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Feedback Control Systems (FCS)
Lecture-34-35
Modern Control Theory
Dr. Imtiaz Hussain
email: [email protected]
URL :http://imtiazhussainkalwar.weebly.com/
Introduction
• The transition from simple approximate models, which are
easy to work with, to more realistic models produces two
effects.
– First, a large number of variables must be included in the models.
– Second, a more realistic model is more likely to contain nonlinearities
and time-varying parameters.
– Previously ignored aspects of the system, such as interactions with
feedback through the environment, are more likely to be included.
Introduction
• Most classical control techniques were developed for linear constant
coefficient systems with one input and one output(perhaps a few
inputs and outputs).
• The language of classical techniques is the Laplace or Z-transform
and transfer functions.
• When nonlinearities and time variations are present, the very basis
for these classical techniques is removed.
• Some successful techniques such as phase-plane methods,
describing functions, and other methods, have been developed to
alleviate this shortcoming.
Introduction
• The state variable approach of modern control theory provides a
uniform and powerful methods of representing systems of arbitrary
order, linear or nonlinear, with time-varying or constant coefficients.
• It provides an ideal formulation for computer implementation and is
responsible for much of the progress in optimization theory.
• The advantages of using matrices when dealing with simultaneous
equations of various kinds have long been appreciated in applied
mathematics.
• The field of linear algebra also contributes heavily to modern control
theory.
Introduction
• Conventional control theory is based on the input–output
relationship, or transfer function approach.
• Modern control theory is based on the description of system
equations in terms of n first-order differential equations, which may
be combined into a first-order vector-matrix differential equation.
• The use of vector-matrix notation greatly simplifies the a
mathematical representation of systems of equations.
• The increase in the number of state variables, the number of inputs,
or the number of outputs does not increase the complexity of the
equations.
State Space Representation
• State of a system: We define the state of a system at time
t0 as the amount of information that must be provided at
time t0, which, together with the input signal u(t) for t  t0,
uniquely determine the output of the system for all t  t0.
• This representation transforms an nth order difference
equation into a set of n 1st order difference equations.
• State Space representation is not unique.
• Provides complete information about all the internal
signals of a system.
6
State Space Representation
• Suitable for both linear and non-linear systems.
• Software/hardware implementation is easy.
• A time domain approach.
• Suitable for systems with non-zero initial conditions.
• Transformation From Time domain to Frequency
domain and Vice Versa is possible.
7
Definitions
• State Variable: The state variables of a dynamic system are
the smallest set of variables that determine the state of the
dynamic system.
• State Vector: If n variables are needed to completely
describe the behaviour of the dynamic system then n
variables can be considered as n components of a vector x,
such a vector is called state vector.
• State Space: The state space is defined as the ndimensional space in which the components of the state
vector represents its coordinate axes.
8
Definitions
• Let x1 and x2 are two states variables that define the state
of the system completely .
dx
dt
x2
State (t=t1)
Velocity
State (t=t1)
State
Vector
x1
Two Dimensional State space
x
Position
State space of a Vehicle
9
State Space Representation
• An electrical network is given in following figure,
find a state-space representation if the output is
the current through the resistor.
State Space Representation
• Step-1: Select the state variables.
v c 
 
iL 
Step-2: Apply network theory, such as Kirchhoff's voltage and current
laws, to obtain ic and vL in terms of the state variables, vc and iL.
iL  iR  iC
Applying KCL at Node-1
iC  iR  iL
C
dvC
v
  C  iL
dt
R
(1)
State Space Representation
Step-2: Apply network theory, such as Kirchhoff's voltage and current
laws, to obtain ic and vL in terms of the state variables, vc and iL.
Applying KVL at input loop
diL
v(t )  L
 vR
dt
diL
L
 vC  v(t )
dt
(2)
Step-3: Write equation (1) & (2) in standard form.
dvC
1
1

vC  i L
dt
RC
C
diL
1
1
  vC  v(t )
dt
L
L
State Equations
State Space Representation
dvC
1
1

vC  i L
dt
RC
C
 1
d vc   RC
  1
dt iL  
 L
 1
vc   RC
    1
 i L  
 L
diL
1
1
  vC  v(t )
dt
L
L
1
0

v


C c   1 v(t )
 
0  iL   L 

1
0

v


C c   1 v(t )
 
0  iL   L 

State Space Representation
Step-4: The output is current through the resistor therefore, the
output equation is
1
iR  vC
R
1
iR  
R
 vc 
0  
 iL 
State Space Representation
 1
vc   RC
    1
 i L  
 L
1
0

v


C c   1 v(t )
 
0  iL   L 

Where,
x(t )  Ax(t )  Bu(t )
1
iR  
R
 vc 
0  
 iL 
y(t )  Cx(t )  Du (t )
x(t) --------------- State Vector
A (nxn) ---------------- System Matrix
B (nxp) ----------------- Input Matrix
u(t) --------------- Input Vector
Where,
y(t) -------------- Output Vector
C (qxn) ---------------- Output Matrix
D ----------------- Feed forward Matrix
Example-1
• Consider RLC Circuit Represent the system in Sate Space and find
(if L=1H, R=3Ω and C=0.5 F):
–
–
–
–
State Vector
System Matrix
Input Matrix & Input Vector
Output Matrix & Output Vector
dvc
C
 u ( t )  iL
dt
iL
Vc
+
+
Vo
-
diL
L
  Ri L  vc
dt
Vo  Ri L
• Choosing vc and iL as state variables
dvc
1
1
  iL  u(t )
dt
C
C
diL 1
R
 vc  iL
dt
L
L
-
Example-1 (cont...)
diL 1
R
 vc  iL
dt
L
L
dvc
1
1
  iL  u(t )
dt
C
C

vc   0
 i    1
 L 
L
1
  v   1 
c
C
  C u(t )



R  iL   
 
0
L
State Equation
Vo  Ri L
Vo  0
vc 
R  
iL 
Output Equation
Example-2
• Consider the following system
K
x(t)
M
f(t)
B
Differential equation of the system is:
M
d 2 x(t )
dt
2
dx(t )
B
 Kx(t )  f (t )
dt
18
Example-2
• As we know
d 2x
dx
v
dt
dt 2
dv

dt
• Choosing x and v as state variables
M
dx
v
dt
d 2 x(t )
dt 2
dx(t )
B
 Kx(t )  f (t )
dt
dv
B
K
1
 v
x
f (t )
dt
M
M
M
 x   0
     K
 v   M
1   x  0 
B      1  f (t )
  v   
M
M 
Example-2
 x   0
     K
 v   M
1   x  0 
B      1  f (t )
  v   
M
M 
• If velocity v is the out of the system then output equation is given as
y (t )  0
 x
1 
v 
Example-3
• Find the state equations of following mechanical translational
system.
• System equations are:
M1
d 2 x1
dt
2
dx1
D
 Kx1  Kx2  0
dt
f (t )  M 2
d 2 x2
dt
2
 Kx2  Kx1
Example-3
• Now
dx1
 v1
dt
d 2 x1
dt 2
dv1

dt
2
dx2
d
x2 dv2
 v2

2
dt
dt
dt
• Choosing x1, v1, x2, v2 as state variables
dx1
 v1
dt
dv1
M1
 Dv1  Kx1  Kx2  0
dt
dx2
 v2
dt
dv2
f (t )  M 2
 Kx2  Kx1
dt
Example-3
• In Standard form
dx1
 v1
dt
dv1
D
K
K

v1 
x1 
x2
dt
M1
M1
M1
dx2
 v2
dt
dv2
K
K
1

x2 
x1 
f (t )
dt
M2
M2
M2
Example-3
dx1
 v1
dt
dv1
D
K
K

v1 
x1 
x2
dt
M1
M1
M1
dx2
 v2
dt
dv2
K
K
1

x2 
x1 
f (t )
dt
M2
M2
M2
• In Vector-Matrix form
 0
 x1   K
   
 v1    M 1
 x 2   0
   K
 v2   M
2

1
D

M1
0
0
0
K
M1
0
K

M2
0
 0 
x


 1
0    0 
v1  



  0  f (t )
1  x2 
 1 



0  v 2  

M2 
Example-3
 0
 x1   K
   
 v1    M 1
 x 2   0
   K
 v2   M
2

1
D

M1
0
0
K
M1
0
K

M2
0
0
 0 
x


 1
0    0 
v1  



  0  f (t )
1  x2 
 1 



0  v 2  

M2 
• If x1 and v2 are the outputs of the system then
1
y (t )  
0
0
0
0
0
 x1 

v
0 
 1
1
  x2 
 
 v2 
Eigenvalues & Eigen Vectors
• The eigenvalues of an nxn matrix A are the roots of the
characteristic equation.
• Consider, for example, the following matrix A:
Eigen Values & Eigen Vectors
Example#4
• Find the eigenvalues if
–K=2
– M=10
– B=3
 x   0
     K
 v   M
1   x  0 
B      1  f (t )
  v   
M
M 
Frequency Domain to time Domain
Conversion
• Transfer Function to State Space
K
f(t)
x(t)
M
B
Differential equation of the system is:
d 2 x(t )
dx(t )
M

B
 Kx(t )  f (t )
2
dt
dt
Taking the Laplace Transform of both sides and ignoring
Initial conditions we get
Ms 2 X ( s )  BSX ( s )  KX ( s )  F ( s )
The transfer function of the system is
X ( s)
1

F ( s ) Ms 2  Bs  K
State Space Representation:
X (s)
 2
F (s) s 
X (s)
 2
F (s) s 
X (s)

F ( s) P( s) 
B
M
1
M
B
M
1
M
B
M
s
K
M
s
K
M
s 2 P ( s )
 2
s P( s)
2
1
M
1
s P( s)
s P ( s )  MK s  2 P ( s )
30
1 2
X (s) 
s P( s)
M
……………………………. (1)
B 1
K 2
F ( s)  P( s) 
s P( s) 
s P( s ) ……………………………. (2)
M
M
From equation (2)
B 1
K 2
P( s)  F ( s)  s P( s)  s P( s)
M
M
……………………………. (3)
Draw a simulation diagram of equation (1) and (3)
-K/M
-B/M
F(s)
P(s)
1/s
1/s
1/M
X(s)
•
•
Let us assume the two state variables are x1 and x2.
These state variables are represented in phase variable form as
given below.
-K/M
x 2
F(s)
P(s)
•
-B/M
1/s
x2  x1
1/s
1/M
x1
X(s)
State equations can be obtained from state diagram.
x1  x2
•
K
B
x 2  F ( s ) 
x1 
x2
M
M
The output equation of the system is
1
x(t )  x1
M
x1  x2
 x1   0
 x    K
 2   M
K
B
x 2  F ( s ) 
x1 
x2
M
M
1   x  0 
1
B       f (t )
   x2  1
M
1
x(t )  x1
M
 1
x (t )  
M
  x1 
0  
  x2 
Example#5
• Obtain the state space representation of the following
Transfer function.
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END OF LECTURES-34-35