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.
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Introduction 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 D i
db k
ll
f
t
t
i
l
t placement
Design state estimator Design state estimator
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Serial compensator Serial compensator
R(z)
Y(z)
ZOH
G(z)
GC(z) C
+
T T
H(z) H( )
State feedback control State feedback control
u(k)
r(k)
y(k)
( (
k k
k )( )( k
ku )( )( k
)1 )1
+ +
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 5
u(k) u(k)
e(k) e(k)
D
Differential term:
tu
)(
)( tde )( td dt
( kTe
)1
T
]
)
Discrete difference:
( kTu
)
)( zE
)( zU
[( ke T 1 )( zEz T
Transfer function of the discrete difference term:
1
)( )( zGD G
1 z zT
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6
e(t) e(t)
u(t) u(t)
Integral I t l
t
Continuous integral: Continuous integral:
)( t tu )(
e
)( d d )(
0
kT
(
k
kT
( kTu
)
e
)( d
Discrete integral: l t
Di
t
i
)1 T )( e d
0
(
0
k
)( d e T )1
kT
=
-
t
u kT ( ( kT
) )
u k [( [( k
1) 1)
T T
] ]
e
dt ( ) ( ) d
=
-
+
-
+
u k [( [( k
1) 1)
T T
] ]
e k [( [( k
1) 1)
T T
] ]
e kT ( ( kT
) )
( (
+ ò + ò
T T 2
-
(
k
1)
T
U
zU )( )(
1 1 zUz )( )( U
T T 2
TF of discrete integral term: TF of discrete integral term:
zGI G )( )(
1 1
zT z 2
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Continuous PID controller:
C ti
PID
ll
t
G
K
)( )( s
PID PID
P P
sK D D
K s
Discrete PID controller:
1
z
G
)( z
K
PID
P
1 1
zTK I 2 z
K D T
z
P
D
I
z
1 1
G
K
z )(
or
PID
TK I
P
z
1
K K D T
z
z
I
D
P
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8
r(k) (k)
e(k)
u(k)
y(k)
+
D/A
G(s)
PID
A/D
1
G
)( )( z
K
PID PID
P P
)( zU )( )( zE
zTK I 2 2 z
1 1 1
zK D T
z
ku )( )( k
ku ( ( k
)1 )
ke ( ( k
)]1 )
keK )([ k )( P
ke )([ )([ ke
ke ( ( ke
)]1 )]1
)1 )1
ke ke ( (
)]2 )]2
ke (2)([ ke (2)([ ke ke
TK I 2
K D T
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9
float PID control(float setpoint float measure) float PID_control(float setpoint, float measure) { ek_2 = ek_1; ek_1 = ek; ek = setpoint – measure; uk_1 = uk; uk = uk 1 + Kp*(ek ek 1) + Ki*T/2*(ek+ek 1) + uk = uk_1 + Kp (ek-ek_1) + Ki T/2 (ek+ek_1) +… Kd/T*(ek – 2ek_1+ek_2);
If uk > umax, uk = umax; If uk < umin, uk = umin; return(uk)
_
;
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Continuous phase lead/lag compensator: Continuous phase lead/lag compensator:
K
)(
sGC
phase lead g phase lag p
b b
a a
as bs bs
Discretization using trapezoidal integral:
K K
zGC )( )( G
( ( (
aT bT
) )2 )2
z z
( ( (
aT bT
) )2 )2
Denote
and
zC
pC
( ( (
)2 )2 )2
aT aT aT
( ( (
)2 )2 )2
bT bT bT
TF f di
d/l
h
t
t
l
TF of discrete phase lead/lag compensator
phase lead
K K
)( zG zG )( C C
C C
z z
phase lag
z C p C
z C z C
p C p C
1Cz 1Cp
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a
Fi
design d i
ti
p
continuous controller, then discretize the controller to have a discrete control system. The performances of the discrete control system The performances of the obtained discrete control system are approximate those of the continuous control system provided y that the sample time is small enough.
Direct design: Directly design discrete controllers
in Z domain.
locus, pole placement, analytical
,
method, …
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K
(
)
zG )( C
C
z C
p C
z z
z C p C
Step 1: Determine the dominant poles from desired
* 2,1z
transient response specification:
(POT)
2 j 1
n
n
* s 2,1
ts
Overshoot g time Settling
n n
*Tse
* z 2,1
*
T n
r
z
e
*
z z
2 nT 1 T 1
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th t th
* 2,1z dominant poles lie on the root locus of the system after compensated: after compensated:
n
m
0
180
arg(
)
arg(
z
)
*
p i
i
* z 1
* z 1
i
i
1
1
where pi and zi are poles and zeros of G(z)
Geometry formula:
0
180
ngles
from
poles
of
zG
*
a
o )( t
* z 1
ngles
from
oserz
of
zG
a
* * 1o)( z t
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 15
* 1z
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 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: Two methods often used for drawing the rays: Bisector method Pole elimination method Pole elimination method
Step 4: Calculate the gain KC using the equation:
zGzG )( )( )( )( zGzG
1 1
C
* zz 1
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R(s)
Y(s)
ZOH
G(s)
GC(z)
+
T
)( sG
sec1.0T
ss (
)5
50
Design the compensator GC(z) so that the compensated system
has dominant poles with
707.0
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Solution: Solution:
The open-loop discrete TF:
)( )( sG
)5
50 ss (
1(
z
)( zG
1 Z )
)( sG s
1(
1 z Z )
2
s
)5
50 s (
5.0
5.0
5.0
z
15.0[(
e 5.0
)]
1
1(10
z
)
5.0
e (5 (5
z z
) )
z e ) 1( 2 z z e e ()1 ()1
aT
aT
aT
aTe
1
aTz (
)
zG zG )( )(
)
e za (
)
(
z
z ) 1( 2 z ()1
e aTe
21.0 z z )(1
18.0 a Z Z 607.0 ) 2 as ( s
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 18
j
The desired poles:
re
* z 2,1
where where
707
.01.0
10
T n
r
e
e
493.0
1
1.0
10
707
.0
707
.01
2
707.0
493.0
je
* z 2,1
375.0
j
320.0
* z 2,1
2 nT
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Im z Im z
The deficiency angle The deficiency angle
0.375+j0.320
+j
180
)
*
( 2 3
1
0
152
9.
1
0
P
125
9.
2
0
2
*
1
6.14
Re z
3
3
0
+1
1
084 * 84 0
A pc
B zc
j j
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20
Determine the pole and the zero of the compensator Determine the pole and the zero of the compensator
using the pole elimination method:
607
.0
Cz
607.0Cz C
OA
OB
AB
pC
607 578
.0OB .0AB
029.0Cp
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 21
i K
t
l
1 1
*
Calculate the gain KC: C l th
zGzG G )( G )( )( )( C
zz
1 1
K K
C
( (
z z
.0 .0
607 029
) ()
)18.0 .0 607
)
z 21.0( z z )(1
z
375.0
j
.0
320
1
KC
320
)1
.0(
375
j
j
.0(21.0[ 0(210[ .0 320
j j ]180)320 375 375 .0 0 ]18.0)320 029 .0 .0)( 375 .0
1
CK
24.1CK
.0 471
.0
702
267 .0
Conclusion: The TF of the lead compensator is:
241)( 24.1)(
zGC G
.0 .0
607 029
z z
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22
Root locus of the uncompensated system t
t d
Root locus of the compensated system t d
t
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K
(
)
sG )( C
C
z C
p C
z z z
z z C p C
C
p Cp1 C 1 z state error requirement:
or
or
K K
K K
K K
P * P
a * a
V * V
Step 2: Chose the zero of the lag compensator:
1Cz
Step 3: Calculate the pole of the compensator: p
1(
z
)
1
C
C
Step 4: Calculate KC satisfying the condition:
)( zGHzG
)(
1
C
* zz
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R(s) R(s)
C(s) C(s)
+
ZOH
G(s)
GC(z)
T
sG )(G )(
sec1.0T 10 T
ss (
)5
50
Design the lag compensator GC(z) so that the compensated system has the velocity constant and the closed
100
* VK
poles are nearly unchanged.
C
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Solution: Solution:
The discrete transfer function of the open-loop system:
sG )(
z
)( zG
1(
ss (
)5
50 50
1( 1(
2
s
)5
)( sG sG )( s 50 50 s (
5.0
5.0
5.0
[( 15.0[( z
e 5.0
)] )]
1 1
z
1(10 1(10
5.0
e (5
z
e e
)
) ) ( z 1( 2 z ()1
1 Z 1 ) 1 1 Z z Z ) ) ) )
zG )( )( G
18.0 607.0
)
(
z
21.0 )(1
z z
aT
aT
aT
aTe
1
aTz (
)
2
)
s
(
e za (
)
a as
z ) 1( 2 z ()1
e aTe
Z Z
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The characteristic equation of the uncompensated system: The characteristic equation of the uncompensated system:
1
zG )(
0
0
1
18.0 607.0
) )
( (
z
z 21.0 )( )(1 z
Poles of the uncompensated system:
699.0
j
.0
547
z 2,1
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Step 1: Determine Step 1: Determine
The velocity constant of the uncompensated system:
1(
1 ) )( zGz
K V
lim 1 z
1 1
z
) )
K K V
9.9VK
lim li 1( 1( z 1
18.0 607.0
)
(
z
1 1 T 1 1.0
z 21.0 z )(1
The desired velocity constant: The desired velocity constant:
100 100
*K VK
Then: Then:
9.9 100
VK V * K V
099,0 099,0
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Step 2: Chose the zero of the lag compensator: Step 2: Chose the zero of the lag compensator:
Chose:
99.0Cz
Step 3: Calculate the pole of the lag compensator:
1(
z
)
1
999
.01
099
)99.01(
p C
C
.0Cp
K
zG )( C
C
z 99,0 s 999,0
Step 4: Determine the gain of the compensator
1
zGzG )( )(
*
C
1
K
C
)18.0 607.0
)
(
21.0( z )(1 z z
z
699.0
j j
.0
547
007.1
1
zz ( )99.0 z ( )999.0 z CK
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Root locus of the uncompensated system uncompensated system
Root locus of the compensated system compensated system
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R(s) R( )
C(s) C( )
ZOH
G(s)
+
GC(z)
T
H(s)
sG )(
sec2T
05.0
)( sH
1
10
10 s
Design the controller GC(z) so that the closed– loop system has the poles with =0.707, n=2 loop system has the poles with =0 707 =2 rad/sec and steady state error to step input is zero.
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 31
The controller to be designed is a PI controller (to meet The controller to be designed is a PI controller (to meet
the requirement of zero error to step input):
K
zG )( )( C
P
1 1 1
zTK zTK I I z 2
The discrete TF of The discrete TF of
the open-loop system: the open-loop system:
)( zGH
z
1(
1(
10 s 10( 10(
05.0 s )1 )1
1z Z )
1 Z )
2.0
1
z
) )
( 1(
z
1 1
(1.0
sHsG )( )( s z 1(05.0 z z )(1 G
K
)
e ) 2.0 20 e z )( PID
P
zTK I 2 z
1 1 1
K D T
z
zGH )( zGH )(
P P
I I
D D
091 .0
)819
(
z
.0
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The characteristic equation of the closed-loop system: The characteristic equation of the closed-loop system:
1
)(
0
zGHzGC )(
K
0
1
P
z
zTK I z 2
1 1
819.0
091.0
2
z
.0(
091
K
.0
091
K
.1
)819
z
.0(
091
K
.0
091
K
.0
)819
0
P
I
P
I
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 33
j j
The desired poles: The desired poles:
re
* z 2,1
where
707.02 02 707
2 2
T T n
r
e
e
059.0
2
1
707
.2
828
22
.01
2 nT
.2
828
.0
059
je
* z 2,1
.0
056
j
.0
018
* z 2,1
The desired characteristic equation: The desired characteristic equation:
(
z
056.0
j
.0
018
)(
z
.0
056
j
.0
0)018
z 2 2
112.0
z
.0
0035
0
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 34
Balancing the coefficients of the system characteristic Balancing the coefficients of the system characteristic
equation and the desired characteristic equation, we have:
091 091
K K
.0 0
091 091
K K
.1 1
819 819
.0 0
112 112
P K 091
.0
.0
I K 091
819.0
.0
0035
P
I
.0 0
P
09.15 13.6
K K
I
Conclusion
09.15
13.6
zGC )(
z z
1 1
2
.0(
091
K
.0
091
K
.1
)819
z
.0(
091
K
.0
091
K
.0
)819
0
z
P
I
P
I
2
z
112.0
z
.0
0035
0
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 35
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)1 )1
k )( )( k
ku )( )( ku
Consider a system:
C
y
( 0)
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 state x(k0) to a arbitrarily final y state x(kf) in a finite time interval k0 k kf . Qualitatively, the system is state controllable if each state variable can be influenced by the input. be influenced by the input
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( (
k k
)1 )1
k )( )( k
ku )( )( k
System:
Controllability matrix
] ]
C C
The necessary and sufficient condition for the
controllability
rank
n
)
( C
Note: we use the term “controllable” instead of
“complete state controllable” for short.
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 38
u(k) u(k)
r(k) r(k)
y(k) y(k)
(
k
)1
k )(
ku )(
+
ku )(
k )(
k
(
The state feedback controller:
ku )(
kr )(
k )(
)(] )(] k
) )1
k
( (
[ [
)( )( kr
d d
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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 closed-
( ) (1)
loop system p y
det[ det[
0] 0]
(2)
0
)
ip
i
1
n )1 ),1
i (i ( ,
Step 2: Write the desired characteristic equation: n z ( are the desired poles th d i d
l
pi
Step 3: Balance the coefficients of the equations (1) and
(2), we can find the state feedback gain K. (2) t t
fi d th
db k
i K
f
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Given the control system: Given the control system:
u(k)
r(k)
y(k)
( (
k
) )1
)( )( k
)( )( ku
+
0
10dC
dB
316.01 01 316 .00 368
092.0 092 0 316.0
the Determined the state feedback gain K so that closed-loop system has a pair of complex poles with =0.707, n=10 rad/sec 0 707 10 rad/sec
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 41
det[
The closed-loop characteristic equation: The closed-loop characteristic equation: 0]
316.01 01 316
.0 0
det
z
k
0
k 1
2
10
.00
316.0
368
01 01
.01
.0
316
.0
092
z
k 1
det
0
.0
316
368
.0
.0
316
z
k 2 k
092 k 1
2
z (
.01
092
368.0
316.0
k
)
z
)(
.0 0
316 316
316 316
.0 0
k 1
2 0( .0(
k k 1
2
z
.0(
316
k
.1
368
)
z
.0
092
k 1
.0(
066
.0
316
k
.0
0
2
k 1
2
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 42
j j
The desired poles: The desired poles:
re
* z 2,1
707.01.0
10
T n
where:
r
e
e
493.0
2
1
1.0
10
707
.0
707
.01
2 nT
707.0 0 707
* * 2,1
z
.0
493
je j
* 2,1 21
375.0 375 0
.0 0
320 320
z z
j j
The desired characteristic equation:
(
z
.0
375
j
.0
320
)(
z
.0
375
j
.0
320
)
0
2 2
z
75.0 750
z
.0 0
243 243
0 0
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Balancing the coefficients of the system characteristic Balancing the coefficients of the system characteristic
equation and the desired characteristic equation, we have:
316.0 316 0
k k
)368.1 1 )368
75.0 750
2
066.0(
.0
316
k
.0
)368
.0
243
k k 1 k 1
2
092.0( 0( 092
12.3
k 1 k
.1
047
2
Conclusion:
047.1
12.3K
2
.0(
092
.0
316
k
.1
368
)
z
.0(
066
.0
316
k
.0
368
)
0
z
k 1
k 1
2
2
2
z
75.0
z
.0
243
0
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Given the control system: Given the control system:
r(k)
u(k)
uR(t)
y(k)
x2
x1
10
ZOH
T=0. T=0
1 s s
1 1s 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, 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 3 Calculate the response of the system to step input with the value of K obtained above. Calculate the POT and settling time.
+
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 45
Solution: Solution: 1. Write the state equations of the discrete open loop system
Step 1: State space equations of open loop continuous system: Step 1: State space equations of open loop continuous system:
sX sX
s )( )( s
tx )( )( tx 1 1
tx )( )( tx 2 2
1 1
sX )( )( sX 2 2
sX )( )(1 sX 1
)1
sUsX )(
)(
( s
tx )( 2
tx )( 2
tu )( R
R
2
sX )(2
)( sX 2 s sU )( R 1s 1 s
0 0 0
tx )( 1 tx )( )( 2
1 1 1
0 1 1
tu )( )( R R
tx )( 1 tx )( )( 2
uR(t)
y( ) y(t)
x2 2
x1 1
10 10
ty )( )( t
10 10
0 0
10 10
tx )( )( t 1
1 s
1 1s
1 tx )( tx )( 2
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 46
Step 2: Transient matrix: Step 2: Transient matrix:
1
1
-
s
)( s
1
s 0
s
0 0 0
1 1
01 10 10
1 1 1
s
1 s s
)1 )1
s )( )(
0
s
1
1 ss ( ( ss 1
1
1
1 s
1 s
)1
1 [
(
s
)]
t )(
L
1L
1
0
0
s
1
1 ( ss 1 1 as
1 ss ( 1 1
)1
te
)
)( )( t t
0
te
1(1
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 47
Step 3: State space equation of the open loop system: Step 3: State space equation of the open loop system:
k )(
ku )(
1.0
)
)(T
d A
dA
e 1.0 1.0
0 0
e
095.01 905.00 00 905
1(1
T
)
d)( )( d B B
e
e
0
e
1.0 0
0
1.0 0
1( e
d d
d d
1.0
1(1 1.0 10
t
)
1
1.0e e 1.0e
0) 1 1 1
e e e
0
e t
e
C
10
0
T
005.0 dB 095.0 1(1 )( t 0 1.0
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 48
2. Calculate the state feedback gain K: 2 Calculate the state feedback gain K:
The closed loop characteristic equation:
det[
0]
det
0
z
k
k 1
2
10
.00
01 01
01 .01
095 095 905
005 005.0 0 095.0
005.01
095.0
005.0
z
k
k 1
det
0
095.0
905
.0
095.0
z
2 k
k 1
2
.01
005
)(
z
.0
905
.0
095
k
905
095
.0
005
k
)
0
.0)
.0(
z (
k 1
k 1
2
2
2
z
k 095.0
)905.1
z
.0(
k 095.0
)905.0
0
k 005.0( 1
2
k 0045 1
2
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 49
j j
The desired dominant poles: Th d i d d
t
l
i
z
re
* 2,1
85.01.0
T n
r r
e e
e e
67.0 67.0
1
2 5.0181.0
693.0
2 nT
693.0
* 2,1
67.0
z
je
* 2,1
z
.0 0
516 516
.0 0
428 428
j j
The desired characteristic equation:
(
z
.0
516
j
.0
428
)(
z
.0
516
j
.0
)428
0
z
03.12
z
.0
448
0
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 50
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 0
k 095 095 k
.1 1
)905 )905
03.1 031
.0( 0(
.0(
2 k 095
.0
.0
)905
.0
448
k 005 005 k 1 k 0045 1
2
0.44
.6
895
k 1 k 2 2
Conclusion:
.60.44K
895
2
z
.0(
005
.0
095
k
.1
)905
z
.0(
0045
.0
095
k
.0
)905
0
k 1
k 1
2
2
03.12
z
z
.0
448
0
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 51
3. Calculate system response and performances: 3 Calculate system response and performances:
k )(
k
(
kr )(
)1
d
Student continuous to calculate the response and performance by themselves following the method performance by themselves following the method presented in the chapter 8.
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 52
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 53
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. 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. from the output measurement
State estimator (or state observer) ) (
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 54
)1 )1
( (
k k
k )( )( k
ku )( )( ku
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 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
state variable x(k) influences the output y(k). (k)
(k) i
i bl
th
fl
t
t
t
t
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 55
)1 )1
( (
k k
k )( )( k
ku ku )( )(
System
)(ˆ kx )(kx
It is require to estimate the state from mathematical It is require to estimate the state from mathematical model of the system and the input-output data.
Observability matrix: Observability matrix:
O O
y
y
rank
n
)
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 56
( (
k k )( )(
ku ku )( )(
k k
)1 )1
Given the system
967 967
.0 0
.0 0
148 148
31dC
dB
where:
dA
.0
297
.0
522
.0 231 231 0 .0 264
Analyze the observability of the system.
Solution: Observability matrix: Solution: Observability matrix:
O
1 077.0 0 077
3 714.1 714 1
O
Because
rank
(
2
det(
484.1)
) O
O
The system is observable
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 57
u(k) u(k)
y(k) y(k)
(
k
k )(
ku )(
)1
+
)(ˆ kx )(kx
)1 )1
(ˆ kx ( kx
1z
+ + + +
)(ˆ ky
)1 )1
)) ))
)(ˆ )(ˆ k k
ku k )( )(
ky k )(( )((
(ˆ (ˆ ky k
State estimator:
where:
[
T ]
l
L
nl
l 1
2
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 58
Requirements: 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.
zI
)
d
0 All the roots of the equation i
t
i
0
) )
0
c c e
Depending on the design of L, we have different state estimator:
g
Luenberger state observer Kalman filter
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 59
Step 1: Write the characteristic equ. of the state observer i ti
f th
t t
th
h
b
t
(1)
det[
0]
Step 1: Write the desired characteristic equation: n
(2)
)
0
ip
z (
i 1
( , (
i i
n )1 ),1 n
are the desired poles of the state estimator are the desired poles of the state estimator
pi p
Step 3: Balance the coefficients of the characteristic ffi
f th
i ti
th
h
t
i
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 60
Problem: Given a system described by the state Problem: Given a system described by the state
equation:
)1 )1
k )( )( k
ku )( )( k
967
.0
148
dA
dB d
31dC
d
.0 0
297 297
.0 0
522 522
.0
231 .0 .0 0 264 264
Assuming that the states of the system cannot be directly 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. 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
Solution Solution
The characteristic equation of the Luenberger state estimator:
det[ det[
0] 0]
967.0
148.0
det
z
0
l 1 l l
10 10
.0 0
297 297
.0 0
522 522
2
01
31
967
z
l 1
det det
0 0
l
.0
.0 297
3 l 1 l 3
z
148.0 .0 522
2
2
2
(1) (1)
z z
l 3 3 l
.1 1
489 489
) )
z z
.1( 1(
.2 2
l 753 753 l
.0 0
549 549
) )
0 0
l ( ( l 1
2
l 413 413 l 1
2
The desired characteristic equation:
(2)
z
z
.0
0468
0
49.02
(
z
)(13.0
z
0)36.0
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 62
Balancing the coefficients of the equations (1) and (2): Balancing the coefficients of the equations (1) and (2):
489.1
49.0 l 753.2
549.0
.0
0468
l 3 2 l 413.1 1
2
l 1
Solve the above set of equations, we have: Solve the above set of equations we have:
653.2
544.1
1l 2l
Conclusion
653
.1
544
.2L
T T
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 63
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 64
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 65
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