Chapter 3 State-Variable Models 1 •State-Variable Models • State equations • State Variables of a Dynamic System • The concept of State • Form of the State Equations • The State Differential Equation • Transfer Function of a State Space Model • The State Transition Matrix • Characteristic Equation and Eigenvalues • Controllability & Observability
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
Chapter 3 State-Variable Models
1
•State-Variable Models• State equations• State Variables of a Dynamic System• The concept of State• Form of the State Equations• The State Differential Equation• Transfer Function of a State Space Model• The State Transition Matrix• Characteristic Equation and Eigenvalues• Controllability & Observability
• The time-domain is the mathematical domain that incorporates the response and description of a system in terms of time, t.
State-Variable Models
Differential equations
Modeling
Nonlinear&Time Varying
LTI Transfer Function
Lapalce Transform
Inverse Lapalce Transform
Physicalsystem
State-VariableModels
State-VariableModels
(cause) (effect)
2
State-Variable Models
3
1 2
1 0 2 0 0
Consider this system shown in the above Figure. A set ofstate variables ( , , , ) for system is a set such that knowledge of the initial values of the state variables[ ( ), ( ), , ( )] at the
n
n
x x x
x t x t x t
…
… 0
1 2 0
initial time t and of the input signals ( ) and ( ) for will be sufficient to determine the future values of the outputs and state variables.
u t u t t t>
System (x)u(t)
Input
x(0) Initial conditions
y(t)
Output
The general form of a dynamic system is schematically shown as
State-Variable Models• State differential equations are an alternative way to describe
a dynamic system (i.e. time-domain method).
• The state-variable model, or state-space model is a particular differential equation model.
• Equations are expressed as n first-order coupled differential equations, but the choice of states is not unique
• All choices of state variables preserve the system’s input-output relationship (that of the transfer function)
4
State Equations
• Any nth order differential equation can easily be converted into a set of 1st order state equations for nonlinear or time-varying systems, the transfer function approach often breaks down.
• Most multivariable and many stochastic design methods are based on state equations
5
• For LTI systems, it is routine to move between state and transfer function representations i.e. between frequencyand time domains
The Concept of State
•Because the choice of states for a given system is not unique, we often choose states that represent physical measurements – voltages, velocity, position, etc.
• However, it may also be convenient to choose states that simplify the mathematical form of the state equations – for example the output and its derivatives – but are not easily measurable.
6
Form of the State EquationsThe state description always consists of two sets of equations, normally written in matrix form. The first set of equations describes the dynamics, and is a set of first order differential equations, each expressing the first derivative of a state as a function of all the (n) states and the (m) inputs (with no derivatives):
))(),(()(or
))(),(),()(),(()(
))(),(),()(),(()())(),(),()(),(()(
m121
m12122
m12111
tutxftx
tututxtxtxftx
tututxtxtxftxtututxtxtxftx
m
nnn
n
n
=
=
==
7
Form of the State Equations
))(),(()())(),(()(
tutxgtytutxftx
==
The second set of equations has no dynamics, and expresses the outputs as a function of the states and inputs:
))(),(()(or
))(),(),()(),(()(
))(),(),()(),(()())(),(),()(),(()(
m121
m12122
m12111
tutxgty
tututxtxtxgty
tututxtxtxgtytututxtxtxgty
npp
n
n
=
=
==
8
Linear Time Invariant Case
In this case, the functionsf and g become linearcombinations of x and ugiving the familiar form:
• If the are n states, m inputs, and p outputs, then A is square (nxn), B is (nxm), C is (pxn) and D is (pxm)
• For a single input, single output system, we have A square (nxn), B=b and is a (nx1) column vector, C=c is a (1xn) row vector, and D=d is a (1x1) scalar (often zero)
))()(()())()(()(
tu,txgtytu,txftx
==
DuCxyBuAxx
+=+=
9
1 11 1 12 2 1 11 1 1
2 21 1 22 2 2 21 1 2
1 1 2 2 1 1
n n m m
n n m m
n n n nn n n nm m
x a x a x a x b u b ux a x a x a x b u b u
x a x a x a x b u b u
= + +… + += + +… + +
= + +… + +
Form of the State Equations
1 2
The state of a linear time in-varying system is described by the set of firstorder differential equations written in terms of the state variables [ ... ]and can be written in general form as:
nx x x
1 11 12 1 111 1 1
2 21 22 2 2
11 2
nm
n
n nm mn n n nn n
x a a a xb b u
x a a a x
b b ux a a a x
= +
or in matrix form as follows:
Form of the State Equations1 11 12 1 1
11 1 12 21 22 2 2
11 2
nm
n
n nm mn n n nn n
x a a a xb b u
x a a a x
b b ux a a a x
= +
n: number of state variables, m: number of inputs.
The column matrix consisting of the state variables is called thestate vector and is written as
1
2
n
xx
x
x
=
Form of the State Equations1 11 12 1 1
11 1 12 21 22 2 2
11 2
nm
n
n nm mn n n nn n
x a a a xb b u
x a a a x
b b ux a a a x
= +
x A x Bu= +
1 11 12 111 1 1
2 21 22 2
11 2
, , ,
nm
n
n nm mn n n nn
x a a ab b u
x a a ax A B u
b b ux a a a
= = = =
Example:
State variables=?
)()()()(2
2tutky
dttdyb
dttydM =++
)()()()(12
2 tutkxtbxdt
tdxM =++
Thus the following two first-order DE’s
13
Mtutx
Mktx
Mb
dttdx
txdt
tdx
)()()()(
)()(
122
21
+−−=
=
1 1
2 2
0 1 0( )1
x xu tk bx x
M M M
= + − −
Example:how about ? )()()()()(
011
1
1 tutyadt
tdyadt
tydadt
tyda n
n
nn
n
n =++++ −
−
−
14
12
1
1 1 0 1
1 ( ( ))
nn
nn n
n
dx xdt
dx xdt
dx a x a x u tdt a
−
− −
=
=
= − + + +
111 2Let , , , .n
ndxdxx y x x
dt dt−= = =
1 1
2 2
0 1 1 1
0 1 00
0 0 0
1// / / n
n n n n n n
x xx xd u
dta
x a a a a a a x−
= +
The State Differential Equation
=
=
=
=
yy
xx
xyy
xx
x
2
1
2
1
)(2 2 tuyyyyy
+−−=
=
ωως
)(10
210
2 tuyy
yy
+
−−
=
ωςω
CxyBuAxx
=+=
)(2 2 tuyyy =++ ωως
[ ]01 10
210
2 =
=
−−
= CBAωςω
• Consider the second-order system:
Its state-space description is
With
15
Series RLC Circuit)()()( )( : 0v i tvtvtvtv CRL ++∑ ==
)()( )( tvRtidtdiLtv C++=
Cti
dtdv
dtdvCti CC )()( =⇔=
cvxix == 21 let
)()()( tvtvRtidtdiL C +−−=
Cti
dtdvC )( =
)(11 211 tv
Lx
Lx
LR
dtdx
+−−=
Cx
dtdx 12 =
ODE:
16
) : that(Notice 12 dt
dxx ≠
Series RLC Circuitcvxix == 21 let
11 2
2 1
1 1 ( ) dx R x x v tdt L L Ldx xdt C
= − − +
=
1 1
2 2
1 1( )
1 00 ux x
BA
Rx xL L v tLx x
C
− − = +
17
[ ] [ ] 12
2
If we let as the output, then = 0 1 , 0 1C C
xv C y x v
x
= = =
[ ] [ ] 11
2
If we let ( ) as the output, then = 1 0 , 1 0 ( ).x
i t C y x i tx
= = =
x Ax Bu= +
An RLC circuit. :0∑ =ii
RivdtdiL
itudt
dvCi
LcL
Lc
c
−=
−== )(
∑ = : 0v i
LcL
Lc
RiLRv
Ldtdi
tuC
iCdt
dv
−=
+−=
1
)(11
Lc ixvx == 21 let 21
2
21
1
)(11
xLRx
Ldtdx
tuC
xCdt
dx
−=
+−=
)( Lc iitu +=
RidtdiLvvv L
LoLc +=+=
18
An RLC circuit.Lc ixvx == 21 let
212
21
1
)(11
xLRx
Ldtdx
tuC
xCdt
dx
−=
+−=
19
1 1
2 2
1 10( )
1 0 ux x
BA
x xC u tCx xR
L L
− = + −
2( ) ( )oy t v t Rx= = ⇒
x Ax Bu= +
[ ]0c R=
[ ] 12 0
2
0 L
xy R Rx Ri v
x
= = = =
An RLC circuit.Lc ixvx == 21 let
212
21
1
)(11
xLRx
Ldtdx
tuC
xCdt
dx
−=
+−=
20
1 1
2 2
1 10( )
1 0 ux x
BA
x xC u tCx xR
L L
− = + −
2If we want ( ) ( )oy t v t Rx= = ⇒
x Ax Bu= +
[ ]0c R=
[ ] 12 0
2
0 L
xy R Rx Ri v
x
= = = =
What happen if is the output?Lv
An RLC circuit.Lc ixvx == 21 let
21
1 1
2 2
1 10( )
1 0 ux x
BA
x xC u tCx xR
L L
− = + −
What happen if is the output?Lv
1 1
2 1 1 2
let
c
L c o L
v x xv x v v x Ri x Rx
∗
∗
= =
= = − = − = −
*1 1
* * *2 1 2 1 2
1 1( ) ( )
x x
x x x x xR R
= ⇔
= + = +
[ ] 11 2 0
2
1 c L
xy R x Rx v v v
x
= − = − = − = ⇒
[ ]1c R= −
State Equation to SFG
LcL
Lc
RiLRv
Ldtdi
tuC
iCdt
dv
−=
+−=
1
)(11
1
2
let
c
L
x vx i
==
1 2
2 1 2
1 1 ( )
1
x x u tC C
Rx x xL L
= − +
= −
Output: .Rxtvty o 21 )()( ==
U(s) Vo(s)1x 2x
1x 2xs1
s1
RC1
C1
−
L1
LR
−
22
Inverted Pendulum on a Cart
State variables?
θMgLT =
23
Presenter
Presentation Notes
The cart must be moved so that mass m is always in an upright position. The state variables must be expressed in terms of the angular rotation ɵ(t) and the position of the cart y(t). The differential equations describing the motion of the system can be obtained by writing the sum of the forces in the horizontal direction and the sum of the moments about pivot point. Assume that M >> m and the angle of rotation ɵ is small so that the equations are linear. u(t) - force on the cart I is the distance from the mass m to the pivot point.
Inverted Pendulum on a Cart
2 2
2 2( ) cosθ ( ) 0d y d θM m mL u tdt dt
+ + − =
The sum of the forces in the horizontal direction:
The sum of the torques about the pivot point:2 2
22 2 0d y d θmL mL mLg θ
dt dt+ − =
dttdxtx
dttdyxtyx )( )( )( )( 4321
θθ ====
mg
2
2
dtdlm θ
2 2
2 2( )d θ d ymLdt dt
← +
2 2
2 2( ) ( ) 0 for small d y d θM m m L u t θdt dt
+ + − =
24
Presenter
Presentation Notes
The differential equations describing the motion of the system can be obtained by writing the sum of the forces in the horizontal direction and the sum of the moments about pivot point. Assume that M >> m and the angle of rotation ɵ is small so that the equations are linear. u(t) - force on the cart I is the distance from the mass m to the pivot point. [Peiveit]
Inverted Pendulum on a Cart
2 4( ) ( ) 0dx dxM m m l u tdt dt
+ + − =
The state variable for the second-order equations are:
The sum of the torques about the pivot point:
0 342 =−+ xg
dtdxl
dtdx
)()( 4321 dtd,,
dtdy,yx,x,x,x θθ=
25
Presenter
Presentation Notes
To obtain the necessary first-order differential equations, we solve for l dx4/dt and substitute into first one.
Inverted Pendulum on a Cart2 4( ) ( ) 0dx dxM m m l u t
dt dt+ + − =
0342 =−+ xg
dtdxl
dtdx
)(32 tuxgm
dtdxM =+
23
1 ( )
1 ( ) (for )
dx m g x u tdt M M
u t M mM
= − +
≈ >>
dtdxgx
dtdxl 2
34 −=
26
(horizontal direction)
(about the pivot point)
2 4( ) ( ) 0dx dxM m m l u tdt dt
+ + − =
Presenter
Presentation Notes
To obtain the necessary first-order differential equations, we solve for l dx4/dt and substitute into first one.
Inverted Pendulum on a Cart2 4( ) ( ) 0dx dxM m m l u t
dt dt+ + − =
0342 =−+ xg
dtdxl
dtdx
)(32 tuxgm
dtdxM =+
23
1 ( )
1 ( ) (for )
dx m g x u tdt M M
u t M mM
= − +
≈ >>
dtdxgx
dtdxl 2
34 −=
27
(horizontal direction)
(about the pivot point)
2 4( ) ( ) 0dx dxM m m l u tdt dt
+ + − =
(horizontal direction)
Presenter
Presentation Notes
To obtain the necessary first-order differential equations, we solve for l dx4/dt and substitute into first one.
Inverted Pendulum on a Cart
0342 =−+ xg
dtdxl
dtdx
)(32 tuxgm
dtdxM =+
2 1 ( ) (for )dx u t M mdt M
≈ >>
28
(horizontal direction)
(about the pivot point)
43
1 ( ) 0dxu t l g xM dt
+ − =
0)(34 =+− tuxgM
dtdxlM
Presenter
Presentation Notes
To obtain the necessary first-order differential equations, we solve for l dx4/dt and substitute into first one.
Inverted Pendulum on a Cart
)t(ulM
xlg
dtdx
xdt
dx
)t(uM
xM
gmdt
dx
xdtdx
1
1
34
43
32
21
−=
=
+−=
=
The four first-order differential equations:
CxyBuAxx
=+=
[ ]0100 =C29
0 1 0 00 0 00 0 0 10 0 0
mgM
gl
A−
=
1
1
0
0M
Ml
B
−
=
Presenter
Presentation Notes
To obtain the necessary first-order differential equations, we solve for l dx4/dt and substitute into first one.
Transfer Function of a State Space Model
)()(A)( sBUsXssX +=
)()(C)()()(A)(
tDutxtytButxtx
+=+=
[ ] )()( sBUsXsI-A =
Taking the Laplace transform (assume zero initial conditions)
So,
[ ] 1( ) ( )X s sI A BU s−= −
)()(C)( sBUsXsY +=
30
)( ωσ js +=
Transfer Function of a State Space Model[ ] )()( 1 sBUsI-AsX −=
)()(C)( sBUsXsY +=Substituting into the output equation
Therefore, for single variable case, the transfer function of the system is
For multivariable case,
yields [ ] )()()(}{)( 1 sUsGsUDBsI-ACsY =+= −
)()()( sU/sYsG =
)()()( sU/sYsG jiij =
31
TF from State Space Model Example
3410
)()()( 2 ++
==sssU
sYsG
(a)
?)( =sG
A,B,C and D √
[ ] )()()(}{)( 1 sUsGsUDBsI-ACsY =+= −
[ ]xy
uxx
01010
4310
=
+
−−
=
32
TF from State Space Model Example
uxxx +−−= 212 43
110xy =
uyyy 1034 =++
10/10/
10/
12
12
1
yxxyxx
yx
====
=
3410
)()()( 2 ++
==sssU
sYsG
21 xx =
)(10)()34()(10)(3)(4)(
2
2
sUsYsssUsYssYsYs
=++
=++
(b)
?)( =sG[ ]xy
uxx
01010
4310
=
+
−−
=
Dynamics equation:
33
uxxx +−−= 212 43
TF from State Space Model Example
11
1
1)()()(
asasasUsYsG n
nn
n ++== −
+
11 1( ) ( ) ( )n n
n na s a s a Y s U s−+ + + =
L.F:
)()()()()(121
11 tutya
dttdya
dttyda
dttyda n
nnn
nn =++++ −
−
+
[ ]
+
==
− equation State
eqaution dyanmic:LF)()(
)(1 DBsI-ACsUsY
sG
34
TF from State Space Model Example
1
2 1
3 2
1
1
Let
n
n n
x yx x yx x y
d yx xdt
−
−
== == =
= =
1 2
1 1 2 21
1
1 ( );n n nn
x x
x a x a x a x ua
y x+
=
= − − − +
=
−−−
=
+++ 11211 ///100000010
nnnn aaaaaa
A
=
+1/1
00
na
B
[ ]001 =C D=?
? )()()()()(121
1
1 tutyadt
tdyadt
tydadt
tyda n
n
nn
n
n =++++ −
−
+
[ ] 1( )( )( )
Y sG s C sI - A B DU s
−= = + ⇒35
11 1
1 ( ) n nn n
G sa s a s a−
+
=+ + +
Realization of a Transfer Function,x Ax Bu y Cx Du= + = +• A state description is a
realization of G(s) if 1[ ] ( )C sI A B D G s−− + =
• A realization of G(s) is minimal if there exists no realization of G(s) of less order
• An LTI systems is observable if the initial state x(0) can be uniquely deduced from knowledge of u(t) and y(t) fort Є [0 T]
• An LTI system is controllable if for every x(t0) and every T >0, there exist u(t0+t), 0<t ≤ T such that x(t0+T) =0
36
From Transfer Function to State Space• For a given transfer function, there is no unique state
space realization
• Engineering dictates the use of a realization of leastorder, a minimal realization (A realization of G(s) is minimal if there exists no realization of G(s) of less order)
• A minimal realization is both controllable and observable
• All the possible A matrices for the different space realizations should have the same eigenvalues
37
[ ])()()( 1
sDsNDBsI-ACsG =+= −
0) Det( 0)( :Pole =⇒= sI-AsD
From Transfer Function to State Space
3)(1)( )()(1
)(63
6131
1
6)( ==⇔+
=+
=+
= sH,s
sGsGsH
sGs
s
ssT
2)(1)( )()(1
)(2
1121
1
)( ==⇔+
=+
=+
= sH,s
sGsGsH
sGs
s
ssT
Why & How?
1 2 1 2
15 5Δ 5( 1) 1 1( ) 5, 5 ,Δ Δ 1,Δ 1 51Δ 51 5
k kkP ssT s P P
s s ss
+ += = = = = = = = +
++
∑
1
1
5155
5)1(5)( −
−
+
+=
++
=ss
sssGc
38
From Transfer Function to State Space
1
1
5155
5)1(5)( −
−
+
+=
++
=ss
sssGc
39
1
1
12 1 2
ss s
−
−=+ +
1
1
6 63 1 3
ss s
−
−=+ +
From Transfer Function to State Space
1x
1x2x
2x
3x
3x
3211 063 xxxx ++−=
[ ]Xy
trxxx
xxx
X
001
)(150
5002020063
3
2
1
3
2
1
=
+
−−−
−=
=
State-variable differential equation:
]5)([5520 33212 xtrxxxx −++−=
)(500 3213 trxxxx +−−=
1xy =40
From Transfer Function to State Space
))()(()(
)3)(2)(5()1(30)(
)()(
321 sssssssq
sssssT
sRsY
−−−=
++++
==)3()2()5(
)()()( 321
++
++
+==
sk
sk
sksT
sRsY
1x
2x
3x
30 10 20 321 =−=−= kkk
State-variable differential equation:
)(111
300020005
3
2
1
3
2
1
trxxx
xxx
+
−−
−=
diagonal canonical form[ ]
=
3
2
1
3010-20- xxx
y41
The State Transition Matrix Φ(t)
How to compute this matrix?
nRtx ∈)(
42
The State Transition Matrix (u=0)
0)0()()(
)(
000
1
==−=∫=∫
==∈=⇒=
txetxttAx/xlnAdtx
dx
Adtx
dxAx
dt/dxRxifAxxAxx
At
• The state transition matrix satisfies the homogenous (i.e. zero-input) state equation.
• It represents the evolution of the system’s free response to non-zero initial conditions:
“zero input response”
x(0) is initial condition. Hence it is a constant.
Atet =)(φ
=?
)()( tAxtx =
)0((t))( xtx φ=
BuAxx += u=0
43
The State Transition Matrix (u=0)
• If , there is another way to find x(t) using Taylor series expansion
• Successive differentiation of gives:
0 At time
)(
32)3(
==
==
txAx
xAxAx
kk
nRtx ∈)(
Axx =
44
xAxAx 2==
)0()()( xttx φ=
The State Transition Matrix
+
+
++++=
++++=
kk
kk
txA!k
txA!
tAxx
tx!k
tx!
txxtx
)0(1)0(21)0()0(
)0(1)0(21)0()0()(
22
)(2
This series converges for all finite t. It is called the matrix exponential
)0()121( 22 xtA
!ktA
!AtI kk
+++++=
+++++= kkAt tA!k
tA!
AtIe 121 22
+++++= kkAt tA!k
tA!
AtIe 121 22
)0()( xetx At=45
The Matrix Exponential
122121 )( AtAtAtAtttA eeeee ==+
IeA =0
AtAt ee −− =1)( i.e. the matrix exponentialbehaves very much like thefamiliar scalar exponentialfunction. Note that A mustbe square.
T)( AttA eeT
=
AeAe AtAt =
AtAt Aeedtd
=
46
State Transition Matrix (u≠0) Φ(t)BuAxx +=
Buexedtd AtAt −− =)(
=∫ −t A dxe
dtd
0)( ττ
BueAxxe AtAt −− =− )(
∫ −+=∫+= −−tt tAAt dButxtBudexetx00
)( )()Φ()0()Φ()0()( τττττ
Atet =)Φ( 47
BuAxx =−
)0()( 0xetxe AAt −− − ∫= −t A dBue0
)( τττ
∫ −+=t
dButxttx0
)()Φ()0()Φ()( τττ
State transition Matrix (u≠0) –Φ(s)
Which compares with the time domain solution:
Let
BuAxx +=)()()0()( sBUsAXxssX +=−
)()()0()()( 11 sBUAsIxAsIsX -- −+−=
)()0()()( sBUxsXAsI +=−
1)()( -AsIs −=Φ)()()0()()( sBUsxssX ΦΦ +=
∫ −+=t
dτButxttx0
)()()0()()( ττΦΦ
Atet =)Φ( 48
)( ωσ js +=Can it exist?
State Transition Matrix
• Note that the system response has twocomponents:
• Natural response – “zero input response”due to initial conditions
• Forced response – “zero state response”due to input
• Overall response is the sum of the two components
BuAxx +=
∫ −+=t
dτButxttx0
)()()0()()( ττΦΦAtet =)Φ(
1)()( -AsIs −=Φ
49
Example
The time-domain state transition matrix can be obtained using the inverse Laplace transform
−−
=3120
A
+−
=−31
2 then
ss
AsI
−+=−=
ss
sAsIsΦ -
123
)Δ(1][ )( 1
)3)(1(232)3()Δ( with 2 ++=++=++= sssssss
50
Example
0,3 2211 == ααa=1,b=2
( ) ( 1)( 2)s s s∆ = + +
−+=−=
ss
sAsIsΦ -
123
)Δ(1][ )( 1
{ }
+−−+−−
== −−−−
−−−−
tttt
tttt-
eeeeeeeest 22
221
2222)( L)( Φφ
is response free the then11
)0( conditions inital Assuming
=x
)0()()( 2
2
== −
−
t
t
eexttx φ ttttt
ttttt
eeeeeeeeee
222
222
2222
−−−−−
−−−−−
=+−−
=+−−51
Example
Note that for otherinitial conditions,the phase planeplot will not be astraight line.
52
Characteristic Equation and Eigenvalues• Recall that, for a transfer function G(s)=N(s)/D(s),
the roots of the characteristic equation D(s)=0 are the poles of the system.
• Recall that the denominator of the transfer functionof a state-space representation is det(sI-A)
• The characteristic equation is then det(sI-A)=0• The roots of this equation are the eigenvalues of
the matrix A.
[ ] )()()(}{)( 1 sUsGsUDBsI-ACsY =+= −
For the stable system, the real parts of all the eigenvalues must be negative.
For the stable system, what must the egenvialues be ?
53
)( ωσ js +=
Controllability
DuCxyBuAxx
+=+= Theorem: This system is controllable if
and only if the following controllability matrix has rank n:
Note that for a single input system S will be an [n × n] square matrix and the rank test is that
][ 12 BABAABBS n−=
∆[S]≠0
54
Controllability Example
1 20 0
−?
=
−
−=
01
10
12BA
[ ]
:matrixility controllab
== ABBS
det S= 055
Observability
DuCxyBuAxx
+=+=
Theorem: This system is observable if and only if the following observability matrix has rank n:
TnCACACACV ][ 12 −=
56
Observability Example
1 02 0−
det V=
?
[ ]01 01
10
02=
=
−
−= CBA
:matrixity observabil
=
=CAC
V
057
Uncontrollable System
The state x2 cannot be affected by input u and henceis uncontrollable
58
Unobservable System
The state x2 does not affecte y and hence is unobservable
59
Controllability and Observability
60
Basic Idea of State Feedback
Consider the state feedback controller where is a constant feedback gain matrix
Then one can write
Whereas the poles of the open-loop system are given by the eigenvalues of A, the poles of the closed-loop systemare given by the eigenvalues of (A-BK).
BuAxx +=
rKxu +−=
BrxBKAr-KxBAxx
+−=++=
)( )(
61
State Feedback Design
-
Control system with state feedback
•The poles of the closed-loop system can be arbitrarily assigned if and only if the system is controllable
BrxBKAr-KxBAxx
+−=++=
)( )(
62
State Feedback Design•Often, states are not all measurable. Hence, it is
necessary to design an observer to construct them from the output vector.
• Such an observer can be designed if and only if the system is observable.
• The observed state is then used instead of the true state to generate the feedback.
63
State Variable Feedback
• Choice of feedback matrix gains allows the eigenvalues (or poles) to be assigned as we choose.
• Note that we have not addressed (yet) the issue of a reference input r.
• We shall be returning to state variable design later.