state space model: linear: or in some text: where: u: input y: output
DESCRIPTION
State space model: linear: or in some text: where: u: input y: output x: state vector A, B, C, D are const matrices. Example. State transition, matrix exponential. State transition matrix: e At. e At = I nxn +At+ A 2 t 2 + A 3 t 3 + - PowerPoint PPT PresentationTRANSCRIPT
State space model:
linear:
or in some text:
where: u: input y: output x: state vectorA, B, C, D are const matrices
),( :eqOutput
),( :eq State
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Example
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State transition, matrix exponential
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State transition matrix: eAt
• eAt = Inxn+At+ A2t2+ A3t3 +
• eAt is an nxn matrix • eAt =ℒ-1((sI-A)-1), or ℒ (eAt)=(sI-A)-1
• eAt= AeAt= eAtA
• eAt is invertible: (eAt)-1= e(-A)t
• eA0=I• eAt1 eAt2= eA(t1+t2)
dt
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Example
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Example
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I/O model to state space• Infinite many solutions, all equivalent.
• Controller canonical form:
1 1
1 1 0 1 1 01 1
0 1 1
0 1 1
0 1 0 0 0
0 0 1 0 0
0
0 0 0 1 0
1
[0]
n n n
n nn n n
n
n
d d d d dy a y a y a y b u b u b u
dt dt dt dt dt
x x u
a a a
y b b b x u
I/O model to state space• Controller canonical form is not unique
• This is also controller canonical form
1
1 1 01
1
1 1 01
1 2 1 0
1 2 1 0
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1 0 0 0 0
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0 1 0 0 0
0 0 1 0 0
[0]
n n
nn n
n
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n n
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d d dy a y a y a y
dt dt dtd d
b u b u b udt dta a a a
x x u
y b b b b x u
Example
t
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n=4 a3 a2 a1 a0 b1 b0=b2=b3=0
xy
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0010
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>> n=[1 2 3];d=[1 4 5 6];>> [A,B,C,D]=tf2ss(n,d)
A = -4 -5 -6 1 0 0 0 1 0B = 1 0 0C = 1 2 3D = 0
>> tf(n,d) Transfer function: s^2 + 2 s + 3---------------------s^3 + 4 s^2 + 5 s + 6
DuCxy
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Characteristic values• Char. eq of a system is
det(sI-A)=0the polynomial det(sI-A) is called char. pol.the roots of char. eq. are char. valuesthey are also the eigen-values of A
e.g.
∴ (s+1)(s+2)2 is the char. pol. (s+1)(s+2)2=0 is the char. eq.
s1=-1,s2=-2,s3=-2 are char. values or eigenvalues
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Solution of state space model
Recall: sX(s)-x(0)=AX(s)+BU(s)
(sI-A)X(s)=BU(s)+x(0)
X(s)=(sI-A)-1BU(s)+(sI-A)-1x(0)
x(t)=(ℒ-1(sI-A)-1))*Bu(t)+ ℒ-1(sI-A)-1) x(0)
x(t)= eA(t-τ)Bu(τ)d τ+eAtx(0)
y(t)= CeA(t-τ)Bu(τ)d τ+CeAtx(0)+Du(t)
DuCxy
BuAxx
t
0
t
0
But don’t use those for hand calculation
use:X(s)=(sI-A)-1BU(s)+(sI-A)-1x(0)
x(t)=ℒ-1{(sI-A)-1BU(s)}+{ℒ-1 (sI-A)-1} x(0)
& Y(s)=C(sI-A)-1BU(s)+DU(s)+C(sI-A)-1x(0)
y(t)= ℒ-1{C(sI-A)-1BU(s)+DU(s)}+C{ℒ-1 (sI-A)-1} x(0)
e.g.
xy
uxx
01
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1
20
01 If u= unit step
1
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Note: T.F.=D+ C(sI-A)-1B
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Eigenvalues, eigenvectors
Given a nxn square matrix A, p is an eigenvector of A if Ap p∝i.e. λ s.t. Ap= λpλ is an eigenvalue of A
Example: ,
Let ,
∴p1 is an e-vector, & the e-value=1
Let ,
∴p2 is also an e-vector, assoc. with the λ =-2
20
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In Matlab
>> A=[2 0 1; 0 2 1; 1 1 4];
>> [P,D]=eig(A)
P = 0.6280 0.7071 0.3251
0.6280 -0.7071 0.3251
-0.4597 -0.0000 0.8881
p1 p2 p3
D =1.2679 0 0
0 2.0000 0
0 0 4.7321
λ1
λ2
λ3
If we use [P,D]=eig(A) get approximate but wrong answerShould use:
>>[P,J]=jordan(A)P =
0.3750 0 1 0.625 0 8 4 0 -0.375 0 0 0.375 0 16 9 0
J=
-8 0 0 0 0 -16 1 0 0 0 -16 1 0 0 0 -16
a 3x3 Jordan block assoc. w/. λ=-16
• In general, if λ, P is an e-pair for A,AP= λP λP-AP=0 λIP-AP=0 (λI-A)P=0
∵ P≠0 det(∴ λI-A)=0 ∴ λ is a sol. of char. eq of A
• char. pol. of nxn A has deg=n ∴ A has n eigen-values.e.g. A= , det(λI-A)=(λ-1)(λ+2)=0
⇒ λ1=1, λ2=-2
20
01
• If λ1 ≠λ2 ≠λ3⋯then the corresponding P1, P2, will be linearly ⋯independent, i.e., the matrix
P=[P1 P⋮ 2 P⋮ ⋯ n] will be invertible.AP1= λ1P1
AP2= λ2P2
⋮A[P1 P⋮ 2 ]=[AP⋮ ⋯ 1 AP⋮ 2 ]⋮ ⋯
=[λ P1 ⋮ λ P2 ]⋮ ⋯
=[P1 P2 ]⋯
0
0
0
0
2
1
• ∴ AP=PΛ P-1AP= Λ=diag(λ1, λ2, ⋯)
∴If A has n lin. ind. Eigenvectors then A can be diagonalized.
Note: Not all square matrices can be diagonalized.
If A does not have n lin. ind. e-vectors(some of the eigenvalues are identical),then A can not be diagonalized
E.g. A=
det(λI-A)= λ4+56λ3+1152λ2+10240λ+32768
λ1=-8λ2=-16λ3=-16λ4=-16
by solving (λI-A)P=0
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