Bài giảng Fundamentals of Control Systems - Chapter 6: Design of control systems - Huỳnh Thái Hoàng

 (cont’)
 The desired characteristic equation: 
0)2)(( 22  nnssas 
0)648)(( 2  ssas
064)648()8( 23  asasas (2)
 Balancing the coefficients of the equations (1) and (2), we have:
  aK 810010  25156




aK
aKP
D
64100
648100100

 

541
14,12
.
P
K
K
a

I  ,D
Conclusion: G 5411006412)( 
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s
s
sC ,,
Manual tuning of PID controllers
 Effect of increasing a parameter of PID controller independently 
on closed-loop performance:
Para-
meter Rise time POT
Settling 
time
Steady-
state
error
Stability
KP Decrease Increase
Small
change Decrease Degrade
KI Decrease Increase Increase Eliminate Degrade
KD
Minor
change Decrease Decrease No effect
Improve if
K smallD
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A d f l t i f PID t ll
Manual tuning of PID controllers (cont.)
 proce ure or manua un ng o con ro ers:
1. Set KI and KD to 0, gradually increase KP to the critical 
i K (i th i k th l d l tga n cr .e. e ga n ma es e c ose - oop sys em 
oscilate)
2. Set KP Kcr /2
3. Gradually increase KI until the steady-state error is 
eliminated in a sufficient time for the process (Note that 
too much KI will cause instability).
4. Increase KD if needed to reduce POT and settling time 
(Note that too much KD will cause excessive response 
and overshoot)
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Control systems design in state-space
using pole placement method
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Controllability
  )()()( tutt BAxxC id t
  )()( tty Cx
 ons er a sys em:
 The system is complete state controllable if there exists an
unconstrained control law u(t) that can drive the system from
an initial state x(t0) to a arbitrarily final state x(tf) in a finite
time interval t0  t tf . Qualitatively, the system is state
controllable if each state variable can be influenced by the
input.
y(t)
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Signal flow graph of an incomplete state controllable system
Controllability condition
  )()()( ttt BAxx
 

)()( tty
u
Cx
 System:
 Controllability matrix
][ 12 BABAABB  nC 
 The necessary and sufficient condition for the controllability is: 
nrank )(C
N t th t “ t ll bl ” i t d f “ l t o e: we use e erm con ro a e ns ea o comp e e 
state controllable” for short.
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Controllability – Example
  )()()( tutt BAxx
  )()( tty Cx
 Consider a system
where:  10 5  32A 

2
B  31C
E l h ll bili f h
 Solution: Controllability matrix:
va uate t e contro a ty o t e system.
 ABBC
 Because:



 16
2
2
5
C
84)det( C  2)( Crank
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 The system is controllable
State feedback control
r(t) u(t) y(t)x(t)
+ C)()()( tutt BAxx 
K
 Consider a system described by the state equations:




)()(
)()()(
tty
tutt
Cx
BAxx
 The state feedback controller: )()()( ttrtu Kx
  )()(][)( trtt BxBKAx
 The state equations of the closed-loop system:
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  )() ty(t Cx
Pole placement method
If the system is controllable then it is possible to determine,
the feedback gain K so that the closed-loop system has the
poles at any location.
 Step 1: Write the characteristic equation of the closed-loop
system 0]det[  BKAIs (1)
 Step 2: Write the desired characteristic equation:
n
0)(
1

i
ips
)1( nip th d i d l
(2)
 Step 3: Balance the coefficients of the equations (1) and (2),
, ,i  are e es re po es
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we can find the state feedback gain K.
Pole placement method – Example
 Problem: Given a system described by the state state - 
equation: 
  )()()( tutt BAxx  )()( tty Cx
 010 0  100C



 100A



 3B
  374 1
 Determine the state feedback controller )()()( ttrtu Kx 
so that the closed-loop system has complex poles with 
and the third pole at 20. 10;6,0  n
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Pole placement method – Example (cont’)
 Solution
 The characteristic equation of the closed-loop system:
0]det[  BKAIs 
  03
0
100
010
010
001
d t 



 kkk
1374100
e 321  







 




s
 The desired characteristic equation:
(1)0)12104()211037()33( 313212323  kkskkkskks
0)2)(20( 22  nnsss 
(2)23
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0200034032  sss
Pole placement method – Example (cont’)
 B l th ffi i t f th ti (1) d (2) ha ance e coe c en s o e equa ons an , we ave:
  3233 32 kk




200012104
340211037
21
321
kk
kkk
 Solve the above set of equations, we have:
 578220k

 

48217
839,3
,
2
1
k
k
 ,3
 482178393578220K Conclusion:
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,,,
D i f t t ti tes gn o s a e es ma ors
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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 applications 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)
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Observability
  )()()( tutt BAxx
  )()( tty Cx
 Consider a system:
 The system is complete state observable if given the control 
law u(t) and the output signal y(t) in a finite time interval 
t0  t tf , it is possible to determine the initial states x(t0). 
Qualitatively the system is state observable if all state variable, 
x(t) influences the output y(t).
y(t)
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Signal flow graph of an incomplete state observable system
Observability condition
  )()()( tutt BAxxS t
  )()( tty Cx
 ys em
It is necessary to estimate the state from mathematical )(ˆ tx
 Observability matrix:  C
model of the system and the input-output data. 




 2CA
CA
O




 1nCA

 The necessary and sufficient condition for the observability is: 
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nrank )(O
Observability – Example
  )()()( tutt BAxx
  )()( tty Cx
 Consider the system
 10 1where:  

32
A 

2
B  31C
 Solution: Observability matrix:
Evaluate the observability of the system. 



CA
C
O 


 8
3
6
1
O
 Because 10)det( O  2)( Orank
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 The system is observable
State estimator
u(t) y(t)x(t)r(t) )()()( tutt BAxx  C
+

CB 
L
)(ˆ tx

+
++ )(ˆ ty
A
K




)(ˆ)(ˆ
))(ˆ)(()()(ˆ)(ˆ
tty
tytytutt
xC
LBxAx State estimator:
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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 the dynamic response of the
control loop.
 All the roots of the equation locates 
i th h lf l ft l
 It is required to chose L satisfying:
0)det(  LCAsI
n e a - e s-p ane.
 The roots of the equation are further
from the imaginary axis than the roots of the equation
0)det(  LCAsI
0)det(  BKAsI
 Depending on the design of L, we have different state estimator:
 Luenberger state observer
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 Kalman filter
Procedure for designing the Luenberger state observer
 St 1 W it th h t i ti ti f th t t bep : r e e c arac er s c equa on o e s a e o server
0]det[  LCAIs (1)
 Step 1: Write the desired characteristic equation:
0)( n ips (2)
1i
),1( , nipi  are the desired poles of the state estimator
 Step 3: Balance the coefficients of the characteristic
equations (1) and (2), we can find the gain L.
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Design of state estimators – Example
 Problem: Given a system described by the state equation:

 
)()(
)()()( tutt
C
BAxx
 tty x
 010 0
 001C



 100A



 3B
  374 1
 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 20, 20 and 50.
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Design of state estimators – Example (cont’)
 Solution
 The characteristic equation of the Luenberger state estimator:
0]det[  LCAIs
  0001100
010
010
001
d t
1




 l
l

374100
e
3
2  







 



 l
s
 The desired characteristic equation:
(1)0)457()73()3( 32121
2
1
3  lllsllsls
0)50()20( 2  ss
23
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(2)020000240090  sss
Design of state estimators – Example (cont’)
 B l i th ffi i t f th (1) d (2) l d ta anc ng e coe c en s o e equ. an ea s o:

  9031l
 

20000437
240073
321
21
lll
ll
 Solve the above set of equations, we have:
  87l




12991
21322
1
l
l
3
 T12991213287L Conclusion
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