Control system mathematical modelling of a system

Control System
Control Components - Mathematical Modeling of Systems
Dr. Nilesh Bhaskarrao Bahadure
nbahadure@gmail.com
https://www.sites.google.com/site/nileshbbahadure/home
June 29, 2021
Dr. Nilesh Bhaskarrao Bahadure (Ph.D.) Control System June 29, 2021 1 / 65

Overview I
1 Introduction
2 Translational Motion
Mass
Spring
Dashpot or Damper
3 Rotational Motion
Moment of Inertia
Torsional Spring
Dashpot
Analogous Elements
4 Electrical Elements
Resistor
Inductor
Capacitor
5 Analogous System
Force - Voltage Analogy

Overview II
Torque Voltage Analogy
Force - Current Analogy
6 Steps to solve problems on analogous systems
7 Problems
8 Questions
9 Thank You

What is Mathematical Model
The mathematical model of a system is a mathematical relation which
relates the input, the system, and the output. This relation must be such
as to guarantee that one can determine the system’s output for any given
input.
From the above definition it follows that the mathematical model is not
just any relation, but a very special relation, which offers the capability of
system analysis, i.e., the capability to determine the systems response
under any excitation. Furthermore, the foregoing definition reveals the
basic motive for determining mathematical models. This motive is to have
available appropriate tools that will facilitate the system analysis. It should
be also noted that the mathematical model is useful for other purposes, as
for example to study the systems stability and other properties, to improve
the system’s performance by applying control techniques, etc.

Mathematical Model...
Figure : A system, its input, and its output.
Most of the control systems contain mechanical or electrical or both types
of elements and components. To analyse such systems, it is necessary to
convert such systems into mathematical models based on transfer function
approach. From mathematical angle of view, models of mechanical and
electrical components are exactly analogous to each other.

Analysis of Mechanical System
In mechanical systems, motion can be of different types i.e. Translational,
Rotational, or combination of both. The equations governing such motion
in mechanical systems are often directly or indirectly governed by Newtons
laws of motion.

Translational Motion
Translational Motion
Motion can either be translational, rotational, oscillatory or just random.
It would be translational if there is a change in place, oscillatory when it’s
moving between the two points to and fro and rotational if the body is
spinning. Many illustrations we see in our daily life like a boy kicks a ball,
a bullet fired from the gun, a sportsman throwing the short put. These
show the translational motion. Here in this actions, the movement of
object shifts from one point to another. Translational motion may take
place along the body even in the curved path or straight path. Let’s see
more about translational motion on this page.
The following elements are dominantly involved in the analysis of
translational motion.
1 Mass (inertia) element
2 Spring (elastic) element
3 Dashpot or Damper (frictional) element

Translational Motion...
Translational Motion...
1 Spring stores energy as potential energy
2 Mass stores energy as kinetic energy
3 Damper dissipates energy into heat
If a force is applied to a translational mechanical system, then it is
opposed by opposing forces due to mass, elasticity and friction of the
system. Since the applied force and the opposing forces are in opposite
directions, the algebraic sum of the forces acting on the system is zero.
Let us now see the force opposed by these three elements individually.

Mass
Mass
Mass is the property of a body, which stores kinetic energy. If a force is
applied on a body having mass M, then it is opposed by an opposing force
due to mass. This opposing force is proportional to the acceleration of the
body. Assume elasticity and friction are negligible.
Figure :

Mass...
Fm ∝ a
Fm = M × a = M
d2x
dt2
F = Fm = M
d2x
dt2
where,
F is the applied force
Fm is the opposing force due to mass
M is mass
a is acceleration
x is displacement
Taking Laplace and neglecting initial conditions we can write,
F(s) = Ms2
X(s)

Spring
Spring is an element, which stores potential energy. If a force is applied on
spring K, then it is opposed by an opposing force due to elasticity of
spring. This opposing force is proportional to the displacement of the
spring. Assume mass and friction are negligible.
Figure :

Spring...
f (t) ∝ x(t)
fk = f (t) = Kx(t)
F = Fk = Kx
where,
F is the applied force
Fk is the opposing force due to elasticity of spring
K is spring constant
x is displacement

Dashpot or Damper
If a force is applied on dashpot B, then it is opposed by an opposing force
due to friction of the dashpot. This opposing force is proportional to the
velocity of the body. Assume mass and elasticity are negligible.
Figure :
Consider a mass M as shown in Figure 4 having friction with a support
with a constant B represented by a dashpot. Friction will oppose the
motion of mass M and also opposing force is proportional to velocity of
mass M

Dashpot...
Ffrictional ∝ ν
Ffrictional = B
dx
dt
where,
Ffrictional is the opposing force due to friction of dashpot
B is the frictional coefficient
v is velocity
x is displacement
Ffrictional (s) = BsX(s)

Rotational Motion
Instead of position, rotational motion deals with changes in angle (θ). An
object will always rotate-that is, spin or move in a circle-around an axis of
rotation. The earth spins around on the axis that runs from the North
Pole to the South Pole, creating day and night; figure skaters spin around
on the axis from their head to their feet, creating dizzying gold medals;
Star Fox spins around on the axis from his spaceship’s nose to stern,
creating DO A BARREL ROLLs.

Rotational Motion...
Rotational Motion
Rotational mechanical systems move about a fixed axis. These systems
mainly consist of three basic elements. Those are moment of inertia,
torsional spring and dashpot.
If a torque is applied to a rotational mechanical system, then it is opposed
by opposing torques due to moment of inertia, elasticity and friction of the
system. Since the applied torque and the opposing torques are in opposite
directions, the algebraic sum of torques acting on the system is zero. Let
us now see the torque opposed by these three elements individually.

Moment of Inertia
In translational mechanical system, mass stores kinetic energy. Similarly, in
rotational mechanical system, moment of inertia stores kinetic energy.
If a torque is applied on a body having moment of inertia J, then it is
opposed by an opposing torque due to the moment of inertia. This
opposing torque is proportional to angular acceleration of the body.
Assume elasticity and friction are negligible.
Figure :

Moment of Inertia...
Tdue to inertia ∝ α
Tdue to inertia = Jα = J
d2θ
dt2
where,
T is the applied torque
Tdue to inertia is the opposing torque due to moment of inertia
J is moment of inertia
α is angular acceleration
θ is angular displacement
Tdue to inertia(s) = Js2
θ(s)

Torsional Spring
In translational mechanical system, spring stores potential energy.
Similarly, in rotational mechanical system, torsional spring stores potential
energy.
If a torque is applied on torsional spring K, then it is opposed by an
opposing torque due to the elasticity of torsional spring. This opposing
torque is proportional to the angular displacement of the torsional spring.
Assume that the moment of inertia and friction are negligible.
Figure :

Torsional Spring...
Tk ∝ θ
Tk = Kθ
T = Tk = Kθ
where,
T is the applied torque
Tk is the opposing torque due to elasticity of torsional spring
K is the torsional spring constant
θ is angular displacement

Dashpot
If a torque is applied on dashpot B, then it is opposed by an opposing
torque due to the rotational friction of the dashpot. This opposing torque
is proportional to the angular velocity of the body. Assume the moment of
inertia and elasticity are negligible.
Figure :

Dashpot...
Tb ∝ ω
Tb = Bω = B
dθ
dt
T = Tb = Bω = B
dθ
dt
where,
Tb is the opposing torque due to the rotational friction of the dashpot
B is the rotational friction coefficient
ω is the angular velocity
θ is the angular displacement

Analogous Elements of Translational and Rotational
Motion
Translational Motion Rotational Motion

Motion
Mass (M)
Inertia (J)

Motion
Mass (M)
Inertia (J)
Friction (B)
Friction (B)

Motion
Mass (M)
Inertia (J)
Friction (B)
Friction (B)
Spring (K)
Spring (K)

Motion
Mass (M)
Inertia (J)
Friction (B)
Friction (B)
Spring (K)
Spring (K)
Force (F)
Torque (T)

Motion
Mass (M)
Inertia (J)
Friction (B)
Friction (B)
Spring (K)
Spring (K)
Force (F)
Torque (T)
Displacement (x)
Angular Displacement (θ)

Motion
Mass (M)
Inertia (J)
Friction (B)
Friction (B)
Spring (K)
Spring (K)
Force (F)
Torque (T)
Displacement (x)
Velocity v = dx
dt
Angular velocity ω = dθ
dt

Motion
Mass (M)
Inertia (J)
Friction (B)
Friction (B)
Spring (K)
Spring (K)
Force (F)
Torque (T)
Displacement (x)
Velocity v = dx
dt
Angular velocity ω = dθ
dt
Acceleration a = d2x
dt2
Angular acceleration α = d2θ
dt2
Table : Analogous Elements

Electrical Elements
Similar to the mechanical systems, very commonly used systems are of
electrical type. The behavior of such systems is governed by the rule of
Ohms law. The dominant elements of an electrical system are,
1 Resistor
2 Inductor
3 Capacitor

Resistor
A resistor is an electrical component that limits or regulates the flow of
electrical current in an electronic circuit. Resistors can also be used to
provide a specific voltage for an active device such as a transistor.
Figure :
Consider a resistance carrying current I as shown, then the voltage drop
across it can be written as,
V = IR

Inductor I
An inductor is a passive electronic component that stores energy in the
form of a magnetic field. An inductor, also called a coil or reactor, is a
passive two-terminal electrical component which resists changes in electric
current passing through it. It consists of a conductor such as a wire,
usually wound into a coil. When a current flows through it, energy is
stored temporarily in a magnetic field in the coil. When the current
flowing through an inductor changes, the time-varying magnetic field
induces a voltage in the conductor, according to Faraday’s law of
electromagnetic induction, According to Lenz’s law the direction of
induced e.m.f is always such that it opposes the change in current that
created it. As a result, inductors always oppose a change in current, in the
same way that a flywheel oppose a change in rotational velocity

Inductor II
Figure :
Consider an Inductor carrying current I as shown, then the voltage drop
V = L
di
dt
I =
1
L
Z
Vdt

Capacitor I
Capacitor is an electronic component that stores electric charge. The
capacitor is made of 2 close conductors (usually plates) that are separated
by a dielectric material. The plates accumulate electric charge when
connected to power source. One plate accumulates positive charge and the
other plate accumulates negative charge.
The capacitance is the amount of electric charge that is stored in the
capacitor at voltage of 1 Volt.
The capacitance is measured in units of Farad (F).
The capacitor disconnects current in direct current (DC) circuits and short
circuit in alternating current (AC) circuits.

Capacitor II
Figure :
Consider a Capacitor carrying current I as shown, then the voltage drop
V =
1
C
Z
Idt
I = C
dv
dt

Analogous System
In between electrical and mechanical systems there exists a fixed analogy
and there exists a similarity between their equilibrium equations. Due to
this, it is possible to draw an electrical system which will behave exactly
similar to the given mechanical system, this is called electrical analogous
of given mechanical system and vice versa. It is always advantageous to
obtain electrical analogous of the given mechanical system as we are well
familiar with the methods of analyzing electrical network then mechanical
system.
Two systems are said to be analogous to each other if the following two
conditions are satisfied.
1 The two systems are physically different
2 Differential equation modeling of these two systems are same
Electrical systems and mechanical systems are two physically different
systems. There are two types of electrical analogies of translational
mechanical systems.
1 force voltage analogy i.e. Direct analogy and
2 force current analogy i.e. Indirect analogy

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Mechanical System
In force voltage analogy, the mathematical equations of translational
mechanical system are compared with mesh equations of the electrical
system.
Consider the following translational mechanical system as shown in the
following figure.
Figure :

Force - Voltage Analogy...
The force balanced equation for this system is
F = Fm + Fb + Fk
F = Ma + Bv + Kx
F = M
d2x
dt2
+ B
dx
dt
+ Kx (1)
Taking Laplace,
F(s) = Ms2
X(s) + BsX(s) + KX(S)
This is equilibrium equation for the given system.

Electrical System
In this method, to the force in mechanical system, voltage is assumed to
be analogous one. Accordingly we will try to derive other analogous term.
Consider electric network as shown in Figure 12
Figure :

The equation according to the Kirchhoff’s law can be written as,
V = Ri + L
di
dt
+
1
C
Z
idt
But we cannot compare F and V unless we bring them into same form.
For this we will use current as rate of flow of charge
Therefore, i =
dq
dt
V = R
dq
dt
+ L
d2q
dt2
+
q
C
V = L
d2q
dt2
+ R
dq
dt
+
1
C
q (2)
Taking Laplace,
V (s) = Ls2
Q(s) + RsQ(s) +
1
C
Q(S)

Comparing equations for F and V it is clear that,
1 Inductance L is analogous to mass M
2 Resistance R is analogous to friction or damper B
3 Reciprocal of capacitor i.e. 1
C is analogous to spring of constant K

Torque Voltage Analogy
Mechanical System
In this analogy, the mathematical equations of rotational mechanical
system are compared with mesh equations of the electrical system.
Rotational mechanical system is shown in the following figure.
Figure :

Force - Torque - Voltage Analogy
Translational Rotational Electrical

Force (F)
Torque (T) Voltage (V)

Force (F)
Mass (M)
Inertia (J) Inductance (L)

Force (F)
Mass (M)
Friction Constant (B)
Tortional Friction
Constant (B)
Resistance (R)

Force (F)
Mass (M)
Tortional Friction
Constant (B)
Resistance (R)
Spring Constant K
(N/m)
Torional Spring Con-
stant K (Nm/rad)
Reciprocal of capaci-
tor 1
C

Force (F)
Mass (M)
Tortional Friction
Constant (B)
Resistance (R)
Spring Constant K
(N/m)
stant K (Nm/rad)
tor 1
C
Displacement (X)
θ Charge (q)

Force (F)
Mass (M)
Tortional Friction
Constant (B)
Resistance (R)
Spring Constant K
(N/m)
stant K (Nm/rad)
tor 1
C
Displacement (X)
θ Charge (q)
Velocity x = dx
dt
θ = dθ
dt = ω Current i = dq
dt
Table : Tabular form of Force Torque and Voltage analogy

Force - Current Analogy
Electrical System
In force current analogy, the mathematical equations of the translational
mechanical system are compared with the nodal equations of the electrical
system.
Consider the following electrical system as shown in the following figure.
This circuit consists of current source, resistor, inductor and capacitor. All
these electrical elements are connected in parallel.
Figure :

Force - Current Analogy... I
The equation according to the Kirchhoff’s law can be written as,
I = IR + IL + IC
, so the nodal equation is
I =
V
R
+
1
L
Z
Vdt + C
dV
dt
(3)
But we cannot compare F and V unless we bring them into same form.
For this we will use voltage as rate of flow of flux
Therefore, V =
dφ
dt
where φ = flux

Force - Current Analogy... II
Now, equation 3 becomes,
I =
1
R
dφ
dt
+
1
L
φ + C
d2φ
dt2
I = C
d2φ
dt2
+
1
R
dφ
dt
+
1
L
φ (4)
By taking Laplace,
I(S) = Cs2
φ(S) +
1
R
sφ(S) +
1
L
φ(S)
By comparing Equation 1 and Equation 4 , we will get the analogous
quantities of the translational mechanical system and electrical system.

Force - Current Analogy...
Comparing equations for F and I it is clear that,
1 Capacitor C is analogous to mass M
2 Reciprocal of Resistance 1
R is analogous to friction or damper B
3 Reciprocal of Inductance i.e. 1
L is analogous to spring of constant K

Force - Torque - Current Analogy

Force (F)
Torque (T) Current (I)

Force (F)
Mass (M)
Inertia (J) Capacitance (C)

Force (F)
Mass (M)
Tortional Friction
Constant (B)
1
R

Force (F)
Mass (M)
Tortional Friction
Constant (B)
1
R
Spring Constant K
(N/m)
stant K (Nm/rad)
Reciprocal of Induc-
tance 1
L

Force (F)
Mass (M)
Tortional Friction
Constant (B)
1
R
Spring Constant K
(N/m)
stant K (Nm/rad)
tance 1
L
Displacement (X)
θ φ

Force (F)
Mass (M)
Tortional Friction
Constant (B)
1
R
Spring Constant K
(N/m)
stant K (Nm/rad)
tance 1
L
Displacement (X)
θ φ
Velocity x = dx
dt
θ = dθ
dt = ω Voltage V = dφ
dt
Table : Tabular form of Force Torque and Current analogy

Steps to solve problems on analogous systems I
1 Identify all the displacements due to the applied force.
2 Draw the equivalent mechanical system based on node basis.
3 Write the eqiolibrium equations. At ech node algebraic sum of all the
forces acting at the node is zero.
4 In F-V analogy, use following replacements and rewrite equations
accordingly,
F → V , M → L, B → R, K →
1
C
, x → q, x → i(current)
5 Simulate the equations using loop method. Number of displacement
equal to number of loop currents.
6 In F-I analogy, use following replacements and rewrite equations,
F → I, M → C, B →
1
R
, K →
1
L
, x → φ, x → e(e.m.f .)

Steps to solve problems on analogous systems II
7 Simulate the equations using node basis. Number of displacements
equal to number of node voltages.

Problem 01
Problem 01
Draw the equivalent mechanical system of the given system. Hence write
the set of equilibrium equations for it and obtain electrical analogous
circuits using
1 F - V analogy
2 F - I analogy

Solution 01
Solution 01
The displacement of M1 is x1(t) and as B1 is between M1 and fixed
support hence it is also under the influence of x1(t). While B2 changes the
displacement from x1(t) to x2(t) as it is between two moving points. And
M2 and K are under the displacement x2(t) as K is between mass and
fixed support.
Figure : Equivalent System

Solution 01...
Solution 01

Solution 01...
Solution 01
X
F = 0

Solution 01...
Solution 01
X
F = 0
at node 1, F = M1
d2x1
dt2
+ B1
dx1
dt
+ B2(
dx1
dt
−
dx2
dt
) (5)

Solution 01...
Solution 01
X
F = 0
at node 1, F = M1
d2x1
dt2
+ B1
dx1
dt
+ B2(
dx1
dt
−
dx2
dt
) (5)
at node 1, F = M1s2
X1 + B1sX1 + B2s(X1 − X2) (6)

Solution 01...
Solution 01
X
F = 0
at node 1, F = M1
d2x1
dt2
+ B1
dx1
dt
+ B2(
dx1
dt
−
dx2
dt
) (5)
at node 1, F = M1s2
X1 + B1sX1 + B2s(X1 − X2) (6)
at node 2, 0 = M2
d2x2
dt2
+ Kx2(t) + B2(
dx2
dt
−
dx1
dt
) (7)

Solution 01...
Solution 01
X
F = 0
at node 1, F = M1
d2x1
dt2
+ B1
dx1
dt
+ B2(
dx1
dt
−
dx2
dt
) (5)
at node 1, F = M1s2
X1 + B1sX1 + B2s(X1 − X2) (6)
at node 2, 0 = M2
d2x2
dt2
+ Kx2(t) + B2(
dx2
dt
−
dx1
dt
) (7)
at node 2, 0 = M2s2
X2 + KX2 + B2s(X2 − X1) (8)

Solution 01...
Solution 01
Case - I:

Solution 01...
Solution 01
Case - I:
Now F - V Analogy M→ L, B → R, K → 1
C , x → q
Therefore,

Solution 01...
Solution 01
Case - I:
C , x → q
Therefore,
V (s) = V = L1s2
q1 + R1sq1 + R2s(q1 − q2) (9)

Solution 01...
Solution 01
Case - I:
C , x → q
Therefore,
V (s) = V = L1s2
q1 + R1sq1 + R2s(q1 − q2) (9)
0 = L2s2
q2 +
1
C
q2 + R2s(q2 − q1) (10)
we know that
I(s) = sQ(s)orI(s) = sq(s)
so, q = q(s) =
I(s)
s
So equations 9 and 10 becomes,

Solution 01...
Solution 01

Solution 01...
Solution 01
V (s) = V = L1sI1(s) + R1I1(s) + R2s(I1(s) − I2(s)) → Loop1 (11)

Solution 01...
Solution 01
V (s) = V = L1sI1(s) + R1I1(s) + R2s(I1(s) − I2(s)) → Loop1 (11)
0 = L2sI2(s) +
1
sC
I2(s) + R2s(I2(s) − I1(s)) → Loop2 (12)
Hence,
Figure : Equivalent Electrical System

Solution 01...
IMP

Solution 01...
IMP
Number of loop currents equal to number of displacements

Solution 01...
Solution 01
Case - II:

Solution 01...
Solution 01
Case - II:
Now F - I Analogy M→ C, B → 1
R , K → 1
L, x →φ
Therefore,

Solution 01...
Solution 01
Case - II:
R , K → 1
L, x →φ
Therefore,
I(s) = I = C1s2
φ1 +
1
R1
sφ1 +
1
R2
s(φ1 − φ2) (13)

Solution 01...
Solution 01
Case - II:
R , K → 1
L, x →φ
Therefore,
I(s) = I = C1s2
φ1 +
1
R1
sφ1 +
1
R2
s(φ1 − φ2) (13)
0 = C2s2
φ2 +
1
L
φ2 +
1
R2
s(φ2 − φ1) (14)
we know that
V (s) = sφ(s)

Solution 01...
Solution 01

Solution 01...
Solution 01
I(s) = I = C1sV1(s) +
V1(s)
R1
+
1
R2
(V1(s) − V2(s)) → Node1 (15)

Solution 01...
Solution 01
I(s) = I = C1sV1(s) +
V1(s)
R1
+
1
R2
(V1(s) − V2(s)) → Node1 (15)
0 = C2sV2(s) +
1
sL
V2(s) +
1
R2
(V2(s) − V1(s)) → Node2 (16)
Hence,

Solution 01...
IMP

Solution 01...
IMP
Number of node voltages equal to number of displacements

Problem 02
Problem 02
Draw the equivalent mechanical system and analogous systems based on
F-V and F-I methods for the given system

Solution 02
Solution 02

Solution 02
Solution 02
No element under x1(t) alone as force is directly applied to a spring K1. So
it will store energy and hence is the cause to change the force applied to
M2. Hence displacement of M2 is x2 and as B2 and K2 are connected to
fixed supports both are under x2(t) only as shown in the equivalent system.

Solution 02
Solution 02
No element under x1(t) alone as force is directly applied to a spring K1. So
it will store energy and hence is the cause to change the force applied to
M2. Hence displacement of M2 is x2 and as B2 and K2 are connected to
fixed supports both are under x2(t) only as shown in the equivalent system.
Figure : Equivalent System

Solution 02...
Solution 01

Solution 02...
Solution 01
at node 1, F = K1(X1 − X2) (17)

Solution 02...
Solution 01
at node 1, F = K1(X1 − X2) (17)
at node 2, 0 = K1(X2 − X1) + M2s2
X2 + K2X2 + B2sX2 (18)

Solution 02...
Solution 01
at node 1, F = K1(X1 − X2) (17)
at node 2, 0 = K1(X2 − X1) + M2s2
X2 + K2X2 + B2sX2 (18)
M2, B2 and K2 are under same displacement

Solution 02...
Solution 02
Case - I:

Solution 02...
Solution 02
Case - I:
C , x → q
Therefore,

Solution 02...
Solution 02
Case - I:
C , x → q
Therefore,
V (s) = V =
1
C1
(q1 − q2) (19)

Solution 02...
Solution 02
Case - I:
C , x → q
Therefore,
V (s) = V =
1
C1
(q1 − q2) (19)
0 =
1
C1
(q2 − q1) + L2s2
q2 +
1
C2
q2 + R2sq2 (20)
we know that
I(s) = sQ(s)or I(s) = sq(s)
so, q = q(s) =
I(s)
s

Solution 02...
Solution 02

Solution 02...
Solution 02
V (s) = V =
1
sC1
(I1 − I2) → Loop1 (21)

Solution 02...
Solution 02
V (s) = V =
1
sC1
(I1 − I2) → Loop1 (21)
0 =
1
sC1
(I2 − I1) + L2sI2 +
1
sC2
I2 + R2I2 → Loop2 (22)
Hence,

Solution 02...
IMP

Solution 02...
IMP
Number of loop currents equal to number of displacements

Solution 02...
Solution 02
Case - II:

Solution 02...
Solution 02
Case - II:
R , K → 1
L, x →φ
Therefore,

Solution 02...
Solution 02
Case - II:
R , K → 1
L, x →φ
Therefore,
I(s) = I =
1
L1
(φ1 − φ2) (23)

Solution 02...
Solution 02
Case - II:
R , K → 1
L, x →φ
Therefore,
I(s) = I =
1
L1
(φ1 − φ2) (23)
0 = C2s2
φ2 +
1
R2
sφ2 +
1
L2
φ2 +
1
L1
(φ2 − φ1) (24)
we know that
V (s) = sφ(s)

Solution 02...
Solution 02

Solution 02...
Solution 02
I(s) = I ==
1
sL1
(V1 − V2) → Node1 (25)

Solution 02...
Solution 02
I(s) = I ==
1
sL1
(V1 − V2) → Node1 (25)
0 = C2sV2 +
1
R2
V2 +
1
sL2
V2 +
1
sL1
(V2 − V1) → Node2 (26)
Hence,

Solution 02...
IMP

Solution 02...
IMP
Number of node voltages equal to number of displacements

Questions
1 What is mathematical model, explain in details
2 Explain mechanical transnational motion system in details
3 Explain mechanical rotational motion system in details
4 Explain analogy between translational and rotational motion of
mechanical system
5 Differentiate between Analogous Elements of Translational and
Rotational Motion
6 Explain electrical elements
7 Explain direct and indirect analogy
8 Explain force-voltage analogy in details
9 Explain force-current analogy in details
10 Give comparison between Force - Torque - Voltage Analogy
11 Give comparison between Force - Torque - Current Analogy

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Control system mathematical modelling of a system

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