Signal Processing and Conditioning

8/12/2019 Signal Processing and Conditioning

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1.0 SIGNAL PROCESSING AND

CONDITIONING.

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1.1. BASICS

Most modern control equipment work on the followingstandard signal ranges.

Electric 4 to 20mA

Pneumatic 0.2 to 1.0 bar

Digital Standards

Older electrical equipment use 0 to 10V.

The advantage of having a standard range is that all equipmentare sold ready calibrated. This means that the minimum signal

(Temperature, speed, force, pressure and so on) isrepresented by 4mA or 0.2 bar and the maximum signal isrepresented by 20mA or 1.0 bar.

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The primary transducer will not produce these standardranges so the purpose of processing and conditioning is

usually to convert the output into the standard range.The vastarray of instrumentation and control equipment availableuses many forms of signal. Here is a summary:

ELECTRICAL voltage, current, digital

MECHANICAL force and movementPNEUMATIC AND HYDRAULICpressure and flow

OPTICAL high speed digital transmission

RADIO analogue and digital transmission

ULTRA VIOLETsimilar application to radio over shortranges.

Processing may do the following things

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Change the level or value of the signal (e.g. Voltage level)

Change the signal from one form to another (e.g. Electrical

to pneumatic and vice-versa)Change the operating characteristic with respect to time

Convert analogue and digital signal from one form to the

other.

Now considering those processes which change the level orvalue of the signal.

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1.2.1 Amplifiers

Amplifiers may amplify VOLTAGE, CURRENT, or BOTH in whichcase it may be a Power Amplifier. Amplifier gain may beexpressed as a ratio or in decibels.

The letter, W indicates that it

refers to power gain.

The gain in dbW is given by

Gain(dbW) = 10 log10 Power Output

Power InputIn practice, an amplifier generates some noise and the inputand output terminals have a resistance that governs the ratioof current to voltage.

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Fig 3.1


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Since electric power into a resistive load is given asP = I R = = V /R

Then

Gain (dbV) = 10log10 .V out/Vin = 20 log10Vout/VinThe letter V indicates it as a voltage gain ( units for voltage gain isdbV)

EXAMPLE

Calculate the gain of a voltage amplifier with an input of 2mV andoutput of 10V

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SOLUTION

Voltage gain, G = 20log1010/0.02

= 73.98dbV

Differential Amplifiers

These have two inputs and the difference between them isamplified. The electronic symbol is shown in fig. 1.4

Voltage=20log10 Vout

V2V1

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EXAMPLE

For the differential amplifier shown below determine the output

voltage if the gain is 15db.

SOLUTION

Input = 5-2 = 3V

G = 15 = 20log10(Vout/3)

15/20 = 0.75 = log10(Vout/3)

Antilog 0.75 = 5.623 = (Vout/3)

Vout= 16.87V

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1.2.2 Attenuators

Sometimes a signal is too big and must be reduced byattenuating it. Electrical signals are attenuated with resistorswhich dissipate the electric power as heat. The gain of anattenuator in db. is negative.

The term amplificationis often used when the level of asignal is increased but not the power. Strictly speaking suchdevices should be called Transformers. For example, an AC.Electrical transformer may increase the voltage but not thepower. We have voltage amplifiers and current amplifiers

which do not necessarily change the power level.

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1.2.3. Transformers

Electrical(Review)Many devices only change the level of a signal without

changing the power. A voltageamplifier is an example. An

electrical transformer foralternating voltages basically

consists of two windings, aprimary and a secondary.

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The coils are wound on a magnetic core. As learnt before, theflux depends upon the number of Turns, T1and the same fluxcuts the secondary. The e.m.f. in the secondary will depend on

the number of turns T2

.It follows that

V1/V2= T1/T2

In an ideal transformer there is no energy loss and so thepower in and power out are equal. V1i1= V2i2. It follows that ifthe voltage is stepped down, the current is stepped up andvice-versa.

MechanicalMechanical transformers are levers and gear boxes whichchange movement, force, speed and torque but not thepower.

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They are used in many instruments (eg. A mechanicalpressure gauge and the nozzle -flapper system, etc).

The gear ratio is in direct proportion to the pitch circlediameters (mean diameters) or number of teeth on eachwheel.

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The lever movements at the ends are in direct proportion to

the length on each side of the fulcrum.

Hydraulic

The hydraulic pressure amplifier shown below increases the

pressure in direct proportion to the area of the pistons. It is

called an intensifier.

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SELF ASSESSMENT EXERCISE No.2

1. An electrical transformer must produce 12 V a.c. output from 240 V a.c.

input. The primary has 2000 turns.How many turns are needed on the secondary? (Answer 100)

2. A pressure intensifier must increase the pressure from 10 bar to 100 bar.

What must be the ratio of the

piston diameters? (Answer 3.162/1)

3. A lever must magnify the movement of a mechanism from 0.1 mm to 2

mm. What must be the ratio of the

lengths either side of the fulcrum? (Answer 20/1)

4. A pair of simple gears must magnify the rotation angle by 4/1 many must

there be on the large gear? (Answer 80 teeth)


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1.3. Signal Converters

Signal converters change the signal from one form to another.Where ever possible, these are the standard inputs and

output ranges. Normally we show them on a block diagram as

a box with an input and output with a label to say what it

does. Figure 3.8 shows some examples. All these examples

have opposite versions, i.e. I/P, P/M, V/f, D/A etc.

Most signal converters have two adjustments.

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NozzleFlapper and Differential PressureCells

The nozzle-flapper system is widely used in DifferentialPressure (D.P.) cells. The form shown below convertsdifferential pressure (e.g. from a differential pressure flowmeter) into a standard pneumatic signal. This is widely used in

the control of air-operated valves

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In operation, the bellows respond to the differential pressureand moves the lever. This moves the flapper towards or awayfrom the nozzle. The air supply passes through a restrictor

(orifice) and leaks out of the nozzle.The output pressure hence depends on how close the flapperis to the end of the nozzle. The range of the instrument isadjusted by moving the pivot and the zero position is adjustedby moving the relative position of the flapper and nozzle.

This system is used in a variety of forms. Instead of bellows, abourdon tube might be used and this is operated by anexpansion type temperature sensor to produce atemperature-pneumatic signal converter.

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REGENT UNIVERSITY COLLEGE OF SCIENCE AND

TECHNOLOGY

WEEKEND SESSIONROQUAH CAMPUS

LEVEL 4003RDTRIMESTER

2011/2012 ACADEMIC SESSION


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COURSE OUTLINE

COURSE TITLE: INDUSTRIAL INSTRUMENTATION AND CONTROL

COURSE CODE: SIEL 4613

Overall Course Objective(s) : The student will be able to understand and

appreciate the principles underlying the

various instrumentation and process control

applications in Industries .

Teaching Method: Power point presentation, quizzes,

assignments and class test.

Contact: Ing.Dr. P.K. AmissahSenior Lecturer/HOD


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COURSE DESCRIPTION

SIEL 4613:- Industrial Instrumentation and Control [ 2 0 3]

Introduction to process control; elements of the process loop;

controller principles: hydraulic, pneumatic, electric and

electronic controllers; digital control principles; final control

elements; control loop characteristics, complex control

system; fundamentals of fluidics, fluidic logic and fluidic

devices; industrial telemetry techniques, pneumatic

indicators, receivers, transmitters, indicating controllers;

electro-pneumatic converters.


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TRIMESTER SCHEDULE

9-10 DECEMBER 2011 WK.1.SCHOOL RE-OPENS

20-21 JANUARY 2012 WK.5.QUIZ

3-4 FEBRUAYRY 2012 ..WK..7/8 ..MID-TRIMESTER EXAMS

10-11 FEBRUARY 201216-17 MARCH 2012 . WK. 13/14/15 END OF TRIM EXAMS

23-24 MARCH 2012 /30-31 .

6-7 APRIL 2012 VACATION PERIOD

13-14 APRIL 2012 .SCHOOL RE-OPENS

Documents

Signal Processing and Conditioning