Lecture "Fundamentals of control systems - Chapter 8: Analysis of discrete control systems" presentation of content: Stability conditions for discrete systems, extension of Routh - Hurwitz criteria, jury criterion, root locus,... Invite you to reference.
Trang 1Lecture 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: hthoang@hcmut.edu.vn
huynhthaihoang@yahoo.com Homepage: www4 hcmut edu vn/ hthoang/
Homepage: www4.hcmut.edu.vn/~hthoang/
Trang 2Chapter 8
ANALYSIS OF DISCRETE CONTROL SYSTEMS
Trang 3 Stability conditions for discrete systems
Content
Stability conditions for discrete systems
Extension of Routh-Hurwitz criteria
Jury criterion
Root locus
St d t t
Steady state error
Performance of discrete systems
Trang 4Stability conditions for discrete systems
Trang 5Stability conditions for discrete systems
A system is defined to be BIBO stable if every bounded
A system is defined to be BIBO stable if every bounded input to the system results in a bounded output.
z
0
The region of stability for a
contin o s s stem is the
The region of stability for a
Trang 6Characteristic equation of discrete systems
Discrete systems described by block diagram:
(
)()
()
1(
k k
y
k r k
k
d
d d
x C
B x
A x
Characteristic equation: det(z I A ) 0
Trang 7Methods for analysis the stability of discrete systems
Algebraic stability criteria
Algebraic stability criteria
The extension of the Routh-Hurwitz criteria
J ’ t bilit it i
Jury’s stability criterion
The root locus method
Trang 8The extension of the
The extension of the Routh Routh Hurwitz criteria Hurwitz criteria
Trang 9The extension of the
The extension of the Routh Routh Hurwitz criteria Hurwitz criteria
Characteristic equation of discrete systems:
Im z
C a acte st c equat o o d sc ete syste s
0
1 1
n
n n
a z
a z
1
Re w
Region of stability
The extension of the Routh-Hurwitz criteria : transform
z w, and then apply the Routh – Hurwitz criteria to the , pp y
characteristic equation of the variable w.
Trang 10The extension of the
The extension of the Routh Routh Hurwitz criteria Hurwitz criteria – – Example Example
Analyze the stability of the following system:
Analyze the stability of the following system:
T
3)
(
e G
G
1
1)
(
1 GH z
Trang 11The extension of the
The extension of the Routh Routh Hurwitz criteria Hurwitz criteria – – Example (cont’) Example (cont’)
G z
3 )
G
s
) (
3 (
3 )
1
s s
s
e
) 1 (
1 )
) )(
)(
1 (
)
( )
1 (
e z
z
B Az
z z
z
0673 0
) 1
( 3 )
1
( e30.5 e0.5
A
) )(
)(
1 (
) (
) )(
(
1
e z e
z z
B Az
z b
s a s
) 1
( )
1 ( 3
0673
0 )
3 1 ( 3
) (
) (
5 0 3 5
0 5
0 5
0
e
e B
A
) 1
( )
1 (
) (
) 1
( )
1 (
e be
e ae
a b ab
e a
e b
A
aT bT
bT aT
bT aT
3 1 ( 3
) (
) 1
( )
1 (
a b ab
e be
e ae
0 )(
223
0 (
z z
GH
Trang 12The extension of the
The extension of the Routh Routh Hurwitz criteria Hurwitz criteria – – Example (cont’) Example (cont’)
The characteristic equation:
The characteristic equation:
0)
(
1 GH z
0)
607
0)(
223
0(
104
0202
11
1 1
1
104
0 202
.
0 )
z GH
0 104
.
0 1
1 202
0 1
1 135
0 1
1 83
0 1
w w
w w
w
) 607
0 )(
223
0 (
G
2 3
Trang 13The extension of the
The extension of the Routh Routh Hurwitz criteria Hurwitz criteria – – Example (cont’) Example (cont’)
The Routh table
The Routh table
Conclusion: The system is stable because all theterms in the first column of the Routh table arepositive
0 611
0 52
1 354
6 648
5 867
.
1 w4 w3 w2 w
positive
Trang 14Jury stability criterion
Trang 15Jury stability criterion
Analyze the stability of the discrete system which has
Analyze the stability of the discrete system which has
the characteristic equation:
0
1
1 1
n n
n n
a z
a z
a z
Jury table: consist of (2n+1) rows.
The first row consists of the coefficients of the
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 (nk+1) terms,
the term at the row i column j defined by:
the term at the row i column j defined by:
3 ,
2 1
, 2
3 ,
1 1
, 1 1
c
Trang 16Jury stability criterion (cont’)
Jury criterion statement: The necessary and
sufficient condition for the discrete system to be
t bl i th t ll th fi t t f th dd f thstable is that all the first terms of the odd rows of the Jury table are positive
Trang 17Jury stability criterion
Jury stability criterion – – Example Example
Analyze the stability of the system which has the characteristic
Solution: Jury table
Analyze the stability of the system which has the characteristic equation:
01
32
Trang 18The root locus of discrete systems
Trang 19The root locus (RL) method
RL is a set of all the roots of the characteristic equation of
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:
)
(z
N
0)
z
N K
)(
)
()
(
0
z D
z
N K z
Denote:
The rules for construction of the RL of continuous system
Assume that G0(z) has n poles and m zeros.
y can be applied to discrete systems, except for the step 8.
Trang 20Rules for construction of the RL of discrete systems
Rule 1: The number of branches of a RL = the order of the
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 K goes to + : m branches of the RL end at m zeros
of G0(z) , the n m remaining branches goes to
approaching the asymptote defined by the rule 5 and
approaching the asymptote defined by the rule 5 and
rule 6
Rule 3: The RL is symmetric with respect to the real axis.
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
total number of poles and zeros of G0(z) to its right is odd.
Trang 21Rules for construction of the RL of discrete system (cont’)
Rule 5: The angles between the asymptotes and the real
Rule 5: The angles between the asymptotes and the real axis are given by:
m n
l
(2 1) (l 0,1,2,)
Rule 6: The intersection between the asymptotes and
the real axis is a point A defined by:
m
n
the real axis is a point A defined by:
z p
p m
n
i i
Rule 7: : Breakaway / break-in points (or
break points for short), if any, are located in 0
d dK
the real axis and are satisfied the equation: dz
Trang 22Rules for construction of the RL of discrete system (cont’)
Rule 8: The intersections of the RL with the unit circle can
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
Rule 9: The departure angle of the RL from a pole p j (of
Rule 9: The departure angle of the RL from a pole p j (of
multiplicity 1) is given by:
0
)(
)(
i j
i
i j
, 1 1
0
)arg(
)arg(
Trang 23The root locus of discrete systems
The root locus of discrete systems – – Example Example
Consider a discrete system described by a block diagram:
Consider a discrete system described by a block diagram:
5)
(
s s
K s
(
1 G z
Trang 24The root locus of discrete systems
The root locus of discrete systems – – Example (cont’) Example (cont’)
5 )
(
s s
K s
5 )
1
s s
( )
1 5 0
[(
) 1
(
5 0 5
0 5
) 1 (
5
)]
5 0 1
( )
1 5 0
[(
) 1
e z
z
e e
z e
z z
K
K z
G
) 607
0 )(
1 (
) (
z z
K z
0 )(
1 (
Poles: p1 1 p2 0 607
) (
) 1 (
)
e z z
a a
Trang 25The root locus of discrete systems
The root locus of discrete systems – – Example (cont’) Example (cont’)
The asymptotes:
The asymptotes:
1 2
) 1 2
( )
1 2
1 2
) 857
0 ( ] 607
0 1 [
OA poles
464
0 021
0 018
0 021
.
z z
0 036
0 021
0 021
0
Trang 26The root locus of discrete systems
The root locus of discrete systems – – Example (cont’) Example (cont’)
The intersection of the root locus with the unit circle:
The intersection of the root locus with the unit circle:
(*) (z 1)(z 0.607) K(0.021z 0.018) 0
(**)
0)
607
0018
.0()
607
1021
.0(
z
Method 1: Apply the extension of Routh – Hurwitz criteria:
Perform the transformation
607
0018
.0(
1)
607
1021
.0(
w
0)
607
0018
.0
(1
)607
1021
.0
Trang 27The root locus of discrete systems
The root locus of discrete systems – – Example (cont’) Example (cont’)
According to the corollary of the Hurwitz criterion the stability conditions are:
0214
3
0036
.0786
0
0
K K
K K
1485
1
8187
05742
Trang 28The root locus of discrete systems
The root locus of discrete systems – – Example (cont’) Example (cont’)
Method 2: S ubstitute z = a + jb into (**) :
Method 2: S ubstitute z = a + jb into (**) :
0 )
607
0 018
0 ( )
)(
607
1 021
0 ( )
6070
0180
0)
607
0018
.0
6070
0180
()
6071
0210
(
2
K
K 1.607) (0.018 0.607) 0021
.0(
z
Trang 29The root locus of discrete systems
The root locus of discrete systems – – Example (cont’) Example (cont’)
Combine wit h a2 + b2 =1 we have the set of equations:
Combine wit h a2 + b2 =1, we have the set of equations:
607
1021
.0(2
2 2
b a
b K
j ab
05742
83.21
K
cr K
Trang 30The root locus of discrete systems
The root locus of discrete systems – – Example (cont’) Example (cont’)
Trang 31Frequency response of discrete systems
Trang 32Frequency response of discrete systems
E act freq enc response: s bstit te z e jT into the
Exact frequency response: substitute into the
transfer function G(z)
e
z
)(e j T
Ex: Transfer function:
)6.0(
10)
(
z z
z G
Frequency response:
)6.0(
10)
(
j T j T
T j
e e
e
Draw the exact Bode diagram of discrete systems:
Difficult
Cannot apply the addition property of the Bode plot
Note: Sampling theorem: f s
Note: Sampling theorem: f s
Trang 33Frequency response of discrete systems
Frequency response of discrete systems – – Example Example
Consider the discrete system:
Consider the discrete system:
T
)3(
12)
(
s s
s
G
1 0
T
Plot the frequency response of the open-loop system
Solution:The transfer function of the open-loop system:
Solution:The transfer function of the open-loop system:
z
G( ) (1 1)Z ( )
7410
7411
0493
00544
0)
Frequency response:
741
0741
.1)
e
Trang 34Frequency response of discrete systems
Frequency response of discrete systems – – Example (cont’) Example (cont’)
Exact Bode diagram (Matlab)
Trang 35The bilinear transformation
2/
1 Tw 2 z 1
Bilinear transformation:
2/1
2/1
Tw
Tw z
z
z T
G( ) G(w) wj
Trang 36Relationship between frequency in w
Relationship between frequency in w plane and continuous frequency plane and continuous frequency
12
e T
z
z
T j
z
z T
Trang 37Procedure for drawing approximate Bode diagram
Step 1: Perform the bilinear transformation:
2/
1 Tw
z
2/
1 Tw
z
Step 2: p Substitute , then apply the procedure for w jw pp y p
drawing the Bode diagram presented in chapter 4
w
Trang 38Stability analysis of discrete systems using frequency response
crossover frequency, remember the relationship:
The stability of discrete systems can be analyzed by using
Bode diagrams as continuous systems
Trang 39Performance of discrete systems
Trang 40Time response of discrete systems
Time response of a discrete system can be calculated by
Time response of a discrete system can be calculated byusing one of the two methods below:
Method 1 : if the discrete system described by a transfer
Method 1 : if the discrete system described by a transfer
function, first we calculate Y(z), and then apply the
inverse z-transform to find y(k) y( )
equations, first we find the solutionq x(k)( ) to the stateequations, then calculate y(k)
Dominant poles of a discrete system are the poles lyingclosest to the unit circle
Trang 41y
y
y and y are the maximum and steady state values of y(k)
y max and y ss are the maximum and steady-state values of y(k)
Settling time: ts ksT
where k s satisfying the condition:
k k
y y
k
y( ) . ss
s
k k
y k
y ,
100
)( ss
k k
y k
Trang 42Transient performances
Method 2
Method 2: Analyzing the transient performances based
Method 2
Method 2: Analyzing the transient performances based
on the dominant poles
The dominant poles: z* re j
2 ,
1
)(ln
)(ln
T n
Trang 43Steady state error
(1
)
()
(
z GH z
G
z
R z
Trang 44Performances of discrete system
Performances of discrete system – – Example 1 Example 1
Y( )
R( )
10 )
T
) 3 )(
2 (
) (
s s
s G
1 Calculate the closed-loop transfer function of the system.p y
2 Calculate the time response of the system to step input
3 Evaluate the performance of the system: POT, settlingtime, steady-state error
Solution:
1 The closed-loop TF of the system:p y G cl (z) G(z)
)(1
)
(
z G
cl
Trang 45Performance of discrete system
Performance of discrete system – – Example 1 (cont’) Example 1 (cont’)
2 (
10 )
s G
2(
10)
1
( 1
s s
s
z Z
))(
)(
1(
)
()
1(
e z
z
B Az
z z
) )(
)(
1 (
) (
) )(
(
1
e z e
z z
B Az
z b
s a s
)(
8190
(
036
0042
0)
z G
) 1
( )
1 (
) (
) 1
( )
1 (
e be
e ae
a b ab
e a
e b
A
aT bT
bT aT
bT aT
0)(
819
0(
)(
z
) (
) 1
( )
1 (
a b ab
e be
e ae
Trang 46Performance of discrete system
Performance of discrete system – – Example 1 (cont’) Example 1 (cont’)
036 0
042
)(1
)
()
(
z G
z
G z
0)(
819
0(
036
0042
z
)741
0)(
819
0(
036
0042
01
))(
z
6430
5181
036
0042
0)
z z
G cl
643
0518
z
Trang 47Performance of discrete system
Performance of discrete system – – Example 1 (cont’) Example 1 (cont’)
036
0 042
.
0 )
z G
) ( ) ( )
(z G z R z
2. The time response of the system to step input
643
0 518
1
z
G k
)
( 643
0 518
1
036
0 042
.
0
z z
0 518
1 1
036
0 042
.
0
2 1
2 1
z
R z
0 518
1
(1 1.518 1 0.643 2) ( ) (0.042 1 0.036 2) ( )
z R z
z z
Y z
y(k) 1 518y(k 1 ) 0 643y(k 2 ) 0 042r(k 1 ) 0 036r(k 2 )
) 2 (
036 0
) 1 (
042 0
) 2 (
643 0
) 1 (
518 1
) (k y k y k r k r k
y
y(k) 1 518y(k 1 ) 0 643y(k 2 ) 0 042r(k 1 ) 0 036r(k 2 )
Trang 48Performance of discrete system
Performance of discrete system – – Example 1 (cont’) Example 1 (cont’)
01
)(k k
;
;
;
;
;
;
;
;
;
;
;
61910
62510
63410
64610
66060
67600
68980
69850
69750
68170
64590
58600
) 2 (
036 0
) 1 (
042 0
) 2 (
643 0
) 1 (
518 1
036
0 )
1 (
042
0 )
2 (
643
0 )
1 (
518
1 )
c
Trang 49Performance of discrete system
Performance of discrete system – – Example 1 (cont’) Example 1 (cont’)
Step Response
0.6 0.7
0
Trang 50036 0
042
Performance of discrete system
Performance of discrete system – – Example 1 (cont’) Example 1 (cont’)
3 Transient performances:
2
1 )
(
643
0 518
1
036
0 042
.
0 )
z z
G k
3. Transient performances:
) ( ) 1
) ( ) 1
1 (
0 518
1
036
0 042
.
0 )
1 (
lim
z z
z
z z
z
624
0 624 0
% 100
624
0 6985
0
% 100
Trang 51Performance of discrete system
Performance of discrete system – – Example 1 (cont’) Example 1 (cont’)
Settling time (5% criterion): 0 624
Settling time (5% criterion):
First, we need to find ks satisfying:
1 (k) 1 k k
05 0
% 5
624
k
( ) 0.655 ,593
.0
From the time response calculated before ks 14
1.0
sec4
.1
;
;
;
;
;
k c
6898 0
6985 0
6975 0
6817 0
6459 0
5860 0
Steady state error:
Since the system is unity negative feedback, we have:
;
;
;
;
;
;
0
ss rss yss 1 0.624