16.513 control systems controllability and observability (chapter 6)
TRANSCRIPT
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16.513 Control Systems
Controllability and Observability(Chapter 6)
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A General Framework in State-Space Approach
(*) Cx y Bu;Axx =+=&Given an LTI system:
The system might be unstable or doesn’t meet the required performance spec. How can we improve the situation?
The main approach: Let u= v- Kx (state feedback), then
Dv-DK)x-(C Bv;BK)x-(A Kx)-D(vCxy Kx);-B(vAxx
=+=+=+=&
The performance of the system is changed by matrix K.Questions:
• Is there a matrix K s.t. A-BK is stable?• Can eig(A-BK) be moved to desired locations?
These issues are related to the controllability of (*)
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Main Result 1: The eigenvalues of A-BK can be movedto any desired locations iff the system (*) is controllable.Another situation: the state x is not completely available. Only a linear combination of x, e.g., y = Cx, can be measured. How can we realize u = v-Kx?
A possible solution: build an observer to estimate x basedon measurement of y. Main result 2: The observer error (difference between thereal x and estimated x) can be made arbitrarily small withinarbitrarily short time period iff (*) is observable.
We will arrive at these conclusions in Chapter 8. Before that, we need to prepare some tools and go through these fundamental problems: controllability and observability.
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Consider an LTI system: DuCxy Bu;Axx +=+=&
Questions:a) Can we move the state from one point in the state space to a
desired location: x0 → xd, by choosing u properly?b) Can we build a controller to stabilize the system?c) Suppose that y contains all the quantities that we
can measure. Can we evaluate the state x from y?
– The first two questions are related to the controllability of the system
– The last one is related to observability.– They describe, in some sense, the potential of a system that
can be explored through feedback design, e.g., stabilizationby output feedback.
Controllability and Observability
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An example: A circuit system
+ − + −
C1C2
x1x2
u
+
−
y
R1 R2
Input: u, a current sourceState: x1,x2, voltages across the two capacitorsOutput: y, voltage between two points
Observation:The input u has no effect on x2
⇒ x2 cannot be altered by u ⇒ x2 is not controllableThe measurement y has no relation to x1, ⇒ any change onx1 is not reflected on y ⇒ x1 is not observable.
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Controllability: Definition
Consider the system.Ru ;R xBu,Axx pn ∈∈+=&
Controllability is a relationship between state and input.
Definition: The system, or the pair (A,B), is said to be controllable if for any initial state x(0)=x0 and any final state xd, there exist a finite time T > 0 and an input u(t), t∈[0,T] such that
(1) x)dτBu(exex(T) dτ)-A(TT
00AT =+= ∫ τ
Comment: There may exist different T and u that satisfy (1).As a result, there may be different trajectories starting from x0 andend at xd. Controllability does not care about the difference.
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Examples: uncontrollable networks.
R4= 1 ΩR3= 1 Ω
R2= 1 ΩR1= 1 Ω
+ −x
+u− 1 Ω1 Ω
+u−
1 F 1 F+x1
−
+x2
−
Observation:• If x(0)=0, then x(t)=0 for
all t > 0. The input u can do nothing about it.
• If the resistance is changedso that R1/R3 ≠ R2/R4, thenyou can bring x to any desiredvalue.
Observation:• If x1(0)=x2(0)=0, then
x1(t)=x2(t) for all t > 0.You cannot bring x(t) to anypoint in the plane.
• This situation can be changedby altering the parameters of the components.
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Theorem 1: The pair (A,B) is controllable if and only if the n×n matrix,
τdeBB'edτeBB'e(t)W τ)(tA't
0τ)A(tτA't
0Aτ
c−−∫∫ ==
is nonsingular for every t > 0.
Main idea of the proof: Suppose Wc(t1)-1 exists. Recall the solution
)dBu(exe)x(t τ)-A(tt
00At
11
11 ττ∫+=
If we choose ]xx)[e(tWeB'u(t) d0At
11
ct)(tA' 11 −−= −−
Then, ]dxx)[e(tWeBB'exe)x(t d0At
11
c)(tA'τ)-A(tt
00At
1111
11 ττ −−= −−∫
The necessity of the condition is shown by contradiction. (p.146)
{ } ]xx)[e(t WdeBB'e xe d0At
11
c)(tA'τ)-A(tt
00At 111
11 −−= −−∫ ττ
dd0At
0At xxxe xe 11 =+−=
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Equivalent conditions: The following are equivalentconditions for the pair (A,B) to be controllable:
τdeBB'edτeBB'e(t)W τ)(tA't
0τ)A(tτA't
0Aτ
c−−∫∫ ==
1) Wc(t) is nonsingular for every t > 0.2) Wc(t) is nonsingular for at least one t > 0.3) For every v∈Rn, v≠0, v’eAt B is not identically zero.4) The matrix Gc = [B AB A2B … An-1B] has full
row rank, i.e., ρ(Gc) = n.5) The matrix M(λ) = [A−λI B] has full row rank at all λ∈C. 6) M(λ) has full row rank at every eigenvalues of A.
Note: M(λ) has full row rank if λ is not an eigenvalue of A.We only need to check the rank of M(λ) at eigenvalues of A.
Note : Of all the conditions, only 4) and 6) can be practically verified.
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dτeBB'e(t)W τA't
0Aτ
c ∫=
Everything starts from the matrix function
We first give some necessary background and tools.
Notice that Wc(t) is a symmetric matrix: Wc’(t)=Wc(t)
Actually it is positive semidefinite. Why? All eigenvalues are non-negative.Wc(t1) is nonsingular iff Wc(t1) > 0.Controllability is related to the positive definiteness.
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Explanations: 1) ⇔ 2) Only need to show 2) ⇒1)Assume that Wc(t1) is nonsingular. Then Wc(t1)>0.⇒ v’Wc(t1)v > 0 for every v∈Rn.⇒ For a fixed v, v’Wc(t)v is analytic in t and
nondecreasing. Also, v’Wc(0)v=0, v’Wc(t1)v > 0 ⇒ v’Wc(t)v > 0, for all t > 0.⇒ For any t>0 and v∈Rn, v’Wc(t)v > 0. ⇒ Wc(t) > 0 for all t > 0.⇒ Wc(t) nonsingular for all t > 0.
dτeBB'e(t)W τA't
0Aτ
c ∫=
1) Wc(t) is nonsingular for any t > 0.2) Wc(t) is nonsingular for at least one t > 0.
Note: If f(t) is analytic and f(t)=0 for t∈[a,b], then f(t)=0 for all t∈R.
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The above relation can also be proven with the result from last lecture
Theorem: Let M(t)∈Rn×n be a symmetric matrix function. Suppose that every element of M(t) is an analytic function, i.e., mij(t) can be differentiated infinitely many times. Then the eigenvalues of M(t) can be arranged as λ1(t),λ2(t),…,λn(t) where each λi(t) is analytic.If M(t) is nondecreasing, i.e., M(t1) ≤ M(t2), for any t1< t2, then λ1(t), λ2(t) ,…, λn(t) are nondecreasing.
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Explanations: 1) ⇔ 3)
dτeBB'e(t)W τA't
0Aτ
c ∫=
Suppose Wc(t1) is singular for certain t1. Exist v∈Rn such that v’Wc(t1) v =0 . Since v’eAτBB’eA’τ v≥0 for all τ, must have⇒ v’eAτBB’eA’τ v=0, v’eAτB=0 for all τ∈[0,t1] ⇒ v’eAtB=0 for all t by analyticity.On the other hand if v’eAtB=0 for all t, ⇒ v’Wc(t) v =0 for all t ⇒ Wc(t) must be singular since it is positive
semi-definite.
1) Wc(t) is nonsingular for any t > 0.3) For every v∈Rn, v≠0, v’eAt B not identically zero.
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3) For every v∈Rn, v≠0, v’eAt B not identically zero.4) The matrix Gc = [B AB A2B … An-1B] has full row
rank i.e., ρ(G) = n.
Explanation: 3) ⇔ 4)
3a) Exists v∈Rn, v≠0, v’eAt B =0 for all t.4a) ρ(G) < n
The opposite of 3) and 4):
4a) ⇒ 3a) Exists v such that v’Gc=0 ⇒ v’B=v’AB=…=v’An-1B=0.Since eAt is a linear combination of {I,A,A2,…An-1},eAtB = k0 (t)B+k1(t) AB+ k2 (t) A2B+…+kn-1(t) An-1B⇒ v’eAtB=0 for all t
3a) ⇒ 4a) v’eAt B =0 for all t ⇒ dj (v’eAtB)/dtj |t=0 = 0 for all j ≥ 0.⇒ v’eAtAjB=0 for all j. ⇒ v’eAtGc=0 ⇒ ρ(Gc) < n.
For the relationship between 4) and 5), or 4) and 6), see page 147.
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Equivalent conditions: The following are equivalentconditions for the pair (A,B) to be controllable:
τdeBB'edτeBB'e(t)W τ)(tA't
0τ)A(tτA't
0Aτ
c−−∫∫ ==
1) Wc(t) is nonsingular for any t > 0.2) Wc(t) is nonsingular for at least one t > 0.3) For every v∈Rn, v≠0, v’eAt B not identically zero.4) The matrix Gc = [B AB A2B … An-1B] has full
row rank, i.e., ρ(Gc) = n.5) The matrix M(λ) = [A−λI B] has full row rank at all λ∈C. 6) M(λ) has full row rank at every eigenvalues of A.
Note: M(λ) has full row rank if λ is not an eigenvalue of A.We only need to check the rank of M(λ) at eigenvalues of A.
Note : Of all the conditions, only 4) and 6) can be practically verified.
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dτeBB'e(t)W τA't
0Aτ
c ∫=
]xx)[e(tWeB'u(t) d0At
11t)(tA' 11 −−= −−
Some observations: Recall from the proof of Theorem 1(slide 8).To bring the state from x(0) = x0 to x(t1) = xd , a particular input is
This is actually the minimal energy control, i.e., if there is anotherinput w(t) to transfer x0 to xd within the same time interval, then
∫∫ ≥ 11
00)()'()()'(
ttduudww ττττττ
If (A,B) is controllable, Wc(t)-1 exists for all t > 0.⇒ The transfer of the state can be accomplished in arbitrarily small
time interval
Note that as t1 decrease, both λmin[Wc(t1)] and λmax[Wc(t1)] decrease.Then ||Wc(t1)-1|| increases. ⇒ larger magnitude of u is required.As t1 → 0, ||Wc(t1)-1|| → ∞ , u(t) → ∞.
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0 1 2 3 4 510-8
10-6
10-4
10-2
100
102
t10 1 2 3 4 5
10-2
100
102
104
106
108
t1
An example:
⎥⎥⎦
⎤
⎢⎢⎣
⎡=
⎥⎥⎦
⎤
⎢⎢⎣
⎡
−−−=+=
111
,321
100010
Bu,Axx BA&
Eigenvalues of Wc(t1) Eigenvalues of Wc(t1)-1
]xx)[e(tWeB'u(t) d0At
11t)(tA' 11 −−= −−
Magnitude of u increases as t1 is decreased.
dτeBB'e(t)W τA't
0Aτ
c ∫=
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Example: Determine the controllability for
⎥⎦⎤
⎢⎣⎡=⎥⎦
⎤⎢⎣⎡
−−=+= b
aBA ,1001 Bu,Axx&
Approach 1: ⎥⎦⎤
⎢⎣⎡== b-b
a-aAB] [BGc
b and a possible allfor 2)( nG =<ρThe system not controllable whatever a and b are.
Approach 2: Check M(λ)=[A-λI B] at λ=-1
⎥⎦⎤
⎢⎣⎡=− b00
a001)M(
ρ(M(-1)) < 2 for all possible a and bSame conclusion on controllability
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Example:
⎥⎦⎤
⎢⎣⎡=⎥⎦
⎤⎢⎣⎡
−−=+= b
aBA ,2001 Bu,Axx&
Approach 1:⎥⎦⎤
⎢⎣⎡== 2b-b
a-aAB] [BGc
0bor 0aeither if 2,)ρ(G
0b and 0a if 2,)ρ(G ab,detG c
cc
⎩⎨⎧
==<≠≠=−=
The system is controllable if a ≠ 0 and b ≠ 0.
Approach 2: Check M(λ)=[A-λI B] at λ=-1
,b00a012)M(,b1-0
a001)M( ⎥⎦⎤
⎢⎣⎡=−⎥⎦
⎤⎢⎣⎡=−
ρ(M(-1))= ρ(M(-2))=2 iff a ≠ 0 and b ≠ 0
Same conclusion on controllability
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A general SI system (diagonalizable)
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
=⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
=+=
nn b
bb
BAM
LMOMM
LL
& 2
1
2
1
,
00
0000
Bu,Axx
λ
λλ
The above system is controllable if and only ifthe eigenvalues are distinct and none of the bi’s is zero
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Example:
⎥⎦⎤
⎢⎣⎡=⎥⎦
⎤⎢⎣⎡ −=+=
2
1, Bu,Axx bbBA αβ
βα&
⎥⎦⎤
⎢⎣⎡
+−==
212
211αbβbbβbαbbAB] [BGc
0or 0either if 2,)ρ(G
0 and 0 if 2,)ρ(G ),b(bdetG 22
21
c
22
21
c22
21
c
⎩⎨⎧
=+=<≠+≠=+=bb
bbβ
ββ
The system is controllable if β≠ 0 and (b1,b2) ≠ (0,0)
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Example: ⎥⎦⎤
⎢⎣⎡=⎥⎦
⎤⎢⎣⎡=
2
1
2
1 ,00
BBBA
AA
Suppose that the eigenvalues of A1 and those of A2 are disjoint. ⇒ (A,B) is controllable iff (A1,B1) is controllable and (A2,B2) is controllable.
Proof: [ ] ⎥⎦⎤
⎢⎣⎡=−=
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11BA-I0B0A-I)M(Consider λ
λλλ BAI
Problem: Show that
)rank(Yrank(X)YZ0X
rank +=⎥⎦
⎤⎢⎣
⎡
)rank(Yrank(X)YZ0X
rank such that, examplean Give +≠⎥⎦
⎤⎢⎣
⎡
If Y has full row rank, then
)rank(Yrank(X)YZ0X
rank +≥⎥⎦
⎤⎢⎣
⎡
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Theorem: Consider the pair
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
=⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
=
mm B
BB
B
A
AA
AM
LOMMLL
2
1
2
1
,
0000000
Suppose that the eigenvalues of Ai and those of Aj are disjoint for i ≠ j . Then (A,B) is controllable iff (Ai,Bi) is controllable for all i.
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Theorem: Let ρ(B) = p. The pair (A,B) is controllable iff
Gcn-p+1: = [B AB A2B … An-pB]
has full row rank. This is equivalent to Gcn-p+1Gc’n-p+1 being
nonsingular, and to Gcn-p+1Gc’n-p+1 > 0 (positive definite.)
Note: The following are equivalent• X has full row rank;• XX’ is nonsingular;• XX’ is positive-definite. • How to prove these?
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Gcn-p+1=[b1 b2 … bp Ab1 Ab2 … Abp …… An-pb1 An-pb2 …An-pbp]
Procedure to determine controllability:
Step 1: form the n×(p×(n-p+1)) matrix
Step 2: From left to right, throw away columns that are dependent on its LHS columns (or LI columns)until n LI columns are detected or until the last column. – Recall: y is dependent on the columns of Z iff ρ(Z) = ρ([Z y])
Step 3:• If the maximal number of LI columns is n, controllable.
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Example:
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡=
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−=
10000100
,0020100020030010
BA
n=4, p=2. ρ(B)=2= p.
[ ]
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12
2121
2c1pn
bA bA Ab Ab b b 400210
021000012001200100
BA AB BG⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−−−−==+−
The first 4 columns are LI. ⇒ ρ(Gcn-p+1)=4 = n
⇒ (A,B) controllable
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Example:
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡=
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡=
10010101
,3000020001200001
BA n=4, p=2. ρ(B)=2.
[ ]
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12
2121
2c1pn
bA bA Ab Ab b b 993311040201080301010101
BA AB BG⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡==+−
The first 3 columns are LI. The 4th is dependent on the first 3.
[ ] rank row full has
9311420183011101
bAAbbb 12
121
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
=
Hence (A,B) is controllable.
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Question:Does similarity transformation retain the controllability property?
Recall that equivalence transformation can make the structurecleaner and simplify analysis.
Theorem: The controllability property is invariant under anyequivalence transformation
Proof: Consider (A,B) with Gc=[B AB A2B …. An-1B]. Let the transformation matrix be P. Then (A,B) ⇔ (PAP-1, PB)
c
1-n
1-n
11-n1-
1nc
PG B]A AB P[B
]BPA PAB [PB PB]PPA PBPAP [PB
]BABAB[G
=====
−
−
L
L
L
L
Since P is nonsingular,
)ρ(G)Gρ( cc =
Effect of equivalence transformation
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Review: Definition of Controllability
Consider the system.Ru ;R xBu,Axx pn ∈∈+=&
Controllability is a relationship between state and input.
Definition: The system, or the pair (A,B), is said to be controllable if for any initial state x(0)=x0 and any final state xd, there exist a finite time T > 0 and an input u(t), t∈[0,T] such that
(1) x)dτBu(exex(T) dτ)-A(TT
00AT =+= ∫ τ
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Next Problem: Observability
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Observability: A dual conceptConsider an n-dimensional, p-input, q-output system:
DuCxy Bu;Axx +=+=&
Definition: The system, is said to be observable if for any unknown initial state x(0), there exists a finite t1> 0 such that x(0) can be exactly evaluated over [0,t1] from the input u and the output y. Otherwise the system is said to be unobservable.
Systemu y
Assume that we know the input and can measure the output, but has no access to the state.
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Examples: Two circuits
1Ω1Ω
1 H
1F
+
x2 = y
−
x1
1Ω1Ω
1 H
1F
+
x2 = y
−
x1
2
2
1
2
1
xy
u,11
xx
1001
xx
=
⎥⎦⎤
⎢⎣⎡+⎥⎦
⎤⎢⎣⎡
⎥⎦⎤
⎢⎣⎡
−−=⎥⎦
⎤⎢⎣⎡&&
2
2
1
2
1
xy
u,11
xx
2110
xx
=
⎥⎦⎤
⎢⎣⎡+⎥⎦
⎤⎢⎣⎡
⎥⎦⎤
⎢⎣⎡
−−=⎥⎦
⎤⎢⎣⎡&&
The output is independent of x1;Or any change on x1 will not be reflected from the output. ⇒ Unobservable.
x1 has some effect on x2= y.We will see later that the circuit is observable.
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Some observations:Recall the state-space solution,
Du(t))dBu(exey(t) τ)-A(tt
00At ++= ∫ ττCC
Since both u and y are known, define Du(t))dBu(ey(t)(t)y τ)-A(tt
0−−= ∫ ττC
xe(t)y 0AtC=
have weAlso known. is (t)yThen
]t[0, intervalan over (t)ygiven detected becan x whether is problem The
1
0
Clearly, the observability depends only on matrices A and C. We can assume that B=0 and D=0.
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Theorem: The system (A,B,C,D) is observable if and only if the n×n matrix
τττ dCeCetW At Ao ')(
0'∫=
is nonsingular for any t > 0.
Proof. Consider the case where B = 0 and D = 0. Then xey(t) 0
AtC= xCeC'ey(t)C'e 0AttA'tA' =
01o00AttA'
0 0AttA'
0tA'
)x(tWxdt CeC'e
dt xCeC'edty(t)C'e1
11
==
=
∫∫∫t
tt
If Wo(t1) is nonsingular, then
dty(t)C'e)(tWx 1
0tA'1
1o0 ∫−=t
and observability is ensured. (Sufficiency proved)
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Take a comparison between Wo and Wc
dτCeC'e(t)W Aτt
0τA'
o ∫= dτeBB'e(t)W τA't
0Aτ
c ∫=Mathematically they have the same structure. So we have the following equivalent conditions:1) Wo(t) is nonsingular for any t > 0.2) Wo(t) is nonsingular for at least one t > 0.3) For every v∈Rn, v≠0, CeAt v not identically zero.
If W0(t) is singular for any t > 0, then there exists v∈Rn, v≠0 such that CeAtv=0 for all t. If we have y(t)= CeAtx0 for some x0, then
y(t)vkCexCekv)(xCe At0
At0
At =+=+This means that different initial conditions will yield the same output. And the difference can never to told. ⇒ unobservable.
Proof continued. (necessity)
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By comparing
dτCeC'e(t)W Aτt
0τA'
o ∫= dτeBB'e(t)W τA't
0Aτ
c ∫=
Duality between controllability and observability
It is straightforward to conclude the following
Theorem of duality: The pair (A,B) is controllable if and only if (A1, C1) = (A’,B’) is observable.
BuAxx +=& zBzCyzAzAz''
1
1
====&Dual systems
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Equivalent conditions for observability:1) The pair (A,C) is observable.2) Wo(t) is nonsingular for some t > 0.3) The observability matrix
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡=
−1n
o
CA
CAC
G M
has full column rank, i.e., ρ(Go) = n.
4) The matrix
⎥⎦⎤
⎢⎣⎡ −= C
λIA)(Mo λ
has full column rank at every eigenvalue of A.
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Theorem: The pair (A,C) is observable if and only if
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡=
−+
qn
o1q-n
CA
CAC
GM
has full column rank, where q=ρ(C).
Theorem: The obserbability property is invariant under any equivalence transformation;
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39
Examples: Two circuits
1Ω1Ω
1 H
1F
+
x2 = y
−
x1
1Ω1Ω
1 H
1F
+
x2 = y
−
x1
[ ] x10y
u,11
xx
1001
xx
2
1
2
1
=
⎥⎦⎤
⎢⎣⎡+⎥⎦
⎤⎢⎣⎡
⎥⎦⎤
⎢⎣⎡
−−=⎥⎦
⎤⎢⎣⎡&&
[ ] x10y
u,11
xx
2110
xx
2
1
2
1
=
⎥⎦⎤
⎢⎣⎡+⎥⎦
⎤⎢⎣⎡
⎥⎦⎤
⎢⎣⎡
−−=⎥⎦
⎤⎢⎣⎡&&
⎥⎦⎤
⎢⎣⎡
−=⎥⎦⎤
⎢⎣⎡= 10
10CACGo
ρ(Go) = 1 < 2
Unobservable
⎥⎦⎤
⎢⎣⎡
−=⎥⎦⎤
⎢⎣⎡= 21
10CACGo
ρ(Go) = 2
Observable
40
Theorem: Consider the pair
[ ]mm
CCCC
A
AA
A L
LOMMLL
212
1
,
0000000
=⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
=
Suppose that the eigenvalues of Ai and those of Aj are disjoint for .i ≠ j . Then (A,C) is observable iff (Ai,Ci) is observable for all i.
21
41
Comments on controllability and observability
Controllability: Recall from the proof of the theorem on slide 8.To bring the state from x(0) = x0 to x(t1) = xd , a particular input is
]xx)[e(tWeB'u(t) d0At
11t)(tA' 11 −−= −−
c
Usually we don’t apply such an input to move the state betweenpoints in the state space. (open-loop control, very sensitive).A more practical problem is to use a simple linear feedback control, such as u = Fx + u0, to bring the state asymptotically toward a certain point. – Such a problem is related to stabilizability.
If a system is controllable, then it is stabilizable. A future result:
42
Observability: Theoretically, the initial state x(0) can be computedfrom the output y(t) by the following formula
dty(t)C'e)(tWx 1
0tA'1
1o0 ∫−=t
And the time interval [0,t1] can be arbitrarily small.
What we will be doing later is to build a LTI system withu and y as input and z as state so that z(t)→x(t) as t → 0.
Actually, we are not only interested in computing x(0) with other available information. Most often, we would like to detect the state for all time.
Again, this approach is complicated and sensitive.
xe
yLTI System
u
State estimator
If the system is observable,then the estimated state canbe made to approach the realstate asymptotically.
22
43
So far, we have learned • Controllability• Observability
Next, we will study• Canonical decomposition: to divide the state space
into controllable/uncontrollable, observable/unobservable subspaces
44
Canonical Decomposition
Consider an LTI system,DuCxyBu,Axx +=+=&
Let z = Px, where P is nonsingular, then
DD,CPCPB,B,PAPA whereuDzCyu,BxAz
11- ====+=+=
−
&
Recall that under an equivalence transformation, all properties,such as stability, controllability and observability are preserved.We also have 1oocc PGG,PGG −==
Next we are going to use equivalence transformation to obtain certain specific structures which reflect controllability and observability.
23
45
Controllability decomposition
Theorem: Suppose that ρ(Gc) = n1 < n. Let Q be a nonsingular matrix whose first n1 columns are LI columns of Gc. Let P=Q-1. Then
Recall Gc=[B AB … An-1B]. Suppose that ρ(Gc) = n1 < n. Then Gc has at most n1 LI columns.They form a basis for the range space of Gc.
[ ]cc
pnc
nnc
c
c
12c1
CCC
RB,RA,0BPBB,
A0AAPAPA 111
=
∈∈⎥⎦⎤
⎢⎣⎡==⎥
⎦
⎤⎢⎣
⎡== ××−
DBA)C(sIDB)A(C and lecontrollab is )B,A(pair theMoreover,
1c
1cc
cc
+−=+− −−sI
See page 159 for the proof.
46
Discussion: After state transformation, the equivalent system is
2c2
c2121c1
zA zuBzAzAz
=++=
&
&
The input u has no effect on z2. This part of state is uncontrollable.The first sub-system is controllable if z2=0. If z2≠0, then
20A
2
212τ)-(tAt
0cτ)-(tAt
010tA
11
ze)(z
)d(zAe )du(Beze)(tzc
1c1
1c1
1c
ττ
ττ
=
++= ∫∫ ττ
ττ de'BBe)(W 'Acc
τAt
01cc
1
∫=tc
Given a desired value for z1, say z1d. If we let
, dzeAe )(v 20A
12τ)-(tAt
01c1c
1 ττ∫=t
]z)(z)[e(tWe'Bu(t) and 1d110tA
11
ct)(t'A
c1c1c −+−= −− tv
Then you can verify that z1(t1)=z1d.
24
47
Example:
⎥⎥⎦
⎤
⎢⎢⎣
⎡=
⎥⎥⎦
⎤
⎢⎢⎣
⎡=
+=
100110
B,110010011
A
BuAxx&n=3, p=2, n-p+1=2.Only need to check Gc
2
[ ]⎥⎥⎦
⎤
⎢⎢⎣
⎡==
111001011110
ABBGc2 ableuncontroll ,32)( 2 <=cGρ
121 QP,
010101110
]qbb[QLet −=⎥⎥⎦
⎤
⎢⎢⎣
⎡== q is picked to make
Q nonsingular
⎥⎦⎤
⎢⎣⎡=
⎥⎥⎦
⎤
⎢⎢⎣
⎡==
⎥⎦
⎤⎢⎣
⎡=
⎥⎥⎦
⎤
⎢⎢⎣
⎡=
⎥⎥⎦
⎤
⎢⎢⎣
⎡
⎥⎥⎦
⎤
⎢⎢⎣
⎡
⎥⎥⎦
⎤
⎢⎢⎣
⎡
−
−==
0B
001001
PBB
,A0AA
100111001
010101110
110010011
101100111
PAQA
c
c
12c
Note: the last column of Q is different from the book (page 161).As a result, Ā12 is different from that in the book, which is 0.
48
Theorem: Suppose that ρ(Go) = n1 < n. Let P be a nonsingular matrix whose first n1 rows are LI rows of Go. Then
[ ] 1
111
noo
pno
nno
o
o
o21
o1
RC ,0CC
RB,RA,BBPBB,
AA0APAPA
×
××−
∈=
∈∈⎥⎦
⎤⎢⎣
⎡==⎥⎦
⎤⎢⎣
⎡==
q
DBA)C(sIDB)A(C and observable is )C,A(pair theMoreover,
1o
1oo
oo
+−=+− −−sI
Observability decomposition (follows from duality)
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡=
−1n
o
CA
CAC
G RecallM
Discussion: After state transformation, the equivalent system is
DuzCy u,BzA zA z
uBzAz
1o
o2o1212
o1o1
+=++=
+=&
& z2 may be affected by z1 but has no effect on y or z1
25
49
Summary for today: • Controllability• Observability• Canonical decomposition
– Controllable/uncontrollable – Observable/unobservable
Next Time:
• Controllability and observability continued– Controllability/observability decomposition– Minimal realization– Conditions for Jordan form conditions– Parallel results for discrete-time systems – Controllability after sampling
• State feedback design (introduction)
50
Problem Set #9
1. Is the following state equation controllable? observable?
[ ]101y ,010
011111110
=⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡+
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−−−−
−= uxx&
If not controllable, reduce it to a controllable one;If not observable, reduce it to an observable one.
2. Is the following state equation controllable? observable?
[ ]101y ,011010
200210201
=⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡+
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡−−
= uxx&
If not controllable, reduce it to a controllable one;If not observable, reduce it to an observable one.
26
51
Bonus Problem: Show that
)rank(Yrank(X)YZ0X
rank +=⎥⎦
⎤⎢⎣
⎡
)rank(Yrank(X)YZ0X
rank such that, examplean Give +≠⎥⎦
⎤⎢⎣
⎡
If Y has full row rank, then
)rank(Yrank(X)YZ0X
rank +≥⎥⎦
⎤⎢⎣
⎡