ee 529 circuits and systems analysis mustafa kemal uyguroğlu lecture 9
TRANSCRIPT
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EE 529 Circuits and Systems Analysis
Mustafa Kemal Uyguroğlu
Lecture 9
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State vectorState vector
a listing of state variables in vector form
(t)x
(t)x
(t)x
(t)x
(t)x
n
1n
2
1
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State equationsState equations
(t)uB(t)xA
(t)x
(t)x
(t)x
(t)x
n
2
1
(t)uD(t)xC(t)y
System dynamics
Measurement
Read-out mapOutput vector
Inpu
t vec
tor
Stat
e ve
ctor
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x:n-vector (state vector)
u:p-vector (input vector)
y:m-vector (output vector)
A:nxn
B:nxp
C:mxn
D:mxp
nn
nn
mm
mm
nn
pp
nn
pp
System matrix
Input (distribution) matrix
Output matrix
Direct-transmission matrix
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Solution of state eq’nsSolution of state eq’ns
Consists of:
Free response Forced sol’n&
(Homogenous sol’n) (particular sol’n)
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Homogenous solutionHomogenous solution
Homogenous equation
xAx has the solution
0xΦ(t)(t)x
State transition matrix X(0)
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State transition matrixState transition matrix
An nxn matrix (t), satisfying
( ) A ( ), (0)
where is identity matrix.
0
t t
n n
I
I
x 0 = x 0
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Determination of Determination of (t):(t): transform methodtransform method
Laplace transform of the differential equation:
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Determination of Determination of (t):(t): transform methodtransform method
11 tt s e
AI AL
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Determination of Determination of (t):(t): time-domain solutiontime-domain solution
( ) . ( )t a t
Scalar case
( ) att e
where
0
12211
k
kkk
at taatate !! .........)(
0
x ax
x t t x
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Determination of (t): time-domain solution
( ) . ( )t t AFor vector case, by analogy
( ) tt e A
where
21 12! !
0
1 ( ) .........t k kk
k
e At At A t
A
Can be verified by substitution.
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Φ(t2-t0)
Properties of TMProperties of TM
(0)=I
-1(t)= (-t)
Ф(t2-t1)Φ(t1-t0)= Φ(t2-t0)
[Φ(t)]k= Φ(kt)
Φ(t)Φ(-t)
Φ(t1-t0) Φ(t2-t1)
t0 t1 t2
Φ(t) Φ(t) Φ(t) Φ(t) Φ(t) Φ(t)Φ(kt)
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General solutionGeneral solutionScalar case
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General solutionGeneral solutionVector case
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General solutionGeneral solution: transform method: transform method
uBxAx L{ }
(s)uB(s)xA(0)x(s)xs ˆˆˆ
1 1ˆ ˆx(s) (sI A) x(0) (sI A) Bu(s)
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Inverse Laplace transform yields:
(t)uB*Φ(t)(0)xΦ(t)(t)x
t
0
)d(uBe(0)xe(t)x )-A(tAt
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t
t0
) )d(uBe)(txe(t)x )-A(t0
t-A(t 0
For initial time at t=t0
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The outputThe output
y(t)=Cx(t)+Du(t)
(t)uD(t)duBeC)(txCe(t)yt
t
)-A(t
0
)t-A(t 0 0
Zero-inputresponse Zero-state response
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ExampleExample
Obtain the state transition matrix (t) of the following system. Obtain also the inverse of the state transition matrix -1(t) .
1
2
0 1
2 3
x x
x x
1
2
For this system
0 1
2 3
A
the state transition matrix (t) is given by
1 1( ) [( ) ]tt e s LA I A
since0 0 1 1
0 2 3 2 3
s ss
s s
I A =
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ExampleExample
2 21
2 2
2( )
2 2 2
t t t tt
t t t t
e e e et e
e e e e
A
1
2 1 1 11 2 1 21
2 2 1 21 2 1 2
3 11( )
2( 1)( 2)
( ) s s s s
s s s s
ss
ss s
s
I A
I A
The inverse (sI-A) is given by
Hence
2 2
2 2
2( )
2 2 2
t t t t
t t t t
e e e et
e e e e
Noting that -1(t)= (-t), we obtain the inverse of transition matrix as:
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Exercise 1Exercise 1
Find x1(t) , x2(t)
The initial condition
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Exercise 1 (Solution)Exercise 1 (Solution)
x = (t)x(0)
11
2 21
2
2 2
( )
4 11 4 11 4 5 4 5
5 4 5 54 5
4 5 4 5
t sI A
ss s s s s ssI A sI As s ss s
s s s s
L
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Example 2Example 2
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Exercise 2Exercise 2
Find x1(t) , x2(t)
The initial condition
Input is Unit Step
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Exercise 2 (Solution)Exercise 2 (Solution)
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Matrix Exponential eMatrix Exponential eAtAt
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Matrix Exponential eMatrix Exponential eAtAt
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P=
1 1 1
1 2
12
22 2
11
21 1
n
n
n nnn
The transformation where
1,2,…,n are distinct eigenvalues of A. This transformation will transform P-1AP into the diagonal matrix
n
2
1
0
0
=APP 1
ˆx = Px
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Example 3Example 3
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Method 2:
1 1
1
2
1 1
2
[( ) ]
0 0 1 1
0 0 2 0 2
1 1 1 1 1 12 1 21 2 2
( )02 11
0022
11 1
2[( ) ]1
02
t
t
t
t
e s
s ss
s s
s s s s s s ss
ss s
ss
ee s
e
L
L
A
A
I A
I A =
I A
I A
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Matrix Exponential eMatrix Exponential eAtAt
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Matrix Exponential eMatrix Exponential eAtAt
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Example 4Example 4
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Laplace TransformLaplace Transform
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