# Z – transform  (Page 2/2)

 Page 2 / 2

$=\left(-1\text{.}2{\right)}^{n}$ $n>0$

## 1.3 z–transform pairs

Table 4.1 gives many useful z-transform pairs . All signals are causal (right-sided), except two which are anticausal (left-sided). Notice that a transform can be expresed equivalently as a function of ${z}^{-1}$ or z , for example

Table 4.1 : Common z-transform pairs

$u\left(n\right)\to X\left(z\right)=$ $\frac{1}{1-{z}^{-1}}$ or $\frac{z}{z-1}$

${a}^{n}u\left(n\right)\to X\left(z\right)=$ $\frac{1}{1-{\text{az}}^{-1}}$ or $\frac{z}{z-a}$

$\left(\text{cos}n{\Omega }_{0}\right)u\left(n\right)\to X\left(z\right)=$ $\frac{1-{z}^{-1}\text{cos}{\Omega }_{0}}{1-{2z}^{-1}\text{cos}{\Omega }_{0}+{z}^{-2}}$ or $\frac{z\left(z-\text{cos}{\Omega }_{0}\right)}{{z}^{2}-2z\text{cos}{\Omega }_{0}+1}$

Which form is more appropriate depending on what we would like to do with the transform. (see sections 4.1.6 , 4.3 and 4.6).

## 1.4 z–transform for systems

The z-transform applies to systems as well as signals because systems are represented by their impulse responses which are functions of index n (time or space …) just like signals. Remember that many other transforms (Laplace, Fourier …) have the same property, and due to this property that the transforms are useful in the analysis and design of systems because signals and systems interact.

Specifically , the z–transform of impulse response h(n) is

$H\left(z\right)=$ $\sum _{n=0}^{\infty }h\left(n\right){z}^{-n}$ (one–sided transform ) (4.8)

or

$H\left(z\right)=$ $\sum _{n=-\infty }^{\infty }h\left(n\right){z}^{-n}$ (two–sided transform )(4.9)

depending on whether the system is causal or noncausal .

$H\left(z\right)$ , the z–transform of $h\left(n\right)$ , is called transfer function or system function of the system .

## Example 4.1.3

A system has impulse respone

h(n) = [1 , 2 , 3 , 4 , 5 , 6]

Find the transfer function .

## Solution

The system is of noncausal FIR type . Its transfer function is

$H\left(z\right)=$ $\sum _{n=-\infty }^{\infty }h\left(n\right){z}^{-n}$ $=\sum _{n=-2}^{3}h\left(n\right){z}^{-n}$

$=h\left(-2\right){z}^{2}+h\left(-1\right){z}^{-1}+h\left(0\right){z}^{0}+h\left(1\right){z}^{-1}+h\left(2\right){z}^{-2}+h\left(3\right){z}^{-3}$

$={z}^{-2}+{2z}^{-1}+3+{4z}^{-1}+{5z}^{-2}+{6z}^{-3}$

On the contrary, if $H\left(z\right)$ is known as above we can easily obtain $h\left(n\right)$ .

## 1.5 eigen-function and eigen-value

We know that if the frequency response of a system is $H\left(\omega \right)$ then for input $x\left(n\right)=$ ${e}^{\text{jn}\omega }$ , the output is $y\left(n\right)=$ ${e}^{\text{jn}\omega }$ $H\left(\omega \right)$ . Because of this , ${e}^{\text{jn}\omega }$ is the eigen-function , and $H\left(\omega \right)$ the eigen-value of the system.

Now , for input

$x\left(n\right)={z}^{n}$ (4.10)

the system output is

$y\left(n\right)=h\left(n\right)\ast x\left(n\right)=$ $\sum _{k=0}^{\infty }h\left(k\right){z}^{n-k}$ $={z}^{n}\left[\sum _{k=0}^{\infty }h\left(k\right){z}^{-k}\right]$

In the brackets is just H(z) , thus

 $y\left(n\right)={z}^{n}H\left(z\right)$ (4.11)

Hence in the z–transform domain , ${z}^{n}$ is eigen-function and $H\left(z\right)$ is eigen-value of the system.

## 1.6 transfer function in terms of filter coefficients

For , Let’s begin with the general filter difference equation which is repeated here

$y\left(n\right)=$ $\sum _{k=1}^{M}{a}_{k}y\left(n-k\right)$ $+\sum _{k=-N}^{N}{b}_{k}x\left(n-k\right)$ (4.12)

where ${a}_{k}$ and ${b}_{k}$ are the filter coefficients (constants) , and the limits M, N can be extended to infinity .

Now we make the replacement $x\left(n\right)={z}^{n}$ and $y\left(n\right)={z}^{n}$ $H\left(z\right)$ to get

${z}^{n}H\left(z\right)=$ $\sum _{k=1}^{M}{a}_{k}{z}^{n-\text{k}}H\left(z\right)$ $+\sum _{k=-N}^{N}{b}_{k}{z}^{n-k}$

From this we derive the expression of $H\left(z\right)$

$H\left(z\right)=$ $\frac{\sum _{k=-N}^{N}{b}_{k}{z}^{-k}}{1-\sum _{k=1}^{M}{a}_{k}{z}^{-k}}$ (4.13)

It is worthwhile to notice that the above transfer function is resulted from the filter equation (4.12) . Other authors write the filter equation differently (for example, all the y terms are on the left side of the equation), leading to a slightly different expression of $H\left(z\right)$ . For nonrecursive fillters the denominator is just 1 and $H\left(z\right)$ becomes that of nonrecursive filters as we know .

The idea here is that when the filter equation is given, we collect its coefficients to place into the expression above of $H\left(z\right)$ without the need to take the z-transform. Vice versa, if we know $H\left(z\right)$ that means we know the filler coefficients , hence the filter equation.

## Example 4.1.4

Given

(a) $H\left(z\right)=$ $\frac{{2z}^{2}-3z}{{z}^{2}+0\text{.}5z-0\text{.}8}$

(b) $H\left(z\right)=$ $\frac{-\text{20}{z}^{2}+5z}{\text{10}{z}^{3}+{5z}^{2}-8z-1}$

Find the filter difference equation.

## Solution

(a) Write $H\left(z\right)$ as a function of ${z}^{-1}$ by multiplying the numerator and denominator with ${z}^{-2}$ :

$H\left(z\right)=$ $\frac{2-{3z}^{-1}}{1+0\text{.}{5z}^{-1}-0\text{.}{8z}^{-2}}=\frac{2-{3z}^{-1}}{1-\left(-0\text{.}{5z}^{-1}+0\text{.}{8z}^{-2}\right)}$

The coefficients are

${a}_{0}=2$ ${a}_{1}=-3$ ${b}_{1}=-0\text{.}5$ ${b}_{2}=0\text{.}8$

Thus the filler equation is

$y\left(n\right)=-0\text{.}5y\left(n-1\right)-0\text{.}8y\left(n-2\right)+2x\left(n\right)-3x\left(n-1\right)$

(b) Multiply the numerator and denominator with $0\text{.}{1z}^{-3}$ to make $\text{10}{z}^{-3}$ in the denominator equal to 1 :

$H\left(z\right)=$ $\frac{-{2z}^{-1}+{5z}^{-2}}{1+0\text{.}{5z}^{-1}-0\text{.}{8z}^{-2}-0\text{.}{1z}^{-3}}$

Collect the coefficients:

${a}_{1}=-2$ ${a}_{2}=5$ ${b}_{1}=-0\text{.}5$ ${b}_{2}=0\text{.}8$ ${b}_{3}=-0\text{.}1$

Thus

$y\left(n\right)=-0\text{.}5y\left(n-1\right)+0\text{.}8y\left(n-2\right)+0\text{.}1y\left(n-3\right)-2x\left(n-1\right)+5x\left(n-2\right)$ y(n)

what are the products of Nano chemistry?
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
learn
da
no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts
Bhagvanji
Preparation and Applications of Nanomaterial for Drug Delivery
revolt
da
Application of nanotechnology in medicine
what is variations in raman spectra for nanomaterials
I only see partial conversation and what's the question here!
what about nanotechnology for water purification
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
Hafiz
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
anyone know any internet site where one can find nanotechnology papers?
research.net
kanaga
sciencedirect big data base
Ernesto
Got questions? Join the online conversation and get instant answers!