Unit-6.pptx of control system and engineering

Unit-6
STATE SPACE ANALYSIS:
Concept of state, state variables and state model,
derivation of state models from block diagrams-
solving time invariant state equations
–state transition matrix and its properties.

Selection of state variables
• The state variables of a system are not unique.
• There are many choices for a given system
Guide lines:
1. For a physical systems, the number of state variables
needed to represent the system must be equal to the
number of energy storing elements present in the system
2. If a system is represented by a linear constant
coefficient differential equation, then the number of state
variables needed to represent the system must be equal
to the order of the differential equation
3. If a system is represented by a transfer function, then the
number of sate variables needed to represent the system
must be equal to the highest power of s in the
denominator of the transfer function.

State space Representation using Physical variables
• In state-space modeling of the systems, the
choice of sate variables is arbitrary.
• One of the possible choice is physical
variables.
• The state equations are obtained from the
differential equations governing the system

State Space Model
Consider the following series of the RLC circuit.
It is having an input voltage vi(t) and the current
flowing through the circuit is i(t).

• There are two storage elements (inductor and
capacitor) in this circuit. So, the number of the
state variables is equal to two.
• These state variables are the current flowing
through the inductor, i(t) and the voltage
across capacitor, vc(t).
• From the circuit, the output voltage, v0(t) is
equal to the voltage across capacitor, vc(t).

Problem
Represent the electrical circuit shown by a state
model

Solution
Since there are three energy storing elements,
choose three state variables to represent the systems
The current through the inductors i1,i2 and voltage
across the capacitor vc are taken as state variables
Let the three sate variables be x1, x2 and x3 be related
to physical quantities as shown
Let, i1 = x1,
i2 = x2,
vc = x3

y(t) = 0 R2
0
y(t) = R2i2(t) = R2x2(t)
x1(t)
x2(t
)
x3(t)
This is output equation

Problem
Obtain the state model for a system represented
by an electrical system as shown in figure

Solution
Since there are two energy storage elements
present in the system, assume two state
variables to describe the system behavior.
Let the two state variables be x1 and x2 be
related to physical quantities as shown
Let v1(t) = x1(t)
v2(t) = x2(t)

State representation using Phase variables
• The phase variables are defined as those particular
state variables which are obtained from one of the
system variables and its derivatives.
• Usually the variables used is the system output and
the remaining state variables are then derivatives of
the output.
• The state model using phase variables can be easily
determined if the system model is already known in
the differential equation or transfer function form.

The block diagram for the state model is

Problem
s2
+7s+
2
Obtain the state model of the system whose
transfer function is given by s3
+9s2
+26s+24
Solution:
Y(s)
s2
+7s+2
=
U(s)
s3
+9s2
+26s+24
Y(s)
Y(s) = x
U(s) C(s)
U(s)
C(s
)
C(s
)
Y(s)
= s2 + 7s +
2
--------(1)
C(s) 1
=
U(s)
s3
+9s2
+26s+24
-------(2)

Problem
A feedback system has a closed-loop transfer
function =
Y(s)
2(s+5)
U(s) (s+2)(s+3)
(s+4)
Solution:
Y(s
) =
U(s) (s+2)(s+3)
(s+4)
2(s+5
)
By partial fraction expansion,
Y(s)
2(s+5)
= =
A
+
B
+
U(s) (s+2)(s+3)(s+4) (s+2) (s+3)
(s+4)
C
Solving for A, B and C
A = 3; B = - 4; C = 1

Solution
From the given system, A =
1
0
1
1
•The solution of state equation is,
•X(t) = L-1 [sI-A]-1 X(0) + L-1 [SI-A]-1 B U(s)
•Here U= 0
•∴ X(t) = L-1 [sI-A]-1 X(0)
[sI – A] = s 1
0 1
0
0 1
-
1 1
=
s − 1 0
−1 s −
1
[sI-A]-1 =
s −
1
1
0
s −
1
1
s−
1
2

[sI-A]-1 = s −
1
0
s −
1
1
s−
1
2
=
1
1
s−
1
1 2
0
1
s−1 s−1
X(t) = L-1 [sI-A]-1 X(0)
= L-1
1
s−
1
1
0
1
s−
1
2
s−1
1
0
= et
te
0
et
0
1 = et
te

Problem
Compute the State transition matrix by infinite
series method A
=
0 1
−1
−2
Solution: For the given system matrix A, the state
transition matrix is,
A
t
∅ (t) = e = I + At
+
A
t2
!
+
A
t
2
3
3
!
+------
A =
0 1
−1
−2
A2 = A. A =
A3 = A2.A =
−1
−2
=
−
3
0 1 . 0 1 −1 −2
−1 −2 −1 −2
=
2 3
−1 −2 .
0 1 2 3
2 3
−4

∅(t) = I + At +
A
t
2
+
A
t
3
3
!
+---
=
1
0
0
1
+
2!
0 1
−1
−2
t +
−1
−2
2 3
t2
2!
+
2 3
−3
−4
t3
3!
+- -
=
1
−
t2
t3
2
2
+ + ⋯ t − t
+
t3
+
⋯
−t + t2 −
t
3
3
2
2
2
2
+ ⋯ 1 − 2t + 3t
+ ⋯
= e−t +
te−t
−te−t
te−t
e−t −
te−t

Problem
Find the state transition matrix by infinite series
method for the system matrix A = 1
1
0
1
Solution: For the given system matrix A, the
state transition matrix is,
A
t
∅ (t) = e = I + At +
A
t2
!
+
A
t
2
3
3
!
+------
A = 1
1
0
1
A2 = A.A = 1
1 .
1 1 =
1
2
0 1 0 1 0
1

A3 = A2. A = 1
2 1 1 1
3
0 1
.
0 1
=
0
1
∅(t) = I + At +
A
t
2 3
3
!
At
+ ---
= +
+
2!
1 0 1
1
0 1 0
1
t +
1
2
0
1
t2
2
!
+
1
3
0
1
t3
3
!
+ ---
=
2
3
1 + t + t
+ t
+
⋯
2!
3!
0
3
t + t2 + t
+
⋯
1 + t +
t
2
2
+
t
3
2!
3!
+.
.
= e
t
0
te
t
et

Unit-6.pptx of control system and engineering

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