CONTROL SYSTEM ANALYSIS

CONTROL
SYSTEM
by - Mohammed Waris Senan

1 Mohammed Waris Senan
Course Outline
 Introduction
Mathematical Modelling, Transfer function models of LTI System, types of control
system, Block diagram technique and Signal flow graph.
 Time Response Analysis
Time response of 1st and 2nd order system, Design specification of 2nd order system and
Concept of stability (Routh-Hurwitz Criteria and Root Locus).
 Frequency Response Analysis
Polar Plot, Bode Plot, Nyquist Criteria etc.
 Introduction to Controller Design
Accuracy, disturbance rejection, sensitivity and robustness of control system, Design
specification in frequency domain, P, I, D, PID controller etc.
 State Variable Analysis
State Space model, Eigen values and stability analysis, concept of controllability and
observability, Difference equation of discrete time systems etc.

2 Mohammed Waris Senan
Introduction
A CONTROL SYTEM is a systematically arranged interconnection of components
which together act to provide a desired response when excited by an input.
OR
A system which controls the output quantity is called a control system.
Examples :
 Automatic Washing Machine.
 Immersion Rod.
 Air Conditioner.
 Fan with regulator.
 Motor pump to fill water in roof tank etc.
Plant (Process)
Input Output

Mohammed Waris Senan3
Introduction
 Command Input:
It is the externally produced input and is independent of the feedback
control system.
 Reference Input:
This produces the standard signals proportional to the command and is
directly used by the system. It is also called as actuating signal.
 Disturbance:
A disturbance is a signal that tends to affect the value of the output of a
system. If a disturbance is created inside the system, it is called internal.
While an external disturbance is generated outside the system.
 Feedback Control:
It is an operation that, in the presence of disturbance tends to reduce the
difference between the output of a system & some reference input.
 Servo Mechanism:
A servo mechanism is a feedback controlled system in which the output is
some mechanical position, velocity or acceleration.

Introduction
 Controlled Variable:
It is the quantity or condition that is measured & controlled.
 Controller:
Controller means measuring the value of the controlled variable of the system
& applying the manipulated variable to the system to correct or to limit the
deviation of the measured value to the desired value.
 Plant:
A plant is a piece of equipment, which is a set of machine parts functioning
together. The purpose of which is to perform a particular operation.
Example: Furnace, a C.R.O., Space craft etc.
 Process:
A progressive continuing operation with gradual changes leading towards a
specified result. Normally plant and process are used in the same sense.
 System:
A system is a combination of components that works together & performs
certain objective.

Control System Types
Control System
Open Loop
System
Closed Loop
System
Open Loop Control System:
 It is a system in which control action is independent of output.
 Control system without feedback.
 Actuating signal has no component of the output
 Examples: Traffic control signal, Toaster, Automatic washing machine etc.
Plant
(Process)
Input Output
Controller
Actuating
Signal

Closed Loop Control System:
 It is a system in which control action is dependent on desired output.
 Control system with feedback.
 Output is compared with the reference input and error signal is produced.
 Examples: Lift in multi storey building, Air Conditioner etc.
Plant
Command
Input
Output
Controller
Feedback
Element
Reference
Input Element
Reference
Input
∓
+
Error
Signal
Feedback
Signal
Error
Detector

Open Loop
Control System
• Construction and design are
simple, hence less expensive.
• Less accurate and less
reliable.
• No Feedback. Output has no
effect on input.
• Sensitive to disturbances and
environmental changes.
• Optimization is not possible
Closed Loop
Control System
• Design is complicated and
hence more expensive.
• Highly accurate and more
reliable.
• Feedback is present. Output
affects the input.
• Almost insensitive to both
disturbances and
environmental changes.
• Optimization is possible.

Design Consideration
Stability Accuracy
Speed of
Response
 For a given input, the response must attain and maintain
some useful value within a reasonable period of time.
 An unstable system produces persistent or violent
fluctuations.
 Unstable system has uncontrollable or unbounded output.
 Unstable systems are useless.
 System must be capable of reducing errors to zero or to a
small tolerable value.
 Accuracy is relative parameter with limits based upon the
particular application.
 As the cost of system increases with the increase of
accuracy, the limits should be made as wide as possible.
 All systems must provide the demand degree of accuracy.
 System must complete its response within an accepted
period of time with reasonable damping.
 If the time required to respond fully is far greater than the
time interval between inputs then the system has no real
value even if the system is stable and accurate.
 A system is to be designed so as to meet all three
requirements of stability, accuracy and speed of response.
 BUT this is not always possible. A stable system is generally
less accurate. A more accurate system is less stable.
 When both the conditions are met, the speed of response
may not be desired.
 Therefore it is necessary to make the effective compromise
between these requirements.

Transfer Function
Laplace transform of output to the Laplace transform
of input with all initial conditions are zero.
System
Input, r(t) Output, c(t)
=
𝐿 {𝑐 𝑡 }
𝐿{𝑟 𝑡 }
Transfer Function (All initial conditions are zero)
𝐺(𝑠) =
𝐶(𝑠)
𝑅(𝑠)

Transfer Function
Systemr(t) c(t)
constants.are''&''Where
)(
)(
......
)()(
)(
)(
......
)()(
011
1
1011
1
1
ba
trb
dt
tdr
b
dt
trd
b
dt
trd
btCa
dt
tdC
a
dt
tCd
a
dt
tCd
a m
m
mm
m
mn
n
nn
n
n  




)()......()()......( 01
1
101
1
1 sRbsbsbsbsCasasasa m
m
m
m
n
n
n
n  



Taking Laplace Transform:
)(
)......(
)......(
)(
)(
,
01
1
1
01
1
1
sG
asasasa
bsbsbsb
sR
sC
So n
n
n
n
m
m
m
m



 



G(s)R(s) C(s)

Transfer Function
Characteristic Equation
Characteristic equation of a linear system can be obtained by
equating the denominator polynomial of transfer function to zero
)......(
)......(
)(
)(
01
1
1
01
1
1
asasasa
bsbsbsb
sR
sC
G(s) n
n
n
n
m
m
m
m


 



0...... 01
1
1  
 asasasa n
n
n
n
So, Characteristic Equation will be:

Transfer Function
)......(
)......(
)(
)(
01
1
1
01
1
1
asasasa
bsbsbsb
sR
sC
G(s) n
n
n
n
m
m
m
m


 



Where, K =
𝑏 𝑚
𝑎 𝑛
is known as the gain factor
and ‘s’is the complex frequency
))()((
))()((
2
21
2
21
CBsAspsps
cbsaszszsK
G(s)



Poles & Zeros

Transfer Function
))()((
))()((
2
21
2
21
CBsAspsps
cbsaszszsK
G(s)



Poles : The poles of G(s) are those values of ‘s’which make G(s) tends to infinity.
For example, s = -p1, s = -p2 and 𝑠 =
−𝐵± 𝐵2 − 4𝐴𝐶
2𝐴
Zeros : The zeros of G(s) are those values of ‘s’which make G(s) tends to zero.
For example, s = -z1, s = -z2 and 𝑠 =
−𝑏± 𝑏2 − 4𝑎𝑐
2𝑎
 Simple Poles (or Zeros) : Poles (or Zeros) are distinct.
 Multiple Poles (or Zeros) : Poles (or Zeros) coincides (repeated).
Poles & Zeros

Transfer Function
Poles & Zeros
2
)2)(3(
)4(50



sss
s
G(s)
Simple Poles : s = 0 and s = -3
Multiple Poles : s = -2, -2
Simple Zero : s = -4
Two case arises if we consider entire s-plane:
1. If Z < P then the value of transfer function becomes zero for s → ∞ . Hence we say
that there are zeros at infinity and the order of such zeros is P – Z.
2. If P < Z then the value of transfer function becomes infinity for s → ∞ . Hence we say
that there are poles at infinity and the order of such poles is Z – P.
Number of poles (P) = 4
Number of zeros (Z) = 1
Therefore, in addition to finite poles and zeros, if we consider poles and zeros at infinity,
then for a rational function total number of poles and zeros are equal.

Transfer Function
Poles & Zeros
Simple Poles : s = 0 and s = -3
Multiple Poles : s = -2, -2
Simple Zero : s = -4
Imaginary (jω)
Real (σ)
-4 -3 -2 -1-5 0
S-plane

Transfer Function
Example 1:
Find the transfer function of the given network.
Solution:
Convert the given network in s-domain.
s
sI
idt
ssI
dt
di
sC
C
sLL
RR
s-DomainnTime Domai
)(
)(
1

(s)
sLR
sL
(s) iVV0 







sLR
sL
(s)
(s)
i 

V
V0
sLR
sL
(s)
(s)
s
i 

V
V
)(G 0
𝑠𝐿
𝑅 + 𝑠𝐿
Vi(s) V0(s) G(s)Vi(s) V0(s)

Transfer Function
Example 2:
Solution:
Convert the given network in s-domain.
(s)
sC
sLR
sC(s) iE
1
1
E0














LCL
R
ss
LC
(s)
(s)
i 1
1
E
E
2
0


LCL
R
ss
LC
(s)
(s)
s
i 1
1
E
E
)(G
2
0


V0(s)Vi(s)
LCL
R
ss
LC
1
1
2


Transfer Function
Question 1:

Mechanical System
1. Translational System
2. Rotational System
Motion takes place in straight line.
Translational System
Forces that resists translational motion are:
 Inertia Force
 Damping Force
 Spring Force
M
x(t)
FM(t)
)()( tMatFM 
dt
tdv
MtFM
)(
)( 
2
2
)(
)(
dt
txd
MtFM 

Mechanical System
x(t)
)()( tBvtFD 
 Damping Force
FD(t)
B
dt
tdx
BtFD
)(
)( 
 Spring Force
)()( tKxtFK 
x(t)
FK(t)
K
Translational System

21
Mechanical System
Motion of a body about a fixed axis.
Rotational System
Torques that resists rotational motion are:
 Inertia Torque
 Damping Torque
 Spring Torque
Mohammed Waris Senan
J
TI(t)
θ(t)
2
2
)(
)(
dt
td
JtTI


 Damping Torque
dt
td
BtTD
)(
)(


 Spring Torque
)()( tKtT  

22
Mechanical System
Translational & Rotational System
)(
)()(
)(
)()(
2
2
2
2
tK
dt
td
B
dt
td
JSystemRotational
tKx
dt
tdx
B
dt
txd
MSystemnalTranslatio


SpringDampingInertiaSystemsMechanical

23
Mechanical System
D’ALEMBERT’S PRINCIPLE
For any body, the algebraic sum of externally applied forces (or
torques) and forces (or torques) resisting motion in any given
direction is zero
K
M
B
F (t)
x(t)
External Force : F(t)
Resisting Forces :
)()(,
)(
)(,
)(
)(,
2
2
tKxtFForceSpring
dt
tdx
BtFForceDamping
dt
txd
MtFForceInertia
K
D
M



3.
2.
1.

24
Mechanical System
M
F (t)
FK(t) FM(t) FD(t)
According to D’Alembert’s Principle :
0)(
)()(
)(
2
2
 tKx
dt
tdx
B
dt
txd
MtF
F(t) + FM(t) + FD(t) + FK(t) = 0
)(
)()(
)(
2
2
tKx
dt
tdx
B
dt
txd
MtF 

25
Mechanical System
External Torque : T(t)
Resisting Torques :
)()(,
)(
)(,
)(
)(,
2
2
tKtTTorqueSpring
dt
td
BtTTorqueDamping
dt
td
JtTTorqueInertia
K
D
I






3.
2.
1.
T(t)J
K
B

26
Mechanical System
T(t)J
K
B
0)(
)()(
)(
2
2
 tK
dt
td
B
dt
td
JtT 

)(
)()(
)(
2
2
tK
dt
td
B
dt
td
JtT 

 J
T(t)
TI(t) TD(t) TK(t)
According to D’Alembert’s Principle :
T(t) + TI(t) + TD(t) + TK(t) = 0

27
Mechanical System
Example 3:
Write differential equations governing the mechanical system as shown.
Solution:
Free body diagram for M1
F (t)
M1
)(1
tFM
)(1
tFB)(1
tFK
)1()}()({
)}()({)(
)( 211
21
12
1
2
1 

 txtxK
dt
txtxd
B
dt
txd
MtF
)()()()( 111
tFtFtFtF KBM 

28
Mechanical System
Solution:
Free body diagram for M2
M2
)(2
tFM
)(1
tFB)(1
tFK
)2()(
)()(
)}()({
)}()({
22
2
22
2
2
2121
12
1 

txK
dt
tdx
B
dt
txd
MtxtxK
dt
txtxd
B
)(2
tFB)(2
tFK
)()()()()( 22211
tFtFtFtFtF KBMKB 

29
Mechanical System
Example 4:
Write differential equations governing the mechanical rotational system as
shown.
Solution:
Free body diagram for J1
)(1
tTJ
)()()()( 111
tTtTtTtT KBM 
J1
T(t)
)(1
tTB )(1
tTK
)1()}()({
)()(
)( 211
1
12
1
2
1  ttK
dt
td
B
dt
td
JtT 


30
Mechanical System
Solution:
Free body diagram for J2
)(2
tTJ
)()()()( 2221
tTtTtTtT KBJK 
J2
)(2
tTB
)(1
tTK
)2()(
)()(
)}()({ 22
2
22
2
2
2121  tK
dt
td
B
dt
td
JttK 


)(2
tTK

31
Mechanical Network
Steps to draw mechanical network:
1. Mark independent nodes for each displacement.
2. Connect each element between the nodes that correspond to the two
displacements at each end of that element.
3. The mass and moment of inertia are always connected from the
reference node to the node representing its displacement.
4. The spring and dashpot elements are connected to the two nodes that
represent the displacement of each end of the element.
5. Force/Torque equation is written for each node by equating the sum of
the forces/torques at each node to zero, a technique similar to the
NODALANALYSIS.

32
Mechanical Network
x(t)
M BKF (t)
0)()()()(  tFtFtFtF KBM )()()()( tFtFtFtF KBM 
)(
)()(
)(
2
2
tKx
dt
tdx
B
dt
txd
MtF 

33
Mechanical Network
x1
M1
BK1F(t)
Example 5:
Draw mechanical network for the mechanical system as shown.
x2
M2
K2
Solution:

34
Mechanical Network
Solution:
)2(0)(
:NodeAt
1222
2
2
2
2
 xxK
dt
xd
M
x
)1(}{)(
:NodeAt
21211
1
12
1
2
1
1
 xxKxK
dt
dx
B
dt
xd
MtF
x

35
Analogous Circuits
1. Force – current (f-i) analogy.
2. Force – voltage (f-v) analogy.
)(
)()(
)(
2
2
tKx
dt
tdx
B
dt
txd
MtF 
MmassofVelocity
dt
tdx
tvLet
)(
)( 
 dttvKtBv
dt
tdv
MtF )()(
)(
)(

36
Analogous Circuits
 dtte
L
te
Rdt
tde
Cti )(
1
)(
1)(
)(

37
Analogous Circuits
 dtte
L
te
Rdt
tde
Cti )(
1
)(
1)(
)(
 dttvKtBv
dt
tdv
MtF )()(
)(
)(

38
Analogous Circuits
L
K
R
B
CM
tetv
titF
1
1
)()(
)()(





)()( ttx 

39
Analogous Circuits
 dttvKtBv
dt
tdv
MtF )()(
)(
)(  dtti
C
tRi
dt
tdi
Lte )(
1
)(
)(
)(

40
Analogous Circuits
C
K
RB
LM
titv
tetF
1
)()(
)()(





)()( tqtx 

41
Analogous Circuits
Example 6:
Draw mechanical network and write differential equation and also draw f-i and
f-v analogy for the mechanical system as shown.

42
Analogous Circuits
x1
M1
B1
K1F(t)
x2
M2K2
B2
B3
Solution:
Mechanical Network :

43
Analogous Circuits
Solution:
)2(0
)(
:NodeAt
12
322
2
22
2
2
2
2



dt
xxd
BxK
dt
dx
B
dt
xd
M
x
)1(
)(
)(
:NodeAt
21
311
1
12
1
2
1
1



dt
xxd
BxK
dt
dx
B
dt
xd
MtF
x

44
Analogous Circuits
Solution:
e1(t)
i (t)
e2(t)
f-i analogy :








1
1
1
B
R








2
2
1
B
R








2
2
1
K
L 2C








1
1
1
K
L 1C








3
3
1
B
R

45
Analogous Circuits
Solution:
f-v analogy :
i1(t) i2(t)3R+
−e (t)
1R 2R2L1L 







1
1
1
K
C 







2
2
1
K
C

46
Analogous Circuits
Question 1:
Draw mechanical network and write differential equation and also draw f-i and
f-v analogy for the mechanical system as shown.

47
Block Diagram
G(s)
H(s)
R(s)
∓
+
E(s)
C(s)

48
Block Diagram
Block diagram reduction technique
1. Blocks in cascade
G2(s)R(s) C(s)G1(s)
R(s) G1(s)×G2(s) C(s)
(s)G(s)G
R(s)
C(s)
G(s) 21 

49
Block Diagram
2. Blocks in parallel
G2(s)R(s) C(s)
G1(s)
G3(s)
+
+
+
R(s) G1(s) + G2(s) + G3(s) C(s)
(s)G(s)G(s)G
R(s)
C(s)
G(s) 321 

G(s)
50
Block Diagram
3. Moving a take off point ahead of a block
R(s) C(s)
C(s)
G(s)
R(s)C(s)
R(s) C(s)G(s) R(s) C(s)G(s)
G(s)


1
𝐺(𝑠)
51
Block Diagram
R(s) C(s)G(s)
R(s)
4. Moving a take off point after a block
R(s) C(s)
R(s) C(s)G(s) R(s) C(s)G(s)
1
𝐺(𝑠)


G(s)
52
Block Diagram
5. Moving a summing point after a block
x C(s)
y
+
−
y
C(s)x G(s)
x C(s)
+
−
G(s)
y G(s)
x
y
+
−
C(s)G(s)


53
Block Diagram
1
𝐺(𝑠)
+
−
C(s)G(s)R(s)
6. Moving a summing point ahead of a block
X(s)
X(s)

+
−
C(s)G(s)
X(s)
R(s) C(s)
+
−
G(s)R(s)
1
𝐺(𝑠)
X(s)

54
Block Diagram
7. Moving a take off point after a summing point
+
−
y
x
x
x - y
x - y
+
y
x
+
−
y
x
x
x - y +
−
y
x
x - y
x - y
+
y
x


55
Block Diagram
8. Moving a take off point ahead of a summing point
+
−
y
x x - y
x - y
y
x
−
x - y
+
−
y
x x - y
x - y
+
−
y
x
x
x - y
y
−
x - y


56
Block Diagram
9. Closed loop transfer function
G(s)
H(s)
R(s)
−
+
E(s)
C(s)
)()(1
)(
sHsG
sG

R(s) C(s)
)()(1
)(
sHsG
sG

R(s) C(s)
Negative feedback Positive feedback

57
Block Diagram
Example 7:
Determine the transfer function using block reduction technique.
G1
R(s)
−
+
C(s)G2
−
+

58
Block Diagram
Example 8:
Determine the transfer function using block reduction technique.
G1R(s) −
+ C(s)
−
+ G2 G3 G4
H2
−
+
H1
AKU, 2019 [6 marks]

59
Signal Flow Graph (SFG)
Signal Flow Graph (SFG) is a graphical representation of system equations
and, in this sense, serves the same purpose as a block diagram
R(s) C(s)
G
−H
1 E(s)

60
R(s) C(s)
−H2
G1
SOME IMPORTANT DEFINITION
1. Forward path and forward path gain:
G4G3G2
G5
−H1
−H3
 A forward path is the path from input node to output node that does not
enter any node more than once and follows the direction of the signal flow.
 The forward path gain is defined as the product of gains in a forward path.
P1 =
G1 G4G3G2P2 =
G5
G1

G3
−H1
G2
−H2
G2
61
R(s) C(s)
2. Loop and loop gain:
G4G3
G5
−H1
−H3
 A loop is a path in a signal flow graph that begins and ends at the same
node without entering any node more than once and following the
direction of the signal flow.
 The loop gain is defined as the product of gains in the branch of the loop.
L1 =
G1L2 =
L3 =
−H2
−H3

G3
−H2
G2
62
R(s) C(s)
3. Nontouching Loop:
G4
G5
−H1
−H3
 Nontouching loops are loops that have no common nodes.
G1
4. Input node:
5. Output node:
It is a node having only outgoing branches.
It is a node having only incoming branches.
L1 = G2−H1
L3 = −H3

63
MASON’S GAIN FORMULA
Where,
𝐶(𝑠)
𝑅(𝑠)
= Transfer function
PK = Gain of Kth forward path.
Δ = Determinant of the graph.
= 1 − ∑ loop gains + ∑ (product of non-touching loop gains taken two at a
time) − ∑ (product of non-touching loop gains taken three at a time) + ……
ΔK = Δ − ∑ loop gain terms in Δ that touches the Kth forward path.
= The part of Δ not touching the Kth forward path.



K
KKP
sR
sC
sT
)(
)(
)(

64
G3
−H2
G2
R(s) C(s)
G4
G5
−H1
−H3
G1
Example 9:
Find
𝐶(𝑠)
𝑅(𝑠)
for the given SFG.

G3
−H2
G2
65
R(s) C(s)
G4
G5
−H1
−H3
G1
P1 = G1 G2 G3 G4
P2 = G1 G5
Solution:
Forward path gain
Loop gain
L1 = −H1 G2
L2 = −H2 G3
L3 = −H3
Δ1 = 1− [0 + 0 + 0] + [0] = 1

G3
−H2
G2
66
R(s) C(s)
G4
G5
−H1
−H3
G1
P1 = G1 G2 G3 G4
P2 = G1 G5
Solution:
Forward path gain
Loop gain
L1 = −H1 G2
L2 = −H2 G3
L3 = −H3
Δ1 = 1− [0 + 0 + 0] + [0] = 1
Δ2 = 1− [L2 + L3 + 0] + [0] = 1 − [−H2 G3 − H3] = 1 + H2 G3 + H3

67
P1 = G1 G2 G3 G4
P2 = G1 G5
Solution:
Forward path gain Loop gain
L1 = −H1 G2
L2 = −H2 G3
L3 = −H3
Δ1 = 1− [0 + 0 + 0] + [0] = 1
Δ2 = 1 + H2 G3 + H3
Δ = 1 − [L1 + L2 + L3] + [L1L3 + 0]
= 1 − [−H1 G2 −H2 G3 −H3] + [− H1 G2 (−H3)]
= 1 + H1 G2 + H2 G3 − H3 + H1 G2 H3
Δ = 1 − ∑ loop gains + ∑ (product of non-touching loop gains taken two at a time)
− ∑ (product of non-touching loop gains taken three at a time) + ……

68
Solution:


 2211
)(
)(
)(
PP
sR
sC
sT



K
KKP
sR
sC
sT
)(
)(
)(
23133221
332514321
1
]1[1
)(
GHHHGHGH
HGHGGGGGG
sT



P1 = G1 G2 G3 G4
P2 = G1 G5
Δ1 = 1− [0 + 0 + 0] + [0] = 1
Δ2 = 1 + H2 G3 + H3
Δ = 1 + H1 G2 + H2 G3 − H3 + H1 G2 H3

69
Example 10:
Find
𝐶(𝑠)
𝑅(𝑠)
for the given SFG.

70
Solution:
G3
−H2
G2
R(s) C(s)
G4
G5
−H1
G1
G6
G8
G7
1 1
P1 = G1 G2 G3
P2 = G4 G5 G6
P3 = G1 G8 G6
Forward Path

71
Solution:
G3
−H2
G2
R(s) C(s)
G4
G5
−H1
G1
G6
G8
G7
1 1
P1 = G1 G2 G3
P2 = G4 G5 G6
P3 = G1 G8 G6
P4 = G4 G7 G3
Forward Path

72
Solution:
G3
−H2
G2
R(s) C(s)
G4
G5
−H1
G1
G6
G8
G7
1 1
P1 = G1 G2 G3
P2 = G4 G5 G6
P3 = G1 G8 G6
P4 = G4 G7 G3
P5 = −H1G4 G7 G8 G6
Forward Path

73
Solution:
G3
−H2
G2
R(s) C(s)
G4
G5
−H1
G1
G6
G8
G7
1 1
P1 = G1 G2 G3
P2 = G4 G5 G6
P3 = G1 G8 G6
P4 = G4 G7 G3
P5 = −H1G4 G7 G8 G6
Forward Path
P6 = −H2G1 G8 G7 G3

74
Solution:
G3
−H2
G2
R(s) C(s)
G4
G5
−H1
G1
G6
G8
G7
1 1
L1 = −H1 G2
L2 = −H2 G5
Loop

75
Solution:
G3
−H2
G2
R(s) C(s)
G4
G5
−H1
G1
G6
G8
G7
1 1
L1 = −H1 G2
L2 = −H2 G5
Loop
L3 = H1 H2 G7 G8

76
Solution:
G3
−H2
G2
R(s) C(s)
G4
G5
−H1
G1
G6
G8
G7
1 1
Δ1 = 1− [0 + L2 + 0] + [0] = 1 + H2G5

77
Solution:
G3
−H2
G2
R(s) C(s)
G4
G5
−H1
G1
G6
G8
G7
1 1
Δ1 = 1− [0 + L2 + 0] + [0] = 1 + H2G5
Δ2 = 1− [0 + L1 + 0] + [0] = 1 + H1G2

78
Solution:
G3
−H2
G2
R(s) C(s)
G4
G5
−H1
G1
G6
G8
G7
1 1
Δ1 = 1− [0 + L2 + 0] + [0] = 1 + H2G5
Δ2 = 1− [0 + L1 + 0] + [0] = 1 + H1G2
Δ3 = 1− [0 + 0 + 0] + [0] = 1
Δ4 = 1− [0 + 0 + 0] + [0] = 1
Δ5 = 1− [0 + 0 + 0] + [0] = 1

79
Solution:
G3
−H2
G2
R(s) C(s)
G4
G5
−H1
G1
G6
G8
G7
1 1
Δ1 = 1− [0 + L2 + 0] + [0] = 1 + H2G5
Δ2 = 1− [0 + L1 + 0] + [0] = 1 + H1G2
Δ3 = 1− [0 + 0 + 0] + [0] = 1
Δ4 = 1− [0 + 0 + 0] + [0] = 1
Δ5 = 1− [0 + 0 + 0] + [0] = 1
Δ6 = 1− [0 + 0 + 0] + [0] = 1

80
Solution:
L1 = −H1 G2
L2 = −H2 G5
L3 = H1 H2 G7 G8
P1 = G1 G2 G3
P2 = G4 G5 G6
P3 = G1 G8 G6
P4 = G4 G7 G3
P5 = −H1G4 G7 G8 G6
P6 = −H2G1 G8 G7 G3
Forward Path gain Loop gainΔ1 = 1− [0 + L2 + 0] + [0] = 1 + H2G5
Δ2 = 1− [0 + L1 + 0] + [0] = 1 + H1G2
Δ3 = 1− [0 + 0 + 0] + [0] = 1
Δ4 = 1− [0 + 0 + 0] + [0] = 1
Δ5 = 1− [0 + 0 + 0] + [0] = 1
Δ6 = 1− [0 + 0 + 0] + [0] = 1
Δ = 1 − [L1 + L2 + L3] + [L1L3 + 0]
= 1 − [−H1 G2 −H2 G5 + H1 H2 G7 G8] + [− H1 G2 H1 H2 G7 G8]
= 1 + H1 G2 + H2 G5 − H1 H2 G7 G8 − H1
2 G2 H2 G7 G8
Δ = 1 − ∑ loop gains + ∑ (product of non-touching loop gains taken two at a time)
− ∑ (product of non-touching loop gains taken three at a time) + ……

81
Solution:


 665544332211
)(
)(
)(
PPPPPP
sR
sC
sT



K
KKP
sR
sC
sT
)(
)(
)(
8722
2
187215221
37812678413746812165452321
1
]1[]1[
)(
GGGHHGGHHGHGH
GGGGHGGGGHGGGGGGGHGGGGHGGG
sT




CONTROL SYSTEM ANALYSIS

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CONTROL SYSTEM ANALYSIS