Transcript Feedback Control Systems (FCS)
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 t 0 as the amount of information that must be provided at time t 0 , which, together with the input signal u(t) for t uniquely determine the output of the system for all t t 0 .
t 0 , • This representation transforms an n th equation into a set of n 1 st order difference order difference equations.
• State Space representation is not unique.
• Provides complete information about all the internal signals of a system.
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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.
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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 such a vector is called state vector.
components of a vector x , • State Space : The state space is defined as the n dimensional space in which the components of the state vector represents its coordinate axes.
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Definitions
• Let
x 1
and
x 2
are two states variables that define the state of the system completely .
x
2 State (t=t 1 ) State Vector
x
1 Velocity
dx dt
State (t=t 1 )
x
Position Two Dimensional State space 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.
i v L c
Step-2: Apply network theory, such as Kirchhoff's voltage and current laws, to obtain i
c
and v
L
in terms of the state variables, v
c
and i
L
.
Applying KCL at Node-1
i L
i R
i C i C
i R
i L C dv C dt
v C R
i L
(1)
State Space Representation
Step-2: Apply network theory, such as Kirchhoff's voltage and current laws, to obtain i
c
and v
L
in terms of the state variables, v
c
and i
L
.
Applying KVL at input loop
v
(
t
)
L di L dt
v R L di L dt
v C
v
(
t
) (2) Step-3: Write equation (1) & (2) in standard form.
dv C dt
1
RC v C
1
C i L di L dt
1
L v C
1
L v
(
t
) State Equations
State Space Representation
dv C dt
1
RC v C
1
C i L d dt
i v c L
1
RC
1
L
i v
c
L
1
RC
1
L di L dt
1
L v C
1
L v
(
t
) 1
C
0
i v c L
0 1
L
v
(
t
) 1
C
0
i v c L
0 1
L
v
(
t
)
State Space Representation
Step-4: The output is current through the resistor therefore, the output equation is
i R
1
R v C i R
1
R
0
i v c L
State Space Representation
i v
c
L
(
t
) 1
RC
1
L Ax
(
t
) 1
C
0
i v c L
Bu
(
t
) 0 1
L
v
(
t
) Where,
x(t)
--------------- State Vector
A (nxn)
---------------- System Matrix
B (nxp) u(t)
----------------- Input Matrix --------------- Input Vector
i R
1
R
0
i v c L
y
(
t
)
Cx
(
t
)
Du
(
t
) Where,
y(t)
-------------- Output Vector
C (qxn) D
---------------- Output Matrix ----------------- 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
i L
– – Input Matrix & Input Vector Output Matrix & Output Vector
V c + V o +
•
C dv dt c
u
(
t
Choosing v
c
) and i
L i L L di L
dt
as state variables
Ri L
v c dv dt c
1
C i L
1
C u
(
t
)
di dt L
1
L v c V
R i L L o
Ri L
Example-1 (cont...)
dv c dt
1
C i L
1
C u
(
t
)
di L dt
1
L v c
R L i L
i v
c
L
0 1
L
1
C R L
i v c L
1
C
0
u
(
t
) State Equation
V o
Ri L V o
0
R
v c i L
Output Equation
Example-2
•
Consider the following system K M
x(t) f(t)
B
Differential equation of the system is:
M d
2
x
(
t
)
dt
2
B dx
(
t
)
dt
Kx
(
t
)
f
(
t
) 18
• •
Example-2
As we know
dx dt
v d
2
x dt
2
dv dt
Choosing
x
and
v
as state variables
dx dt
v M d
2
x
(
t
)
dt
2
B dx
(
t
)
dt
Kx
(
t
)
f
(
t
)
dv dt
B M v
K M x
1
M f
(
t
)
x v
0
K M
1
B M
x v
0 1
M
f
(
t
)
x
v
0
K M
Example-2
1
B M
x v
0 1
M
f
(
t
) • If velocity v is the out of the system then output equation is given as
y
(
t
) 0 1
x v
•
Example-3
Find the state equations of following mechanical translational system.
• System equations are:
M
1
d
2
dt x
2 1
D dx
1
dt
Kx
1
Kx
2 0
f
(
t
)
M
2
d
2
dt x
2 2
Kx
2
Kx
1
Example-3
• • Now
dx dt
1
v
1
d
2
dt x
2 1
dv
1
dt dx dt
2
v
2 Choosing x 1 , v 1 , x 2 , v 2 as state variables
d
2
dt x
2 2
dv
2
dt dx
1
dt
v
1
M
1
dv
1
dt
Dv
1
Kx
1
Kx
2
dx
2
v
2
dt f
(
t
)
M
2
dv
2
dt
Kx
2
Kx
1 0
Example-3
• In Standard form
dx
1
dt
v
1
dv
1
dt
D M
1
v
1
K M
1
x
1
K M
1
x
2
dx
2
dt
v
2
dv
2
dt
K M
2
x
2
K M
2
x
1 1
M
2
f
(
t
)
Example-3
•
dx dt
1
v
1
dv
1
dt
D M
1
v
1
K M
1
x
1
K M
1
x
2
dv dt
2
K M
2 In Vector-Matrix form
x
2
K M
2
x
1 1
M
2
f
(
t
)
dx
2
dt
v
2
v
v
2 1 1 2 0
M
0
K M K
2 1 1
D M
1 0 0 0
K
M
0 1
K M
2 0 0 1 0
x v x v
1 1 2 2 0 0 0 1
M
2
f
(
t
)
Example-3
x v
v
2 1 1 2 0
M
0
K M K
2 1 1
D M
1 0 0 0
K
M
0 1
K M
2 0 0 1 0
x v x v
1 1 2 2 0 0 0 1
M
2
f
(
t
) • If
x
1 and
v
2 are the outputs of the system then
y
(
t
) 1 0 0 0 0 0 0 1
x v x v
2 1 1 2
Eigenvalues & Eigen Vectors
• The eigenvalues of an characteristic equation.
nxn
matrix A are the roots of the • Consider, for example, the following matrix A:
Eigen Values & Eigen Vectors
Example#4
•
Find the eigenvalues if
– K = 2 – M=10 – B=3
x v
0
K M
1
B M
x v
0 1
M
f
(
t
)
Frequency Domain to time Domain Conversion
•
Transfer Function to State Space K M
x(t) f(t)
B
Differential equation of the system is:
M d
2
x
(
t
)
B dx
(
t
)
dt
2
dt
Kx
(
t
)
f
(
t
) 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
)
F
(
s
)
Ms
2 1
Bs
K
State Space Representation:
X
(
s
)
F
(
s
)
s
2
B M
1
M s
K M X
(
s
)
F
(
s
)
s
2
B M
1
M s
K M
s
2
P
(
s
)
s
2
P
(
s
)
X
(
s
)
F
(
s
)
P
(
s
)
B M s
1
M
1
s P
2 (
s P
) (
s
)
K M s
2
P
(
s
) 30
X
(
s
) 1
M F
(
s
)
P
(
s
)
B M s
2
P
(
s
)
s
1
P
(
s
)
K M
……………………………. (1)
s
2
P
(
s
) ……………………………. (2) From equation (2)
P
(
s
)
F
(
s
)
B M s
1
P
(
s
)
K M s
2
P
(
s
) ……………………………. (3) Draw a simulation diagram of equation (1) and (3) F(s) P(s) 1/s
-K/M -B/M
1/s
1/M
X(s)
• • Let us assume the two state variables are
x
1 and
x
2 .
These state variables are represented in phase variable form as given below.
-K/M
F(s) P(s) 2 1/s
x
2
-B/M
x
1 1/s
x
1/M
1 X(s) • • State equations can be obtained from state diagram.
x
1
x
2 2
F
(
s
)
K M
The output equation of the system is
x
1
B M x
2
x
(
t
) 1
M x
1
x
1
x
2 2
F
(
s
)
K M x
1
B M x
2
x
x
2 1 0
K M
1
B M
x x
2 1 0 1
f
(
t
)
x
(
t
) 1
M x
1
x
(
t
) 1
M
0
x
1
x
2
Example#5
• Obtain the state space representation of the following Transfer function.
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