EC2255-_Control_System_Notes_solved_prob(1).pdf

EC2255- Solved Problems in Control System IV Semester ECE
Solved by A.Devasena., Associate Professor., Dept/ECE Page 1
Control Systems
PHYSICAL SYSTEMS:
INTRODUCTION:
 First step towards analysis of a control system is preparation of a mathematical
model which is linear over a satisfactory range operating conditions giving us the
property of linearity and superposition.
 A model may be defined as a representation of the essential aspects of the system
which presents knowledge of the system in a usable manner.
 To be useful, the model must not so complicated that it cannot be understood and
thereby be unsuitable for analysis.
 The components of a control system are diverse in nature and may include
electrical, mechanical, thermal and fluidic devices.
 At the same time, it must not so complicated that it cannot be understood and
thereby be unsuitable for analysis.
 At the same time, it must not be oversimplified and trivial.
 While dealing with control systems, we shall be concerned mostly with dynamic
systems. The behavior of such systems is described in the form of differential
equations.
 Although these will normally be nonlinear, it is customary to linearize them about
an operating point to obtain linear differential equations.
 The components of a control system are diverse in nature and may include
electrical, mechanical, thermal and fluidic devices. The differential equations
relating the input and output quantities for these devices are obtained using the
basic law of physics. These include balancing forces, energy and mass. In
practice, some simplifying assumptions are often made to obtain linear
differential equations. With constant coefficients, although in most cases exact
analysis would lead to nonlinear partial differential equations. For most physical
devices one may classify the variables as either THROUGH or ACROSS
variables, in the sense that the former refer to a point while the latter are measured
between two points.
 The input , output relations of various physical components of a system is
governed by differential equation. The mathematical model of a control system
constitutes a set of differential equation.

SYSTEM THROUGH
VARIABLE
ACROSS
VARIABLE
Electrical Current I Potential difference
or voltage V.
Mechanical
(translational)
Force, F Relative velocity,v
Mechanical
(rotational)
Torque, T Relative angular
velocity ,ω
Thermal Rate of flow of
heat energy,q
Difference in temp.
T
Fluidic Volumetric rate
of fluid flow,Q
Difference in
pressure,P
A mathematical model will be linear if the differential equations describing the
system have constant coefficients. If the coefficients of the differential equations
describing the system are constants, then the model is linear time invariant. If the
coefficients of the differential equations governing the system are functions of time, then
the model is linear time varying.
The differential equations of a linear time invariant system can be reshaped into
different form for the convenience of analysis. One such model for single input and
single output system analysis is transfer function of the system. The transfer function of a
system analysis is transfer function of the system. The transfer function of a system is
defined as the ratio of laplace transfer function of the system. The transfer function of a
defined as the ratio of laplace transfer function of a system is defined as the ratio of
laplace transform of output to the laplace transform of input with zero initial conditions.
An equation describing a physical has integrals and differentials. The response can be
obtained by solving such equations. The steps involved in obtaining the transfer function
are:
 Write differential equation of the system.
 Replace terms involving d/dt by s and integral of dt by 1/s.
Advantages of open loop system.
The advantages of open loop system are
1. Such systems are simple in construction.
2. Very much convenient when input is difficult to measure.
3. Such systems are easy for maintenance point of view.
4. Generally these are not troubled with problems of stability.
5. Such systems are simple to design and hence economical.

Disadvantages of open loop system.
The disadvantages of open loop system are
1. Such systems are inaccurate and unreliable because accuracy of such
systems are totally dependent on the accurate precalibration of the
controller.
2. Such systems give inaccurate results if there are variation in the
external environment i.e. systems cannot sense environmental changes.
3. Similarly they cannot sense internal disturbances in the system, after
the controller stage.
Advantages of closed loop system.
The advantages of closed loop system are:
1.accuracy of such system is always very high because controller modifies
and manipulates the actuating signal such that error in the system will be zero.
2. Such systems senses environmental changes, as well as internal
disturbances and accordingly modifies the error.
1. In such system, there is reduced effect of non-linearites and
distortions.
Comparison between open loop system and closed loop system
OPEN LOOP SYSTEM CLOSED LOOP SYSTEM
Output measurement is not required
for operation of the system.
Output measurement is
necessary.
Highly affected by non-lineariteis Reduced effect of non-
linearities.
Highly sensitive to the disturbances
and environmental changes
Less sensitive to disturbances
and environmental changes.
Feedback element and error detector
are absent
Feedback element and error
detector are absent
Generally stable in nature Stability is the major
consideration while designing.
Linear system.
A system is said to be linear if it obeys the principle of superposition and
homogeneity. The principle of superposition states that the response of the system to a
weighted sum of the responses of the system to each individual input signals.
The system is said to be linear, if it satisfies the following two properties:
 Adaptive property that is for any x and y belonging to the
domain of the function f, we have
F(x+y) = f(x) +f(y)

 Homogeneous property that is for any x belonging to the
domain of the function f and for any scalar constant α
We have
 F(αx) = αf(x)
These two properties together constitute a principle of superposition.
Most of the systems are non-linear in nature because of different non-linearities
such as saturation, friction, dead zone etc. present in the system.
Definition of Transfer Function
Transfer function of a given system is defined as the ratio of the laplace transform
of output variable to laplace transform of input variables at zero input conditions.
Properties of transfer functions.
The properties of transfer function are as follows:
 The transfer function of a system is the laplace transform of its
impulse response. I.e. if the input to a system with transfer
function P(s) is an impulse and all initial conditions are zero, the
transform of the output is P(s).
 The roots of the denominator are the system poles and the roots of
the numerator are system zeros. The system stability can be
described in terms of the location of the roots of the transfer
function.
Advantages of transfer function.
 It helps in the stability analysis of the system.
 It helps in determining the important information about the system
Poles, zeros, characteristic equation etc.
 Once transfer function is known, output response for any type of
reference input can be calculated.
 The system differential equation can be easily obtained by
replacing variable ‘s’ by d/dt.
Disadvantages of transfer function.
The disadvantages of transfer function approach are:
 Only applicable to linear time invariant systems.
 It does not provide any information concerning the physical
structure of the system. From transfer function, physical nature of
the system, whether it is electrical, mechanical, thermal or
hydraulic cannot be judged.

 Effects arising due to initial conditions are totally neglected. Hence
initial conditions loose importance.
Important features of feedback.
 Reduced effects of non-linearities and distortion.
 Increased accuracy.
 Reduced sensitivity of the ratio of the output to input to variations
in system characteristics.
 Tendency toward oscillation or instability.
The basic elements used for modeling mechanical translational system:
The model of mechanical translational system can be obtained by using three
basic elements mass spring and dashpot.
Translational system
Consider a mechanical system in which motion is taking place along a straight
line. Such systems are of translational type. These systems are characterized by
displacement, linear velocity and linear acceleration.
Definition of torque.
This is the motion about a fixed axis. In such systems, the force gets replaced by a
moment about the fixed axis. I.e. {force x distance from fixed axis} which is called
torque.
Definition of friction.
Whenever there is a motion, there exists a friction. Friction may be between
moving element and fixed support or between two moving surfaces. Friction is also non-
linear in nature.
The types of friction.
Friction can be divided into three types. They are
 Viscous friction.
 Static friction.
 Coulomb friction.
The two types of analogies for mechanical system are force-voltage analogy and
force-current analogy.
Analogous systems
Systems whose differential equations are identical form are called analogous
systems.

Two types of variables in physical system.
The two variables in physical systems are through variables and across variable.
Through variables refer to a point, the across variable is measured between two points.
The other names for force voltage analogy and force current analogy are
Force voltage analogy- Loop analysis.
Force current analogy- nodal analysis.
BLOCK DIAGRAM ALGEBRA;
INTRODUCTION:
In block diagram, the system consists of so many components. These components
are linked together to perform a particular function. Each component can be represented
with the help of individual block.
NEED FOR BLOCK DIAGRAM REDUCTION:
Block diagrams of some of the systems turn out to be complex, such that the evaluation
of their performance required simplification (or reduction) of block diagrams which is
carried out by block diagram rearrangements.
DEFINITION:
A block diagram of a system is a pictorial representation of the functions
performed by each component of the system and shows the flow of signals.
ADVANTAGES OF BLOCK DIAGRAM:
 Very simple to construct the block diagram for complicated systems.
 Individual as well as overall performance of the system can be studied by using
transfer functions shown in the block diagram.
 Overall closed loop transfer function can be easily calculated using block diagram
rules.
 The function of the individual element can be visualized from the block diagram.
DISADVANTAGES OF BLOCK DIAGRAM:
The disadvantages of block diagram are:
 Block diagram does not include any information about the physical construction
of the system.
 Source of energy is generally not shown in the block diagram, so block diagram
for a given system is not unique.
The basic components of block diagram are block, branches, summing point, arrows.

BLOCK:
It indicates the function of particular system. R(s) is the reference or controlling
variable.
G(s) is the transfer function of the particular system.C(s) is output or controlled variable.
SUMMING POINT:
It is used to add or subtract one or more signals. + indicates the add or subtract function.
+ indicates the signal is added to reference signal. -indicates the signal is subtracted to
reference signal. It is called negative feedback. The signal which is added or subtracted to
the reference signal is called feedback signal.
TAKE OF POINT
Some of the signal at the top are bypass and it is given to other block for further
performance.
The steps to reduce the block diagram.
 Reduce the series blocks.
 Reduce the parallel blocks.
G(S)
C(S)
R(S)
+
C(S)
R(S)
Summing point
G(s)
C(S)
R(S)
Take off point

 Reduce minor feedback loops.
 As for as possible shift summing point to the left and take-off point to the right.
 Repeat the above steps till canonical form is obtained.
BLOCK DIAGRAM FOR A CANONICAL SYSTEM:
R(s) = reference or controlling variable.
C(S) = Output or controlled variable.
E(S) = Error actuating signal.
B(S) = Feedback signal
H(S) = Feedback element.
G(S) = transfer function for the system.
Rules for reduction of block diagram
Rule 1: If the blocks are in cascade then
G(S) = C(S)
E(S)
Closed loop transfer function is
C(S) .
R(S)
C(S) = G(S) E(S).
Consider a negative feedback signal is applied
Actuating signal E(S) = R(S) – B(S)
Feedback signal C(S)H(S) = B(S)
Put equation (3) in eq.(2)
E(S) = R(S)- C(S)H(S)
Put eq (4) in eq.(1)
C(S) = G(S){R(S) – C(S)H(S)}
(1)
(2)
(3)
(4)
(5)
C(S)[ 1+ G(S) H(S)] = G(S)R(S) (6)
(transfer function for negative feedback signal
C(S) = G(S)______
R(S) 1+G(S) R(S)
(transfer function for positive feedback signal.)
C(S) = G(S)______
R(S) 1-G(S) R(S)

Rule 2: if the blocks are in parallel then, the blocks are added or subtracted
depending on the summing point signal.
Rule 3: Moving the take-off point after the block
Rule 4: moving the take-off point ahead of the block
G1 G2
C(S
)
R(S)
G1G2
=
R(S) C(S)
G1
G2
+
+
+
R(S) C(s)
G1+G2
=
C(S)
R(S)
G
C (S)
R(S) G
1/G
R(S) C(S)
=
G
G
G

Rule 5: Moving summing point after the block, then
Rule 6: Moving the summing point ahead (before) the blockff
Rule 7: Eliminating feedback loop, then

1. Using block diagram reduction techniques find C(s) / R(s) as in fig. Fig1.2.1.1
G1
R(s)
H1
H2
- -
Step 1: eliminating feedback loop I
I
II
G1
1+G1H1
C(s)
H1
-
R(s) C(s)
II
Step2: eliminating feedback loop II
G(s) = G1 / 1 +G1H1 H(s) = H2
C(s) / R(s) =
G1 / 1 +G1H1
1 + (G1 / 1 +G1H1 )(H2)
=
G1
1 +G1H1 + G1H2
Fig1.2.1.1

R(s) C(s)
G1
1 + G1H1+G1H2
Fig1.2.1.2
Answer.
2. Using block diagram reduction technique find closed loop transfer
function C(s) / R(s) shown in fig 1.2.2.1
R(S)
G1 G4 G2
G3
-
-
H1
+
H2
C(S)
Fig 1.2.2.1

R(S)
G1 G4 G2 +G3
-
-
H1
H2
C(S)
Fig 1.2.2.2
I
II
Step 2: Eliminate feedback loopI and combine the blocks (G1G4 / 1+ G1G4H1) &
(G2 + G3) which are in parallel as shown in fig. Fig 1.2.2.3
R(S)
G1G4
1+ G1G4H1
G2+G3
H2
C(S)
II
Step 1: Combine the blocks G1 &G2 which are in cascade and combine the
blocks G2 &G3 which are in parallel as shown in fig 1.2.2.2.

Step 3: eliminate feedback loop II
G1G4
1 +G1G4H1
G2 +G3
1 + G1G4
1 + G1G4H1
G2 +G3
C(s)
R(s)
=
=
G1G4
G2 +G3
1 + G1G4H1 +
G1G4 (G2 +G3)

3.Find the transfer function C(s) / R(s) for the block diagram shown below
as shown in (fig 1.2.3.1) (A.U.2004)
G1(s) G2(s)
-
-
I
II
R(s) C(s)
(fig 1.2.3.1)
G1(s) G2(s)__
1 + G2 (s)
-
II
R(s) C(s)
Step 1: Eliminating feedback loopI (as in fig.1.2.3.2)
(fig 1.2.3.2)

G1(s)G2(s)__
1 + G2 (s)
-
II
R(s) C(s)
Step 2: Combining blocks G1(s) & {G2(s) / 1+ G2(s)} which are in
cascade (as in fig 1.2.3.3)
Fig 1.2.3.3
Step 3: Eliminating feedback loop II (as in fig 1.2.3.4)
C(s) =
R(S)
G1(S) G2(S)
1 + G2(S)
1 + G1(S) G2(S)
1 + G2(S)
G1(S) G2(S)
(1 + G2(S)) + G1(S) G2(S)
C(s) =
R(S)
G1(S) G2(S)
(1 + G2(S)) + G1(S) G2(S)
R(S)
C(S)
Answer.

4. Determine the transfer function C(s) / R(s) of the system shown in
fig.1.2.4.1 by block diagram reduction method. (AU: Dec.2003)
G1
H2
H3
G4
G3
G2
-
+ + +
-
-
H1
R(s)
C(s)
Fig.1.2.4.1.
S1 S3
T1
T2
S2
G1
H2
H
3
G4
G3
G2
-
+ + +
-
-
H1
R(s)
C(s)
Fig.1.2.4.2.
S1 S3
S2
Step:1.
Shifting the summing point S2 before the block G1 and shifting the
take off point T2 after the block G4
1 / G4
1 / G1

G1 G2
H2
H3 / G1G4
G3 G4
+
+
+
-
-
H1
R(s)
C(s)
Fig.1.2.4.3.
S1
S3
S2
Step:2.
Exchange the summing points and take off points using associative law
and combining the series blocks we get
III
II
I
-
G1 G2______
1 + G1G2H1
H3 / G1G4
+
R(s)
C(s)
Fig.1.2.4.4.
S1
Step:3
Eliminating inner feedback loops I, II
III
G3 G4______
1 + G3G4H2
-
+

G1 G2 G3 G4____________
(1 + G1G2H1)( 1 + G3G4H2)
H3 / G1G4
+
R(s)
C(s)
Fig.1.2.4.5.
S1
Step:4
Combine the blocks in series
III
-
+
Step 5:
Eliminate the feedback loop III
C(s) =
R(s)
G1 G2 G3 G4_________
(1 + G1G2H1)( 1 + G3G4H2)
1 + G1 G2 G3 G4____________
(1 + G1G2H1)( 1 + G3G4H2)
__H3___
G1G4
C(s) = ______G1G2G3G4_________________
R(s) (1 + G1G2H1)( 1 + G3G4H2) +G2G3H3
Answer.

5. Using block diagram reduction technique, find the transfer function from
each input to the output C(s) for the system shown in fig.1.2.5.1.
(AU: 2005)
G1 G2 G3
G4
G5
R(s)
X(s)
+
+
+
-
-
-
H5
C(s)
Fig.1.2.5.1.
G1 G2 G3 G5
R(s)
+
+
+
-
-
-
H5
C(s)
Fig.1.2.5.2.
With X(s) = 0, the given block diagram reduces as
I II
III

G1
__G2____
(1 + G2)
G3 G5______
(1 + G5H5)
R(s)
+
-
C(s)
Fig.1.2.5.3.
Step 1: eliminate feedback loops I, II
III
G1
__G2 G3 G5_____
(1 + G2) (1 + G5H5)
R(s)
+
-
C(s)
Fig.1.2.5.4.
Step 2: Combine the blocks in series
III

G1
__G2 G3 G5_________________
{ (1 + G2) (1 + G5H5)}+ {G2G3G5)
R(s)
C(s)
Fig.1.2.5.5.
_ G1G2 G3 G5_________________
{ (1 + G2) (1 + G5H5)}+ {G2G3G5)
R(s)
C(s)
Fig.1.2.5.6.
Step 3: Eliminate feedback loop III
Step 4: Combine the blocks in series.

G2 G3 G5
+
+
-
-
H5
C(s)
Fig.1.2.5.6.
With R(s) = 0, G1 vanishes, but minus sign at summing point must be
considered by introducing block of -1 as shown
I II
-1
G4
X(s)
G2____
( 1+ G2 )
G3 G5_______
( 1+ G5H5)
+
+
C(s)
Fig.1.2.5.7.
Step 5: eliminate feedback loops I, II
-1
G4
X(s)

- G2G3____
( 1+ G2 )
G5_______
( 1+ G5H5)
+
C(s)
Fig.1.2.5.8.
G4
X(s)
+
III

- G2G3____
( 1+ G2 )
G5_______
( 1+ G5H5)
+
C(s)
Fig.1.2.5.8.
G4
X(s)
+
III
G4G5(1 +G2)_____________
(1+ G5H5 )(1 +G2)+ G2G3H5
C(s)
Step 8: eliminate feedback loop III and combine the blocks in series
X(s)
Answer.

6. Reduce the block diagram shown in fig.1.2.6.1. and obtain C(s) / R(s)
(AU: may 2007)
G1
G5
G2 G3
H1
H2
G4
S2
S1 S3
R(s) C(s)
Fig. 1.2.6.1.
+
+
+
+
-
-

Step :1
Shifting the take off point between the blocks G2 and G3 to after the
block G3
G1
G5
G2 G3
H1
H2
G4
S2
S1 S3
R(s) C(s)
Fig. 1.2.6.2
1/G3
-
+ +
-
+
+
Step :2
Combine the blocks G2 and G3, 1/G3 and G5 which are in series
G1 G2 G3
H1
H2
G4
S2
S1 S3
R(s) C(s)
Fig. 1.2.6.3
G5/G3
-
+ +
-
+
+
I

Step :3
Eliminate feed back loop I, and then combine blocks G4 and G5 / G3
which are in parallel
G1
G2 G3_____
1 + G2G3H1
H2
S1
R(s)
C(s)
Fig. 1.2.6.4
G4 + (G5/G3)
-
+
Step :4
Combine the blocks which are in series
(G1G2 G3)( G3G4 + G5)__
G3(1 + G2G3H1)
H2
S1
R(s)
C(s)
Fig. 1.2.6.5
-
+
II

Step :5
Eliminate the feed back loop II
(G1G2 G3)( G3G4 + G5)__
G3(1 + G2G3H1)
1 + G1G2G3(G3G4 + G5 )
G3(1 + G2G3H1)
H2
C(s) =
R(s)
C(s) = G1G2G3 (G3G4 + G5 )___________
R(s) G3(1 +G2G3H1) + [G1G2G3 (G3G4 + G5)] H2
Answer.
7. Determine the overall transfer function of the block diagram shown in
fig. 1.2.7.1. (AU: Nov. 2005)
G1
H1
G3
G2
R(s)
S3
S2
S1
+
-
+
+
-
I
C(s)
Fig.1.2.7.1.

Step :1 eliminate feed back loop I
G1____
1 +G1H1
G3
G2
R(s)
S3
S1
+
-
+
+
C(s)
Fig.1.2.7.2.
Step :2 Shifting the summing point S3 before the block (G1 / 1+G1H1) and
Combine the blocks G2 & (G1 / 1 +G1H1) which are in series.
G3
G2G1___
1 +G1H1
R(s)
S3 S1
+
-
+
+ C(s)
Fig.1.2.7.3.
1 + G1H1
G1
II

Step :3 Eliminate feedback loop II and combine the blocks in parallel
G1G2________
(1 +G1H1) +G1G2
R(s)
C(s)
Fig.1.2.7.4.
{G3(1 + G1H1)}
1 + G1
C(s) = { G1 +G3(1 + G1H1)} ___G1G2________
R(s) G1 1+G1H1 +G1G2
C(s) = { G1 +G3(1 + G1H1)}G2
R(s) 1+G1H1 +G1G2
Answer.

8. Use block diagram reduction technique obtain the equivalent transfer
function C(s) /R(s)
G1
G2
G3
G4
G5
R(s) C(s)
H1
H2
I
II
-
-
+
+
Fig.1.2.8.1
-
+
G1____
1 +G1H1
(G2+G3)
G4____
1 +G4H2
G5
R(s) C(s)
Fig.1.2.8.2
-
+
Step 1:
Eliminate the feed back loops I ,II and combine the blocks G2 and G3
which are in parallel
III

R(s) C(s)
Fig.1.2.8.3
-
+
Step 2:
Combine the blocks in series
III
G1(G2+G3) G4 G5___
( 1 +G4H2) (1 +G1H1)
Step 2:
Eliminate the feed back loop III
G1(G2+G3) G4 G5___
( 1 +G4H2) (1 +G1H1)
1 + __G1(G2+G3) G4 G5___
( 1 +G4H2) (1 +G1H1)
C(s) =
R(s)
C(s) = ______ G1(G2+G3) G4 G5_________________
R(s) ( 1 +G4H2) (1 +G1H1) + G1(G2+G3) G4 G5
Answer.

Signal flow graph:
The graphical representation of the variables of a set of linear algebraic equations
representing the system is called signal flow graph.
The properties of signal flow graph
 The signal in the system flows along the branches and along the arrows associated
with the branches.
 The value of variable represented by any node is an algebraic sum of all the
signals entering at the node.
 The signals gets multiplied by the branch gain or branch transmittance when it
travels along it.
 Applicable only to linear time invariant systems.
Need for signal flow graph:
Block diagrams are very successful for representing control systems, but for
complicated systems, the block diagram reduction process is tedious and time consuming.
So signal flow graphs are needed which does not require any reduction process because
of availability of a flow graph formula, which relates the input and output system
variables.
The transmittance is the gain acquired by the signal when travels from one node
to another node in the signal flow graph.
Node represents a system variable, which is equal to the sum of all incoming
signals at the node, outgoing signals from the node do not affect the value of the node
variable.
A signal travels along a branch from one node to another in the direction
indicated by the branch arrow and in the process gets multiplied by the gain or
transmittance of the branch.
A node having incoming and outgoing branches is known as chain node.
A feedback loop consisting of only one node is called self loop.
The product of all the gains of the branches forming a loop is called loop gain.
A path from the input to output node is defined as forward path.

1..Using Mason’s gain formula obtain C(s)/R(s) for the signal flow graph shown in
fig1.3.1
By Mason’s Gain formula
C(S)
R(S)
=
K
∑ PK ∆K
1
∆
Here K = 2
Forward paths
P1 = G1G2
P2 = G1G3
Individual loops
L1 = - G1H1
L2 = - G1G2H2
L3 = - G1G3H2
There are no two non touching loops.
∆ = 1 – (L1 +L2+L3) = 1 – (- G1H1 – G1G2H2 - G1G3H2)
= 1 + G1H1+ G1G2H2 + G1G3H2

C(S)
R(S)
=
P1 ∆1 + P2∆2
∆
=
∆1 = 1
∆2 = 1
(G1G2)(1) + (G1G3)
= 1 + G1H1+ G1G2H2 + G1G3H2
Ans.

2. Using Mason’s gain formula, determine C(S) / R(S) for the figure 1.3.5
2.Using Mason’s gain formula obtain C(s)/R(s) for the signal flow graph
shown in fig1.3.2
K
∑ PK ∆K
1
∆
=
C(S)
R(S)
Here K =1
Forward paths
P1 = G1G2G3
Individual loops
L1 = - G1H2
L2 = - G2G3H3
L3= - G1G2H1
∆ = 1 – (L1 +L2+L3) = 1 – (- G1H2 - G2G3H3 - G1G2H1)
∆1 = 1
C(S) = P1∆1 = ____ (G1G2G3)(1)______________
R(S) ∆ 1 + G1H2 + G2G3H3 + G1G2H1
Ans.

K
∑ PK ∆K
1
=
C(S)
R(S) ∆
Here K =2
Individual loops
L1 = - G2H1
L2 = G1G2H1
Forward paths
P1 = G1G2G3
P2 = G4
There are no two non touching loops
∆ = 1 – (L1 +L2+L3) = 1 – (- G1H2 + G1G2H1)
= 1 + G1H2 - G1G2H1
∆1 = 1
∆2 = 1

C(S)
R(S)
=
P1 ∆1 + P2∆2
∆
=
(G1G2G3)(1) + (G4)(1)
1 + G1H2 - G1G2H1
Ans.
4. Using Mason’s gain formula, determine C(S) / R(S) for the
figure 1.3.4

C(S)
R(S)
=
K
∑ PK ∆K
1
∆
Forward paths
P1 = G1G2G3G4
P2 = G5G6G7G8
Here K =2
Individual loops
L1 = G2H2
L2 = G3H3
L3 = G6H6
L4 = G7H7
Two pairs of two non touching loops are
there. They are
L1 L3 = G2H2 G6H6
L2 L4 = G3H3 G7H7
∆ = 1 – (L1 +L2+L3 + L4 ) +( L1 L3 + L2 L4)
= 1 – (G2H2 + G3H3 + G6H6 + G7H7) + (G2H2 G6H6+ G3H3 G7H7)
∆1 = 1 – (G6H6 + G7H7) = 1 - G6H6 -G7H7
∆2 = 1 – (G2H2+ G3H3) = 1 – G2H2- G3H3
C(S)
R(S)
=
P1 ∆1 + P2∆2
∆
(G1G2G3G4)( 1 –G6H6 -G7H7) + (G5G6G7G8)( 1 – G2H2- G3H3)
=
1 – (G2H2 + G3H3 + G6H6 + G7H7) + (G2H2 G6H6+ G3H3 G7H7)
Ans.

Fig.1.3.5
C(S)
R(S)
=
K
∑ PK ∆K
1
∆
Here K = 2
Forward paths
P1 = G1G2G4G5
P2 = G1G2G3
Individual loops
L1 = - G4 G5H2
L2 = - G2G4G5H1
L3 = - G1 G2G4 G5
L4 = - G1 G2G3
L5 = - G2G3H1
L6 = - G3H2
There are no two non touching loops
∆ = 1 - ( L1 + L2 + L3 + L4 + L5+ L6 )
= 1 – (- G4 G5H2 - G2G4G5H1 - G1 G2G4 G5 - G1 G2G3 - G2G3H1 - G3H2)
= 1 + G4 G5H2 + G2G4G5H1 + G1 G2G4 G5 + G1 G2G3 + G2G3H1 + G3H2)
∆1 = 1
∆2 = 1

C(S)
R(S)
=
P1 ∆1 + P2∆2
∆
=
(G1G2G4G5)(1) + (G1G2G3)(1)
1 + (G4 G5H2 + G2G4G5H1 + G1 G2G4 G5 + G1 G2G3 + G2G3H1
+ G3H2)
G3H2)
Ans.
6.Find the transfer function for the signal flow graph as shown in fig 1.3.6

C(S)
R(S)
=
K
∑ PK ∆K
1
∆
Forward paths
P1 = G1G2G3G4 G5G6
P2 = G1G2G7G6
P3 = G1G2G3G4 G8
Here K =3
Individual loops
L1 = - G4H4
L2 = - G5 G6H1
L3 = -G2G3G4 G5G6 H2
L4 = - G1G2G3G4 G5G6 H3
L5 = - G8H1
L6 = - G1G2G7 G6H3
L7 = -G1G2G3 G4G8 H3
L8 = -G2G3 G4G8 H2
L9 = - G2G7 G6H2
∆ = 1 – (L1 +L2+L3 + L4 + L5 + L6 + L7 +L8 + L9) + (L1L6 + L1L9)
= 1 –( - G4H4 - G5 G6H1 - G2G3G4 G5G6 H2 - G1G2G3G4 G5G6 H3 - G8H1
- G1G2G7 G6H3 - G1G2G3 G4G8 H3 - G2G3 G4G8 H2 - G2G7 G6H2)
+( G1G2 G4G6G7H3H4 + G4 G2G6G7H2H4)
There are two pairs of two non touching loops. They are
L1L6 = G1G2 G4G6G7H3H4
L1L9 = G4 G2G6G7H2H4
∆1 = 1
∆2 = 1 – ( - G4H4)
∆3= 1

7.Determine the transfer function C(S)/R(S) for the signal flow graph
shown in fig.1.3.7.
C(S)
R(S)
=
P1∆ 1 +P1∆2 + P3∆3
∆
= (G1G2G3G4 G5G6)(1) + (G1G2G7G6)( 1 + G4H4)+ (G1G2G3G4 G8)(1)
1 + G4H4 + G5 G6H1 +G2G3G4 G5G6 H2 + G1G2G3G4 G5G6 H3
+G8H1 + G1G2G7 G6H3 +G1G2G3 G4G8 H3 + G2G3 G4G8 H2
+G2G7 G6H2 + G1G2 G4G6G7H3H4 + G4 G2G6G7H2H4
Ans.

C(S)
R(S)
=
K
∑ PK ∆K
1
∆
Forward paths
P1 = G1G2G3
P2 = G1G4
Here K =2
Individual loops
L1 = - G1 G2H1
L2 = - G2 G3H2
L3 = -G1G2G3
L4 = - G4 H2
L5 = - G1G4
∆ = 1 – (L1 +L2+L3 + L4 + L5 )
= 1 –( - G1 G2H1 - G2 G3H2- G1G2G3 - G4 H2- G1G4)
= 1 + G1 G2H1 + G2 G3H2+ G1G2G3 + G4 H2+G1G4
∆1 = 1
∆2 = 1

C(S)
R(S)
=
P1∆ 1 +P1∆2
∆
= (G1G2G3)(1) + (G1G4)( 1 )
1 + G1 G2H1 + G2 G3H2+ G1G2G3 + G4 H2+G1G4
Ans.
8.Find C(S)/R(S) for the following system using mason’s gain formula
for the signal flow graph shown in fig.1.3.8.

C(S)
R(S)
=
K
∑ PK ∆K
1
∆
Forward paths
P1 = G1G2
P2 = G4
P3= G7G8
P4 = G1G5 G8
P5 = G7G6 G2
Here K =5
Individual loops
L1 = G9
L2 = G3
L3 = G5G6
∆ = 1 – (L1 +L2+L3 ) + (L1L2)
= 1 – ( G9 + G3 + G5G6) + (G9 G3)
= 1 - G9 - G3 - G5G6 + (G9 G3)
There is one pair of two non touching loops. It is
L1L2 = G9 G3
∆1 = 1 – L1 = 1 – G9
∆2 = 1 – (L1 +L2+L3 ) + (L1L2) = 1 - G9 - G3 - G5G6) + (G9 G3)
∆3 = 1 – L2 = 1 – G3
∆4 = 1
∆5 = 1

C(S)
R(S)
=
P1∆ 1 +P1∆2 +P3∆ 3 +P4∆4 +P5∆ 5
∆
(G1G2)( 1 – G9) + (G4)(( 1 - G9 - G3 - G5G6) + (G9 G3) ) + (G7G8)
(1 – G3) + (G1G5 G8)(1) + (G7G6 G2)(1)
1 - G9 - G3 - G5G6 + (G9 G3)
Ans.
=
9. Determine the transfer function C(s)/ R(s) for the signal flow
graph shown in fig. 1.3.9.

C(S)
R(S)
=
K
∑ PK ∆K
1
∆
Forward paths
P1 = G1G2 G3
P2 = G4
Here K =2
Individual loops
L1 = - G1G2H1
L2 = - G2G3H2
L3 = - G1G2 G3
L4 = - G4
L5 = G2G4 H1 H2
∆ = 1 – (L1 +L2+L3 L4 +L5)
= 1 – (- G1G2H1 - G2G3H2- G1G2 G3 - G4 + G2G4 H1 H2)
= 1 + G1G2H1 + G2G3H2 + G1G2 G3 + G4 - G2G4 H1 H2
∆1 = 1
∆2 = 1

C(S)
R(S)
=
P1∆ 1 +P1∆2
∆
(G1G2 G3)(1) + (G4) (1)
1 + G1G2H1 + G2G3H2 + G1G2 G3 + G4 - G2G4 H1 H2
Ans.
10.Determine the transfer function C(S) /R(S) using Mason’s gain formula
for the signal flow graph shown in fig.1.3.10

C(S)
R(S)
=
K
∑ PK ∆K
1
∆
Forward paths
P1 = G1G2 G3 G4
P2 = G1 H3 G4
Here K =2
Individual loops
L1 = - G1G2 G3 G4H1
L2 = - G3G4H2
L3 = - G1G4 H3 H1
L4 = - G3G4
∆ = 1 – (L1 +L2+L3 L4)
= 1 – (- G1G2 G3 G4H1 - G3G4H2 - G1G4 H3 H1 - G3G4 )
= 1 + G1G2 G3 G4H1 + G3G4H2 + G1G4 H3 H1 + G3G4
∆1 = 1
∆2 = 1

C(S)
R(S)
=
P1∆ 1 +P1∆2
∆
= (G1G2 G3 G4)(1) + (G1 H3 G4)( 1 )
1 + G1G2 G3 G4H1 + G3G4H2 + G1G4 H3 H1 + G3G4
Ans.

13.For the signal flow graph given in fig.1.3.13. Valuate the closed loop
transfer function of the system. (A.U.April.2006)

C(S)
R(S)
=
K
∑ PK ∆K
1
∆
Forward paths
P1 = a12a23a34a45
P2 = a12a23a35
Here K =2
Individual loops
L1 = a23a32
L2 = a23a34a45 a52
L3= a23a34a45 a42
L4 = a44
L5= a23a35
∆ = 1 – (L1 +L2+L3 L4 + L5 )
= 1 – (a23a32 + a23a34a45 a52 + a23a34a45 a42 + a44 + a23a35)
= 1 – a23a32 - a23a34a45 a52 - a23a34a45 a42 - a44 - a23a35
∆1 = 1
∆2 = 1 – L4 = 1- a44

C(S)
R(S)
=
P1∆ 1 +P1∆2
∆
= (a12a23a34a45)(1) + (a12a23a35)( 1- a44 )
1 – a23a32 - a23a34a45 a52 - a23a34a45 a42 - a44 - a23a35
Ans.
14. Using Mason’s gain formula find C(S) /R(S) for the fig. 1.3.14.

C(S)
R(S)
=
K
∑ PK ∆K
1
∆
Forward paths
P1 = G1G2 G3 G4 G5
P2 = G1 G2 G3 G4 G6
Here K =2
Individual loops
L1 = H1
L2 = G3H2
L3 = G4 H3
L4 = H4
The combination of two non touching loops:
L1 L2 = H1G3H2
L1 L3 = H1 G4 H3
L1 L4 = H1H4
L2 L4 = G3H2 H4
The combination of three non touching loops:
L1 L3 L4 = H1 G4 H3 H4
L1 L2 L4 = H1G3H2 H4

C(S)
R(S)
=
P1∆ 1 +P1∆2
∆
= (G1G2 G3 G4 G5)(1) + (G1 G2 G3 G4 G6 )(1)
)(1) + (a12a23a35)( 1- a44 )
1 – ( H1 + G3H2+ G4 H3 + H4 )
+ (H1G3H2 + H1 G4 H3 + H1H4 + G3H2 H4)
– (H1 G4 H3 H4 + H1G3H2 H4 )
Ans.
∆ = 1 – (L1 +L2+L3 L4) + (L1 L2 + L1 L3 + L1 L4 + L3 L4) – (L1 L3 L4 + L1 L2 L4)
= 1 – ( H1 + G3H2+ G4 H3 + H4 ) + (H1G3H2 + H1 G4 H3 + H1H4 + G3H2 H4)
– (H1 G4 H3 H4 + H1G3H2 H4 )
∆1 = 1
∆2 = 1

1. Obtain the analogous electrical network for the system shown in fig.5. (AU:Nov./Dec.-
2007)
The Mass M1 is under the displacement x1(t).
The friction B1 is responsible to change the displacement from x1(t) to x2(t)
The Mass M2 is under the displacement x2(t).
The friction B2 and spring K1 are responsible to change the displacement from x2(t) to
x3(t)
The Mass M3 and spring K2 are under the influence of displacement x3(t).
The equivalent Mechanical system is shown in fig.5.a.

The equilibrium equations are
F(t) = M1(d2
x1(t)/ dt2
) + B1d(x1(t)- x2(t))/dt ------------------------------------------------(1)
0 = B1d(x2(t)- x1(t))/dt + M2(d2
x2(t)/ dt2
) + K1(x2(t)- x3(t)) + B2d(x2(t)- x3(t))/dt --------(2)
0 = K1(x3(t)- x2(t)) + B2d(x3(t)- x2(t))/dt + M3(d2
x3(t)/ dt2
) + K3x3(t) ---------------------(3)
Using force- voltage analogy ,
Mass is replaced by inductance, friction or dashpot is replaced by resistance, spring is
replaced by reciprocal of capacitance, displacement is replaced by charge. Rate of change
of displacement is replaced by current, force is replaced by voltage.
V(t) = L1di1(t)/dt + R1 (i1(t) – i2(t) ) ----------------------------------------------------------------
(4)
0 = R1 (i2(t) – i1(t) ) + L2di2(t)/dt + 1/C1∫(i2(t) – i3(t))dt + R2 (i2(t) – i3(t) )---------------------
(5)
0 = 1/C1∫(i3(t) – i2(t)) dt + R2 (i3(t) – i2 +(t) ) + L3di3(t)/dt + 1/C2∫(i3(t)dt.----------------------
(6)
The analogous electrical network is shown in fig.5.c.

2. Draw the equivalent mechanical system of the system shown in fig.6. write the set of
equilibrium equations for it and obtain electrical analogous circuits using i) F-V analogy
ii)F-I analogy.
( AU: May-
2009)
As shown in fig.6. M1,K1,and B1 are under the displacement x1 as K1 and B1 are with
respect to rigid support. K2 is between x1 and x2 as it is responsible for the change in
displacement. While M2 , K3 and B2 are under the displacement x2. Hence the equivalent
mechanical system is as shown in fig.6.a.

The equilibrium equations are
F(t) = M1(d2
x1(t)/ dt2
) + B1dx1(t)/dt + K1x1(t) + K2(x1(t)- x2(t)) ----------------------
--------(1)
0 = M2(d2
x2(t)/ dt2
) + B2dx2(t)/ dt + K2(x2(t)- x1(t)) + K3x2(t) -----------------------
-------(2)
Using force- voltage analogy ,
Mass is replaced by inductance, friction or dashpot is replaced by resistance, spring is
replaced by reciprocal of capacitance, displacement is replaced by charge. Rate of change
of displacement is replaced by current, force is replaced by voltage.
V(t) = L1di1(t)/dt + R1 i1(t) + 1/C1∫(i1(t) dt + 1/C2∫(i1(t) – i2(t)) dt --------------------
----------(3)
0 = L2di2(t)/dt + R1 i2(t) +1/C2∫(i2(t) – i1(t))dt + 1/C3∫(i2(t) dt---------------------(
4)
The analogous system for force voltage analogy is shown in fig.6.c.

Using force- current analogy ,
Mass is replaced by capacitance, friction or dashpot is replaced by reciprocal of
resistance, spring is replaced by reciprocal of inductance, displacement is replaced by
flux. Rate of change of displacement is replaced by voltage, force is replaced by current.
The analogous system for force current analogy is shown in fig.6.d.

I(t) = C1dV1(t)/dt + 1/R1 V1(t) + 1/L1∫(V1(t) dt + 1/L2∫(V1(t) – V2(t)) dt -------------
--------------(5)
0 = C2dV2(t)/dt + 1/R2 V2(t)+1/L2∫(V2(t) – V1(t))dt + 1/L3∫(V2(t) dt-----------------
--------------(6)
3.Write the equations for mechanical system shown in figure 7.a.
( AU: April 2005, Dec 2005)
Due to force f(t) applied to M1, it will displace by the displacement x1(t).
As K1 and B1 are between M1 and fixed support, both are under same
displacement x1(t)
Due to friction B3, the force transferred to M2 is different than f(t), hence M2 will
displace by the displacement x2(t).
As K2 and B2are between M2 and fixed support, both are under same displacement
x2(t)
The equivalent system is shown in fig.7.b.
The node equations for the system are
At node x1
f(t) = M1(d2
x1(t)/ dt2
) + B1dx1(t)/dt + K1x1(t) + B3d(x1(t)- x2(t))/dt ------------------
-----(1)
At node x2

0 = M2(d2
x2(t)/ dt2
) + B2dx2(t)/dt + K2x2(t) + B3d(x2(t)- x1(t))/dt --------------------
---(2)
Taking laplace transforms of equations(1) and (2) we get
F(s) =M1s2
X1(s) + B1sX1(s) + K2X1(s) + B3s(X1(s) – X2(s)) -----------------------(3)
0 =M2s2
X1(s) + B2sX2(s) + K2X2(s) + B3s(X2(s) – X1(s)) -----------------------(4)
4.Obtain the mathematical model of the following mechanical system shown
in figure 8.a
( AU:April 2004)
The displacement is x(t) as shown if fig.8.a.
Both B1 and B2 are between mass M and fixed support. Hence under the
influence of x(t). the spring k between mass M and fixed support. Hence under the
influence of x(t).
The equivalent mechanical model is shown in fig.8.b.
f(t) = M(d2
x(t)/ dt2
) + (B1 + B2)dx1(t)/dt + Kx (t) -------------------(1)
This equation represents the mathematical model.

5. Obtain the differential equations of the following mechanical systems as
shown in fig.9.a.
The equivalent nodal diagram for the given mechanical system is shown in
fig.9.b.
At node x1
0 = M2(d2
x1(t)/ dt2
) + K1x1(t) + K1x1(t) + K2(x1(t)- x2(t)) -----------------------------
-(1)
F(t) = M1(d2
x2(t)/ dt2
) + K2(x2(t)- x1(t)) ----------------------------(2)
Taking Laplace transforms of equations(1) and (2) we get
0 =M2s2
X1(s) + K1X1(s) + K2(X1(s) – X2(s)) -----------------------(3)
F(s) =M2s2
X1(s) + K2 (X2(s) – X1(s)) -----------------------(4)

6. For the spring, damper and mass system shown in fig.10.a. Obtain the
differential equations governing the system. F(t) is the force applied
(Au:2005).
The equivalent nodal diagram for the given mechanical system is shown in
fig.10.b.
The differential equations governing the system are
f(t) = M2(d2
x2(t)/ dt2
) + B2d(x1(t)- x2(t))/dt + K1(x2(t)- x1(t)) + B1d(x2(t)- x1(t))/dt ---(1)
0 = M1(d2
x1(t)/ dt2
) + B1d(x1(t)- x2(t))/dt +K1(x1(t)- x2(t))

7. Show that the systems shown in fig.11.a and fig.b are analogous system. (AU:
April-2007)
There are three displacements xi(t) xo(t) and xy(t). The input is xi(t) and output is
xo(t). So transfer function of the mechanical system is X0(s) / Xi(s). The
equilibrium equations are
B1d(xo(t)- xi(t))/dt + K1(xo(t)- xi(t)) + B1d(xo(t)- xy(t))/dt = 0 ---------(1)
B2d(xy(t)- xo(t))/dt + K2xy(t) = 0 ------------------------------------ (2)
Taking Laplace transform of both equations, neglecting initial conditions,
B1sXo(s) - B1sXi(s) + K1Xo(s) - K1Xi(s) + B2sXo(s) – B2sXy(s) = 0
Therefore Xo(s)[sB1 + k1 +s B2] + Xi(s)[sB1 + k1] – B2sXy(s) = 0 -------------------
--- (3)
B2sXy(s) - B2sXo(s) + K2Xy(s) = 0
Therefore Xy(s)[sB2 + k2 ] = B2s Xo(s)
Therefore Xy(s) = {B2s / [sB2 + k2 ]} Xo(s)} -------------------------------------------
(4)
Substitute equation 4 in equation 3,
Xo(s)[sB1 + k1 +s B2] - Xi(s)[sB1 + k1] - B2s{B2s / [sB2 + k2 ]} Xo(s)}
Therefore Xo(s)[ sB1 + k1 +s B2 – s2
B2
2
/ (sB2 + k2) = Xi(s)[sB1 + k1]
Xo(s) / Xi (s) = [sB1 + k1][ sB2 + k2] / {s2
B1B2 + sB1K2 + sK1B2 + K1K2 + s2
B2
2
+
sK2B2
– s2
B2
2
}
Xo(s) / Xi (s) = ___________ k2K1 ( 1 + B1/K1) ( 1 + B2/K2) ___________
k2K1 [( 1 + s2
B1B2/K1K2 +s B1/K1 + s B2/K2) + s B2/K1
Xo(s) = __________ ( 1 + B1/K1) ( 1 + B2/K2) ____ -------------
Required transfer function.
Xi (s) ( 1 + B1/K1) ( 1 + B2/K2) + s B2/K1
The s domain network for the given electric network is

Now I (s) = ____ Ei(s) ________
Z(s) + R1 + 1/sC1
And Eo(s) = I(s)[ R1 + 1 /sC1] with Z(s) = R2 parallel 1 /sC2
Therefore Eo(s) = Ei(s) [ R1 + 1 /sC1]
Z(s) + R1 + 1/sC1
Eo(s) = Ei(s)( 1 + sR1C1)
sC1Z(s) + sC1R1 +1
substituting Z(s) = R2 x 1/sC2
R2 + 1/sC2
Eo(s) = __________ ( 1 + sR1C1) ( 1 + sR2C2) ____
Ei(s) ( 1 + sR1C1) ( 1 + sR2C2) + sC1 R2
As spring is replaced by reciprocal of capacitance, friction is replaced by
Resistance, the two transfer functions are identical , hence the two systems are
analogous in nature.

22. Find the impulse response of the second order system whose open loop transfer
function
G(S)= ____9_____
S2
+4S+9 (AU may 2009)
Comparing the denominator term with S2
+2ZωnS+ωn
2
ωn
2
=9 2Zωn = 4.
ωn=3 Z= __4_ _ = ___4__ = 0.667
2ωn 2×3
The impulse response for Z<1 is given by
C(t)= _ωn___ e-Z ω
n t sin(ωn√1-Z2
)t
√1-Z2
= __3_____ e-2t
sin(3√1-(0.667)2
)t
√1-(0.667)2
=4.035 e-2t
sin2.235t
Obtain the unit step response of a second order system [under damped condition]
[AU: nov/dec
06]
Response of the second order system
__C(S) = ___ωn
2
_______
R(S) S2
+2ZωnS+ωn
2
For unit step input R(S)=1
For under damped systems Z<1
S2
+2ZωnS+ωn
2
=0 has two roots
S1,2 = - Zωn ± jωn√1-Z2
Now let Zωn=α
And ωn√1-Z2
= ωd

S1,2= α+j ωd
For unit step input R(S)=1/s
C(S)= R(S) [ ___ωn
2
_______ ] = [ ___ωn
2
_______ ]
S2
+2ZωnS+ωn
2
S(S2
+2ZωnS+ωn
2
)
C(S)=A + __BS+C____
S S2
+2ZωnS+ωn
2
___ωn
2
_______ =A(S2
+2ZωnS+ωn
2
)+(BS+C)S
S2
+2ZωnS+ωn
2
S2
+2ZωnS+ωn
2
ωn
2
= AS2
+2AZωnS+Aωn
2
+BS2
+CS
ωn
2
=S2
(A+B)+S( 2AZωn+C)+A ωn
2
equvating constant term
ωn
2
= A ωn
2
A=1
Equvating S term
0= 2AZωn +C
C=2AZωn
C=2Zωn
Equvating S2
term
0=A+B
0=1+B
B=-1
Zωn=α
C(S)= 1 + ___-S-2α___
S S2
+2αS+ωn
2
C(S)= 1 - { ___S+2α___}
S S2
+2αS+ωn
2
Now consider S2
+2αS+ωn
2
Add and subtract α2
term to the above eq
S2
+2αS+ α2
+ ωn
2
- α2
(Zωn=α
=(S+ α2
)+ ωn
2
-Z2
ωn
2
α2
= Z2
ωn
2
=(S+ α2
)+ ωn
2
(1 -Z2
)
WKT

ωd = ωn√1-Z2
ωd
2
= ωn(1-Z2
)
= (S+ α) 2
+ ωd
2
―> put this expression in the denominator ofC(S) we get
C(S)= 1 - { ___S+2α___}
S (S+α)2
+ωd
2
Now
L-1
{ ___S+α___} = e-at
cosωt
(S+α)2
+ω2
L-1
{ ___ω___ } = e-at
sinωt
(S+α)2
+ω2
Adjusting for C(s) we get
C(S)= 1 - { ___S+α_ _ + ___α_ ____ }
S (S+α)2
+ωd
2
(S+α)2
+ωd
2
Multiply and divide by ωd to the last term
C(S)= 1 - { ___S+α_ _ + _α _ _ ωd ___ }
S (S+α)2
+ωd
2
ωd (S+α)2
+ωd
2
Taking inverse laplace
C(t)=1- e-αt
cosωd t- α e-αt
sinωd t
ωd
Now put
Zωn=α
And ωn√1-Z2
= ωd
C(t)=1- e-αt
cosωd t- α e-αt
sinωd t
ωd
C(t)=1- e- Zωn t
cosωd t+ e-Zωn t
sinωd t
ωn√1-Z2
C(t)=1- e-Zωn t
(√1-Z2
ωd t+Zsin ωd t)

ωn√1-Z2
by trigonometry
sin ( ωd t +θ)= sin ( ωd t)cos θ+cos + cos ( ωd t) sinθ
compare this equation with the above expression,
we can write sin θ= √1-Z2
cos θ =Z
sinθ = tanθ= √1-Z2
cosθ Z
θ= tan-1
√1-Z2
radians.
Z
Hence using this in the expression
C(t)=1- e-Zωn t
(sin (ωd t+θ)
ωn√1-Z2

Find kp , kv , ka and steady state error for a system with open loop transfer function as
G(S)H(S)= 15(S+4)(S+7)__
S(S+3)(S+6)(S+8)
Where the input is r(t)=4+t+t2
The given open loop transfer function in time constant form
G(S)H(S)= 15(S+4)(S+7)__
S(S+3)(S+6)(S+8)
=15× 4× 7(1+S) (1+s)
4 7
S ×3× 6× 8(1+s_) (1+s_) (1+s_)
3 6 8
= 35(1+0.25s)(1+0.14s)______
12s(1+0.3s)(1+0.16s)(1+0.125s)
=2.9(1+0.25S)(1+0.14S)
s(1+0.3s)(1+0.16s)(1+0.125s)
Now kp= Lt G(S)H(S)=∞
S->0
KV= Lt S G(S)H(S)= 2.9(1+0.25S)(1+0.14S)
S-> 0 s(1+0.3s)(1+0.16s)(1+0.125s)
KV=2.9
Ka= = Lt S2
G(S)H(S)=0
S->0
Input is =4+t+t2
=4+t+2.t2
2
Now input combination of three standard inputs
A1=4 step of 4
A2=1 ramp of 1
A3=2, parabolic input of 2.
a) For step 3 the error is
Ess 1=A1___ = 4___ = 0
1+kp 1+∞

b) For ramp of 1 the error is
Ess 2=A2___ = 1___
KV 2.9
c) For parabolic of 2, the error is
Ess 3=A3___ = 2___ = ∞
Ka 0
Then the steady state error is
Ess=e ss1+ess2+ess3
=0+1___ +∞
2.9
=∞

Steady state analysis:
Steady state error is an important measure of the accuracy of a control systems.
Basically these errors occur from the nature of inputs, non-lineraities present in the
system
Etc. the steady state error ess is the difference between the input (or desired value )and
the
Output of a closed loop system input as t->∞
Mathematically
Ess =Lt e(t) =Lt [ r(t)-c(t)]
t->∞ t->∞
by using the final value theorem
ess = Lt e(t) =Lt SE(S)
t->∞ s->0
This is valid provided that SE(S) has no poles on the jω axis,
No multiple poles on the jω axix and is in the right half of S-plane.
Consider is a closed loop signal shown below
E(s)=error signal
B(S)=feed back signal
Closed loop transfer function = C(S) = G(S)___
R(S) 1+G(S)H(S)
C(S) = G(S)___. R(S)
1+G(S)H(S)
E(S)=R(S)-B(S)----- >1
WKT
B(S)=R(S)-B(S)------ >2
Put eq 2 in eq1
E(S)=R(S)-C(S)H(S)------ >3
WKT
C(S)=E(S)G(S)------- >4
Put 4 in eq 3
E(S)=R(S)-E(S)G(S)H(S)
E(S)+E(S)G(S)H(S)=R(S)

[1+G(S)H(S)]E(S) =R(S)
E(S)=___R(S)___ for non unity feed back
1+G(S)H(S)
E(S)=___R(S)__ For unity feed back
1+G(S)
ESS= Lt SR(S)_____
s->0 1+G(S)H(S)
Steady state error deoends on
(i) R(S) i.e reference i/p its type and magnitude.
(ii) G(S)H(S) i.e open loop transfer function.
(iii) Dominant non linearities if any.
Effect of input on steady state error:
(static error co efficient method)
(a) Reference i/p is step of magnitude A
R(S) = __A__
S
ESS= Lt SR(S)_____
s->0 1+G(S)H(S)
= Lt S.A/S_____
s->0 1+G(S)H(S)
= Lt A_____
s->0 1+G(S)H(S)
ESS=_____A___
1+ Lt G(S)H(S)
s->0
For a system selected Lt G(S)H(S) is constant and called

s->0
positional error co_efficient of the system denoted as Kp.
Kp= Lt G(S)H(S) =positional error co-efficient
s->0
and corresponding error is
ess=__A__
1+kp
Reference i/p is ramp of magnitude A
R(S)=A/S2
.
ESS= Lt SR(S)_____
s->0 1+G(S)H(S)
= Lt S.A/S2
_____
s->0 1+G(S)H(S)
= Lt A_____
s->0 S[1+G(S)H(S)]
ESS =_____A___
Lt S G(S)H(S)
s->0
for a seleted system Lt S G(S)H(S) is constant an called velocity error co efficient as
Kv.
s->0
ESS=A/Kv
Kv= Lt S G(S)H(S) =velocity error effect.
s->0
reference input is parabolic of magnitude A
R(t)=__A__ t2
2
R(S)= _A__

S3
ESS= Lt SR(S)_____
s->0 1+G(S)H(S)
= Lt S.A/S3
_____
s->0 1+G(S)H(S)
= Lt A_____
s->0 S2
[1+G(S)H(S)]
ESS =_____A___
Lt S2
G(S)H(S)
s->0
for a seleted system Lt S2
G(S)H(S) is constant an called velocity error co efficient as
Ka.
s->0
ESS=A/Ka
Ka= Lt S2
G(S)H(S) =acceleration error effect.
s->0
types of feed back control systems:
The open loop transfer function G(S) of a unity feed back system can be
written in two standard forms namely the time constant form and polr zero form.
G(S)=K(TZ1S+1) (TZ2S+1) (TZ3S+1)…….. (TZmS+1)
Sn
(TP1S+1) (TP2S+1) (TP3S+1)…….. (TPmS+1) (TIME CONSTANT FORM)
= _K’
(S+Z1) (S+Z2) (S+Z3)…….. (S+Zm)
Sn
(S+P1) (S+P2) (S+P3)…….. (S+Pm) (POLE ZERO FORM)
Where the relation between K and K’
is

K=K’
(product of all zeros)
Product of all poles.
In both respresentatiom, the terms Sn
in the denominator represent the n
poles at the origin, which represent the type number of the system. If n=0,
the system is known as type 0 system and if n=1 the system is known as
type 1 system and so on.
The type number of a system is defined as the number of poles of G(S) at
the origin or number of integrators present in the systems.
For example G(S)=(S+5)____ is a type 0 system
(S+4)(S+8)
G(S)=(S+9)___ is a type 1 system
S(s+3)(s+7)
G(s)=(s+10)____ is a type 2 system
S2
(S+2)(S+9)

Steady state errors of type 0 unity feedback system.
Let G(S) be an open loop transfer function of a system with no poles at the
origin (n=0).the steady state errors for standard inputs are as follows.
Positional error essp(t)= Lt s.1/s__
s->0 1+G(S)
= ____1____
1+ Lt G(S)
s->0
____1__ = __1___
1+kp 1+k
Velocity error essv(t) = Lt s.1/s2
__
s->0 1+G(S)
= ____1____ = 1/0=∞
1+ Lt s G(S)
s->0
acceleratiom error essa(t) = Lt s.1/s3
__
s->0 1+G(S)
= ____1____ = 1/0=∞
1+ Lt s2
G(S)
s->0
steady state error of type1 unity feed back system:
Let G(S) be an open loop transfer function of a system with one poles at the
origin (n=1).the steady state errors for standard inputs are as follows.
Positional error essp(t)= Lt s.1/s__
s->0 1+G(S)
= ____1____
1+ Lt G(S)
s->0
____1__ = 0
1+∞

__
s->0 1+G(S)
= ____1____ = ____1____ = 1/k
Lt s+ s G(S) Lt s G(S)
s->0 s->0
__
s->0 1+G(S)
= ____1____ = 1/0=∞
Lt s2
+s2
G(S)
s->0
for tupe 1 system
G(S)=k(Tz1S+1) (Tz2S+1)…….
S(Tp1S+1) (Tp2S+1)…….
Lt G(S) =∞=kp
s->0
Lt s G(S) =k=kv
s->0
Lt s2
G(S) =0=ka
s->0
steady state errors of type 2 unity feedback system.
G(S) be an open loop transfer function of a system with two poles at the origin (n=2).the
steady state errors for standard inputs are as follows.
Positional error essp(t)= Lt s.1/s2
__
s->0 1+G(S)
= ____1____
1+ Lt G(S)
s->0
____1__ = 0
1+∞

__
s->0 1+G(S)
= ____1____ = 1/∞=0
Lt s G(S)
s->0
__
s->0 1+G(S)
= ____1____ = 1/k
Lt s2
+s2
G(S)
s->0
for tupe 2 system
G(S)=k(Tz1S+1) (Tz2S+1)…….
S2
(Tp1S+1) (Tp2S+1)…….
Lt G(S) =∞=kp
s->0
Lt s G(S) =∞=kv
s->0
Lt s2
G(S) =k=ka
s->0
generalized error coefficient method (or dynamic error co efficients)
E(S)=___R(S)__
1+G(S)H(S)
Let us assume that is the product of two polynomial of s
E(S)=F1(S)F2(S)
Where f1(s)=__1___ . F2(S) =R(S)
1+G(S)H(S)
F(S)=F1(S).F2(S) then using convolution integral
t
L-1
{ F(S)}= F(t)= ∫F1(ι) F2(t-ι)dι
0
t t
Similarly e(t)= ∫F1(ι) F2(t-ι)dι = ∫F1(ι) R(t-ι)dι

0 0
R(t-ι) can be expanded by using taylor series form as
R(t-ι)=R(t)-ιR’(t)+ι2
R”(t)+ι2
R’”(t)+…..
2! 3!
Substituting e(t)= t
∫F1(ι) [R(t)-ιR’(t)+ι2
R”(t)+ ι2
R’”(t)+…..]dι
0 2! 3!
t t
= ∫F1(ι) R(t) dι - ∫F1(ι) ιR’(t)dι+…….
0 0
Ess=lt e(t)
t->∞
= lt t t
t->∞ [ ∫F1(ι) R(t) dι - ∫F1(ι) ιR’(t)dι+…]
0 0
∞ ∞ ∞
= R(t) ∫F1(ι) dι - R’(t) ∫F1(ι) ι dι+ R”(t) ∫F1(ι) ι2
dι]
0 0 0 2!
∞
Where ko= = ∫F1(ι) dι
0
∞
K1= - ∫F1(ι) ι dι
0
∞
K2= - ∫F1(ι) ι2
dι
0
Substituting these values we have
Ess= k0R(t)+k1R’(t)+k2 R”(t)+…….
2!
Where ko, k1,k2…… are called dynamic error coeff
To calculate these values co efficient use the following method
According to the definition of laplace transform

∞
F(s)= ∫ F1(ι)e-sι
dι
0
∞
Now k0=∫ F1(ι)dι
0
Multiplying by e-sι
to both sides
∞
e-sι
k0=∫ F1(ι) e-sι
dι = F1(s)
0
Taking limit as s->0 of both sides
Lt k0 e-sι
= lt F1(s)
s->0 s->0
where F(S)= ___1__
1+G(S)H(S)
TAKING derivative of k0 e-sι
w.r.t ‘s’ we get
-ι k0 e-sι
= dF1(S)
ds
substituting ∞
k0 = ∫ F1(ι) dι
0
∞
∫ F1(ι) e-sι
dι= dF1(S)
0 ds
Taking limit as s->0 of both side
K1= lt dF1(S)
s->0 ds
in general
Kn= lt dn
F1(S)

s->0 ds
Introduction to PID controllers:
A controller is a device which when introduced in feed back or forward path of system,
controls the steady state and transient response as per the requirement. This controller
converts the applied input to some other form of error which is proportional to the error due to
which steady state and transient response gets improved. The output of the controller is
proportional to the amount of error generated by that device.
The performance of this controlling phenomena may be done by means of electrical,
mechanical, pneumatic or hydraulic medium.
Classification of controllers:
1.ON-OFF controller
2. proportional controller
3. integral controller
4. derivative controller
5. proportional plus integral controller
6. proportional plus derivative controller
7. proportional +integral+derivativecontroller
ON-OFF controller:
This type of controller is the simplest and cheapest typehere the actuating device
(controller)
Is the capable of assuming only one two positions, with either zero(or) maximum input to the
process.
Proportional controller:
It is defined as the action of a controller in which the output signal m(t) is proportional to
the measured actuating error signal e(t).
M(t) α e(t)
M(t) =kpe(t)
Where kp is proportional sensitivity or the gain
Taking laplace transform we get
M(s)=kpE(S)

Kp= M(S) = change in controller output
E(S) change in derivation
So the proportional controller is essentially an amplifier with an adjustable gain.
Integral controller:
It is denoted by the term ‘I’ in the PID controller. In a controller with integral action,
the value of the output m(t) is propprtional to the measured actuating error signal e(t).
dm(t) = kie(t)
ds
t
m(t) = ki ∫e(t) dt
0
Where ki is an adjustable constant the transfer function of the integral controller is
M(S) = ki
E(S) s
The integral controller increases the type number of system by one. This integral control
overcomes the drawback of proportional control by reducing the steady state error to zero
without the use of excessive large control gain. The integral control action is also called as reset
control action.
Derivative controller:
Derivative controller do not affect state error but effect transient response. The output of
derivative controller depends on the rat e of change of error signal. The main drawback of
derivative controller is that it amplifies the noise signal. Therefore it is not possible for us to use
the derivative control action alone. We can use this derivative controller along with integral or
proportional controller
In time domain m(t)= kd dl(t)
Dt
In laplace domain m(S)=kds E(S)
The output of the derivative controller is zero when the error signal constant. That is the
derivative control has no input when it is acting on constant signal.
Proportional plus integral controller (PI controller)
In PI controller the controller output m(t) is proportional to a linear combination of
actuating signal e(t) plus integral of the error signal.

M(t) =kpe(t) +ki ∫e(t) dt.
Taking laplace transform
M(s) = kpE(S)+kI E(S)
S
=E(S)[kp+ki_]
S
Assuming kp=1 we can write
G(S)=(1+ki ) ωn
2
= (ki+s) ωn
2
___s_ s2
(s+s2z ωn)
S(s+2z ωn)
i.e the system becomes type 2 in nature
and C(S) = (ki+s) ωn
2
R(S) s3
+2z ωns2
+s ωn
2
+ki ωn
2
i.e it becomes third order
that is it integral is included along with proportional controller means the system
relatively becomes less stable as ki must be designed in such a way that system will remain in
stable condition. Second order system will always stable.
The transient response of of system gets affected badly if PI controller is not designed
properly. By including ‘I’ controller along with ‘p’ controller, steady state the system gets
improved and we can get the accurate output from the system.
In general PI controller improves steady state part but it affects transient part of the
system.
Proportional plus derivative controller (PD controller)
A controller in a forward path which changes the controller output corresponding to
the proportional plus derivative of error signal is known as PD controller
M(T) =kpe(t) +kpTd de(t)
Dt
Taking laplace transform
M(S) = kpE(S) +kpTd s E(S) =E(S)[kp+TdS]
Assuming kp=1
G(S) = (1+sTd) ωn
2
S(s+2z ωn)

C(S) = (1+sTd) ωn
2
R(S) s2
+2[2zωn+ ωn
2
Td] + ωn
2
C0mparing the denominator with standard form s2
+2z ωns+ ωn
2
ωn
2
term is same as in the standard form
but 2z ωn =2z ωn+ ωn
2
Td
z= 2z ωn + ωn
2
Td
z ωn z ωn
because of this derivative controller the damping ratio increases by a factor ωnTd
2
Kp= lt G(S)H(S) =∞
S->0
KV= LT S G(S)H(S) = ωn
s->0 2z
as there is no change in co efficient, the error will also remain same.
PD controller has following effects on the system.
1. It reduces settling time
2. It reduces peak overshoot
3. Steady state error remains unchanged.
4. Type of the system remains unchanged
5. It increases damping ratio
PD controller improves transient part without affecting steady state.
Proportional+integral +derivative controller(PID controller)
A PID controller is the combination of proportional control action and derivative
control action with an adjustable gain for each action.
t
M(t)= kpe(t)+kpTdde(t) + kp ∫ e(t) dt
dt TI 0
taking laplace transform
M(S)= kpE(S)+kpTdSE(S) + kp E(S)
TiS
M(S)= kpE(S)[1+Td + kp ]
Ti S

Pd controller improves transient portion, pi controller improve steady state portion combination
of two may be used to improve overall time response of the system.
G(s) =(K+STd+KI ) ωn
2
S____
S(S+2Z ωn)
C(S) = (K+STd+KI ) ωn
2
/ S(S+2Z ωn)
R(s) S_________
1+ (K+STd+KI ) ωn
2
S____
S(S+2Z ωn)
Transient response and steady state response.
The time response of a system consists of two parts.
Transient response
Steady state response
This can be denoted by equation as
Total time response =transient response +steady state response
C(t)=ctr(t)+css(t)
Transient response of the system is the portion of total time response during which output
Changes from one value to another value. In other words, it is the response before the output
reaches
the steady state value.
Steady state response of the system for a given input after a very long time. In steady
state, the output response settles to its final steady state value or steady oscillations.
Derivation of steady state error:
Definition: steady state error : it is the difference between the actual output and the desired
output.
Consider a simple closed loop system using negative feedback us shown.
Where E(s)=error signal and B(s)=feed back signal.
Now E(S)=R(S)-B(S)
BUT B(S)=C(S)H(S)

E(S)=R(S)-C(S)H(S)
C(S)=E(S)G(S)
E(S)=R(S)-E(S)G(S)H(S)
E(S)+E(S)G(S)H(S)=R(S).
E(S)=R(S)___
1+G(S)H(S) for non unity feed back .
E(S)=R(S)___
1+G(S) for unity feed back.
This E(S) is the error in laplace domain and is expression in ‘s’. we want to calculate the error
value. In the time domain corresponding error will be e(t). now steady state of the system is that
state. Which remains as t-> ∞.
Steady state error ess=Lt e(t)
t->∞
standard test signals:
the impulse signal has zero amplitude every where expect at the origin as shown in fig.
mathematically the impulse signal can be represented by
A δ(t)=0 for t#0 and
ε
∫ A δ(t) dt= A
-ε
Where ε tends to zero. Here the value a represents the area of the signal or energy content of
the signal and the laplace transform of the impulse is given by.
L[ A δ(t)] =A
Unit impulse signal:
If A=1, then the signal is called unit impulse signal that is for a unit impulse signal.
δ(t)=0 for t#0 and
ε
∫ A δ(t)dt = 1
-ε
The laplace transform of the unit impulse signal is

L[ δ(t)] =1.
Step signal:
Step signal of size A is a signal that changes from zero level to another level A is a zero
time and stays there for ever as shown fig. the step signal is applied to the system to study the
behavior of the system for a sudden change in input.
Mathematically
r(t) = A for t> 0
= 0 for t< 0
The laplace transform of the the step signal is R(S)=L[r(t)]=A/s.
Unit step signal:
If the the magnitude a of the step signal is unity, then the step signal is shown and is
denoted bt u(t). that is
U(t) =1 for t > 0
=0 for < 0.
The laplace transform of the unit step msignal u(t) is
L[u(t)] =1/s.
Ramp signal:
The ramp signal increases linearly with time from an initial value of zero at t=0.
Mathematically representation of the ramp signal is given by
R(t) = At for t> 0
= 0 for t=0
Where A represents the slope of the line
The laplace transform of the ramp signal is known as unit ramp signal.
Parabolic signal:
The instantaneous value of a parabolic signal varies as square the time from an initial
value of zero at t=0.
Mathematically representation of the

Parabolic signal is
r(t)=At2
t>0.
= 0 t<0.
The laplace transform of the parabolic signal is given by
R(s)= L[At2
] =2A/s3
To have a convinent form for laplace transform of parabolic signal is A/s3
. the parabolic signal
in time domain is often defined as At2
/2.
Steady state erroer and error constants
The stady state error ess is the difference between the input (or desired value ) and the
output of a closed loop system for a known input as t->∞.
Mathematically
Ess = Lt e(t)
t->∞
= Lt [r(t)-c(t)]
t->0
bu using final value theorem
Ess = Lt e(t) = Lt s E(S)
t->∞ s->0
this is valid provided that SE(S) has no poles on the jω axis no multiple poles on the jω axis and
is the rignt half of s plane. The steady state error is a measure of system accuracy. With the help
of step ramp and paroboic inputs it is passible for us to judge te steady state behavior of the
system.
Analysis of first order system:
Order: order of system is the highest power of s in the denominator of a closed loop
transfer funttion. Consider a simple system shown in fig. find v0(t)
i.e response if it is excited by unit step input.
Vi(t)=1 t>0
=0 t<0
Vi(s)=1/s.

Now first calculate system T.F using laplace T.f
Vi(t) = R i(t)+ 1 ∫ i(t) dt
C
V0(t) = 1 ∫ i(t) dt
C
Vi(s) = R I(s)+ 1 I(s)
Cs
V0(s) = 1 I(s)
Cs
I(S)=CSV0(S)
Vi(S)=I(S)[R + 1 ]
SC
Vi(S)=CSV0(S)[R + 1 ]
SC
V0(S) = ___1__
VI(S) 1+SRC
V0(S) = ___1__
VI(S) 1+TS where T=RC
The step response of first order system .
Let input applied vi(t) is the unit step voltage substituting vi(s)=1/s is the transfer function
V0(S) = ___1__ = A + __b__
S( 1+SRC) s 1+SRC
1= A(1+SRC) +BS
1=A+ASRC+BS
1=A+S(B+ARC)
Equate constant term
1=A
Equate s term
0=S(B+ARC)
Put A=1

B+RC=0
B=-RC
V0(S) = 1 + __-RC__
S 1+SRC
= 1 + __-RC__
S RC (S+1/RC)
= 1 + __-1_____
S (S+1/RC)
Taking laplace inverse
V0(t) =1- e-t/RC
=css+ct(t)
Closed loop transfer function of a system is given by
C(S) = G(S)___
R(S) 1±G(S)R(S)
The equation which gives poles of system is called characteristics equation which is
1+G(S)H(S)=0
For the first order system this equation is also first order is general of the form
1+TS=0
AS closed loop poles are the roots of characteristics equation so for first order system there is
only one closed loop pole i.e s=-1/T
The time taken for the step response of a system to reach 63.2% of the final value is known as
the time constant of the system.
Unit ramp response of a first order system:
The transfer function of a first order system is given by
C(S) = ___K__
R(S) (TS+1)
For a ramp input r(t)=t; t>0; R(S)=1/S2
. There fore
C(S) = ___K__
S2
(TS+1)

C(S)= K [A +B + __C__]
S S2
TS+1
__K__ = K[AS(TS+S)+BTS+B+CS2
]
S2
(TS+1) S2
(TS+1)
__K__ = K[AS(TS+S)+BTS+B+CS2
]
1=AS(TS+S)+BTS+B+CS2
Equate constant term equate s term equate s2
term
A=1 A+BT=0 AT+C=0
1+BT=0 C=-AT
BT=-1 C=-T
B=-1/T
C(S)= K [1 +-1/T + __-T__]
S S2
TS+1
Taking inverse laplace transform
C(s)=k [-T+t+e=t/T
]; t>0
22) A unity feed back heat treatment system has open loop transfer function
G(s)= ____10000______ . the output of set point is 5000
c. what is the steady state
temperature.
(1+S)(1+0.5S)(1+0.02S)
Here h(s)=1 R(S)=500/s
Steady state error =Lt S.R(S)
S->0 1+G(S)
=Lt S.500/S
S->0 1+ ____10000______ .
(1+S)(1+0.5S)(1+0.02S)
500/1+10000 =0.040
STEADY STATE ERROR ESS=0.040
C
STEADY state temperature c(t)=r(t)-ess
=5000
c- 0.040c
= 499.960
c

(
1. Determine the stability of the following system using Routh’s criterion.
(AU:May/ June 2008)
(a) G(s)H(s) =______1______
(s +2)(s +4)
(b) G(s)H(s) =______9______
s2
(s +2)

1.a.)G(s)H(s) =______1______
(s +2)(s +4)
The characteristic equation is
1 + G(s)H(s) = 0
1 + ___1______ = 0
(s +2)(s +4)
( s + 2 ) ( s+4 ) + 1 =0
s2
+ 6s + 9 = 0
a0 =1 a1 = 6 a2 =9
Routh’s array
s2
1 9
s1
6 0
s0
9
No sign change in the first column of the Routh’s array, hence the
system is stable.

2. Utilize the Routh table to determine the number of roots of the following
polynomials in the right half of the s-plane. Comment about the stability of the
system. (AU:May/June 2006)
( i ) s5
+ 2s4
+ 3s3
+ 6s2
+ 10s + 15
( ii ) s5
+ 6s4
+ 15s3
+ 30s2
+ 44s + 24
1.b.)G(s)H(s) =_____9______
s2
(s +2)
The characteristic equation is
1 + G(s)H(s) = 0
1 + ___9______ = 0
s2
(s +2)
s2
(s +2) + 9 =0
s3
+ 2s2
+9 = 0
a0 =1 a1 = 2 a2 =0 a3 =9
Routh’s array
s3
1 0
s2
2 9
s1
- 4.5
S0
9
b1 = 2 x 0 – 1x 9 = - 4.5
2
b2 = 0
c1 = - 4.5 x 9 = 9
- 4.5
There are two sign changes in the first column of the Routh’s array. Hence
the system is unstable.

2) i) s5
+ 2s4
+ 3s3
+ 6s2
+ 10s + 15 = 0
a0 =1 a1 = 2 a2 = 3 a3 = 6 a4 =10 a5 = 15
Routh’s array:
s5
1 3 10
s4
2 6 15
s3
0 2.5
s3
ε 2.5
s2
6 ε - 5 15
ε
Replace the 0
by small
positive
number ε
s1
6 ε – 5 2.5
____ε _________ - 15 ε
6 ε - 5
ε
s0
15
as ε tends to zero, the routh’s array becomes
s5
1 3 10
s4
2 6 15
s3
0 2.5
s2
-∞ 15
s1
2.5 0
s0
15
There are two sign
changes in the first
column of the routh’s
array. There fore there
are two roots on the
right half of the s- plane.

2) ii ) s5
+ 6s4
+ 15s3
+ 30s2
+ 44s + 24 = 0
a0 =1 a1 = 6 a2 = 15 a3 = 30 a4 =44 a5 = 24
Routh’s array:
s5
1 15 44
s4
6 30 24
s3
10 40
s2
6 24
s1
0
Replace the 0
by small
positive
number ε
s0
24
as ε tends to zero, the routh’s array becomes
s5
1 15 44
s4
6 30 24
s3
10 40
s2
6 24
s1
0
s0
24
There is no sign change in
the first column of the
routh’s array.
No roots of the
polynomial are in the
right half of the s- plane.
Hence the system is
stable.
s1
ε

3. for a system with, F(s) = s4
+ 22s3
+ 10 s2
+ s + K = 0. obtain the marginal value
of K, and the frequency of oscillations of that value of K (AU May/june 2007)
F(s) = s4
+ 22s3
+ 10 s2
+ s + K = 0
a0 = 1 a1 = 22 a2 = 10 a3 = 1 a4 = K
Routh’s array
s4
1 10 K
s3
22 1 0
s2
9.95 K
s1
9.95 – 22K
9.95
s0
K
For a stable system, the first column of the routh’s array must be
positive.
Therefore K > 0 , And 9.95 – 22K > 0
0 < K < 0.452.
The marginal value of K is 0.452.
Consider 9.95s2
+ K = 0. when K = 0.452, to find the frequency of
oscillations s= + j 0.913 or jω = + j 0.913. ω = 0.213 rad/sec.

4. Using Routh’s criterion for stability, discuss the stability of the system whose
characteristic equation is s3
+ 10s2
+ 50s + 500 = 0 (AU: April/ may 2005)
s3
+ 10s2
+ 50s + 500 = 0
a0 = 1 a1 = 10 a2 = 50 a3 = 500
Routh’s array
s3
1 50 0
s2
10 500
s1
0 0
by special case 2, replace zero row by the derivative of the
auxiliary equation ( the auxiliary equation is the row, just above the
zero row)
Auxilary equation is
A(s) = 10s2
+ 500
dA(s) = 20s
ds
s3
1 50 0
s2
10 500
s1
20 0
s0
500
No sign change in the first column of the Routh’s array. But due to
special case 2, system , may not be stable. For sufficient conditions
solve A(s) = 0.

5. Using Routh –Hurwitz criterion, determine the relation between K and T so that unity
feedback control system whose open loop transfer function given below is stable.
10s2
+ 500 = 0
s2
= - 50
s = + j √ 50.
as dominant roots are on imaginary axis, and no root in right half of s-plane
because no sign change in first column of routh’s array. the system is
marginally stable. It oscillates with the frequency √ 50 rad./ sec.

1.Sketch the root locus diagram of the following open loop transfer function
G(s)H(s) =________K_____ (AU: Nov/DEC:2005)
s(s+2)(s+5)
Step 1:
Number of poles P = 3
At s
Number of Zeros Z= 0
0,-2,-5
Step 2:
Number of separate root loci (branches)= Number of poles (P) = 3
Step 3:
The starting point of the root locus branches are from poles 0,-2,-5.
Terminating point of the root locus branches are at ZEROs. But here we have
no ZEROs. So The root locus branches terminates at infinity.
P1
P2
P3
+jω
-jω
+σ
-σ
Consider P1. to the R.H.S. of P1 the sum of poles and zeros are
odd. So P1 is the part of root locus. Consider P2. to the R.H.S. of
P2 the sum of poles and zeros are even. So P2 is not the part of
root locus. Consider P3. to the R.H.S. of P3 the sum of poles and
zeros are odd. So P3 is the part of root locus.
-2
-5 0

Step 4:
Number of asymptotes Na = P-Z = 3- 0 = 3
Angle of asymptotes : θn = (2n+1)π ( n = 0,1,2,…….P-Z-1)
P – Z
Here n = Na = P-Z = 3-0 = 3 n = 0,1,2.
For n =0 θ0 = (2(0)+1)π = 60ο
3 – 0
For n =1 θ1 = (2(1)+1)π = 180ο
3 – 0
For n =2 θ2 = (2(2)+1)π = 300ο
3 – 0
Step 5:
Centroid (σ )
= (∑ real parts of poles) - (∑ real parts of zeros)
P – Z
= (0 -2-5) – (0)
3 – 0
= -7 = -2.3
3
Step 6:
Breakaway point:
1 + G(s)H(s) = 0
1 + K_________ = 0
s(s+2)(s+5)
s(s2
+ 7s + 10) +K= 0
K = - s(s2
+ 7s + 10)
K = - (s3
+ 7s2
+ 10s)
to find breakaway point, differentiate k with respect to s and equate to
zero.
dk = 0
ds
dk = - 3s2
- 14s – 10 = 0
ds
i.e. 3s2
+ 14s +10 = 0
now solve the above quadratic equation, we get s = - 0.88 , -3.78.
The valid breakaway point is – 0.88. because breakaway point must be
a part of root locus.

Step 7:
Intersection of root locus with imaginary axis (jω)
Characteristic equation = 1 + G(s)H(s) = 0
s3
+ 7s2
+ 10s + k = 0
a0 = 1 a1 = 7 a2 = 10 a3 = k
Routh’s array
s3
1 10
s2
7 K
s1 7(10) – K 0
7
s0
K
To have intersection on imaginary axis, any one row should be zero.
Let us make s1
row = 0
70 – K = 0
7
70- K = 0
K = 70
Now , consider auxiliary equation
7s2
+K = 0
Put K = 70 in above equation.
7s2
+70 = 0
s2
+10 = 0
s = + j √10 = + j 3.2
The complete root locus is shown in fig.4.1

fig. 4.1

2.Sketch the root locus diagram of the following open loop transfer function
G(s)H(s) =________K_____ (AU: Nov/DEC:2006)
s(s+1)(s+3)
Step 1:
At s
0,-1,-3
Step 2:
Step 3:
The starting point of the root locus branches are from poles 0,-1,-3
no ZEROs. So The root locus branches terminates at infinity.
P1
P2
P3
+jω
-jω
+σ
-σ
P2the sum of poles and zeros are even. So P2 is not the part of
-1
-3 0

Step 4:
P – Z
Here n = Na = P-Z = 3-0 = 3 n = 0,1,2.
For n =0 θ0 = (2(0)+1)π = 60ο
3 – 0
For n =1 θ1 = (2(1)+1)π = 180ο
3 – 0
For n =2 θ2 = (2(2)+1)π = 300ο
3 – 0
Step 5:
Centroid (σ )
P – Z
= (0 -1-3) – (0)
3 – 0
= -4 = -1.33
3
Step 6:
Breakaway point:
1 + G(s)H(s) = 0
1 + K__________ = 0
s(s+1)(s+3)
s(s+1)(s+3) +k = 0
K = - s(s2
+ 4s + 3)
K = - (s3
+ 4s2
+ 3s)
zero.
dk = 0
ds
dk = - 3s2
- 8s – 3 = 0
ds
i.e. 3s2
+ 8s +3 = 0
now solve the above quadratic equation, we get s = - 0.451 , -2.22
The valid breakaway point is – - 0.451. because breakaway point must
be a part of root locus.

Step 7:
s3
+ 4s2
+ 3s + k = 0
a0 = 1 a1 = 4 a2 = 3 a3 = k
Routh’s array
s3
1 3
s2
4 K
s1 12 – K 0
4
s0
K
Let us make s1
row = 0
12-K = 0
4
12 - K = 0
K = 12
4s2
+K = 0
Put K =12 in above equation.
4s2
+12 = 0
s2
+3 = 0
s = + j √3= + j 1.714

fig. 4.2

Step 1:
At s
0,-3+j4, -3 –j4
Step 2:
Step 3:
The starting point of the root locus are from poles 0, -3+j4, -3 –j4. the
terminating point of the root locus is at ZERO. Here we have no
ZEROs. So the root locus terminates at infinity.
P1
+jω
-jω
+σ
-σ
Consider P1. to the R.H.S. of P1 the sum of poles and zeros are odd.
So P1 is the part of root locus.
0
-3+j4
-3-j4
3.Obtain the root locus for a unity feedback system with open loop transfer
function G(s)H(s) =________K_____ (AU: may/june:2006 )
s(s2
+6s+25)

Step 4:
P – Z
Here n = Na = P-Z = 3-0 = 3 n = 0,1,2.
For n =0 θ0 = (2(0)+1)π = 60ο
3 – 0
For n =1 θ1 = (2(1)+1)π = 180ο
3 – 0
For n =2 θ2 = (2(2)+1)π = 300ο
3 – 0
Step 5:
Centroid (σ )
P – Z
= (0 -3-3) – (0) = -2
3 – 0
Step 6:
Breakaway point:
1 + G(s)H(s) = 0
1 + K__________ = 0
s(s2
+6s+25)
s(s2
+6s+25)+K = 0
K = - s(s2
+6s+25)
K = - (s3
+ 6s2
+ 25s)
zero.
dk = 0
ds
dk = - 3s2
- 12s – 25 = 0
ds
i.e. 3s2
+ 12s +25 = 0
now solve the above quadratic equation, we get
s = - 2 + j0.5√39
this point is not on the root locus. Therefore there is no breakaway
point.

Step 7:
1 + k__________ = 0
s(s2
+6s+25)
s3
+ 6s2
+ 25s + K = 0
a0 = 1 a1 = 6 a2 = 25 a3 = K.
Routh’s array
s3
1 25
s2
6 K
s1 150 – K 0
6
s0
K
Let us make s1
row = 0
150-K = 0
6
150 - K = 0
K = 150
6s2
+K = 0
6s2
+150= 0
s2
+25 = 0
s = + j 5

Φp1
Φp2
Step 8:
We have complex poles. So angle of departure has to be calculated.
Angle of Departure calculation diagram.
P3
P2
Consider point P2.
Φp1 = -180ο
– tan-
(3/4)
= - 143.13ο
Φp3= -90ο
∑ Φp = Φp1 + Φp3 = - 143.13ο
- 90ο
= -233.13ο
∑ Φz = 0 ο
Φ = ∑ Φp - ∑ Φz = -233.13ο
- 0 ο
= -233.13ο
Angle of departure = 180ο
- -233.13ο
= - 53.13 ο
Similarly for point P3. the angle of departure is - 53.13 ο
The complete root locus is shown in fig.4.3.
+j4
-j4
-3

Fig.4.3

Step 1:
At s
0, -2+j4, -2 –j4 , -4
Step 2:
Step 3:
The starting point of the root locus are from poles 0, -2+j4, -2 –j4 , -4 .
The terminating point of the root locus is at ZERO. Here we have no
P1
+jω
-jω
+σ
-σ
So P1 is the part of root locus. Consider P2.To the R.H.S of the P2
the sum of poles and zeros are even. So p2 is not the part of root
locus.
0
-2+j4
-2-j4
-4
P2
4.Determine the root locus of the system whose open loop gain is
G(s)H(s) =________K_______ (AU: Nov/Dec:2006 )
S(s+4)(s2
+4s+20)

Step 4:
P – Z
Here n = Na = P-Z = 4-0 = 4 n = 0,1,2,3
For n =0 θ0 = (2(0)+1)π = 45ο
4– 0
For n =1 θ1 = (2(1)+1)π = 135ο
4 – 0
For n =2 θ2 = (2(2)+1)π = 225ο
4 – 0
For n =3 θ3 = (2(3)+1)π = 315ο
4 – 0
Step 5:
Centroid (σ )
P – Z
= (0 -4-2 -2) – (0)
4 – 0
= -8 = - 4
4
Step 6:
Breakaway point:
1 + G(s)H(s) = 0
1 + k__________ = 0
s(s+4) (s2
+4s+20)
s(s+4) (s2
+4s+20) + K = 0
k = - (s 2
+4s)( (s2
+4s+20)
k = - (s4
+8s3
+ 36s2
+80s)
zero.
dk = 0
ds
dk = - ( 4s3
+24s2
+72s +80) = 0
ds
i.e. s3
+6s2
+18s +20 = 0

When s= 0, dk/ds =80 (+v)
When s =-4, dk/ds = -80 (-ve)
When s=0, dk/ds = 0. the breakaway point is -2.
The roots of s2
+ 4s +10 = 0 are
-2 + j2.45
The other break away points are -2 + j2.45
1 6 18 20
0 -2 -8 -20
0
1 4 10
-2
Step 7:
1 + k__________ = 0
s(s+4)(s2
+4s+20)
s(s+4) (s2
+4s+20) + K = 0
s4
+8s3
+ 36s2
+80s + K = 0
a0 = 1 a1 = 8 a2 = 36 a3 = 80 a4 = K .

Routh’s array
s4
1 36 K
s3
8 80 0
s2
26 K
s1
2080 –8 K
26
s0
K
To have intersection on jω axis, any one row should be zero.
So make s1
row to be zero.
2080 –8K = 0
K = 260.
26s2
+K = 0
26s2
+260= 0
s2
= - 10
s = + j √10 = + j 3.16

Step 8:
Angle of departure:
Фp4
Фp1
Фp2
Фp3
Angle of departure Фd = 180ο
+ ∑ Фp
Consider point P3
∑ Фp = - Фp1 - Фp2 -Фp4
= - (180ο
– tan-1
4/2 ) - tan-1
4/2 - 180ο
= - 270ο
+ ∑ Фp
= 180ο
- 270ο
= - 90ο
The complete root locus is shown in fig.4.4.

Fig.4.4.

5. Sketch the root locus for the unity feedback system whose open loop
transfer function is
G(s)H(s) =________K_____
s(s+3)(s+8)

Step 1:
At s
0,-3,-8
Step 2:
Step 3:
The starting point of the root locus branches are from poles 0,-3,-8.
no ZEROs. So the root locus branches terminates at infinity.
P1
P2
P3
+jω
-jω
+σ
-σ
-3
-8 0

Step 4:
P – Z
Here n = Na = P-Z = 3-0 = 3 n = 0,1,2.
For n =0 θ0 = (2(0)+1)π = 60ο
3 – 0
For n =1 θ1 = (2(1)+1)π = 180ο
3 – 0
For n =2 θ2 = (2(2)+1)π = 300ο
3 – 0
Step 5:
Centroid (σ )
P – Z
= (0 -3-8) – (0)
3 – 0
= -11 = -3.66
3
Step 6:
Breakaway point:
1 + G(s)H(s) = 0
1 + K__________ = 0
s(s+3)(s+8)
s(s2
+ 11s + 24) +k = 0
K = - s(s2
+ 11s + 24)
K = - (s3
+ 11s2
+ 24s)
zero.
dk = 0
ds
dk = - 3s2
- 22s – 24 = 0
ds
i.e. 3s2
+ 22s +24 = 0
now solve the above quadratic equation, we get s = - 1.3, - 6.
The valid breakaway point is – 1.3. Because breakaway point must be
a part of root locus. But -6 is not the part of the root locus.

Step 7:
s3
+ 11s2
+ 24s + K = 0
a0 = 1 a1 = 11 a2 = 24 a3 = K
Routh’s array
s3
1 24
s2
11 K
s1 11(24) – K ) 0
11
s0
K
Let us make s1
row = 0
264 – K = 0
11
264 – K = 0
K = 264
11s2
+K = 0
11s2
+264 = 0
s2
+24 = 0
s = + j √24 = + j 4.9

Fig.4.5.

6.Determine the root locus of the system whose open loop gain is
G(s)H(s) =________K_______ ;
S(s+4)(s2
+8s+32)
k > 0. Sketch the root locus of the system. Hence find the value of K so that
system has a damping factor of 0.707

Step 1:
At s
0, -4+j4, -4–j4 , -4
Step 2:
The starting point of the root locus are from poles 0, -4+j4, -4–j4 , -4
The terminating point of the root locus is at ZERO. Here we have no
P1
+jω
-jω
+σ
-σ
So P1 is the part of root locus. Consider P2.Tothe R.H.S of the P2
the sum of poles and zeros are even. So p2 is not the part of root
locus.
0
-4+j4
-4-j4
-4
P2

Step 4:
P – Z
Here n = Na = P-Z = 4-0 = 4 n = 0,1,2,3
For n =0 θ0 = (2(0)+1)π = 45ο
4– 0
For n =1 θ1 = (2(1)+1)π = 135ο
4 – 0
For n =2 θ2 = (2(2)+1)π = 225ο
4 – 0
For n =3 θ3 = (2(3)+1)π = 315ο
4 – 0
Step 5:
Centroid (σ )
P – Z
= (0 -4-4 -4) – (0)
4 – 0
= -12 = - 3
4
Step 6:
Breakaway point:
1 + G(s)H(s) = 0
1 + K__________ = 0
s(s+4) (s2
+8s+32)
s(s+4) (s2
+8s+32) + K = 0
K = - (s 2
+4s)( (s2
+8s+32)
K = - (s4
+12s3
+ 64s2
+128s)
zero.
dk = 0
ds
dk = - ( 4s3
+36s2
+128s +128) = 0
ds
i.e. s3
+9s2
+32s +32 = 0

If s =-1.57, then the value of k is nearly zero. So s=-1.57 is the
valid breakaway point.
1 9 32 32
0 -1.57 -11.25 -31.92
0.08
1 7.43 20.35
-2
Step 7:
1 + k__________ = 0
s(s+4)(s2
+8s+32)
s(s+4) (s2
+8s+32) + K = 0
s4
+12s3
+ 64s2
+128s + K = 0
a0 = 1 a1 = 12 a2 = 64 a3 = 128 a4 = K .

Routh’s array
s4
1 64 K
s3
12 128 0
s2
53.3 K
s1
6826.7 –12K
53.3
s0
K
So make s1
row to be zero.
6826.7 -12K= 0
53.3
K = 568.9
53.3s2
+K = 0
Put K =568.9 in above equation.
53.3s2
+568.9= 0
s2
= - 568.9
53.3
s = + j 3.26

Step 8 :
Angle of Departure
-4
-4+4j
P3
P4
-4-4j
Фp1
+ ∑ Фp
Consider point P3
∑ Фp = - Фp1 - Фp2 -Фp4
= - (135 ο
+ 90 ο
+ 90 ο
)
= - 335ο
+ ∑ Фp
= 180ο
- 335ο
= - 135ο
Similarly at P4
Фd = + 135ο
P2
Фp4
Фp2
The complete root locus is shown in fig. 4.6.

G(S)H(S) =________K_______
S(s+4)(s2
+8s+32)
Fig.4.6.

Calculation of the value of k at damping factor ζ = 0.707
δ = cos-1
ζ = cos-1
0.707 = 45ο
K = __│Mp1_ Mp2 Mp3 Mp4│____
1
Where Mp1 = magnitude of pole P1 i.e. the distance of pole P1 from K
+j4
-j4
-4
K = 2.1 * 3 *3.6 *6.2
1
K = 140.62

7.Sketch the root locus diagram of the following open loop transfer
function
G(s)H(s) =________K_(s+4)____
s(s2
+8s + 13)

Step 1:
At s
0,-3+2j,-3-2j
Step 2:
Step 3:
The starting point of the root locus branch are from poles 0,-3+2j,-3-2j
. the terminating point of the root locus at ZEROs. Here we have one
ZERO. So one branch terminates at ZERO. The other poles terminate
at infinity.
P1
+jω
-jω
+σ
-σ
root locus.
-4
0
s -4
- 3+ 2j
- 3 - 2j
P2

Step 4:
P – Z
Here n = Na = P-Z = 3-1 = 2 n = 0,1,
For n =0 θ0 = (2(0)+1)π = 90ο
3 – 1
For n =1 θ1 = (2(1)+1)π = 270ο
3 – 1
Step 5:
Centroid (σ )
P – Z
= (0 -3-3) – (-4)
3 – 1
= -2 = -1 2
Step 6:
Breakaway point:
In this problem, no poles are adjacent to each other. The pole at origin
terminates at ZERO s = -5. the complex poles move to infinity along 90ο
And 270ο
asymptotes. Hence there is no valid breakaway point.
Step 7:
1 + K_(s+4)____ = 0
s(s2
+6s + 13)
s3
+6s2
+ 13s + K(s+4) = 0
s3
+6s2
+ (13 + K)s + 4K = 0
a0 = 1 a1 = 6 a2 = 13 + k a3 =4K

Routh’s array
s3
1 13 + K
s2
6 4K
s1
78 + 2K 0
8
s0
4K
So make s1
row to be zero.
78 + 2 K = 0
6
K = -39.
K value is negative. But to have intersection on jω axis, k
value should be positive. So root locus does not intersect on
jω axis.

EC2255-_Control_System_Notes_solved_prob(1).pdf

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