english cl 2
TRANSCRIPT
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2. State equations
State equations
Solution of the state equations
Assumption: We assume that all the Laplace transforms involved in the
following reasonings exist.
x(t) = A x(t) + Bu(t)
y(t) = C x(t) + Du(t)
L
x(s) = (sIn A)
1x(0) + (sIn A)1Bu(s)
y(s) = C(sIn A)1x(0) + [C(sIn A)1B + D]u(s)
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2. State equations
Back to the time domain
x(s) = (sIn A)1x(0) + (sIn A)1Bu(s)
y(s) = C(sIn A)1x(0) + [C(sIn A)1B + D]u(s)
L1
x(t) = L1{(sIn A)1} x(0) + L1{(sIn A)1}B u(t)
y(t) = CL1{(sIn A)1} x(0 ) + [CL1{(sIn A)1}B + D] u(t)
L1{(sIn A)1} = ?
(sInA)1 is a rational matrix function that is (strictly) proper. Its
Laplace transform can be computed componentwise.
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2. State equations
Example:
ComputeL1{(sIn A)1} for A=
1 2
2 1
.
(sIA)1 = s1(s1)2+4 2(s1)2+4
2(s1)2+4
s1(s1)2+4
L1{(sIA)1}= et
cos 2t sin 2t
sin 2t cos 2t
(check this!)
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2. State equations
General expression for L1{(sIn A)1}
(sIA)1 = s1I+k=1
Aksk1
L1{(sIn A)1} = L1{s1}I+
k=1
AkL1{sk1}
= I+k=1
AkL1{dk1s
dsk(1)k
k!}
=k=0
Aktk
k!=: eAt
this notation is chosen by analogy with the scalar case
L1{ 1sa
}= eat =
k=0aktk
k!
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2. State equations
General form of x(t) and y(t)
x(t) = L1{(sIn A)1} x(0) + L1{(sIn A)1}B u(t)
y(t) = CL1{(sIn A)1} x(0 ) + [CL1{(sIn A)1}B + D] u(t)
x(t) = e
At
x(0) + eAt
B u(t)y(t) = CeAt x(0 ) + [CeAtB + D] u(t)
x(t) = eAtx(0) +
t
0 eA(t)B u()d
y(t) = CeAtx(0) +t0
CeA(t)B u()d+Du(t)
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2. State equations
x(t) = eAtx0 +t0
eA(t)B u()d is a solution ofx= Ax + Bu.
It is the only solution ofx= Ax+ Bu such that x(0) = x0, for a fixed
given u. [Prove this fact using the following theorem.]
Theorem - existence and uniqueness of solution
Consider a first order differential equation x= F(x) with initial condition x(t0) = x0 (IVP
- initial value problem). IfF is a lipschitzian, then the (IVP) has a unique solution. This
solution is of class C1, i.e. is continuously differentiable.
x(t) is defined for all inputs u(t) that guarantee the existence of theintegral
t0
eA(t)B u()d. Here we take as admissible the inputs u(t)
which are piecewise continuous.
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2. State equations
The solutions of the state eqautions can also be written as follows (if
the initial conditions are given at time t0):
x(t) = eA(tt0)x(t0) +tt0
eA(t)B u()d
y(t) = CeA(tt0)x(t0) +tt0
CeA(t)B u()d +Du(t)
Check this!
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2. State equations
Zero-input evolution/response- state and output evolution for zero input
Zero-state evolution/response- state and output evolution for zero initial
state
x(t) = eAtx(0) xl(t)
+ t
0eA(t)B u()d
xf(t)
y(t) = CeAtx(0) yl(t)
+ t
0
CeA(t)B u()d+Du(t)
yf(t)
zero-input evolution zero-state evolution
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2. State equations
Impulse response and transfer function
yf(t) =t0
CeA(t)B u()d+Du(t)
Impulse response
yf(t) = [CeAtB + D] u(t)L1 L
yf(s) = [C(sInA)1B+D] u(s)
Transfer function
Impulse = Dirac -function; ui = yf = CeAtB + Di.
ui - i-th component of u Di - i-th colunm of D
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2. State equations
Discretization
Discretization starting at time t0, with discretization interval .
xd(k) := x(t0+k); analogous definitions for ud e yd.
Process 1
Approximate x(t) x(t+)x(t)
.
This leads to:
xd(k+ 1) =(I+A)xd(k) +Bud(k)yd(k) = Cxd(k) +Dud(k) Check!
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2. State equations
Process 2
Suppose u(t) constant in each interval [t0+k t0+ (k+ 1))
Compute the state trajectories of the continuous system at the
discretization instants:
x(t0+ (k+ 1)) =
eA((t0+(
k+1)
(t0+
k))x(t0 + k) +
t0+(k+1)t0+k e
A(t0+(
k+1)
)Bu()d
This leads to:
xd(k+ 1) = eAxd(k) +
0 eABd
ud(k)
yd(k) = Cxd(k) +Dud(k) Check this!
Exercise: Compare the discrete systems obtained by the two different
processes.
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2. State equations
DiscretizaoExacta
EXEMPLO:
Considere-se o sistema compartimental contnuo,
1 0 1 0
0 0 0 1
0 1 1 0
x x u
= +
& (1)
( ) ( ) ( )
0.05 0.05 0.05 0.05
0.05 0.05 0.05
1 1.05 0.05 1.95 2.05
1 0 1 0 0.05
0 1 0.95
e e e e
x k x k u k
e e e
+
+ = + +
E o sistema compartimental discretizadocorrespondente, para h=0.05seg:
(2)
Assim
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2. State equations
0 50
2
4
6
Tempo (seg)
Amplitude
0 50
2
4
6
Tempo (seg)
Amplitude
0 50
2
4
6RESPOSTA AO DEGRAU UNITRIO
Tempo (seg)
Amplitude
DiscretizaoExacta
Respostas ao degrau do sistema contnuo e da sua discretizaoexacta
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2. State equations
DiscretizaoExacta
Respostas foradas do sistema contnuo e da sua discretizao exacta
nos intervalos de discretizao
0 50
5
10
15
Tempo (seg)
Amplitude
0 50
2
4
6
Tempo (seg)
Amplitude
0 50
5
10RESPOSTA FORADA
Tempo (seg)
Amplitude
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S
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2. State equations
DiscretizaoAproximada
EXEMPLO:
Considere-se novamente o sistema compartimental contnuo (1):
( ) ( ) ( )
0.95 0 0.05 0
1 0 1 0 0.05
0 0.05 0.95 0
x k x k u k
+ = +
E o sistema compartimental discretizado aproximadamente correspondente, para
h=0.05seg:
Assim
(3)
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2. State equations
DiscretizaoAproximada
Respostas ao degrau do sistema contnuo e da sua discretizaoaproximada
0 50
2
4
6
Tempo (seg)
Am
plitude
0 50
2
4
6
Tempo (seg)
Am
plitude
0 50
2
4
6RESPOSTA AO DEGRAU UNITRIO
Tempo (seg)
Am
plitude
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2. State equations
General solution for discrete state equations
x(k+ 1) = Ax(k) +Bu(k)
y(k) = Cx(k) +Du(k)
x(k) = Akx(0) +k1
l=0 Ak1lBu(l)
y(k) = CAkx(0) +l=0CAk1Ak1lBu(l) + Du(k)
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2. State equations
Invertible transformations (isomorphisms) in the state space
State transformation: x x= Sx
S invertible(i.e., x x= Sx is an isomorphism)
Question: What are the evolution equations for x(t)?
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2. State equations
x = Ax + Buy = Cx + Du
Sx = SAS1Sx + SBuy = CS1Sx + Du
Replacing Sx by x, yields:
x =
A SAS1 x +
BSB u
y = CS1 Cx + Du
x = Ax + Bu
y = Cx + Du
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2. State equations
Thus:
x = Ax + Bu
y = Cx + Du
x = Ax + Bu
y = Cx + DuTransformation S
(A , B , C , D) (A, B, C, D) =
= (SAS1
, SB, CS
1
, D)
(A , B , C , D) Algebraically equivalent
e systems
(A, B, C, D) invertible matrix Ssuch that
A= SAS1, B = SB
C= CS1, D = D
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2. State equations
Prove That:
Proposition: If two systems are algebraically equivalent then they have the
same transfer function.
Remark: Two systems with the same transfer function are called
zero-state equivalent.
So, the previous proposition states thatwhenever two systems are
algebraically equivalent systems, they are also zero-state equivalent.
However: there are systems that are zero-state equivalent, but not
algebraically equivalent.
Give an example!
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