Characteristic values

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Characteristic values

• Char. eq of a system is
• det(sI-A)0
• the polynomial det(sI-A) is called char. pol.
• the roots of char. eq. are char. values
• they are also the eigen-values of A
• e.g.
• ? (s1)(s2)2 is the char. pol.
• (s1)(s2)20 is the char. eq.
• s1-1,s2-2,s3-2 are char. values or eigenvalues

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gtgt MySysss(A,B,C,D) a x1 x2 x3 x1
-1 0 0 x2 0 -2 1 x3 0 0 -2 b
u1 x1 1 x2 0 x3 1 c
x1 x2 x3 y1 1 1 1 d u1
y1 0 Continuous-time model.
gtgt A-1 0 0 0 -2 1 0 0 -2 A -1 0
0 0 -2 1 0 0 -2 gtgt
B101 B 1 0 1 gtgt C1 1
1 C 1 1 1 gtgt D0 D 0
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gtgt tf(MySys) Transfer function 2 s2 8 s
7 --------------------- s3 5 s2 8 s 4 gtgt
zpk(MySys) Zero/pole/gain 2 (s2.707)
(s1.293) --------------------- (s1)
(s2)2 gtgt roots(2 8 7) ans -2.7071
-1.2929 gtgt roots(1 5 8 4) ans -2.0000
-2.0000 -1.0000
gtgt ssym('s') s s gtgt det(seye(3)-A) ans
(s1)(s2)2 gtgt eig(A) ans -1 -2
-2 gtgt DCinv(seye(3)-A)B ans
1/(s1)1/(s2)21/(s2) gtgt simplify(ans) ans
(2s28s7)/(s1)/(s2)2
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gtgt inv(seye(3)-A) ans 1/(s1), 0,
0 0,
1/(s2), 1/(s2)2 0,
0, 1/(s2) gtgt ilaplace(ans) ans
exp(-t), 0, 0
0, exp(-2t),
texp(-2t) 0, 0,
exp(-2t) gtgt tsym('t') t t gtgt
expm(At) ans exp(-t), 0,
0 0,
exp(-2t), texp(-2t) 0,
0, exp(-2t)
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gtgt expm(At)A ans -exp(-t),
0, 0
0,
-2exp(-2t), exp(-2t)-2texp(-2t)
0, 0,
-2exp(-2t) gtgt
Aexpm(At) ans -exp(-t),
0, 0
0,
-2exp(-2t), exp(-2t)-2texp(-2t)
0, 0,
-2exp(-2t) gtgt
Aexpm(At)-expm(At)A ans 0, 0, 0 0,
0, 0 0, 0, 0
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can
?
set t0
?No
can
?
v
at t0
?
v
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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-t)Bu(t)d teAtx(0)
• y(t) CeA(t-t)Bu(t)d tCeAtx(0)Du(t)

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S.S to T.F.

• X(s)(sI-A)-1BU(s)
• Y(s)C(sI-A)-1BU(s)DU(s)
• (D C(sI-A)-1B)U(s)
• ? T.F. H(s) D C(sI-A)-1B
• In matlab ss2tf
• eig
• roots
• poly
• use help to find out how to use these

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• In Matlab
• gtgt A0 1-2 -3
• gtgt B01
• gtgt C1 3
• gtgt D0
• gtgt n,dss2tf(A,B,C,D)
• n
• 0 3.0000 1.0000
• d
• 1 3 2
• gtgttf(n,d)

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But dont use those for hand calculation

• useX(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-1C(sI-A)-1BU(s)DU(s)CL-1
  (sI-A)-1 x(0)
• e.g.

u unit step
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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. ? s.t. Ap ?p
• ?is an eigenvalue of A
• Example ,
• Let ,
• ?p1 is an e-vector, the e-value1
• Let ,
• ?p2 is also an e-vector, assoc. with the ? -2

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• In general, if ?, p is an e-pair for A,
• Ap ?p
• ?p-Ap0
• ?Ip-Ap0
• (?I-A)p0
• ? p?0 ? det(?I-A)0
• ? ? is a sol. of char. eq of A
• char. pol. of nxn A has degn
• ? A has n eigen-values.
• e.g. A , det(?I-A)(?-1)(?2)0
• ? ?11, ?2-2

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• If ?1 ??2 ??3?
• then the corresponding p1, p2, ? will be
  linearly independent, i.e., the matrix
• Pp1?p2 ? ?pn will be invertible.
• Ap1 ?1p1
• Ap2 ?2p2
• ?
• Ap1?p2 ? ?Ap1?Ap2 ? ?
• ?p1? ?p2 ? ?
• p1 p2 ?

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• ? APP?
• P-1AP ?diag(?1, ?2, ?)
• ?If A has n lin. ind. Eigenvectors then A can be
  diagonalized.
• Note Not all square matrices can be diagonalized.

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Example
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• In Matlab
• gtgt A2 0 1
• 0 2 1
• 1 1 4
• gtgt P,Deig(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

?1
?2
?3
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• 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(?I-A) ?456?31152?210240?32768
• ?1-8
• ?2-16
• ?3-16
• ?4-16
• by solving (?I-A)P0

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• If we use P,Deig(A)
• get approximate but wrong answer
• Should use gtgtP,Jjordan(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/. ?-16
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More Matlab Examples

• gtgt ssym('s')
• gtgt A0 1-2 -3
• gtgt det(seye(2)-A)
• ans
• s23s2
• gtgt factor(ans)
• ans
• (s2)(s1)

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• gtgt P,Deig(A)
• P
• 0.7071 -0.4472
• -0.7071 0.8944
• D
• -1 0
• 0 -2
• gtgt P,Djordan(A)
• P
• 2 -1
• -2 2
• D
• -1 0
• 0 -2

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• A 0 1
• -2 -3
• gtgt exp(A)
• ans
• 1.0000 2.7183
• 0.1353 0.0498
• gtgt expm(A)
• ans
• 0.6004 0.2325
• -0.4651 -0.0972
• gtgt tsym('t')
• gtgt expm(At)
• ans
• -exp(-2t)2exp(-t),
  exp(-t)-exp(-2t)
• -2exp(-t)2exp(-2t), 2exp(-2t)-exp(-t)

?
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v
v
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Similarity transformation
same system as()
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Example
diagonalized
decoupled
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Invariance
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Controllability
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Example
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• In Matlab
• gtgt Sctrb(A,B)
• gtgt rrank(S)
• If S is square (when B is nx1)
• gtgt det(S)

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Observability
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Example
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• In Matlab
• gtgt Vobsv(C,A)
• gtgt rrank(V)
• rank must n
• Or if single output (ie V is square), can use
• gtgt det(V)
• det must be nonzero

Lookfor controllability Lookfor observability
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• Recall linear transformation
• Controllabilitybeing able to use u(t) to drive
  any state to origin in finite time
• Observabilitybeing able to computer any x(0)
  from observed y(t)
• After transformation, eigenvalues, char. poly,
  char. eq, char. values, T.F., poles, zeros
  un-changed, but eigenvector changed

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• Controllability is invariant under transf.

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• Observability invariant under transf.

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State Feedback
D
1 s
r
u
x



y
B
C


-
A
K
feedback from state x to control u
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• In Matlab
• Given A,B,C,D
• ?Compute QCctrb(A,B)
• ?Check rank(QC)
• If it is n, then
• ?Select any n eigenvalues(must be in complex
  conjugate pairs)
• ev?1 ?2 ?3 ?n
• ?Compute
• Kplace(A,B,ev)

ABk will have eigenvalues at
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• Thm Controllability is unchanged after state
  feedback.
• But observability may change!