Automatics Control Notes

Automatic Control Has Application In many Areas:
# Mechanical Engg.
* Power Plant
* Engines
* Production
* process Plant
* transportation
* Defense Applications
# Electrical Engg.
# Textile Engg.
# Aerospace Engg.
11/18/2020 DINESH PANCHAL

Main component involved in automatic control may be electrical, mechanical,
hydraulic, pneumatic depending on the application.
Types of control Systems :
Control system controls the energy inputs to a system and gives the desired
output. Control system may be classified as:
1. Open or closed Loop System
2. Analog or Digital Control System
3. Regulators and Servomechanisms
4. Sequence Control

Open Loop System:
In this system output of the system is no measured and output is not
compared with the input value. In this system output value varies as the input
is varied.

Closed Loop System:
In this system , the actual value of the variable being controlled is
measured and compared with the desired value and action is taken to eliminate
the error. This system is also called feedback systems.

Analog Control System:
In this system , an analog signal varies continuously and can have any
value in the given range.

Digital Control System:
In this type of control system the signal have discrete values. In this
system a digital computer is used for comparing the actual and desired values
of the controlled variables by using specified control algorithm .It is used for
complex control systems e.g. boiler control system. In this system the control
devices are of analog type but these analog signal are converted in to the digital
signals and vice versa in digital control system.

Digital Control System:

Regulators:
Regulators are the control systems in which the controlled variable in
the system is fixed but in some cases it is not changed much as in voltage
control. Another case is the temp. control system where the temp is to be
maintained constant.
Servomechanism:
This type of control system is employed where the controlled variable
is changing time to time e.g. in case of radar or gun control the position is often
changed and the target may occupy any popsition.

Sequence Control:
it is the type of control system in which the set of operation are to be
performed in a given sequence.
e.g. In a cloth washing machine
- Filling of water in the tub
- Washing
- Draining the tub
- Rinsing
- Spin Drying
In this each operation is controlled by the timer
.

another example of this type of control system is used in production machines
- Job in position
- Guard in position
- tool in position
- tool motion
- Tool withdrawal
- Job withdrawal
This type of control system uses logic control devices.
.

S.NO. OPEN LOOP CLOSED LOOP
1 These are not reliable These are reliable
2 It is easier to build It is difficult to build
3 If calibration is good, they perform
accurately
they are accurately
4 They are more stable They are less stable
5 Optimization is not possible Optimization is possible
6 They are simple They are complex
7 Less maintenance high maintenance

Typical simplified block diagram of closed loop system
is shown in fig.
r : reference input or desired input or set point
c : controlled variable or actual value of controlled
variable or output
. 11/18/2020 DINESH PANCHAL

m : manipulated variable, which is manipulated to control the system
b : disturbance or load input
e : error which is the difference of two signals.
Comparator :
Comparison element like potentiometer or digital computer
Control Elements :
These are the functional elements which modifies the error signals like
differentiating or integrating or amplifying ect. They may also include amplifier
and final control elements.
.

Amplifier:
It magnify the input signals by using external energy. E.g. Hydraulic,
Pneumatic or electrical amplifier.
Final Control Elements :
Elements which are adjacent to the system being controlled e.g.
control valve, servomotor, drive unit
.

System or Process or Plant:
The main element which require control of its variables,
e.g. rotating system, needs speed control, inertia system needs
position control.
Feedback Elements:
Measuring elements or sensor ment for measure the actual
value of controlled variable, for comparison with the desired
input value.
.

 -To find out the performance of control system one has to determine the
error over a period of time to different inputs.
 - Error is the difference between the desired value and the actual value.
 As the system is dynamic in nature so frequency or time domain analysis
is done.
 If the feedback system is not designed properly , it may become unstable
under certain conditions. So system is to be checked for the stability.
 So Routh, Nyquist, Bode, Root Locus techniques are used to check
dynamic performance of system.

 As we know that hardware used in the control system may be electrical,
hydraulic or pneumatic or combination of these.
 The first aim is to find out the governing equitation's or transfer functions
of various element
 After that the overall transfer function of the system is to be found out.
 The overall equitation is to be used to find out the performance of the
system.

 Draw the Block Diagram of the system given below.

1. Mathematical Modeling:
By using the physical laws and governing equations between input
and outputs of a system is derived .The elements in the system may be
electrical, mechanical, pneumatic.
If a function Y is dependent on X which is independent variable

Y=F(X)
A small variation ΔY from operating point O is
If the function Y depends on independent variable
Y= F( )
X
dX
dY
Y  0
.........,, ,321 XXX
.........,, ,321 XXX
.................20
2
10
1






 X
X
Y
X
X
Y
Y

2. Block Diagram Representation:
(A) Transfer Function
y= Ax
A= y/x
= Output/Input Where A is the transfer
Function

(B) Summing Junction:
Summing Junction is used to represent the addition or
subtraction of signals
c = a + d – b
Sum of variable leaving the Junction = Some of variable entering the
Junction

( C) Comparison Element:
This is used to compare the desired output to the actual output
e = Input – Output = r - c

Representation of Spring Mass System:
Which may be written as D
D
dt
d
isdotBut
tfkxxcxm
tfkxxcxm



).(
)(
)(
...
...

 
   
Kcsmssf
sx
KcDmDtf
tx
tfKcDmDtx
tftKxtcDxtxmD
tfKxcDxxmD







2
2
2
2
2
1
)(
)(
1
)(
)(
sbyreplacedisDtionTransformaLaplaceAfter
)(
)()()()(
)(
isFunctiontransferDomainTime

1. Liquid, Gas, Thermal System:
These systems are analogous to electrical system
Level, Pressure, Temp. = Potential
liquid or gas flow rate, heat flow rate = current
Tank =Capacitance
These analogy is useful for driving the governing equations.

Suppose a tank of capacity C to which there is a inflow of liquid at a rate qi
at a pressure pi and outflow rate q0 from the tank. There is a pressure
drop at inlet due to resistance R .If the level of the liquid in the tank is h

Pressure at the bottom of tank is
As per Electrical Analogy
As
……………(1)
As CV=Q As per Electrical Analogy
…………….(2)
ghp 
i
i
q
pp
R


I
VV
R 21 

0
0So
qrateflow
toanalogusiscurrentandpressuretoanalogusisVAs
Currenti
dt
dQ
QCV
qqCDp
qq
dt
dp
C
dt
dV
C
dt
dV
C
CVQ
i
i





Now Substituting the value of qi from eq. 1 to eq. 2
………………………(3)
)1(
)1(
0
0
0
0
0
0











RCD
Rqp
p
RqpRCDp
RqppRCDp
ppRqRCDp
R
pp
qCDp
q
R
pp
CDp
i
i
i
i
i
i

Vol. of liquid accumulated in the tank is
This increase in vol. increase the height (h) of water in the tank of area A
Increase in Vol. =
So ………………………..(4)
And p=hρg …………………………………..(5)
0qqi 
dt
dh
A
oi qq
dt
dh
A 

=
4
128
RisResistanceBut
eq.aboveinhofvaluePut this
g
p
h
4and2eq.From
d
L
g
A
C
g
ADp
CDp
ghp
ADh
dt
dh
ACDp












Gas Pressure System:
11/18/2020 DINESH PANCHAL)1(1(
)1(
)1(
)(
)(
1noeq.inofvalueput theNow
andSoVBut
1...................................qrateflowso
CDVcurrent
0
0
0
0
0
0
21
i
RCD
Rq
RCD
p
p
RCD
Rqp
p
RCDpRqp
pRCDpRqp
RCDpRqpp
CDpq
R
pp
q
R
pp
q
RqppiRV
CDpq
i
dt
dv
C
dt
dq
CVq
i
i
i
i
i
i
i
i
i
ii
o



















C is the capacitance of the system
As we know gas eq. pV=m R T
From eq. no. 1
=dm/dt Rate of accumulation of mass
=Dm
So
From eq. p V= m R T , Now put the value of
m/p in above eq.
And
CDpqqi  0
p
m
C
Cpm


RT
V
p
m

RT
V
C 
52
8
d
fl
R



Mechanical Rotating System:
Here we have to determine the governing equation of the
rotating system
Torque developed by Motor = u
Load Torque =
Speed of Rotation= ω
Moment of Inertia of Rotating Mass = I
Viscous damping Coefficient = B
Total Torque=Load Torque + Torque for Inertia +
Damping Torque
So Governing Eq. is
Lu
 BIDuu L 

Geared System:
Torque developed by Motor =
Load Torque =
Inertia =
Speed of Rotation=
Gears are used between motor and load for different speed and torque. So
torque and rotation are different at the two ends.
MU
LU
21, JJ
21,

























1
2
...
1
2
21
2
2
2
1
1
22
2
2
1
1
2
2
2
2
22
2
11
2
2
2
22
2
11
eq
3...................................................
Similarly
2.................................................
bysidebothDivide
1.....................................
2
1
2
1
2
1
JissystemtheofInertiaTotal











LMeqeqM
eq
EQ
EQ
EQ
EQ
UBJU
BBB
JJJ
JJJ
JJJ
JJJ

Fig. shows hydraulic control system. This is used
to controlling heavy loads, with application of
small force. It may be used in machine tools,
governing of turbine, ship steering system . High
pressure pump supplies oil to hydraulic unit.
Input motion given to system = x
Output motion given by system is = y
When x motion is given to the end A then upper
port is uncovered and high pressure oil flow
through upper port and moves the power piston
in downward direction by an amount of y. The
piston valve moves by an amount e
From the diagram
…………………………….1
21 eee 

From similar triangle ACX and Bce
From similar triangle ACy and Bcy
Now put the value if e1 and e2 in eq .1
………..2
ba
xb
e
AC
xBC
e
AC
x
BC
e




1
1
1
ba
ya
e
AC
yAB
e
AC
y
AB
e




2
2
2
2
yx
e
bifa
ba
ay
ba
bx
e








diagramblockdrawcanweequatationtheFrom
abovetheFrom
pistontheofareatheisAWhereAq
tCoefficienPortCeWherCqSo
movementpistonthetoalproportiondirectlyisflowoilofRate
a
bx
y
2no.eq.fromThen.conditionstatesteadyunderissystemthen0eWhen
1
1
11
11
11
DA
eC
y
DyAeC
Dy
e







A control valve is the final control element in a fluid
Process system and control the flow rate of the
system
From the fig pressure applied to the diaphragm is p
Spring constant of Diaphragm = K
Moving mass of valve = m
Damping Constant = B
Force Applied by Diaphragm on Valve= F = p(t) A…..1
For a spring Mass system eq. is given as
 
   
KBDmD
A
tp
tx
KBDmDtxF
tKxtBDxtxmDF
KxxBxmF





2
2
2
)(
)(
2and1no.eq.From
2.......................)(
)()()(


Comparison elements or comparator find out the difference between the input
and actual value of the controlled variable and gives a signal proportional
to the error or difference
1. Potentiometer type :
This is the simplest type of error detector, which produces a signal voltage
which us proportional to the error signal.
Potentiometer may be rotary or linear
Suppose Input =
Output =
Two identical rotary potentiometer are
Connected as shown in fig and the
voltage e(t) depend on the difference
Of position i.e.
And it will be zero if position of these
Two are same.
r
c
crand cr  

Block Diagram and Transfer Function Representation
ABC
r
q
ABC
r
q
r
p
p
m
m
q
r
q
m
q
C
p
m
B
r
p
A




**
**
FunctionTransferOverall
,,
FunctionTransfer

From the above given Fig.
Input = r
Output = c
Disturbance = b
Transfer Function are A, G1, G2 and H
 bcHrAGGc
mGc
bcHrAGm
bmm
cHrAGm
eGm
cHrAe







)(
)(
)(
12
12
11
1
1
1

 
 
b
HGG
G
r
HGG
AGG
c
HGG
bGrAGG
c
bGrAGGHGGc
bGrAGGcHGGc
bGcHGGrAGGc
bcHGrAGGc
bcHrAGGc
)1()1(
)1(
)1(
)(
12
2
12
12
12
212
21212
21212
21212
112
12













Fig. shows the temp. control
System Thermal capacitance = C
Actual value of temp. at any time =ϴ
The disturbance is due to the heat
loss to the surrounding and temp
is controlled by transferring the heat
To the chamber .
Heat Inflow =
Suppose to set the temp of chamber
by moving the input lever by z upward .
As the lever Pivots about P .So the
movement Of piston valve by z/2 The
piston covers both ports under Steady
state condition. Due to upward motion
of piston power piston moves y amount
in
inq
r

This apply force on rheostat contact point and rheostat put the smaller
resistance. Hence voltage increases across the heater and gives more heat
and increase the temp of chamber.Due to increase in temp. liquid in
feedback bellow expand and moves the lever by x. Due to x point p moves
downward. Hence piston moves x/2 in downward.
Lever input Z is proportional to
Flow rate of oil through port = be = Ady ……………………… 3
Where b is the port coefficient A is the area of the piston
Heat input is proportional to the y
net heat accumulate in the chamber and increase the temp of
chamber
1............................................
22
xz
e 
r
2................................................1 rCz 
4.........................................2 yCqin 
oin qq 

So
C is the thermal capacitance of the chamber and is the rate of heat loss to
the surrounding
Heat Loss =
Where R is the resistance to heat loss
X is directly proportional to the temp of the chamber
From eq. 5 and 6 eliminate
5..................................CDqq oin 
oq
6...........................
R
q a
o
 


7..........................................4Cx 
oq
8............................
1
1
















 













R
CDR
R
q
R
CD
R
q
R
CD
R
q
CD
RR
q
CD
R
q
a
in
a
in
a
in
a
in
a
in

From the above eq.































 






 






 






 












 






 






 

AD
bCC
R
RCD
RAD
bCC
AD
bCC
R
RCD
RAD
bCC
R
RCD
RAD
bCC
AD
bCC
R
RCD
RAD
bxC
AD
bzC
R
RCD
R
xz
AD
bC
R
RCD
RAD
be
C
R
RCD
R
yC
R
RCD
R
q
ar
ar
ar
a
a
a
a
a
in
2
1
2
2
1
2
1
22
1
22
1
22
1
1
1
4212
4212
4212
22
2
2
2

    bCRCRCDAD
AD
bCRCRCDAD
RbCC
bCRCARCDAD
AD
bCRCARCDAD
RbCC
bCRCARCDAD
ADR
RbCRCARCDAD
ADR
AD
bCC
bCRCARCDAD
ADR
RbCRCARCDAD
ADR
AD
bCC
ADR
bCRCARCDAD
RAD
bCC
AD
bCC
R
RCD
RAD
bCC
ar
ar
ar
ar
ar
ar
4242
12
42
2
42
2
12
42
2
42
2
12
42
2
42
2
12
42
2
12
4212
12
2
12
22
2
22
22
2
*
22
2
*
2
22
2
*
22
2
*
2
2
22
2
2
1
2





















 























It is the another way to represent the system by network. As we are representing the
system by block diagrams.
# In signal flow graph junction is called Node
# Node are connected by path called Branches.
# Direction of signal is shown by an arrow.
# Value of variable is indicated at nodes.
# Gain is shown in the center of path.
Fro the fig.
12
112
3 xGx
xGx



Fig.
Fig.
22113 xHxHx 
23214 xGGGx 

Some Important points.
# Input node have only outgoing branches
# output node have only incoming branch.
# Forward path is a path from input to output node.
# Feedback is a path which start and ends at the same node .
Fig.

PROBLEM : Draw signal flow graph for following equitations.
56
345
324
423
512
xx
gxfxx
exdxx
cxbxx
axxx






PROBLEM : Draw signal flow graph for following equitation.
4453355
4443344
2233
5524423321122
xaxax
xaxax
xax
xaxaxaxax





Meson’s Formula gives the overall gain M of the system, which is the ratio of output
to input
# = 1- (sum of all individual loop gain)+(sum of product of gains of all
possible combinations of two non touching loops) – (sum of
product of gains of all possible combinations of three non
touching loops) + …………………..
# = value of for that part of the signal flow graph not touching the
forward path.



k
kkM
M

k 
th
k

A signal flow graph shown for a system (a) Derive overall transfer function c/r using
Mason’s formula and draw block diagram
Sol. : There is only one forward path
total gain of forward path=
No. of feed back loop system = 3 with gains =
There is one combination of two non – touching loops
Product of gains of these non touching loops=
There is no combination of three or more non touching loops
cxxrx 321
43211 YYYYM 
342321 ,, YYYYYY 
3412 YYYY




k
kkM
M
# = 1- (sum of all individual loop gain)+(sum of product of gains of all possible
combinations of two non touching loops) – (sum of product of gains of all possible
combinations of three non touching loops) + …………………..
# = value of for that part of the signal flow graph not touching the
forward path.
= 1+
= 1 Since there is no part of graph not touching the first forward path
So M =c/r =
For Block diagram governing eq. are
k


 3412432321 YYYYYYYYYY 
1
4231432321
4231
1 YYYYYYYYYY
YYYY

 
 
  323
2312
121
Ycxx
Yxxx
Yxrx




 Block Diagram:

For the block diagram shown in fig. draw signal flow graph and overall
transfer function
Sol. :
Signal Flow Graph

From signal flow graph r is input and c is output
There are two forward path and
Gain of 1st forward path =
Gain of 2nd Forward Path =
There are three feed back loop with gains
There are no non touching loop and all loops touch both forward path
cGGrG 321 cGGrG 421
321 GGG
421 GGG
2421232121 ,,1 HGGGHGGGHGG 
24212321121
421321
2211
24212321121
21
1
1
1
HGGGHGGGHGG
GGGGGG
MM
r
c
M
HGGGHGGGHGG









 For given block diagram draw signal flow graph and find out overall
transfer function

#Controller may be hydraulic, pneumatic or electrical.
1. Hydraulic Controller
# they modify the error signal to suitable form
# Hydraulic control system needs high pressure fluid and may be used in ships,
machines tools, aircraft control
#hydraulic control system gives higher torque/inertia ratio, high power to size ratio
# They can withstand shock and vibrations
# They have sealing difficulties
# Hydraulic fluids are generally inflammable.
2. Pneumatic Controller :
# these are suitable where the compressed air is easily available
# They are suitable where fire – hazard are important.
# Proper sealing is required

3. Electrical Controller :
# these are good for remote positioning. Where signals are transmitted
through wires or microwaves
# They have a drawback of fire – hazards .
# Generally they are used for position and speed control.

1. Proportional Control :
# this type of control action is used in level process
# controlled variable is level of the liquid
# x is the change in the level or error.
# y is the movement of the valve.
We cannot use any types of controllers at anywhere, with each type controller,
there are certain conditions that must be fulfilled. With proportional controllers
there are two conditions and these are written below:
(a) Deviation should not be large, it means there should be less deviation
between the input and output.
(b)Deviation should not be sudden.
If the level of the tank increases to the set level then due to lever valve moves y
distance in the downward direction and closes the valve and reduces the flow
rate . Hence decrease in the level of liquid starts and vice versa.
y=Kx……………………………1.
Where K is the constant of proportionality

Block diagram for the system for the given system is below
Fig of proportional control system.

Where is the reference input and is the controlled variable or output
# only changes only when if error x changes.
# for the sensity of the controller term Proportional band term is used (P.B.)
# P.B. Is defined as the % change in variable x require to give 100% change in
controller output y. If full scale value of input and output value are
Then P.B. =
Advantages of Proportional Controller
Now let us discuss some advantages of proportional controller. Proportional
controller helps in reducing the steady state error, thus makes the system more
stable.
Slow response of the over damped system can be made faster with the help of
these controllers.
Disadvantages of Proportional Controller
Now there are some serious disadvantages of these controllers and these are
written as follows: Due to presence of these controllers we some offsets in the
system. Proportional controllers also increase the maximum overshoot of the
system.
rx cx
inq
0,0 yx
0
0
y
y
x
x

2. Integral Control System :
As the name suggests in integral controllers the output (also called the actuating
signal) is directly proportional to the integral of the error signal. Now let us
analyze integral controller mathematically. As we know in an integral controller
output is directly proportional to the integration of the error signal, writing this
mathematically we have,
`
Advantages of Integral Controller
Due to their unique ability they can return the controlled variable back to the
exact set point following a disturbance that’s why these are known as reset
controllers.
Disadvantages of Integral Controller
It tends to make the system unstable because it responds slowly towards the
produced error.

Fig. shows the integral type control system
e=c x where c is the constant
Flow rate of high pressure oil= be = A dy/dt Where b is port constant
and A is the area of the piston.
b c x = ADy Where D is the d/dt

From the above eq.
Where
In case of level changes x , controller output y keeps on increasing with time t till
the error is brought to zero..Thus there is no steady state error in this type of
control action. But system may be unstable under certain conditions due to
friction of moving component. Usually it is combined with the proportional
action controller called PI controller.
D
x
K
A
bcxdt
y
A
bcxdt
dy
A
bx
dt
dy
A
bcx
Dy
AD
bcx
y
AD
bc
x
y
 





1
A
bc
K 1

Due to combined action output will be
Graphs.
D
x
KKxy 1

Derivative Control :
# In this type of control y is proportional to the dx/dt
# Control action is proportional to the rate of change of controlled variable.
When the level rises x then the piston will move in the downward with a
velocity. It will produces a force proportional to it due to the viscosity of
the fluid in the cylinder. As cylinder is mounted on the springs, springs
deflects y is proportional to the force developed.

# In this type of controller, controller output is more if rate of change of x is
more
P-D Controller :
it is the combination of proportional and derivative control system.
Dxk
dt
dx
ky
dt
dx
y
22 

xDKy
DxKKxy
dt
dx
KKxy
yyy
d
d
)1(
2
21







On – Off Controller :
This type of control action control valve have two position
either fully open or fully closed. As per the fig. when level rises to a certain level
i.e. highest set point, the switch is off and the solenoid is de- energized, closing
the valve. Thus the level starts reducing till it reaches the lowest set point, the
switch is on , the solenoid is energized and the valve is fully open.

# Proportional, derivative and integral control action can be obtained by using
hydraulic servo motor suitably
# Transfer function of hydraulic servo motor shown in fig. 2 have eq.
is of integral type and fig. 1 have P-D Control action
Fig. 1 Fig 2
 e
A
C
DA
eC
y
1
1
1
1

 
 
1...........................
0
0
fWhere0SummationForce
.
fcDfk
kz
y
kzfcDfky
fcDykzfky
cDy
f
z
kky
a
l
yc
f
z
yk













 
 
 
 
 
   
   
   
   
   
bk
fxbcDbk
bk
zbkADfkfcAD
bk
xbfcDbfk
bk
zbkfcADADfk
xbfcDbfkzbkfcADADfk
xbfcDbfkzbkADfcDADfk
bxfcDbxfkbkzADzfcDADzfk
bkzbxfcDbxfkADzfcDADzfk
bkzbxfcDbxfkADzfcDADzfk
fcDfk
bkzfcDfkbx
ADz
fcDfk
bkz
bxADz
fcDfk
kz
xbADz

























2
2
2
2no.eq.inyofvaluePut the
bADzOilPressureHighofRateFlow

   
 
fx
k
cD
z
fx
k
cD
z
fx
k
cD
z
b
fAD
kb
cfAD
bk
fxbcDbk
bk
zbkADfkfcAD






























1
1100
largeveryis
k
b
andsmallvery veryisAIf
11
Now
2
2

Electronic Controller :
Electronic controller are easily available and used in low and medium
power servomotor as proportional controller
From the fig ( a)
If I is the current then (Voltage across Capacitor)
As we know q = C v
= Voltage Across Capictor and put this value in eq. 1
  1..............................1vvKv io 
1viRvo 
DC
i
v
DCvi
dt
dCv
dt
dq



1
1
1

then
From Eq. 1 Put the value of
1
1
1
1
1
1
2............................Then
1
1
1









RCD
DC
i
iRv
v
DC
i
iRv
v
DC
iiR
DC
i
v
v
DC
i
iRv
o
o
o
o
1KvKvv io 
RCDv
v
Kv
RCD
K
v
RCD
K
If
Kv
RCD
K
v
Kv
RCD
v
Kv
RCD
v
KKvv
i
o
io
io
i
o
o
o
io
























1
1
1
1
1
1
1
1
1

 Then
RCDv
v
Kv
RCD
K
v
i
o
io









1
1
1

D
i
qDqi
dt
dq
i  ,,

Derive the expression for the overall transfer function and show that it is of
PDI type of controller
From the fig. as we know
Put the value of q in eq. 1
Also V=iR11/18/2020 DINESH PANCHAL
2..........
1..............
D
i
qDq
dt
dq
i
C
q
VCVq


DC
q
V 

The eq. for the circuit having
And for the circuit having
00andCR
  0
0
1
0
0
000010
0
11
e
DC
i
R
DC
ieiRii
DC







110 , andRCC
 
 
DC
DC
i
iRi
DC
i
DC
i
iRi
DCDC
i
DC
i
iRi
DCDC
i
iRi
DCDC
i
DC
i
iRi
DC
ii
DC
iRi
DC
ii
DC
0
0
1
111
1
0
0
1
111
10
0
0
1
111
10
0
111
10
0
0
1
111
1
01
0
11
1
01
0
1
1
1
1
11
0
11













Now put the value of in eq.
0i 0
0
1
0
0
0
1
e
DC
i
R
DC
i 






 
 
 
  0
0
00
10
11010
1
0
0
00
0
1
1
1
0
0
00
0
1
1
1
0
0
1
00
0
1
1
1
0
0
1
00
0
0
0
1
1
1
0
0
1
0
0
0
0
1
1
1
0
0
1
0
0
0
0
1
11
1
1
0
0
1
0
0
0
0
1
11
1
1
1
1
1
11
1
1
11
1
11
11
111
1
11
e
DC
DRC
DCC
CRDCCC
i
e
DC
DRC
DC
R
DC
i
e
DC
DRC
DC
R
DC
i
e
DC
i
DRC
DC
R
DC
i
e
DC
i
DRC
DC
DC
DC
R
DC
i
e
DC
i
R
DC
DC
DC
R
DC
i
e
DC
i
R
DC
DC
DC
i
iR
DC
i
e
DC
i
R
DC
DC
DC
i
iRi
DC











 






















































































 
)1(
)(
)(
)(
eq.Fron
But
1
1Now
0101010011
11
0
1
001000110001010
101
0
1
10010001100011010
101
0
1
100100011000110101
1011
0
1
10
10010001100011010
11
0
1
111
0
10
100100110001110
1
0
0
00
10
1110
1
DRCDRDRCCDRCDRC
DCR
e
e
DRCCDRCCDRCDRCCRDCCC
DCCR
e
e
CDRCCDRCCDRCDRCCCRDCCC
DCCR
e
e
CDRCCDRCCDRCDRCCCRDCCCi
DCCRi
e
e
DCC
CDRCCDRCCDRCDRCCCRDCCC
Ri
e
e
eRi
e
DCC
CDRCCDRCDRCDRCCCDRCC
i
e
DC
DRC
DCC
CDRCC
i






















 











 

Now
Further
From above two eq.
  )1(
)1(
0101
2
010011
11
0
1
0101010011
11
0
1
RRCCDRCRCRCD
DCR
e
e
DRCDRDRCCDRCDRC
DCR
e
e




2
3
12
02
R
R
ee
andKeeei


 
2
3
01
2
3
10
0
2
3
1
R
R
Keee
R
R
eKee
Ke
R
R
ee
i
i
i




Now put the value of e1 in eq………
 
 
 
 
 
 
 
 
  
 
 
  30101
2
010011211
30101
2
0100110
30101
2
010011
30101
2
010011211
0
30101
2
010011
211
0
3
2
0101
2
010011
11
0
3
2
0101
2
010011
11
0
0
0
3
2
0101
2
010011
11
0
0
0101
2
010011
11
0
2
3
0
0101
2
010011
11
0
1
)1(
)1(
)1(
)1(
)1(
)1(
)1(
)1(
)1(
)1(
RRRCCDRCRCRCDKDRCR
RRRCCDRCRCRCD
e
e
RRRCCDRCRCRCD
RRRCCDRCRCRCDKDRCR
e
e
K
RRRCCDRCRCRCD
DRCR
e
e
K
R
R
RRCCDRCRCRCD
DCR
e
e
R
R
RRCCDRCRCRCD
DCR
e
Ke
e
e
R
R
RRCCDRCRCRCD
DCR
e
Kee
RRCCDRCRCRCD
DCR
e
R
R
Kee
RRCCDRCRCRCD
DCR
e
e
i
i
i
i
i
i
i























The basic component of a pneumatic controller is a flapper nozzle system
motion e of the flapper is the input and output of the controller is
pressure p0 which is applied to the final control element. The block
diagram is shown in fig.
Proportional Controller :

if is constant it is proportional controller
The fig shows the pneumatic proportional controller.
Supply Pressure =Constant=
Pressure in the Chamber= variable= which changes with motion e
Capital letters are used for absolute value and small letters are used for
operating values
When there is no gap between nozzle and flapper
If gap is very large
Changes in the pressure in the chamber operate a relay whose output
pressure varies from supply pressure to atm. Pressure, depending on
motion y of elastic end chamber. The relay is ment to amplify the pressure
changes
Transfer function between and e may be obtained by analysis of
1. Flapper Nozzle element 2. Chamber 3. Relay 4. Feedback Element
e
p0
SP
1P
SPP 1
.1 atmPP 
1p
SP
1p
0p

Mass flow rate of air through orifice depend on pressure difference
But is constant.
So
Or Where
Mass flow rate of air out of the nozzle depend on
Mass flow rate of air in the chamber = (Mass flow rate through orifice) –
(mass flow rate through nozzle outside)
)1( PPS 
SP
 1PFMin 
D
m
D
m
m
pam
inin
in
in


flowMass
)1.......(..........11
01
1
dP
dM
a 
 1,0
XPFM 
andXP1
 
 3..............................NozzletheoutsideflowMass
and
2..................................
NozzletheoutsideRateflowMass
3120
0
0
0
3
01
0
2
3120
0
0
1
01
0
0
D
xapa
D
m
m
X
M
a
P
M
aWhere
xapam
X
X
M
P
P
M
m


















Now
 
 
 
 
 7......................................................................
p
y
p
bellowofbalanceforceFor
6......................................................................................v
bellowyofntDisplacemeXchamberinbellowofareachamberofin volumeChange
5................................................................................................
a
va
s
constantsareaawherevasa
,PSo
chambertheinsidevol.andmasson thedependchambertheinsidePressure
constantgaslUniverasatheisChamber,airin theof
masstheisair,oftempabsolutetheischamber,inairofvol.theisWhere
LawGastoaccordingBut
4..........................................
NozzletheoutideFlowMass-NozzlethroughflowMasschambertheinsideflowMass
11
11
11
4
151
c
5415c41
11
11
111
3
1211
12113
31211
31211
K
A
KyA
yA
p
andp
VSF
R
STV
TRSVP
a
Dspapa
xDspapaxa
xapapaDs
D
xapapa
s
C
C
C
c
c
cc
















Now from eq. 1 and 2
 
    
  
  
yofput valueNow
Aaaa
ap
vofput valueNow
vaaa
ap
vaaa
ap
a
va
p
eq.aboveinofvalueput theNow
p
p
151412411
4311
1
151412411
4311
151412411
4311
4
151
1211
311
1211
311
3
1211
11
ypDpapaK
aA
x
y
pDpapaK
aA
x
y
pDpapaK
aA
p
DpapaK
aA
x
y
s
DspapaK
aA
a
Dspapa
K
A
x
y
c
cc











 







  
 
 
 2142
1
5214
431
2
1
5214
431
2
1
54241
431
1
154241
431
11
151412411
4311
151412411
4311
abyDivide
a1a
a
a1a
a
a1aa
a
Aa1aa
a
]
Aaaa
ap
Aaaa
ap
aa
K
A
Daa
a
K
A
x
y
K
A
Daa
a
K
A
K
A
DaaK
aA
x
y
K
A
DaaK
aA
x
y
K
Ap
pDpapaK
aA
x
y
ypDpapaK
aA
x
y






















































































 
 
 
 
 
 
   
 
 
 
   
 
 10.........................................................................................
2
flapperofmiddleat theisnozzleif
9...........................................................................................
YmotionondependPpressureSince
8....................................................................................Cy
CthensmallisIf
1
C-Where
11
a
a
a
a1
eWher
a
a1
1
a
a
60
0
1
11
21
321
3
1214
43
1
214
2
1
5
214
2
1
5
214
43
1
ze
x
yap
x
K
A
K
A
x
y
D
C
K
A
x
y
aa
a
D
aa
a
K
A
x
y
D
aa
a
K
A
x
y
aa
K
A
aa
K
A
D
aa
a
K
A
x
y









































































 
lyrespectiveareaandstiffnesstheareandWhere
11...............................................................................
bellowsfeedbackforeq.balanceForce
0
ff
ff
AK
pAzK 

FIG.
)(pondependbellowfeedbackthemrateflowmassin theChange
belowase''and'p'betweenrelationthederiveTo.derivation
previousin theassameremainsx''andpbetweenrelationThe
.x''amountbyincreasesnozzlenearflapperofdistancethen
givenise''inputIf.z''feedbacknoistheremeat that tichangesande''
motioninputany timeatifSobelolow.fedbackrelay tofromairof
entrythebeforejustintroducedisresistancepneumaticsystemin this
0f
0
0
fp

 
 
.
bellowofstiffnessandareatheareKandAWhere
bellowofbalanceforceFor
bellowofvol.in theChange
constantstheareaandaWhere
,
Si.e.bellowinairofmassoffunctionisbellowin thePPressure
bellowairin thechangeMass
ff
f
87
87
ff
6
6
006
f
ff
fff
f
ffffff
f
f
ff
ff
ff
K
pA
zzKpA
zAv
vasapVSFP
D
s
s
dtss
a
mpa
pppam









 
 





 







 












7
8
6
6
7
8
6
6
0
6
6
6
60
6
6
6
60
eq.From
a
zvAap
DpaK
apA
a
vap
DpaK
apA
H
p
z
DspaK
apA
a
mpa
K
pA
H
p
z
mpaK
apA
a
mpa
K
pA
H
p
z
ff
ff
ff
ff
ff
ff
f
fff
ff
ff
f
ff
f
fff
ff
ff
f
ff
f

 
67
2
8
67
2
8
7
2
8
6
6
0
7
8
6
6
7
8
6
6
0
1
Where
1
1
1
1
1
aa
K
vAa
DK
A
aa
K
vAa
DK
A
a
K
vAa
DaK
aA
H
p
z
a
K
A
vAa
DaK
aA
a
K
pA
vAap
DpaK
apA
H
p
z
f
f
f
ff
f
f
f
f
f
f
f
f
f
f
f
f
f
f
f
f
ff
ff
ff
ff
f











































































Fig.
 
 
 
f
ff
ff
f
ff
f
f
f
A
DK
DK
Ae
p
DK
A
G
H
G
GH
G
e
p














1
1
1
largevery veryisGIf
1
1
1
1
1
1
diagramblockaboveFron the
0
0

 
 
D
ppa
ppa
e
p
109
1
109
0
0
0
s
2bellowhethrough tflowMass
2bellowhethrough trateflowMass
sameremainwill
pandbetween xrelationbut the
derivedbewillzandpbetween
relationbelowTheasderivedbe
mayfig.inshowncontroller
for thefunctionTransfer




 
 
 
   
 z
z
z
z
Aisbelloweachofarea
andsameare2and1bellowofStiffness
eq.balanceForce
2bellowofareatheisWherez-z-
constsnts.areandand2bellowinchangevol.theisWhere
vandsoffunctiontheispaspFurther
11111910091010
11111091010
1111
109
10
10111111010
111111010
1
1110
11
11101
1111111101
ff
f
f
ff
ffff
ff
ff
ff
DK
AADapaapaa
DK
zDK
K
Ap
DK
AADappaa
z
K
Ap
K
AAa
D
ppa
a
z
K
Ap
z
K
AAasa
K
Ap
z
K
Avasa
K
Ap
z
K
Ap
K
Ap
AAAv
aav
vasa















 










   
  
 
 
   
 
 
 
910
2
111
910
1
0
910
2
111
910
1
910
2
111
910
1
0
910
2
111910
910
1
2
111910
1
111910
191019101
0
1119101910191010
1111910109101091010
109101091011191010
1111109101091010
1111
1
10
910091010
11111910091011111910091010
1
FandEWhere
1
1
1
1
z
z
zz
aa
K
Aa
Kaa
A
FD
ED
p
z
aa
K
Aa
D
Kaa
DA
Kaa
AaKD
Kaa
DA
p
z
Kaa
ADaKaaDK
Kaa
DA
ADaKaaDK
DA
AADaKaaDK
AaaAaaDA
p
z
AADaKaaDKzAaaAaaDAp
AADaKaaDKzApaaApaaADp
ApaaApaaAADaKaaDKzADp
AADazKApaaApaazDKADp
AADa
A
zKAp
aapaazDKADp
DK
AADapaapaazDK
DK
AADapaapaa
DK
zDK
K
Ap
f
f
f
f
f
f
f
ff
f
fffff
fff
ff
fff
ff
f
f
f
ff
f
f














































 






EDE
F
He
p
H
G
GH
G
e
p
1
ED
FD1
0
1
largevery veryisGIf
1
1
1
1
0
1
1
0










TRANSIENT AND STEADY RESPONSE
# Transient and steady response of feedback control system response is
analyzed for various inputs
# Governing eq. is analyzed by the Laplace transformation
# Solution of higher order equation may be find out by using the digital
computer

# fig. shows transfer function representation between torque variable u(t) and
rotary function Ð(t) in time domain
 
  aDID 
 2
1
tu
t
G(t)functionTransfer


 
 
 
   
 
 
 
 
 
 
 
 
 
    trtc
tr
tc
1
tGofvalueput theNow
11tr
tc
1tHBut
1tr
tc
edistrubancnowithsystemeabovfor thefunctionTransfer
2
2
2
KKaDID
KaDID
K
aDID
K
tG
K
tKG
tKG
So
tHtKG
tKG













For the general case having disturbance:
    


for toutputstatesteadyandc(t)outputresponsetransientthegivesThis
domainin timesolvedbecantscoefficienlinearorder withsecondofiseq.aboveThe
trtc 2
KKaDID
        
     
     
     
        tbtrKtG
tutG
tbtrK
tutu
tctrKteKtu





tcK-
tc
tcK-
tb
2
12
1

          
             
             
            
     
  
   
  
     
  1
0
existb(t)edistrubanconlyand0r(t)If
b(t)cedistrubaanandr(t)inputthetorelatedisc(t)Output
11
K1
tcK
tcK-
tcK-
tGK
tbtG
tc
tGK
tbtG
tGK
trtGK
tc
tbtGtrtGKtGtc
tbtGtrKtGtGtc
tbtGtGtrKtGtc
tbtrKtGtc












For second order system the eq. Become where
     
  1
0
tGK
tbtG
tc


     
  
 
 
 
 
 
    
responsetransientfor thesolvediseq.aboveThe
1
G(t)
1
1
2
2
2
tbKaDIDtc
KaDIDtb
tc
KaDID
tb
K
tb
tGK
tbtG
tc










 
aDID
tG

 2
1

  
    
      
        
2
1
0
1
0
22
1
1
0
R(s)TransformLaplaceitsThen
ar(t)inputrampFor
R(s)TransformLaplaceitsThen
Constantar(t)inputstepFor
00
0
And
aswrittenLaplaceofInverse
)()(And
)(isTransformLaplaceitsfunctiongeneralanyFor
s
a
dttea
t
s
a
dtae
fsfsFssFDL
fssFsDFL
tfsFL
L
tfLdttfesF
sFf(t)
st
a
st
st




















FINAL VALUE THEOREM :
It gives the steady state value of any function as given below. The
steady state value of function f(t) is given by
   sFtf LtLt st 0


     
  
   
  
b(t)loadandr(t)inputtodueresponseofsumtheisoutputWhere
11
underasistopicpreviousinderivedasoutputofequatationThe
controlalproportioninconstantiselementcontroltheofKfunctionTransfer
tGK
tbtG
tGK
trtGK
tc





Transient response due to reference input r(t):
When load does not exist and only input exist i.e. R(t) exist and b(t)=0
     
  
   
  
     
  
   
  
 
     
  
 
 
 
 
   
 
   
       
111
SystemofFrequencyNatural
I
K
or
I
K
But
1
KbyDeno.andNum.theDivide
11
1
tGwheresystemordersecondofcasethetakeweIf
1
0
1
become
11
eq.The
2
2
2
2
2
2
2
2
2
2
2

























































KI
aD
K
ID
tr
IKK
IaDD
tr
KK
aDD
tr
tc
K
aD
K
ID
tr
tc
KaDID
trK
tc
KaDID
trK
K
tG
trK
tGK
trtGK
tc
aDID
tGK
trtGK
tGK
trtGK
tc
tGK
tbtG
tGK
trtGK
tc
nnn
nn



     
     
   
 
   
  1
tanWhere1sin
1
1tc
alsoiseq.aboveofSol.
1sin
1
1cos1tc
belowgiveneq.isaldifferentithisofSolution1
2
bewilleq.Thedr(t)puti.e.inputstepFor1
2
1
2
1
1
21
DampingCritical
systemofDamping
C
C
RatioDamping
2
But
1
1
1
2
12
2
2
2
2
2
2
2
2
2
2
2
2
c
2
2
2
2

































































































t
e
d
ted
dtc
DD
trtc
DD
DD
tr
DD
tr
tc
KI
a
KI
aDD
tr
KI
aD
K
ID
tr
tc
n
t
nn
t
nn
nn
nnnn
nnn
n
n

The plot of the below eq., Which is the transient response for a second order
system having proportional control and having step input
   
 










2
12
2
2
2
2
1
tanWhere1sin
1
1tc
alsoiseq.aboveofSol.
1sin
1
1cos1tc



























t
e
d
ted
n
t
nn
t
n
n

IMPORTANT POINTS:
# Graph has been plot for different damping ratio .
# If The there is a peak value of response initially after that response c(t)
oscillate
about the reference input value r(t) =d till the steady state is reached.
# For there is no oscillation.

1
1

(A) Rise Time (tr) : It is time taken by the system to reach the value of
output equal to input first time. It corresponds to the point P. Lower the
value of lower the value of
2
2
1
1
1
tan











 


n
rt

rt

(B) Time (tp) : for peak value and peak overshoot: It is time taken by the
system to reach the max. value of output first time. It compounds to the
point the point .
It is the time when first overshoot take place .It can be found out by
putting
dc/dt =0
2P
 
   




























2max
p2
2
2
2
1
1C
first timeoutputofvalueMax.ofvalueget thewe
c(t)foreq.in thetofvaluetheputtingBy
1
01sin
1
1cos1tc
dt
d
0







ed
t
ted
tc
dt
d
n
p
nn
tn

( C) Percentage Overshoot: is the percentage by which the peak response
exceeds the value of step input
2
1
-
e100OversshootPeak% 




0.0 1.57 3.15 100.0
0.2 1.80 3.20 53.0
0.4 2.20 3.42 25.4
0.6 2.80 3.93 9.5
0.8 4.50 5.24 1.5
 rnt pnt OvershootPeak%

(D) Settling Time: It is the time taken system when the output c(t) remain
with in the range of . It corresponds to the point . From the eq.
settling time can be find out when x is given the put c(t) =x
(E) Steady State : When the output of the system become equal to the reference
input and does not change with time. This will become possible when
.
%x 3P
X (Percentage 1 2 3 4 5
4.6 3.9 3.5 3.2 3.0  sn t
t
 











2
1
1


te
dxtc
n

PROBLEM : For first order system shown in fig. derive the sol. For the output
for unit step input=1
SOLUTION:
 
 
 
 
 
  bD
K
KaD
K
K
aDt
t
G
G
t
t
GH
G
t
t
i
















1
1
aD
K
GBut
1
G
1
1
1
1Hfig.giventheFrom
1
asbelowgivenssystemsystemclosedtheofFunctionTransferOverall
i
0
i
0
i
0







 
 
 
     
      
 
   
   
 
 
   3no.eq.incond.boundaryput thisNow0t0at tSo
.zeroissystemofoutputthestartingAt
3...........................AeAe
solutionParticularsolutionryComplementt
bygiven1.iseq.ofsolutionSo
K
eq.ofsolutionParticular
-br
00
0
2eq.intofvaluethePut
teq.ialdifferientthisofsolutionryComplement
2.................................
1.................................
Given1InputBut
0
bt-rt
0
00
0
i
i
0






















b
K
b
K
b
brAebr
bAerAe
Ae
KtbtD
KtbD
t
bD
K
t
t
rt
rtrt
rt

 
 
 
graphinshownisformgraphicalinOutput
1
b
K
e
b
-outputSo
b
K
-A
b
K
A0
Ae
rt
0
rt
0
bt
e
b
KK
t
b
K
t








STEADY STATE RESPONSE FOR UNIT INPUT AND NO LOAD:
To find out the steady state response and steady state error of the system
to any input r(t) or any load b(t), we have to apply the limit t→∞ to the
solution or apply s→0 using final value theorem in Laplace
Transformation eq.
We can also find out the steady state error by using the Final Value Theorem
and Laplace Transformation
   
 
0d-dc-rerrorstateSo
Outputc
tstatesteadyAt
1sin
1
1cos1tc 2
2
2













 
d
ted nn
tn




     
  
   
  
   
   
 KaDD
Then
,0tbif
11
2
2







I
trK
tc
aDIDtG
tGK
tbtG
tGK
trtGK
tc

   
 
   
 
 
     
   
 
 
1
K
I
K
I
1
K
I
1
1
1
K
I
sRsEError
1
K
IKass
TranformLaplaceTaking
KaDD
2
2
2
2
2
2
2


















































s
K
a
s
s
K
a
s
sR
s
K
a
s
sR
s
K
a
s
sR
sR
sC
s
K
a
s
sR
I
sRK
sC
I
trK
tc

   
 
domaintimefromobtainedassameisWhich
0
100
00
d
1
K
I
K
I
d
1
K
I
K
I
0EErrorStateSteady
TheoremValueFinaltoAccording
1
K
I
K
I
So
s
d
R(s)Laplaceitsanddr(t)inputstepFor
1
K
I
K
I
2
2
2
2
0
0
2
2
2
2









































































s
K
a
s
s
K
a
s
s
K
a
s
s
K
a
s
s
d
s
s
s
K
a
s
s
K
a
s
s
d
sE
s
K
a
s
s
K
a
s
sRsE
s
s

   
 
largebeshouldgainandpossibleassmallasbeshouldconstantdampingerrortheminimiseTo
and
2
Where
2
10
0
1
K
I
K
I
1
K
I
K
I
1
1
K
I
K
I
1
0EErrorStateSteady
TheoremValueFinaltoAccording
1
K
I
K
I
1
So
s
1
R(s)Laplaceitsandtr(t)inputrampunitFor
1
K
I
K
I
n
22
2
2
2
20
0
2
2
2
2
2
2
I
K
KI
a
K
a
o
K
a
s
K
a
s
K
a
s
s
K
a
s
s
K
a
s
ss
K
a
s
s
K
a
s
s
s
s
s
K
a
s
s
K
a
s
s
sE
s
K
a
s
s
K
a
s
sRsE
n
s
s



































































































TRANSIENT RESPONSE TO LOAD INPUT AND r(t)=0 :
   
 
 
   
 
 
 
 
   
       
We
tbtcKaDID
KaDID
tb
K
tG
tb
tKG
tbtG
tc
aDID
tKG
tbtG
tc
t-ctc-0tc-trerrorstateSteady
tany timeatoutputofvaluethefindtosolvediseq.aboveThe
)(
11
)(
1
tGtakeSystemOrdersecondFor
1
)(
2
2
2













# This analysis is simple in nature and useful technique
# It is used to study the behavior of the system
# In this technique behavior of the output is analyzed when harmonic signal
applied at input
# the frequency of the output signal is ω, it is changed from low value to high
value
# For the linear system frequency of the o/p and i/p remain same and ratio of
their magnitude is one and phase difference between the o/p and i/p signal
depend on ω
# Frequency response analysis may be carried out experimentally or
analytically

Diagram
Input =r(t)=Harmonic in nature=
Otput = e(t) = Harmonic in nature =
D=d/dt = jωt
tj
er 
tj
ec 

 








jaI
jG
jSoD
tcjecjtDc
ecjtDcec
dt
d
tc
dt
d
ectc
asIs
sG
aDID
tj
tjtj
tj










2
2
2
1
)(
)()(
)()(
)(
1
)(
1
GFunctionTransferIf

 
GH
G
r
c
GrGHcGrcGHc
cGHGrccHrGc
Gec
cHrcre






1
functiontransferloopclosedFor1
)()(
2eq.in to1eq.fromeofvalueput theNow
2.......................................
1....................'

FG
r
c
FG
r
c
FGH
r
c
G
G
r
c






'
'
1*
systemfeedbackunitFor
1
Then1HIf

# It is the plot of magnitude and phase of the Transfer Function in the
polar co-ordinate system as is changed from 0 to
FIG. 6.7
iM i
i 

(A) Polar Plot of First Order System :-
Fig 6.10
But H = 1
So But
Now
GH
G
r
c
aD
K
G



1
G
G
G
r
c
1
1
1
1 



aD
K
G 
aD
KG
r
c
1
1
1
1
1
1



 1.................................
1
1
1
1
TD
K
aDr
c





Where T = a/K
Now put jω in place of D In eq. no. 1
Compare with complex no. a+ jb
2222
22
11
1
1
1
1
1
*
1
1
1
1
1
1









T
jT
Tr
c
T
jT
Tj
Tj
Tjr
c
TjTDr
c















2
22
2
22
22
22
22
22
11
1
1
1
1


























T
T
T
baM
baMMagnitude
T
T
b
and
T
a

Next
Phase Angle
     2222
22
2
22
2
22
11
1
1
11
1





TT
T
M
T
T
T
M

























T
T
T
T
T
a
b
1
22
22
1
tan
1
1
1tan
tan









ωt M Ф
0 1 0
0.5 0.89 -27
1 0.707 -45
2 0.45 -63
4 0.42 -76
5 0.196 -79
0 -90

Diagram 6.11

POLAR PLOT OF SECOUND ORDER SYSTEM :-
Fig. 6.12
But For feedback system overall transfer function is
So
asIs
K
sG

 2
)(
1)(
)()(1
)(
)(
)(



sButH
sHsG
sG
sR
sC
1
)(
1
1
)(
)(
)(
1
)(
1
1
)(1
)(
)(
)(
2








sG
sR
sC
asIs
K
sButG
sG
sG
sG
sR
sC

22
2
22
2
2
2
2
22
2)(
)(
jsputNow
2)(
)(
So
2
Let
1
)(
)(
Now
nn
n
nn
n
n
jsR
sC
ss
I
K
I
as
s
I
K
sR
sC
KI
a
I
K
I
K
I
as
s
I
K
KasIs
K
K
asIssR
sC























       222
2
222
2
2
2
22
2
2
2
22
2
2
2
22
2
21
21
21
21
)(
)(
21
21
*
21
1
21
1
)(
)(
Put
2
1
1
2)(
)(
bydenumandnum.Divide
2)(
)(
RR
RjR
RjR
RjR
jr
jc
RjR
RjR
RjRRjRjr
jc
R
jjjr
jc
jjR
jC
n
nnn
nn
n
n
n
nn
n















































           
   
   
       
 
    
 
  
     222224
224
224
2
222
224
2
222
2
222
2
22
222
222
2
222222
2
222
2
21
1
221
1
221
221
21
221
21
2
21
1
21
2
21
1
jbano.complexwithCompare
21
2
21
1
21
21
)(
)(
RRRRR
M
RRR
RRR
RR
RRR
M
RR
R
RR
R
baM
RR
R
b
RR
R
a
RR
Rj
RR
R
RR
RjR
jr
jc































































Now
M will be maximum when
   
   
   222
2
1
2
222
2
222
21
1
1
2
tan
1
2
21
1
21
2
tan
RR
M
R
R
R
R
RR
R
RR
R
a
b




















2
2
2
12
1
21
21







peak
n
M
and
Or
R

DIAGRAM 6.13

Draw the polar plot for open loop frequency response of fig. 6.12
when a =2, K =20 , I = 1
Fig. 6.12

Solution :-
 
           
222 2222
2
22
2
2
2
2
2
2
22
22
2
2
)(
*)(
*)(
)()(
)()()(
)()()(
functiontransferoverallfunctiontransferloopopenFor
1)(
)(

















jaI
jKa
jaI
KI
jaI
jaIK
jF
jaI
jaI
jaI
K
jF
jaI
jaI
jaI
K
jaI
K
jF
jaI
K
jajI
K
jHjGjF
asIs
K
sHsGsF
sH
asIs
K
sG




























Now
       
       
 
 
     





























2
tan
4
20
4
40
tan
4
120
4
400
4
4400
4
1600400
4
40
4
20
4
40
4
20
)(
*2**1
*2*20*
*2**1
*1*20
)(
)(
1
24
2
24
1
224224
24
224
242
24
2
24
2
2424
2
2222
2
2222
2
22
22











































j
M
M
j
jF
j
j
j
jF
jaI
jKa
jaI
KI
jF

Fig. 6.14
 



2
tan
4
120
1
2



M
ω F(Jω) Ф
0 -90
0.5 19.4 256
1 8.9 243
2 3.5 225
3 1.84 213.6
4 1.12 207
5 0.74 201.8
6 0.53 198
0 180

# It is the plot of magnitude and phase angle of transfer function against
in linear or logthrmic fashion they are also known as Bode Plots.
1.First order System :
FIG. NO. 6.10

22
2
22
2
22
22
22
22
2222
222222
1
1
11
1
1
1
1
jbawitheq.abovetheCompare
11
1
)(
)(
11
1
1
1
1
1
*
11
1
)(
)(
1
1
1
)(
1
1
)(1
)(
)(
)(
frequencyloopclosedFor
1
)(
TT
T
T
baMMagnitude
T
T
b
T
a
T
Tj
Tjr
jc
T
Tj
TT
Tj
Tj
Tj
Tjjr
jc
Tj
jG
jG
jG
jr
jc
K
a
WhereT
Tjja
K
jG
aD
K
G







































































.
tvsandtvs.MplotNow
tan
1
1
1tantan
AnglePhase
1
22
22
11







T
T
T
T
a
b 






ωT M
0 1
1 0.707
2 0.4472
3 0.3162
4 0.2425
5 0.1961
6 0.1644
7 0.1414
8 0.124
0 1 2 3 4 5 6 7
Ф 0 -45 -63.43 -71.56 -75.96 -78.69 -80.83 -81.86
-90
-80
-70
-60
-50
-40
-30
-20
-10
0
AxisTitle
Mag. Vs. ωT

-90
-80
-70
-60
-50
-40
-30
-20
-10
0
0 1 2 3 4 5 6 7 8
Ф Vs ωT
Ф
ФT Ф
0 0
1 -45
2 -63.4
3 -71.6
4 -76
5 -78.7
6 -80.8
7 -81.9
∞ 90

POLAR PLOT FOR SECOUND ORDER SYSTEM:
From Fig. 6.12
   222
2
1
21
1
1
2
tan
RR
M
R
R






 

2
tantan-90and
41
20
M
2)1)(ss(s
20
G(s)(iii)
tan-90and
1
10
M
1)s(s
10
G(s)(ii)
tan-90and
1
1
M
1)s(s
1
G(s)(i)
forplotpolarDraw.1
110
22
10
2
10
2


























1. Determine the transfer function of the system
2. Rewrite the Transfer Function both numerator and denominator in stranded form as
shown in fig.
))()((
))()()((
)(
111
1111
pspspss
zszszszsK
sH



polesarepandzerosarezWhere
111
1111
)(
111
1111
)(
321
4321
1
321
321
4321
4321


























































































p
s
p
s
p
s
s
z
s
z
s
z
s
z
s
K
sH
p
s
p
s
p
s
ppsp
z
s
z
s
z
s
z
s
zzzKz
sH

3. The transfer function contains:
1.

# Positive Gain margin means system is stable and negative gain
margin means system is unstable.
# For min. phase system both phase margin and gain margin
must be positive.

PROBLEM: Sketch the Bodes plot for transfer function
Determine the (a) P.M. (b) G.M. (c) Stability of system (d) Phase crossover
frequency (e) Gain crossover frequency
PROBLEM : 2 Sketch the Bodes plot for transfer function
PROBLEM : 3 Sketch the Bodes plot for transfer function
)001.01()1.01(
1000
)(
ss
sS


)1.01)(125.01(
)5.01(16
)( 2
sss
s
sG



)1.01)(25.01(
50
)(
sss
sG



PROBLEM -1

PROBLEMS : 2

PROBLEM: 3

# A system is said to be stable if it return to equilibrium position after being
disturbed.
# If the output response shown in fig 7.1 (A) or (b) then system is unstable
# While the system is stable if the output as per the dia. 7.2 (a) and (b)

# Transfer function for the given fig. is
equatationsticcharacteritheis0GH1i.e.
equatationsticcharacterithegiveswhichzerotoequal
c
ofrdenominatoPut
element.feedbacktheisH
lementseforwardoffunctiontransfercombinedtheisGWhere
1



r
GH
G
r
c

   
 AAAc(t)
AAAc(t)Then
,If
AAAc(t)
responseTransient
prootrealthebeingandconjugateomplexare,and,,,areeq.abovetheofrootsIf
0eq.sticCharacteri
1
r
c
order.thirdofeq.sticcharacteriithwithsystemSuppose
unstable.issystemThen thepart.realpositiveorpositiveiseq.sticcharacteritheofroottheofanyIf#
system.theofresponsetransientthegiveseq.abovetheofrootThe
0ID
equatationsticCharacteri
0rDenominatoputequatationsticCharacteriFor
ID
K
1
ID
1
1
1
1
1r
c
1Hfeedbackand
ID
K
GwithsystemordersecondFor
321
321
32
321
1323232
23
23
2
2
2
2
1
1
321
jytjytxttp
tjyxtjyxtp
tptptp
eeee
eee
jyxpp
eee
pjyxpppppp
dcDbDaD
dcDbDaD
w
KaD
KaD
K
aD
G
GH
G
aD

























# Routh’s criterion is used to find the with positive real part
# It is applied to the characteristics eq. of the system.
# If any a’s are zero or negative then the eq. definitely has roots either
pure imaginary or have positive real part
# If all a’s are positive we use Routh’s criterion if any of root is positive.
....ee
....dd
ccc
..........bbb
.....aaaa
.....aaaa
asarrayanintcoefficientheArrange
0..................
21
21
321
321
7531
6420
1
2
2
3
3
2
2
1
10  

nnn
nnnn
asasasasasasa

b
ba-ba
b
b-aba
a
aa-aa
a
aa-aa
a
aa-aa
Where
....ee
....dd
ccc
..........bbb
.....aaaa
.....aaaa
1
3113
2
1
2112
1
1
3113
2
1
2112
1
1
3115
2
1
2113
1
1
7016
3
1
5014
2
1
3012
1
21
21
321
321
7531
6420
d
dcdc
e
d
dcdc
e
c
cbcb
d
c
cbcb
d
cc
bbb











# The above process is continued in each row and column , until zero are
obtained for additional coefficients in the array
# According to the routh’s criterion , the no. of change of sign in left hand
column of the array is equal to the no. of roots of the eq. with positive real
part. If there is any root with positive real part, then the system is
unstable.

  
.
unstable.issystemSopart.realpositivewithrootstwohaseq.sticcharacterithe.Thus
2.5to4-and4-to1i.e.signinchangestwoaretheretop,fromstartingleftin thecolumnfirstInthe
0d
............6
5.2
046*5.2
c
..........b0b2.5
4-
6-1*4-
b
.....0a0a6a-4a
.....0a0a1a1a
0..................
belowgiveneq.atandredwitheq.abobetheCompare:SOLUTION
06s4ss
partrealpositiveany withifrootsofno.thefindeq.sticscharacteriwithasystemFor:PROBLEM
1
1
321
7531
6420
1
2
2
3
3
2
2
1
10
23












nnn
nnnn
asasasasasasa

 
 
0
28
4K
2s5
1
0sG1eq.sticcharacteriFor
11r
c
28
4K
2s5
28
18.0
4.015sG
:SOLUTION
stable.issystemfor theKofvaluetheFindsystemcontrolFor:
2
i
2
i
2
i


























ss
s
G
G
GH
G
ss
s
sss
K
s
PROBLEM
i

0.625to0isKofrangeSo
0.625K0
4
32K-20
changesignnoFor
0
0
4
32K-20
4K4
58
arraytheArranging
4K5s48
04K2s5s28
0
28
4K2s5s
1
0
28
4K
2s5
1
i
i
i
i
i
i
23
i
223
23
i
2
2
i
 










ss
ss
ss
ss
s

Special Case :
0
02
017/8
216
41-
282
531
Array
025s8s3s2ss
becomeEq.
2)(sbyeq.hemultiply tSo2.e.g.no.realarbitryanisAwhere
)(sbyeq.hemultiply tsituationsuchInrow.oneinzeroistcoefficienfirstisThere
0
0
020
131
arraytheArranging
012s3ss
eq.sticscharacterigivenFor the
2345
24





A

There are two changes of the sign in the first column at left.
Thus there are two roots with positive real part. So the system
is unstable.

(unstable)01322ss(vii)
(unstable)0123s6s(vi)
(Stable)05.15.35.4(v)
(unstable)05432s(iv)
(Stable)012552s(iii)
(unstable)02322s(ii)
(Stable)01462s(i)
(Stable)01462s(i)
(Stable)01462s(i)
eq.ticscharactersthehavingsystemtheofstabilitytheFind
2345
2345
23
234
234
234
234
234
234









sss
sss
sss
sss
sss
sss
sss
sss
sss

0)95s(s
0
)95s(s
s)-Ke(1
1
0H(s)G(s)1eq.ticscharactersSo
0rdenominatoPut
H(s)G(s)
G(s)
FunctionTransfer
)95s(s
s)-Ke(1
)1(efrequencylowFor
:Ans
stableissystemforKofvaluemax.theFind
)95s(s
Ke
H(s)G(s)
functiontransferloopopenanhassystemfeedbackA#
0.73K04)2(2(ii)
unstableissystemsorootscomplexgivesIt01510520s(i)
stableissystemfor theKofvaluetheFind#
2
2
2
s-
2
s-
23
234













KsKs
s
s
s
s
sKKss
ssKs


# By using the Routh’s criterion we can find out the stability of the system.
But it is impossible to find at a glance how unstable or stable the system
i.e. whether by changing the system parameter , it may change its state of
stability or instability or vice-versa.
This is possible by using the Nyquist Criterion which is graphical plot.
Zero’s and Poles of a Transfer Function:
 
 
 
 
      
    
zerosthearez......z,z,z
polestheare......pp,p,p
b
a
Where
.....
.......
rDenominato
Numerator
......
........
FunctionTransferofFormGeneral
m321
n321
n
m
21
21
2
210
2
210
K
pspspss
zszszsK
sF
sD
sN
sbsbsbbs
sasasaa
sF
n
q
m
n
n
q
m
m









system1typecalledisit1qif
system0typecalledisit0qIf
complexorrealbemayrootsandgaincalledisK



 Roots of characterstics eq. of control system affects the response and
stability of system
 So plot of roots for varying system parameter are important.
 The locus of each root as the gain K varies known as Root Locus.

The block diagram is shown in fig for second order system
FIG. 8.2
 
2I
a
-s
2I
a
-s
4
KFor
2I
a
-sj
2I
a-
sKFor
I
a
-s0s0KFor
to0fromKfor varingplottedbecansandslociofThe
22
s
eq.abovetheofrootsThe
001
issystemtheofeq.ticsCharacters
21
2
21
21
21
2
1,2
2












I
a
j
I
K
I
a
I
a
KassIsGH

# Thus there are two root loci
# First root loci starting from origin and till –a/2I on the real axis
# then moving parallel to the imaginary axis in +ve direction
# The second root loci starts from –a/I and till –a/2I is on the real axis and
then goes parallel to the imaginary axis in –ve direction
# the root loci meet at x=-a/2I when K=
# Each point on the locus corresponds to the particular value of K
I
a
4
2

The root of previous eq. Can be easily calculated and root locus were drawn.
For higher order system we need digital computer to find the roots.
     
    v21
u321
00
00
p-s.................p-sp-s
z-s..................z-sz-sz-sK
GH(s)F(s)If
360180critrionAngular1GH
iscreitreonMagnitude......eq.from.....0,1,2,3,4.mWhere
3601801GHF
-1GH(s)or0(s)G1
issystemabovetheofeq.ticsCharactersThe





mGH
m
H

   
        mpszssGH
pspsps
zszszsK
H
u
i
i
u
i
i
u
360180sF
eq.ofcriterionangularThe
1
........
.........
sGsF
bewouldcriterionmagnitudeThe
zero.thearep..............p,p,pspointsWhere
zero.thearez..............z,z,zspointsWhere
11
111
11
v321
v321







 

RULE NO. 1 :
No. of separate root loci = No. of roots of characteristics eq.
RULE NO. 2 :
  
  
polesare...,pandzerosare....,zereWh
0
......p-sp-s
......z-sz-sK
1
eq.ticscharactersFor
or0eitheratterminateandF(s)
functiontransferloopopenofpolesatstartlocirootthatstateIt
2121
21
21
pz



RULE NO. 3 :
Root loci are symmetrical with respect to real axis. This is due to
the fact that complex roots appear as complex conjugates
RULE NO. 4 :
The real axis loci : A point on the real axis lies on the root locus if
the total number of real poles and zeros to its right , is an odd no.

A system with transfer function Draw the root
loci.
Solution :
According to rule 2, the loci start at poles
and ends at zero or infinity. One root
locus starts at pole 0 and ends at zero -3
, another root s starts at pole -6 and ends
at infinity.
 
 6
32
)()(



ss
sK
sGHsF
3zatareZeros
6pand0patarePoles
GH(s)abovetheFrom
1
21



Problem: For a open loop system for the given transfer function find the root
loci and find the value of K for which system is stable
Solution :
 
 20
10
)( 2



ss
sK
sGH
 



atends
andandfromstartssandslocirootTwo#
zzero
atendsandfromstartsslocusrootThus#
permittednotiswhichnos.,
inevenareofsiderighton thepointanyfrom
asinfinityrdsright towatogonotcanitFrom#
and,atstartlociRoot
10i.e.1zerosofNo.
20and0i.e.3polesofNo.
2121
1
33
3
3
321
1
321
pp
p
p
p
ppp
zN
pppN
Z
P

   
Kofvaluesallforstableissystemthe
plane,s''ofhalfrightenter thelocitheofnoneSince
5
2
10
1-3
1020-00
270,90
13
360180
3.....2,1,0,,
360180
1 1
00
















  
zp
v
i
u
i
ii
zp
NN
zp
f
m
m
NN
m


RULE NO. 6 :
Problem : Find the root loci for the system with
Solution :
0
11
belowgivenusedformulaaxisrealon thepointthefindTo
:pointsawayBreak
11


  
u
i i
v
i i zxx-p
  32
)(


sss
K
sGH
no.inoddarezerosand
polesofno.totalpoleofrighton thepointanyfromas
touprighttowardsgocanandfromstartsslociRoot#
and,atstartlociRoot
ili.e.0zerosofNo.
3and20i.e.3polesofNo.
3
33
321
321
p
p
ppp
nN
pppN
Z
P




   
     
0
3
1
2
1
0
1
0
11
:pointawayBreak
67.1
3
5
13
0032
300,180,60
03
360m180
N
360m180
Asymptotes#
axisrealaboutlsymmetricabewilland
axisrealfromawaybreakand2-and0frostartslocirootSo#
4no.ruletodue3-and2-betwwenpossibleislocusrootNo#
.atendsandandfromstartssandslociRoot#
11
1 1
000
p
2121



























 

 
xxx
zxx-p
NN
zp
f
N
pp
u
i i
v
i i
zp
v
i
u
i
ii
z


     
        
   
        
     
     
 
planes''ofhalfrightenter thelocitheaspointthisbeyondunstablebewillsystemThe
8no.ruleusingbyoutfoundbecan
axisimagenoryononintersectithetoingcorrospondKandbofvalueandaxis.realat
0.785-at xsepratelocusroottheSo.2-and0betweenlies-0.785xofvalueThe
785.0-,54.2
6
71.4
,
6
29.15
6
29.510
6
7210010
3*2
6*3*410010
2
4bb-
eq.abveofRoot
06103
02365
023623
02332
0
320
203032
0
3
1
2
1
0
1
2
2
22
22
2
2

























a
ac
xx
xxxxxx
xxxxxxx
xxxxxx
xxx
xxxxxx
xxx

RULE NO. 7 :
RULE NO. 8:
u
v
zszszszs
pspspsps



.....
......
K
:locusrootonpointanytoingcorrospondlocusrooton thKofvaluuethefindTo
321
321
stability.andyinstabilitofborderon theispointThis#
K.ofvalue
ingcorrospondfindandbforsolveandeq.sticcharacteriin thejbsPut#
itintersectlocusrootat whichaxis,imagenoryonbdistancefindTo#
:axisimagenorywithlocirootofonIntersecti


Problem : Find the root loci for the system with
Solution :
  32
)(


sss
K
sGH
  
  
   
 
 
 
30KforunstablebewillSystem
30K02.45*5-KNow
2.45b
0b-60b-6b
5bK05b-K
imagenorypart withimagenoryandrealpart withrealCompare
0j065b-K0j0K5b-6jbjb-
0j0K6jb5b-jb-032jjb
eq.abovein thejbsPut
032ss0
32ss
K
1
0GH(s)1
problemgivenfor theeq.sticsCharacteri
2
22
22
3223
23












bbj
Ksjbb
Ks
s

RULE NO. 9 :
 
  
   
    1321
1321
11
1
21
21
22
3
1
360180departureofAngle
360180
setisfiedbetoiscriterionangularSo
locus.rooton theissand,saypolecomplex
thenear toveryspointtrialaby taking
andpolethefromdepartorstartlocus
rootin whichdirectionthefindorder toIn#
andatconjugates
complexbemaypolestwoabovetheFrom#
2ps
zsK
GH(s)If
:polecomplexafromdepartureofAngle








m
m
p
pp
pp
ss nn

PROBLEM:
SOLUTION: fig. 8.12
 
  1251s
5sK
GH(s)withsystemFor the 2



ss
   
  0
1
0
2
01
1
01
3
321
11
321
1
2,12,11
124412290180
90
44
5.2
4.2
tan
122
5.1
4.2
tan180
360180
fromdepartureofanglethefindTo
and,atstartlociRoot
5i.e.1zerosofNo.
4.25.2
2
48255
1
i.e.3polesofNo.


















m
p
ppp
zN
jppp
N
Z
P

5.0
13
55.25.21
180,270,90
13
360180360180 000














 
ZP
ZP
NN
zp
f
m
NN
m


Problem:

0
2
21
2
1,2
22
45then0.707Suppose
foundbecan,knownisIf
1
tan
fig.theFrom
shownasaresandsroot
theofpartimagenoryandrealiffig.theFrom
1casedunderdampefortrueisThis
1s
eq.aboveofrootsThe
02s
issystemordersecoundofeq.ticsCharacterssystem.theof
ratiodampingbyinfluencedgreatlyisresponseits
systemordersecoundofresposetransientFor the















n
n
nn
nn
j
s

 
3.1
3.91.33.9t
5.3secperAsvaluestatesteadyfor2%xwithinbetoresponse
for the3sec.isinputsteptosystemoprdersecoundfor thetimesettingtheSuppose#
specifiedbecan
locirootfor thezonethenspecifiedisx%bandwidthgivenafortimesettingtheif#
response.transientthedominatepolesdominantThese#
poles.
dominantcalledarepolesuchzeros,andpolesotherthanaxisimagenorytoclose
planes''inpoleconjugatecomplexanyisthereiforder,higherofissystemtheIf#
poles.conjugatecomplextwohasitsystemordersecoundFor the#
area.shadedbyshownislocirootsorrootshein which tzoneThe#
45beshould
thenresponsetransientdesiredfor0.707thanlessbeshouldthatrequiredisitIf#
n
nsn
0











 
    
  
3j4-
2
100648
ppolestwoareThere
0zi.e.zerooneisThere
0
258s
10
258s
258s
258s
0
258s
258s
aswrittenbemayeq.Above
to0fromiesvarkofethat valurequiredisIt
0258s
iseq.sticcharacteriIf
Kofluecertain vatoscorrospond
locuson thepointEachK.gainofvaluedifferentforplotslociThe#
....
....
sGHreWhe
0G1eq.sticscharacterifor thelocirootdrawn thehaveWe
1,2
1
2
1
2
1
2
22
1
2
1
1
2
21
21



















s
sk
s
sks
ss
sks
sks
pspss
zszsK
sH
n

 
 
0
23
0
23
0
0654
321
2
23
23
A2ss
3Bs
sGH
0
A2ss
3Bs
1
aswrittenbecaneq.Now
Bofionvariat
forlociofstsrtfor thesuitablyselectedaofvaluetheisA#
AAwhenBofariationareforthevsand,slociRoot#
0BwhenAofariationareforthevsand,slociRoot#
fig.inshowndrawnasbecanabovetheoflocirootThe#
0
2s
1
02s
0BtakeFirst we
032s
below
oneeq.likesticcharacteriin theBabdA
parameterstwoofeffecttheifFurther













s
s
s
A
As
ABss

# Several control system are being used in the machine control like
hydraulic, pneumatic, electrical
# CNC and DNC system are also used to control the machines.
1. Hydraulic control : There are two type of systems
(a) Valve Control :
(b) Pump Controlled:
(a) Valve Control :
It involve the use of spool valve with constant pressure supply and
response is fast and efficiency is somehow low. There is chances of
leakage and contamination of dirt which affect the performance of
system

# Fig shows the valve controlled copying unit as applied in the machine unit.
# It uses a template and stylus is kept pressed in contact
# Stylus is connected to the piston valve
# The complete unit i.e. shaded part can move to and fro on the fixed rod.
# Machine tool slide is attached to the assembly.

# Tool slide follow the motion of stylus.
# High pressure oil at cont. pressure is supplied through the middle port.
# if stylus moves to left, High pressure oil goes through the port 2 and
assembly moves to the right, following the stylus motion
# By tool slide follow the profile on which the stylus moves.

(b) Pump Controlled: It has slow response .Its efficiency is high for large
power units this system is used.
# It is automatic positioning system for machine tools
# The fig. of pump controlled system is shown in fig.
# It consist axial type piston pump with control lever , in which discharge is
controlled by motion of control lever.
p

# This controls the travel of pistons located in rotating cylinder.
# The control lever can be operated by small torque motor, which is
operated depending on the error between desired position and actual
position
# The two electric potentiometers convert mechanical position in to voltage
signals.

2. NC/DNC/CNC System:
(a) NC SYSTEM:
# NC System for machine came in early 1950 for commercial use.
# They are of two type (i) open loop (ii) closed loop

# Fig. 12.3 shows the open loop NC system in which stepper moor is used to
control the slide having a motor rotating a screw shaft with fixed nut on
the slide
# he stepper motor moves as per the pluses generated in Machine
Controlled Unit
# MCU consist tape recorder, buffer storage and decoder.
# Tape has no. of holes depending on commands or program
# Pulses are generated by the decoder depending on desired output.
# For the turning on lathe machine two motion are required in x and y
direction . For each direction one stepper motor is required.

FIG. no. 12.4
# In NC closed loop system motion is measured by the sensor
# output is compared with the desired output (input command) and error
signal is generated
# This error signal after amplification is converted digital to analog signals
by DA convertor .
# Then these signal fed to the servo motor which rotate the shaft having
screw.
# Which moves the slide
cx

(b) DNC SYSTEM:
# It was evolved in late sixties
# It uses a central computer to communicate with several machine tools.
# It eliminate the tape reader for each machine tools
# All program are stored in the central computer and program for each
machine tool can be downloaded when required.
# There is two way communication between machine and computer.

(b) CNC SYSTEM:
# It uses micro computer for each machine tool separately.
# A large hard wired controllers are used in NC machine tool have been
replaced by computers
# Computer does the function of comparator and store the program.
# The program can be stored in floppy disc or CD can be edited at site.
# CNC is useful in FMS

# Engine are the prime mover and drive the loads but load may vary with
time
# So with the change of load the speed of the engine will also changes .In
most of cases the speed of the engine must be constant as in the case of
electric generator
# With change in speed the frequency of A.C. changes which is not desirable.
# So maintain the speed of the engine speed governor is attached to the
engine
# Governor may be of mechanical, hydraulic, pneumatic or electronic.
# Now a days microprocessor have application in the engine control.
1. Mechanical Governor
2. Hydraulic Governor
3. Pneumatic Governor
4. Electronic Governor

1. Mechanical Governor :
# Principle of mech. Governor is
centrifugal force
# For the fixed setting of the
adjusting lever the speed of the
engine is constant
# If the load on the engine decrease
then speed of the engine increase
and spindle of the governor is
rotated by the engine increase its speed.
# With increase in speed centrifugal force on the ball increases and they
move in outward direction , which uplift the sleeve.
# Due to the movement of the sleeve fuel control valve move in such a way
so that reduce the supply of fuel to the engine

# Thus the speed of the engine get reduced and reach to the equilibrium state.
# An increase in the load will have opposite effect.
# In case we want to change the equilibrium speed, we can change by
adjusting the lever which adjust the fuel valve corresponding to the
desired speed
# In Automotive application , the speed of the engine is protected against the
over speed.
# They are also protected against the idea ling speed.
# Between the min. speed and max. speed s, the control on the speed is
exercised by the driver through accelerator pedal.
# For max. high speed , the rising of the sleeve has to rise against the high
stiff spring.
# Min speed another spring of small stiffness is used.

2. Hydraulic Controller:
# In large unit like power plants ,
mechanical governor is used to
sense the speed and fuel valve is
controlled by means of hydraulic
servo.
# For the given setting of the speed
lever, if the speed increases due
to decrease in the load .The
outward movement of the balls
uplift the sleeve which gives the
downward movement to the piston
valve.
# The high pressure oil enters through
the bottom port of the hydraulic servo and move the power piston which
reduces the supply of the fuel by closing the fuel valve.
# Thus reducing the speed of the engine
# If the speed decrease then opposite action is taken by the system.

3. Pneumatic Governor:
# Due to change in speed , there is change in the air pressure in the intake
manifold.
# A spring loaded diaphragm is connected through a lever to the fuel
control valve which control the supply of the fuel to the engine.
# At equilibrium position , the spring force equal to the force due to the
pressure difference on two side of the diaphragm.
# If the speed of the engine increases , due to reduction of the load, the inlet
manifold pressure reduces and moving the fuel control rod to the right,
thus reducing the supply of fuel to the engine. So that engine speed return
to the desired value.

# Reverse action is taken by the system when speed of engine decreases due
to increase in the load.
# The butterfly valve is nearly closed at idling.
# As it is opened manifold pressure increases, moving the diaphragm and
fuel control valve to the left and increases the fuel supply up to speed
which is desired.
# Pneumatic governor are good for low speed.
# Pneumatic governor performance is not so good for high speed.

4. Electronic Governor :
# Electronic governor use electrical and electronic component
# They give efficient control of speed
# Speed is sensed by the electric sensor which give feedback signals the
sensor may be of electromagnetic type
# The desired speed is obtained in firm of electrical signal and compared with
sensor speed.
# Error voltage after the comparison is amplified and applied to the motor
actuator system , which control the fuel supply of fuel injection system
# The control system works till steady state is reached.
# With the development of computer tech. micro processor bases control
system are being applied for efficient control in engine.11/18/2020 DINESH PANCHAL

CARBURETTOR CONTROL :
# Development of electronic control carburetor for petrol engine is going
on so that engine run using the optimum use of fuel air mixture and min
exhaust emission
# A microprocessor based system is shown in fig.
# The main input variables are engine speed, switch position, engine temp.,
intake manifold temp., and main throttle angle.

# The operating conditions are determined by the above variables and may
be idling, accelerating, decelerating with engine cut off, starting and
warming and constant speed running.
# By using the suitable algorithms , the processing unit regulate the fuel
supply with the help of carburetor and chock valve.
# Suitable sensors are used for temp. and position measurement.

DIESEL FUEL IGNITION CONTROL :
# Now a days electronic control system has been employed to control the
injection timing and fuel control in a diesel engine.

# The timing of injection is controlled by the cam ring positing , while the
amount of fuel injected is controlled by the metering valve and valve is
controlled by the fueling actuator.
# Fig shows feedback control system for injection timing and fuel metering
, which is dependent on the output of the speed sensor.
# The desired timing is determined by speed and load while fuel control is
related to the speed.
# Feedback control system is applied to control injection timing which is
manipulated by the cam actuator system.
# The injection pump output is controlled depending on the need for fuel

Automatics Control Notes

Recommended

Recommended

More Related Content

What's hot

What's hot (20)

Similar to Automatics Control Notes

Similar to Automatics Control Notes (20)

More from Dinesh Panchal

More from Dinesh Panchal (13)

Recently uploaded

Recently uploaded (20)

Automatics Control Notes