Bài giảng Fundamentals of Control Systems - Chapter 7: Mathematical model of continuous systems - Huỳnh Thái Hoàng

kekeku
)(2)(10)( 1 zEzzEzU  
1210)()(  zzUzGC
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 27
)(zE
Calculate TF from block diagram – Example 3 (cont’)
 sG )( 20 s


s
zzG )1()( 1 Z
  2.05e s
2
.5)(
s
esG


2 )1()20( zz
)1(10 z

  31)1( sz Z 3
11
)1(2
.)1(5 

z
zz
2)1(
.)(  zzzG
 sHsG )()(


s
zzGH )1()( 1 Z 1.0)( sH
 sG )()1(10 1
z
zzT
s 3
2
3 )1(2
)1(1




Z

s
z. Z
)1(01.0)(  zzGH
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 28
sTs eez 2.02)1( zz
Calculate TF from block diagram – Example 3 (cont’)
Th l d l t f f ti e c ose - oop rans er unc on:
 

 
2
)1(1.0.210 zz
)()(1
)()()(
zGHzG
zGzGzG
C
C
k 
 

 
 
2
)1(01.0.2101
)1(
zz
zzz
  )1(zzz
2.08.02  zz
1
)1(1.0)(
210)(

 
zG
zzGC

02.008.01.12
)( 234  zzzzzGk
2
)1(01.0)(
)1(


zzGH
zz
z
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 29
2)1( zz
State-space model of discrete system
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 30
The discrete state space (SS) equation
 The state-space model of a discrete system is a set of first-
order difference equations of the form:
  )()()1( krkk BxAx
  )()( kky d
dd
xC
h
 n
aaa  11211 b1
w ere:
 )(1 kx

n
d
aaa

 22221A 



d b
2B



 )()( 2 kxk x  nnnn aaa 21  nb
 C
 )(kxn
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 31
nd ccc 21
Derive SS equation from difference equation
 Case 1: The right hand side of the difference equation-
does not involve the differences of the input:
)()()1()1()( kubkyakyankyankya  ... 0110 nn  
 Define the state variables:
The first state variable is the 
t t f th t
)()(1  kykx
ou pu o e sys em;
The ith state variable (i=2..n) 
is set to be one sample time- )1()(
)1()(
23
12


kxkx
kxkx
advanced of the (i1)th state 
variable. )1()( 1   kxkx nn

6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 32
Derive SS equation from difference equation
Case 1 (cont’)
  )()()1( kukk dd BxAx
 The state equations:
  )()( kky d xC
where:
 0010  0



 0100






 0




)(
)(
2
1
kx
kx




121
1000
aaaa
d 
A




0
0
b
dB




)(
)(
kx
k x

 
0000 aaaa
nnn   0a
 
n
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 33
0001 dC
Derive SS equation from difference equation – Case 1 example
 Write the state equations of the system described by:
  )()( kykx
)(3)(4)1(5)2()3(2 kukykykyky 




)1()(
)1()( 12
1
kxkx
kxkx Define the state variables:
23
 The state equations:   )()()1( krkk dd BxAx  )()( kky d xC
 010010 



 0
0
0
0
dB
where:




 



50522
100 100 
123 aaa
dA  5.1
0
0
a
b
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 34
..
000 aaa  001dC
Derive SS equation from difference equation
 Case 2: The right-hand side of the difference equation
involve the differences of the input:
 )()1()1()( 110 kyakyankyankya
 Define the state variable:
... nn
)()1(...)2()1( 1210 kubkubnkubnkub nn  
The first state variable is 
the output of the system; )()(1 kykx 
The ith state variable (i=2..n) 
is set to be one sample 
time ad anced of the (i 1)th
)()1()(
)()1()(
223
112
kukxkx
kukxkx




- v 
state variable minus a 
quantity proportional to the )()1()( 11 kukxkx nnn   

6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 35
input
Derive SS equation from difference equation
Case 2 (cont’)

 
)()(
)()()1(
kk
kukk dd
C
BxAx
 The state equation: y d x
where:
 0010



 0100









2
1




)(
)(
2
1
kx
kx




121
1000
aaaa nnn
d 
A




n
d
 1
B




)(
)(
kx
k
n
x

 
0000 aaaa
  n
 
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 36
0001 dC
Derive SS equation from difference equation
Case 2 (cont’) 
The coefficient i in the vector Bd are defined as:
0
0
1 a
b 
0
111
2 a
ab  
0
12212
3 a
aab  
1122111 aaab nnnn    

6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 37
0a
n
Derive SS equation from difference equation – Case 2 example
 Write the state equations of the system described by:
  )()( kykx
)(3)2()(4)1(5)2()3(2 kukukykykyky 




)()1()(
)()1()( 112
1
krkxkx
krkxkx

 Define the state variables:
 223
 The state equations:  )()()1( kukk dd BxAx   )()( kky d xC
 010010 
 1

B
where:






50522
100 100 
123 aaa
dA 





3
2

d
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 38
 ..
000 aaa  001dC
Derive SS from difference equation – Case 2 example (cont’)
 Th ffi i t  i th t B l l t de coe c en i n e vec or d are ca cu a e as:
  5010b
  25.05.010
.
2
111
2
0
1
ab
a




 375.0
2
5.05)25.0(13
2
12212
3
0
aab
a

0a
 50






3750
25.0
.
dB
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 39
.
Formulation of SS from block diagram
y(t)r(t) e (t)e(kT)e(t)
+ G(s)ZOHT
R
 Step 1: Write the state space equations of the open-loop
ti tcon nuous sys em:
y(t)
G(s)
eR(t)

 
)()(
)()()(
tt
tett R
Cx
BAxx
y
 Step 2: Calculate the transient matrix:
)]([)( 1 st  L
  1
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 40
)( -ss AI where
Formulation of SS equations from block diagram (cont’)
 Step 3: Discretizing the open-loop continuous SS equation:
G(s)ZOH
e(kT) y(kT)
  )()(])1[( kTekTTk Rdd BxAx 
 

A
T
d T )(
with  )()( kTkTy d xC




CC
B
d
d Bd
0
)( 
 Step 4: Write the closed-loop discrete state equations (which
has input signal r(kT))
   )()(])1[( kTrkTTk dddd BxCBAx
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 41
  )()( kTkTy d xC
Formulation of SS equations from block diagram – Example
F l t th SS ti d ibi th t ormu a e e equa ons escr ng e sys em:
y(t)r(t) e (t)e(kT)e(t) x x
+ ZOHT
R
s
1
as 
1 K2 1
where a = 2, T = 0.5, K = 10
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 42
Formulation of state equations from block diagram – Example (cont’)
 Solution:
y(t)eR(t)
s
1
2
1
s 10
x2 x1 Step 1:
s
sXsX )()( 21  )()( 21 sXssX  )()( 21 txtx 
2
)()(2  s
sEsX R )()()2( 2 sEsXs R )()(2)( 22 tetxtx R

)(
1
0
)(
)(
20
10
)(
)(
2
1
2
1 te
tx
tx
tx
tx
R











 BA
   )(010)(10)( 11 txtxty
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 43
 )(2 txC
Formulation of state equations from block diagram – Example (cont’)
 Step 2: Calculate the transient matrix
 
11
1
20
1
20
10
10
01
)(



















s
s
sss -AI






 
1
)2(
11
0
12
)2(
1 ssss
 

2
0
s
sss
   1111 11 LL





 



 













10
)2(
2
10
)2()]([)(
1
11 ssssssst
L
LL
 2ss



 
 tet
2 )1(
2
11)(
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 44
  te 20
Formulation of state equations from block diagram – Example (cont’)
 Step 3: Discretizing the open   )()(])1[( kTekTTk BxAx-
loop continuous state equations:   )()( kTkTy d
Rdd
xC
 11














368.00
316.01
0
)1(
2
1
0
)1(
2
1)(
5.02
5.02
2
2
e
e
e
eT
Tt
t
t
dA
 








 












 



 TTT
d d
e
ed
e
ed
0 2
2
0 2
2
0
)1(
2
1
1
0
0
)1(
2
11)( 




BB






 







 

092.02
1
22
5.0
22 22
5.02
2
2 ee
T


 



 
  316.0
2
1
22
5.02
0
2 ee 
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 45
 010 CCd
Formulation of state equations from block diagram – Example (cont’)
 Step 4: The closed-loop discrete state equations:
 

 
)()(
)()(])1[(
kTkT
kTrkTTk dddd
C
BxCBAx
y d x
     316.0080.0010092.0316.01CBAwhere
 Conclusion: The closed-loop state equation is:
 368.0160.3316.0368.00ddd
)(
092.0)(316.0080.0)1( 11 kr
kxkx


316.0)(368.0160.3)1( 22 kxkx  
   )(010)( 1 kxky
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 46
 )(
.
2 kx
Calculate transfer function from state equation
 Given the state equation
  )()()1( kukk dd BxAx  )()( kky d xC
Th di t f f ti i e correspon ng rans er unc on s:
zzYzG BAIC 1)()()(  dddzU )(
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 47
Calculate transfer function from state equation - Example
 Calculate the TF of the system described by the SS equation:




)()(
)()()1(
kky
kukk
d
dd
xC
BxAx



1070
10
dA 


2
0
dB  01dC
 Solution: The transfer function is:
 ..
ddd zzG BAIC
1)()( 
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