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MODERN CONTROL SYSTEMS Emam Fathy Department of Electrical and Control Engineering email: [email protected] http ://www.aast.edu/cv.php?disp_unit=346&ser=68525 Lecture 1 State Space Representation 1

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Modern Control Systems (MCS)email: [email protected]
– MIMO systems.
Conventional control theory is applicable to:
– SISO systems.
3
Example
Consider the mechanical system shown in figure. We assume that the system is linear. The external force u(t) is the input to the system, and the displacement y(t) of the mass is the output. The displacement y(t) is measured from the equilibrium position in the absence of the external force. This system is a single-input, single-output system.
From the diagram, the system equation is
() + () + () = ()
This system is of second order. This means that the system involves two integrators. Let us define state variables 1() and 2() as
1 = ()
Example
Then we obtain
1 = 2()
2 = −

2 −
)()()( tButAxtx
)()( tCxty
Example(summary)
• The system equation is () + () + () = ()
• Let 1 = () 2 = ()
1 = 2()
2 = −

State Equation
Output Equation
)()()( tButAxtx
)()()( tDutCxty
Where, x(t) ------------ State Vector A(nxn) ------ System Matrix B(nxp) ------- Input Matrix u(t) ----------- Input Vector
y(t) ----------- Output Vector C(qxn) ------ Output Matrix D -------------- Feed forward Matrix
Canonical Forms Canonical forms are the standard forms of state space models.
Each of these canonical form has specific advantages which makes it convenient for use in particular design technique.
There are several canonical forms of state space models
– Phase variable canonical form
– Controllable Canonical form
– Observable Canonical form
– Diagonal Canonical form
• Jordan Canonical Form
It is interesting to note that the dynamics properties of system remain unchanged whichever the type of representation is used.
• Obtain the state equation in phase variable form for the following differential equation, where u(t) is input and y(t) is output.
• The differential equation is third order, thus there are three state variables:
• And their derivatives are (i.e state equations)
2 3
3 + 4
1 = 2
2 = 3
Phase Variable Canonical form
Phase Variable Canonical form
• In vector matrix form
2 = 3


































• where u is the input and y is the output.
• This equation can also be written as
• We will present state-space representations of the system defined by above equations in controllable canonical form and observable canonical form.
ububububyayayay nn
Controllable Canonical Form ()
u
x
x
x
x
aaaax
x
x
x
n
n
nnnn
n
() =
u
b
b
b
b
x
x
x
x
a
a
a
a
x
x
x
x
n
n
n
n
n
n
n
n
−3 −1 −2 +
17
= −2 1 0 −1 0 1 −3 0 0
+ 0 3 3
Diagonal Canonical Form
+ 1 + 2 ………( + )
= + 1
+ 1 +
ub
x
x
x
State Space to T.F
• Now Let us convert a space model to a transfer function model.
• Taking Laplace transform of equation (1) and (2) considering initial conditions to zero.
• From equation (3)
)()()( 1 sBUAsIsX (5)
Transfer Matrix (State Space to T.F) • Substituting equation (5) into equation (4) yields
)()()()( 1 sDUsBUAsICsY
)()()( 1 sUDBAsICsY
DBAsIC sU
sY 1)(
Example 3
Convert the following State Space Model to Transfer Function Model if K=3, B=1 and M=10;
)(tf
Example 3
)(
Obtain the transfer function T(s) from following state space representation.
Answer
State Controllability A system is completely controllable if there exists an
unconstrained control u(t) that can transfer any initial state x(to) to any other desired location x(t) in a finite time, to ≤ t ≤ T.
controllable
uncontrollable
System is said to be state controllable if
BABAABBC n T
xy
uxx
21
0
1
30
01
• Thus
state controllable.
1 A B
State Observability A system is completely observable if and only if there exists a
finite time T such that the initial state x(0) can be determined from the observation history y(t) given the control u(t), 0≤ t ≤ T.
observable
unobservable







OT is obtained as
• Since () ≠ therefore system is not completely state
observable.
Check the state controllability, state observability of the following system
10, 1
Eigenvalues & Eigen Vectors
Eigenvalues & Eigen Vectors
• The eigenvalues of an nxn matrix A are the roots of the characteristic equation.
• Consider, for example, the following matrix A:
46
48
Similarity Transformations • It is desirable to have a means of transforming one state-space
representation into another.
• Consider state space model
• Along with this, consider another state space model of the same plant
• Here the state vector , say, represents the physical state relative to some other reference, or even a mathematical coordinate vector.
)()()( tButAxtx
)()()( tDutCxty
)()()( tuBtxAtx
)()()( tuDtxCty
Similarity Transformations • When one set of coordinates are transformed into another
set of coordinates of the same dimension using an algebraic coordinate transformation, such transformation is known as similarity transformation.
• In mathematical form the change of variables is written as,
• Where T is a nonsingular nxn transformation matrix.
• The transformed state () is written as
)( )( txTtx
• Taking time derivative of above equation
)( )( 1 txTtx
• Substituting = −1() in above equation
• Since output of the system remain unchanged [i.e. = ()] therefore above equation is compared with = + () that yields
)()()( tuDtxCty
CTC DD
Transformation to DCF
• In linear algebra, a square matrix A is called diagonalizable, if there exists an invertible matrix P such that P-1AP is a diagonal matrix.
• n-by-n square matrix A is called invertible (also nonsingular) if there exists an n-by-n square matrix B such that AB=BA=I
• Diagonalizable matrices are o easy to handle
o their eigenvalues and eigenvectors are known
o can raise a diagonal matrix to a power by simply raising the diagonal entries to that same power
o the determinant of a diagonal matrix is simply the product of all diagonal entries.
Example
() + 1 0 ()
= 1 1 ()
Find Eigen values, Eigen vectors, and diagonal form of this system
− = 0
= − 1 − 4 + 2 = 2 − 5 + 6
1 = 2 2 = 3 68
Eigen vectors
1 2 −1 −4
−11 + 4 12 = 2 11
Let 11 = 1 , so 12 = 1
2 , 1 =
1 2 −1 −4
−21 + 4 22 = 3 22
Let 21 = −1 , so 22 = −1 , 2 = −1 −1
, ≠ 0
= 1 −1 1
1 = 11() + 1() = 11()
1 = −1 = 1 −1 1
2 −1
2 −1
2 −1
1() + 2 1 ()