applications of zapplications of z--transform...

Applications of zApplications of z--TransformTransform

S Wongsa

��

S Wongsa

Dept. of Control Systems and Instrumentation Engineering,

KMUTT

Summary from previous class

� �� ndzzzXnx

1)(1

][

��

� ��C

ndzzzX

jnx

1)(2

][�

residualz, iztrans

� Applications of z-Transform

• Solution of Linear Difference Equations

• Discrete-Time LTI Systems

• Characterisation of Discrete-Time LTI Systems

• Frequency Response Analysis

Today’s goal

��

Applications of z-Transform

Given the following difference equation

where y and x are the output and the input variables, respectively.

� Solution of linear difference equations

Mnxbn-xbnxb

Nnyanyanyany

M

N

][...]1[][

...][...]2[]1[][

10

21

��

��

��

i) Taking z-transform to both sides and rearranging gives

)()(

)()(

...1

...)(

1

1

1

10 zXzA

zBzX

zaza

zbzbbzY

N

N

M

M ��

� ��

��

ii) The solution of the difference equation is

y[n] = Z-1{Y(z)}


� Solution of linear difference equations by MATLAB

For N � M, the output of the system described by

can be found using the MATLAB command ‘filter’.

y=filter(b,a,x,zi) .

Mnxbn-xbnxb

Nnyanyanyany

M

N

][...]1[][

...][...]2[]1[][

10

21

��

��

��

y=filter(b,a,x,zi) .

b = [b0 b1 … bm]

a = [1 a1 a2 … an ]

x is the input vector.

zi is the initial condition vector = [zi(1) zi(2) … zi(N)] and

where

)1()(

)1(...)2()1()2(

)(...)2()1()1(

32

21

��

��

��

yaNzi

nyayayazi

nyayayazi

N

N

N

�


EXAMPLE : Zero initial conditions

]1[][]2[1.0]1[7.0][ �� nununynyny

If u[n] is the unit step and suppose that y[n]=0, n<0, compute the output

response y[n].

��2

)()()( zUzGzY �

��

��

��

��

11.07.0)(

2

2

z

z

zz

zzzY

][))2.0(3

2)5.0(

3

5(][ nuny nn ��

2.03

2

5.03

5)(

��

��

z

z

z

zzY


0.5

0.6

0.7

0.8

0.9

1

y[n

]

b=[1 -1];

a=[1 -0.7 0.01];

n=0:20;

x=ones(1,length(n));

y=filter(b,a,x);

Filter command:

��

0 2 4 6 8 10 12 14 16 18 200

0.1

0.2

0.3

0.4

n

y[n

]

Analytical solution:

for ii=0:length(u)-1

y1(ii+1)=5/3*(0.5^ii)-2/3*(0.2^ii);

end


EXAMPLE : Non-zero initial conditions

]1[][]2[1.0]1[7.0][ �� nununynyny

If u[n] is the unit step and suppose that y[-2]=0,y[-1]=1, compute the output

response y[n].

]1[]2[][1.0]1[7.0]2[ �� nununynyny

• Replacing n with n+2:

��

��

��

��

��

��

2.08.0

5.05.2

1.07.0

)1.07.1()(

2z

z

z

z

zz

zzzY

• With y[0]=1.7 and y[1]=1.09,

][))2.0(8.0)5.0(5.2(][ nuny nn ��


b=[1 -1];

a=[1 -0.7 0.1];

n=0:20;

x=ones(1,length(n));

zi(1)=-a(2)*1;

zi(2)=-a(3);

y=filter(b,a,x,zi);

Filter command:

0.8

1

1.2

1.4

1.6

1.8

y[n

]

y=filter(b,a,x,zi);

Analytical solution:

for ii=0:length(u)-1

y1(ii+1)=2.5*(0.5^ii)-0.8*(0.2^ii);

end

0 2 4 6 8 10 12 14 16 18 200

0.2

0.4

0.6

0.8

n

Discrete-Time LTI Systems

��

h[n] is the impulse response function. The z-transform of h[n] is referred to

as the transfer function.

)(

)()(

zX

zYzH �

Difference Equation to Transfer Function

• Given a difference equation

Mnxbn-xbnxb

Nnyanyanyany

M

N

][...]1[][

...][...]2[]1[][

10

21

��

��

��

If all initial conditions are zero, we find the transfer function to be

)(

)(

...1

...)(

1

1

1

10

zA

zB

zaza

zbzbbzH

N

N

M

M ��

� ��

��

EXAMPLE For the unit sample response

� ][])5.0(5.0[][ nunhnn ��

, find the transfer function H(z).

� ]}[)5.0(][5.0{)(1

nunuZzHnn ��

Solution

��

� ]}[)5.0(][5.0{)( nunuZzH ��

25.0

2

5.05.0)(

2

2

��

��

��

z

z

z

z

z

zzH

Gain, Poles & Zeros

))...()((

))...()(()(

21

21

p

z

n

n

pzpzpz

zzzzzzKzH

��

��

A transfer function can be factored into

• K is called the system gain.

• zi, i=1,…,nz is called the system zeros.

��

• zi, i=1,…,nz is called the system zeros.

• pi, i=1,…,np is called the system poles.

x

5.0)(

��

z

zzH

Characterisation of Discrete-Time LTI Systems

� Causality

• A system is causal if the output sequence value at n=n0 depends only on the

input sequence values for n� n0, for every choice of n0, i.e. the output

sequence value is only dependent on past or current inputs.

• For a causal discrete-time LTI system, we have

��

0,0][ �� nnh


� Stability

• A system is stable in the bounded-input, bounded-output (BIBO) sense if

and only if every bounded input sequence produces a bounded output

sequence.

• It can be shown that a discrete-time LTI system is BIB0 stable if its

impulse response is absolutely summable, that is,

��

impulse response is absolutely summable, that is,

��

��n

nh ][

� Stability

A discrete-time system is stable, if and only if

pi nip ,...,1for 1|| ��

• Stability condition


��

pnppp ,...,, 21where are the poles of H(z).

� Marginal Stability

A discrete-time system is marginally stable if and only if

and poles dnonrepeate allfor 1|| �ip

• Marginal Stability Condition


��

poles. repeated allfor 1|| �i

i

p

Given a first-order system with a pole at z = a,

az

zzH

��)( n

anh �][

EXAMPLE


��

0 5 100

0.2

0.4

0.6

0.8

1

n

a=1.0

X

Stable Marginally Stable Unstable

• The pulse response for various pole locations


��


� Relationship between ZT & DTFT

��

��

��n

njenxX ][)( �

�

��

��n

nznxzX ][)(

DTFT ZT

• In going from the DTFT to the ZT we replace by .�j

e z

��

• This evaluation is equivalent to evaluating the z-transform on the unit circle in

the complex plane.

• Also, replacing with , ZT will become DTFT.z �je

• Therefore, if unit circle is in the ROC � DTFT exists.


� Frequency Response Analysis

Consider a DT transfer function H(z), the discrete frequency response function

(FRF) is

�� jez

jzHeHH |)()()(

��

• is the magnitude or gain of the FRF. |)(| �H

• is the phase of the FRF. )(��H

where � is the discrete frequency in rad/sample.


� Response to a sinusoidal input

If a DT system is stable with transfer function H(z), then in steady-state

Anx �][ )0(][ AHny �

��

nAnx 0sin][ �� ))(sin(|)(|][ 000 �� HnHAny

nAnx 0cos][ �� ))(cos(|)(|][ 000 �� HnHAny


• EXAMPLE

5.0)(

��

z

zzH

5.0)(

��

�

j

j

e

eH

��

5.0��je

��

��sin)5.0(cos

sincos)(

j

jH


• EXAMPLE

��

��sin)5.0(cos

sincos)(

j

jH

• If a sinusoidal input is appliednnx3

sin][�

�

1.1547,|)3/(| ��H �30)3/( �� H

��

)303

1.1547sin(][ �� nny�

6 8 10 12 14 16 18 20 22 24-1.5

-1

-0.5

0

0.5

1

1.5

n

u[n]

y[n]


� Bode plot

A Bode plot in the discrete time is a graph of |H(�)| and �H(�) plotted

as a function of �, where � is usually ranging from 0 to �.

��

��sin)5.0(cos

sincos)(

j

jH

5.0)(

��

z

zzH

num = 1; % numerator

den = [1 -0.5]; % denominator

freqz(num,den); % DT frequency response

��

�� sin)5.0(cos j

0 0.2 0.4 0.6 0.8 1-30

-20

-10

0

Normalized Frequency (�� rad/sample)

Phase (

degre

es)

0 0.2 0.4 0.6 0.8 1-5

0

5

10

Normalized Frequency (�� rad/sample)

Magnitude (

dB

)

freqz(num,den); % DT frequency response

Summary

��

applications of zapplications of z--transform...

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