discrete-time system analysis using the z...

Discrete-Time System Analysis Using the z-Transform

S Wongsa

11

S Wongsa

Dept. of Control Systems and Instrumentation Engineering,

KMUTT

Overview

� z-Transform

� Definition

� Some properties

� The Inverse z-Transform

� Partial fraction expansion

22

� Applications of z-Transform

� Solution of Linear Difference Equations

� Characterisation of Discrete-Time LTI Systems

- Transfer function, zero-pole-gain

- Causality, Stability

z-Transform

Given a DT signal x[n], its z-transform is defined as

∑∞

−∞=

−=n

nznxzX ][)(

where z is a complex variable , called the z-transform variable.

∑∞

=

−=0

][)(n

nznxzX

One-sided / Unilateral z-TransformTwo-sided / Bilateral z-Transform

ωjrez =

33

� The notation

)(][ zXnx ⇔

� If

))...()((

))...()(()(

21

21

p

z

n

n

pzpzpz

zzzzzzzX

−−−

−−−=

- zi, i=1,…,nz, are called the zeros of X(z).

- pi, i=1,…,np, are called the poles of X(z).

The Region of Convergence

• The set of values of z for which the sum converges and X(z) exists is called

the region of convergence or ROC.

EXAMPLE :

,][z

nuan ⇔ |||| az >

Causal exponential

44

,][az

znuan

−⇔ |||| az >

|||| az <

Left-sided exponential

,]1[az

znuan

−⇔−−−

Some Common z-Transform Pairs

55

The Inverse z-Transform

∫ −=C

n dzzzXj

nx 1)(2

1][

π

where C is a closed contour that includes all poles of X(z).

Definition

66

� If X(z) is in the rational form, i.e. X(z)=B(z)/A(z), we can compute the inverse

z-transform using the partial fraction expansion.

� But doing it requires integration in the complex plane and it is rarely used in engineering

practice.


• Partial Fraction Expansion

Given

))...()((

))...()(()(

21

21

p

z

n

n

pzpzpz

zzzzzzzX

−−−

−−−=

• It is assumed that nz ≤≤≤≤ np (proper form) before carrying out the next step.

If this is not the case, carry out the long division until it is.

77

i) If the poles are distinct and are all nonzero:

expand X(z) into a sum of terms

p

p

n

n

pz

zc

pz

zcczX

−++

−+= ...)(

1

10

where

00 |)( == zzXc

ipzi

i zXz

pzc =

−= |)(

)(for pni ≤≤1



EXAMPLE : Distinct poles

2

102)(

2

2

−−+

=zz

zzX

345)( ++−=zz

zX ][)2(3][)1(4][5][ nununnx nn +−+−= δ

88

23

145)(

−+

++−=

z

z

z

zzX ][)2(3][)1(4][5][ nununnx nn +−+−= δ

...20]3[,16]2[,2]1[,2]0[ ==== xxxx



EXAMPLE : Complex poles

)2)(1(

1

2

1)(

2

3

23

3

−+++

=−−−

+=

zzz

z

zzz

zzX

• The poles of X(z) are . 2,866.05.0 j±−

99

• Expanding X(z) gives

2

643.0

866.05.0

)0825.0429.0(

866.05.0

)0825.0429.0(5.0)(

−+

−+−

+++

++−=

z

z

jz

zj

jz

zjzX

• The inverse z-transform of X(z) is

][)2(643.0][89.103

4cos874.0][5.0][ nununnnx n+

++−= oπδ



ii) Repeated poles:

If the pole p1 is repeated r times and the other np-r poles are distinct,

pnr

r

r

pz

zc

pz

zc

pz

zc

pz

zc

pz

zcczX

−++

−+

−++

−+

−+= + ...

)(...

)()( 1

2

210

1010

pnr

r pzpzpzpzpzczX

−++

−+

−++

−+

−+=

+

...)(

...)(

)(11

2

11

0

where

1

)()(

!

11

pz

r

i

i

irz

zXpz

dz

d

ic

=

−

−= for 1,...,0 −= ri

• The constants are computed in the same way as

in the distinct pole case.pnrr cccc ,..., and 2,10 ++



EXAMPLE : Repeated poles

)1()1(

26

1

26)(

2

23

23

23

+−−+

=+−−−+

=zz

zzz

zzz

zzzzX

75.05.325.5 zzz

1111

1

75.0

)1(

5.3

1

25.5)(

2 ++

−+

−=

z

z

z

z

z

zzX

][)1(75.0][5.3][25.5][ nunnununx n−++=

Some properties of the z-Transform

Properties Sequence1 z-transform

Linearity

Time shifting

)()( zbYzaX +][][ nbynax +

][ knx −

][ knx + ∑−

=

−−1

0

][)(k

i

ikk zixzXz

∑=

−− −+k

i

kik zixzXz1

][)(

1212

Scaling

Differentiation

Convolution

Initial value

Final value

theorem

1All sequences are causal.

][nxan )/( azX

∑∞

−∞=

−l

lnylx ][][ )()( zYzX

][nnx )(zXz &−

exists.limit theif ),(lim]0[ zXx z ∞→=

circle.unit theinside are )()1( of poles theif

),()1(limor )()1(lim][lim 1

11

zXz

zXzzXznx zzn

−

−−= −→→∞→


EXAMPLE : Revisit the distinct poles example

)2.0)(1.0(

1.02

02.01.0

1.02)(

2

2

2

−++

=−−

+=

zz

z

zz

zzX

][)2.0(3][)1.0(4][5][ nununnx nn +−+−= δ

1313


EXAMPLE : Initial Value & Final Value

Find x[0] and x[∞] given

� ][∞x

�

2)(lim]0[ == ∞→ zXx z

]0[x

)2.0)(1.0(

1.02

02.01.0

1.02)(

2

2

2

−++

=−−

+=

zz

z

zz

zzX

1414

�

0

)()1(lim][lim 1

=

−= →∞→ zXznx zn

][∞x

][)2.0(3][)1.0(4][5][ nununnx nn +−+−= δNB:


� Z-transform Partial Fraction Expansion Using MATLAB

Z-transform partial-fraction expansion can be computed using the MATLAB

command ‘residuez’ (X(z) in ascending powers of z-1)

02.01.0

1.02)(

2

2

−−+

=zz

zzX

21

2

02.01.01

1.02)( −−

−

−−+

=zz

zzX

1515

>> num=[2 0 0.1];

>> den=[1 -0.1 -0.02];

>> [r,p,k]=residuez(num,den)

r = 3

4

p = 0.2000

-0.1000

k = -5

11 2.01

13

1.01

145)( −− −

++

+−=zz

zX

2.03

1.045)(

−+

++−=

z

z

z

zzX

By calculation:


• The Inverse z-Transform Using MATLAB

x =iztrans(X) is the inverse Z-transform of the scalar sym X with default

independent variable z.

02.01.0

1.02)(

2

2

−−+

=zz

zzX ][)2.0(3][)1.0(4][5][ nununnx nn +−+−= δ

1616

>> syms X x z

>> X=(2*z^2+0.1)/(z^2-0.1*z-0.02);

>> x=iztrans(X)

x =

3*(1/5)^n + 4*((-1/10))^n - 5*kroneckerDelta(n, 0)

Overview

� z-Transform

� Definition

� Some properties



1717






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

−+++

+−−−−−−−=

1818

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)}

(If all initial conditions are zero)


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[)(1.0]1[)(7.0)( 1121 −+−=−+−++−+− −−−− uzUzzUyzyzYzyzYzzY

1919NB: ][ knx − ∑

=

−− −+k

i

kik zixzXz1

][)(

( ) 111

11.07.0)()1.07.01(

1

1

1

121 =−

−−

=+−++−−

−

−−−−

z

z

zzzYzz

121 1.07.1)()1.07.01( −−− −=+− zzYzz


EXAMPLE : Non-zero initial conditions

−=

−= 8.05.2

)1.07.1()(

zzzzzY

121 1.07.1)()1.07.01( −−− −=+− zzYzz

2020

−

−−

=+−

=2.0

8.05.0

5.21.07.0

)(2 zzzz

zY

][))2.0(8.0)5.0(5.2(][ nuny nn −=


• 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

−+++

+−−−−−−−=

2121

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

M


• NB

The initial condition vector can be formulated by using the command ‘filtic’

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

b = [b0 b1 … bM]

a = [1 a a … aN ]

where

2222

a = [1 a1 a2 … aN ]

The vectors x and y contain the most recent input or output first, and oldest input

or output last.

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

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

Nyyyy

Mxxxx

−−−=

−−−=


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[-2]=0,y[-1]=1, )1()(

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

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

32

21

−−=

+−−−−−−−=

−−−−−−−=

yaNzi

Nyayayazi

Nyayayazi

N

N

N

M

]1[][]2[1.0]1[7.0][ −−=−+−− nununynyny

2323

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

Analytical solution:

for ii=0:length(x)-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

zi = filtic(b,a,[1 0]);

NB: zi can also be obtained by

][))2.0(8.0)5.0(5.2(][ nuny nn −=

Overview

� z-Transform

� Definition

� Some properties



2424






Discrete-Time LTI (Linear-Time Invariant) Systems

� Linearity : defined by the principle of superposition

LTI x1[n] y1[n] LTI x2[n] y2[n]

2525

LTI ax1[n]+bx2[n] ay1[n]+by2[n]

Scaling & additivity

properties

Discrete-Time LTI (Linear-Time Invariant) Systems

� Time Invariant

LTI x[n] y[n] LTI x[n-k] y[n-k]

2626

Discrete-Time LTI Systems

2727

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

transfer function.

)(

)()(

zX

zYzH =

EXAMPLE For the unit impulse response

( ) ][])5.0(5.0[][ nunh nn ⋅−+=

, find the transfer function H(z).

( ) ]}[)5.0(][5.0{)( nunuZzH nn ⋅−+⋅=

Solution

2828

( ) ]}[)5.0(][5.0{)( nunuZzH ⋅−+⋅=

25.0

2

5.05.0)(

2

2

−=

++

−=

z

z

z

z

z

zzH

Difference Equation to Transfer Function

• Given a difference equation

Mnxbn-xbnxb

Nnyanyanyany

M

N

][...]1[][

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

10

21

−+++

+−−−−−−−=

2929

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 :

]1[][]2[1.0]1[7.0][ −−=−+−− nununynyny

Find the transfer function of the difference equation above.

1.07.0

)1(

1.07.01

1

)(

)()(

)()()(1.0)(7.0)(

221

1

121

+−−

=+−

−==

−=+−

−−

−

−−−

zz

zz

zz

z

zU

zYzH

zUzzUzYzzYzzY

3030

1.07.01.07.01)( 221 +−+− −− zzzzzU

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.

3131

• 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 output at time n depends only on present and past inputs but

not on future.

� For real-time processing the system needs to be causal.

EXAMPLE

3232

]1[][][ −+= nxnxny

]1[][][ ++= nxnxny

Causal system

Noncausal system


� Causality condition

For a causal discrete-time LTI system, we have

0,0][ <= nnh

3333Pictures from Steven W. Smith, The Scientist and Engineer's Guide to Digital Signal Processing.


� 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.

When |x[n]| < B , if |y[n]|<∞, the system is stable.

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

3434

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

impulse response is absolutely summable, that is,

∞<∑∞

−∞=n

nh ][

Proof: Stability

We get

2121 zzzz +≤+

zzzz =

3535

using property that

If the impulse response is absolutely summable, i.e.

Then , i.e. y[n] is bounded.∞<][ny

2121 zzzz =

� Stability

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

pi nip ,...,1for 1|| =<

� Stability condition


3636

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


3737

poles. repeated allfor 1|| <i

i

p

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

az

zzH

−=)( nanh =][

EXAMPLE


3838

0 5 100

0.2

0.4

0.6

0.8

1

n

a=1.0

X

Stable Marginally Stable Unstable

discrete-time system analysis using the z...

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