{- 1} z) = \frac{1}{1 - (a^{- 1} z)^{- 1}} \\ = \frac{1}{1 - a z^{- 1}} R_{y} : | z | > | a | \end{array}

Verifying this result by applying the ZT to the sequence

y [n] = a^{n} u [n]

leads indeed to the same result

Y (z)

with the same ROC as follows from the following equations:

\begin{array}{rcl} via ZT of y [n] & \circ - \circ & Y (z) = \sum_{n = 0}^{\infty} a^{n} z^{- n} \overset{| z | > | a |}{=} \frac{1}{1 - a z^{- 1}} R_{y} : | z | > | a | \end{array}

Finally, as a special case note that if

x [n]

is multiplied by a complex exponent

e^{j n θ_{0}}

e^{j n θ_{0}} x [n] \circ - \circ X (e^{- j θ_{0}} z)

which corresponds to a rotation of the

z

-plane

Property: Convolution Theorem

Perhaps the most important ZT property is the convolution theorem, which states that convolution in time domain is mapped to multiplication in $z$ domain.

$x [n] ⋆ y [n] \circ - \circ X (z) \cdot Y (z) with R_{w} = R_{x} \cap R_{y}$

The proof of this property is shown in the following steps: First write out the convolution sum and transform to the $z$ -domain.

$\begin{array}{rcl} w [n] = x [n] * y [n] & \circ - \circ & W (z) = \sum_{n = - \infty}^{\infty} (\sum_{p = - \infty}^{\infty} x [p] y [n - p]) z^{- n} \end{array}$

In the next step change the order of the two summations and then add an extra exponent $z^{- p}$ in the first summation which has to be compensated this in the second summation:

$\begin{array}{rcl} W (z) & = & \sum_{p = - \infty}^{\infty} x [p] {\sum_{n = - \infty}^{\infty} y [n - p] z^{- n}} \\ = & \sum_{p = - \infty}^{\infty} x [p] z^{- p} \cdot {\sum_{n = - \infty}^{\infty} y [n - p] z^{- (n - p)}} \end{array}$

The first summation is the ZT of the sequence $x [n]$ and the second summation between large brackets is the ZT of the sequence $y [n]$ , which finalizes the proof.

$\begin{array}{rcl} W (z) & = & X (z) \cdot Y (z) with R_{w} = R_{x} \cap R_{y} \end{array}$

The ROC of the result $W (z)$ of the convolution property includes the intersection of the ROC's of $X (z)$ and $Y (z)$ . However, the ROC $R_{w}$ may be larger, if there is a pole-zero cancellation in the product $X (z) \cdot Y (z)$ .

Example

Calculate the convolution result of the two finite length sequences:

\begin{array}{rcl} x [n] = δ [n] + 2 δ [n - 1] & and & y [n] = δ [n] + δ [n - 1] \end{array}

Calculate first the result via the convolution procedure and verify this result via the convolution property of the ZT.

Both

x

and

y

are finite length sequences and the convolution result

w [n]

is a finite length sequence which consists of 3 delta pulse with coefficients

1

3

and

2

respectively.

\begin{array}{rcl} w [n] = x [n] * y [n] & = & = 1 δ [n] + 3 δ [n - 1] + 2 δ [n - 3] \end{array}

The ZT of the two sequences is as follows:

\begin{array}{rcl} x [n] = δ [n] + 2 δ [n - 1] & \circ - \circ & X (z) = 1 + 2 z^{- 1} \\ y [n] = δ [n] + δ [n - 1] & \circ - \circ & Y (z) = 1 + z^{- 1} \end{array}

Multiplying these two polynomials results in a finite length polynomial with coefficients

1

3

and

2

, which is indeed the ZT of the convolution result

w [n]

\begin{array}{rcl} W (z) = X (z) \cdot Y (z) & = & = (1 + 2 z^{- 1}) \cdot (1 + z^{- 1}) = 1 + 3 z^{- 1} + 2 z^{- 2} \end{array}

Finally the ROC of the result

W (z)

of the convolution theorem includes the intersection of the ROC's of

X (z)

and

Y (z)

Example

Calculate the ROC of

x [n] ⋆ y [n]

with

x [n] = a^{n} u [n]

and

y [n] = δ [n] - a δ [n - 1]

\begin{array}{rcl} x [n] = a^{n} u [n] & \circ - \circ & X (z) = \frac{1}{1 - a z^{- 1}} with R_{x} : | z | > | a | \\ y [n] = δ [n] - a δ [n - 1] & \circ - \circ & Y (z) = 1 - a z^{- 1} with R_{y} : | z | > 0 \end{array}

In this example the ZT

X (z)

of the infinite length exponential decaying sequence

x [n]

has one pole at

z = a

and the ROC contains all complex variables

z

outside a circle with radius

a

. The ZT

Y (z)

of the finite length sequence

y [n]

has one zero at

z = a

and the ROC contains all complex variables

z

except

z = 0

\begin{array}{rcl} W (z) = X (z) \cdot Y (z) = \frac{1}{1 - a z^{- 1}} \cdot (1 - a z^{- 1}) = 1 with R_{w} : all z \end{array}

The ZT of the convolution of

x [n]

with

y [n]

W (z) = X (z) \cdot Y (z)

, which, due to the pole-zero cancellation, has a ROC that is the entire

z

-plane.

Property: Conjugation

$x^{*} [n] \circ - \circ X^{*} (z^{*}); R_{x}$ As a corollary, note that if $x [n]$ is real-valued, $y [n] = x^{*} [n] = x [n]$ , then $X (z) = X^{*} (z^{*})$ is such as case.

The proof of the conjugation property goes as follows:

The ZT of the complex conjugate of the sequence $x [n]$ can be rewritten by taking the complex conjugation operation outside the brackets and interchange in the exponents of the variable $z$ which leads to the final result. $\begin{array}{rcl} y [n] = x^{*} [n] \circ - \circ & Y (z) & = \sum_{n = - \infty}^{\infty} x^{*} [n] z^{- n} = {(\sum_{n = - \infty}^{\infty} x [n] {z^{- n}}^{*})}^{*} \\ = {(\sum_{n = - \infty}^{\infty} x [n] {z^{*}}^{- n})}^{*} = X^{*} (z^{*}); R_{y} = R_{x} \end{array}$ The ROC is not influenced by the complex operation.

Example

Calculate the ZT of the sequence

{(e^{- j θ_{0}})}^{n} u [n]

From the table with ZT-pairs it follows that

a^{n} u [n] \circ - \circ 1 / (1 - a z^{- 1})

\to

\begin{array}{rcl} x [n] = {(e^{j θ_{0}})}^{n} u [n] & \circ - \circ & X (z) = \frac{1}{1 - e^{j θ_{0}} z^{- 1}} \end{array}

With

y [n] = x^{*} [n]

,we can find the result

Y (z)

by using the conjugation property as follows:

\begin{array}{rcl} Y (z) = X^{*} (z^{*}) & = & {\frac{1}{1 - e^{j θ_{0}} (z^{*})^{- 1}}}^{*} = {\frac{1}{1 - e^{j θ_{0}} (z^{- 1})^{*}}}^{*} \\ = & \frac{1}{1 - e^{- j θ_{0}} z^{- 1}} \end{array}

Property: Derivative

$n \cdot x [n] \circ - \circ - z \frac{d}{d z} X (z); R_{x}$

The proof of this property goes as follows:

The ZT of the sequence $n \cdot x [n]$ can be pre-multiplied with $z$ , which has to be compensated inside the summation: $\begin{array}{rcl} y [n] = n \cdot x [n] & \circ - \circ & Y (z) = \sum_{n = - \infty}^{\infty} n \cdot x [n] z^{- n} = z \sum_{n = - \infty}^{\infty} {n \cdot x [n] z^{- n - 1}} \end{array}$ The function in between brackets can be rewritten as a differentiation, which can be taken outside the summation, because of linearity, which leads to the final result: $\begin{array}{rcl} Y (z) & = & = z \sum_{n = - \infty}^{\infty} {- \frac{d}{d z} x [n] z^{- n}} = - z \frac{d}{d z} (\sum_{n = - \infty}^{\infty} x [n] z^{- n}) = - z \frac{d}{d z} X (z) \end{array}$ The ROC is not influenced.

Example

Calculate the ZT of

y [n] = n a^{n} u [n]

The sequence

y [n]

n

times the exponential decaying sequence

a^{n} u [n]

. From the derivative property it follows that we can find the ZT

Y (z)

y [n]

as follows:

\begin{array}{rcl} Y (z) & = & - z \frac{d}{d z} (\frac{1}{1 - a z^{- 1}}) = - z {\frac{- a z^{- 2}}{(1 - a z^{- 1})^{2}}} = \frac{a z^{- 1}}{(1 - a z^{- 1})^{2}} \end{array}

Property: Initial value

When the sequence $x [n] = 0$ for $n < 0$ , the initial value $x [0]$ may be found from the ZT $X (z)$ by taking the limit for $z$ to infinity: $x [0] = lim_{z \to \infty} {X (z)}$ The proof goes as follows with $x [n] = 0$ for $n < 0$ : $\begin{array}{rcl} lim_{z \to \infty} {X (z)} & = & lim_{z \to \infty} {\sum_{n = - \infty}^{\infty} x [n] z^{- n}} \\ = & lim_{z \to \infty} {\dots x [- 1] z + x [0] + x [1] z^{- 1} + \dots} \\ = & lim_{z \to \infty} {0 + x [0] + x [1] z^{- 1} + \dots} = x [0] \end{array}$ This initial value property can be shown in the following example:

Example

Calculate the initial value

x [0]

of the sequence

x [n] = a^{n} u [n]

via the initial value property of the ZT.

The given ZT

X (z)

is related to the exponential decaying function

x [n]

from which it is clear that the initial value

x [0] = 1

\begin{array}{rcl} X (z) = \frac{1}{1 - a z^{- 1}} & \circ - \circ & x [n] = a^{n} u [n] \to x [0] = 1 \end{array}

The same result can be found via the initial value property as follows:

lim_{z \to \infty} {X (z)} = lim_{z \to \infty} {\frac{1}{1 - a z^{- 1}}} = 1

One sided Z-Transform

The ZT $X (z)$ of the sequence $x [n]$ , as discussed before, is the two-sided ZT. The one-sided ZT $\tilde{X} (z)$ is defined by the given equation in which the summation index $n$ does not start from $n = - \infty$ but from $n = 0$ . $\tilde{X} (z) = \sum_{n = 0}^{\infty} x [n] z^{- n}$ Most of the properties of the one-sided ZT are the same as the properties of the two-sided ZT. One that is different, however, is the shift property. Specifically, if the sequence $x [n]$ has a one-sided ZT $\tilde{X} (z)$ , the difference of the one-sided ZT of the shifted sequence $x [n - k]$ is represented in red in the following shift property of the one-sided ZT: $x [n - k] \circ - \circ \sum_{p = - k}^{- 1} x [p] z^{- (p + k)} + z^{- k} \cdot \tilde{X} (z)$ For the proof of this property we apply the one-sided ZT to the shifted sequence $x [n - k]$ and then substitute a new variable $p$ for $n - k$ , which finalizes the proof: $\begin{array}{rcl} x [n - k] & \circ - \circ & \sum_{n = 0}^{\infty} x [n - k] z^{- n} \overset{p = n - k}{= = = =} \sum_{p = - k}^{\infty} x [p] z^{- (p + k)} \\ = \sum_{p = - k}^{- 1} x [p] z^{- (p + k)} + z^{- k} \cdot (\sum_{p = 0}^{\infty} x [p] z^{- p}) \end{array}$ It is this shifting property that makes the one-sided ZT useful for solving Difference Equations with initial conditions, as can be shown by the following example.

Example

Consider the linear constant Difference Equation

y [n] = \frac{1}{4} y [n - 2] + x [n]

. Find the solution for this equation assuming

x [n] = δ [n - 1]

. First use the ZT for the case

y [n] = 0

for

n < 0

and then use the one sided ZT shift property for the case

y [- 2] = y [- 1] = 1

∙ \underline{Case y [n] = 0 for n < 0} \to Use ZT

In case the initial conditions are zero, we can use the usual ZT. Together with the ZT of

x [n] = δ [n - 1]

this leads to the given fraction. This fraction can be expanded as a sum of two terms which are related to the given sum of two exponential decaying sequences.

\begin{array}{rcl} Y (z) & = & \frac{1}{4} z^{- 2} \cdot Y (z) + X (z) with X (z) = z^{- 1} \to \end{array}

\begin{array}{rcl} Y (z) = \frac{z^{- 1}}{1 - \frac{1}{4} z^{- 2}} = \frac{1}{1 - \frac{1}{2} z^{- 1}} - \frac{1}{1 + \frac{1}{2} z^{- 1}} \\ \circ - \circ y [n] = ({(\frac{1}{2})}^{n} - {(- \frac{1}{2})}^{n}) \cdot u [n] \end{array}

∙ \underline{Case y [- 2] = y [- 1] = 1} \to Use one-sided ZT:

In case the initial conditions are nonzero, for example

y [- 2] = y [- 1] = 1

we have to use the one-sided ZT to solve the Difference Equation. The one-sided ZT

\tilde{X} (z)

x [n] = δ [n - 1]

is the same as the ZT

X (z)

which is equal to

Z^{- 1}

. Using this together with the shift property of the one-sided ZT results in the given fraction, which can be expanded as a sum of two terms which are related to the given sum of two weighted exponential decaying sequences.

\begin{array}{rcl} \tilde{Y} (z) & = & \frac{1}{4} \cdot (y [- 2] + y [- 1] z^{- 1} + z^{- 2} \cdot \tilde{Y} (z)) + \tilde{X} (z) \end{array}

\begin{array}{rcl} \to \tilde{Y} (z) & = & \frac{\frac{1}{4} + \frac{5}{4} z^{- 1}}{1 - \frac{1}{4} z^{- 2}} = \frac{\frac{11}{8}}{1 - \frac{1}{2} z^{- 1}} - \frac{\frac{9}{8}}{1 + \frac{1}{2} z^{- 1}} \\ \tilde{Y} (z) & \circ - \circ & y [n] = (\frac{11}{8} {(\frac{1}{2})}^{n} - \frac{9}{8} {(- \frac{1}{2})}^{n}) \cdot u [n] \end{array}

The influence of the initial conditions is reflected by the two weights denoted in red.

Last updated on May 30, 2020

Properties of the Z-transform

Screencast video [⯈]

Property: Linearity

Example

Property: Shifting

Property: Time reversal

Property: Multiply by anan

Example

Property: Convolution Theorem

Example

Example

Property: Conjugation

Example

Property: Derivative

Example

Property: Initial value

Example

One sided Z-Transform

Example

Property: Multiply by $a^{n}$