Lecture "Fundamentals of control systems - Chapter 9: Design of discrete control systems" presentation of content: Introduction, discrete lead - lag compensator and PID controller, design discrete systems in the Z domain,.... 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
Trang 2Chapter 9
DESIGN OF DISCRETE CONTROL SYSTEMS
Trang 3 Introduction
Content
Introduction
Discrete lead – lag compensator and PID controller
Design discrete systems in the Z domain
Design discrete systems in the Z domain
Controllability and observability of discrete systems
Design state feedback controller using pole
placement
Design state estimator
Design state estimator
Trang 4Discrete lead lag compensators
and PID controllers
Trang 5 State feedback control
State feedback control
+
C
) ( )
( )
1
+ x(k 1 ) A d x(k) B d u(k) C d
Trang 6Transfer function of discrete difference term
u(k)
e(k)
D
u(k) e(k)
e kT
e kT
u ( ) ( ) [( 1 ) ]
T
z E z
z
E z
z
GD ( )
Trang 7Transfer function of discrete integral term
t
( ))
(
Continuous integral:u t e d
0
)()
T k
)()
Trang 8Transfer function of discrete PID controller
Continuous PID controller:
s K
K K
z
z T
K z
z T
K K
z
1
1 2
K z
z T
K K
Trang 9Digital PID controller
z T
K K
z
U z
1
1 2
) (
)
( )
) (
) ( )
( )
z T
z z
PID
1 2
) (
) (
)]
1 (
) ( [ )
1 (
K
k e k
e K
k u k
u
D I
P
Trang 10Digital PID control programming
float PID control(float setpoint float measure)
Note: Kp, Ki, Kd, uk, uk_1, ek, ek_1, ek_2 must be declared as
global variables; uk 1, ek 1 and ek e must be initialized
to be zero; umax and umin are constants
Trang 11TF
TF of discrete phase lead/lag compensator of discrete phase lead/lag compensator
Continuous phase lead/lag compensator:
a
s K s
GC
)
) 2
( )
K
G
) 2 (
) 2 (
) (
)
( )
bT
K z
GC
) 2
) 2 (
) 2
) 2
Trang 12Approaches to design discrete controllers
Indirect design: First design a continuous controller, then discretize the controller to have a discrete control system The performances of the obtained discrete control system are approximate those of the continuous control system provided y p that the sample time is small enough.
Direct design: Directly design discrete controllers
in Z domain.
Methods: root locus, pole placement, analytical
method, … ,
Trang 13D i di t t ll i th Z d i
Design discrete controllers in the Z domain
Trang 14Procedure for designing discrete lead compensator using the RL
) (
)
C
C C
p z
z
z K
Step 1: Determine the dominant poles from desired
transient response specification:
* 2 , 1
(POT) Overshoot
*
Ts
e z
e z
z T n 1
Trang 15Procedure for designing discrete lead compensator using the RL
Step 2: Determine the deficiency angle so that the
dominant poles lie on the root locus of the system after compensated:
* 2 , 1
) arg(
poles from
ngles
Trang 16Procedure for designing discrete lead compensator using the RL
Step 3: Determine the pole & zero of the lead compensator
Step 3: Determine the pole & zero of the lead compensator
Draw 2 arbitrarily rays starting from the dominant pole such that the angle between the two rays equal to * The
* 1
z
intersection between the two rays and the real axis are the
positions of the pole and the zero of the lead compensator
Two methods often used for drawing the rays:
Bisector method
Pole elimination method
Step 4: Calculate the gain K C using the equation:
1 )
( ) ( z G z
1
z z
C z G z G
Trang 17Design discrete lead compensator using RL
Design discrete lead compensator using RL – – Example Example
50 )
(
s s
s
Design the compensator G C (z) so that the compensated system
has dominant poles with 0 707, (rad/sec)n 10
Trang 18Design discrete lead compensator using RL
Design discrete lead compensator using RL – – Example (cont’) Example (cont’)
Solution:
50 )
z
) (
s s
50 )
1
s s
) 1 (
5
)]
5 0 1
( )
1 5
0
[(
) 1
(
5 0 5
0 5
0 1
e z
z
e e
z e
z z
) 1 (
)
e z z
a a
0 )(
1 (
) (
z z
G
Trang 19Design discrete lead compensator using RL
Design discrete lead compensator using RL – – Example (cont’) Example (cont’)
The desired poles: z* re j
2 , 1
where
493
0
10 707 0 1
e e
r Tn
where
707
0 707
0 1
10 1
0
1 0 493 e j
320
0 375
0
* 2 ,
Trang 20Design discrete lead compensator using RL
Design discrete lead compensator using RL – – Example (cont’) Example (cont’)
The deficiency angle Im z
3 2
1
*
)(
Trang 21Design discrete lead compensator using RL
Design discrete lead compensator using RL – – Example (cont’) Example (cont’)
Determine the pole and the zero of the compensator
Determine the pole and the zero of the compensatorusing the pole elimination method:
607
0
z C
607
OA
p C
607
0
OB
578
0
AB
029
Trang 22Design discrete lead compensator using RL
Design discrete lead compensator using RL – – Example (cont’) Example (cont’)
C l l t th i K G ( G ) ( ) 1
Calculate the gain K C: GC ( z ) G ( z ) z z* 1
1
)18.021
.0()607
0)(
1(
)029
z K
]180)
3200
3750
(210
1)
1320
.0375
.0)(
029
0320
.0375
.0(
]18.0)
320
0375
.0(21.0[
j
K C
1702
.0471
0
267
024
1)
G
Conclusion: The TF of the lead compensator is:
029
0
24.1)
(
z z
G C
Trang 23Design discrete lead compensator using RL
Design discrete lead compensator using RL – – Example (cont’) Example (cont’)
Root locus of the Root locus of the
Trang 24Procedure for designing discrete lag compensator using the RL
z
z
)(
)
C
C C
p z
z
z K s
The discrete lag compensator:
Step 1: Denote 1 p C Determine to meet the steady
Step 1: Denote Determine to meet the state error requirement:
Step 2: Chose the zero of the lag compensator: zC 1
Step 3: Calculate the pole of the compensator:
)1
()
( z*
C z GH z G
Trang 25Design discrete lag compensator using RL
Design discrete lag compensator using RL – – Example Example
(
Design the lag compensator G C (z) so that the compensated
) 5 (
) (
s s
Trang 26Design discrete lag compensator using RL
Design discrete lag compensator using RL – – Example (cont’) Example (cont’)
( 1 Z
) 5 (
50 )
(
s s
s G
50 )
1
s s
( 10
5 0 5
0 5
0
) 1 (
5
)]
( )
[(
) 1
(
e z z
0 )(
1 (
) (
z z
z G
) (
) 1 (
)
e z z
a a
Trang 27Design discrete lag compensator using RL
Design discrete lag compensator using RL – – Example (cont’) Example (cont’)
The characteristic equation of the uncompensated system:
The characteristic equation of the uncompensated system:
0 )
(
1 G z
)607
0)(
1(
18.021
z
))(
(
Poles of the uncompensated system:
547
0699
.0
2 ,
z
Trang 28Design discrete lag compensator using RL
Design discrete lag compensator using RL – – Example (cont’) Example (cont’)
(lim
18.021
0)
1(li
0)(
1(
)1
(
lim1
.0
0
0,099
Trang 29Design discrete lag compensator using RL
Design discrete lag compensator using RL – – Example (cont’) Example (cont’)
Step 2: Chose the zero of the lag compensator:
Step 2: Chose the zero of the lag compensator:
Chose: z C 0.99
Step 3: Calculate the pole of the lag compensator:
)1
0
99,
0)
z
G C C
Step 4: Determine the gain of the compensator
1)
()( z*
z
C z G z G
1)
607
0)(
1(
)18.021
.0()999
0(
)99.0(
z z
z K
Trang 30Design discrete lag compensator using RL
Design discrete lag compensator using RL – – Example (cont’) Example (cont’)
Root locus of the uncompensated system
Root locus of the compensated system uncompensated system compensated system
Trang 31Design discrete PID controller using analytical method
10)
Trang 32Design discrete PID controller using analytical method
The controller to be designed is a PI controller (to meet
The controller to be designed is a PI controller (to meetthe requirement of zero error to step input):
1)
( K I T z
1
12
)(
K K
z
G C P I
The discrete TF of the open-loop system:
The discrete TF of the open-loop system:
(
05.0
10)
1( z 1 Z
11
)1
(05
0)
1
2
0 1
K z
z T
K K
z
1
12
1(
1.0
0)
0(
)(
z
z GH
Trang 33Design discrete PID controller using analytical method
The characteristic equation of the closed-loop system:
The characteristic equation of the closed-loop system:
0 )
( )
(
1 G C z GH z
0 819
0
091 0 1
1 2
z T K
K P I
0 )
819 0 091
0 091
0 ( )
819
1 091
0 091
0 (
Trang 34Design discrete PID controller using analytical method
The desired poles: z1,2 re j
2 707 0 2
T
where
059
0
2 707 0
2707
.012
0056
.0
*
2 ,
z
The desired characteristic equation:
The desired characteristic equation:
0)
018
0056
.0)(
018
0056
.0
2
z2 0.112z 0.0035 0
Trang 35Design discrete PID controller using analytical method
Balancing the coefficients of the system characteristic
Balancing the coefficients of the system characteristic
equation and the desired characteristic equation, we have:
0819
.0091
.0091
.0
112
0819
.1091
.0091
.0
I P
I P
K K
K K
09.15
I
P
K K
1
113
.609
.15)
G C
Conclusion
0 )
819 0 091
0 091
0 ( )
819 1 091
0 091
0 (
2 K P K I z K P K I
z
Trang 36Design of discrete control systems es g o d sc ete co t o syste s
in state space domain
Trang 37) ( )
( )
1 (
k k
y
k u k
k
d
d d
x C
B x
A
x
The system is complete state controllable if there exists
The system is complete state controllable if there exists
an unconstrained control law u(k) that can drive the
system from an initial statey x(k( 00)) to a arbitrarily final y
state x(k f) in a finite time interval k0 k k f Qualitatively, the system is state controllable if each state variable can
be influenced by the input
be influenced by the input
Trang 38) ( )
( )
1 (
k k
y
k u k
k
d
d d
x C
B x
Note: we use the term “controllable” instead of
“complete state controllable” for short.
Trang 39Discrete state feedback control
( )
1 (k A d x k B d u k
(
) ( )
( )
1 (
k k
k u k
C
B x
A x
Consider a system described by the state-space equation:
The state feedback controller: u ( k ) r ( k ) Kx ( k )
y ( k ) Cd x ( k )
x ( k 1 ) [ A B K ] x ( k ) B r ( k )
The state equation of the closed-loop system:
) ( )
( ] [
) (
Trang 40Pole placement method
If the system is controllable then it is possible to
If the system is controllable, then it is possible to
determine the feedback gain K so that the closed-loop
system has poles at any location
Step 1: Write the characteristic equation of the loop systemp y det[ det[ z I I A A B B K K ] ] 0 0 (1)
closed-d d
Step 2: Write the desired characteristic equation:
0 )
) 1
pi are the desired poles
(2), we can find the state feedback gain K
Trang 41Discrete pole placement design
Discrete pole placement design – – Example 1 Example 1
Given the control system:
Given the control system:
+
r(k) u(k)
C y(k)
) ( )
( )
1 (k A d x k B d u k
K
) ( )
( )
0 0
316
0
092
Trang 42Discrete pole placement design
Discrete pole placement design – – Example 1 (cont’) Example 1 (cont’)
The closed-loop characteristic equation:
The closed-loop characteristic equation:
0 ]
d d
z
092 0
316 0
1 0
316
0
092
0 368
0 0
316
0 1
1 0
z
0 316
0 368
0 316
0
092
0 316
0 092
0
1 det
2 1
k
k k
092 0
316 0
( 316 0
) 316
0 368
0 )(
092
0 1
k z
k z
368
0 0
d
B
A
) 368
1 316
0 092
0
092
0 316
0 ( 316
0
d
B
0 )
368
0 316
0 066
0 (
Trang 43
Discrete pole placement design
Discrete pole placement design – – Example 1 (cont’) Example 1 (cont’)
The desired poles: * j
The desired poles: z1,2 re j
493
0
10 707 0 1
e e
r Tn
where:
707
0 707
0 1
10 1
0
707 0
493
02
320
0 375
0 )(
320
0 375
0
Trang 44Discrete pole placement design
Discrete pole placement design – – Example 1 (cont’) Example 1 (cont’)
Balancing the coefficients of the system characteristic
Balancing the coefficients of the system characteristic
equation and the desired characteristic equation, we have:
0 )
368
0 316
0 066
0 (
75 0 )
368
1 316
0 092
0 (
2 1
2 1
k k
k k
1
12 3
2
1
k k
Conclusion: K 3 12 1 047
0)
368
0316
.0066
.0()
368
1316
.0092
.0
z
0 243
0 75
0
2 z
z
Trang 45Discrete pole placement design
Discrete pole placement design – – Example 2 Example 2
Given the control system:
Given the control system:
1 Write the state equations of the discrete open loop system
2 Determine the state feedback gain K = [k1 k2] so that the closed loop system has a pair of complex poles wit h =0 5 closed loop system has a pair of complex poles wit h =0.5 ,
n=8 rad/sec.
Trang 46Discrete pole placement design
Discrete pole placement design – – Example 2 (cont’) Example 2 (cont’)
Solution:
Solution:
1 Write the state equations of the discrete open loop system
Step 1: State space equations of open loop continuous system:
s
X s
(
2 s
s
U s
) ( )
1 s X s
sX x1(t) x2(t)
) ( )
( )
1 (s X2 s U R s
0 )
(
) ( 1
0
1 0
) (
)
1
t u
t x t
) ( 1
0 )
( 10 )
0 10 )
( 10 )
t x
t x t
1
Trang 47Discrete pole placement design
Discrete pole placement design – – Example 2 (cont’) Example 2 (cont’)
Step 2: Transient matrix:
Step 2: Transient matrix:
) (s s I A -
1
1 0
1 0
1 0
) 1 (
) (
s
s s s
s
)]
( [ )
1 1
1
1
s s
( 1
Trang 48Discrete pole placement design
Discrete pole placement design – – Example 2 (cont’) Example 2 (cont’)
Step 3: State space equation of the open loop system:
Step 3: State space equation of the open loop system:
(
) ( )
( )
1 (
k k
c
k u k
k
d
d d
x C
B x
A x
1 0
0
) 1
0
095
0 1
0
) 1
(
1 )
t t
0
1 0
Trang 49Discrete pole placement design
Discrete pole placement design – – Example 2 (cont’) Example 2 (cont’)
2 Calculate the state feedback gain K:
2 Calculate the state feedback gain K:
The closed loop characteristic equation:
0 ]
det[ I A B K
d d
z
005 0
095 0
1 0
0
005
0 905
0 0
095
0 1
1 0
z
0 095
0 905
0 095
0
005
0 095
0 005
0
1 det
2 1
k
k k
z
0 )
005
0 095
0 ( 905
0 )
095
0 905
0 )(
005
0 1
Trang 50Discrete pole placement design
Discrete pole placement design – – Example 2 (cont’) Example 2 (cont’)
Th d i d d i t l * j
The desired dominant poles: z1 , 2 re j
67 0
8 5 0 1
e e
r Tn
67 0
e e
r
693
0 5
0 1
8 1
0
02
516 0
0 516
.
02
428
0 516
0 )(
428
0 516
0
0 448
0 03
1
z
Trang 51Discrete pole placement design
Discrete pole placement design – – Example 2 (cont’) Example 2 (cont’)
Balancing the coefficients of the closed loop characteristic
Balancing the coefficients of the closed loop characteristic
equation and the desired characteristic equation, we have:
0 )
905
0 095
0 0045
0 (
03 1 )
905
1 095
0 005
0 (
2 1
2 1
k k
k k
6
0 44
905
0 095
0 0045
0 ( )
905
1 095
0 005
0
2 k k z k k
z
Trang 52Discrete pole placement design
Discrete pole placement design – – Example 2 (cont’) Example 2 (cont’)
3 Calculate system response and performances:
State space equation of the closed-loop system:
(
) ( )
( )
1 (
k k
c
k r k
k
d
d d
d
x C
B x
K B
A x
Student continuous to calculate the response and
performance by themselves following the method
performance by themselves following the method
presented in the chapter 8
Trang 53D i f di t t t ti t
Design of discrete state estimators
Trang 54The concept of state estimation
To be able to implement state feedback control
To be able to implement state feedback control
system, it is required to measure all the states of the system
from the output measurement.
State estimator (or state observer) ( )