<< Chapter < Page Chapter >> Page >

= ( 1 . 2 ) n size 12{ {}= \( - 1 "." 2 \) rSup { size 8{n} } } {} n > 0 size 12{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 size 12{z rSup { size 8{ - 1} } } {} or z , for example

Table 4.1 : Common z-transform pairs

u ( n ) X ( z ) = size 12{u \( n \) rightarrow X \( z \) ={}} {} 1 1 z 1 size 12{ { {1} over {1 - z rSup { size 8{ - 1} } } } } {} or z z 1 size 12{ { {z} over {z - 1} } } {}

a n u ( n ) X ( z ) = size 12{a rSup { size 8{n} } u \( n \) rightarrow X \( z \) ={}} {} 1 1 az 1 size 12{ { {1} over {1 - ital "az" rSup { size 8{ - 1} } } } } {} or z z a size 12{ { {z} over {z - a} } } {}

( cos n Ω 0 ) u ( n ) X ( z ) = size 12{ \( "cos"n %OMEGA rSub { size 8{0} } \) u \( n \) rightarrow X \( z \) ={}} {} 1 z 1 cos Ω 0 1 2z 1 cos Ω 0 + z 2 size 12{ { {1 - z rSup { size 8{ - 1} } "cos" %OMEGA rSub { size 8{0} } } over {1 - 2z rSup { size 8{ - 1} } "cos" %OMEGA rSub { size 8{0} } +z rSup { size 8{ - 2} } } } } {} or z ( z cos Ω 0 ) z 2 2z cos Ω 0 + 1 size 12{ { {z \( z - "cos" %OMEGA rSub { size 8{0} } \) } over {z rSup { size 8{2} } - 2z"cos" %OMEGA rSub { size 8{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 ( z ) = size 12{H \( z \) ={}} {} n = 0 h ( n ) z n size 12{ Sum cSub { size 8{n=0} } cSup { size 8{ infinity } } {h \( n \) z rSup { size 8{ - n} } } } {} (one–sided transform ) (4.8)

or

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} } } } {} (two–sided transform )(4.9)

depending on whether the system is causal or noncausal .

H ( z ) size 12{H \( z \) } {} , the z–transform of h ( n ) size 12{h \( n \) } {} , 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 ( 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} } } } {} = n = 2 3 h ( n ) z n size 12{ {}= Sum cSub { size 8{n= - 2} } cSup { size 8{3} } {h \( n \) z rSup { size 8{ - n} } } } {}

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

= z 2 + 2z 1 + 3 + 4z 1 + 5z 2 + 6z 3 size 12{ {}=z rSup { size 8{ - 2} } +2z rSup { size 8{ - 1} } +3+4z rSup { size 8{ - 1} } +5z rSup { size 8{ - 2} } +6z rSup { size 8{ - 3} } } {}

On the contrary, if H ( z ) size 12{H \( z \) } {} is known as above we can easily obtain h ( n ) size 12{h \( n \) } {} .

1.5 eigen-function and eigen-value

We know that if the frequency response of a system is H ( ω ) size 12{H \( ω \) } {} then for input x ( n ) = size 12{x \( n \) ={}} {} e jn ω size 12{e rSup { size 8{ ital "jn"ω} } } {} , the output is y ( n ) = size 12{y \( n \) ={}} {} e jn ω size 12{e rSup { size 8{ ital "jn"ω} } } {} H ( ω ) size 12{H \( ω \) } {} . Because of this , e jn ω size 12{e rSup { size 8{ ital "jn"ω} } } {} is the eigen-function , and H ( ω ) size 12{H \( ω \) } {} the eigen-value of the system.

Now , for input

x ( n ) = z n size 12{x \( n \) =z rSup { size 8{n} } } {} (4.10)

the system output is

y ( n ) = h ( n ) x ( n ) = size 12{y \( n \) =h \( n \) *x \( n \) ={}} {} k = 0 h ( k ) z n k size 12{ Sum cSub { size 8{k=0} } cSup { size 8{ infinity } } {h \( k \) z rSup { size 8{n - k} } } } {} = z n k = 0 h ( k ) z k size 12{ {}=z rSup { size 8{n} } left [ Sum cSub { size 8{k=0} } cSup { size 8{ infinity } } {h \( k \) z rSup { size 8{ - k} } } right ]} {}

In the brackets is just H(z) , thus

{} y ( n ) = z n H ( z ) size 12{y \( n \) =z rSup { size 8{n} } H \( z \) } {} (4.11)

Hence in the z–transform domain , z n size 12{z rSup { size 8{n} } } {} is eigen-function and H ( z ) size 12{H \( z \) } {} 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 ( n ) = size 12{y \( n \) ={}} {} k = 1 M a k y ( n k ) size 12{ Sum cSub { size 8{k=1} } cSup { size 8{M} } {a rSub { size 8{k} } y \( n - k \) } } {} + k = N N b k x ( n k ) size 12{+ Sum cSub { size 8{k= - N} } cSup { size 8{N} } {b rSub { size 8{k} } x \( n - k \) } } {} (4.12)

where a k size 12{a rSub { size 8{k} } } {} and b k size 12{b rSub { size 8{k} } } {} are the filter coefficients (constants) , and the limits M, N can be extended to infinity .

Now we make the replacement x ( n ) = z n size 12{x \( n \) =z rSup { size 8{n} } } {} and y ( n ) = z n size 12{y \( n \) =z rSup { size 8{n} } } {} H ( z ) size 12{H \( z \) } {} to get

z n H ( z ) = size 12{z rSup { size 8{n} } H \( z \) ={}} {} k = 1 M a k z n k H ( z ) size 12{ Sum cSub { size 8{k=1} } cSup { size 8{M} } {a rSub { size 8{k} } z rSup { size 8{n - "k "} } H \( z \) } } {} + k = N N b k z n k size 12{+ Sum cSub { size 8{k= - N} } cSup { size 8{N} } {b rSub { size 8{k} } z rSup { size 8{n - k} } } } {}

From this we derive the expression of H ( z ) size 12{H \( z \) } {}

H ( z ) = size 12{H \( z \) ={}} {} k = N N b k z k 1 k = 1 M a k z k size 12{ { { Sum cSub { size 8{k= - N} } cSup { size 8{N} } {b rSub { size 8{k} } z rSup { size 8{ - k} } } } over {1 - Sum cSub { size 8{k=1} } cSup { size 8{M} } {a rSub { size 8{k} } z rSup { size 8{ - 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 ( z ) size 12{H \( z \) } {} . For nonrecursive fillters the denominator is just 1 and H ( z ) size 12{H \( z \) } {} 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 ( z ) size 12{H \( z \) } {} without the need to take the z-transform. Vice versa, if we know H ( z ) size 12{H \( z \) } {} that means we know the filler coefficients , hence the filter equation.

Example 4.1.4

Given

(a) H ( z ) = size 12{H \( z \) ={}} {} 2z 2 3z z 2 + 0 . 5z 0 . 8 size 12{ { {2z rSup { size 8{2} } - 3z} over {z rSup { size 8{2 } } +0 "." 5z - 0 "." 8} } } {}

(b) H ( z ) = size 12{H \( z \) ={}} {} 20 z 2 + 5z 10 z 3 + 5z 2 8z 1 size 12{ { {-"20"z rSup { size 8{2} } +5z} over {"10"z rSup { size 8{3} } + 5z rSup { size 8{2} } -8z - 1} } } {}

Find the filter difference equation.

Solution

(a) Write H ( z ) size 12{H \( z \) } {} as a function of z 1 size 12{z rSup { size 8{ - 1} } } {} by multiplying the numerator and denominator with z 2 size 12{z rSup { size 8{ - 2} } } {} :

H ( z ) = size 12{H \( z \) ={}} {} 2 3z 1 1 + 0 . 5z 1 0 . 8z 2 = 2 3z 1 1 ( 0 . 5z 1 + 0 . 8z 2 ) size 12{ { {2-3z rSup { size 8{ - 1} } } over {1+0 "." 5z rSup { size 8{ - 1} } - 0 "." 8z rSup { size 8{ - 2} } } } = { {2 - 3z rSup { size 8{ - 1} } } over {1 - \( - 0 "." 5z rSup { size 8{ - 1} } +0 "." 8z rSup { size 8{ - 2} } \) } } } {}

The coefficients are

a 0 = 2 size 12{a rSub { size 8{0} } =2} {} a 1 = 3 size 12{a rSub { size 8{1} } = - 3} {} b 1 = 0 . 5 size 12{b rSub { size 8{1} } = - 0 "." 5} {} b 2 = 0 . 8 size 12{b rSub { size 8{2} } =0 "." 8} {}

Thus the filler equation is

y ( n ) = 0 . 5y ( n 1 ) 0 . 8y ( n 2 ) + 2x ( n ) 3x ( n 1 ) size 12{y \( n \) = - 0 "." 5y \( n - 1 \) - 0 "." 8y \( n - 2 \) +2x \( n \) - 3x \( n - 1 \) } {}

(b) Multiply the numerator and denominator with 0 . 1z 3 size 12{0 "." 1z rSup { size 8{ - 3} } } {} to make 10 z 3 size 12{"10"z rSup { size 8{ - 3} } } {} in the denominator equal to 1 :

H ( z ) = size 12{H \( z \) ={}} {} 2z 1 + 5z 2 1 + 0 . 5z 1 0 . 8z 2 0 . 1z 3 size 12{ { { - 2z rSup { size 8{ - 1} } +5z rSup { size 8{ - 2} } } over {1+0 "." 5z rSup { size 8{ - 1} } - 0 "." 8z rSup { size 8{ - 2} } - 0 "." 1z rSup { size 8{ - 3} } } } } {}

Collect the coefficients:

a 1 = 2 size 12{a rSub { size 8{1} } = - 2} {} a 2 = 5 size 12{a rSub { size 8{2} } =5} {} b 1 = 0 . 5 size 12{b rSub { size 8{1} } = - 0 "." 5} {} b 2 = 0 . 8 size 12{b rSub { size 8{2} } =0 "." 8} {} b 3 = 0 . 1 size 12{b rSub { size 8{3} } = - 0 "." 1} {}

Thus

y ( n ) = 0 . 5y ( n 1 ) + 0 . 8y ( n 2 ) + 0 . 1y ( n 3 ) 2x ( n 1 ) + 5x ( n 2 ) size 12{y \( n \) = - 0 "." 5y \( n - 1 \) +0 "." 8y \( n - 2 \) +0 "." 1y \( n - 3 \) - 2x \( n - 1 \) +5x \( n - 2 \) } {} y(n)

Questions & Answers

what does nano mean?
Anassong Reply
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
Damian Reply
absolutely yes
Daniel
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
characteristics of micro business
Abigail
for teaching engĺish at school how nano technology help us
Anassong
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.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
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
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
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