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Control System Design-II
.
Department of Electrical EngineeringPakistan Institute of Engineering and Applied Sciences
P.O. Nilore Islamabad Pakistan
url: www.pieas.edu.pk/aqayyumEmail: [email protected]
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Preliminaries of MCT ctd….. The central concept on which the modern control theory is based,
is STATE.
State: The state of dynamic system is the smallest set of variables(referred to as state variables) such that the knowledge of thesevariables at together with the knowledge of input for0t t= 0t t>
Control System Design-II by Dr. A.Q. Khan
completely determines the behavior of the process; that is,
The concept of state is not limited to physical/engineeringsystems but also biological systems, economic systems, socialsystems etc.
Deter mines
0 t t 0 0t t t tx(t) , and u(t) x(t) →
>= >
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State variables: The smallest set of variables that determinesthe behavior of the system
State vector:Vector of state variables. For ‘n’ state variables, astate vector of n-components to be constructed. Let ‘x’ be a
Preliminaries of MCT ctd…..
Control System Design-II by Dr. A.Q. Khan
State space: It is an ‘n’ dimensional space whose coordinatesare state variables.
[ ]1 2 3 4x x x x . . . x=
1 2 3 nx ,x , x , , xK
Any state can be represented by a point in state space
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State space equation: It is a first order differential equation(or set of first order differential equations).
It generally involves three variables of interest.
State variables
Preliminaries of MCT ctd…..
Control System Design-II by Dr. A.Q. Khan
Output variables
Let us consider
x x u,y x
= +=
&
State variables
Output equation
Input variablesState equation
Outputvariables
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There exists many kinds of state space representation.However the no. of states (state variables) are the same.
A higher order differential equation can be represented inthe form of set of first order differential equations. Hence
Preliminaries of MCT ctd…..
Control System Design-II by Dr. A.Q. Khan
Let us consider a 2nd order dynamic system
y by cy u+ + =&& &
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The preceding example shows
Preliminaries of MCT ctd…..
1
1 2
x yx y x
== =& &
& && &
Control System Design-II by Dr. A.Q. Khan
Note that the states and are the outputs of integrator.
Remember! The state space model for a system is a set of1st order differential equations.
2 2 1x y u y cy u x cx= = − − = − −
1x
2x
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Let us consider
Preliminaries of MCT ctd…..
1 1 1 2 n 1 2 p
2 2 1 2 n 1 2 p
3 3 1 2 n 1 2 p
x f (x , x , , x ,u , u , , u , t)
x f (x , x , , x , u ,u , , u , t)
x f (x , x , , x ,u , u , , u , t)
=
= =
& L L
& L L
& L L
Control System Design-II by Dr. A.Q. Khan
n n 1 2 n 1 2 px f (x , x , , x , u ,u , ,u , t)
=
MM
& L L
a e equa ons
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1 1 1 2 n 1 2 p
2 2 1 2 n 1 2 p
3 3 1 2 n 1 2 p
y h (x , x , , x , u , u , , u , t)
y h (x , x , , x ,u ,u , , u , t)
y h (x , x , , x , u ,u , , u , t)
=
= =
L L
L L
L L
Preliminaries of MCT ctd…..
Control System Design-II by Dr. A.Q. Khan
m m 1 2 n 1 2 p
y h (x , x , , x ,u ,u , , u , t)
=
M
M
L L
u pu equa on
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Combining the two
Preliminaries of MCT ctd…..
11 1
22 2
ux yux y
x(t) , y(t) , u(t)
= = = MM M
State Vector Output Vector Input Vector
Control System Design-II by Dr. A.Q. Khan
pn m
1 1 2 n 1 2 p
2 1 2 n 1 2 p
n 1 2
ux yf (x ,x , ,x ,u ,u , ,u ,t)
f (x ,x , ,x ,u ,u , ,u ,t)
f(x)
f (x ,x
=
L L
L L
MM
1 1 2 n 1 2 p
2 1 2 n 1 2 p
n 1 2 p m 1 2 n 1 2 p
h (x ,x , ,x ,u ,u , ,u ,t)
h (x ,x , ,x ,u ,u , ,u ,t)
, h(x) =
, ,x ,u ,u , ,u ,t h (x ,x , ,x ,u ,u , ,u ,t)
L L
L L
MM
L L L L
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The resultant state space model is
Preliminaries of MCT ctd…..
x f (x , u , t)
y h (x , u , t)==
&
Control System Design-II by Dr. A.Q. Khan
It may represents Linear System
Nonlinear System
Linear Time varying system
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The LTV system is represented by
Preliminaries of MCT ctd…..
( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )
x t A t x t B t u t
y t C t x t D t u t
= +
= +
&
Control System Design-II by Dr. A.Q. Khan
The LTI system is represented by
( ) ( ) ( )( ) ( ) ( )
x t Ax t Bu t
y t Cx t Du t
= +
= +
&This is the focusof this course
11
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Transfer function V/S State Space Two important Questions
Given a Transfer function G(s) = Y(s)/U(s), how to obtain thestate space representation?
Given the state space representation G: (A,B,C,D), how toobtain Transfer function G(s) = Y(s)/U(s)?
Control System Design-II by Dr. A.Q. Khan
Let us consider a system G with the following SS-form
( ) ( ) ( )
( ) ( ) ( )
: x t Ax t Bu t
G
y t Cx t Du t
= +
= +
&
12
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Laplace Transform of the state equation is
( ) ( )
( ) ( ) ( )1
( ) (0)
(0)
0
sX s x AX s BU s
sI A X s x BU s
X s sI A x BU s−
− = + ⇒
− = + ⇒
= − +
State Space Transfer function
Control System Design-II by Dr. A.Q. Khan
Laplace Transform of the output equation is
Substituting X(s) in the above equation
Note that for T/F, x(0) = 0, then
( ) ( ) ( )Y s CX s DU s= +
( ) ( ) ( )( )1 (0) ( )Y s C sI A x BU s DU s−= − + +
( )( )
1
( )
Y sC sI A B D
U s
−= − +
13
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Note that A,B,C,D are constant matrices
As
characteristics polynomial of G(s)
Correlation between transfer function
and state space model Ctd……
( )1 ( )
( ) adj sI A
G s C sI A B D C B DsI A
− −= − + = +
−
Control System Design-II by Dr. A.Q. Khan
Poles of G(s) are the eigenvalues of A. Hence G(s) can be written as
where Q(s) is a polynomial in s. If A,B,C,D are time varying or uncertain, then it is difficult to derive
equivalent transfer function. For a nonlinear system, transfer function representation is also
difficult to obtain.
Q(s)G(s)
sI A=
−
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Some important facts: Recall that the characteristic equation is obtained by equating
the denominator of the transfer function to ZERO.
In SS-representation, (sI-A) is very important. Thecharacteristic equation in this case is |sI-A| = 0.
Control System Design-II by Dr. A.Q. Khan
- .some properties
If the coefficients of A are real, its eigenvalues are either real orin complex conjugate pairs
The eig(A) = eig(transpose(A))
15
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An Example Example 2: Obtain the transfer function from the state-
space model derived in Example 1.
Solution: Given that
1 1 0
A , B = , C 1 0 , D 0k b 1
= = =
Control System Design-II by Dr. A.Q. Khan
[ ]
[ ]
1
1
22
m m mG (s ) C (s I A ) B D
s 1 0
1 0 0k b 1s
m m m
b0s 1
1 1m = 1 0 1
b k k m s b s k s s s m
m m m
−
−
− −
= − +
− = + +
+
= + + + + −
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State-representation for MIMO system Consider a multiple input multiple output (MIMO) system
with ‘p’ inputs and ‘m’ outputs; i.e.;
11
22
uy
uy u
= =
Control System Design-II by Dr. A.Q. Khan
The transfer function matrix is given aspm
uy
( ) ( )
1 m pY(s)
G s C sI A B DU(s)
− ×
= = − + ∈ ℜ
Note that the Eigenvalues of A does not necessarily imply
the closed loop poles
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Transfer function State SpaceTransfer function State SpaceTransfer function State SpaceTransfer function State Space
Transform T/F into n-th order differential equation
Get the SS- representation then.
Consider an n-th order differential equation
Control System Design-II by Dr. A.Q. Khan
Let1
n n
n y a y a y u
−
+ + + =LLLLLL
1
2 1
3 2
1n
n
x y
x y x
x y x
x y
−
=
= =
= =
=
& &
&& &
M
18
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1 2
2 3
1 1 2 1
1 1
2 2
0 1 0 0 0
0 0 1 0 0
n n n n
x x
x x
x u a x a x a x
x x
x xu
−
=
=
= − − −
= +
&
&M
& L
& L
& L
Control System Design-II by Dr. A.Q. Khan
Let the output of this system be1 1 1n nn na a a x x−
− − − & L
[ ]
1
2
1 1 0 0
n
x
x y x
x
= =
LM
19
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In compact form
where
0 1 0 0 0
x Ax Bu
y Cx Du
= +
= +
&
L
L
Control System Design-II by Dr. A.Q. Khan
[ ]
1 1
,
1
1 0 0 , 0
n n
A B
a a a
C D
−
= =
− − −
= =
M O O M M
L
L
20
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Case: When derivatives are present in
ControlConsider an n-th order differential equation with in n
derivatives in control channel1 1 .
1 0 1 1
Let
n n n n
n n n y a y a y b u b u b u b u
− −
−+ + + = + + + +L L
Control System Design-II by Dr. A.Q. Khan
1 0
2 0 1 1 1
3 0 1 2 2 2
1 1 2
0 1 2 1 1 1
n n n
n n n n n
x y ß u
x y ß u ß u x ß u
x y ß u ß u ß u x ß u
x y ß u ß u ß u ß u x ß u− − −
− − − −
= −= − − = −
= − − − = −
= − − − − − = −
& & &
&& && & &
M
& &L21
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0 1
0 0
1 1 1 0
2 2 1 1 2 0
1 1 1 1 0
w h e r e , , a r e d e t e r m i n e d f r o m
n
n n n n n
ß ß ß
ß b
ß b a ß
ß b a ß a ß
ß b a ß a ß a ß− −
=
= −
= − −
= − − − −
L
M
L L
Control System Design-II by Dr. A.Q. Khan
W i t h t h i s c h o i c e o f s t a t e v a r i a b le s , t h ee x i s t a n c e a n d u n i q u n e s s o f s ta t e e q u a t i o n
i s g u a r a n t e e d . L e t u s c
1 2 1
2 3 2
o n s t r u c t
s t a t e s p a c e e q u a t i o n
x x ß u
x x ß u
= +
= +
&
&
M22
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1 1
1 1 2 1
n n n
n n n n n
x x ß u
x a x a x a x ß u− −
−
= +
= − − + +
M
&
& L
1 1 1
2 2 2
Combining the above set of equations
0 1 0 00 0 1 0
x x x x
u
β β
= +
& L& L
M O O M MM M
Control System Design-II by Dr. A.Q. Khan
[ ]
1 1
1
2
1 01 0 0
n n nn n
n
a a a x x
x
x
y x ß u
x
β −− − −
= = +
& L
L M
Note that
•The derivatives of the control
only affect B matrix
•A and C matrices are unchanged
•Matrix D is non-zero in this case
23
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Decomposition of Transfer functionConsider the following transfer function
( )( )
n n n n n n n n
n
Y s b s b s b s b s b U s s a s a s a
− − −
− −+ + + + +=
+ + + +
1 2 3
0 1 2 3
1 2
1 2
⋯
⋯
Control System Design-II by Dr. A.Q. Khan24
Steps:1. Express the transfer function in –ve power of s. Multiply
numerator and denominator by n s −
( )( )
n n n
n
Y s b b s b s b s b s U s a s a s a s
− − − −
− − −+ + + + +=
+ + + +
1 2 3
0 1 2 3
1 2
1 21
⋯
⋯
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2. Multiply numerator and denominator by a dummy variableX(s)
Decomposition of Transfer function ctd…
( )( )
( )( )
n
n n
n
Y s X s b b s b s b s b s
U s a s a s a s X s
− − − −
− − −
+ + + + +=
+ + + +
1 2 3
0 1 2 3
1 2
1 21
⋯
⋯
Control System Design-II by Dr. A.Q. Khan25
3. Rewrite numerator and denominator as
( ) ( )
( )
( ) ( ) ( )
n
n
n
n
Y s b b s b s b s X s
U s a s a s a s X s
− − −
− − −
= + + + +
= + + + +
1 2
0 1 2
1 2
1 21
⋯
⋯
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4. Construct the state diagram in the above two equations.Re-arrange the last equation as
5. Substituting as( ) ( ) ( ) ( )n n X s U s a s a s a s X s − − −= − + + +1 21 2 ⋯
Decomposition of Transfer function ctd…
Control System Design-II by Dr. A.Q. Khan26
( ) ( )( ) ( )
( )
,
,
n n
n
n n
n
n n n n n
n n n n n
s X s x s X s x x
s X s x x s X s x x
X s x x a x a x a x a x u
y b x b x b x b x b x
− − +
− + −−
− −
− −
→ → =← ← → →= =← ←
→← = − − − − − +
= + + + + +
1
1 2 1
2 1
3 2 1
1 1 2 2 3 1
1 1 2 2 3 1 0
ℓ ℓ
ℓ
ɺ
ɺ ɺ⋯
ɺɺ ⋯
ɺ⋯
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substituting
into n n n n n
n n n n n
x a x a x a x a x u
y b x b x b x b x b x
− −
− −
= − − − − − +
= + + + + +
1 1 2 2 3 1
1 1 2 2 3 1 0
ɺ ⋯
ɺ⋯
Decomposition of Transfer function ctd…
Control System Design-II by Dr. A.Q. Khan27
( )( ) ( )
( )
n n n n
n n n n
n n n n
n
As
y b x b x b x b x
b a x a x a x a x u b b a x b b a x
b b a x b u
− −
− −
− −
= + + + +
+ − − − − − += − + − + +
− +
1 1 2 2 3 1
0 1 1 2 2 3 1
0 1 1 0 1 2
1 0 1 0
⋯
⋯
⋯
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The resultant state space model is
where
x Ax Bu
y Cx Du
= +
= +
ɺ
Decomposition of Transfer function ctd…
Control System Design-II by Dr. A.Q. Khan28 ( ) ( ) ( )
, ,
n n n
n n n n
A B D b
a a a a
C b b a b b a b b a
− −
− −
= = =
− − − −
= − − −
0
1 2 1
0 1 0 1 1 0 1
0 1 0 0 00 0 1 0 0
0 0 0 0 0
1
⋯⋯
⋯
⋮ ⋮ ⋮ ⋯ ⋮ ⋮
⋯
⋯
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Matlab functions ss2tf: Transforms state-space to transfer function
[num,den] = ss2tf(A,B,C,D)
G(s) = num/den = Y(s)/U(s) For multi-input systems, the above function still works; that is, [num,den] = ss2tf(A,B,C,D,iu)
Control System Design-II by Dr. A.Q. Khan
iu: ith input and should be an integer value, that is, 1,2,…. For example [num,den] = ss2tf(A,B,C,D,1) returns transfer
function from first input.
tf2ss: Transforms the transfer function to state-space [A,B,C,D]=tf2ss(num,den) This function is only for SISO
29
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State space model of Inverted PendulumState space model of Inverted PendulumState space model of Inverted PendulumState space model of Inverted Pendulum
Control SystemControl SystemControl SystemControl System
Control System Design-II by Dr. A.Q. Khan30
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State space model of Inverted PendulumState space model of Inverted PendulumState space model of Inverted PendulumState space model of Inverted Pendulum
Control SystemControl SystemControl SystemControl System----Free body diagramFree body diagramFree body diagramFree body diagram
Control System Design-II by Dr. A.Q. Khan31
ɺɺ
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( )
( )
( )
M m x m u
I m m x mg or
M M m g u
Mx u mg
Let
θ
θ θ
θ θ
θ
+ + =
+ + =
= + −
= −
2
ɺɺɺɺ ℓ
ɺɺ ɺɺℓ ℓ ℓ
ɺɺℓ
ɺɺ
ɺ ɺ
Control System Design-II by Dr. A.Q. Khan32
, , , ,
x x
M m x gx u
M M
x x
m x gx u
M M
= = = =
=
+= −
=
= − +
1 2 3 4
1 2
2 1
3 4
4 1
1
1
ɺ
ɺℓ ℓ
ɺ
ɺ
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x x M m g
x x M M
u x x
x x m g
+ −
= + −
1 1
2 2
3 3
4 4
0 1 0 0 0
10 0 0
0 0 0 1 0
10 0 0
ɺ
ɺℓ ℓ
ɺ
ɺ
Control System Design-II by Dr. A.Q. Khan33
x
x y
x
x
=
1
2
3
4
1 0 0 0
0 0 1 0
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Canonical forms 4 canonical forms are important
Controllable canonical form
Observable canonical forms
Diagonal canonical form
Control System Design-II by Dr. A.Q. Khan
Given a nth order linear differential equation
and the transfer function
1 1 .
1 0 1 1
n n n n
n n n y a y a y b u b u b u b u− −
−+ + + = + + + +L L
( )
1
0 1 1
1
1 1
n n
n n
n n
n n
b s b s b s bG s
s a s a s a
−
−−
−
+ + + +=
+ + + +
L
L34
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Controllable Canonical forms The controllable canonical form for the system shown is
{
1 1
2 2
1 1
1 2 1
0 1 0 0 0
0 0 1 0 0
0 0 1 0
1
n n
n n nn n
x x
x x
u x x
a a a a x x
− −
− −
= + − − − −
& L
& L
M M O O O M M M& L L
& L144444 444443
Control System Design-II by Dr. A.Q. Khan
[ ]0 1 1 0 1 1 0
cc
c
B A
n n n n
C
y b a b b a b b a b− −= − − −L L1444444442444444 3 {
1
2
0
1 c D
n
n
x
x
b u
x x
−
+
M44
•Very important in design of state-feedback
controller
•Pole placement35
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Observable Canonical forms The observable canonical form for the system shown is1 1 0
2 2 1 1 01
1 1 2 2 02
0 0 0
1 0 0
0 0
n nn
n nn
n n
x x b a ba
x x b a ba
u
x x b a ba
− −−
− −
−− −−
= +
−−
& L
& L
M M MM O O O M
& L L
&
Control System Design-II by Dr. A.Q. Khan
[ ]
1 1 01
0 0 1
o b o b
o b
n n
A B
C
aa x x
y
−−
=
L
1 4 4 442 4 4 4 43 1 442 4 43
L L1 4 442
1
2
0
1n
n
x
x
b u
x
x
−
+
M4 4 43
•Very important in design of
state observers
•
Pole placement•Note also that
,
,
T T
ob c ob c
T
ob c ob c
A A B C
C B D D
= =
= =
36
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Diagonal canonical form (DCF) Consider a transfer function having distinct poles, that is,
The diagonal canonical form is given as
( )
( ) ( )( ) ( ) ( ) ( ) ( )
1
0 1 1 1 10
1 2 1 1
n n
n n n
n n
Y s b s b s b s b cc cb
U s s p s p s p s p s p s p
−
−+ + + += = + + + ++ + + + + +
LL
L
1 11 0 0 0 1
0 0 0 1
x x p
x x
− −
& L
& L
Control System Design-II by Dr. A.Q. Khan
{
[ ]
1 1
1
2
1 2 0
1
0 0 0 1
0 0 0 1
d d
d
n n
nn n
B A
n
C n
n
u x x
p x x
x
x
y c c c b u
x
x
− −
−
= +
−
= +
M M O O O M M M& L L
& L1 4 4 4 442 4 4 4 4 43
L L M1 4 4 4 2 4 4 4 3
37
J d C i l f
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Jordan Canonical form
Suppose a system transfer function has multiple poles ofmultiplicity m, the proceeding diagonal can be modified. Thismodified representation is known as Jordan Canonical form.
( )( ) ( ) ( ) ( )
1
0 1 1
3
1 2
31 2 40 3 2
n n
n n
n
n
Y s b s b s b s bU s s p s p s p
c cc c cb
−
−+ + + +=+ + +
= + + + + + +
LL
L
Control System Design-II by Dr. A.Q. Khan
1 21 1 ns p s p s ps p s p+ +
[ ]
1
1
1
4
5
6
1 2 2 n 0
p 1 0 0 0 0 0
0 p 1 0 0 0 0
0 0 p 0 0 0 0x x u
0 0 0 p 0 0 1
0 0 0 0 p 0
0 0 0 0 0 p 1
y c c c c x b u
− − −
= + −
−
−
= +
&
M
L L
38
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An Example( )
( )1 2
2 2
1 2
0 1 2 0 1 2
Y s b s bs 3
U s s 3s 2 s a s a
Obtain CCF, OCF, DCF
Here b 0, b 1, b 3; a 1,a 3,a 2
0 1 0CCF : A , B ,C 3 1 , D 0
++= =
+ + + +
= = = = = =
= = = =
Control System Design-II by Dr. A.Q. Khan
[ ]T T T0 c 0 c 0 c 0
0 2 3OCF : A A , B C ,C 0 1 B , D 0
1 3 1
− −
− = = = = = = = −
[ ]D D D D1 0 1
DCF : A , B ,C 2 1 , D 00 2 1
− = = = − = −
39
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Similarity TransformationSimilarity TransformationSimilarity TransformationSimilarity Transformation
State-space representation is not unique Often desirable to work with some especial form of state-
space models
Control System Design-II by Dr. A.Q. Khan40
A transformation exists from one canonical form to anothercanonical form
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Similarity TransformationSimilarity TransformationSimilarity TransformationSimilarity Transformation
x Ax Bu
y Cx Du
= +
= +
ɺConsider
Letx Pz =
Where is non-sin ular matrix.n n ×
Control System Design-II by Dr. A.Q. Khan41
z P x −= 1
The transformed equation is
z P x P Ax P Bu
P APz P Bu
CPz Du η
− − −
− −
= = +
= +
= +
1 1 1
1 1
ɺ ɺ
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Similarity TransformationSimilarity TransformationSimilarity TransformationSimilarity Transformation
z Az Bu
Cz Du η
= +
= +
ɺ
where, ,A P AP B PB C CP D D −= = = =1
Control System Design-II by Dr. A.Q. Khan42
Invariance Properties:• Characteristic equation• Eigenvalues•
Eigenvectors• Transfer function/Transfer function matrix (MIMO Systems)
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Controllability canonical formConsider the dynamic system
to get the CCF, we choose
where
x Ax Bu
y Cx Du
= +
= +
ɺ
P MW =
n −2 1
Control System Design-II by Dr. A.Q. Khan43
n n
n n
a a a
a a
W
a
− −
− −
=
1 2 1
2 3
1
1
1 0
1 0 0
1 0 0 0
⋯
⋯
⋯
⋮ ⋮ ⋯ ⋮ ⋮
⋯
⋯
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and are found as
Then
s a
n n n n n sI A s a s a s a s a
− −−− = + + + + =
1 2
1 2 1 0⋯
0 1 0 0 0
0 0 0 0 0
⋯
⋯
Control System Design-II by Dr. A.Q. Khan44
,
,
n n n
A P AP B PB
a a a a
C CP D D
−
− −
= = = = − − − −
= =
1 2 1
0
0 0 0 1
1
⋮ ⋮ ⋯ ⋮ ⋮
⋯ ⋮
⋯
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Observable canonical formConsider the dynamic system
to get the OCF, we choose
where
x Ax Bu
y Cx Du
= +
= +
ɺ
( )P WN −
= 1
Control System Design-II by Dr. A.Q. Khan45
,
n n
n n
n
C a a a CA a a
N W CA
a
CA
− −
− −
−
= =
1 2 1
2 3
2
1
1
1
1 0
1 0 0
1 0 0 0
⋯
⋯
⋮ ⋮ ⋯ ⋮ ⋮
⋮ ⋯
⋯
s
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and are found as
Then
s a
n n n n n sI A s a s a s a s a
− −−− = + + + + =
1 2
1 2 1 0⋯
n
n
a
a −−
− − = = −
1
1
0 0 0
1 0 0
⋯
⋯
⋯
Control System Design-II by Dr. A.Q. Khan46
[ ]
n
a
C CP
−
−
= =
10 0 0
0 0 0 1
⋮ ⋮ ⋯ ⋮ ⋮
⋯
⋯
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Diagonal canonical form
In DCF, the system matrix A is transformed into Diagonalmatrix,
The entries on the diagonal are the eigenvalues of the A
The transformation matrix is obtained as
Control System Design-II by Dr. A.Q. Khan47
where are the eigenvectors corresponding to theeigenvalues
If the matrix A is in CCF and has distinct eigenvalues, Then Pmatrix has especial structure.
n n
e e e e −
=1 2 1
⋯
i λi e
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Diagonalization of AAAA Let a matrix A has distinct eigenvalues and is
represented as1 2 n, ,λ λ λL
0 1 0 0
0 0 1 0
A
=
L
L
L L L L L
L L L L L
Control System Design-II by Dr. A.Q. Khan
To diagonalize A, let us use some transformation; that is,
With
n n 1 1a a a−
− − − L L x Pz=
1 2 3 n
2 2 2 21 2 3 n
n 1 n 1 n 1 n 1
1 2 3 n
1 1 1 1
P
− − − −
λ λ λ λ
= λ λ λ λ λ λ λ λ
M
M
M
M M M M M
M
48
Vandermondematrix
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Then
Case: If A has multi le ei envalues, then the dia onalization
Diagonalization of AAAA1
2
1
3
n
0 0 0
0 0 0
P AP 0 0 0
0 0 0
−
λ λ = λ λ
L
L
L
L L L L LL
Control System Design-II by Dr. A.Q. Khan
is not straight forward. For instance A has eigenvaluesThen P can be constructed as
1 2λ =λ =λ
-11 1
P with P does not exists.
= λ λ This implies that diagonalization cannot be achieved this way
49
Is there any way possible??
An ExampleAn ExampleAn ExampleAn Example
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An ExampleAn ExampleAn ExampleAn Example
Diagonalise the following system dynamics
----------- (a)
[ ]
0 1 0 0
x 0 0 1 x 0 u
6 11 6 6
y 1 0 0 x
= +
− − − =
&
Control System Design-II by Dr. A.Q. Khan
Step-1: Find eigenvalues of A . In this case the eigenvaluesare
Step-2: Form matrix P1 2 3
1, 2, 3λ = − λ = − λ = −
1 2 3
2 2 2
1 2 3
1 1 1 1 1 1P 1 2 3
1 4 9
= − − − = λ λ λ λ λ λ
50
Apply transformation to (a)x Pz=
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pp y ( )
[ ]
1 1
0 1 0 0 0 1 0 0
Pz 0 0 1 Pz 0 u z P 0 0 1 Pz P 0 u
6 11 6 6 6 11 6 6
y 1 0 0 Pz
1 0 0 3
− − = + ⇒ = + − − − − − −
=
−
& &
Control System Design-II by Dr. A.Q. Khan
[ ]z 0 2 0 z 6 u, y 1 1 1 x
0 0 3 3
⇒ = − + − = −
&
• Transformation matrix P modifies the coefficient matrix of z into
diagonal form• All the states are decoupled from one another• The eigenvalues remain unchanged
51
Fact: The eigenvalues are unchanged during similarity
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transformation.
Proof:
( )
1 1 1
1
1
Consider
I P AP P P P AP
= P I A P
= P I A P
− − −
−
−
λ − = λ −
⇒ λ −
⇒ λ −
Control System Design-II by Dr. A.Q. Khan
( )
( )
( )
1
1
= P P I A
= P P I A
= I A
−
−
⇒ λ −
⇒ λ −
⇒ λ −Hence proved that the eigenvalues are invariant under linear transformation
52
Example
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Example Diagonalize the following
Step 1: The eigenvalues are -1,-3,-3[ ]
0 1 0 0
x 0 0 1 x 0 u
9 15 7 3y 1 0 0 x
= +
− − − =
&
Control System Design-II by Dr. A.Q. Khan53
Step 2: Form P matrix1
1 2 3
2 2 2
1 2 3
1 1 1 1 1 1
P 1 3 3 P does not exist.
1 9 9
−
= λ λ λ = − − − ⇒ λ λ λ
Diagonalization ?
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Generalized Eigenvalues
A has multiple eigenvalues The formula for computing eigenvectors
does not hold to com ute all ei envectors( )i i I A e λ − = 0
Control System Design-II by Dr. A.Q. Khan54
Let a q be distinct eigenvalues among n eigenvalues of A. Theequation hold for computing qeigenvectors.
( )i i I A e λ − = 0
Let be the eigenvalues of mth order i e; Thejλ m n q≤ −
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Let be the eigenvalues of mth order. i.e; . The
corresponding eigenvectors are called generalized eigenvectorsand can be determined using the following formula
j λ m n q ≤
( )
( )
j n q
j n q n q
I A e
I A e e
λ
λ
− +
− + − +
− =
− =−
− =−
1
2 1
0
Control System Design-II by Dr. A.Q. Khan55
( )
n q n q
j n q m n q m I A e e λ
− + − +
− + − + −− =−
2
1
⋮
An Example
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An Example
2 3
1
Diagnolize
; ; ; ;
:
Eigenvector associated to is
A
Solution
λ λ λ
λ
− = = = =
=
−
1
0 6 5
1 0 2 2 1 1
3 2 4
2
Control System Design-II by Dr. A.Q. Khan56
( ) ( )1
I A e I A e e e
e e e
λ − = − = − − = ⇒ − − −
1 21 21 21
31 31 31
2 1 2 2 0
3 2 4
( ) ( )2
e e e
I A e I A e e e
e e e
λ
= − − − − = − = − − = ⇒ = −
− − − −
11 12 12
2 21 22 22
31 32 32
1
2
11 6 5
31 1 1 2 0
73 2 35
7
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( ) ( )2
;
e e
I A e I A e e e e e
e
e e
λ
−
− = − = − − = − − − −
⇒ = − −
11 12
3 21 22 2
31 32
13
23
33
1 6 5
1 1 1 2
3 2 3
1
22
49
46
49
Control System Design-II by Dr. A.Q. Khan57
;P
A P AP −
= − − − − − −
= 1
2 1 1
3 221
7 49
5 462
7 49
=
2 0 0
0 1 1
0 0 1
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Transfer function State space-Decomposition of transfer functionsDecomposition of transfer functionsDecomposition of transfer functionsDecomposition of transfer functions Direct decomposition :
Applied to a transfer function which is not in factored form. It canbe conducted in two ways1. Direct decomposition to CCF
Control System Design-II by Dr. A.Q. Khan58
2. rect ecompos t on to
Cascade decompositionApplied to transfer functions that are written as product of simplefirst order or second order components
Parallel decompositionApplied to transfer functions whose denominator is in factoredform and partial fraction expansion can be obtained
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Direct decomposition to CCFConsider the following transfer function
( )( )
n n n
n n n n
n
Y s b s b s b s b U s s a s a s a
− − −
− −+ + + += + + + +
1 2 3
1 2 3
1 2
1 2
⋯⋯
Control System Design-II by Dr. A.Q. Khan59
Steps:
1. Express the transfer function in –ve power of s. Multiplynumerator and denominator by n s −
( )( )
n
n n
n
Y s b s b s b s b s U s a s a s a s
− − − −
− − −+ + + += + + + +
1 2 3
1 2 3
1 2
1 21
⋯⋯
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2. Multiply numerator and denominator by a dummy variableX(s)
Direct decomposition to CCF ctd…
( )( )
( )( )
n n
n
n
Y s X s b s b s b s b s
U s a s a s a s X s
− − − −
− − −
+ + + +=
+ + + +
1 2 31 2 3
1 2
1 21
⋯
⋯
Control System Design-II by Dr. A.Q. Khan60
3. Rewrite numerator and denominator as
( ) ( ) ( )
( ) ( ) ( )
n
n
n
n
Y s b s b s b s b s X s
U s a s a s a s X s
− − − −
− − −
= + + + +
= + + + +
1 2 3
1 2 3
1 2
1 21
⋯
⋯
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Direct decomposition to CCF ctd…4. Construct the state diagram in the above two equations.
Re-arrange the last equation as
5. Substituting as( ) ( ) ( ) ( )n n X s U s a s a s a s X s − − −= − + + +1 21 2 ⋯
Control System Design-II by Dr. A.Q. Khan61
( ) ( )( ) ( )
( )
,,
n n
n
n n
n
n n n n n
n n n n
s X s x s X s x x s X s x x s X s x x
X s x x a x a x a x a x u
y b x b x b x b x
− − +
− + −−
− −
− −
→ → =← ←
→ →= =← ←
→←
= − − − − − +
= + + + +
1
1 2 1
2 1
3 2 1
1 1 2 2 3 1
1 1 2 2 3 1
ℓ ℓ
ℓ
ɺ
ɺ ɺ⋯
ɺ
ɺ ⋯
⋯
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The resultant state space model is
where
Direct decomposition to CCF ctd…
x Ax Bu
y Cx Du
= +
= +
ɺ
Control System Design-II by Dr. A.Q. Khan62 [ ]
,
,
n n n
n n n
A B
a a a a
C b b b b D
− −
− −
= = − − − −
= =
1 2 1
1 2 1
0 1 0 0 0
0 0 1 0 0
0 0 0 0 0
1
0
⋯
⋯
⋯
⋮ ⋮ ⋮ ⋯ ⋮ ⋮
⋯
⋯
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Direct decomposition to OCFConsider the following transfer function
( )( )
n n n
n n n n
n
Y s b s b s b s b U s s a s a s a
− − −
− −+ + + += + + + +
1 2 3
1 2 31 2
1 2
⋯
⋯
Control System Design-II by Dr. A.Q. Khan63
Steps:
1. Express the transfer function in –ve power of s. Multiplynumerator and denominator by n s −
( )( )
n
n n
n
Y s b s b s b s b s U s a s a s a s
− − − −
− − −+ + + += + + + +
1 2 3
1 2 3
1 2
1 21
⋯⋯
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2. Expand the equation asDirect decomposition to OCF ctd…
( ) ( )
( ) ( )
n
n
n
n
a s a s a s Y s
b s b s b s b s U s
− − −
− − − −
+ + + +
= + + + +
1 2
1 2
1 2 3
1 2 3
1 ⋯
⋯
Control System Design-II by Dr. A.Q. Khan64
or
( ) ( ) ( )
( ) ( )
n
n
n
n
Y s a s a s a s Y s
b s b s b s b s U s
− − −
− − − −
= − + + +
+ + + + +
1 2
1 2
1 2 3
1 2 3
⋯
⋯
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The output of the integrators are assigned as state variable
Finally we get
Direct decomposition to OCF ctd…
x A x B u
y C x D u
= +
= +
ɺ
Control System Design-II by Dr. A.Q. Khan65
where
[ ]
,
,
n n
n n
n n
a b
a b
A B a b
a b
C D
− −
− −
= = −
= =
1 1
2 3
1 1
0 0 0
1 0 0
0 1 0
0 0 1
0 0 0 1 0
⋯
⋯
⋯
⋮ ⋮ ⋮ ⋯ ⋮ ⋮
⋯
⋯
Cascade Decomposition
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p
( )( )
1 2
1 2
1 2 1 2
where, , , and are real constant. Each of the first
+ += + +
Y s s b s bK U s s a s a
a a b b
Control System Design-II by Dr. A.Q. Khan66
1 1 2
2
decomposition.
The state space model is
b= − −
&&
x a a x
[ ]
2 1
2 2
1 2 2 2
+0
b b
−
= − − +
x K ua x K
y a a x Ku
Parallel Decomposition
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p
( )
( )
( )
( ) ( )
( )
1 2
1 2
w h e r e
i s a p o l y n o m i a l o f o r d e r l e s s t h a n 2
a n d , a r e r e a l c o n s t a n t .
=+ +
Y s Q s
U s s a s a
Q s
a a
Control System Design-II by Dr. A.Q. Khan67
( )( ) ( ) ( )
1 2
1 2
1
2
T h e s t a t e s p a c e m o d e l i s g i v e n a s
0 10 1
= ++ +
−
= + − &
Y s K K U s s a s a
a x xa
[ ]1 2
=
u
y K K x
Relationship b/w various methodRelationship b/w various methodRelationship b/w various methodRelationship b/w various method
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Control System Design-II by Dr. A.Q. Khan68
Solving TimeSolving TimeSolving TimeSolving Time----Invariant state equationInvariant state equationInvariant state equationInvariant state equation---- A reviewA reviewA reviewA review
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Solving TimeSolving TimeSolving TimeSolving Time Invariant state equationInvariant state equationInvariant state equationInvariant state equation A reviewA reviewA reviewA review
( )
( ) 2 k 0 1 2 k
Consider
x Ax Bu, (1) x t ???
Let us consider a homogenous state equation
x ax (2)
Let
x t b b t b t b t (3)
= + =
=
= + + + + +
&
&
L L
Control System Design-II by Dr. A.Q. Khan69
1 2 k
k 1 2 k
1 2 k 0 1 2 k
1 0
2 2
2 1 0 2 0
3
3
x t t t
subsitituting (3) and (4) into (2)
b 2b t kb t a (b b t b t b t ) (5)
we have
b ab
12b ab a b b a b , Similarly
2
1b a
6
−
−
= + + +
+ + + = + + + + +
=
= = ⇒ =
=
& L
L L L
3 k
0 0 k 0
1 1b a b , b a b
3 k = =LL
( ) 2 0
As
1x t 1 at at b
= + + + L
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( )
( )
( ) ( ) ( )
0
0
2 2 at
x t 1 at at b2
At t 0, equation (3) im plies
b x 0 , therefore
1x t 1 at a t x 0 e x 0
2Now consider
x Ax (6)
+ + +
=
=
= + + + =
=
L
&
Control System Design-II by Dr. A.Q. Khan70
( ) 20 1 2
Analogy with scalar, we have
x t b b t b t= + + +
( )
( ) ( ) ( ) ( )
k k
k 1
1 2 k
2 2 At
b t (7)
x t b 2b t kb t (8)
proceeding in similar fashion, we
1x t 1 At A t x 0 e x 0 = (t)x 0 (9)2
where (t ) is state transition
−
+ +
= + + +
= + + + = φ
φ
L L
& L
L
matrix.
Referring to equation (9), the term is of particulari t t
Ate
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interest.
It is referred to as matrix exponential and analogously, it can berepresented as
This matrix exponential for an converges absolutely forall finite time.
k k At
k 0
A te
k!
∞
=
= ∑
n n×
Control System Design-II by Dr. A.Q. Khan71
( )
( )
At At
A B t At Bt
A B t At Bt
At At
d e Aedx
e e e if AB BA
e e e if AB BA e e = I
+
+
−
• =
• = =
• ≠ ≠•
C id i (6) d ki L l T f
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( ) ( ) ( )
( ) ( ) ( ) ( ) ( )
( ) ( ) ( )
1
11 At
C onsider equation (6) and taking Laplace T ransform
sX (s) - x (0) AX s X s sI A x (0)
x t sI A x 0 x t e x 0
orx t t x 0
where
−
−−
= ⇒ = −
= − ⇒ =
= φ
L
Control System Design-II by Dr. A.Q. Khan72
( )t : n n m atrix and is the unique solution
: State transition m atr
φ ×
( )
( ) ( )-1 -At
ix. It has all the inform ation abou t
the free motion o f the system define by x Ax
0 I
Also note t =e = t
=
φ =
φ φ −
&
Case-I: If be distinct eigenvalues of A, thenwill contain exponentials
1 2 n, ,λ λ λL ( )tφ
1 2 nt t te ,e , eλ λ λL
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will contain exponentials
Case-II: If the matrix A is diagonal with , then
e ,e , e
1 2 n, ,λ λ λL
( )
1
2
n
t
t
t
e 0 0
0 e 0t
0 0 e
λ
λ
λ
φ =
L
L
M O O M
L
Control System Design-II by Dr. A.Q. Khan73
Case-III: if there is multiplicity of eigenvalues, then will contain terms like in addition to1 1 1 2 n, , , ,λ λ λ λ λL
( )tφ 1 1t t2te , t eλ λ
1 2 nt t te ,e , eλ λ λL
Properties of ( )tφ
( ) AtAs t eφ =
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( )
( )
( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( )
( ) ( )
1 2 1 2
A(0)
1 1At At 1
A t t A t A t1 2 1 2
n
As t e
0 e I
t e e t or t t
t t e e e t t
t nt
− −− −
+
φ =
• φ = =
• φ = = = φ − φ = φ −
• φ + = = = φ φ
• φ = φ
Control System Design-II by Dr. A.Q. Khan74
An Example
( ) ( ) ( )2 1 1 0 2 0 t t t t t t• φ − φ − = φ −
0 1For x Ax, where A=
2 3
= − −
&
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( ) ( )( ) ( ){ }
( )
-1
1At
1
Obtain t and t
Solution : As e t sI A
s 1Note that sI A
2 s 3
s 3 11sI A
−
−
φ φ
= φ = −
− − =
+ +
− = −
-1L
Control System Design-II by Dr. A.Q. Khan75
( )( ) ( )
Using partial fraction expansion
1t
s 1 s 2φ =
+ +
-1L
( )-t 2 t -t 2 t t 2 t t 2 t
-t 2 t -2t t t 2 t 2t t
2 1 1 1
s 3 1 s 1 s 2 s 1 s 2
2 s 2 2 2 1
s 1 s 2 s 2 s 1
2e e e e 2e e e e, t
2e 2e 2e e 2e 2e 2e e
− −
− −
− − + + + + +
= − − − + + + +
− − − −= ⇒ φ − =
− − − −
-1L
Solution to Homogenous State equationsConsider
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n 1
n 1
n n
n p
x
ux Ax Bu (1)
A
B
Now
x Ax Bu
×
×
×
×
∈ ℜ∈ ℜ
= + − − − − − − − − − ∈ ℜ
∈ ℜ
− =
&
&
Convolution integral
Control System Design-II by Dr. A.Q. Khan76
( )
( ) ( ) ( )
At At At
t t
A At A
0 0
e x Ax e Bu multiply by e both side
Integrating both sides
de x Ax d e Bu d e x
d
− − −
− τ − − τ
− =
− τ = τ τ ⇒ ττ∫ ∫
&
& ( )
( ) ( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( )
t t
A
0 0
t t
A tAt A At
0 0
t
0
d e Bu d
e x t x 0 e Bu d x t e x 0 e Bu d
x t t x 0 t Bu d (2)
− τ
− τ− − τ
τ = τ τ
⇒ − = τ τ ⇒ = + τ τ
⇒ = φ + φ − τ τ τ − − − − − − − −
∫ ∫
∫ ∫
∫
Laplace Transform ApproachLaplace Transform ApproachLaplace Transform ApproachLaplace Transform Approach
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( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )1
Consider
x Ax Bu
The Laplace transform of the above system is
sX s x 0 AX s BU s sI A X s x 0 BU s
X s sI A x 0 BU s−
= +
− = + ⇒ − = + ⇒
= − +
&
Control System Design-II by Dr. A.Q. Khan77
( ) ( ) ( ) ( )
( ) ( )
( ) ( )
( )i
i
t
A tAt
0
0 i
tA t t A t
t
x t e x 0 e Bu d
For initial time t t 0, then
x t e x 0 e Bu d
− τ
− −τ
⇒ = + τ τ
= ≠
= + τ τ
∫
∫
An ExampleAn ExampleAn ExampleAn ExampleConsider
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( ) ( )
( )
x Ax Bu,
where
0 1 0 0 0 for t
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Laplace Transform method:
Method of Diagonalization of A:
( ) ( ){ }1At
t e sI A −
φ = = −-1LDiscussed in
Previous slides
1if D P A P then−=
Control System Design-II by Dr. A.Q. Khan79
Caley-Hamilton Theorem and minimal polynomial theorem
Minimal polynomials of A involves distinct roots Minimal polynomials of A involves repeated roots
A t 1 D t
e P e P
−
=