2_16_mark_mechatranics

flounderconvoyElectronics - Devices

Nov 15, 2013 (3 years and 8 months ago)

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2 mark and 16 marks question and answer mechatronics

ME1402


UNIT
-
3

1.What are the characteristics of negative feed back?(n/d
-
2007,2009)

1.

Reduce the noise and distortion,etc..

2.

180degree phase shif t to be provided.

2.write the analogous electrical elemen
ts in force voltage
analogy for the elements of mechanical translational
system? (n/d
-
2007,2009)


It represents the types of forces experienced when
object is moved against frictional forces or when the
object is pushed through a fluid.


3.Derive the
equation for a translational mechanical
system model with spring and mass.(N/D
-
2008)


It represents the stiffness of a system.The stiffness is
given by the releationship between the force (F)used to
extend or compress a spring and the resulting exten
sion or
compression(X).


Fαx, F=Kx

4.Give an example for two step mode control unit. (N/D
-
2008)


Room Heating system is an example for two step mode
control unit.

5.What are the various control modes used in
mechatronics control system?(A/M
-
2010)


Propotional control mode


Integral control mode


Derivative control mode


ON
-
OFF Mode


Propotional Integral derivative control
mode(PID)

6.What are the building blocks in electrical system?(N/D
-
2010)


Resistor


Inductor


Capacitor

7.Give some examples for electro mechanical systems?(
N/
D
-
2010)


Potentiometers,motors,generators.

8. What are the building blocks in mechanical or
translational system?(N/D
-
2011)


Springs


Dashpots and


Masses.

9.Define adaptive control.(N/D
-
2011)


The control system which can adapt changes and it can
chang
e its parameters depending on the situation is known
as adaptive contol system and the method is called
adaptive control.

10.What is stack point register?(M/J
-
2012)


It is a reserved area of the memory in the RAM whose
temporary information may be stored
.A 16 BIT Stack
pointer is used to hold the addressof the most recent stack
entry.



Unit 4

1.what are the criteria that need to be considered for
selecting a PLC?

Ans: The criteria that need to be considered for selecting a
PLC is

i.system definition,

ii.choosing the input and output hardware,

iii.analog input/output module,

iv. input and output timing consideration.

2.draw the ladder rungs to represent : two switches are
normally open and both have to be closed for a motor to
operate
.

------
[ ]
--------
------
[ ]
----------------
( )


Key Switch 1 Key Switch 2 Door Motor

The above realizes the function: Door Motor = Key
Switch 1
AND

Key Switch 2

This circuit shows two key sw
itches that security guards
might use to activate an electric motor on a bank vault
door. When the normally open contacts of both switches
close, electricity is able to flow to the motor which opens
the door.


3.
design a PLC timing circuit that will switch

a output on
for 10 seconds and then switch it off.

INDUSTRIAL STOP/STAR
T
[
EDIT
]

In common industrial latching start/stop log
ic we have a
"start" button to turn on a motor contactor, and a "stop"
button to turn off the contactor.

When the "start" button is pushed the input goes true, via
the "stop" button NC contact. When the "run" input
becomes true the seal
-
in "run" NO contact

in parallel with
the "start" NO contact will close maintaining the input
logic true (latched or sealed
-
in). After the circuit is
latched the "stop" button may be pushed causing its NC
contact to open and consequently the input to go false.
The "run" NO co
ntact then opens and the circuit logic
returns to its quiescent state.


--
+
----
[ ]
--
+
----
[
\
]
----
( )


| start | stop run


| |


+
----
[ ]
--
+


run





-------
[ ]
--------------
( )


run motor

The above realizes the function: run = ( start
OR

run )
AND

(
NOT

stop )

Note the use of parenthesis to group the logical OR
function before evaluating the logical AND function
(which has a higher order of operation priority). Also
n
ote the use of
NOT

to represent the "stop" NC
contact logic.

This
latch

configuration is a common
idiom

in

ladder
logic. In ladder logic it is referred to as seal
-
in logic. The
key to understanding the latch is in recognizing that
"start" switch is a momentary switch (once the user
releases the button, the switch is open again). As soon as
the "run" solenoid e
ngages, it closes the "run" NO
contact, which latches the solenoid on. The "start" switch
opening up then has no effect.


4.what are the logic functions that can be obtained by
using switches in series?

Ans: The logic functions that can be obtained by usin
g
switches in series AND ,NOT, NAND

5.what do you mean by electronic counter?

Ans :
Ever since electronics have been used in battle in an
attempt to gain superiority over the enemy, effort has
been spent on techniques to reduce the effectiveness of
those e
lectronics. More recently, sensors and weapons are
being modified to deal with this threat. One of the most
common types of
ECM

is
radar jamming

or
spoofing
.

6.why are PLC’s considered suitable for shop floor?

A
Programmable Logic Controller
,
PLC

or
Programmable Controller

is a
digital computer

used for
automation

of
electromechanical

processes, such as
control of machinery on factory
assembly lines
,
amusement rides
, or
light fixtures
. The abbreviation
"PLC" and the term "Programmable Logic Controller" are
registered trademarks of the
Allen
-
Bradley

Company
(
Rockwell Automation
). PLCs are used in many
industries and machines.
Unlike general
-
purpose
computers, the PLC is designed for multiple inputs and
output arrangements, extended temperature ranges,
immunity to electrical noise, and resistance to vibration
and impact. Programs to control machine operation are
typically stored

in battery
-
backed
-
up or
non
-
volatile
memory
. A PLC is an example of a
hard

rea
l time

system
since output results must be produced in response to input
conditions within a limited time, otherwise unintended
operation will result.


11. what is shift register
? How many data is required for
a shift register?

Shift
register

c
an be used

where sequence of op
e
rations is
required or movement or track th
e flow of parts and
information

It requires 3 inputs: (i) file


address of the bit array


(ii) Control


Address of control
structure



(iii) Bid address
-

Address of source
bit


(iv) Length


Number of bits in bit
array

Shift register c

12.Draw the ladder rungs to represent

the following: a
motor is switched on by pressing spring return push
button sw
itch,and the mortor remains on until another
spring return push button switch is pressed.


I N


----

----

M

I___ ___________I I_________ M_______I

I___I/I
-----------------
I

N
-

I

13.obtain a NOR logi
c function using ladder program?

Ans :
If we have an OR gate by a NOT gate will invert the
outputs of the OR gate as illustrated in the truth table
M
M
M
M

.therefore the combination of OR and not gates is termed
a NOR gate

14. what

is meant by ladder diagram?

Ans :ladder diagrams is the most commonly used
programming method evolved from electrical relay
circuits and is in the form of graphical language

Or

Ladder diagram is a network of contacts and coils are
arranged on rungs betwee
n two vertical lines called rails
used.

15.what are the advantage of programmable logic
controller over their relay system?


(i) Rewiring should be easily done in PLC.


(ii) No vertical connections are allowed.


(iii) In PLC ,there must always be one outpu
t on each
line.

17.give typical specification of programmable logic
controller.


(i) Central processing unit


(ii) Input / output modules


(iii) Programmer / Monitor.



18.draw the ladder diagram for delay off timer.


When the contact IN1 is closed, the c
ontact will
energise the timer T1 and holds the output lamp ON for
specified set value of 10 sec .The action of an OFF delay
timer is to delay setting the lamp OFF.

19.b
rief on EEPROM.

20.write short notes on basic arrangement of PLC system.

(i) Central
processing unit

(ii) Input / output modules

(iii) Programmer / Monitor.



Unit 5:


1.differntiate between traditional and mechatronics
designs with example.

Sl no

Traditional
design

Mechatronics design

1

It is based on
the traditional
mechanical
elements

such
as gears
,spring
,levers,etc…

It is based on the
mechanical,electronics,computer
technology control engineering.

2

It involves
more
complicated
mechanisms
and moving
components

It involves less complicated
mechanisms and moving
components.


2.how
a operated switch is replaced with PLC? Show the
ladder diagram.


In PLC system,the time duration can be easily
adjusted by changing the timer preset values in the
program whereas the trational system requires various
size of cams.

3. list down the various

mechatronics elements in
an
automatic camera.

i) auto foucssing mechanism control

ii)
Aperture drive

iii) Shutter drive

iv) mirror drive

5. mention any four statements in problem definition

of
mechatronics system design.


i) Need for design


ii) Analysis
of problem


iii) Preparation of specification


iv) Gerneration of possiple solution

6.what are the various movements of robots?


i) cw and acw rotation of the robot on its base


ii) linear movement of the arm horizontally


iii) up and down movement of the

arm


iv) open and close movement of the gripper.

7.identify the sensor , signal conditioner and display
elements in the bourden preasure gauge.


Sensor


Burdon tube acting as sensor


Signal conditioner


ratchet & pinion mechanism is
acting as signal con
ditioner


Display elements


Pointer is the display element.

8.list the advantages of mecharonics design

over
traditional design.


i) Mechatronics system serves the purpose effectively
with high dimensional accuracy requirements


ii) It provides increased
productivity in the industry


iii) it facilitation automation in the production
,assembly and quality control.

9. what are the basic principle involved in mechatronics
design?


Need of design,analysis of problems,preparation of
specification,generation of
possible solution,selection of
suitable solution,production of detailed
solution,production of working drawing,implementation
of design.

10.what are embedded systems?


The hardware which is designed for particular
applications is known as embedded system.

11.what is timed switch
?


It is a device which is used to switch on a motor for
some prescribed time.



12.list the seven modules of mechatronics design
approach.

Need of design,analysis of problems,preparation of
specification,generation of possible solut
ion,selection of
suitable solution,production of detailed
solution,production of working drawing,implementation
of design.

16.what are the factors to be considered in the design of
mechatronics system?


Objects to be handled,actuators,power source,range
of

gripping
force,positioning,maintenance,environment,temp
protection,materials.

18.name any four impartant s
e
nsor used in the p
i
ck and
place robot.


TRIAC,LED,Solenoid,M68HC11

19
.mention the basic components of
a
ny industrial robot.


Locomotion,sensors,perc
eption,knowledge,planning,
autonomy,collaboration

20.denote on intelligent mechatronics system.


Mechanical linkages,drives


Hydraulic and pneumatic actuators


Electrical motors,switches

16 marks

Unit 3

1.a.i.how does a microcontroller differ from

a
micropr
osser?

Difference between microprocessor and
microcontroller

Microprocessor is an IC which has only the CPU inside
them i.e. only the processing powers such as Intel’s
Pentium 1,2,3,4, core 2 duo, i3, i5 etc. These
microprocessors don’t have RAM, ROM, and
other
peripheral on the chip. A system designer has to add them
externally to make them functional. Application of
microprocessor includes Desktop PC’s, Laptops, notepads
etc.



But this is not the case with
Microcontrollers
.
Microcontroller has a CPU, in
addition with a fixed
amount of RAM, ROM and other
peripherals

all
embedded on a single chip. At times it is also termed as a
mini computer or a computer on a single chip. Today
different manufacturers produce microcontrollers with a
wide range of features

available in different versions.
Some manufacturers are ATMEL, Microchip, TI,
Freescale
, Philips, Motorola etc.





Microcontrollers are designed to perform specific tasks.
Specific means applications where the relationship of
input and output is defined.

Depending on the input, some
processing needs to be done and output is delivered. For
example, keyboards, mouse, washing machine, digicam,
pendrive, remote, microwave, cars, bikes, telephone,
mobiles, watches, etc. Since the applications are very
specific
, they need small resources like RAM, ROM, I/O
ports etc and hence can be embedded on a single chip.
This in turn reduces the size and the cost.



Microprocessor find applications where tasks are
unspecific like developing software, games, websites,
photo

editing
, creating documents etc. In such cases the
relationship between input and output is not defined. They
need high amount of resources like RAM, ROM, I/O
ports etc.



The clock speed of the Microprocessor is quite high as
compared to the microcontrol
ler. Whereas the
microcontrollers operate from a few MHz to 30 to 50
MHz, today’s microprocessor operate above 1GHz as
they perform complex tasks. Read more about
what is
microcontroller.



Com
paring microcontroller and microprocessor

in
terms of cost is not justified. Undoubtedly a
microcontroller is far
cheaper

than a microprocessor.
However microcontroller cannot be used in place of
microprocessor and using a microprocessor is not advised
in
place of a microcontroller as it makes the application
quite costly. Microprocessor cannot be used stand alone.
They need other peripherals like RAM, ROM,
buffer
, I/O
ports etc and hence a system designed around a
microprocessor is quite costly.


ii.draw
a block diagram of a basic microcontroller and
explain the function of each subsystem.

LOCK DIAGRAM OF 8051

MICROCONTROLLER:

Microcontroller 8051 block diagram is shown below.
Let’s have a closer look at each & every fraction or block
of this design:


8051 MICROCONTROLLER

BLOCK DIAGRAM
EXPLANATION:

CPU (CENTRAL PROCESS
OR UNIT):

As you may be familiar that Central Pro
cessor Unit or
CPU is the mind of any processing machine. It scrutinizes
and manages all processes that are carried out in the
Microcontroller. User has no power over the functioning
of CPU. It interprets program
printed

in storage space
(ROM) and carries
out all of them and do the projected
duty.

INTERRUPTS:

As the heading put forward, Interrupt is a sub
-
routine call
that reads the Microcontroller’s key function or job and
helps it to perform some other program which is extra
important at that point of tim
e. The characteristic of
Interrupt is extremely constructive as it aids in emergency
cases. Interrupts provides us a method to postpone or
delay the current process, carry out a sub
-
routine task and
then all over again restart standard program
implementati
on.

The Micro
-
controller 8051 can be assembled in such a
manner that it momentarily stops or break the core
program at the happening of
interrupt
. When sub
-
routine
task is finished then the implementation of core program
initiates automatically as usual. T
here are 5 interrupt
supplies in 8051 Microcontroller, two out of five are
peripheral interrupts, two are timer interrupts and one is
serial port interrupt.

MEMORY:

Micro
-
controller needs a program which is a set of
commands. This program enlightens Microc
ontroller to
perform precise tasks. These programs need a storage
space on which they can be
accumulated

and
interpret

by
Microcontroller to act upon any specific process. The
memory which is brought into play to accumulate the
program of Microcontroller i
s recognized as Program
memory or code memory. In common language it’s also
known as Read Only Memory or ROM.

Micro
-
controller also needs a memory to amass data or
operands for the short term. The storage space which is
employed to momentarily data storage

for functioning is
acknowledged as Data Memory and we employ Random
Access Memory or RAM for this principle reason.
Microcontroller 8051 contains code memory or program
memory 4K so that is has 4KB Rom and it also comprise
of data memory (RAM) of 128 byte
s.

BUS:

Fundamentally Bus is a group of wires which functions as
a communication canal or mean for the transfer Data.
These buses comprise of 8, 16 or more cables. As a result,
a bus can bear 8 bits, 16 bits all together. There are two
types of buses:

1.

Addr
ess Bus:

Microcontroller 8051 consists of 16 bit
address bus. It is brought into play to address
memory positions. It is also utilized to transmit the
address from Central Processing Unit to Memory.

2.

Data Bus:

Microcontroller 8051 comprise of 8 bits
data bu
s. It is employed to cart data.

OSCILLATOR:

As we all make out that Microcontroller is a digital circuit
piece of equipment, thus it needs timer for its function.
For this function, Microcontroller 8051 consists of an on
-
chip oscillator which toils as a ti
me source for CPU
(Central Processing Unit). As the productivity thumps of
oscillator are steady as a result, it facilitates harmonized
employment of all pieces of 8051 Microcontroller.
Input/output Port: As we are acquainted with that
Microcontroller is e
mployed in
embedded

systems to
manage the functions of devices. Thus to gather it to other
machinery, gadgets or peripherals we need I/O
(input/output) interfacing ports in Micro
-
controller. For
this function Micro
-
controller 8051 consists of 4
input/outpu
t ports to unite it to other peripherals.

Timers/Counters: Micro
-
controller 8051 is incorporated
with two 16 bit counters & timers. The counters are
separated into 8 bit registers. The timers are utilized for
measuring the intervals, to find out pulse widt
h etc.

EXPLANATION ON 8051
PIN DIAGRAM:


For explaining the pin diagram and pin configuration of
microcontroller 8051, we are t
aking into deliberation a 40
pin Dual inline package (DIP). Now let’s study through
pin configuration in brief:
-

Pins 1


8:
-

recognized as Port 1. Different from other
ports, this port doesn’t provide any other purpose. Port 1
is a domestically pulled up,

quasi bi directional
Input/output port.

Pin 9:
-

As made clear previously RESET pin is utilized to
set the micro
-
controller 8051 to its primary values,
whereas the micro
-
controller is functioning or at the early
beginning of application. The RESET pin has
to be set
elevated for two machine rotations.

Pins 10


17:
-

recognized as Port 3. This port also
supplies a number of other functions such as timer input,
interrupts, serial communication indicators TxD & RxD,
control indicators for outside memory interfa
cing WR &
RD, etc. This is a domestic pull up port with quasi bi
directional port within.

Pins 18 and 19:
-

These are employed for interfacing an
outer crystal to give system clock.

Pin 20:
-

Titled as Vss


it symbolizes ground (0 V)
association.

Pins
-

21
-
2
8:
-

recognized as Port 2 (P 2.0


P 2.7)


other
than serving as Input/output port, senior order address bus
indicators are multiplexed with this quasi bi directional
port.

Pin
-

29:
-

Program Store Enable or PSEN is employed to
interpret sign from outer pro
gram memory.

Pin
-
30:
-

External Access or EA input is employed to
permit or prohibit outer memory interfacing. If there is no
outer memory need, this pin is dragged high by linking it
to Vcc.

Pin
-
31:
-

Aka Address Latch Enable or ALE is brought
into play to
de
-
multiplex the address data indication of
port 0 (for outer memory interfacing). Two ALE throbs
are obtainable for every machine rotation.

Pins 32
-
39:

recognized as Port 0 (P0.0 to P0.7)


other
than serving as Input/output port, low order data &
address

bus signals are multiplexed with this port (to
provide the use of outer memory interfacing). This pin is
a bi directional Input/output port (the single one in
microcontroller 8051) and outer pull up resistors are
necessary to utilize this port as Input/ou
tput.

Pin
-
40:

termed as Vcc is the chief power supply. By and
large it is +5V DC.


Or

b.
draw the block diagram of 8085 microproccesor and
explain the function of each element.

ARCHITECHTURE or FUNCTIONAL BLOCK
DIAGRAM OF 8085

The functional block diagram o
r architechture of 8085
Microprocessor is very important as it gives the
complete details about a Microprocessor. Fig. shows
the Block diagram of a Microprocessor.


8085 Bus Structure:


Address Bus:



The address bus is a group of 16 lines generally
identified as A0 to A15.



The address bus is unidirectional: bits flow in one
direction
-
from the MPU to peripheral devices.



The MPU uses the address bus to perform t
he first
function: identifying a peripheral or a memory
location.


Data Bus:



The data bus is a group of eight lines used for data
flow.



These lines are bi
-
directional
-

data flow in both
directions between the MPU and memory and
peripheral devices.



The MPU uses the data bus to perform the second
function: transferring binary information.



The eight data lines enable the MPU to
manipulate 8
-
bit data ranging fr
om 00 to FF (28 =
256 numbers).



The largest number that can appear on the data
bus is 11111111.


Control Bus:



The control bus carries synchronization signals
and providing timing signals.



The MPU generates specific control signals for
every operation it pe
rforms. These signals are
used to identify a device type with which the MPU
wants to communicate.


Registers of 8085:




The 8085 have six general
-
purpose registers to
store 8
-
bit data during program execution.



These registers are identified as B, C, D, E,
H, and
L.



They can be combined as register pairs
-
BC, DE,
and HL
-
to perform some 16
-
bit operations.



Accumulator (A):



The accumulator is an 8
-
bit register tha
t is part of
the arithmetic/logic unit (ALU).



This register is used to store 8
-
bit data and to
perform arithmetic and logical operations.



The result of an operation is stored in the
accumulator.


Flags:



The ALU includes five flip
-
flops that are set or
r
eset according to the result of an operation.



The microprocessor uses the flags for testing the
data conditions.



They are Zero (Z), Carry (CY), Sign (S), Parity
(P), and Auxiliary Carry (AC) flags. The most
commonly used flags are Sign, Zero, and Carry.


The bit position for the flags in flag register is,



1.Sign Flag (S):








After execution of any arithmetic and logical
operation, if D7 of the result is 1, t
he sign

















flag
is set. Otherwise it is reset.








D7 is reserved for indicating the sign; the
remaining is the magnitude of number.








If D7 is 1, the number will be viewed as negative
number. If D7 is 0, the number will be















viewed
as positive number.


2.Zero Flag (z):








If the result of arithmetic and logical operation is
zero, then zero flag is set otherwise it is reset.


3.Auxiliary Carry Flag (AC):








If D3 generates any carry when doing any


arithmetic an
d logical operation, this flag is set.








Otherwise it is reset.


4.Parity Flag (P):








If the result of arithmetic and logical operation
contains even number of 1's then this flag will be







set and if it is odd number of 1's it will be reset
.


5.Carry Flag (CY):








If any arithmetic and logical operation result any
carry then carry flag is set otherwise it is















reset.


2.a.i.derive the mathematical model for a machine
mounted on the ground to study the effects of ground
dist
urbances on the machine bed displacement.

ii.compare the control system performance for a system
with proportional
control and a system with integral
control.

Often control systems are designed using Proportional
Control.


In this control method, the contr
ol system acts in
a way that the
control effort is proportional to the error
.


You should not forget that phrase.


The control effort is
proportional to the error in a proportional control system,
and that's what makes it a proportional control system.


If

it doesn't have that property, it isn't a proportional control
systems.



Here’s a block diagram of such a system.


In this
lesson we will examine how a proportional control system
works.


An integral controller has one very good quality.


An
integral controller will normally ensure zero SSE in a
control system
-

for step (constant) inputs.



An integral controller is not particularl
y difficult to
implement.



In an analog system, an integral control system
integrates the error signal to generate the control
signal.


If the error signal is a voltage, and the control
signal is also a voltage, then a proportional controller
is just an ana
log integrator.



In a digital control system, an integral control system
computes the error from measured output and user
input to a program, and integrates the error using
some standard integration algorithm, then generates
an output/control signal from th
at integration.

Integral controllers have these properties.



The controller integrates the error as shown in the
block diagram of an example system below.

o

The integral controller has a transfer function of
K
i
/s



So, the actuating signal (the input to the sy
stem being
controlled) is proportional to the integral of the error.




We can examine some of the features of integral
control using an
integral controller.


Or

b.i.derive the differential equation governing the
mechanical system of an electric motor .

ii.explain the charecteristics of PID controller.

A
proportional
-
integral
-
derivative controller

(
PID
controller
) is a generic
control loop

feedback mechanism

(
controller
) widely used in
industrial control systems
. A
PID controller calculates an "error" value as the
differen
ce between a
measured

process variable

and a
desired
setpoint
. The

controller attempts to minimize the
error by adjusting the process control inputs.

The PID controller calculation
algorithm

involves three
separate
constant

parameters, and is according
ly
sometimes
called

three
-
term control
: the
proportional
,
the
integral

and

derivative

values, denoted
P,

I,

and
D.

Simply put, these values can be interpreted in terms of
time:
P

depends on the
present

error,
I

on the
accumulation of
past

errors, and
D

is a
prediction of
future

errors, based on current rate of change.
[1]

The
weighted sum of these three actions is used to adjust the
process via a control element such as the position o
f a
control valve
, a
damper
, or the power supplied to a
heating element.

In the absence of knowl
edge of the underlying process, a
PID controller has historically been
considered

to be the
best controller.
[2]

By tuning the three parameters in the
PID controller algor
ithm, the controller can
provide

control action designed for specific process requirements.
The response of the controller can be described in terms
of the responsiveness of the controller to an error, the
degree

to which the controller
overshoots

the setpoint,
and the degree of system oscillation. Note that the use of
the PID algorithm for control does not guarantee
optimal
control

of the system or system stability.

Some applications may require using only one or two
actions to provide the appropriate system control. This is
achieved by setting the other parameters to zero. A PID
contro
ller will be called a PI, PD, P or I controller in the
absence of the
respective

control actions. PI controllers
are fairly common, since derivative action is sensitive to
measurement

noise, whereas the absence of an
integral

term may prevent the system fr
om reaching its target
value due to the control action.


3.a.i.with a block diagram,explain the use of
microcontroller for the house hold application.

ii.list various applications of microcontroller.

APPLICATIONS OF MICR
OCONTROLLERS:

Microcontrollers are m
ostly used in
following

electronic
equipments :



Mobile Phones



Auto Mobiles



CD/DVD Players



Washing Machines



Cameras



In Computers
-
> Modems and Keyboard Controllers



Security Alarms



Electronic Measurement Instruments.



Microwave Oven.

-

See more at:
http://www.
wikiforu.com/2012/10/applications
-
of
-
microcontroller.html#sthash.RtHMxsgi.dpuf


Or

b.i.write the program to divide two 8
-
bit numbers and to
store result in memory again by microprocessor using
8085.

MICROPROCESSOR

-

PROGRAM TO DIVIDE TWO
NUMBERS

OF 8 BIT D
ATA

AIM: To write an assembly language program to
divide

two numbers of 8 bit data.

APPARATUS REQUIRED: Microprocessor 8085 Trainer
Kit.

PROCEDURE:

PROBLEM ANALYSIS:

¸The division in 8085 is performed as repeated
subtraction. The dividend is stored in A re
gister and
divisor in
B
-
register.

¸The initial value of quotient is assumed as Zero.

¸Subtraction should be performed only when the dividend
is greater than divisor.

¸Subtraction is continued until dividend is lesser than the
divisor.

¸For each subtraction

quotient is incremented by one.

¸Then store the quotient and remainder in memory.

ALGORITHM:

1.Load the divisor in the accumulator and move it to B
register.

2.Load the dividend in accumulator.

3.Clear C register to account for quotient.

4.Check whether d
ivisor is less than dividend.

5.If divisor is less than the dividend go to step 8,
otherwise go to next step.

6.Increment the contents of C register( quotient)

7.Go to step 4.

8.Store the content of accumulator (remainder) in
memory.

9.Move the content of
C register (quotient) to
accumulator and store in memory.

10.Stop

TABULATION:

INPUT

INPUT

OUTPUT

Decimal

Hexa

Decimal

Hexa

Quotient

Remainder














ASSEMBLY LANGUAGE PROGRAM
: DIVIDE TWO
NUMBERS OF 8


BIT DATA

ADDRESS

OPCODE

MNEMONICS

DESCRIPTION









8100

3A

LDA 8201H

LOAD THE

8101

01



ACCUMULATOR

8102

82





8103

47

MOV A,B

GET THE
DIVISOR IN B







REGISTER

8104

3A

LDA 8200H

GET THE
DIVIDEND IN A

8105

00



REGISTER

8106

82





8107

0E

MVI C,00H

CLEAR C
REGISTER FOR

8108

00



QUO
TIENT

8109

B8

AGAIN
: CMP
B

COMPARE

810A

DA

JC
STORE

IF THE DIVISOR
IS LESS

810B

12



THAN THE
DIVIDEND GO

810C

81



TO STORE

810D

90

SUB B

SUBTRACT THE
DIVISOR







FROM THE
DIVIDEND

810E

0C

INR C

INCREMENT
QUOTIENT







BY ONE FOR
EACH







SU
BTRACTION

810F

C3

JMP
AGAIN

GO TO AGAIN

8110

09





8111

81





8112

32

STORE:
STA
8203H

STORE THE
REMAINDER

8113

03



IN MEMORY

8114

82





8115

79

MOV A, C

MOVE THE
CONTENTS OF







C REGISTER TO







ACCUMULATOR

8116

32

STA 8202H

STORE THE
Q
UOTIENT IN

8117

02



MEMORY

8118

82





8119

76

HLT

END

SAMPLE DATA
:





DATA



RESULT



8200

C9 (Dividend)

8202

14 (Quotient)

8201

0A (Divisor)

8203

01 (Remainder)

RESULT: The Microprocessor Division program is
verified using 8085 Trainer Kit.




ii.discuss the working of microprocessor controller traffic
signal system

Basic principle of traffic light control system

Some lights don't have any sort of detectors. For example,
in a large city, the traffic lights may simplyoperate on
timers
--

no mat
ter what time of day it is, there is going to
be a lot of traffic. In the suburbsand on country roads,
however, detectors are common. They may detect when a
car arrives at anintersection, when too many cars are
stacked up at an intersection (to control the

length of the
light),or when cars have entered a turn lane (in order to
activate the arrow light).There are all sorts of
technologies for detecting cars
--

everything from lasers
to rubber hoses filledwith air! By far the most common
technique is the indu
ctive loop. An inductive loop is
simply a coilof wire embedded in the road's surface. To
install the loop, they lay the asphalt and then come
back

and cut a groove in the asphalt with a saw. The wire
is placed in the groove and sealed with a
rubberycompoun
d. You can often see these big
rectangular loops cut in the pavement because the
compoundis obvious.


Inductive loops work by detecting a change of inductance.
To understand the process, let's first look atwhat
inductance is. This figure is helpful:We see
here is a
battery, a light bulb, a coil of wire around a piece of iron
(yellow), and a switch. Thecoil of wire is an inductor. If
we have read How Electromagnets Work, we will also
recognize thatthe inductor is an electromagnet. If we were
to take the indu
ctor out of this circuit, then what wehave
is a normal flashlight. We close the switch and the bulb
lights up. With the inductor in the circuitas shown, the
behavior is completely different. The light bulb is a
resistor (the resistance creates heatto make
the filament in
the bulb glow). The wire in the coil has much lower
resistance (it's just wire),so what you would expect when
you turn on the switch is for the bulb to glow very dimly.
Most of thecurrent should follow the low
-
resistance path
through the lo
op. What happens instead is that whenyou
close the switch, the bulb burns brightly and then gets
dimmer. When we open the switch, the

bulb burns very
brightly and then quickly goes out.

Microprocessor as traffic light control system:

A traffic light contro
l and information transmission
device comprising: a microprocessor locatedinside the
traffic control box and to control all the circuitries; a
traffic light controller connected toand controlled by said
microprocessor to send out the stop, go and direction

signals; an electronicdisplay board connected to and
controlled by said microprocessor to display characters,
patterns andgraphic images; a video camera connected to
and controlled by said microprocessor to monitor
thetraffic flow; a compression circuitry

connected to said
microprocessor and said video camera, saidcompression
circuitry compresses the image data captured by said
video camera and sends the data tosaid microprocessor;
an I/O interface connected to said microprocessor to
receive, transmit data

andcontrol signals; a traffic flow
detector connected to said I/O interface and gathering the
traffic flowinformation; the traffic flow information is
input to said microprocessor; a DSL connected to said
I/Ointerface 6 and receiving, transmitting data an
d control
signals; a broadband network linking sai


4.a.a hot object with capacitance
C and temperature T,
cools in a large room at a temperature T
a.
if the thermal
system has a resistance R,derive an equation describing
how the temperature of the hot obje
ct changes with the
time and give an electrical analogue of the system.

Or

b.i.propose a model for a stepped shaft used to rotate a
mass and derive an equation relating the input

torque and
the angular rotation .neglect damping effect.

ii.describe and comp
are the charecteristics of

1) proportional controller
.

2) proportional plus integral controller
.

5.a.i.explain the mathematical model for rotating mass on
the end of a shaft and for torsional spring
mass
system
.

ii.distinguish between continous and discre
et process
controllers.


Or

b.i.discuss the pr
inciple and operation of a PID controller
.

ii.

explain the principle and operation of self
-
tuning
circuit.

6.

a.i.explain the mathematical model for resistor,
inductor, capacitor system using kirchoffs’s law.

i
i.compare the control system performance for a system
with proportional control and a system with integral
control

or

B.i.a motor is used to rotate a load. Design a
mathematical model and draw a mechanical model and
equivalent model for the same.

ii.enumer
ate the essential charecteristics of the PID
controller

PID CONTROLLER THEOR
Y

This section describes the parallel or non
-
interacting form of the PID controller. For other
forms please see the section
Alternative nomenclature and PID forms
.

The PID control scheme is named after its three correcting terms, whose sum constitutes the
manipulated variable (MV). The proportional, integral, and derivative terms

are summed to
calculate the output of the PID controller. Defining
as the controller output, the final form
of the PID algorithm is:


where

: Proportional gain,

a tuning parameter

: Integral gain, a tuning parameter

: Derivative gain, a tuning parameter

: Error

: Time or instantaneous time (the present)

: Variable of integration; takes on values from time 0 to the present
.

PROPORTIONAL TERM



Plot of PV vs time, for three values of K
p

(K
i

and K
d

held consta
nt)

The proportional term produces an output value that is proportional to the current error value.
The proportional response can be adjusted by multiplying the error by a constant
K
p
, called the
proportional gain constant.

The proportional term is given b
y:


A high proportional gain results in a large change in the output for a given change in the error. If
the proportional gain is too high,

the system can become unstable (see
the section on loop
tuning
). In contrast, a small gain results in a small output response to a large input error, and a
less responsive or les
s sensitive controller. If the proportional gain is too low, the control action
may be too small when responding to system disturbances. Tuning theory and industrial practice
indicate that the proportional term should contribute the bulk of the output chan
ge.
[
citation needed
]

DROOP

Because a non
-
zero error is required to drive it, a proportional controller generally operates with
a steady
-
state error, refer
red to as
droop
.
[note 1]

Droop is proportional to the process gain and
inversely proportional to proportional gain. Droop may be mitigated by adding a compensating
bias term

to the setpoint or output, or corrected dynamically by adding an integral term.

INTEGRAL TERM



Plot of PV vs time, for three values of K
i

(K
p

and K
d

held constant)

The contribution from the integral term is proportional to both the magnitude of the error and the
duration of the error. The
integral

in a PID controller is the sum of the instantaneous error over
time and gives the accumulated offset that should have been corrected previously. The
accumulated error is then multip
lied by the integral gain (
) and added to the controller output.

The integral term is given by:


The integral term accelerates the movement of the process towards setpoint and eliminates the
residual steady
-
state error that occurs with a pure proportional controller. However, s
ince the
integral term responds to accumulated errors from the past, it can cause the present value to
overshoot

the setpoint value (see
the section on loop tuning
).

DERIVATIVE TERM



P
lot of PV vs time, for three values of K
d

(K
p

and K
i

held constant)

The
derivative

of the process error is calculated by determining the slope of the error over time
and multiplying th
is rate of change by the derivative gain
K
d
. The magnitude of the contribution
of the derivative term to the overall control action is termed the derivative gain,
K
d
.

The derivative term is given by:


Derivative action predicts system behavior and thus improves settling time and stability of the
system.
[7]
[8]

Derivative action, however, is seldom used in practice
[
citation needed
]

because of i
ts
inherent sensitivity to measurement noise.
[9]

If this noise is severe enough, the derivative action
will be erratic and actually degrade control performance. Large
, sudden changes in the measured
error (which typically occur when the set point is changed) cause a sudden, large control action
stemming from the derivative term, which goes under the name of
derivative kick
. This problem
can be ameliorated to a degree i
f the measured error is passed through a linear
low
-
pass filter

or a
nonlinear but simple
median
filter
.
[9]

LOOP TUNING

Tuning

a control loop is the adjustment of its control parameters (proportional band/gain, integral
gain/reset, derivative gain/rate) to the o
ptimum values for the desired control response. Stability
(bounded oscillation) is a basic requirement, but beyond that, different systems have different
behavior, different applications have different requirements, and requirements may conflict with
one a
nother.

PID tuning is a difficult problem, even though there are only three parameters and in principle is
simple to describe, because it must satisfy complex criteria within the
limitations of PID control
.
There are accordingly various methods for loop tuning, and more sophisticated techniques are
the subject of patents; this section describes some traditional manual methods for loop tuning.

Designing and tuning a P
ID controller appears to be conceptually intuitive, but can be hard in
practice, if multiple (and often conflicting) objectives such as short transient and high stability
are to be achieved. Usually, initial designs need to be adjusted repeatedly through c
omputer
simulations until the closed
-
loop system performs or compromises as desired.

Some processes have a degree of
nonlinearity

and so parameters that work well at full
-
l
oad
conditions don't work when the process is starting up from no
-
load; this can be corrected by
gain
scheduling

(using different parameters in different operating regions).
PID controllers often
provide acceptable control using default tunings, but performance can generally be improved by
careful tuning, and performance may be unacceptable with poor tuning.

STABILITY

If the PID controller parameters (the gains of the proporti
onal, integral and derivative terms) are
chosen incorrectly, the controlled process input can be unstable, i.e., its output
diverges
, with or
without
oscillation
, and is limited only by saturation or mechanical breakage. Instability is
caused by
excess

gain, particularly in the presence of significant
lag
.

Generally, stabilization of response is required and the process must not oscillate for any
combination of process conditions and setpoints, though sometimes
marginal stability

(bounded
oscillation) is acceptable or desired.
[
citation needed
]

OPTIMUM BEHAVIOR

The optimum behavior
on a process change or setpoint change varies depending on the
application.

Two basic requirements are
regulation

(disturbance rejection


staying at a given setpoint) and
command tracking

(implementing setpoint changes)


these refer to how well the contr
olled
variable tracks the desired value. Specific criteria for command tracking include
rise time

and
settlin
g time
. Some processes must not allow an overshoot of the process variable beyond the
setpoint if, for example, this would be unsafe. Other processes must minimize the energy
expended in reaching a new setpoint.

OVERVIEW OF METHODS

There are several metho
ds for tuning a PID loop. The most effective methods generally involve
the development of some form of process model, then choosing P, I, and D based on the dynamic
model parameters. Manual tuning methods can be relatively inefficient, particularly if the
loops
have response times on the order of minutes or longer.
[
citation needed
]

The choice of method will depend largely on whether or not the loop can be t
aken "offline" for
tuning, and on the response time of the system. If the system can be taken offline, the best tuning
method often involves subjecting the system to a step change in input, measuring the output as a
function of time, and using this respons
e to determine the control parameters


7. a.explain the construction of PID controller. Also write
mathematical equation for this.

MODIFICATIONS TO THE

PID ALGORITHM

The basic PID algorithm presents some challenges in control applications that have been
ad
dressed by minor modifications to the PID form.

INTEGRAL WINDUP

For more details on this topic, see
Integral windup
.

One common problem resulting from the ideal PID implement
ations is
integral windup
, where a
large change in setpoint occurs (say a positive change) and the integral term accumulates an error
larger than the maximal value for the re
gulation variable (windup), thus the system overshoots
and continues to increase as this accumulated error is unwound. This problem can be addressed
by:



Initializing the controller integral to a desired value



Increasing the setpoint in a suitable ramp



Disa
bling the integral function until the PV has entered the controllable region



Preventing the integral term from accumulating above or below pre
-
determined bounds

OVERSHOOTING FROM KN
OWN DISTURBANCES

For example, a PID loop is used to control the temperature

of an electric resistance furnace
where the system has stabilized. Now when the door is opened and something cold is put into the
furnace the temperature drops below the setpoint. The integral function of the controller tends to
compensate this error by i
ntroducing another error in the positive direction. This overshoot can
be avoided by freezing of the integral function after the opening of the door for the time the
control loop typically needs to reheat the furnace.

REPLACING THE INTEGR
AL FUNCTION BY A M
ODEL BASED PART

Often the time
-
response of the system is approximately known. Then it is an advantage to
simulate this time
-
response with a model and to calculate some unknown parameter from the
actual response of the system. If for instance the system is
an electrical furnace the response of
the difference between furnace temperature and ambient temperature to changes of the electrical
power will be similar to that of a simple RC low
-
pass filter multiplied by an unknown
proportional coefficient. The actual

electrical power supplied to the furnace is delayed by a low
-
pass filter to simulate the response of the temperature of the furnace and then the actual
temperature minus the ambient temperature is divided by this low
-
pass filtered electrical power.
Then,
the result is stabilized by another low
-
pass filter leading to an estimation of the
proportional coefficient. With this estimation, it is possible to calculate the required electrical
power by dividing the setpoint of the temperature minus the ambient temp
erature by this
coefficient. The result can then be used instead of the integral function. This also achieves a
control error of zero in the steady
-
state, but avoids integral windup and can give a significantly
improved control action compared to an optimi
zed PID controller. This type of controller does
work properly in an open loop situation which causes integral windup with an integral function.
This is an advantage if, for example, the heating of a furnace has to be reduced for some time
because of the f
ailure of a heating element, or if the controller is used as an advisory system to a
human operator who may not switch it to closed
-
loop operation. It may also be useful if the
controller is inside a branch of a complex control system that may be temporari
ly inactive.

PI CONTROLLER



Basic block of a PI controller

A
PI Controller

(proportional
-
integral controll
er) is a special case of the PID controller in which
the derivative (D) of the error is not used.

The controller output is given by


where
is the error or deviation of actual measured value (
PV
) from the setpoint (
SP
).

.

A PI controller can be modelled easily in software such as
Simulink

or
Xcos

using a "flow chart"
box involving
Laplace

operators:


where

= proportional gain

= integral gain

Setting a value for
is often a trade off between decreasing overshoot and increasing settling
time.

The lack
of derivative action may make the system more steady in the steady state in the case of
noisy data. This is because derivative action is more sensitive to higher
-
frequency terms in the
inputs.

Without derivative action, a PI
-
controlled system is less respo
nsive to real (non
-
noise) and
relatively fast alterations in state and so the system will be slower to reach setpoint and slower to
respond to perturbations than a well
-
tuned PID system may be.

DEADBAND

Many PID loops control a mechanical device (for examp
le, a valve). Mechanical maintenance
can be a major cost and wear leads to control degradation in the form of either
stiction

or a
deadband

in the mechanical response to an input signal. The rate of mechanical wear is mainly a
function of how often a device is activated to make a change. Where wear is a significant
concern, the PID loop may have an output
deadband

to reduce the frequency of activation of the
output (valve). This is accomplished by modifying the controller to hold its output steady if the
change would be small (within the defined deadband ran
ge). The calculated output must leave
the deadband before the actual output will change.

SETPOINT STEP CHANGE

The proportional and derivative terms can produce excessive movement in the output when a
system is subjected to an instantaneous step increase in

the error, such as a large setpoint change.
In the case of the derivative term, this is due to taking the derivative of the error, which is very
large in the case of an instantaneous step change. As a result, some PID algorithms incorporate
the following
modifications:

Derivative of the process variable

In this case the PID controller measures the derivative of the measured
process variable

(PV),
rather than the derivative
of the error. This quantity is always continuous (i.e., never has a
step change as a result of changed setpoint). For this technique to be effective, the derivative
of the PV must have the opposite sign of the derivative of the error, in the case of negati
ve
feedback control.

Setpoint ramping

In this modification, the setpoint is gradually moved from its old value to a newly specified
value using a linear or first order differential ramp function. This avoids the
discontinuity

present in a simple step change.

Setpoint weighting

Setpoint weighting uses different multipliers for the error depending on which element of the
controller it is used in. The
error in the integral term must be the true control error to avoid
steady
-
state control errors. This affects the controller's setpoint response. These parameters
do not affect the response to load disturbances and measurement noise.



Or

b. Explain the ad
aptive control system in detail.


Adaptive control

is the control method used by a
controller

which must adapt to a
controlled system with parameters which vary, or are initially uncertain. For example, as an
aircraft flies, its mass will slowly decrease a
s a result of fuel consumption; a control law is
needed that adapts itself to such changing conditions. Adaptive control is
different

from
robust
control

in that it does not ne
ed
a priori

information about the bounds on these uncertain or time
-
varying parameters;
robust

control guarantees that if the changes are within given bounds the
control law need not be changed, while adaptive control is concerned with control law changing

themselves.


PARAMETER ESTIMATION

The
foundation

of adaptive control is
parameter estimation
.
Common

methods of estimation
include
recursive least squares

and
gradient descent
. Both of these methods provide update laws
which are used to

modify estimates in real time (i.e., as the system operates).
Lyapunov stability

is used to derive these update laws and show convergence criterion (typically persiste
nt
excitation
).
Projection (mathematics)

and normalization are commonly used to improve the
robustness of estimation algorithms.


8.explain the working
and constructional details ,
advantages and
disadvantages of a servo motor.


A
servomotor

is a
rotary actuator

that allows for precise control of angular position.
[1]

It
consists of a motor coupled to a sensor for position feedback, through a
reduction gearbox
. It
also r
equires a relatively sophisticated controller, often a dedicated module designed specifically
for use with servomotors.

Servomotors are used in applications such as
robotics
,
CNC

machinery

or
automated

manufacturing.

MECHANISM

As the name suggests, a servomotor is a
servomechanism
. More specifically, it i
s a
closed
-
loop

servomechanism that uses position feedback to control its motion and final position. The input to
its control is some signal, either
analogue

or

digital, representing the position commanded for the
output shaft.

The motor is paired with some type of
encoder

to provide position and speed feedback. In the
simplest

case, only the posit
ion is measured. The measured position of the output is compared to
the command position, the
external

input to the controller. If the output position differs from that
required, an
error signal

is generated which then causes the motor to rotate in either direction, as
needed to bring the output shaft to the
appropriate

position. As the positions approach, the error
signal red
uces to zero and the motor stops.

The very simplest servomotors use position
-
only sensing via a
potentiometer

and
bang
-
bang
control

of their motor; the motor always rotates at full speed (or is stopped). This type of
servomotor is not widely used in industrial
motion c
ontrol
, but they form the basis of the
simple

and cheap
servos

used for
radio
-
controlled models
.

More sophisticated servomotors measure both the position and also the speed of the output shaft.
They may also control the speed of their motor, rather than always running at full speed. Both of
these enhancements, usual
ly in combination with a
PID control

algorithm, allow the servomotor
to be brought to its commanded position more quickly and more precisely, with less
overshooting
.

SERVOMOTORS VS. STEP
PER MOTORS

Servomotors are generally used as a high performance alternative to the
stepper motor
. Stepper
motors have some inherent ability to control position, as they have built
-
in output steps. This
often allows them to be used as an open
-
loop position control, without any feedback encoder, as
their drive signal specifies the n
umber of steps of
movement

to rotate. This lack of feedback
though
limits

their performance, as the stepper motor can only drive a load that is well within its
capacity, otherwise missed steps under load may lead to positioning errors. The encoder and
cont
roller of a servomotor are an additional cost, but they optimise the performance of the
overall system (for all of speed, power and accuracy) relative to the capacity of the basic motor.
With larger systems, where a powerful motor represents an increasing
proportion of the system
cost, servomotors have the advantage.

Many applications, such as
laser cutting

machines, may be offered in two ranges, the low
-
priced
range using stepper

motors and the high
-
performance range using servomotors.
[2]

ENCODERS

The first servomotors were developed with
synchr
os

as their encoders.
[3]

Much work was done
with these systems in the development of
radar

and
anti
-
aircraft artillery

during
World War II
.

Simple servomotors may use resistive potentiometers as their p
osition encoder. These are only
used at the very simplest and cheapest level, and are in close competition with stepper motors.
They suffer from wear and
electrical

noise in the potentiometer track. Although it would be
possible to
electrically differentiate

their position signal to obtain a speed signal, PID controllers
that can make use of such a speed signal generally warrant a more precise encoder.

Modern servomotors use
optical encoders
, either
absolute

or
incremental
. Absolute encoders can
determine their position at power
-
on, but are more complicated and expensive. Incremental
encoders are simpler, cheaper and work at faster speeds. Incremental systems, like stepper
motors
, often combine their inherent ability to measure intervals of rotation with a simple zero
-
position sensor to set their position at start
-
up.

Many servomotors are rotary, but are used for ultimate control of a linear motion. In some of
these cases, a linea
r encoder is used.
[4]

These servomotors avoid inaccuracies in the drivetrain
between the motor and linear carriage, but their design is made more complicated as they are no
longer a p
re
-
packaged factory
-
made system.

MOTORS

The type of motor is not critical to a servomotor and different types may be used. At the
simplest, brushed permanent magnet DC motors are used, owing to their simplicity and low cost.
Small industrial servomotors ar
e typically electronically
-
commutated brushless motors.
[5]

For
large industrial servomotors, AC induction motors are typically used, often with
variable
frequency drives

to allow control of their speed. For ultimate performance in a compact package,
brushless AC motors with permanent magnet fields are used, effectively large versions of
Brushless DC electric motors
.
[6]

Drive modules for servomotors are a standard

industrial component. Their design is a branch of
power electronics
, usually based on a three
-
phase MOSFET
H bridge
. These standard modules
accept a single direction and pulse count (rotation distance) as input. They may also include
over
-
temperature monitoring, over
-
torque and stall detection features.
[7]

As the encoder type,
gearhead ratio and overall system dynamics are application specific, it is more difficult to
produce the overall controller as an off
-
the
-
shelf module and so these are often implemented as
part of the main contr
oller.

CONTROL

Most modern servomotors are designed and supplied around a dedicated controller module from
the same manufacturer. Controllers may also be developed around
micr
ocontrollers
, but this is
rarely worth the time and trouble, compared to buying off
-
the
-
shelf.


Or

9.explain the various mechanical actuation system with
neat sketches.

Diaphragm actuator works on the principle of gauge pressure i e difference between inp
ut
pressure signal and atmospheric pressure (Pressure = force per unit area)







































































Where,























= Pressure difference (P
ascal)













































A = Diaphragm area (
)















































F = Force (N)

Construction

• Diaphragm actuator consists of a diaphragm made of rubber and is sandwiched between two
circular steel discs. The diaphragm is mounted in upper chamber where the varying pressu
re
from controller is applied.

• The bottom chamber contains a spring that forcing the diaphragm and connected to shaft
down against the spring force. Fig. shows the construction of actuator diaphragm.
























































Fig. Pneumatic diaphragm actuator

Working:

• In upper chamber on ap
plying the varying air pressure results in a varying force on the top
of the diaphragm. Initially, with no air pressure, the spring (S) forces the diaphragm upward
and holds the valve fully open. The pressure on opposite side of diaphragm is maintained at
atmospheric pressure by open whole H.

• The effect of changes in input air pressure applies a force on the diaphragm this forces
diaphragm down against the sprig force Fig shows the condition of diaphragm after applying
pressure and maximum travel of shaft

(high pressure state).






































Fig Pn
eumatic diaphragm actuator under high pressure

• The pressure and force are linearly related by equation (1). As the compression of a spring is
linearly related to force hence the shaft position is linearly related to applied control pressure.
It is given
by





























Where,























= Shaft travel (meters)




































= Differential pressure (Pa)






































A =Diaphragm are
a







































K = Spring constant (N/m).

1
1. compare the power transmission by gear,belt and
chain drives.

1.2.1 POWER TRANSMIS
SION USES

Power transmission machines use either chains, gears, or belts.
Tabl
e 1.1

provides a comparison
of typical applications.

Usually, chain is an
economical

part of power transmission machines for low speeds and large
loads. However, it is also possible to use chain in high
-
speed conditions like automobile engine
camshaft dri
ves. This is accomplished by devising a method of operation and lubrication.

Basically, there are lower limits of fatigue strength in the gear and the chain, but not in the belt.
Furthermore, if a gear tooth breaks, the gear will stop at the next tooth. Th
erefore, the order is
gear > chain > belt in the aspect of reliability.

In most cases:

1.

An increase in gear noise indicates that the end of the service life is near.

2.

You will know that the chain is almost at the end of its life by wear elongation or an incr
ease in
vibration caused by wear elongation.

3.

It is difficult to detect toothed
-
belt life without stopping the machine and inspecting the belt
carefully.

It is possible to decrease gear noise by adjusting the gears precisely or by adapting the drive to a
he
lical or double helical gear. Both of these are expensive, and thrust load may occur with the
use of helical gears.

Chain is more suitable to long
-
term continuous running and power transmission with limited
torque fluctuation. Gears are more fit to reversi
ng or intermittent drives.

The greater the shaft center distance, the more practical the use of chain and belt, rather than
gears.

Table 1.1

comparison Table

Type

Roller Chain

Tooth Belt

V Belt

Spur Gear

Synchronization





Transmission
Efficiency





Anti
-
Shock





Noise/Vibrati
on





Surrounding
Condition

Avoid Water,
Dust

Avoid Heat, Oil,
Water, Dust

Avoid Heat, Oil,
Water, Dust

Avoid Water, Dust

Space Saving

(High Speed/ Low
Load)





Space Saving

(Low Speed/ High
Load)


Compact


Heavy Pulley


Wider Pulley


Less Durability Due to Less
Engagement

Lubrication


Required


No Lube


No Lube


Required

Layout Flexibility





Excess Load onto
Bearing





Excellent
Good
Fair
Poor

Generally, under the same transmission conditions, the cost of toothed belts and pulleys is much
higher than the cost of
chains and sprockets.

See the following features and points of notice about roller chain transmission.

Features of Chain Drives:

1.

Speed reduction/increase of up to seven to one can be easily accommodated.

2.

Chain can accommodate long shaft
-
center distances (l
ess than 4 m), and is more versatile.

3.

It is possible to use chain with multiple shafts or drives with both sides of the chain.

4.

Standardization of chains under the American National Standards Institute (
ANSI
), the
International Standardization Organization
(
ISO
), and the Japanese Industrial Standards (
JIS
)
allow ease of selection.

5.

It is easy to cut and connect chains.

6.

The sprocket diameter for a chain system may be smaller than a belt pulley, while transmitting
the same torque.

7.

Sprockets are subject to less
wear than gears because sprockets distribute the loading over
their many teeth.


12.a.discuss about the design of an enhanced nonlinear
PID controller.

An enhanced nonlinear PID (EN
-
PID) controller that exhibits the improved performance than the
convention
al linear fixed
-
gain PID controller is
proposed

in this paper, by incorporating a sector
-
bounded nonlinear gain in cascade with a conventional PID control architecture. To achieve the high
robustness against noise, two nonlinear tracking differentiators ar
e used to
select

high
-
quality
differential

signal in the presence of measurement noise. The criterion to determine the nonlinear gain
to retain the stability of the proposed EN
-
PID control system is addressed, by using the Popov stability
criterion. The ma
in advantages of the proposed EN
-
PID controller lie in its high robustness against noise
and easy of implementation. Simulation results performed on a robot manipulator are presented to
demonstrate the better performance of the developed EN
-
PID controller
than the conventional fixed
-
gain PID controller.

Or

b.i.discuss about the
hardware

of microprocessor
.

A
microprocessor

(
microscopic processor

or
processor on a chip
) is the miniaturized circuitry
of a computer processor
--

the CPU, the part that processes,

or manipulates, data into
information. It is the most important part of personal computers and workstations. It controls
logic of almost all digital devices, for example: mobile devices, clocks and computers.
Microprocessors are different on the basis of
instructions set, bandwidth and clock speed. It is
very tiny in size but it can perform various operations within a matter of few seconds. It is like a
engine that goes on when the computer starts. Computers cannot perform any operations without
a micropro
cessor. Microprocessors take the data as an input, translate it into a machine language
and display the result as an output. We use so many things in our daily life which work with
embedded systems and microprocessors are a part of it. Microprocessors
tech
nology

has taken a
once bulky series of switches and miniaturized
powerful

electronic circuitry to be able to fit in
personal computers, cell phones, pda's, and other popular devices. When referring to
microprocessors, people are usually talking about the
CPU, which is made up of at least one
microprocessor that
handles

the chief "brain" functions.

According to scientific law, the speed of microprocessors will increase rapidly from year to year,
and you can see evidence of this as current technology
becomes

outdated very quickly.
Microprocessors are necessary in almost any engineering project.
Circuit

board
design

may
depend on a specific microprocessor construction that can be supplied through contract
manufacturing, and outsourced services.

WORKSTATION

The

Work Station Computer is an expensive, powerful personal computer, usually used for the
complex scientific mathematical and engineering
Calculations

and for computer aided design
and manufacturing. These computers are commonly used by engineers and other
scientific
people who would need the extra power in a computer for many many , calculations. The
workstation offer many capabilities similar to the midsized main frames, such as designing car
engines

and intricate mechanical functions. The workstations hav
e incredible graphic capabilities
witch has helped catch the
public's

eye. The capabilities of a low end workstation overlap those
of a high end desktop microcomputer.



ii.what are the type of control used in MIMO systems?
Discuss.

MULTI
-
INPUT, MULTI
-
OUTP
UT

Systems with more than one input and/or more than one output are known as
Multi
-
Input
Multi
-
Output

systems, or they are frequently known by the
abbreviation

MIMO
. This is in
contrast to systems that have only a
single

input and a single output (SISO), l
ike we have been
discussing previously.


13.a.i.explain the system modeling of a chamber filled
with fluid

PURE FLUID CAPACITOR


A physical element in which the energy stored is a function of fluid pressure may be thought of as a
PURE FLUID CAPACITOR
. An
IDEAL FLUID CAPACITO
R

is a pure fluid capacitor defined by the linear
constituency relation

V
(
t
) =
C
f

p
(
t
)

(24)



or


( 25)



where
V
(
t
) is the volume of the fluid in the capacitor,
Q
(
t
) is the fluid volume flow rate into the
capacitor,
p
(
t
) is the fluid pressure in the capacitor, and
C
f

[m
3
.Pa
-
1
] is the constant parameter called
the
F
LUID CAPACITANCE

of the capacitor.

As example of a subsystem which may behave approximately like an ideal fluid capacitor is a
closed
PRESSURIZED CHAMBER

with rigid walls completely filled with fluid shown in
Fig.

40
a
. The energy is stored in the chamber because of the compressibility of the fluid, and the
variation of pressure in the chamber occurs as a result of the amount of fluid that has been forced
into it.



Figure 40:

(
a
) Chamber containing compressed fluid. (
b
) Symbol of pure fluid capacitor. (
c
) Model of
the pressurized chamber. (
d
) Open fluid reservoir.


According to the law of conserv
ation of mass the time rate of mass entering the chamber through
the inlet
A

must equal to the increase of mass in the chamber, or


(26)



where
[kg/m
3
] is the fluid density and
[m
3
] is the constant chamber volume.

To employ
chamber

pressure
p

rather tha
n the mass density
as a variable, we can utilize the
equation of state of the liquid. For most liquids, a reasonably good first
-
order approximation for
their equation of st
ate is


(27)



where
[N/m
2
] is the bulk modulus of elasticity of the liquid. Substi
tuting (
27
) into (
26
) yields


(28)



where
p
A

is the gage pressure measured at the chamber inlet
A
.

Comparing this result with (
25
) indicates that the chamber behaves like an ideal fluid capacitor
wi
th the capacitance


(29)



In the above derivation we have tacitly made the following assumptions:



the influence of gravity on the fluid is negligible



fluid inertia a
nd frictional effects can be neglected



the pressure of the fluid which has been forced into the chamber is uniform (but not
necessarily constant) throughout the tank



the pressure and temperature variations of the bulk modulus of elasticity are negligible


Under the above assumptions the pressurized chamber can be represented in the dynamic
diagram of a system model by the symbol of the pure fluid capacitor given in Fig.

40
b
. The
sy
mbol consists of a square representing the capacitor volume and pin representing the capacitor
inlet. The capacitor is associated with two variables, the capacitor pressure
p

measured with
respect to a constant pressure reference, and the capacitor flow ra
te
Q
. The assumed positive
polarity of
p