cs ppt good one as well containg everyting .ppt

Control
Systems
1
Department of Electronics & Communication Engineering,
Madan Mohan Malaviya University of Technology,
Gorakhpur
Subject Code: BEC-
26
Third Year
ECE
Unit-I
Shadab A.
Siddique
Assistant Professor

UNIT- I
❖ Electrical Networks and Mechanical Systems
❖ Force-Voltage Analogy
❖ Force-Current Analogy 2
UNIT-I: Basic Components of a control system, Feedback and its effect, Types of
feedback control Systems, Block diagrams: representation and reduction, Signal
Flow Graphs, Modeling of Physical Systems: Electrical Networks and Mechanical
Systems, Force-voltage analogy, Force-current analogy.
➢Introduction to Control Systems
❖ Control System – Definition and Practical Examples
❖ Basic Components of a Control System
➢Feedback Control Systems:
❖ Feedback and its Effect
❖ Types of Feedback Control Systems
❖ Transfer Function
➢Block Diagrams:
❖ Representation and reduction
❖ Signal Flow Graphs
➢Modeling of Physical Systems:
Shadab. A. Siddique

Course Assessment methods:
Continuous assessment through tutorials, attendance, home
assignments, quizzes, practical work, record, viva voce and Three Minor
tests and One Major Theory & Practical Examination
Course Outcomes:
The students are expected to be able to demonstrate the following knowledge,
skills and attitudes after completing this course,
✓Describe the response characteristic and differentiate between the open loop
and closed loop of a control system.
3
Shadab. A. Siddique

UNIT-
I
➢ Introduction to Control Systems
❖ Control System – Definition and Practical
Examples
➢ Feedback Control Systems:
➢ Block Diagrams:
➢ Modeling of Physical Systems:
❖ Force-Current Analogy
4
Shadab. A. Siddique

Input
➢ The stimulus or
excitation applied to a control
system from an external
source in order to produce the
output is called input
Input
Output
➢ The actual response obtained from
a system is called as output.
Output
Input
5
Shadab. A. Siddique

Control
➢ It means
command a system so that
the
to regulate, direct
or
desired objective is
attained
Syste
m
Input
(May or may not
be desired)
➢ It is a
combination
or
arrangement
of different
physical
components
connected
or related in
such a
manner so
as to form
an entire unit
Output
System
Control System:
➢ A control system is an arrangement of different physical elements connected in such a
manner so as to regulate, direct, command itself or some other system to achieve a certain
objective.
Combining above definitions
System + Control = Control
System
Input
Desired
Output
Control System
6
Shadab. A. Siddique

Difference between System and Control System
➢ An example :
Fan
230V/50Hz
AC Supply Air Flow
Input Output
➢ A Fan: Can't Say System: A Fan without blades cannot be a
“SYSTEM”
Because it cannot provide a desired/proper output. i.e. airflow
Input Output
Fan
(System)
230V/50Hz
AC Supply
No
Airflow
(No Proper/
Desired
Output)
7
Shadab. A. Siddique

A Fan: Can be a
System
➢ A Fan with blades but without regulator can be a “SYSTEM” Because it
can
provide a proper output. i.e. airflow
230V/50Hz
AC Supply
A Fan: Can be a Control
System
Airflow
(Proper
Output)
Input Output
➢ A Fan with blades and with regulator can be a “CONTROL SYSTEM”
Because it can provide a Desired output. i.e. Controlled airflow
230V/50Hz
AC Supply Controlled
Airflow
Input Output
Control
Element
(Desired
Outpu t)
Shadab. A. Siddique
➢ Butit cannot
be
a “Control System” Because it cannot
provide desired output i.e. controlled airflow

UNIT-
I
➢ Introduction to Control Systems
Examples
➢ Block Diagrams:
9
Shadab. A. Siddique

Basic Components of Control System:
Feedback Path
✓Basic components in the control
systems are shown in the above block
diagram.
✓Disturbances can be external or
Reference
Transducer Controlle
r
Plant/Process
Feedback
Transducer
Command
I/p
Reference
I/p
Feedback
Signal
Actuating
Signal
Error
Signal
Controlled
O/p
c(t)
r(t) e(t)
b(t) c(t)
u(t)
Forward
Path
∓
Disturbances
10
Shadab. A. Siddique

➢ Plant/Process:- The portion of a system which is to be controlled or regulated
is called a plant or process.
➢ Controller:- The element of a system itself or external to the system which
controls the plant/process is called controller. It consists of error
detector to control logic element.
➢ Error detector:- It received the measured signal (feedback) & compare
it with reference input and determine the error signal.
comparison with the input.
➢ Input:- It is applied signal to a control system from an external energy
sources
in order to produce specific output.
➢ Output:- It is a particular signal of interest or the actual response
obtained from the control system when input is applied.
➢ Disturbances:- It is a signal which tends to adversely effect the value of the
output of a system. It may be external disturbances or internal disturbances.
Basic Components of Control System:
11
Shadab. A. Siddique
𝑒 𝑡 = 𝑟 𝑡 ∓
𝑏(𝑡)
➢ Feedback:- It is used to
fed
back the
o/p
signal to error
detector
for

Classification of Control System:
Control system can be broadly classified as-
1.Natural control system e.g: Respiratory system, Biological systems of human
body
2.Man-made control system e.g: Vehicle
3.Combination control system e.g: Driving a car
4.Time variant and Invariant control system
5.Linear and Nonlinear control system
6.Continuous time and Discrete time control system
7.Deterministic (o/p is predictable) and stochastic (o/p is unpredictable) control
system
8.Lumped parameter and Distributed parameter control system
9.SISO (Serial input serial output) and MIMO (Multiple input and multiple
output) control system
10.Open loop and Closed loop control system
Controller Plant/
process
o/p i/p
i/p
Open Loop Control System
Controller Plant/
process
o/p
Feedback
∓
Closed Loop Control System
Shadab. A. Siddique
12

Classification of Control
System
(Depending on control action)
Open Loop Control
System
Closed Loop Control
System
Open Loop Control System
Definition: “A system in which the control action is totally independent of the
output
of the system is called as open loop system”
Fig. Block Diagram of Open loop Control
System
Process
Reference
I/p
r(t)
u(t)
Controlled o/p
c(t)
13
Shadab. A. Siddique
Controller

OLCS
Examples
➢ Electric Hand Drier:- Hot
long as you keep your hand
under the
air (output) comes out as
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.
14
Shadab. A. Siddique

➢ Automatic Tea/Coffee Vending
Machine:-
These
OLCS
Examples
machines also function for pre adjusted time only.
➢Light Switch:- lamps glow whenever light
switch is
on irrespective of light is required or
not.
➢Volume on Stereo
System:- Volume
is
adjusted
manually irrespective of output volume
level.
Advantages of OLCS
❖Simple in construction and design
❖Economical
❖Easy to maintain
❖Generally stable
❖Convenient to use as output is difficult to
measure
Disadvantages of OLCS
❖They are inaccurate
❖They are unreliable output
cannot be corrected
❖Any change in automatically.
15
Shadab. A. Siddique

Closed Loop
System
Definition:- A system in which the control action is somehow dependent
on the
output is called as closed loop system
Forward Path
Reference
Transducer Controlle
r
Plant
Feedback
Transducer
Command
I/p
Reference
I/p
r(t
)
Feedback
Signal
Manipulated
Signal
Error
Sign
al
e(t)
Controlled
O/p
c(t)
b(t) c(t)
m(t)
Feedback
Path
±
16
Shadab. A. Siddique

CLCS
Examples
➢ Automatic
Electric
Iron:- Heatin
g
element
s
are
controlled
by
output
temperature of the
iron.
➢ Servo voltage stabilizer:- Voltage
controller
operates depending upon output
voltage of the
system.
17
Shadab. A. Siddique

Advantages of
CLCS
➢ Closed loop control systems are more accurate even in the presence of non-
linearity
➢ Highly accurate as any error arising is corrected due to presence of feedback
signal
➢ Bandwidth range is large
➢ Facilitates automation
➢ The sensitivity of system may be made small to make system more stable
➢ This system is less affected by noise
Disadvantages of CLCS
➢They are costlier
➢They are complicated to design
➢Required more maintenance
➢Feedback leads to oscillatory response
➢Overall gain is reduced due to presence of feedback
➢Stability is the major problem and more care is
needed
closed loop system
to design a
stable
18
Shadab. A. Siddique

Difference Between OLCS &
CLCS
Open Loop Control System Closed Loop Control System
1.The closed loop systems are complex and
costlier
2.They consume more power
to 3. The CL systems are not
easy to construct of because more
number of components
required
4.The closed loop systems are accurate
& more reliable
5.Stability is a major problem in closed
loop systems & more care is needed to
design a stable closed loop system
6.Large bandwidth
7.Feedback element is present
8.Output measurement is necessary
9.The changes in the output due to external
disturbances are not corrected
automatically. So they are more sensitive to
noise and other disturbances.
10. Examples: Guided Missile, Temp control
19
of oven, Servo Voltage Stablizer etc.
1. The open loop systems are simple &
economical.
2.They consume less
power
3.The OL
systems
are construct
because less
easier
number
components required
4.The open loop systems are
inaccurate
& unreliable
5.Stability is not a major problem in
OL control systems. Generally
OL systems are stable
6.Small Bandwidth
7.Feedback element is absent
8.Output measurement is not
necessary
9.The changes in the output due
to external disturbances are
not corrected automatically. So they
are more sensitive to noise and
other disturbances.
Shadab. A. Siddique

UNIT-
I
(15
Marks)
➢ Introduction to Control system
Examples
➢ Block Diagrams:
❖ Force-Current Analogy 20
Shadab. A. Siddique

Feedback and its Effect:
The effect of parameter variation on the system performance can be analyzed by
sensitivity of the system.
➢Effect of Parameter variation in open loop control system:
Consider an open loop control system whose transfer function will be
G(s)
R(s) C(s) C(s
)
R(s)
C s =
G s
. R s
= G(s) = T(s)
Let ∆G(s) be the changes in G(s) due to parameter variation, the corresponding
change in the output be ∆C(s)
C(s) + ∆C(s) = [G(s) + ∆G(s)] R(s)
= G(s) . R(s) + ∆G(s) . R(s)
C(s) + ∆C(s) = C(s) + ∆G(s) . R(s)
Therefore, ∆C(s) = ∆G(s) . R(s)
➢Effect of Parameter variation in closed loop control system:
R(s) C(s)
G(s)
H(s)
B(s)
+
±
E(s)
Consider a closed loop control system whose
transfer function is given by,
TF = T(s) =
C (s) = G(s)
R(s) 1 ± G(s).H(s) 21
Shadab. A. Siddique

C(s) + ∆C(s)
R(s)
=
G(s) + ∆G(s)
1 ± [G(s) + ∆G(s)] . H(s)
C(s) + ∆C(s =
G(s) . R(s) + ∆G(s) . R(s)
1 ± [G(s) + ∆G(s)] .
H(s)
∆C(s) =
G(s) . R(s) + ∆G(s) . R(s)
1 ± [G(s) + ∆G(s)] .
H(s)
 C(s)
Assumption: G(s) . H(s) ≈ [G(s) + ∆G(s)] .
H(s)
∆C(s) =
G(s) . R(s) + ∆G(s) . R(s)
1 ± G(s) .
H(s)
 C(s)
∆C(s =
G(s) . R(s)
1 ± G(s) . H(s) + 1 ± G(s) . H(s)  C(s)
∆G(s) . R(s)
∆C(s = C
s
+
∆G(s) . R(s)
1 ± G(s) . H(s)  C(s)
Therefore,
|1 ± G(s) . H(s)| ≥
1
Let ∆G(s) be the changes in G(s) due to
parameter variation in system.
The corresponding change in the output is
∆C(s)
➢ Sensitivity of a control system:
An effect in the system parameters due
to parameter variation can be
studied mathematically by defining
the term sensitivity of a control system.
Let the variable in a system is varying by T
due to the variation in parameters K of the
system. The sensitivity of system
parameters K is expressed by,
% Change in
T
S =
% Change in K
𝐾 𝑐𝑎𝑛 𝑏𝑒 𝑜𝑝𝑒𝑛 𝑙𝑜𝑜𝑝 𝑔𝑎𝑖𝑛
𝑜𝑟 𝑓𝑒𝑒𝑑𝑏𝑎𝑐𝑘
S = d ln K
= ∆K/K
=
d ln T ∆T/T 𝝏T/T
𝝏K/
K
𝑆𝐾
=
�
�
K
𝝏T
𝑇
𝝏K
∆C(s) =
∆G(s) . R(s)
22
factor.
Shadab. A. Siddique
1 ± G(s) .
H(s)
A symbolic representation
𝑆𝐾
�
�
represents
the sensitivity of a variable T with respect
to the variation in the parameter K.
T may be the transfer function of
output variable & K may be the gain or
feedback

SG =
𝑇(𝑠)
𝜕𝐾
For open
loop control
system:
T G(s)
𝜕𝑇
T(s) = C(s)
=
G(s) R(s)
Forward path transfer function
= Gain = G(s)
S
=
G
T G(s)
𝜕𝐺(𝑠)
𝐺(𝑠)
𝜕𝐺(𝑠)
=
1
---------------- (1)
SG =
1
T
For closed loop control system:
T(s) = C(s)
= G(s) =
R(s)
G(s)
1 + G(s) .
H(s)
Forward path transfer function = Gain = G(s)
SG =
T G s
G(s)
1 + G(s) .
H(s)
.
𝜕[
}
G(s)
1 + G(s) .
H(s)
𝜕𝐺(𝑠)
SG = [1 + G(s) .
H(s)] .
T 1 + G(s)H(s)
−G(s)H(s)
[1 + G(s)H(s)]2
ST
=
G
1
1 + G(s)H(s)
---------------- (2)
Comparing equation (1) and (2), it can be
observed that due to feedback the sensitivity
function get reduces
by
the
factor
open
loop
1
1 + G(s)H(s)
compared to an
transfer function.
➢ Sensitivity of T(s) with H(s):
Let K be the gain ( forward path transfer function) & T be the overall transfer function of
the control system.
23
Maj. G. S. Tripathi
Shadab. A. Siddique Shadab. A. Siddique

➢ Sensitivity of T(s) with H(s):
Let us calculate the sensitivity function which indicates the sensitivity of overall transfer
function T(s) with respect to feedback path transfer function H(s). Such a function can be
expressed as,
SH =
𝑇(𝑠)
𝜕𝐻(𝑠)
For closed loop
control system,
T H(s)
𝜕𝑇(𝑠)
T(s) = C(s)
= G(s) =
R(s)
G(s)
1 +
G(s)H(s)
Feedback path transfer function = H(s)
SH =
T H s 1 +
G(s)H(s)
G(s)
1 + G(s)H(s)
H(s) . [1 +
G(s)H(s)]
G(s)
.
𝜕[
}
G(s)
𝜕𝐻(𝑠
)
SH
=
T
. [ [1 + G(s)H(s)]2
. G(s) ]
G(s)
SH
=
T  G(s)H(s)
1 + G(s)H(s)
---------------- (3)
It can be observed from equation (2) and (3) that the closed loop system is more sensitive
to variation in feedback path parameter than the variation in forward path parameter i.e.
24
gain.
Shadab. A. Siddique

Problem 1:
A position control system is shown in the figure, calculate the sensitivity ST
and
ST
k a
R(s)
+
k
S(S + a)
C(s)
-
Solution:
To find the sensitivity first we need to
calculate transfer function of system shown,
T(s) =
C(s) s (s+a)
=
R(s)
k
1 + k
s
(s+a)
=
. 1 s (s
+ a) + k
k
S = . =
T 𝜕k
k
T k 𝜕T k
k
s (s + a) +
k
𝜕
k
s (s + a) +
k 𝜕k
= s (s + a) + k .
s (s + a) + k − k.
1 (s (s + a) + k)2
a k
s (s + a) +
k
S = . =
T 𝜕a 𝜕a
T a 𝜕T a s (s + a) + k
a
𝜕
k
= . s (s + a) +
k
k
. (s (s + a) +
k)2
− k . s
S
k
T
= s (s + a) +
k
s (s + a)
T − as
= s (s + a) + k
Shadab. A. Siddique
S
a
25

Problem 2:
In the system shown in the figure, calculate the system
sensitivity
i ST
(ii) ST
and (ii) S
G1 G2 H
T
R(s) C(s)
+
-
G1 G1
H
1
s +
1
1
s +
2
G1
=
, G2
=
and H =
s
Solution:
[G1(s) . R(s) + H(s) . C(s)] . G2(s) = C(s)
G1 . R(s) . G2(s) = C(s) . [1 - H(s) . G2(s)]
T(s) = C(s) G1(s) . G2(s)
R(s) 1 + H(s) .
G2(s)
=
(i)
S
G
1
T = . =
T 𝜕G1
G1 𝜕T G1
G1G2
1 + G2H
𝜕
G1G
2
1 +
G2H
𝜕G1 =
1 +
G2H
G2
.
G2 (1 +
G2H)
(1 + HG2)2
= 1
(ii)
S
G
2
T = . =
T 𝜕G2
G2 𝜕T G2
G1G2
1 + G2H
𝜕
G1G
2
1 +
G2H
𝜕G2 =
1 +
G2H
G1
.
G1 (1 + G2H) −
G1G2H
(1 + HG2)2
= 1 + G2H
1
(iii) S =
H
T H 𝜕T
.
=
T
𝜕H
H
G1G2 1
+ G2H
𝜕
G1G2
1 +
G2H
𝜕H =
H (1 +
G2H)
G1G
2
.
− G1 G2 − HG2
(1 + HG2)2
. G2 = 1 + G2H
T − HG2 − s . s + 2
1
− s
SG
2
T
=
1
1 +
G2H
=
1
1 + s . 1
s +
2
=
s + 2
2s +
2
= 1 + G2H =
1 + s . 1 = 2s +
S
H
26
s + Shadab. A. Siddique

Effect of feedback on time constant of a control system:
Less the time constant, faster is the response.
Consider an open loop system with overall transfer function as G(s)
=
1
k
1 +
sT
.
Let the input be unit step, r(t) = u(t) » R(s)
=
s
By using partial fraction
method
C(s) = A
+
S 1 +
sT
C(s) = A
B
1 + sT +
Bs
s (1 +
sT)
C(s) =
K
1 + sT 1 + sT s
. R(s) = k
.
1
Therefore, k = A(1 + sT) +
Bs For s = 0, A = k
For s = -1/T, B = -Tk
C(s) = k
-
S 1 +
sT
k
kT
C(s) = k
-
S s +
1/T
Taking Inverse Laplace
transform
−t
e(t) = k u(t) – k e ൗT u(t)
e(t) = k (1 – e T ) u(t)
−t
ൗ
+
-
G s
=
1 +
sT
C(s)
T(s) =
C(s)
=
R(s)
Where T = time constant ------------- (4)
Consider an open loop system
Here feedback H(s) = h(t) is introduced in the
system k
H(s) = h(t)
R(s)
k
1+sT
1 + k
1+sT
.
H(s)
T(s) =
C(s
)
=
R(s) 1 +
sT+k H(s)
k
27
Shadab. A. Siddique

Let us consider H(s) is a unity value function,
i.e. H(s) = 1
T(s) =
C(s) k
=
R(s) sT
+ k +1
C(s)
=
k (
1
)
T
s + k
+1
T
. R(s)
for unit step
input, r t
k
= u t » R(s) = 1
s
C(s)
=
T
.
1
s + k
+1
T
s
taking inverse laplace transform
for C(s)
c(t)
=
k
1
T
L−
1
s
.
s +
k +
1
T
1
s . s + k
+1
T
= s
+
A B
s + k
+1
T
c(t)
=
k
T L−
1
A
s
+
B
s +
k +
1
T
c(t) =
L
T
k −1
T
k +
1 s
−
T
k +
1
s
+
k +
1
T
k
c(t) =
u(t) –
e T k
+ 1 k
+ 1
T T − k+1
tൗ
T u(t
)
k
c(t) =
k + 1
1 –
e
− k+1
tൗ
T u(t
)
Where,
T
--------- (5)
It can be observed that the new time
constant due to unity feedback system
isT
k+
1
. Thus for positive value of k > 1, the
time constant
T
k+
1
is less than T.
Thus it can be concluded from
equation
(4) and (5) that the time constant
of closed loop system is less than open
loop system. We know that lesser the
time constant faster is the response.
Hence feedback improves the time
system.
For s = 0, A =
k+1
T
For s = - , B =
k+1
T
– T
k+1 28
Shadab. A. Siddique
time constant =
k+
1

Effect of feedback on overall gain of a control
system:
Consider an open loop system with overall transfer function = G(s) = overall gain of
the system.
If the feedback H(s) is introduced in such a system, the overall gain
becomes
G(s)
1 ± G(s) . H(s)
.
The +ve and –ve sign in the denominator gets decide by the sign of feedback. For a
–ve
feedback the gain G(s) is reduced by
G(s)
1 + G(s) .
H(s)
,
so due to –ve feedback overall gain of the system
reduces.
Effect of feedback on stability of a control system:
Consider an open loop system with overall transfer function = G(s) =
is located at s = -T.
k
s +
T
then open loop pole
T. F =
C(s)
R(s) = 1 + G(s) . H(s) =
1 + k
G(s)
k
s +
T
s + T .
1
= s + T +
k
k
Now, let the unity feedback (H(s) = 1) is introduced in the system. The overall
transfer
function of closed loop becomes
k
s + T +
k
The closed loop poles is now located at s = - (T +
k) 29
Shadab. A. Siddique

Imaginary Imaginary
Real Real
×
S = -T
×
S = - (T +
k)
For open loop
For closed loop
The stability of the system depends on the location of poles in
s- plane. Thus it can be
concluded that the feedback effect the stability of the system.
There are two main types of feedback control systems: negative feedback and
positive feedback. In a positive feedback control system the setpoint and output values are
added. In a negative feedback control the setpoint and output values are subtracted.
As a rule negative feedback systems are more stable than positive feedback
systems. Negative feedback also makes systems more immune to random variations in
component values and
Types of feedback control
systems:
inputs 30
Shadab. A. Siddique

Transfer Function:
It is a mathematical expression relating output or response of the system to the input. It
is
denoted by T s
=
C(s
)
R(s)
❖ Open Loop control system:- G(s)
R(s) C(s)
C s = G s . R
s
C(s)
= G(s) = T(s)
OLTF
R(s)
❖ Closed Loop control system:-
R(s)
Inpu
t
G(s)
C(s)
Output
H(s)
B(s)
Feedb
ack
Signa
l
+
Error
Signal
E(s)
-
Error signal is given by;
E(s) = R(s)  B(s) --------------------- (1)
R(s) = E(s) + B(s)
Gain of feedback network is given by;
H(s) = B(s) ,
C(s
)
B(s) = H(s) . C(s) --------
(2)
Gain for CL system is given
by;G(s) = C(s)
E(s)
C(s) = G(s).E(s) ---------------
(3)
Substitute value of E(s) from
eq. 1 to 3
C(s) = G(s) . (R(s)  B(s))
C(s) =G(s) . R(s)  G(s) .
B(s) --------(4)
Substitute value of B(s) from
eq. 2 to 4
C ( s ) = G( s) R(s)  G( s).
H( s). C( s)
G( s). R( s) = C(s) + G( s).
C(s) G(s)
=
T.F.= = T(s)
R(s) 1 +
Shadab. A. Siddique

UNIT-
I
(15
Marks)
Examples
➢ Block Diagrams:
32
Shadab. A. Siddique

Block Diagram of a Control System:
✓ If the system is simple & has limited parameters then it is easy to analyze such
systems using the methods discussed earlier i.e transfer function, if the system is
complicated and also have number of parameters then it is very difficult to analyze it.
✓ To overcome this problem block diagram representation method is used.
✓ It is a simple way to represent any practically complicated system. In this
each component of the system is represented by a separate block known as functional
block.
✓ These blocks are interconnected in a proper sequence.
✓ The block diagram has following five basic elements associated with it
a) Blocks
b) T.F of elements shown inside the block
c) Summing point
d) Take off points
e) Arrow
✓ For a closed loop systems, the functions of comparing the different signals is
indicated by the summing point, while a point from which a signal is taken for
feedback purpose is indicated by take off point in block diagram.
✓ The signal can travel along the arrow only. 33
Shadab. A. Siddique

Advantages of block diagram:
a) Very simple to construct the block diagram for complicated systems.
b) The function of individual element can be visualised from the block diagram.
c) Individual as well as the overall performance of the system can be studied by
using transfer function.
d) Overall closed loop transfer function can be easily calculated
using block diagram reduction rule.
Disadvantages of block diagram:
a) Block diagram does not includes about physical construction of the system.
b) Source of energy is generally not shown in the block diagram.
Basic elements of block diagram:
Blocks:
-
It is shorthand, pictorial representation of the cause and effect
relationship
between input and output of a physical system.
Input Block Output
Output:- The value of the input is multiplied to the value of block gain to get the
output.
Input 2s
X(s)
Output
Y(s)
34
Output, Y(s) = 2s .

Output =x+y-z
+
x
-
output
Summing Point:- Two or more signals can be added/ subtracted at summing
point.
Basic elements of block
diagram:
z
Take off Point:- The output signal can be applied to two or more points from a take off
point.
Take off point
Forward path
y
Forward Path:- The direction
of flow of signal is from input
to output.
G1 G2
H1
-
R(s) + C(s)
Feedback
path
Feedback Path:- The direction
of flow of signal is from output
to input.
35
Shadab. A. Siddique

Rule 1:- For blocks in cascade:- Gain of blocks connected in cascade gets
multiplied with each other.
Block Diagram Reduction Techniques
G1 G2
R(s)
R1(s)
C(s)
R1(s) = G1R(s)
C(s) = G2R1(s) = G1G2R(s)
C(s) = G1G2R(s)
G1G2
R(s) C(s)
R(s) C(s)
Find
Equivalent
R(s) C(s)
R(s) C(s)
Find
Equivalent
G1 G2 G3
C1(s)
G1G2G3
R(s) C(s)
G1G2
R(s) C(s)
G3
36
C1(s)
Shadab. A. Siddique
G1G2G3
G1 G2 G3

C(s)= (G1-G2+G3)
R(s)
G1-G2+G3
R(s) C(s)
G1
G2
R(s) C(s)
G3
R1(s)
R3(s)
+
+
R2(s)
C(s)= R1(s)-R2(s)+R3(s)
= G1R(s)-G2R(s)+G3R(s)
C(s)=(G1-G2+G3) R(s)
Rule 2:- For blocks in parallel:- Gain of blocks connected in parallel gets
added
algebraically by considering the sign.
-
Rule 3:- Eliminate feedback loop:- Feedback loop can be either +ve or -
ve C(s)
G
H
R(s)
+
+
-
R(s)
B(s)
E(s)
- Sign is of –ve feedback
+ sign is for +ve feedback
37
Shadab. A. Siddique
G
1 ±GH
C(s)

From Shown Figure, E(s)  R(s) B(s) and
C(s) G.E(s)  G[R(s)  B(s)]  GR(s)
GB(s)
But, B(s)  H . C(s)
C(s) G . R(s) G . H . C(s)
C(s)  G.H.C(s)  GR(s)
C(s)[1G.H(s)] G. R(s)
For Negative
Feedback
C(s)
G
H
R(s)

B(s)
E(s)
For Positive
Feedback
C(s)
G
H
R(s)
+
+
B(s)
E(s)
From Shown Figure, E(s)  R(s) B(s) and
C(s) G.E(s)  G[R(s)  B(s)]  GR(s)
+GB(s)
But, B(s)  H . C(s)
C(s) G . R(s) G . H . C(s)
C(s)  G.H.C(s)  GR(s)
C(s)[1 G.H(s)] G. R(s)
+
C(s)
G(𝑠)
R(s) =
1+G(s)H(s)
C(s) G(𝑠)
R(s) = 1−
38
Shadab. A. Siddique

C(s)
B1
B2
X = R(s) - B1
C(s) = X - B2
C(s) = R(s)-B1-B2
R(s) + X +
-
C(s)
B1
R(s) + X +
-
B2
X = R(s) - B2
C(s) = X - B1
C(s) = R(s) - B2 -
B1
Rule 4:- Associative law for summing point:- It hold god for summing point which are
directly connected to each other i.e. there is no any summing point or take off point or
block in between summing points.
The order of summing points can be changed if two or more summing points are in
series.
39
Shadab. A. Siddique

Rule 5:- Shift summing point before block:-
R(s) C(s)
X
+
G
C(s) = R(s)G + X C(s) = G [R(s) + X/G]
= GR(s) + X
+
C(s)
R(s) +
G
1/G
X
+
R(s) + C(s)
G
X
C(s) = G [R(s)+X]
= GR(s)+GX
C(s) = GR(s)
+XG
= GR(s)+XG
+
R(s) C(s)
X
+
G
G
+
Rule 6:- Shift summing point after block:-
49
Shadab. A. Siddique

R(s) C(s)
G
C(s) = GR(s) and
X = C(s) = GR(s)
C(s) = GR(s) and
X = GR(s)
X
R(s) C(s)
X
G
G
Rule 7:- Shift a take-off point before the
block:-
Rule 8:- Shift a take-off point after the
block:-
R(s) C(s)
G
C(s) = GR(s) and
X = R(s)
C(s) = GR(s) and
X = C(s).{1/G}
= GR(s).{1/G}
= R(s)
X
R(s) C(s)
X
G
1/G
41
Shadab. A. Siddique

✓ While solving block diagram for getting single block equivalent, the said rules
need to be applied. After each simplification a decision needs to be taken. For
each decision we suggest preferences as
✓ First Choice:-
First preference
Second preference
Third preference
: Rule 1 (for Series)
: Rule 2 (for Parallel)
: Rule 3 (for Feedback
Loop)
✓ Second Choice:- equal preferences to
all Rule 4
Rule 5/6
Rule7/8
: Adjusting summing order
: Shifting summing point before/after the
block
: Shifting take off point before/after block
42
Shadab. A. Siddique

G4
R(s) C(s)
Problem 1:
Determine transfer function of the system shown in the
figure.
Solution:
✓Rule 1 cannot be used as there are no immediate series blocks.
✓Hence Rule 2 can be applied to G4, G3, G5 in parallel to get an equivalent of G3+G4+G5
G1 G2 G3 G6
+ +
G5
H1
H2
+
+
+
-
-
✓ Apply Rule 2 Blocks in
Parallel
G4
R(s) C(s)
G1 G2 G3 G6
+ +
G5
H1
+
+
+
-
-
43
H2
Shadab. A. Siddique

R(s) C(s)
G1 G2 G3+G4+G5 G6
+ +
H1
H2
-
-
✓ Apply Rule 1 Block in Series
✓ Apply Rule 3 Elimination of feedback
R(s) C(s)
G1 G2 (G3+G4+G5) G6
+ +
H1
H2
-
-
R(s) C(s)
G1
1 + G1H1
G2 (G3+G4+G5) G6
+
-
H2
Shadab. A. Siddique
44

✓ Apply Rule 3 Elimination of feedback loop
R(s) C(s)
G1G2 (G3+G4+G5)
1 + G1H1
G6
+
H2
-
R(s) C(s)
G1G2 (G3+G4+G5)
1 + G1H1 + G1G2H 2(G 3 + G 4 + G
5)
G6
R(s) C(s)
G1G2G6 (G3+G4+G5)
1 + G1H1 + G1G2H 2(G 3 + G 4 + G
5)
C(s) G1G2G6 (G3+G4+G5)
R(s)
=
1 + G1H1 + G1G2H 2(G 3 + G 4 + G
5)
45
Shadab. A. Siddique

R(s) C(s)
Problem 2:
Determine transfer function of the system shown in the figure.
-
G1 G2
+ +
H2
H1
-
-
+
Solution:
-
R(s) C(s)
G1 G2
+ +
H2
H1
-
-
+
R(s) C(s)
G1
+ +
H1
-
-
G
2
1 +
G2H2
46
Shadab. A. Siddique

✓ Now Rule 1, 2 or 3 cannot be used
directly.
✓ There are possible ways of going ahead.
✓ Use Rule 4 & interchange order of summing so that Rule 3 can be
used on G.H1 loop.
✓ Shift take off point
after
G2
1 + G2H2
block reduce by Rule 1, followed by Rule
3.
Which option we have to use????
✓Apply Rule 4 Exchange summing
point
R(s) C(s)
G1
+ +
H1
-
-
G
2
1 +
G2H2
✓ Apply Rule 4 Elimination of feedback loop
R(s) C(s)
G1
+ +
H1
-
-
G
2
1 +
G2H2
2 1
1 2
47
Shadab. A. Siddique

✓ Apply Rule 1 Block in series
R(s) C(s)
+
-
G
2
1 +
G2H2
2
G
1
1 + G1H1
✓ Apply Rule 1 Block in series
R(s) C(s)
+
-
G1G
2
(1 + G1H1 + G2H2 + G1G2H1H2)
✓Now which Rule will be applied
-------It is blocks in parallel OR
-------It is feed back loop
✓Let us rearrange the block diagram to
understand
✓Apply Rule 3 Elimination of feed back
loop
R(s) C(s)
+
-
G1G
2
(1 + G1H1 + G2H2 + G1G2H1H2)
1
48
Shadab. A. Siddique

R(s) C(s)
G1G2
(1 + G1H1 + G2H2 + G1G2H1H2 + G1G2)
C(s) G1G2
R(s)
=
1 + G1H1 + G2H2 + G1G2H1H2 + G1G2
Note 1: According to Rule 4
✓By corollary, one can split a summing point to two summing point and sum in any order
B B
G
H
+
-
R(s)
C(s) R(s) C(s)
G
H
+ +
-
+
+
49
Shadab. A. Siddique

UNIT-
I
(15
Marks)
Examples
➢ Block Diagrams:
50
Shadab. A. Siddique

Signal Flow Graph (SFG) Representation:
1. SFG is a graphical representation of variables of a set of linear algebraic equations
representing the system.
2. Variables are represented by small circles called nodes.
3. The line joining the nodes are called branches which is associated with a T.F and
an arrow.
4. Eg: V = IR
5. Let V & I are the variables in which I is the input & V is the output
R
V = IR
I
Properties of SFG:
1.SFG is only applicable to LTI system.
2.The signal gets multiplied by branch gain when it travels along it.
3.The value of variable is represented by any node is the algebraic sum of the signals
entering at the node.
4. The no. of branches leaving a node doesn’t affect the value of variable represented
by
51
that node. Shadab. A. Siddique

Introduction
✓SFG is a graphical representation of variables of a
set of linear algebraic equations
representing the system.
✓Variables are represented by small circles called nodes.
✓The directed lines joining the nodes are called branches which is associated with a T.F and
an arrow.
✓Alternative method to block diagram representation, developed by Samuel Jefferson Mason.
✓Advantage: the availability of a flow graph gain formula, also called Mason’s gain formula.
✓It depicts the flow of signals from one point of a
system to another and gives the relationships among the signals.
Fundamentals of Signal Flow Graphs
• Consider a simple equation below and draw its signal flow
graph:
•
The signal flow graph of the equation is shown below;
• Every variable in a signal flow graph is represented by a Node.
• Every transmission function in a signal flow graph is represented by a
Branch.
• Branches are always unidirectional.
• The arrow in the branch denotes the direction of the signal flow.
y 
ax
x y
a
52
Shadab. A. Siddique

Signal-Flow Graph Model 1:
Y1(s) G1 1(s)R1(s)  G1 2(s)R2(s)
Y2(s) G2 1(s)R1(s)  G2 2(s)R2(s)
𝑅1(𝑠
)
𝑅2(𝑠
)
𝑌1(
𝑠)
𝑌2(
𝑠)
r1 and r2 are inputs and x1 and x2 are outputs
a1 1x1  a1 2x2  r1 x1
a2 1x1  a2 2x2  r2 x2
�
�
1
𝑟
2
𝑥
1
𝑥
2
53
Shadab. A. Siddique

b
x4
x3
x2 g
x1
x0
h
e
d
a
xo is input and x4 is output
x1  ax0  bx1 
cx2 x2  dx1  ex3
x3  fx0  gx2 x4
 hx3
Terminologies of
SFG:
f
c
• An input node or source contain only the outgoing branches. i.e., X1
• An output node or sink contain only the incoming branches. i.e., X4
• A path is a continuous, unidirectional succession of branches along which no node is
passed more than ones. i.e.,
X1 to X2 to X3 to X4 X1 to X2 to X4 X2 to X3 to X4
• A forward path is a path from the input node to the output node. i.e.,
X1 to X2 to X3 to X4 , and X1 to X2 to X4 , are forward paths.
Shadab. A. Siddique
54

• A feedback path or feedback loop is a path which originates and terminates on the
same node. i.e.; X2 to X3 and back to X2 is a feedback path.
• A self-loop is a feedback loop consisting of a single branch. i.e.; A33 is a self loop.
• The gain of a branch is the transmission function of that branch.
• The path gain is the product of branch gains encountered in traversing a path. i.e. the
gain of forwards path X1 to X2 to X3 to X4 is A21A32A43
Cntd… Terminologies of
SFG
55
Shadab. A. Siddique

• The loop gain is the product of the branch gains of the loop. i.e., the loop gain of the
feedback loop from X2 to X3 and back to X2 is A32A23.
• Two loops, paths, or loop and a path are said to be non-touching if they have no nodes in
common.
Consider the signal flow graph below and identify the following:
Cntd… Terminologies of
SFG
a) Input node.
b) Output node.
c) No. of Forward paths.
d) No. of Feedback paths (loops).
e) Determine the loop gains of the feedback
loops.
f) Determine the path gains of the forward paths.
g) Non-touching loops
56
Shadab. A. Siddique

• There are two forward path gains:
Consider the signal flow graph below and identify the following
• There are four loops:
57
Shadab. A. Siddique

Consider the signal flow graph below and identify the following
• Nontouching loop gains:
a) Input node:
b) Output node:
c)
d)
e)
R(s)
C(s)
No of Forward paths: two
No. of Feedback paths (loops): four
Determine the loop gains of the feedback
loops: 1. G2(s)H1(s)
2. G4(s)H2(s)
3. G4(s)G5(s)H3(s)
4. G4(s)G6(s)H3(s)
f)Determine the path gains of the
forward paths:
1. G1(s) G2(s)
G3(s) G4(s) G5(s) G7(s)
2. G1(s) G2(s)
G3(s) G4(s) G6(s) G7(s)
g)Non-touching loops:
3. [G2(s)H1(s)][G4(s)G6(s)H3(s)]
Shadab. A. Siddique
58

Consider the signal flow graph below and identify the followin
g:
a) Input node.
b) Output node.
c) Forward paths.
d) Feedback paths.
e) Self loop.
f) Determine the loop gains of the feedback loops.
g) Determine the path gains of the forward paths.
• Input and output Nodes
a) Input node
a) Output node
59
Shadab. A. Siddique

(c) Forward Paths
(d) Feedback Paths or Loops
69
Shadab. A. Siddique

(d) Feedback Paths or Loops
61
Shadab. A. Siddique

(e) Self Loop(s)
(f) Loop Gains of the Feedback Loops
(g) Path Gains of the Forward Paths
62
Shadab. A. Siddique

Mason’s Rule (Mason, 1953)
n = number of forward
paths.
Pi = the i th forward-path gain.
∆ = Determinant of the system
∆i = Determinant of the ith forward path
• ∆ is called the signal flow graph determinant or characteristic function. Since ∆=0 is
the
• The block diagram reduction technique requires successive application of fundamental
relationships in order to arrive at the system transfer function.
• On the other hand, Mason’s rule for reducing a signal-flow graph to a single transfer
function requires the application of one formula.
• The formula was derived by S. J. Mason when he related the signal-flow graph to the
simultaneous equations that can be written from the graph.
Mason’s Gain Formula:
• The transfer function, C(s)/R(s), of a system represented by a signal-flow graph
is;
n
 Pi i
Where,
i 1


R(s)
C(s)
system characteristic equation. 63
Shadab. A. Siddique

• ∆ = 1- (sum of all individual loop gains) + (sum of the products of the gains of
all possible two loops that do not touch each other) – (sum of the products of the gains of
all possible three loops that do not touch each other) + … and so forth with sums of
higher number of non-touching loop gains.
• ∆i = value of Δ for the part of the block diagram that does not touch the i-th forward path
(Δi = 1 if there are no non-touching loops to the i-th path.)
Systematic approach for problems:
1. Calculate forward path gain Pi for each forward path i.
2. Calculate all loop transfer functions
3. Consider non-touching loops 2 at a time
4. Consider non-touching loops 3 at a time
5. etc
6. Calculate Δ from steps 2,3,4 and 5
7. Calculate Δi as portion of Δ not touching forward path i
 Pi i i
1


n
R(s)
C(s)
64
Shadab. A. Siddique

Example#1: Apply Mason’s Rule to calculate the transfer function of the system
represented by following Signal Flow Graph
Therefore,
C

P11  P2 2
R 
L3  G1G3G4 H 2
There are three feedback loops
L1  G1G4 H1 , L2  G1G2G4 H 2 ,
65
There are two forward paths,
Solution:
There are no non-touching loops, therefore
∆ = 1- (sum of all individual loop gains)
  1  L1  L2  L3 
  1  G1G4 H1  G1G2G4 H 2  G1G3G4 H 2 
Eliminate forward path-1
∆1 = 1- (sum of all individual loop gains)+...
∆1 = 1 ∆2 = 1- (sum of all individual loop gains)+...
∆ = 1
2
Shadab. A. Siddique

Example#1: Continue
Example#2: Apply Mason’s Rule to calculate the transfer function of the
system represented by following Signal Flow Graph
P1 P2
Solution:
1.
Calculate forward path gains
for each forward path.
P1  G1G2G3G4 (path
2. Calculate all loop gains. L4  G7 H 7
L3  G6 H 6 ,
L1  G2 H 2 , L2  H 3G3 , 66
Shadab. A. Siddique

Example#2: Continue
3. Consider two non-touching loops.
L1L3
L2L4
L1L4
L2L3
4. Consider three non-touching loops: None.
5. Calculate Δ from steps 2,3,4
  1  L1  L2  L3  L4  L1L3  L1L4  L2 L3  L2 L4 
  1  G2 H 2  H 3G3  G6 H 6  G7 H 7  
G2 H 2G6 H 6
Eliminate
forward path-1
 H 3G3G6 H 6  H 3G3G7 H 7 
Eliminate forward path-2
 G2 H 2G7 H 7
1  1  L3  L4 
1  1  G6 H 6  G7 H 7   2  1  G2 H 2  G3 H 3 
 2  1  L1  L2 
67
Shadab. A. Siddique

Example#2: Continue
Y ( s )

P11  P2  2 R( s) 
Y (s)

G1G2G3G4 1  G6 H6  G7 H7  G5G6G7G8 1  G2 H2  G3H3 
R(s) 1  G2 H2  H3G3  G6 H6  G7 H7  G2 H2G6 H6  G2 H2G7 H7  H3G3G6 H6  H3G3G7 H7 
Example#3: Find the transfer function, C(s)/R(s), for the signal-flow graph in
figure below.
Solution:
•There is only one forward Path
P1  G1(s)G2 (s)G3 (s)G4 (s)G5 (s)
68
Shadab. A. Siddique

Example#3: Continue
•There are four feedback loops.
• Non-touching loops taken two at a time.
• Non-touching loops taken three at a
time.
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Shadab. A. Siddique

• Eliminate forward path-1
Example#3: Continue
79
Shadab. A. Siddique

E(s) G1 G4
G3
Example#5: From Block Diagram to Signal-Flow Graph Models
－
－
－
C(s)
R(s
)
G1 G2
H2
H1
G4
G3
H3
E(s) X1
X2
X3
R(s) 1 C(s)
－
H2
－
H1
－
H3
X1
G2 X2 X3
  1  (G1G2G3G4 H 3  G2G3 H 2  G3G4 H1 )
P1  G1G2G3G4 ; 1  1 R( s) 1  G1G2G3G4 H 3  G2G3 H 2  G3G4 H1
 1 2 3 4
C ( s) G G G G
G 
71
Shadab. A. Siddique

Example#4: Find the control ratio C/R for the system given below.
Solution:
•The signal flow graph is shown in
the figure.
•The two forward path gains are
•The five feedback loop gains are
• All feedback loops touches the two
forward paths, hence
• Hence the control ratio
• There are no non-touching loops,
hence
T = 72
Shadab. A. Siddique

G1
G2
+
— X
+
－
－
－
+ C(s)
R(s) E(s)
Y2
Y1
X1
2
－
1
-1
1
-1
-1
-1
-1
1
1
G1
G2
1
R(s) E(s) C(s)
X1
X2
Y2
Y1
Example#6: Find the control ratio C/R for the system given below.
Solution:
73
Shadab. A. Siddique

7 loops:
3 ‘2 non-touching
loops’ :
Δ  1  2G2  4G1G2
p1  (  1)  G1  1
Then:
4 forward
paths:
Δ1  1  G2
Δ2  1
p2  (  1)  G1  (  1)  G2  1
Δ3  1  G1
p3  1  G2  1
Δ4  1
p4  1  G2  1  G1  1
We have
1  2G2  4G1G2

G2  G1  2G1G2
 p k  k
C( s )
R( s )


Example#6: Continue
74
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UNIT-
I
(15
Marks)
Examples
➢ Block Diagrams:
S. A. Siddique
75

Mathematic Modeling of Dynamical
Systems
• The set of mathematical equation describing the dynamic characteristics of a system is
called mathematical model of a system.
• Dynamics/ mathematical equation of many systems can
be written in terms of differential equations
– Mechanical, thermal, electrical, economic, biological systems etc.
• We said that these D.E.’s can be derived using basic physical laws
• All systems we will study will be ‘causal’, i.e. the system’s response at any
depends only on past and not future inputs
• Recall transfer functions:
time ‘t’
– It is the ratio of Laplace Transform of output to Laplace Transform of input, when
initial conditions are zero.
– We assume
• Zero initial conditions
• Linearity
• Time Invariance
76
Maj. G. S. Tripathi
Shadab. A. Siddique

Similarities in Mechanical and Electrical
Systems
3 basic components in mechanical systems:
–Mass
–Spring
–Damper
✓ 3 basic components in electrical systems:
– Resistance
– Capacitor
– Inductor
• Basic form of differential equations is the same.
• Therefore, learning to model one type of system easily leads to modeling method for
the other.
• Also, electrical and mechanical systems can be easily cascaded in block diagrams due to this
similarity.
• In fact, many other types of systems have similar forms
• We will begin with electrical systems. This will make modeling mech easier!
Modelling Electrical Systems (Nise)
• Current (i) is the rate of flow of charge (q)
• Taking Laplace
transform
• Impedance (complex resistance) is defined as
• It’s mathematical inverse is called admittance.
i t
=
dq(t
) d
t
⇒ q t = න i(t)
dt
I s =
sQ(s)
⇒
Q S =
I(s)
s
1
z
=
v
i I(s)
⇒ Z s
=
V(s)
)
77
Shadab. A. Siddique

Passive Electric
Components
• We will combine these elements in complex networks using Kirchoff’s
Laws
– Current and Voltages in a loop sum to zeros
78
Shadab. A. Siddique

Single Loop RLC Circuit
• Find transfer function of Vc(s) to input
V(s)
⇒
79
Shadab. A. Siddique

Simplifying the Procedure
• Let us look at this in another
way.
• Let’s define impedance (similar to resistance) as
• Unlike resistance, impedance is also applicable to capacitors &
inductors.
• It represents information about dynamic behavior of components.
c
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Solving Multi-Loop Electric
Circuits
• For multiple loops and loads, use the following recipe.
– Replace passive element values with their impedances.
– Replace all sources and time variables with their Laplace
transform.
– Assume a transform current and a current direction in each mesh.
– Write Kirchhoff’s voltage law around each mesh.
– Solve the simultaneous equations for the output.
– Form the transfer function.
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Multi-loop
Example
• Find the transfer function I2(s) / V
(s)
• Solving for Loop 1 and Loop 2
(1)
(2)
• There are various ways to solve this.
• This will yield the following transfer
function
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Summarizing the
Method
• Let us look at the pattern in the last
example
• This form will help us write such equations rapidly
• Mechanical equations of motion (covered next) have the same form. So, this
form is very useful!
83
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Class
Quiz
84
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Mechanical Systems (Translational)
• Many concepts applied to electrical networks can also be
applied to mechanical systems via analogies.
• This will also allow us to model hydraulic/pneumatic/thermal systems.
85

Electrical/Mechanical
Analogies
• Mechanical systems, like electrical networks, have 3 passive, linear
components:
– Two of them (spring and mass) are energy-storage elements; one of them,
the viscous damper, dissipates energy.
– The two energy-storage elements are analogous to the two electrical energy-
storage elements, the inductor and capacitor. The energy dissipater is
analogous to electrical resistance.
• Displacement ‘x’ is analogous to current I
• Force ‘f’ is analogous to voltage ‘v’
• Impedance (Z=V/I) is therefore Z=F/X
• Since, [Sum of Impedances] I(s) = [Sum of applied voltages]
• Hence, [Sum of Impedances] X(s) = [Sum of applied forces]
86
Shadab. A. Siddique

Electric Mechanical
Analogy
87
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Spring Mass Damper System
88
Shadab. A. Siddique

Transfer
Function??
• System has two degrees of freedom, since each mass can be moved in the
horizontal direction while the other is held still.
• 2 simultaneous equations of motion will be required to describe system.
• The two equations come from free-body diagrams of each mass.
• Forces on M1 are due to (a) its own motion and (b) motion of M2 transmitted to M1
through the system.
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Force Analysis
Forces on M1
a.Hold M2
still, move
M1 to right
b.Hold M1
still, move M2
to right
c.combined
Forces on M2
a.Hold M1
still, move
M2 to right
b.Hold M2
still, move M1
to right
c.combined
90
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Use
Analogy
91
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Homework
92
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Mechanical Systems
(Rotational)
• Torque replaces force; angular displacement replaces translational
displacement.
93
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Example#1:
Physical
system
Schematic of
the
system
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Homework
97
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UNIT-
I
(15
Marks)
Examples
➢ Block Diagrams:
98
S. A. Siddique

Force to Voltage Analogy:
Electrical Analogies of Mechanical
Systems
99
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Force to Current Analogy:
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