Lecture Fundamentals of control systems: Chapter 8 - TS. Huỳnh Thái Hoàng

Lecture Notes

Fundamentals of Control Systems
Instructor: Assoc. Prof. Dr. Huynh Thai Hoang
Department of Automatic Control
Faculty of Electrical & Electronics Engineering
Ho Chi Minh City University of Technology
Email:

Homepage: www4.hcmut.edu.vn/~hthoang/
www4 hcmut edu vn/ hthoang/
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Chapter 8

ANALYSIS OF
DISCRETE CONTROL SYSTEMS

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Content

Stability conditions for discrete systems
 Extension of Routh-Hurwitz criteria
 Jury
J
criterion
it i
 Root locus
 Steady
St d state
t t error
 Performance of discrete systems


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Stability conditions for discrete systems

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Stability conditions for discrete systems



A system is defined to be BIBO stable if every bounded
input to the system results in a bounded output.

I s
Im

Stable

Res  0

I z
Im

Re s

| z | 1

Re z
1

z  eTs

The region of stability for a
contin o s system
continuous
s stem is the
left-half s-plane
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Stable

The region of stability for a
di
discrete
t system
t
iis th
the
interior of the unit circle

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Characteristic equation of discrete systems


Discrete systems described by block diagram:
R(s)
+
GC(z)
ZOH
G(s)

T

Y(s)

H(s)
 Characteristic equation: 1  GC ( z )GH ( z )  0


Discrete systems described by the state equation

 x( k  1)  Ad x( k )  Bd r ( k )

 y ( k )  Cd x( k )
 Characteristic equation: det( zI  Ad )  0
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Methods for analysis the stability of discrete systems


Algebraic stability criteria
 The extension of the Routh-Hurwitz criteria
 Jury’s
J ’ stability
t bilit criterion
it i



The root locus method

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The extension of the RouthRouth-Hurwitz criteria

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The extension of the Routh
Routh--Hurwitz criteria


Characteristic
C
a acte st c equat
equation
o o
of d
discrete
sc ete syste

systems:
s

a0 z n  a1 z n 1    an  0
Im z

Im w

Region
R
i
off
stability

Re z
1



Region of
stability

1 w
z
1 w

Re w

The extension of the Routh-Hurwitz criteria: transform
zw,, and then apply

pp y the Routh – Hurwitz criteria to the
characteristic equation of the variable w.

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The extension of the Routh
Routh--Hurwitz criteria – Example


Analyze the stability of the following system:
R(s)

+



T  0.5

ZOH

G(s)

Y(s)

H(s)

3e  s
Gi
Given
that:
th
t G( s) 
s3


1
H ( s) 
s 1

Solution:
Sol
tion
The characteristic equation of the system:
1  GH ( z )  0

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The extension of the RouthRouth-Hurwitz criteria – Example (cont’)
s
G
(

s
)
H
(
)
s
3
e


 GH ( z )  (1  z 1 )Z 
G ( s) 

s
( s  3)


s


3e
3
e
1
1
 (1  z )Z 

H (s) 
( s  1)
 s ( s  3)( s  1) 

z ( Az  B)
1  2
 3(1  z ) z
( z  1)( z  e 30.5 )( z  e 10.5 )
(1  e 30.5 )  3(1  e 0.5 )
A
 0.0673


1
z ( Az  B)
3(1  3)

Z



 s3(s0.5 a)( s  b) 

( z  1)( z  e aT )( z  e bT )

3e 30.5 (1  e 0.5 )  e 0.5 (1  e
) aT
B
b(1  e  0).0346
a(1  e bT )
A
3(1  3)
ab(b  a)
0.202 z  0.104


aeaT (1  e bT )  be bT (1  e aT )
GH ( z )  2
z ( z  0.223)( zB  0.607) ab(b  a)

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The extension of the RouthRouth-Hurwitz criteria – Example (cont’)

 The characteristic equation:

1  GH ( z )  0





0.202 z  0.104
1 2
0
z ( z  0.223)( z  0.607)



z  0.83z  0.135z  0.202 z  0.104  0

4

3

2

1 w
Perform the transformation: z 
1 w

1 w 
1 w 
1 w 
1 w 
  0.83
  0.135
  0.202
  0.104  0
1 w 
1 w 
1 w 
 1 z w
0.202
 0.104
GH ( z )  2
G
z ( z  0.223)( z  0.607)
4
3
2


4

3

2



1.867 w  5.648w  6.354 w  1.52 w  0.611  0

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The extension of the RouthRouth-Hurwitz criteria – Example (cont’)



The Routh table



Conclusion: The system is stable because all the
terms in the first column of the Routh table are
positive
positive.

4
3
2

1.867 w  5.648w  6.354 w  1.52 w  0.611  0

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Jury stability criterion

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Jury stability criterion


Analyze the stability of the discrete system which has
the characteristic equation:

a0 z n  a1 z n 1    an 1 z  an  0


Jury table: consist of (2n+1) rows.
 The first row consists of the coefficients of the
characteristic polynomial in the increasing index order.
 The even row (any) consists of the coefficients of the
previous row in the reverse order.
 The odd row i = 2k+1 (k1) consists (nk+1) terms,
the term at the row i column j defined by:

1 ci 2,1 ci 2,n  j k 3
cij 
ci 2,1 ci 1,1 ci 1,n  j k 3
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Jury stability criterion (cont’)


Jury criterion statement: The necessary and
sufficient condition for the discrete system to be
stable
t bl is
i that
th t allll the
th first
fi t tterms off th

the odd
dd rows off th
the
Jury table are positive.

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Jury stability criterion – Example


Analyze the stability of the system which has the characteristic
equation:
3
2

5 z  2 z  3z  1  0



Solution: Juryy table
Row 1
Row 2
R
Row
3

Row 4
Row 5
Row 6
Row 7



Since all the first terms of the odd rows are positive, the system is stable.
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The root locus of discrete systems

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The root locus (RL) method


RL is a set of all the roots of the characteristic equation of
a system when a real parameter changing from 0  +.



Consider a discrete system which has the characteristic
equation:

N ((zz )
1 K
0
D( z )
Denote: G0 ( z )  K

N ( z)
D( z )

Assume that G0(z) has n poles and m zeros.


The rules for construction of the RL of continuous system
y
can be applied to discrete systems, except for the step 8.

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Rules for construction of the RL of discrete systems


Rule 1: The number of branches of a RL = the order of the
characteristic equation = number of poles of G0(z) = n.



Rule 2:
 For K = 0: the RL begin at the poles of G0(z).
 As
A K goes to
t + : m branches
b
h off th
the RL end
d att m zeros
of G0(z), the nm remaining branches goes to 
approaching the asymptote defined by the rule 5 and
rule 6.



Rule 3: The RL is symmetric with respect to the real axis.

 Rule

4: A point on the real axis belongs to the RL if the
t t l number
total
b off poles
l and

d zeros off G0(z)
( ) to
t its
it right
i ht is
i odd.
dd

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Rules for construction of the RL of discrete system (cont’)


Rule 5: The angles between the asymptotes and the real
axis are given by:
( 2l  1)





nm

Rule 6: The intersection between the asymptotes and
the real axis is a point A defined by:

poles   zeros

OA 

nm



(l  0,1,2,)

n

m

i 1

i 1

(pi and zi are
p

z
 i i

nm

poles
l and
d zeros

of G0(z) )

Rule 7: : Breakaway / break-in points (or
break points for short), if any, are located in
the real axis and are satisfied the equation:

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dK
0
d
dz
21


Rules for construction of the RL of discrete system (cont’)


Rule 8: The intersections of the RL with the unit circle can
be determined by using the extension of the Routh-Hurwitz
criteria or by substituting z=a+jb (a2+b2 =1) into the
characteristic equation.
9: The departure angle of the RL from a pole pj (of
multiplicity 1) is given by:

 Rule

m

 j  1800   arg(( p j  zi ) 
i 1

n

 arg(( p

i 1,i  j

j

 pi )

The geometric
Th
t i form
f
off the
th above
b
formula
f
l is
i
j = 1800 + (angle from zi (i=1..m) to pj )
 (angle
(
l pi (i=1..m,
(i 1

i≠j) to
t pj )
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The root locus of discrete systems – Example


Consider a discrete system described by a block diagram:
R(s)

+



T  0.1

ZOH

G(s)
( )

Y(s)

5K
G( s) 

s( s  5)


Sketch the RL of the system when K=0+. Determine
the critical gain Kcr



Solution:The characteristic equation of the system:

1  G( z)  0
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The root locus of discrete systems – Example (cont’)
 G(s) 
5K
5
K
 G ( z )  (1  z )Z 
G ( s) 

 s 
s ( s  5)
 5K 
1

 (1  z )Z  2

 s ( s  5) 
0 . 5
) z  (1  e 0.5  0.5e 0.5 )] 
1  z[(0.5  1  e

 K (1  z )
 0 .5
2
5( z  1) ( z  e )


0.021z  0.018
 G( z)  K
( z  1)( z  0.607)
1



Th characteristic
The
h
t i ti equation
ti :

0.021z  0.018
1 K
 0 (*)
( z  1)( z  0.607)



 aT
 aT
 aT


a
z
(
aT

1

e
)
z

(
1

e

aTe
)
 Poles: p1  1 p2 Z 0.607

 2


2
 aT
s
(
s

a
)
a
(
z

1
)
(
z

e
)

 Zeros: z1  0.857 

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

The root locus of discrete systems – Example (cont’)


The asymptotes:

(2l  1) (2l  1)
  


2 1
nm
poles   zeros [1  0.607]  ( 0.857)

 OA  2.464

OA 
nm
2 1


The breakaway/break-in points:
((*))

Then

( z  1)( z  0.607)
z 2  1.607 z  0.607
 K 


0.021z  0.018
0.021z  0.018
dK
0.021z 2  0.036 z  0.042


dz
(0.021z  0.018) 2
dK
0
dz

6 December 2013



 z1  2.506

 z2  0.792

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