EC8391-CONTROL SYSTEMS

ENGINERRING

By CT.Vijay Nagaraj, ME.,PhD*

INTRODUCTION –CONTROL SYSTEM
Control system deals with the control of engineering systems
that are governed by the laws of physics and are therefore
called “physical systems”
The word “control” means to regulate ,to direct or to
command .
The word “system” means a combination of devices and
components connected together to perform a certain function .
Control system used in many applications. Eg: Control of
temperature ,velocity ,flow ,pressure ,acceleration etc….
 INTRODUCTION –CONTROL SYSTEM (contd…)
Configuration Of A Control System

COMMAND MANIPULATED DISTURBANCE

INPUT VARIABLE INPUT
CONTROLLER PROCESS
CONTROLLE
OUTPUT

Example : Automatic driving System:

DIRECTION ACTUATOR HEADING
SPEED BRAIN FEET VEHICLE
SPEED

(CONTROLLER) (PROCESS)
It has two inputs and two controlled outputs. Command inputs are
direction of highway and speed limit with traffic signals .
 INTRODUCTION –CONTROL SYSTEM (Contd..)
Manual And Automatic Control:
“Control” is a process of causing a system variable such as
temperature or position to conform to some desired value or
trajectory, called reference value or trajectory.
For example, driving a car implies controlling the vehicle to
follow the desired path to arrive safely at a planned destination.
i. If you are driving the car yourself, you are performing manual control of
the car.

ii. If you use design a machine, or use a computer to do it, then you have built
an automatic control system.
 1.1 CONTROL SYSTEM TERMINOLOGY & ITS
BASIC STRUCTURE
Control System Terminology:
(i)Reference input :
It provides input signal for desired output.
(ii)Error Detector :
It is an element in which one system variable is
subtracted from another variable to obtain third variable also
called comparator .
(iii)Feedback element :
It measures the controlled output ,convert or
transforms to a suitable value for comparison with reference
input.
(iv) Error signal:
It is an algebraic sum of reference input and
feedback .
 1.1 CONTROL SYSTEM TERMINOLOGY & ITS
BASIC STRUCTURE (Contd…)
(v)Controller :
(i) It is an element that is required to generate
the appropriate control signal.
(ii) The controller operates until the error
between controlled output and desired output is reduced to
zero .
(vi)Controlled System :
It is a body , a plant , a process or a machine of
which a particular condition is to be controlled .
Eg: Room heating System , space craft , reactor boiler
(vii) Controlled Output :
It is produced by actuating signal
as input to the controller .controlled output is made equal to
desired output with help of feedback system.
 1.1 CONTROL SYSTEM TERMINOLOGY & STRUCTURE
OF CONTROL SYSTEM
Basic Structure Of Control System:
Elements of control system
Plant
Actuator
Sensor
Controller
 1.1 CONTROL SYSTEM TERMINOLOGY & ITS
BASIC STRUCTURE (contd..)

Basic Structure Of Control System

The Plant is the system or process through which a particular
quantity or condition is controlled.
The Control elements are components needed to generate the
appropriate control signal applied to the plant.
The Feedback elements are components needed to identify the
functional relationship between the feedback signal and the
controlled output.
The Reference point is an external signal applied to the
summing point of the control system to cause the plant to
produce a specified action.
An Actuator is a type of motor that is responsible for moving or
controlling a mechanism or system
1.1 CONTROL SYSTEM TERMINOLOGY & STRUCTURE OF
CONTROL SYSTEM

Types Of Control System:

Control system consists of Two types :
 Open Loop System
Closed Loop System

OPEN LOOP SYSTEM:

DISTURBANCE INPUT
COMMAND MANIPULAED
INPUT CONTROLLER VARIABLE PROCESS

CONTROLLED
OUTPUT
 OPEN LOOP CONTROL SYSTEM
• In open loop control system we have a process which we have to
control and some input to change the process and out put. We have an
example of a tank level control

Hi level
Start/s switch
top

lo level
switch
 1.1 CONTROL SYSTEM TERMINOLOGY & STRUCTURE
OF CONTROL SYSTEM(Contd..)

Open Loop System

Advantages
Simple and economical.
Easier to construct.
Always stable.

Drawbacks
They are inaccurate and unreliable.
The changes in the output due to external disturbances are not
corrected automatically.
 1.1 CONTROL SYSTEM TERMINOLOGY &
STRUCTURE OF CONTROL SYSTEM(Contd..)
Practical Examples of Open Loop Control System
 Electric Hand Drier – Hot air (output) comes out as long as
you keep your hand under the machine, irrespective of how
much your hand is dried.
 Automatic Washing Machine – This machine runs according
to the pre-set time irrespective of washing is completed or
not.
 Bread Toaster - This machine runs as per adjusted time
irrespective of toasting is completed or not.
 Automatic Tea/Coffee Maker – These machines also function
for pre adjusted time only.
 1.1 CONTROL SYSTEM TERMINOLOGY & STRUCTURE OF
CONTROL SYSTEM(Contd..)
Closed Loop System:
DISTURBANCE INPUT

COMMAND MANIPULATED
INPUT CONTROLLER VARIABLE PROCESS

CONTROL O/P
FEEDBACK
MEASUREMENT
SIGNAL

ERRROR DETECTOR (OR)

REF ERROR
CONTROLLE
I/P SIGNAL CONTROLLER
DSYSTEM
CONTROLLED
FEEDBACK
OUTPUT
Closed loop control
• Closed loop control system have information
about the change in process with respect to
change in input. Now consider the previous
example in open loop control system .

Start/s
Hi level
top
switch
lo level
switch
Contro
l
system
 1.1 CONTROL SYSTEM TERMINOLOGY &
STRUCTURE OF CONTROL SYSTEM(Contd..)
Closed Loop System:
Advantages
Systems are accurate even in the presence of non-linearities.
Less affected by noise.
Less sensitivity provides more stable systems.

Drawbacks
Complex and costlier systems.
Feedback may lead to oscillatory response.
Feedback reduces the overall gain of the system.
More care is needed to design a stable system.
1.1 CONTROL SYSTEM TERMINOLOGY &
STRUCTURE OF CONTROL SYSTEM(Contd..)
Practical examples of Closed loop System:

Automatic Electric Iron - Heating elements are controlled

by output temperature of the iron.
Servo Voltage Stabilizer – Voltage controller operates
depending upon output voltage of the system.
Water Level Controller– Input water is controlled by water
level of the reservoir.
An Air Conditioner – An air conditioner functions
depending upon the temperature of the room.
Cooling System in Car – It operates depending upon the
temperature which it controls.
1.1 CONTROL SYSTEM TERMINOLOGY & STRUCTURE
OF CONTROL SYSTEM(Contd..)
Comparison of Closed Loop And Open Loop Control System

S.NO OPEN LOOP CONTROL CLOSED LOOP CONTROL

SYSTEM SYSTEM
1 The Feedback Element Is Absent The Feedback Element Is Always
Present
2 An Error Detector Is Not Present An Error Detector Is Always Present
3 It Is Stable One It May Become Unstable
4 Easy To Construct Complicated Construction
5 It Is An Economical It Is Costly
6 Having Small Bandwidth Having Large Bandwidth

7 It Is Inaccurate It Is Accurate
8 Less Maintenance More Maintenance
9 It Is Unreliable It Is Reliable
 1.2 FEEDBACK AND FEEDFORWARD CONTROL
THEORY
FEEDBACK:
 A feedback loop is a common and powerful tool
when designing a control system .
 Feedback loop takes the system output into
consideration ,which enables the system to adjust its
performance to meet a desired output response.
 1.2 FEEDBACK AND FEEDFORWARD CONTROL
THEORY(contd..)
Feedforward:
Feed forward controller detects the disturbance directly
and takes an appropriate control action in order to eliminate its
effect on the process output.
 1.3 DIFFERENCE BETWEEN FEEDBACK AND
FEEDFORWARD CONTROL

Feedback Feed forward

 The main objective is to error self  The main objective is to control or

nulling minimize transient response .
 It compensates the any disturbance which  It compensates any disturbance before they
affecting the controlled variable . affect the controlled variable .
 It does not include feed forward structure .  it includes feedback structure .

 Once disturbance enters a process ,it must  Disturbance are measured before they enter
propage through process and force the the process and required value of
controlled variable to deviate before manipulated variable to maintain desired
correctiv actions are taken controlled variable
 Example:  Example:
 Residential heating system  Heat Exchanger
 1.3.ELECTRICAL AND MECHANICAL
TRANSFER FUNCTION MODELS
 In control theory, transfer function are commonly used to
characterize the input -output relationship of components or systems that
can be described by linear time –invariant differential equation .

Input output
Transfer function
R(S) C(S)
 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.
 It is not applicable to Non-Linear systems.
 The transfer function of a system is independent of input and it
depends only on system parameters, but the output of a system depends on
input.

Transfer Function = Laplace transform of output (C(S)

Laplace transform of input (R(S)with zero intial
condition
 1.3 ELECTRICAL AND MECHANICAL TRANSFER
FUNCTION MODELS
Let us consider a linear time invariant system defined by following
differential equation

Where,
X - input of the System.
Y - Output of the System.
1.3 ELECTRICAL AND MECHANICAL TRANSFER FUNCTION
MODELS (Contd..)
Calculation Of The Transfer Function:
Consider the following ODE where y(t) is input of the system and
x(t) is the output. d 2 x(t ) dy(t ) dx(t )
A 2
C B
dt dt dt

or
Ax' ' (t )  Cy' (t )  Bx' (t )

Taking the Laplace transform on either sides

A[ s 2 X ( s )  sx(0)  x' (0)]  C[ sY ( s )  y(0)]  B[ sX ( s )  x(0)]
 Calculation of the Transfer Function (Contd..)

A[ s 2 X ( s )  sx(0)  x' (0)]  C[ sY ( s )  y(0)]  B[ sX ( s )  x(0)]

• Considering Initial conditions to zero in order to find the transfer

function of the system

As 2 X ( s )  CsY ( s )  BsX ( s )
• Rearranging the above equation

As 2 X ( s )  BsX ( s )  CsY ( s )
X ( s )[ As 2  Bs ]  CsY ( s )
X ( s) Cs C
 
Y ( s ) As  Bs As  B
2
Stability of Control System:

• Roots of denominator polynomial of a transfer function are

called ‘poles’.

• The roots of numerator polynomials of a transfer function are

called ‘zeros’.
1.3.1 Translational and rotational mechanical
systems
Mechanical Translational system
The system model can be obtained by using the
following basic elements.
1. Mass (M) – Weight of the mechanical system.
2. Spring (K) – Elastic deformation of the body.
3. Dash-pot (B) – Friction existing in the mechanical system.

Newton’s second law of motion for translational

system
The sum of forces acting on a body is zero.
Mechanical Translational system ( cont…)
List of symbols used
X = Displacement, m.
V = dx/dt = Velocity, m/sec.
a = dv/dt = Acceleration, m/sec2
M = Mass, Kg.
K = Stiffness of spring, N/m.
B = Viscous friction co-efficient, N-sec/m.
f = Applied force, N.
fm = Opposing force offered by mass of the body, N.
fb = Opposing force offered by friction (dash-pot) of the body, N.
fk = Opposing force offered by elasticity (spring) of the body, N.
Force Balance Equations Of Idealized Elements
Ideal Mass Element
Consider f=Applied Force
fm=Opposing force due to mass

d 2x
fm  2
dt
d 2x
fm  M
dt 2
By Newton’s second law,

d 2x
f  fm M
dt 2
Force Balance Equations Of Idealized Elements
Ideal Dashpot with one end fixed to reference
Consider f=Applied Force
fb=Opposing force due to friction

dx
fb 
dt
dx
fb  B
dt
By Newton’s second law,

dx
f  fb  B
dt
Force Balance Equations Of Idealized Elements
Ideal Spring with one end fixed to reference
Consider f=Applied Force
fK=Opposing force due to elasticity

fk  x
f k  Kx
By Newton’s second law,

f  f k  Kx
Problem 1 : Mechanical translational system
Write the differential equations governing the mechanical system
shown in figure below and determine the transfer function.
 Problem 2 : Mechanical translational system
cont…
Determine the transfer function of the given system shown in the
figure given below.
1.3.2 Mechanical Rotational system:
The system model can be obtained by using the following basic
elements.
1. Moment of Inertia (J) – Weight of the rotational mechanical system.
2. Spring (K) – Elastic deformation of the body.
3. Dash-pot (B) – Friction existing in the mechanical system.

Newton’s second law of motion for rotational system

The sum of torque acting on a body is zero.
 Mechanical Rotational system cont…
List of symbols used
θ = Angular displacement, rad.
dθ/dt = Angular Velocity, rad/sec.
d2θ/dt2 = Angular acceleration, rad/sec2
J = Moment of Inertia, N-m.
K = Stiffness of spring, N-m/rad.
B = Rotational friction co-efficient, N-m/(rad/sec).
T = Applied torque, N-m.
Tj = Opposing torque due to moment of inertia of the body, N-m.
Tb = Opposing force offered by friction (dash-pot) of the body, N-m.
Tk = Opposing force offered by elasticity (spring) of the body, N.
Torque Balance Equations Of Idealized Elements
Ideal Mass Element
Consider T=Applied torque
Tj =Opposing torque due to moment of inertia

d 2
Tj 
d 2
d 2
Tj  J
d 2
By Newton’s second law,

d 2
T  Tj  J
d 2
Torque Balance Equations Of Idealized Elements
Ideal Dash-pot with one end fixed to reference
Consider T=Applied torque
Tb =Opposing torque due to friction

d
Tb 
d
d
Tb  B
d
By Newton’s second law,

d
T  Tb  B
d
Torque Balance Equations Of Idealized Elements

Ideal Spring with one end fixed to reference

Consider T=Applied torque
Tb =Opposing torque due to elesticity
Tk  
Tk  K

By Newton’s second law,

T  Tk  K
Problem 1: Mechanical Rotational system
Write the differential equations governing the mechanical
rotational system shown in figure given below. Also obtain the
transfer function of the system.
1.3.3 Modeling of Electric systems
The models of electrical systems can be obtained by using
Resistor, Capacitor and Inductor.
For modeling electrical systems the electrical network is formed
by R, L, C and voltage or current network.
The differential equations can be formed by writing Kirchoff’s
current law (KCL) by choosing various nodes in the network or
Kirchoff’s voltage law (KVL) by choosing various closed path in
the network.
 Current – Voltage relation of Resistor, Inductor
and Capacitor

Elements Voltage across the elements Current through the elements

v(t )
Resistor V(t) = R i(t) i (t ) 
R

1
Inductor v(t )  L
di (t ) i (t )   v(t )dt
dt L

1
Capacitor v(t )   i (t )dt i (t )  C
dv(t )
C dt
Problem 1: Modeling of Electric systems
Obtain the transfer function of the electrical system
shown in figure given below.
1.3.4 Electrical Analogous of Mechanical system
It is easier to construct electrical models of any kind of
mechanical system.
It will helps us to analyze any kind of the system.
The electrical system has two type of inputs.
Voltage source
Current source
Therefore we have four types of analogous circuit
Force-voltage analogy
Force-current analogy
Torque-voltage analogy
Torque-current analogy
1.3.4 (a) Electrical Analogous of Mechanical
Translational System
Force – Voltage Analogous Circuit
The electrical analogous elements are,

F(t) – Force
M– Mass
f ( t )  e (t )
B – Dash – pot M L
K – Elasticity
BR
x – Displacement
L – Inductance K  1/ C
R – Resistance x  i (t )
C – Capacitance
1.3.4 Electrical Analogous of Mechanical
Translational System
Force – Current Analogous Circuit
The electrical analogous elements are,

F(t) – Force
f (t )  i (t )
M– Mass
B – Dash – pot M C
K – Elasticity B  1/ R
x – Displacement
K  1/ L
L – Inductance
R – Resistance x  v (t )
C – Capacitance
 Problem 1: Electrical Analogous of Mechanical
Translational System
Write the differential equations governing the
mechanical system shown in fig. Draw the force –
voltage and force – current analogous circuits and
verify by writing Mesh and Node equations.
 1.3.4(b)Electrical Analogous of Mechanical
Rotational System

Torque – Voltage Analogous Circuit

T – Torque
T  v (t )
J – Moment of Inertia
B – Dash – pot JL
K – Elasticity BR
Θ – Angular displacement
K  1/ C
L – Inductance
R – Resistance   i (t )
C – Capacitance
 Electrical Analogous of Mechanical Rotational
System
Torque – Current Analogous Circuit

T – Torque
T  i (t )
J – Moment of Inertia
B – Dash – pot J C
K – Elasticity B  1/ R
Θ – Angular displacement
K  1/ L
L – Inductance
R – Resistance   v(t )
C – Capacitance
 Problem 1: Electrical Analogous of Mechanical
Rotational System
Write the differential equations governing the mechanical
rotational system shown in figure below. Draw the torque-
voltage and torque-current electrical analogous circuits and
verify by writing mesh and node equations.
Problem 1: Electrical Analogous of Mechanical
Rotational System
Torque – voltage analogous electrical network
Problem 1: Electrical Analogous of Mechanical
Rotational System
Torque-current analogous electrical network
1.4 BLOCK DIAGRAM REDUCTION
TECHNIQUES
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.
Basic elements of Block diagram are
Block
Branch point
Summing point
Example - Block Diagram

Block
Summing point

Branch point
Block Diagram Reduction Techniques
The block diagram can be reduced to find the overall transfer
function of the system.
The rules are framed such that any modification made on the
diagram does not alter the input-output relation.
Rules of Block Diagram Algebra
1. Combining the blocks in cascade

2. Combining parallel blocks

Rules of Block Diagram Algebra cont..
3. Moving the branch point ahead of the block

4. Moving the branch point before the block

Rules of Block Diagram Algebra cont..
5. Moving the summing point ahead of the block

6. Moving the summing point before the block

Rules of Block Diagram Algebra cont..
7. Interchanging summing point

8. Splitting summing points

Rules of Block Diagram Algebra cont..
9. Combining summing points

10. Elimination of feedback loop

Problem 1 : Block Diagram Reduction
Reduce the block diagram shown in figure and find the transfer
function.
Problem 2: Block Diagram Reduction
Determine the overall transfer function C(s)/R(s) for the
system shown in figure below.
Problem 3 : Block Diagram Reduction
For the system represented by the block diagram shown in figure
below. Evaluate the closed loop transfer function when he input
R is (i) at station I and (ii) at station II.
 1.5 SIGNAL FLOW GRAPH
It is used to represent the control system in graphical
manner.
It is a diagram that represents a set of simultaneous
linear algebraic equations.
The block diagram approach and signal flow graph
approach gives the same information.
The overall gain can be find easily by using Mason’s
gain formula.
1.5 Signal flow graph (cont…)
Mason’s gain formula

Overall gain,
1
T   PK  K
 K
Where, T = T(s) = Transfer function of the system.
PK = Forward path gain of Kth forward path.
Δ = 1 – {sum of individual loop gains} + {sum of gain products
of all possible combinations of two non-touching
loops} – {sum of gain products of all possible
combinations of three non touching loops} + ……………
ΔK = Δ for that part of the graph which is not touching Kth
forward path
 1.5 Signal flow graph (cont…)
Properties
It is applicable to linear systems only.
The algebraic equations which are used to construct signal
flow graph must be in the form of cause and effect
relationship.
A node represents the variable or signal.
A node adds the signals of all incoming branches and
transmits the sum to all outgoing branches.
A branch indicates functional dependence of one signal on the
other.
The signal travel along branches only in the marked direction.
The signal flow graph of a system is not unique.
Problem 1: Signal flow graph
Find the overall transfer function of the system whose
signal flow graph is shown in figure below.
Problem 2: Signal flow graph

Convert the block diagram to signal flow graph and

determine the transfer function C(s)/R(s).
 1.6 SERVO MOTOR
Servomotor: The servo system is the one, in which the
output is some mechanical variable such as position,
velocity or acceleration. The motors used in the servo
systems are called servomotors.
These motors are usually coupled to the output shaft for
power matching. There are two types of servo motors.
1. DC Servomotors
2. AC Servomotors
 1.6 SERVO MOTOR(contd..)
DC servo motor:

 D.C. servomotors are separately excited or permanent

magnet d.c servomotor .The armature of d.c servomotor has a
large resistance, therefore torque speed characteristics is linear.
 DC servo motors are controlled by DC command signals
applied directly to coils. Time constant for field circuit is large,
due to large time constant, the Response is slow and therefore
they are not commonly used.
 1.6 SERVO MOTOR(contd..)

DC servo motor:
 The magnetic fields that are formed interact with permanent
magnets and cause the rotating member to turn.
 One type of PM uses a wound armature and brushes like a
conventional DC motor, but uses magnets as pole pieces
 Another type uses wound field coils and a permanent magnet
rotor
 1.6 SERVO MOTOR(contd..)

AC Servomotors:

These motor having two parts namely stator and rotor. A.C. Servomotors
are two phase induction motor.
The stator has two distributed windings. These windings are displayed
from each other by 900.
 1.6 SERVO MOTOR(contd..)

AC Servomotors:
 One winding is called main winding or reference winding.
 The reference winding is excited by constant a.c. voltage. Other winding
is called control winding, these winding is excited by variable control
voltage of the same frequency as the reference winding but having a phase
displacement of 900 electrical.
 The variable control voltage for control winding is obtained from a
servo amplifier.
 The rotor of a.c. servomotors are of two types (a)squirrel cage rotor
(b)drag cup type rotor .
 The squirrel cage rotor having large length and small diameter, so its
resistance is very high. The air gap of squirrel cage is kept small. In drag
cup type there are two air gaps.
•
 1.6 SERVO MOTOR(contd..)
AC Servomotors:
For the rotor a cup of nonmagnetic conducting material is used. A
stationary iron core is placed between the conducting cup to complete the
magnetic circuit .
The resistance of drag cup type is high and having high starting torque
.The schematic diagram of two phase a.c. servomotor and the two types
of rotor Controlled by AC command signals applied to the coils.
 AC Brushless Servo Motor Operates on the same principle as
single‐phase induction motor.
 1.7 SYNCHROS
A system consisting of a generator and a motor so connected that
the motor will assume the same relative position as the generator;
the generator and the motor are synchronized
The synchro is a type of rotary electrical transformer that is used
for measuring the angle of a rotating machine.
The term synchro is a generic name for a family of inductive
devices which works on the principle of a rotating transformer
(Induction motor).
 The trade names for synchronous are Selsyn, Autosyn and
Telesyn.
Basically they are electro mechanical devices or electromagnetic
transducer which produces an output voltage depending upon
angular position of the rotor.
 1.7 SYNCHROS (Contd..)
Synchros
A Synchro system is formed by interconnection of the devices
called the synchro transmitter and the synchro control
transformer. They are also called as synchro pair.
The synchro pair measures and compares two angular
displacements and its output voltage is approximately linear with
angular difference of the axis of both the shafts. They can be
used in the following two ways.
i. To control the angular position of load from a remote
place / long distance.
ii. For automatic correction of changes due to disturbance
in the angular position of the load.
1.7 SYNCHROS (Contd..)
Constructional features of synchro transmitter
• Transmitter
1.7 SYNCHROS (Contd..)
Schematic symbol of a synchro transmitter
 1.7 SYNCHROS (Contd..)
Synchro Transmitter
• The constructional features, electrical circuit and a schematic
symbol of synchro transmitter .The two major parts of
synchro transmitters are stator and rotor.
• The stators identical to the stator of three phase alternator. It is
made of laminated silicon steel and slotted on the inner
periphery to accommodate a balance three phase winding.
• The stator winding is concentric type with the axis of the
three coil 120° apart. The stator winding is star connected(Y -
connection).
• The rotor is of dumb bell construction with a single winding.
The ends of the rotor winding are terminated on two slip
rings. A single phase AC excitation voltage is applied to the
rotor through the slip rings.
1.7 SYNCHROS (Contd..)
Working Principles of synchros transmitter:
• When the rotor is excited by AC voltage, the rotor current flows,
and a magnetic field is produced. The rotor magnetic field
induces an emf in the stator coil by transformer action.
• The effective voltage induced in any stator coil depends upon the
angular position of the coils axis with respect to rotor axis
 1.7 SYNCHROS (Contd..)
Schematic control of synchro control transformer
 1.7 SYNCHROS (Contd..)
Synchro Control Transformer:

• The constructional features of synchro control

transformer are similar to that of synchro transmitter,
except the shape of rotor.
• The rotor of the control transformer is made cylindrical
so that the air gap is practically uniform. This feature of
the control transformer minimizes the changes in the
rotor impedance with the rotation of the shaft.
1.7 SYNCHROS (Contd..)
Working of synchros control transformer :
• The generated emf of the synchro transmitter is applied as input
to the stator coils of control transformer. The rotor shaft is
connected to the load whose position has to be maintained at the
desired value.
• Depending on the current position of the rotor and the applied
emf on the stator, an emf is induced on the rotor winding. This
emf can be measured and used to drive a motor so that the
position of the load is corrected.
 1.8 MULTIVARIABLE CONTROL
SYSTEM
Control System

SISO MIMO
SISO:
 One input affect primarily one output and has only weak
effect on other ouputs .
 It is possible to ignore weak interactions (coupling)and
design controllers under the assumption that one input affect only
one output.
 1.8 MULTIVARIABLE CONTROL SYSTEM
MIMO:
It consists of appropriate number of separate SISO systems.
Coupling effects are considered as disturbance and may not cause
Significant degradation in their performance if the coupling is weak .
If it have strong interaction (coupling)if one input affect more than one
output appreciably .
There are two approaches for design of controllers for each system.
Design a decoupling controller to cancel the interactions inherent in
the system .
Design a single controller for multivariable system ,taking into
account .
Examples of multivariable control system are
(i)Automobile driving system
(ii)Antenna stabilization System
 1.8 MULTIVARIABLE CONTROL SYSTEM
(contd..)
(i)AUTOMOBILE DRIVING SYSTEM:

The system to be controlled has two inputs (steering and

Acceleration /braking) and has two controlled outputs (heading and Speed)
Automobile driving system is MIMO System .
It can decouple this system into two SISO System for the purpose of
design.
 1.8 MULTIVARIABLE CONTROL SYSTEM
(contd..)
Antenna Stabilization System (or) Servomechanism for steering of
Antenna :
 1.8 MULTIVARIABLE CONTROL SYSTEM
(contd..)
 The control system has for steering of an antenna can be
treated as two independent System.
(a) Azimuth angle servomechanism
(b) Elevation angle Servomechanism
This is because interaction affect are usually small.
 The occurrence of Azimuth angle causes an error signal to
pass through amplifier, which increases the angular velocity of the
servomotor is a direction towards an error reduction.
 The main disturbance input is the deviation of load from the
nominal estimated value
Example : Effect of Wind Power:
Here the wind power is the disturbance in controlled
system.