<< Chapter < Page Chapter >> Page >

1.1 definition: the z-transform x(z) of a causal discrete – time signal x(n) is defined as

X ( z ) = size 12{X \( z \) ={}} {} n = 0 x ( n ) z n size 12{ Sum cSub { size 8{n=0} } cSup { size 8{ infinity } } { size 11{x \( n \) }} z rSup { size 8{ - n} } } {} ( 4 . 1 ) size 12{ \( 4 "." 1 \) } {}

z is a complex variable of the transform domain and can be considered as the complex frequency. Remember index n can be time or space or some other thing, but is usually taken as time. As defined above , X ( z ) size 12{X \( z \) } {} is an integer power series of z 1 size 12{z rSup { size 8{ - 1} } } {} with corresponding x ( n ) size 12{x \( n \) } {} as coefficients. Let’s expand X ( z ) size 12{X \( z \) } {} :

X ( z ) = size 12{X \( z \) ={}} {} n = x ( n ) z n size 12{ Sum cSub { size 8{n= - infinity } } cSup { size 8{ infinity } } {x \( n \) z rSup { size 8{ - n} } } } {} = size 12{ {}={}} {} x ( 0 ) + x ( 1 ) z 1 + x ( 2 ) z 2 + . . . size 12{x \( 0 \) +x \( 1 \) z rSup { size 8{ - 1} } +x \( 2 \) z rSup { size 8{ - 2} } + "." "." "." } {} (4.2)

In general one writes

X ( z ) = size 12{X \( z \) ={}} {} Z [ x ( n ) ] size 12{Z \[ x \( n \) \] } {} (4.3)

In Eq.(4.1) the summation is taken from n = 0 size 12{n=0} {} to size 12{ infinity } {} , ie , X ( z ) size 12{X \( z \) } {} is not at all related to the past history of x ( n ) size 12{x \( n \) } {} . This is one–sided or unilateral z-transform . Sometime the one–sided z-transform has to take into account the initial conditions of x ( n ) size 12{x \( n \) } {} (see section 4.7).

In general , signals exist at all time , and the two-sided or bilateral z–transform is defined as

H ( z ) = size 12{H \( z \) ={}} {} n = h ( n ) z n size 12{ Sum cSub { size 8{n= - infinity } } cSup { size 8{ infinity } } {h \( n \) z rSup { size 8{ - n} } } } {}

= x ( 2 ) z 2 + x ( 1 ) z + x ( 0 ) + x ( 1 ) z 1 + x ( 2 ) z 2 + . . . size 12{ {}=x \( - 2 \) z rSup { size 8{2} } +x \( - 1 \) z+x \( 0 \) +x \( 1 \) z rSup { size 8{ - 1} } +x \( 2 \) z rSup { size 8{ - 2} } + "." "." "." } {} (4.4)

Because X ( z ) size 12{X \( z \) } {} is an infinite power series of z 1 size 12{z rSup { size 8{ - 1} } } {} , the transform only exists at values where the series converges (i.e. goes to zero as n size 12{n rightarrow infinity } {} or - size 12{ infinity } {} ). Thus the z-transform is accompanied with its region of convergence (ROC) where it is finite (see section 4.4).

A number of authors denote X + ( z ) size 12{X rSup { size 8{+{}} } \( z \) } {} for one-side z-transform.

Example 4.1.1

Find the z–transform of the two signals of Fig.4.1

Solution

(a) Notice the signal is causal and monotically decreasing and its value is just 0 . 8 n size 12{0 "." 8 rSup { size 8{n} } } {} for n 0 size 12{n>= 0} {} . So we write

x ( n ) = 0 . 8 n u ( n ) size 12{x \( n \) =0 "." 8 rSup { size 8{n} } u \( n \) } {}

and use the transform ( 4 . 1 ) size 12{ \( 4 "." 1 \) } {}

X ( z ) = size 12{X \( z \) ={}} {} n = 0 x ( n ) z n size 12{ Sum cSub { size 8{n=0} } cSup { size 8{ infinity } } {x \( n \) z rSup { size 8{ - n} } } } {}

= 1 + 0 . 8z 1 + 0 . 64 z 2 + 0 . 512 z 3 + . . . size 12{ {}=1+0 "." 8z rSup { size 8{ - 1} } +0 "." "64"z rSup { size 8{ - 2} } +0 "." "512"z rSup { size 8{ - 3} } + "." "." "." } {}

= 1 + ( 0 . 8z 1 ) + ( 0 . 8z 1 ) 2 + ( 0 . 8z 1 ) 3 + . . . size 12{ {}=1+ \( 0 "." 8z rSup { size 8{ - 1} } \) + \( 0 "." 8z rSup { size 8{ - 1} } \) rSup { size 8{2} } + \( 0 "." 8z rSup { size 8{ - 1} } \) rSup { size 8{3} } + "." "." "." } {}

Applying the formula of infinite geometric series which is repeated here

1 + a + a 2 + a 3 + . . . = size 12{1+a+a rSup { size 8{2} } +a rSup { size 8{3} } + "." "." "." ={}} {} n = 0 a n size 12{ Sum cSub { size 8{n=0} } cSup { size 8{ infinity } } {a rSup { size 8{n} } } } {} = 1 1 a size 12{ {}= { {1} over {1 - a} } } {} a < 1 size 12{ \lline a \lline<1} {} (4.5)

to obtain

X ( z ) = size 12{X \( z \) ={}} {} 1 1 0 . 8z 1 size 12{ { {1} over {1 - 0 "." 8z rSup { size 8{ - 1} } } } } {} = z z 0 . 8 size 12{ {}= { {z} over {z - 0 "." 8} } } {}

The result can be left in either of the two forms .

(b) The signal is alternatively positive and negative with increasing value .The signal is divergent . We can put the signal in the form

x ( n ) = ( 1 . 2 ) n 1 u ( n 1 ) size 12{x \( n \) = \( - 1 "." 2 \) rSup { size 8{n - 1} } u \( n - 1 \) } {}

which is ( 1 . 2 ) n u ( n ) size 12{ \( - 1 "." 2 \) rSup { size 8{n} } u \( n \) } {} delayed one index(sample) . Let’s use the transform ( 4 . 1 ) size 12{ \( 4 "." 1 \) } {}

X ( z ) = size 12{X \( z \) ={}} {} n = 0 x ( n ) z n size 12{ Sum cSub { size 8{n=0} } cSup { size 8{ infinity } } {x \( n \) z rSup { size 8{ - n} } } } {}

= 0 + 1 . 0 ( z 1 ) 1 . 2 ( z 1 ) 2 + 1 . 44 ( z 1 ) 3 1 . 718 ( z 1 ) 4 + . . . size 12{ {}=0+1 "." 0 \( z rSup { size 8{ - 1} } \) - 1 "." 2 \( z rSup { size 8{ - 1} } \) rSup { size 8{2} } +1 "." "44" \( z rSup { size 8{ - 1} } \) rSup { size 8{3} } - 1 "." "718" \( z rSup { size 8{ - 1} } \) rSup { size 8{4} } + "." "." "." } {}

= z 1 [ 1 + ( 1 . 2z 1 ) + ( 1 . 2z 1 ) 2 + ( 1 . 2z 1 ) 3 + . . . ] size 12{ {}=z rSup { size 8{ - 1} } \[ 1+ \( - 1 "." 2z rSup { size 8{ - 1} } \) + \( - 1 "." 2z rSup { size 8{ - 1} } \) rSup { size 8{2} } + \( - 1 "." 2z rSup { size 8{ - 1} } \) rSup { size 8{3} } + "." "." "." \] } {}

= z 1 1 1 + 1 . 2z 1 = z 1 1 + 1 . 2z 1 = 1 z + 1 . 2 size 12{ {}=z rSup { size 8{ - 1} } { {1} over {1+1 "." 2z rSup { size 8{ - 1} } } } = { {z rSup { size 8{ - 1} } } over {1+1 "." 2z rSup { size 8{ - 1} } } } = { {1} over {z+1 "." 2} } } {}

1.2 the inverse z-transform

The inverse z-transform is denoted by Z 1 size 12{Z rSup { size 8{ - 1} } } {} :

x ( n ) = Z 1 [ X ( z ) ] size 12{x \( n \) =Z rSup { size 8{ - 1} } \[ X \( z \) \] } {} (4.6)

The signal x ( n ) size 12{x \( n \) } {} and its transform constitutes a transform pair

X ( n ) Z ( z ) size 12{X \( n \) ↔Z \( z \) } {} (4.7)

One way to find the inverse transform , whenever possible , is to utilize just the z-transform definition. General methods of the inverse z-transform are discursed in section 4.5 and 4.6

Example 4.1.2

Find the inverse z-transform of the following

  1. X ( z ) = size 12{X \( z \) ={}} {} z z 0 . 8 size 12{ { {z} over {z - 0 "." 8} } } {}
  2. 1 z + 1 . 2 size 12{ { {1} over {z+1 "." 2} } } {}

Solution

(a) Let’s write

X ( z ) = size 12{X \( z \) ={}} {} z z- 0 . 8 size 12{ { {z} over {"z-"0 "." 8} } } {} = size 12{ {}={}} {} 1 1 0 . 8z 1 size 12{ { {1} over {1-0 "." 8z rSup { size 8{-1} } } } } {}

= 1 + ( 0 . 8z 1 ) + ( 0 . 8z 1 ) 2 + ( 0 . 8z 1 ) 3 + . . . size 12{ {}=1+ \( 0 "." 8z rSup { size 8{ - 1} } \) + \( 0 "." 8z rSup { size 8{ - 1} } \) rSup { size 8{2} } + \( 0 "." 8z rSup { size 8{ - 1} } \) rSup { size 8{3} } + "." "." "." } {}

= 1 + 0 . 8z 1 + 0 . 64 z 2 + 0 . 512 z 3 + . . . size 12{ {}=1+0 "." 8z rSup { size 8{ - 1} } +0 "." "64"z rSup { size 8{ - 2} } +0 "." "512"z rSup { size 8{ - 3} } + "." "." "." } {}

By comparing term by term with Equation ( 4 . 2 ) size 12{ \( 4 "." 2 \) } {} we get

x ( n ) = [ 1,0 . 8,0 . 64 , 0 . 512 ; . . . ] size 12{x \( n \) = \[ 1,0 "." 8,0 "." "64",0 "." "512"; "." "." "." \] } {}

or

x ( n ) = size 12{x \( n \) ={}} {} 0 . 8 n u ( n ) size 12{0 "." 8 rSup { size 8{n} } u \( n \) } {}

(b) Let’s write

X ( z ) = size 12{X \( z \) ={}} {} 1 z + 1,2 size 12{ { {1} over {z+1,2} } } {} = z 1 1 + 1,2 z 1 size 12{ {}= { {z rSup { size 8{ - 1} } } over {1+1,2z rSup { size 8{ - 1} } } } } {} = z 1 1 1 + 1,2 z 1 size 12{ {}=z rSup { size 8{ - 1} } { {1} over {1+1,2z rSup { size 8{ - 1} } } } } {}

Next , let’s expand X ( z ) size 12{X \( z \) } {} :

X ( z ) size 12{X \( z \) } {} = z 1 [ 1 + ( 1 . 2z 1 ) + ( 1 . 2z 1 ) 2 + ( 1 . 2z 1 ) 3 + . . . ] size 12{ {}=z rSup { size 8{ - 1} } \[ 1+ \( - 1 "." 2z rSup { size 8{ - 1} } \) + \( - 1 "." 2z rSup { size 8{ - 1} } \) rSup { size 8{2} } + \( - 1 "." 2z rSup { size 8{ - 1} } \) rSup { size 8{3} } + "." "." "." \] } {}

= 0 + 1 . 0z 1 1 . 2z 2 + 1 . 44 z 3 1 . 728 z 4 + . . . size 12{ {}=0+1 "." 0z rSup { size 8{ - 1} } - 1 "." 2z rSup { size 8{ - 2} } +1 "." "44"z rSup { size 8{ - 3} } - 1 "." "728"z rSup { size 8{ - 4} } + "." "." "." } {}

Thus

x ( n ) = [ 0,1 . 0, 1 . 2,1 . 44 , 1 . 728 , . . . ] size 12{x \( n \) = \[ 0,1 "." 0, - 1 "." 2,1 "." "44", - 1 "." "728", "." "." "." \] } {}

or

x ( n ) = ( 1 . 2 ) n 1 u ( n 1 ) size 12{x \( n \) = \( - 1 "." 2 \) rSup { size 8{n - 1} } u \( n - 1 \) } {}

That is

x ( n ) = 0 size 12{x \( n \) =0} {} n 0 size 12{n<= 0} {}

Questions & Answers

how to know photocatalytic properties of tio2 nanoparticles...what to do now
Akash Reply
it is a goid question and i want to know the answer as well
Maciej
Do somebody tell me a best nano engineering book for beginners?
s. Reply
what is fullerene does it is used to make bukky balls
Devang Reply
are you nano engineer ?
s.
what is the Synthesis, properties,and applications of carbon nano chemistry
Abhijith Reply
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
so some one know about replacing silicon atom with phosphorous in semiconductors device?
s. Reply
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
SUYASH Reply
for screen printed electrodes ?
SUYASH
What is lattice structure?
s. Reply
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
Sanket Reply
what's the easiest and fastest way to the synthesize AgNP?
Damian Reply
China
Cied
types of nano material
abeetha Reply
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials​ and their applications of sensors.
Ramkumar Reply
what is nano technology
Sravani Reply
what is system testing?
AMJAD
preparation of nanomaterial
Victor Reply
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
Himanshu Reply
good afternoon madam
AMJAD
what is system testing
AMJAD
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
Hello
Uday
I'm interested in Nanotube
Uday
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
Prasenjit
can nanotechnology change the direction of the face of the world
Prasenjit Reply
how did you get the value of 2000N.What calculations are needed to arrive at it
Smarajit Reply
Privacy Information Security Software Version 1.1a
Good
Berger describes sociologists as concerned with
Mueller Reply
Got questions? Join the online conversation and get instant answers!
QuizOver.com Reply

Get the best Algebra and trigonometry course in your pocket!





Source:  OpenStax, Z-transform. OpenStax CNX. Jul 29, 2009 Download for free at http://cnx.org/content/col10798/1.1
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Z-transform' conversation and receive update notifications?

Ask