Cse UNIT 2

MECH NOTES
DEPARTMENT OF MECHANICAL ENGINEERING
Subject Name: Control System Engineering Subject Code: MET65
UNIT-2
Modeling of Physical Systems – Mechanical, Thermal, Fluid and Electrical systems –
Block Diagram reduction Techniques – Introduction to Time Domain Analysis – Standard Test
Signals
Part –A (2 Marks)
1. What are the three basics elements in electrical and mechanical system?
(Or)
What are the basic elements in control systems?
The basic elements of Electrical system:

▷ Resistor,
▷ Inductor and
▷ Capacitor
The basic elements of mechanical systems are
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▷ In Translational mechanical system

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✓ Mass,
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✓ Spring and
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✓ Dashpot.
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▷ In Rotational mechanical system

✓ Mass with moment of inertia I,
✓ Dash pot with rotational coefficient B and
✓ Torsional spring with stiffness K
2. What are the analogous systems?
The systems for which the differential equations have similar forms are known as analogous
system.
▷ Force-voltage analogy or direct analogy.
▷ Force-current analogy or inverse analogy.
3. What is electrical element analogous to mass in mechanical system?
▷ Inductance is element for mass of force voltage analogy.
▷ Capacitance is the element for mass of force current analogy.
4. What is block diagram? What are the basic components of block diagram?
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.
The basic elements of block diagram are
▷ Block,
▷ Branch,
▷ Point and
▷ Summing point.
5. Give the statement of mason’s gain formula?
Mason’s gain formula states that the overall gain of the system (transfer function) as follows
CONTROL SYSTEM ENGINEERING – MECH DEPT.

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T= T(s) = Transfer function of the system
K= Number of forward path
Pk = Forward path gain of k forward path
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6. What are the two assumptions to be made while deriving transfer function of electrical
systems?
▷ It is assumed that there is no loading, ie. No power us drawn at the output of the system. If
the system has more than one non loading element in tandem, then the transfer function of
each element can be determined independently and the overall transfer function of the
physical system is determining by multiplying the individual transfer functions.
▷ The system should be approximated by a linear lumped constant parameters model by
making suitable assumptions.
7. Define signal flow graph.
A signal flow graph is diagram that represents a set of simultaneous linear algebraic equations.
By taking Laplace transform the time domain differential equations governing a control system
can be transferred to a set of algebraic equations in s-domain. This signal flow graph of the
system can be constructed using these equations.
8. What are the advantages of SFG approach of determining transfer function?
▷ SFG does not require any reduction process because of availability of a flow graph gain
formula which relates the input and output system variables.
▷ A signal flow graph is graphical representation of the relationships between the variable of a
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set of linear algebraic equations.

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▷ SFG method is simple and less time consuming

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▷ Instead of feedback various feedback loops are considered for analysis.

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9. What is meant by robust control?

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Robust control is a branch of control system which deals with the uncertainty because of
disturbance and other factors in its approach to the controller design. Its aim is to achieve robust
performance and stability in the presence of the small modelling errors. It measures
performance changes of a control system with changing system parameters for designing
proper controllers.
10. Explain the disadvantages and advantages of block diagram reduction process over signal
flow graph?
Advantages:
▷ Block diagrams can be easily visualized.
▷ It is identical to physical system.
▷ It consists of blocks and each block represents a function of each components of a system.
Disadvantages:
▷ Block diagram reduction is not a generalized procedure like signal flow graph.
▷ Block diagram reduction is difficult to simplify than signal flow graph if the system has complex
structures.
▷ Many rules block diagram reduction depends upon the complexity of the system on the other
hand signal flow graph has systematic approach.
11. Write the force balance equations of ideal dashpot and ideal spring?

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Part –B
1. Discuss about the mathematical model of physical system in control system engineering?
Mathematical Models of Physical Systems

Mathematical models of physical systems are used to design and analyze control systems.
Mathematical models are described by ordinary differential equations. If the coefficients of the
describing differential equations are function of time, then the mathematical model is linear
time-varying. On the other hand, if the coefficients describing differential equations are constants,
the model is linear time-invariant. The differential equations describing a LTI system can be
reshaped into different forms for the convenience of analysis. For transient response or frequency
response analysis of single-input-single-output linear systems, the transfer function representation is
convenient. On the other hand, when the system has multiple inputs and outputs, the vector-matrix
notation may be more convenient.
Powerful mathematical tools like Fourier and Laplace transforms are available for linear
systems. Unfortunately no physical system in nature is perfectly linear. Certain assumptions must
always be made to get a linear model. In the presence of strong nonlinearity or in presence of
distributive effects it is not possible to obtain linear models. A commonly adopted approach is
to build a simplified linear model by ignoring certain nonlinearities and other physical properties that
may be present in a system and thereby get an approximate idea of the dynamic response of the
system. A more complete model is then built for more complete analysis.
Transfer function
The transfer function of an LTI system is the ratio of Laplace transform of the output variable
to the Laplace transform of the input variable assuming zero initial conditions. Following are some
examples of how transfer functions can be determined for some dynamic system elements.
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The insertion of an isolation amplifier between the two RC-circuits will produce no loading effect.
A. Transfer Function of Armature Controlled DC Motor

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In Laplace domain,
Neglecting , ;
where, and
and are called the motor gain and time constant respectively. These two parameters are
usually supplied by the
manufacturer.
The block diagram model is,
B. Transfer Function of a Field-controlled DC Motor

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We obtain,
2. Discuss about the basics of mechanical translational system in details and also state the
procedure to obtain transfer function
MECHANICAL TRANSLATIONAL SYSTEM
The basic elements of mechanical translational system in control system is given three
elements are
✓ Mass,
✓ Spring and
✓ Dash-pot

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List of symbols used
X= Displacement (m)
f= applied force (Newton (N))
fm= opposing force offered by mass of the body (N)
fk= opposing force offered by the elasticity of the body (spring), (N)
fb= opposing force offered by the friction of the body (dash-pot), (N)
M= Mass, kg
K= stiffness of spring, (NM)
B= Viscous friction co-efficient, (N-sec/m)
Force-balance equations of idealized elements
1. Mass
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Let a force be applied on the mass. The mass will offer an opposing force which is
proportional to acceleration of the body.
Let f=applied force
fm= opposing force due to mass
By Newton’s second law,
2. Dash-pot
Let a force be applied on dash-pot. The dash-pot will offer an opposing force which is
proportional to velocity of the body.
Let f=applied force
fm= opposing force due to friction

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If the dash-pot has displacement at both ends, the opposing force is proportional to differential
velocity.
3. Spring
Let a force be applied on spring. The spring will offer an opposing force which is proportional
to displacement of the body.
Let f=applied force
fm= opposing force due to elasticity

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If the spring has displacement at both ends, the opposing force is proportional to differential
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velocity.
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Procedure to determine the transfer function of mechanical translational system

The differential equations governing the system is obtained by writing force balance equations at
the nodes in the system
▷ Mass elements are considered as nodes in the system. In some cases nodes may be with
mass element (spring and dash-pot combination)
▷ Let as assign the linear displacement of the masses be x1, x2,… and assign displacement to
each mass node.
▷ The first derivative of the displacement is velocity and the second derivative of the
displacement is acceleration.
▷ Draw the free body diagrams of the system.
▷ The free body diagram is obtained by drawing each mass separately and then marking all the
forces acting on that mass (node).
▷ The mass has to move in the direction of applied force and then the opposing force acts in a
direction opposite to applied force.
▷ The displacement, velocity and acceleration of the mass will be in the direction of the applied
force.
▷ If there is no applied force, the displacement, velocity and acceleration of the mass in a
direction opposite to that of applied force.
▷ For each free body diagram, write on differential equation by equating the sum of applied
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forces to the sum of opposing forces.
▷ Take Laplace transform of differential equations to convert them to algebraic equations.
Then rearrange the s-domain equations to eliminate the unwanted variables and obtain ratio
between output variable and input variable (transfer function).
3. Write the differential equations governing the mechanical system shown in Figure below. And
determine the transfer function?
In the given system, applied force f(t) is the input and displacement X is the output.
Let, Laplace transfer of f(t)=
Laplace transfer of x =
Laplace transfer of X1=
Hence the required transfer function is
Free body diagram-1

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+
On taking Laplace transform of above equation with zero initial conditions we get,
=0
Free body diagram-2

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;
+ =
Substituting X1(s) from equation (1) in equation (2) we get,
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4. Determine the transfer function of the system shown below.

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Let, Laplace transfer of f(t) =

Laplace transfer of Y1=
Laplace transfer of Y2=
The system has two nodes and they are masses M1 and M2. The
differential equations governing the system are the force balance
equations at these nodes.
Free body diagram-1

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Free body diagram-2
Substituting for Y1(S) from equation (3) in equation (2) we get,

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5. Discuss about the basics of mechanical rotational system in details and also state the
procedure to obtain transfer function
MECHANICAL ROTATIONAL SYSTEMS

The model of rotational mechanical systems can be obtained by using three elements
▷ Moment of inertia of mass (J)
▷ Dash-pot with frictional coefficient (B)
▷ Torsional spring stiffness (K)
List of symbols used in mechanical rotational system
θ- Angular displacement (rad)
- Angular velocity (rad/sec)
–Angular acceleration, (rad/sec )

2
T-Applied torque, Nm
J-Moment of inertia, Kgm /rad
2
B- Rotational frictional co-efficient, Nm/rad/sec

K- Stiffness of the spring, Nm/rad
Force-balance equations of idealized elements

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1. Mass
Consider an ideal mass element which has negligible friction and elasticity. The opposing
torque due to moment of inertia is proportional to the angular acceleration.
Let T=applied torque

Tj= opposing torque due to moment of inertia of the body
2. Dash-pot
Consider an ideal frictional element dashpot which has negligible moment of inertia and
elasticity. Let a torque be applied on it. The dashpot will offer an opposing force which is
proportional to the angular velocity of the body.
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Tb= opposing torque due to friction

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When the dashpot has angular displacement at both ends, then the opposing force is proportional to
differential angular velocity.
3. Spring
Consider an ideal frictional element dashpot which has negligible moment of inertia and
friction. Let a torque be applied on it. The torsional spring will offer an opposing force which is
proportional to the angular displacement of the body.

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Tk= opposing torque due to elasticity
When the spring has angular displacement at both ends, then the opposing force is proportional to
differential angular displacement.
6. Write the differential equations governing the mechanical rotational system as shown in fig,
obtain the transfer function of the system.
In the given system, applied force f(t) is the input and displacement X is the output.
Let, Laplace transfer of T =
Laplace transfer of x =
Laplace transfer of X1=
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The system has two nodes and they are mass J1and J2, the differential equations governing
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the system are given by torques balance equations at these nodes. Let the displacement of mass J1
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be 1 . The free body diagram of J1 is shown in fig. the opposing forces acting on J1are marked as Tj,
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and Tk.
Free body diagram-1
On taking Laplace transform of above equation we get,
Free body diagram-2

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Substitute from equation 2 in equation 1 we get,
7. Discuss the basic of thermal system in detail
THERMAL SYSTEMS
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Thermal systems are those systems in which heat transfer takes place from one substance
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to another. They can be characterised by thermal resistance and capacitance, analogous to

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electrical resistance and capacitance. Thermal system is usually a non linear system and since the
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temperature of a substance is not uniform throughout the body, it is a distributed system. But for
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simplicity of analysis, the system is assumed to be linear and is represented by lumped parameters.
(a) Thermal Resistance
There are two types of heat flow through conductors: Conduction or Convection and
Radiation
For conduction of heat flow through a specific conductor, according to Fourier law,
Where,
q= Heat flow, Joules/Sec
K= Thermal conductivity, J/sec/m/deg k
A= Area normal to heat flow, m 2
∆X= Thickness of conductor, m

θ= Temperature in K0
For convection heat transfer,

q= HA (θ1-θ2)
Where,
H=Convection coefficient, J/m /sec/deg k
2

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The thermal resistance is defined by,
The unit of R is deg sec / J

For radiation heat transfer, the heat flow is governed by Stefan-Boltzmann law for a surface
receiving heat radiation from a black body:
Where,
σ is a constant, (5.6697 x 10 -8 j/sec/m /K )
2 4
K is a constant
E is emissivity
A is surface in m 2
The radiation resistance is given by
Where, θa is the average temperature of radiator and receiver.

(b) Thermal Capacitance
Thermal capacitance is the ability to store thermal energy. If heat is supplied to a body, its
internal energy raises. For the system shown in below,
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Where,
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C is the thermal Capacitance

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C=WCp
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Where,
W is weight of the block in kg
Cp is specific heat at constant pressure in J/ deg/kg
8. Find the transfer function of the thermal system shown in the below figure. Heat is
supplied by convection to a copper rod of diameter D.
The thermal resistance of the copper rod,
Here A is the surface area of the rod

Hence A = πDL
Where L is the length of the rod
The thermal capacitance of the rod, is given by

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C= WCP
Where, Cp is the specific heat of copper

ρ is the density of copper
We have, q = HA (u - c)
Combining these two equations,
Where RC = T is the time constant of the system
But
Thus the transfer function of the system is,

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9. Discuss the basic of fluid system in detail

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FLUID SYSTEMS
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Fluid systems are those systems in which liquid or gas filled tanks are connected through
pipes, tubes, orifices, valves and other flow restricting devices. Compressibility of a fluid is an
important property which influences the performance of fluid systems. If the velocity of sound in
fluids is very high, compared to the fluid velocity, the compressibility can be disregarded. Hence
compressibility effects are neglected in liquid systems. However compressibility plays an important
role in gas systems. The type of fluid flow, laminar or turbulent, is another important parameter in
fluid systems. If the Reynolds number is greater than 4000, the flow is said to be turbulent and if the
Reynolds number is less than 2000, it is said to be laminar flow. For turbulent flow through pipes,
orifices, valves and other flow restricting devices, the flow is found from Bernoulli's law and is given
by
Where, q = liquid flow rate, m3/ sec

K = flow constant
A= area of restriction, m2
g= acceleration due to gravity, m/sec2
h= head of liquid, m
The turbulent resistance is found from
It can be seen that the flow resistance depends on hand q and therefore it is non linear. It has
to be linearised around the operating point and used over a small range around this point. The
laminar flow resistance is found from the Poisseuille-Hagen law:

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Where, h= head, m
L= length of the tube, m
D = inside diameter of the pipe, m
q= liquid flow rate, m3/sec
m= absolute viscosity, kg-sec/m2
γ = fluid density kg/m3
Since the flow is laminar, head is directly proportional to the flow rate and hence, laminar flow
resistance is given by
Liquid storage tanks are characterized by the capacitance and is defined by
Where, v = volume of the liquid tank in m3. Hence the capacitance of a tank is given by its area of
cross section at a given liquid surface.
10. Obtain the transfer function for the liquid level system shown in the below figure
The capacitance of the vessel is C=A

Where A is the area of cross section of the vessel
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The outflow q is equal to

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Where R is the laminar flow resistance of the valve
=
11. Briefly explain about the basics of electrical system used in control system
ELECTRICAL SYSTEMS
The basic elements of electrical system used in control system are
▷ Resistor,
▷ Capacitor and
▷ Inductor
The electrical systems can be modelled by using resistor, inductor, capacitors and voltage or current
source. Kirchoff’s voltage and current laws are used to frame the differential equations for the
electrical systems.
Current voltage relation of R, L and C
Element Voltage across the element Current through the element

Resistor V(t)= Ri(t)

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Inductor
Capacitor
12. Obtain the transfer function of the electrical network shown below
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In the given network, input is e(t) and V2(t) is the output.

Let, Laplace transform of e(t)=
Laplace transform of V2(t) =

Transform the voltage source in series with resistance R1 into equivalent current source. The
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network has two nodes. Let the node voltages be V1 and V2. The Laplace transform of node voltages
V1 and V2are V1(s) and V2(s) respectively.
At node 1, by KCL
At node 2, by KCL
On substituting V1(s) in equation (1)

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13. Obtain the Transfer function of armature controlled DC motor.
Ra = Armature resistance, Ω
La = Armature Inductance, H
Ia = Armature current, A
Va = armature voltage, V
Eb = back emf, V,Kt= Torque constant, N-m/A
T = Torque developed by motor, N-m
θ = Angular displacement of shaft, rad
J = Moment of inertia of motor and load, Kg-m2
B = Frictional coefficient of motor and load, N-m/(rad/sec)
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Kb = Back emf constant, V/(rad/sec)
By Kirchhoff’s voltage law we can write
Torque of the DC motor is proportional to the product of flux and current. Since flux is constant, the
torque is proportional to ia alone.
The mechanical system is shown in figure below
The differential equation governing the mechanical system is given by
The back EMF of dc motor is proportional to speed of the shaft.

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The Laplace transform of various time domain signals involved in this system are
Let Laplace transform of Va(t) =
Let Laplace transform of eb(t) =
Let Laplace transform of T=
Let Laplace transform of ia(t) =
Let Laplace transform of θ =

The differential equation governing the armature controlled DC motor speed control system is

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On equating (6) and (7)
Equation (5) can be written as
Substituting (8) and (9) in (10)

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The require transfer function is
The transfer function of armature controlled dc motor can be expressed in another form

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Where
14. Derive the transfer function of field controlled DC motor.

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Let
Rf = Field resistance, Ω
Lf = Field inductance, H
If = Field current, A
Vf = Field voltage, V
T = Torque developed by motor, N-m
Ktf = Torque constant, N-m/A
J = Moment of inertia of rotor and load, Kg-m /rad

2
B = Frictional coefficient of rotor and load, N-m/(rad/sec)

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The equivalent circuit of the field circuit is shown above
By KVL,
Torque of the dc motor is proportional to the product of flux and current. Since armature current is
constant, the torque is proportional to flux alone, but flux is proportional to field current.
The mechanical system is shown in figure below

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The differential equation governing the mechanical system is given by

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The Laplace transform of various time domain signals involved in this system are
Let Laplace transform of if(t) =
Let Laplace transform of T=
Let Laplace transform of vf(t) =
Let Laplace transform of θ =
The differential equation governing the armature controlled DC motor speed control system is

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On equating (5) and (6)
Equation (5) can be written as
Substituting (7) in (8)

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Where
15. Brief in detail about the force voltage and force current analogous system in detail
Electrical analogous of mechanical translational systems
The three basic elements mass, dash-pot and spring that are used in modelling mechanical
translational systems are analogous to resistance, inductance and capacitance of electrical
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systems. The input force in mechanical system is analogous to either voltage source or current
source in electrical systems. The output velocity in mechanical systems is analogous to either
current or voltage in electrical systems. The electrical system has two types of inputs either voltage
or current source, there are two types of analogies:
▷ force-voltage analogy and
▷ force-current analogy.
Guidelines to obtain Force- voltage analogy
1. The elements in series in electrical systems have same current, while the elements in
mechanical systems having same velocity are said to be in series.
2. The elements having same velocity in mechanical systems should have analogous same current
in electrical analogous systems
3. Each node in the mechanical system corresponds to a closed loop in electrical system. A mass
is considered to be node.
4. The number of meshes in electrical system is same as that of the number of masses in
mechanical system. Hence the number of mesh currents and system equation will be same as
that of the number of velocities of nodes in mechanical system.
5. The mechanical driving sources (forces) and passive elements connected to the node (mass) in
mechanical system should be represented by analogous elements in a closed loop in analogous
electrical system
6. The elements connected between two (nodes) masses in mechanical system is represented as a
common element between two meshes in electrical analogous system
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Analogous elements in force-voltage analogy
Mechanical systems Electrical system

Input: Force Input: voltage source
Output: Velocity Output: current through the element

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Analogous quantities in force-voltage analogy
Item Mechanical system Electrical system

Independent variable Force, f Voltage e,v
(input)
Dependent variable Velocity, v Current, i
(output)
Dissipative element Frictional coefficient of dashpot, Resistance, R
B
Storage element Mass, M Inductance, L
Stiffness of spring, K Inverse of capacitance, 1/C
Physical law
Newton’s second law Kirchoff’s voltage law
Changing the level of Lever Transformer
independent variable
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Analogous elements in force-current analogy

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Input: Force Input: Current source

Output: Velocity Output: Voltage across the element

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Analogous quantities in force-current analogy

Independent variable Force, f Current, i
(input)
Dependent variable Velocity, v Voltage, v
(output) Displacement, x Flux, ϕ
Dissipative element Frictional coefficient of Conductance G= 1/R
dashpot, B
Storage element Mass, M Capacitance, C
Stiffness of spring, K Inverse of inductance, 1/L
Physical law
Newton’s second law Kirchoff’s current law
Guidelines to obtain Force- current analogy
1. The elements in parallel in electrical systems have same voltage, while the elements in
mechanical systems having same force are said to be in parallel.
2. The elements having same velocity in mechanical systems should have analogous same voltage
3. Each node in the mechanical system corresponds to a node in electrical system. A mass is
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considered to be node.
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4. The number of meshes in electrical system is same as that of the number of nodes(masses) in
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mechanical system. Hence the number of node voltage and system equation will be same as
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that of the number of velocities of nodes in mechanical system.
5. The mechanical driving sources (forces) and passive elements connected to the node (mass) in
mechanical system should be represented by analogous elements connected to a node in
electrical system.
6. The element connected between two nodes (masses) in mechanical system is represented as a
common element between two meshes in electrical analogous system.
16. For the mechanical system shown in fig .Draw the mechanical network diagram and hence
write the differential equations describing the behaviour of the system. Draw the force-voltage
and force-current analogous electrical circuits.
For the given diagram, writing the differential equation for mass M1
Force voltage analogous circuit

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The electrical analogous elements for the elements of mechanical system is given by
f(t) = e(t) M1 → L1 B1 → R1 K1 → 1/C1
v1 = i1 M2 → L2 B2 → R2 K2 → 1/C2 B12 → R12
The mesh basis equations using Kirchhoff’s voltage law for the circuit shown is
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Force current analogous circuit

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The electrical analogous elements for the elements of mechanical system is given by
f(t) = i(t) M1 → C1 B1 → 1/R1 K1 → 1/L1
v1 → v1 M2 → C2 B2 → 1/R2 K2 → 1/L2
v2 → v2 B12 → 1/R12

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The node basis equations using Kirchhoff’s current law for the circuit is
7.
17. Write the differential equations for the mechanical system in fig. Also obtain an analogues
electrical circuit based on force-current and force voltages analogous
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Free body diagram-1

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+
- =
Free body diagram-2

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-
=
Substituting from equation 2 in equation 1 we get,

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From equation 2 we get,
Substitute equation3 in equation1 we get,
- =
The different equations of two nodes are
On replacing the displacements by velocity in the differential equations

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Force-voltage analogous circuit
The electrical analogous elements for the elements of mechanical system are given below.
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The mesh basis equations using Kirchhoff’s voltage law for the circuit shown in fig are given
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below .
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It is observed that the mesh basis equations are similar to the differerential equations governing
the mechanical system.
Force –current analogous circuit

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The node basis equations using Kirchhoff’s current law for the circuit shown in fig are given
below .
It is observed that the node basis equations are similar to the differential equations governing
the mechanical system.
18. Explain in detail about the electrical analogous of mechanical rotational system
ELECTRICAL ANALOGOUS OF MECHANICAL ROTATIONAL SYSTEMS

Mechanical rotational system basic elements
▷ Moment of inertia,
▷ Dashpot and
▷ Torsional spring
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Electrical analogous to these elements-

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▷ resistance,
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▷ inductance and
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▷ capacitance
Input torque- voltage source or current source
The angular velocity- current or voltage in an element
Two types of analogous-

✓ toque- voltage analogy and
✓ torque- current analogy
Torque -voltage analogy

Analogous elements in force-voltage analogy

Input: Force Input: voltage source
Output: Velocity Output: current through the element

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Guidelines to obtain Force- voltage analogy
1. The elements in series in electrical systems have same current, while the elements in
mechanical systems having same velocity are said to be in series.
2. The elements having same velocity in mechanical systems should have analogous same current
3. Each node in the mechanical system corresponds to a closed loop in electrical system. The
moment of inertia of mass is considered to be node.
4. The number of meshes in electrical system is same as that of the number of nodes (moment of
inertia of masses) in mechanical system. Hence the number of mesh currents and system
equation will be same as that of the number of velocities of nodes in mechanical system.
5. The mechanical driving sources (Torque) and passive elements connected to the node (moment
of inertia of mass) in mechanical system should be represented by analogous elements in a
closed loop in analogous electrical system
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6. The elements connected between two nodes (moment if inertia of masses) in mechanical
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system is represented as a common element between two meshes in electrical analogous

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system
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Analogous quantities in force-voltage analogy

Independent variable (input) Torque, T Voltage e,v
Dependent variable (output) Angular Velocity, ω Current, i
Angular displacement, θ Charge, q
Dissipative element Rotational coefficient of Resistance, R
dashpot, B
Storage element Moment of inertia, J Inductance, L
Stiffness of spring, K Inverse of capacitance, 1/C
Physical law Newton’s second law Kirchoff’s voltage law

Analogous elements in force-current analogy

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Input: Force Input: current source
Output: Velocity Output: voltage across the element
Analogous quantities in force-current analogy

Independent variable Torque, T Current, i
(input)
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Dependent variable Angular Velocity, ω Voltage, v

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(output) Angular displacement, θ Flux, ϕ

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Dissipative element Rotational coefficient of Conductance G= 1/R

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dashpot, B
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Storage element Moment of inertia, J Capacitance, C

Stiffness of spring, K Inverse of inductance, 1/L
Physical law
Newton’s second law Kirchoff’s current law
Guidelines to obtain torque-current analogy
1. The elements in parallel in electrical systems have same voltage, while the elements in
mechanical systems having same force are said to be in parallel.
2. The elements having same angular velocity in mechanical systems should have analogous same
voltage in electrical analogous systems
3. Each node in the mechanical system corresponds to a node in electrical system. The moment of
inertia of mass is considered to be node.
4. The number of nodes in electrical system is same as that of the number of nodes(moment of
inertia of mass) in mechanical system. Hence the number of node voltage and system equation
will be same as that of the number of velocities of nodes in mechanical system.
5. The mechanical driving sources (torque) and passive elements connected to the node (mass) in
mechanical system should be represented by analogous elements connected to a node in
32
electrical system.
6. The element connected between two nodes (moment of inertia of mass) in mechanical system
is represented as a common element between two meshes in electrical analogous system.
19. Write the different equations governing the mechanical rotational system shown in fig draw
the both the electrical analogues circuits.
Free body diagram-1
By Newton’s second law, +

es
Free body diagram-2

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By Newton’s second law, +
On replacing angular displacement by angular velocity in the differential equations 1 and 2
Torque- Voltage Analogous Circuit:

33
The mesh basis equations using Kirchhoff’s voltage law for the circuit shown in fig are given below.
It is observed that the mesh basis equations 5 and 6 are similar to the differential equations 3 and 4
governing the mechanical system.
Torque- Current Analogous Circuit:
es
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The node basis equations using Kirchhoff’s current law for the circuit shown in fig are given below
It is observed that the node basis equations 7 and 8 are similar to the differential equations 3 and 4
governing the mechanical system.
20. Describe the rules followed in the block diagram reduction
RULES OF BLOCK DIAGRAM ALGEBRA

Rule-1: Combining the blocks in cascade
Rule-2: Combining the blocks in parallel

34
Rule-3: Moving the branch point after the block
Rule-4: Moving the branch point before the block
Rule-5: Moving summing point after the block
Rule-6: Moving summing point before the block

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Rule-7: Interchanging summing point
Rule-8: Splitting summing point
Rule-9: Combining summing points

35
Rule-10: Elimination of negative feedback loop
Rule-11: Elimination of positive feedback loop

es
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21. Using block diagram reduction technique find closed loop transfer function of the system
whose block diagram is shown in figure.
Step 1: Moving the branch point before the block
Step 2: Combining the blocks

in cascade and eliminating parallel blocks

36
Step 3: Moving summing point before the block
Step 4: Interchanging summing points and modifying branch points

es
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Step 5: Eliminating the feedback path and combining blocks in cascade
Step 6: Eliminating the feedback path

37
22. Obtain the closed loop transfer function C(s)/R(s) of the system whose block diagram is
shown in figure.
es
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Step 1: Splitting the summing point and rearranging the branch points
Step 2: Eliminating feedback path

38
Step 3: Moving the branch point after the block
Step 4: Combining the blocks in cascade and eliminating feedback path

es
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Step 5:Combining the blocks in

cascade and eliminating
feedback path

39
Step7: Combining the blocks in cascade
23. Explain about time domain specification with neat diagram
TIME DOMAIN SPECIFICATIONS

The desired performance characteristics of control systems are specified in terms of time
domain specifications. Systems with energy storage elements cannot respond instantaneously and
will exhibit transient responses whenever they are subjected to inputs or disturbances. The desired
performance characteristics of a system of any order may be specified in terms of the transient
response to a unit step input signal. The response of a second order system for unit step input for
various values of damping ration is shown in fig
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The transient response of the practical control system often exhibits damped oscillation before
reaching steady state. A typical damped oscillatory response of the system is shown in the
following
The transient response characteristics of a control system to a unit step input is specified in terms
of the following time domain specifications
▷ Delay time td
▷ Rise time tr
▷ Peak time tp
40
▷ Maximum overshoot Mp
▷ Settling time ts
The time domain specifications are defined as follows:
1.Delay time (td): It is the time taken to reach 50% of the final value for the very first time
2. Rise time (tr): It is the time taken to reach 0 to 100% of the final value for the very first time.
In under damped systems, the rise time is calculated from 0 to 100%. But
for over damped system it is the time taken by the response to raise form
10% to 90%. For critically damped system it is the time taken for response
to raise from 5% to 95%
3.Peak time (tp) : It is the time taken to reach peak value for the very first time or It is the time
taken for the response to reach the peak overshoot (Mp).
4.Maximum overshoot (Mp): It is otherwise called as peak overshoot. It is defined as the ratio
of the maximum peak values to the final value where the maximum peak value is measured
form final value.
Let c ( = final value of c(t)

c (tp) = maximum value of c(t)
Now peak overshoot =
% peak overshoot % = 100

5.Settling time (ts): It is defined as time taken by the response to reach and stay within a
specified error. It is usually expressed as % of final value. The usual
tolerable error is 2% or 5% of the final value.
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24. Derive the expressions for time domain specifications

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RISE TIME ( )
M
The unit step response of second order system for under damped case is given by C(t) = 1 -
At t = tr , c(t) = c( =1
c( =1- =1
=0
Since 0, the term =0
When = 0, ....... =0
= =
Rise time =
Here = ; damped frequency of oscillation,

41
Rise time = in sec.
or should be measured in radians
PEAK TIME ( )
To find the expression for peak time differentiate c(t) with respect to t and equate to 0. i.e,
c(t) =0
The unit step response of under damped second order system is given by
c(t) = 1 -
Differentiating c(t) with respect to t
c(t) = ( +(
Put
c(t) = ( +(
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= [ -
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= [
= [ ]=
at t= , c(t) = 0 [on constructing right angle triangle with and
we get =
= ]
=0
Since 0, the term =0
When = 0, ....... =0
Peak time (tp) =
The damped frequency of oscillation,
Peak time (tp) =

42
PEAK OVERSHOOT (MP)
% peak overshoot % = 100
Where c( = final steady state value
c(tp) = peak response at t =

The unit step response of second order system is given by
c(t) = 1 -
At t = , c(t) = c( = 1- = 1-0 =1
At t= c(t) = c(tp) = 1-
= 1- [(tp) =
= 1- = 1+ [
= ]
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=1+ = 1+
M
% of peak overshoot % = 100
= 100 = 100
SETTLING TIME ( )
The response of second order system has two components. They are,
1. Decaying exponential component
2. Sinusoidal component,
In this the decaying exponential term dampens or reduces the oscillations produced buy sinusoidal
hence the settling time is decided by the exponential component. The settling time can be by
equating exponential component to percentage tolerance errors.
2% tolerance error band at t = , = 0.02
For least values of , = 0.02

On taking natural logarithm we get,
43
= ln(0.02) = -4 =
For the second order system the time constant, T =
Settling time (ts) = = 4T (for 2% error)
For 5% error, = 0.05

On taking natural logarithm we get,
= ln(0.05) = -3 =
Settling time (ts) = = 3T (for 5% error)

In general for a specified percentage error settling time can be evaluated using equating
Settling time (ts) = =

25. Discuss about the test standard test signals used in control system
TEST SIGNALS
The test signal is nothing but the input signal which is required to predict the response of a
system. The characteristics of actual input signals are a sudden change in magnitude, a constant
velocity and a constant acceleration.
The commonly used test input signals are:
es
▷ Impulse
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▷ Step
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▷ Ramp
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▷ Acceleration and
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▷ Sinusoidal signals
1. STEP SIGNAL:
The step signal is a signal whose value changes from zero to A ( a constant value ) at t=0
and remains constant at A for t>0.
Mathematical representation of the step signal is,
✓ R(t) = A; t≥0
✓ R(t) = 0; t<0
2. RAMP SIGNAL:
The ramp signal is a signal whose value increases linearly with time from an initial value of
zero at t=0.
Mathematical representation of the ramp signal is,
✓ R(t) = At; t≥0
✓ R(t) = 0; t<0
3. PARABOLIC SIGNAL:
The parabolic signal is one, in which the instantaneous value varies as square of the time an
44
initial value of zero at t=0. The sketch of the signal with respect to time resembles a parabola.
The mathematical representation of the parabolic signal is,
✓ R(t) = (At )/2; t≥0
2
✓ R(t) = 0; t<0
4. IMPULSE SIGNAL:
This signal is available for a very short duration is called impulse signal.
The mathematical representation is given as,
✓ δ(t) = ∞; t=0
✓ δ (t) = 0; t≠0
STANDARD TEST SIGNALS:

Name of the signal Time domain equation of Laplace transform of the signal,
signal, r(t) R(s)
Step A A/s
Unit step 1 1/s
Ramp At A/s 2
Unit ramp t 1/s2

Parabolic At /2
2 A/s 3
Unit parabolic t2/2 1/s3

es
Impulse δ(t) 1
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Reference:
▷ N.C. Jagan – “Control Systems” – BS Publications (2008).

▷ R. AnandaNatrajan and P. Ramesh Babu, “Control Systems Engineering”, fourth edition, Sci Tech
Publications (India) Pvt. Limited, Chennai, 2013.
▷ I.J.Nagrath and M.Gopal, “Control Systems Engineering”, fifth edition, New age international (p)
Limited, New Delhi, 2007.
▷ A.Nagoor kani, Control System Engineering” First edition, RBA publications, Chennai

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MECH NOTES

DEPARTMENT OF MECHANICAL ENGINEERING

Subject Name: Control System Engineering Subject Code: MET65

The basic elements of Electrical system:

▷ In Translational mechanical system

▷ In Rotational mechanical system

CONTROL SYSTEM ENGINEERING – MECH DEPT.

set of linear algebraic equations.

▷ SFG method is simple and less time consuming

▷ Instead of feedback various feedback loops are considered for analysis.

9. What is meant by robust control?

CONTROL SYSTEM ENGINEERING – MECH DEPT.

Mathematical Models of Physical Systems

A. Transfer Function of Armature Controlled DC Motor

CONTROL SYSTEM ENGINEERING – MECH DEPT.

The block diagram model is,

B. Transfer Function of a Field-controlled DC Motor

CONTROL SYSTEM ENGINEERING – MECH DEPT.

f= applied force (Newton (N))

fm= opposing force offered by mass of the body (N)

K= stiffness of spring, (NM)

B= Viscous friction co-efﬁcient, (N-sec/m)

Force-balance equations of idealized elements

By Newton’s second law,

By Newton’s second law,

CONTROL SYSTEM ENGINEERING – MECH DEPT.

By Newton’s second law,

By Newton’s second law,

By Newton’s second law,

Procedure to determine the transfer function of mechanical translational system

Hence the required transfer function is

Free body diagram-1

By Newton’s second law,

Free body diagram-2

CONTROL SYSTEM ENGINEERING – MECH DEPT.

Substituting X1(s) from equation (1) in equation (2) we get,

4. Determine the transfer function of the system shown below.

Let, Laplace transfer of f(t) =

By Newton’s second law,

CONTROL SYSTEM ENGINEERING – MECH DEPT.

Free body diagram-2

By Newton’s second law,

Substituting for Y1(S) from equation (3) in equation (2) we get,

MECHANICAL ROTATIONAL SYSTEMS

- Angular velocity (rad/sec)

–Angular acceleration, (rad/sec )

B- Rotational frictional co-efﬁcient, Nm/rad/sec

CONTROL SYSTEM ENGINEERING – MECH DEPT.

Let T=applied torque

By Newton’s second law,

Tb= opposing torque due to friction

By Newton’s second law,

By Newton’s second law,

Let T=applied torque

By Newton’s second law,

Hence the required transfer function is

By Newton’s second law,

On taking Laplace transform of above equation we get,

Free body diagram-2