Fundamentals of Control Systems - Chapter 7: Mathematical model of discrete time Control System

( −−== zzUzGC⇒
16 November 2011 © 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⇒
16 November 2011 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 28
sTs eez 2.0==2)1( −zz
Calculate TF from block diagram – Example 3 (cont’)
‘ The closed-loop transfer function:
⎥⎤⎢⎡ +⎥⎦
⎤⎢⎣
⎡ −
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
16 November 2011 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 29
2)1( −zz
State-space model of discrete system
16 November 2011 © 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 K 11211 ⎤⎡b1
w ere:
⎤⎡ )(1 kx
⎥⎥⎢⎢=
n
d
aaa
MMM
K 22221A ⎥⎥
⎥
⎢⎢
⎢
=d bM
2B⎥⎥
⎥
⎢⎢
⎢
= )()( 2 kxk Mx ⎥⎦⎢⎣ nnnn aaa K21 ⎥⎦⎢⎣ nb
[ ]C
⎥⎦⎢⎣ )(kxn
16 November 2011 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 31
nd ccc K21=
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()( kbkkkk ... 0110 uyayanyanya nn =++++−+++ −
‘ Define the state variables:
ŽThe first state variable is the 
f h
)()(1 = kykx
output o t e system;
ŽThe 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
M
16 November 2011 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 32
Derive SS equation from difference equation
Case 1 (cont’)
⎨⎧ +=+ )()()1( krkk dd BxAx
‘ The state equations:
⎩ = )()( kky d xC
where:
⎤⎡ 0010 ⎤⎡ 0
⎥⎥
⎥
⎢⎢
⎢ 0100
MMMM
K
K
⎥⎥
⎥
⎢⎢
⎢ 0
M⎥⎥
⎤
⎢⎢
⎡
)(
)(
2
1
kx
kx
⎥⎥
⎥
⎢⎢
⎢=
121
1000
aaaa
d
K
A
⎥⎥
⎥
⎢⎢
⎢=
0
0
b
dB
⎥⎥⎦⎢
⎢
⎣
=
)(
)(
kx
k Mx
⎥⎦⎢⎣
−−−− −−
0000 aaaa
nnn K ⎥⎦⎢⎣ 0a
[ ]
n
16 November 2011 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 33
0001 K=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
16 November 2011 © 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
‘ D fi th t t i bl
−... nn
)()1(...)2()1( 1210 kubkubnkubnkub nn −− ++++−++−+
e ne e s a e var a e:
ŽThe first state variable is the 
output of the system;
)()(1 kykx = 
ŽThe ith state variable (i=2..n) 
is set to be one sample time- )()1()(
)()1()(
223
112
krkxkx
krkxkx
−+=
−+=
β
β
advanced of the (i−1)th state 
variable minus a quantity 
i l h i
)()1()( 11 krkxkx nnn −− −+= β
M
16 November 2011 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 35
proport ona to t e nput
Derive SS equation from difference equation
Case 2 (cont’)
⎩⎨
⎧ +=+
)()(
)()()1(
kk
krkk dd
C
BxAx
‘ The state equation: =y d x
where:
⎤⎡ 0010
⎥⎥
⎥
⎢⎢
⎢ 0100
MMMM
K
K
⎥⎥
⎤
⎢⎢
⎡
β
β
2
1
⎥⎥
⎤
⎢⎢
⎡
)(
)(
2
1
kx
kx
⎥⎥
⎥
⎢⎢
⎢=
121
1000
aaaa nnn
d
K
A
⎥⎥
⎥
⎢⎢
⎢=
−n
d
β 1
MB
⎥⎥⎦⎢
⎢
⎣
=
)(
)(
kx
k
n
Mx
⎥⎦⎢⎣
−−−− −−
0000 aaaa
K ⎥⎦⎢⎣ nβ
[ ]
16 November 2011 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 36
0001 K=dC
Derive SS equation from difference equation
C 2 ( t’)ase con
The coefficient βi in the vector Bd are defined as:
0
0
1 a
bβ =
0
111
2 a
ab ββ −=
0
12212
3 a
aab βββ −−=
1122111 aaab nnnn ββββ −−−− −−−−= K
M
16 November 2011 © 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
β
β‘ Define the state variables:
− 223
‘ The state equations: ⎨⎧ +=+ )()()1( kukk dd BxAx ⎩ = )()( kky d xC
⎤⎡ 010010 ⎥
⎤⎢⎡ 1β
β
B
where:
⎥⎥
⎥
⎦⎢
⎢⎢
⎣
==
50522
100 100 
123 aaa
dA ⎥
⎥
⎦⎢
⎢
⎣
=
3
2
β
d
16 November 2011 © 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 β i th t B l l t de coe c en i n e vec or d are ca cu a e as:
⎪⎧ === 5010bβ
⎪⎪⎨ −=×−=−= 25.05.010
.
2
111
2
0
1
ab
a
ββ
⎪⎪
⎪
⎩
=×−−×−=−−= 375.0
2
5.05)25.0(13
2
12212
3
0
aab
a
βββ
0a
⎤⎡ 50
⎥⎥
⎥
⎦⎢
⎢⎢
⎣
−=
3750
25.0
.
dB⇒
16 November 2011 © 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
continuous system:
y(t)
G(s)
eR(t) ⎩⎨
⎧ +=
)()(
)()()(
tt
tett R
Cx
BAxx&
=y
‘ Step 2: Calculate the transient matrix:
)]([)( 1 st Φ=Φ −L
( ) 1
16 November 2011 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 40
)( -ss AI −=Φwhere
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 ⎪⎪
⎧ Φ=
∫
A
T
d T )(
with⎩ = )()( kTkTy d xC
⎪⎪⎩
⎨
=
Φ=
CC
B
d
d Bd
0
)( ττ
‘ Step 4: Write the closed-loop discrete state equations
( hi h h i t i l (kT))w c as npu s gna r
[ ]⎨⎧ +−=+ )()(])1[( kTrkTTk dddd BxCBAx
16 November 2011 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 41
⎩ = )()( kTkTy d xC
Formulation of SS equations from block diagram – Example
l h SS i d ibi h‘ Formu ate t e equat ons escr ng t e system:
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
16 November 2011 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 42
Formulation of state equations from block diagram – Example (cont’)
‘ S l ti
y(t)eR(t)
s
1
2
1
+s 10
x2 x1‘ Step 1:
o u on:
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⎥⎦
⎤⎢⎣
⎡+⎥⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡
−=⎥⎦
⎤⎢⎣
⎡
43421&
&
⇒ BA
[ ] ⎥⎤⎢⎡== )(010)(10)( 11 txtxty
16 November 2011 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 43
⎦⎣ )(2 tx321C
Formulation of state equations from block diagram – Example (cont’)
‘ Step 2: Calculate the transient matrix
( )
11
1
20
1
20
10
10
01
)(
−−
⎟⎟⎠
⎞
⎜⎜⎝
⎛
⎥⎦
⎤⎢⎣
⎡
+
−=⎟⎟⎠
⎞
⎜⎜⎝
⎛
⎥⎦
⎤⎢⎣
⎡
−−⎥⎦
⎤⎢⎣
⎡=−=Φ
s
s
sss -AI
⎥⎥
⎤
⎢⎢
⎡
+=⎥⎦
⎤⎢⎣
⎡ +=
1
)2(
11
0
12
)2(
1 ssss
⎥⎥⎦⎢⎢⎣ +
+
2
0
s
sss
⎤⎡ ⎬⎫⎨⎧⎬⎫⎨⎧⎫⎧ ⎤⎡ 1111 11 LL
⎥⎥
⎥⎥
⎦⎢
⎢⎢
⎢
⎣ ⎭⎬
⎫
⎩⎨
⎧
⎭⎩ +⎭⎩=
⎪⎪⎭
⎪⎪⎬
⎪⎪⎩
⎪⎪⎨
⎥⎥
⎥⎥
⎦⎢⎢
⎢⎢
⎣
+=Φ=Φ
−
−−
−−
10
)2(
2
10
)2()]([)(
1
11 ssssssst
L
LL
++ 2ss
⎥⎥
⎤
⎢⎢
⎡ −=Φ
− tet
2 )1(
2
11)(⇒
16 November 2011 © 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 loop ⎧ +=+ )()(])1[( kTekTTk BxAx-
continuous state equations: ⎩⎨ = )()( kTkTy d
Rdd
xC
⎤⎡⎤⎡ 11
⎥⎦
⎤⎢⎣
⎡=⎥⎥⎦⎢
⎢
⎣
−=⎥⎥⎦⎢
⎢
⎣
−=Φ=
×−
×−
=
−
−
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
∫∫∫ ⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧
⎥⎥⎦
⎤
⎢⎢⎣
⎡ −=
⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧
⎥⎦
⎤⎢⎣
⎡
⎥⎥⎦
⎤
⎢⎢⎣
⎡ −=Φ=
−
−
−
− TTT
d d
e
ed
e
ed
0 2
2
0 2
2
0
)1(
2
1
1
0
0
)1(
2
11)( ττττ
τ
τ
τ
τ
BB
⎥⎤⎢⎡⎥
⎥⎤⎢⎢
⎡
⎟⎟⎠
⎞
⎜⎜⎝
⎛ −+
⎥⎥
⎤
⎢⎢
⎡
⎟⎟⎠
⎞
⎜⎜⎝
⎛ +
×−−
092.02
1
22
5.0
22 22
5.02
2
2 ee
Tττ
⎦⎣=⎥⎥⎦⎢
⎢
⎣ +−
=
⎥⎥⎦⎢
⎢
⎣ −
= ×−− 316.0
2
1
22
5.02
0
2 ee τ
16 November 2011 © 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
16 November 2011 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 46
⎦⎣ )(
.
2 kx
Calculate transfer function from state equation
‘ Gi th t t tiven e s a e equa on
⎨⎧ +=+ )()()1( kukk dd BxAx⎩ = )()( kky d xC
Th di f f i i‘ e correspon ng trans er unct on s:
zzYzG BAIC 1)()()( −−== dddzU )(
16 November 2011 © 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)()( −−=
[ ] ⎥⎤⎢⎡⎟⎟⎞⎜⎜⎛ ⎥⎤⎢⎡⎥⎤⎢⎡=
− 01001
01
1
z ⎦⎣⎠⎝ ⎦⎣ −−
−⎦⎣ 21.07.010
2)(G⇒
16 November 2011 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 48
7.01.02 ++= zzz