Bài giảng Fundamentals of Control Systems - Chapter 9: Design of discrete control system - Huỳnh Thái Hoàng

g/ 42
 316.0d 0)368.0316.0066.0( 21  kk
Discrete pole placement design – Example 1 (cont’)
 The desired poles: j* rez 2,1
493.010707.01.0   eer nTwhere:
707.0707.01101.01 22   nT
7070* j .493.02,1 ez 
 32003750*21 jz  ..,
 The desired characteristic equation:
0)320.0375.0)(320.0375.0(  jzjz
024307502
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 43
 ..  zz
Discrete pole placement design – Example 1 (cont’)
 Balancing the coefficients of the system characteristic 
equation and the desired characteristic equation, we have:
  750)368131600920( kk
  243.0)368.0316.0066.0(
....
21
21
kk
 



047.1
12.3
2
1
k
k
Conclusion:  047.112.3K
0)368.0316.0066.0()368.1316.0092.0( 2121
2  kkzkkz
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 44
0243.075.02  zz
Discrete pole placement design – Example 2
 Given the control system:
r(k) y(k)+ ZOHT=0
u(k) uR(t)
10x1x2
1
1
s s
1
.
1
++ k2
k1
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 with =0 5. ,
n=8 rad/sec.
3 Calculate the response of the system to step input with the
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 45
.
value of K obtained above. Calculate the POT and settling time.
Discrete pole placement design – Example 2 (cont’)
 Solution:
1. Write the state equations of the discrete open loop system
Step 1: State space equations of open loop continuous system:
sXsX )()( 21  )()( 21 sXssX  )()( 21 txtx s
1
)()(2  s
sUsX R )()()1( 2 sUsXs R )()()( 22 tutxtx R
)(
1
0
)(
)(
10
10
)(
)( 11 tu
txtx
R







 
22 txtx 
   )(010)(10)( 1 txtt y(t)uR(t) 10x1x21 1
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 46


)(2
1 tx
xy
1s s
Discrete pole placement design – Example 2 (cont’)
Step 2: Transient matrix:
  1)( -ss AI 
1
10
1 









s
s1
10
10
10
01 









 s
 



 )1(
11
)( sss
 

1
10
s
s
)]([)( 1 st  L









   sss
1
)1(
11
1L 














1
)1(
11 11
sss
LL
   as0   

1
0 1
s
L
 
tet )1(1)(
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 47

 te0
Discrete pole placement design – Example 2 (cont’)
Step 3: State space equation of the open loop system:




)()(
)()()1(
kkc
kukk
d
dd
xC
BxAx
)(Td A 

  

1.0
1.0
0
)1(1 e



90500
095.01
dA

T
d d)(  BB
e .
   
1.0 0)1(1 

de    
1.0 )1( 

de
0
  1.0   e
  0 10 e   0 e
    110 1.0e  005.0
0
 
 e
0
)1(1)( 

  

e
et t
t 
  1
.
1.0e
  095.0dB
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 48
 010 CCd 1.0T
Discrete pole placement design – Example 2 (cont’)
2 Calculate the state feedback gain K:.
The closed loop characteristic equation:
0]det[  KBAI ddz
 00500950101   0
095.0
.
905.00
.
10
det 21  
 kkz

0
095.0905.0095.0
005.0095.0005.01
det
21
21 








kzk
kkz
0)005.0095.0(905.0)095.0905.0)(005.01( 2121  kkkzkz
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 49
0)905.0095.00045.0()905.1095.0005.0( 2121
2  kkzkkz
Discrete pole placement design – Example 2 (cont’)
Th d i d d i t l j*e es re om nan po es: rez 2,1
67.085.01.0   eer nT
693.05.0181.01 22   nT
 693.0* 67.02,1 jez 
42805160* j
The desired characteristic equation:
..2,1z 
0)428.0516.0)(428.0516.0(  jzjz
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 50
0448.003.12  zz
Discrete pole placement design – Example 2 (cont’)
Balancing the coefficients of the closed loop characteristic 
equation and the desired characteristic equation, we have:
 031)905109500050( kk
 

448.0)905.0095.00045.0(
....
21
21
kk





895.6
0.44
2
1
k
k
 895.60.44KConclusion:
0)905.0095.00045.0()905.1095.0005.0( 2121
2  kkzkkz
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 51
0448.003.12  zz
Discrete pole placement design – Example 2 (cont’)
3 Calculate system response and performances:.
State space equation of the closed-loop system:
 
 

)()(
)()()1(
kkc
krkk
d
ddd
xC
BxKBAx
Student continuous to calculate the response and 
performance by themselves following the method 
presented in the chapter 8.
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 52
D i f di t t t ti tes gn o scre e s a e es ma ors
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 53
The concept of state estimation
 To be able to implement state feedback control 
system, it is required to measure all the states of the 
system.
 However, in some application, we can only measure 
the output, but cannot measure the states of the 
system.
 The problem is to estimate the states of the system 
from the output measurement .
 State estimator (or state observer)
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 54
Observability
  )()()1( kukk BxAx
  )()( kky d
dd
xC
 Consider the system:
 The system is complete state observable if given the 
control law u(k) and the output signal y(k) in a finite 
time interval k0  k kf , it is possible to determine the 
initial states x(k0).
 Qualitatively, the system is state observable if all 
t t i bl (k) i fl th t t (k)s a e var a e x n uences e ou pu y .
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 55
Observability condition
  )()()1( kukk dd BxAx
  )()( kky d xC System
It is require to estimate the state from mathematical)(ˆ kx
 C
model of the system and the input-output data. 
 Observability matrix: 



2
d
d
AC
AC
O 




1
dd

 nddAC
 The necessary and sufficient condition for the observability: 
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 56
nrank )(O
Observability – Example
  )()()1( kukk BxAx
  )()( kky d
dd
xC
 Given the system
 14809670  2310
where: 

522.0297.0
..
dA  264.0
.
dB  31dC
 Solution: Observability matrix:
Analyze the observability of the system. 


 d
AC
C
O 


7141
3
0770
1
O
dd ..
 Because 484.1)det( O  2)( Orank
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 57
 The system is observable
State estimator
u(k) y(k)x(k)
)()()1( kukk dd BxAx  Cd
L
)(ˆ kx
+
+ )1(ˆ kx Cd
)(ˆ ky
Bd 1z++
 ))(ˆ)(()()(ˆ)1(ˆ kkkkk LBA
Ad
 

)(ˆ)(ˆ kky
yyu
d
dd
xC
xx State estimator:
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 58
T
nlll ][ 21 Lwhere:
Design of state estimators
 Requirements: 
 The state estimator must be stable, estimation error should 
approach to zero.
 Dynamic response of the state estimator should be fast 
enough in comparison with that of the control loop.
 All the roots of the equation 
l t i id th it i l i th l
 It is required to chose L satisfying:
0)det(  ddzI LCA
oca es ns e e un c rc e n e z-p ane.
 The roots of are further from the 
unit circle than the roots of 0)det(  KBA ddzI
0)det(  ddzI LCA
 Depending on the design of L, we have different state estimator:
 Luenberger state observer
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 59
 Kalman filter
Procedure for designing the Luenberger state observer
St 1 W it th h t i ti f th t t b ep : r e e c arac er s c equ. o e s a e o server
0]det[  ddz LCAI (1)
 Step 1: Write the desired characteristic equation:
n
0)(
1

i
ipz
)1( nip  are the desired poles of the state estimator
(2)
St 3 B l th ffi i t f th h t i ti
, ,i 
 ep : a ance e coe c en s o e c arac er s c 
equations (1) and (2), we can find the gain L. 
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 60
Design of state estimators – Example
 Problem: Given a system described by the state 
equation: 
 )()()1( kkk BA
 

)()( kky
u
d
dd
xC
xx



52202970
148.0967.0
dA 


2640
231.0
dB  31dC
 Assuming that the states of the system cannot be directly
 .. .
measured. Design the Luenberger state estimator so that 
the poles of the state estimator lying at 0 13 and 0 36
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 61
 . . .
Design of state estimators – Example (cont’)
 Solution
 The characteristic equation of the Luenberger state estimator:
0]det[ z LCAI dd
  031
52202970
148.0967.0
10
01
det 1 










l
l
z
.. 2  
 03148.0967.0det 11 




 llz
(1)0)549075324131()48913(2  llzllz
3522.0297.0 22

   lzl
 The desired characteristic equation:
.... 2121
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 62
0)36.0)(13.0(  zz (2)00468.049.02  zz
Design of state estimators – Example (cont’)
 Balancing the coefficients of the equations (1) and (2): 
  49.0489.13 21 ll  0468.0549.0753.2413.1 21 ll
 Solve the above set of equations we have: , 
  653.21l  544.12l
 T544.1653.2L Conclusion
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 63
Simulation of discrete state estimator
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 64
State estimation simulation result
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 65