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8/18/2019 CDS-II (3)
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Control System Design-II
.
Department of Electrical EngineeringPakistan Institute of Engineering and Applied Sciences
P.O. Nilore Islamabad Pakistan
url: www.pieas.edu.pk/aqayyumEmail: [email protected]
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Solving TimeSolving TimeSolving TimeSolving Time----Invariant state equationInvariant state equationInvariant state equationInvariant state equation---- A reviewA reviewA reviewA review
( )
( ) 2 k 0 1 2 k
Consider
x Ax Bu, (1) x t ???
Let us consider a homogenous state equation
x ax (2)
Let
x t b b t b t b t (3)
= + =
=
= + + + + +
&
&
L L
Control System Design-II by Dr. A.Q. Khan2
1 2 k
k 1 2 k
1 2 k 0 1 2 k
1 0
2 2
2 1 0 2 0
3
3
x t t t
subsitituting (3) and (4) into (2)
b 2b t kb t a (b b t b t b t ) (5)
we have
b ab1
2b ab a b b a b , Similarly2
1b a
6
−
−
= + + +
+ + + = + + + + +
=
= = ⇒ =
=
& L
L L L
3 k
0 0 k 0
1 1b a b , b a b
3 k = =L L
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( )
( )
( ) ( ) ( )
2
0
0
2 2 at
As
1
x t 1 at at b2
At t 0, equation (3) im plies
b x 0 , therefore
1x t 1 at a t x 0 e x 0
2
Now consider
x Ax (6)
= + + +
=
=
= + + + =
=
L
L
&
Control System Design-II by Dr. A.Q. Khan3
( ) 20 1 2Analogy with scalar, we havex t b b t b t= + + +
( )
( ) ( ) ( ) ( )
k
k
k 1
1 2 k
2 2 At
b t (7)
x t b 2b t kb t (8)
proceeding in similar fashion, we
1x t 1 At A t x 0 e x 0 = (t)x 0 (9)
2
where (t ) is state transition
−
+ +
= + + +
= + + + = φ
φ
L L
& L
L
matrix.
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Referring to equation (9), the term is of particular
interest. It is referred to as matrix exponential and analogously, it can be
represented as
This matrix exponential for an converges absolutely forall finite time.
Ate
k k At
k 0
A te
k!
∞
=
= ∑
n n×
Control System Design-II by Dr. A.Q. Khan4
( )
( )
At At
A B t At Bt
A B t At Bt
At At
d e Ae
dx
e e e if AB BA
e e e if AB BA
e e = I
+
+
−
• =
• = =
• ≠ ≠
•
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( ) ( ) ( )
( ) ( ) ( ) ( ) ( )
1
11 At
C onsider equation (6) and taking Laplace T ransform
sX (s) - x (0) AX s X s sI A x (0)
x t sI A x 0 x t e x 0
or
−
−−
= ⇒ = −
= − ⇒ =
L
Laplace Transform ApproachLaplace Transform ApproachLaplace Transform ApproachLaplace Transform Approach
Control System Design-II by Dr. A.Q. Khan5
( )
where
t : n n m atrix and is the unique solution
: State transition m atr
φ ×
( )
( ) ( )-1 -At
ix. It has all the inform ation about
the free motion o f the sys tem define by x Ax
0 I
Also note t =e = t
=
φ =
φ φ −
&
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Case-I: If be distinct eigenvalues of A, then
will contain exponentials Case-II: If the matrix A is diagonal with , then
1 2 n, ,λ λ λL ( )tφ
1 2 nt t t
e ,e , eλ λ λ
L
1 2 n, ,λ λ λL
( )
1
2
n
t
t
t
e 0 0
0 e 0t
0 0 e
λ
λ
λ
φ =
L
L
M O O M
L
Control System Design-II by Dr. A.Q. Khan6
Case-III: if there is multiplicity of eigenvalues, then will contain terms like in addition to
1 1 1 2 n, , , ,λ λ λ λ λL
( )tφ 1 1t t2te , t eλ λ
1 2 nt t te ,e , eλ λ λ
L
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Properties of ( )tφ
( )( )
( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( )
( ) ( )
1 2 1 2
At
A(0)
1 1At At 1
A t t A t A t
1 2 1 2
n
As t e
0 e I
t e e t or t t
t t e e e t t
t nt
− −− −
+
φ =
• φ = =
• φ = = = φ − φ = φ −
• φ + = = = φ φ
• φ = φ
Control System Design-II by Dr. A.Q. Khan7
An Example
( ) ( ) ( )2 1 1 0 2 0 t t t t t t• φ − φ − = φ −
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( ) ( )
( ) ( ){ }
( )
-1
1At
1
0 1For x Ax, where A=
2 3
Obtain t and t
Solution : As e t sI A
s 1Note that sI A2 s 3
s 3 11sI A
−
−
= − −
φ φ
= φ = −
−
− = +
+ − = −
&
-1L
Control System Design-II by Dr. A.Q. Khan8
( )
( )( )
Using partial fraction expansion
1t
s 1 s 2
φ =
+ +
-1L
( )-t 2 t -t 2 t t 2 t t 2 t
-t 2 t -2t t t 2 t 2t t
2 1 1 1
s 3 1 s 1 s 2 s 1 s 2
2 s 2 2 2 1s 1 s 2 s 2 s 1
2e e e e 2e e e e, t
2e 2e 2e e 2e 2e 2e e
− −
− −
− − + + + + +
= − − − + + + +
− − − −= ⇒ φ − =
− − − −
-1L
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Solution to Homogenous State equations
n 1
n 1
n n
n p
Consider
x
ux Ax Bu (1)
A
B
Now
x Ax Bu
×
×
×
×
∈ ℜ
∈ ℜ= + − − − − − − − − −
∈ ℜ
∈ ℜ
− =
&
&
Convolution integral
Control System Design-II by Dr. A.Q. Khan9
( )
( ) ( ) ( )
At At At
t t
A At A
0 0
e x Ax e Bu multiply by e both sideIntegrating both sides
de x Ax d e Bu d e x
d
− − −
− τ − − τ
− =
− τ = τ τ ⇒ ττ∫ ∫
&
& ( )
( ) ( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( )
t t
A
0 0
t tA tAt A At
0 0
t
0
d e Bu d
e x t x 0 e Bu d x t e x 0 e Bu d
x t t x 0 t Bu d (2)
− τ
− τ− − τ
τ = τ τ
⇒ − = τ τ ⇒ = + τ τ
⇒ = φ + φ − τ τ τ − − − − − − − −
∫ ∫
∫ ∫
∫
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Laplace Transform ApproachLaplace Transform ApproachLaplace Transform ApproachLaplace Transform Approach
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )1
Consider
x Ax Bu
The Laplace transform of the above system is
sX s x 0 AX s BU s sI A X s x 0 BU s
X s sI A x 0 BU s−
= +
− = + ⇒ − = + ⇒
= − +
&
Control System Design-II by Dr. A.Q. Khan10
( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( )i
i
tA tAt
0
0 i
tA t t A t
t
x t e x 0 e Bu d
For initial time t t 0, then
x t e x 0 e Bu d
− τ
− −τ
⇒ = + τ τ
= ≠
= + τ τ
∫
∫
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An ExampleAn ExampleAn ExampleAn Example
( ) ( )( )
Consider
x Ax Bu,
where
0 1 0 0 0 for t
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Laplace Transform method:
Method of Diagonalization of A:
Computation of=?:
Some Further resultsAt
e
( ) ( ){ }1At
t e sI A −
φ = = −-1LDiscussed in
Previous slides
1if D P A P then−=
Control System Design-II by Dr. A.Q. Khan12
Caley-Hamilton Theorem and minimal polynomial theorem
Minimal polynomials of A involves distinct roots
Minimal polynomials of A involves repeated roots
A t D t 1
e P e P −=
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Cayley-Hamilton Theorem Very useful theorem in linear algebra
after the names of two mathematicians: Arthur Cayley andWilliam Hamilton
It states, “Every square matrix over commutative ring (Real or”
Control System Design-II by Dr. A.Q. Khan13
For example if A is an square matrix withcharacteristics polynomial
---------- (1)
Then
-------- (2)
n n×
( )
n n 1
1 n 1 n
a a a 0−
−
φ λ = λ + λ + + λ + =L
( ) n n 11 n 1 nA A a A a A a I 0−
−φ = + + + + =L
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While the CH-Theorem shows that , there may be a
polynomial having lower degree than , for which. The monic polynomial of lowest degree for which
is referred to as minimal ol nomial of A.
( ) 0φ λ =
( )ϕ λ ( )φ λ
( )A 0ϕ =
A 0=
Control System Design-II by Dr. A.Q. Khan14
The minimal polynomial plays important role in thecomputation of polynomials of n×n matrix.
CH-Theorem together with minimal polynomials helps in the
computation of function of matrix, e.g. exp(At), Cos(At) etc.
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Computation of
Case-1: Minimal polynomials of A involves only
distinct roots
Let us assume that the minimal polynomial of A is m. Then
can be obtained by solving the following determinant equation
( ) AtA eφ =
( )Aφ
1
2
t2 m 1
1 1 1
t2 m 1
1 eλ−
λ−
λ λ λL
L
Control System Design-II by Dr. A.Q. Khan15
------------ (3)
Expanding this determinant equation about the last column, weobtain
----- (4)
m
2 2 2
tm 1 m 1 m 1
m m m
2 m 1 At
0
1 e
1 A A A e
λ− − −
−
=
λ λ λ
M M M M M M
M
L
( ) ( ) ( ) ( )At 2 m 10 1 2 m 1e t I t A t A t A −
−= α + α + α + + αL
Sylvester’s interpolation
formula
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and
------ (5)
can be solved using the equations (5). Then can beobtained by substituting the computed
( ) ( ) ( ) ( )( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
1
2
m
t 2 m 10 1 1 2 1 m 1 1
t 2 m 1
0 1 2 2 2 m 1 2
t 2 m 1
0 1 m 2 m m 1 m
e t t t t
e t t t t
e t t t t
λ −−
λ −
−
λ −
−
= α + α λ + α λ + + α λ
= α + α λ + α λ + + α λ
= α + α λ + α λ + + α λ
L
L
M
L
'sα A te'sα
Control System Design-II by Dr. A.Q. Khan16
• If A is an n×n matrix and has distinct eigenvalues, then thenumber of the numbers of to be determined is m=n
• If A has multiple eigenvalues, but its minimal has only simpleroots then m < n.
( )k tα
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An Example
1
2
1
2
Find
:
The eigenvalues are , .From the theory just discussed, we have
1 e
e
I A e
At
t
t
At
e
A
Solution
λ
λ
λ λ
λ
λ
= −
= = −
=
−
1 2
0 1
0 2
0 2
1 0
Control System Design-II by Dr. A.Q. Khan17
,
1
e
I A e
Expanding
t
At
−
−
− =
1 2
2
0 1
1 2 0
( )
( )
the determinant, we obtain
At t At t
t
At
t
e A I Ae e A I Ae
e e
e
− −
−
−
− + + − = ⇒ = + − −
=
2 2
2
2
12 2 0 22
11 1
2
0
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( ) ( )
( ) ( )( ) ( )
1 2
:
,
where
Since , ; Then
At
t
t
Alternative
As
e t I t A
t t e t t e
λ
λ
α α
α α λα α λ
λ λ
= +
+ =+ =
= = −
1
2
0 1
0 1 1
0 1 2
0 2
Control System Design-II by Dr. A.Q. Khan18
( )
( ) ( ) ( ) ( )
( ) ( )
t t
t At t
t
t
t t e t e
Therefore
e e I e A
e
α
α α α− −
−−
−
=− = ⇒ = −
−
= + − =
0
2 2
0 1 1
22
2
11
2 12
11 11
1 22
0
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Computation of
Case-2: Minimal polynomials of A involves multiple
roots
Let us assume that the minimal polynomial of A involves three
equal roots and the remaining aredistinct. Then using Sylvester's criterion, can be obtainedas
( ) AtA eφ =
( )1 2 3λ = λ = λ ( )4 5 m, ,λ λ λL
( )Aφ
Control System Design-II by Dr. A.Q. Khan19
(6)
( )( )
( )
1
1
1
m
2 tm 3
1 1
t2 m 2
1 1 1
t2 3 m 1
1 1 1 1
t2 3 m 1
m m m m
2 3 m 1 At
m 1 m 2 t0 0 1 3 e2 2
0 1 2 3 m 1 te
1 e 0
1 e
1 A A A A e
λ−
λ−
λ−
λ−
−
− −λ λ
λ λ − λ
λ λ λ λ =
λ λ λ λ
L
L
L
M M M M L M M
L
L
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Equation (6) can be expanded in similar fashion as before
(7)and
( ) ( ) ( ) ( )
( )
( ) ( ) ( ) ( )
1
1
1
2t m 3
2 3 1 m 1 1
t 2 m 1
1 2 1 3 1 m 1 1
t 2 m 1
m 1 m 2te t t t2 2
te t 2 t 3 t t
λ −
−
λ −
−
λ −
− −= α + α λ + + α λ
= α + α λ + α λ + + α λ
L
L
( ) ( ) ( ) ( )
At 2 m 1
0 1 2 m 1e t I t A t A t A
−
−= α + α + α + + αL
Control System Design-II by Dr. A.Q. Khan20
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
2
m
1 2 1 3 1 m 1 1
t 2 m 1
1 2 2 3 2 m 1 2
t 2 m 1
0 1 m 2 m m 1 m
e t t t t
e t t t t
−
λ −
−
λ −
−
= α + α λ + α λ + + α λ
= α + α λ + α λ + + α λ
L
M
L
Extend the above formulation to other cases; e.g; two or
more sets of multiple roots. Try at least two examples
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( ) ( ) ( )( ) ( ) ( ) ( )
( ) ( ) ( )
Example: Find
:
The eigenvalues are , .From the theory just discussed, we have
and
At
At
t t
t
e
A
Solution
e t I t A t Ate t t te t t
e t t t e
λ
λ
λ λ λ
α α αα α λ α α
α α λ α λ
=
= = =
= + += + ⇒ = +
= + + ⇒
1
2
1 2 3
2
0 1 2
2
1 2 1 1 2
2
0 1 2 2 2
2 1 4
0 2 00 3 1
2 1
2 4
( ) ( ) ( )t
t t
t t t
λ
α α α= + +3
2
0 1 2
2
2 4
Control System Design-II by Dr. A.Q. Khan21
( ) ( )
( )
( ) ( ) ( )
Solving the above set of equations, we have
,
Finally
t t t t t t
t t t
t t t t
At
t e e te t e e te
t e e te
e e e te
e t I t A t A
α α
α
α α α
= − + = − + −
= − +
− + −= + + =
0 1 3 2 3 0 1 2
2 2 2 2
0 1
2 2
2
2 2 2
2
0 1 2
4 3 2 4 4 3
3
12 12 13 t t
t
t t t
e e
e
e e e
+ − +
2
2
2
4 4
0 0
0 3 3
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Controllability and ObservabilityControllability and ObservabilityControllability and ObservabilityControllability and Observability
( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( )( ) ( )
0 0
n n p
0 0
m
Consider a dynamical system described byx t A t x t B t u t , x t x
y t C t x t
where
x t : State, x t : State at initial time t , u t : Control input,
y t : output/measurement vector. The solution x t is g
= + =
=
∈ ℜ ∈ ℜ ∈ ℜ∈ ℜ
ɺ
( ) ( ) ( ) ( ) ( ) ( )t
0 0
iven as
x t = t,t x t + t, B u dφ φ τ τ τ τ
Control System Design-II by Dr. A.Q. Khan22
( )
( ) ( ) ( ) ( )
( ) ( ) ( )
0t
0
0 0 0 0 n
where t,t is the transition matrix and it is the solution of
t,t =A t t,t , t ,t =I
In this course, we shall study LTI systems of the following form
x t Ax t Bu t ,
φ
φ φ φ
= +
ɺ
ɺ
( )( ) ( )
( ) ( )0
0 0
A t-t
0
x t xy t Cx t
The transition m atrix is
t, t =e
==
φ
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A continuous time LTI system is said to be
reachable at time if it is possible by means ofany unconstrained control input vector to transferthe system from initial state to any other state in afinite interval of time.
A continuous time LTI system is said to becontrollable (to the zero state) at time if it is
0
t
0t
Control System Design-II by Dr. A.Q. Khan23
input vector to transfer the system from initialstate to origin.
State Controllability
Output Controllability
( )0x t
A system is said to controllable at time if it is possible by means ofan unconstrained control vector to transfer from any initial state
to any other state in a finite interval of time
0t
( )0x t
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( )( )
( )
T h e o r e m : A n e c e s s a ry a n d s u ff ic ie n t c o n d i t io n s
fo r s y s t e m re a c h a b ility [c o n tro lla b ility ] a t t im e is
th a t W ,t is p o s it iv e d e f in ite fo r s o m e t
w h e re W .,. is a n n × n G ra m m ia n m a tr ix d e fin e d b y
W , t
τ
τ τ
τ φ τ
≥
= ( ) ( ) ( ) ( )
t
T T , B B t, d τ
λ λ λ φ λ λ∫
Control System Design-II by Dr. A.Q. Khan24
Uniform Reachability/Uniform ControllabilityStrong notion
It refers to the fact that the property is independent of initial
time.
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Theorem: For LTI systems given in before, the pair (A,B) issaid to be completely state controllable if and only if
n 1rank B AB A B n− = ⋯
( ) ( ) ( ) ( ) ( ) ( )t
0
0
Proof : The solution is
x t = t x 0 + t, B u d , t 0
Apply the definition of state controllability,
φ φ τ τ τ τ =∫
Control System Design-II by Dr. A.Q. Khan25
( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( )
1
1
1
t
1 1
0
t
0
At 2
0 1 2 n 1
.e., x t 0, t ere ore
0= t x 0 + t B u d
x 0 B u d
using e a I a A a A a− −
=
φ φ − τ τ τ τ ⇒
= − φ −τ τ τ τ
φ −τ = = τ + τ + τ + + + τ
∫
∫ ⋯
( )
n 1
n 1k
kk 0
A
a A
−
−
=
= τ
∑
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----- (1)
( ) ( ) ( ) ( ) ( ) ( )
( ) ( )
( )
1 1
1
t tn 1 n 1k k
k kk 0 k 00 0
t
k k
0
1
n 12k n 1
k
Now
x 0 a A B u d A B a u d
Let a u d
x 0 A B B AB A B
− −
= =
−−
= − τ τ τ τ = − τ τ τ
β = τ τ τ
β β = − β = − = Μβ
∑ ∑∫ ∫
∫
∑ ⋯
Controllability matrix
Control System Design-II by Dr. A.Q. Khan26
For system to be completely state controllable, equationmust holds. This requires that matrix M must have n linearly
independent row/column vectors. Rank (M)= n For , the dim of M is n×np. The Rank (M) should be n
for CSS.
k 0
n
=
β
pu ∈ ℜ
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[ ] ( )
1 1
2 2
x x1 1 1u
x 0 1 x 0
1 1B AB Rank 1 2
0 0
Hence the system is not completely state cont
M
rollable
M
= + −
= = ⇒ = ≠
ɺ
ɺ
Example-I
Example-II
Control System Design-II by Dr. A.Q. Khan27
[ ] ( )
1 1
2 2
x x1 1 0u
x 0 1 x 1
0 1B AB Rank 21 1
The system is completely state controlla
M
ble
M
= + −
= = ⇒ = −
ɺ
ɺ
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Controllability conditions in s-plane
Necessary and sufficient conditions: No cancellationof poles and zeros
The system losses its controllability duringcancellation of poles and zeros
An Example
Control System Design-II by Dr. A.Q. Khan28
This system is not CSC.
Check it in state space??
( )( )
( )( )( )
X s s 2.5U s s 2.5 s 1
+= + −
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1 1 2
2 1 2
Controllab
3
2 1.5ility ?
4
An Example
???
x x x u
x x x u
= − + +
= − + +
&
&
Control System Design-II by Dr. A.Q. Khan29
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1 1 2
2 1 2
1
A n E x a m p le
Appa ran t lyC o n tro llab ili ty ???? Y E S
S o lu t ion
32 4
:
1 .5 x x x u x x x u
y x
= − + += − + +
=
⇒
&
&
Control System Design-II by Dr. A.Q. Khan30
[ ]
C h eck the con tro llab i liy m a trix . H e re
3 1 1,
2 1 .5 41 1
4 4
A B
B A M B
− = =
−
= =
( ), R a n k 1 M =
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1 1 2
2 1 2
An Example
Apparantly
Let us proceed as follows
32 1.5 4
Controllability ???? YES
Solution:
x x x u x x x u
= − + += − +
⇒
+
&
&
Control System Design-II by Dr. A.Q. Khan31
[ ]
Check the controllabiliy matrix. Here3 1 1
,2 1.5 4
1 1,
44
A B
B A B M
− = = −
= =
( )Rank 1 M =
Interesting
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( ) ( ) ( ) ( )
( )
( )
( )
( )
( )
( ) ( )
1 1 1
2
2
F r o m t h e s t a t e - s p a c e m o d e l , w e h a v e
1 . 5 2 . 5 2 . 5
1 . 5 2 . 5 2 . 5
2 . 5 2 . 5
2 . 5 11 . 5 2 . 5
x x x u u
s s Y s s U s
Y s s s
U s s ss s
+ − = + →+ − = +
+ +
= = + −+ −
&& & &
L
Remarks
Control System Design-II by Dr. A.Q. Khan32
The system can not be controllable if it is not reducible to controllablecanonical form B should not be in the eigen-space of A.Alternatively B should not be in the null space of A
ConclusionActuator position is very important
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1 1 2 1 2
2 2
3 3 1
1
A n o th e r E x a m p
4 2
2
l
3
e
x x x u u
x x
x x u
y x
= − + + +
= −
= − +
=
&
&
&
Control System Design-II by Dr. A.Q. Khan33
C o n tro lla b ility ????
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Output Controllability
This can also be defined in the manner similar to statecontrollability.
The system is said to be completely output controllable if it ispossible to construct an unconstrained control vector u(t)that will transfer the output y(t) to origin from any initialcondition.
Let us consider
Control System Design-II by Dr. A.Q. Khan34
The output controllability matrix
should have a rank m
( ) ( ) ( ) ( )( ) ( )
0 0x t Ax t Bu t , x t xy t Cx t
= + ==
ɺ
n 1
OM CB CAB CA B−
= ⋯
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Stabilizability For partially controllable systems, if the controllable modes
are stable and the unstable modes are controllable, thesystem is said to be stabilizable.
Remember: Controllability for continuous-time LTI system
implies Reachability. However for discrete-time LTI system this
Control System Design-II by Dr. A.Q. Khan35
does not hold.
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Observability and Reconstructability Definition: The system is said to be observable, if every state x(t0 ) can
be determined from the observation y(t) over finite interval of timet0≤t ≤t1.
In completely observable system, every transition in stateaffects output (or every element of the output vector in caseof multiple output systems)
Control System Design-II by Dr. A.Q. Khan36
e cu ty, n pract ce, ar se w t t e use o state ee accontrol when all the system states are not available forfeedback.
The observability concept is useful in solving the problem of
reconstructing the un-measurable states of the system fromthe measurable states in shortest possible length of time.
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Consider
Definition: Let y(t;t0,x,u) be the output of the above system to the
initial state x0. The present state x0 is unobservable if the (future)output y(t;t0,x,0)=0 for all t ≥ t0
Definition: The present state x0 is unconstructible if the (past) output
( ) ( ) ( )
( ) ( )0 0x t Ax t , x t x
y t Cx t
= =
=
ɺ
Control System Design-II by Dr. A.Q. Khan37
( )y ; ,x,σ τ σ τ = ∀ ≤0 0
The LTI system shown above or the pair (A,C) is
completely observable if and only if it is completely state
reconstructable
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Complete Observability and ReconstructabilityConsider
( ) ( ) ( )
( ) ( )0 0x t Ax t , x t x
y t Cx t
= =
=
ɺ
Control System Design-II by Dr. A.Q. Khan38
The above system (the pair (C,A)) is said to be observable ifand only if the following observability matrix is full rank
( )n
OM C A C A C −∗ ∗ ∗ ∗ ∗ =
1
⋮ ⋮ ⋯ ⋮
Consider the system describe before, the output is
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( ) ( )
( )
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )
At
n
At k k
k
n k n
k n
k
y , p
Recall that , therefore
y t Ce x
e t A
y t C t A x t Cx t CAx t CA x
C
CA
t t t
α
α α α α
α α α
−
=
−−
−=
−
=
=
= = + + +
=
∑
∑
1
0
11
0 1 1
0
0
0
0 0 0 ⋯
⋯
⋯ ⋯ OM α
=
Control System Design-II by Dr. A.Q. Khan39
n CA −
1
⋮
⋯
( )
nm n
n
For Complete observability, the observability matrix s
Observability matrix
Sometimes in literature
OM
OM C A
is used as obser
hould be Full Rank
vability matri
C A C
.
×
−∗ ∗ ∗ ∗ ∗
∈ ℜ
=
=
1
⋮ ⋮ ⋯ ⋮
x
An Example:
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[ ]
( )
Here
OM C A C
x Ax Bu,
An
R
Example:
where
A ,
y Cx
an
B ,C
k OM ∗ ∗ ∗
= = = − −
= = ⇒ =
= + =
1 1 01 0
2
1
1
1 1
1
20
ɺ
Control System Design-II by Dr. A.Q. Khan40
Hence the system is completely state observable
[ ]
Now Find the obserbabiliy for the following ???
x x u, y x
= + = − − −
0 1 0 0
0 0 1 4 5 10
6 11 6 1
ɺ
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Analogous to Controllability, the necessary and sufficient
condition for complete observability is that no cancellationoccur in the transfer function or transfer matrix.
If there is some cancellation of Poles/Zeros, the
observability will be lost.
0 1 0 0
An Example
Control System Design-II by Dr. A.Q. Khan41
[ ]
( ) ( )
x x u
y x
Rank OM C A C A C ∗ ∗ ∗ ∗
= + − − − =
= = = 2
0 0 1 06 11 6 1
4 5 1
2
ɺ
Rank of the observability matrix
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Let us see what happens to Poles and zeros cancellation.
( ) ( ) ( ) ( )
( ) ( )( )
( )
x x x
Y s Y s s s X s s s
X s X s
U s
= + + ⇒
= + + = = + +
1
1 1 1
2
1
1
4 5
4 5 1 4
ɺ ɺɺConsider the output equation
Also Computing from stat
y
e equations, we have
Control System Design-II by Dr. A.Q. Khan42
( )( ) ( )( )( )
( )
( )
( )
( )
( )
( )
( )( )
( )( )( )( )
( )( )
X s
U s s s s s s s
Y s Y s X s s s
U s X s U s s s s
s
s s
= =+ + + + + +
+ +
= = + + ++
=+ +
1
3 2
1
1
1 1
6 11 6 1 2 3
1 4
1 2 3
4
2 3 Poles and Zero Cancellation
occurs
h h l d ll
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Let us see what happens to Poles and zeros cancellation.
( ) ( ) ( ) ( )
( ) ( )( )
( )
x x x
Y s Y s s s X s s s
X s X s
U s
= + + ⇒
= + + = = + +
1
1 1 1
2
1
1
4 5
4 5 1 4
ɺ ɺɺConsider the output equation
Also Computing from stat
y
e equations, we have
Control System Design-II by Dr. A.Q. Khan43
( )( ) ( )( )( )
( )
( )
( )
( )
( )
( )
( )( )
( )( )( )( )
( )( )
X s
U s s s s s s s
Y s Y s X s s s
U s X s U s s s s
s
s s
= =+ + + + + +
+ +
= = + + ++
=+ +
1
3 2
1
1
1 1
6 11 6 1 2 3
1 4
1 2 3
4
2 3 Poles and Zero Cancellation
occurs
A CSC l b CSO d
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A CSC system may not necessarily be CSO and vice versa
[ ]
Illusrativde Example0 1 0
0.4 1.3 1
0.8 1
0 1
x x u
y x
= + − −
=
&
Control System Design-II by Dr. A.Q. Khan44
( )
1 1.30.8
Show if there is Poles Zero Cancellation or not??????
0.4; Rank 1
1 0.5OM OM NOT CSO
− −
= = ⇒ −
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T r y th i s o n e
2 0 0
0 2 00 3 1
x x
=
&
Control System Design-II by Dr. A.Q. Khan45
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Duality properties
The Linear time-invariant system
x Ax Bu
y Cx
= +=
ɺ
Symmetry properties existsObservability (Reconstructability) Controllability (Reachability)
Control System Design-II by Dr. A.Q. Khan46
* *
T T T
1
s t e ua o t e sy
z A z Bv
y C x
stem an v ce versa
where A A ,B C ,C B =
= +
=
= =1
1
1
1
1ɺ
The principle of duality is useful in that it allows the statereconstruction problem can be casted as dual control problem and
vice versa.
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Controllability and Observability indicesControllability and Observability indicesControllability and Observability indicesControllability and Observability indices
−
− = n 1
1
Consi
der the controlability matrix
If there exists a least integer q such that
M B AB A B⋯
Controllability index
Control System Design-II by Dr. A.Q. Khan47
= µ
Then this least integer refferred to as is calle controlability indexd
⋯
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( ) − = *
n 1* * * * *
Consider the observability matrix
If there exists a least integer v, such that
OM C C A C A⋯
Observability index
Controllability and Observability indicesControllability and Observability indicesControllability and Observability indicesControllability and Observability indices
Control System Design-II by Dr. A.Q. Khan48
( ) −
= ≤ ≤µ
n 1* * * * *
Then this least integer refferred to
C C
as
Rank( ) n
is call
ed
1 v n
obs
A C A
erva
⋯
bility index
•Very powerful concept• Quite often used in minimum order observer design