Signal Conditioningdragon.ccut.edu.tw/~mejwc1/e-mea/PDF/em_7.pdf · Signal Conditioning ... balance, the ratio of resistance of any two ... strain gauge cemented to it to detect diaphragm

Signal Conditioning

Modulated and UnmodulatedSignal

Input CircuitryResonant Circuits

Electronic Amplifiers

Signal Conditioning

• Dynamic mechanical quantities– Large amplifications– Good transient response

• Mechanical amplification limited– Undesirable signal loading

• Backlash, elastic deformations,…– Frequency response

Advantages of Electrical Signal Conditioning

• Converting resistance changes to voltage changes

• Subtracting offset voltages• Increasing signal voltages • Removing unwanted frequency

components• Power amplification

– Provide a greater output than input• Easily Recording procedures


Amplitude Modulation (AM)The carrier frequency is held constant and its

amplitude is varied by the measurand


Amplitude Modulation (AM)






Amplitude Demodulation




Frequency Modulation (FM)The carrier amplitude is held constant and its

Frequency is varied by the measurand


• AM is the more common form– Nearly any mechanical signal from a passive pickup

can be transduced into an analogous form– Sensors based on either inductance or capacitance

require an ac excitation

• Demodulated– Extracting the signal information from the modulated

carrier, by rectification and filtering– Using an oscilloscope or oscillograph, and then to

read the result from the envelope of the carrier – FM demodulation is more complex operation

Input Circuitry• Detector-Transducers type

– Passive, those requiring an auxiliary source of energy in order to produce a signal

– Active, those that are self-powering• The most common form of input circuits

– Simple current-sensitive circuits – Ballast circuits– Voltage-dividing circuits– Bridge circuits– Resonant circuits– Amplifier input circuits

The Simple Current-Sensitive Circuit

Current indicatoror recorderSensing outputcurrent, i0

Rm

kRtResistance-typetransducer

ei


• Simple circuit– The transducer may use any one of the

various form of variable-resistance elements– Transducer resistance kRt

• k : percentage factor (0%~100%)• Rt : the maximum transducer resistance value

– The remaining circuit resistance Rm


max 0

law sOhm' using

iReik

RkRei

m

io

mt

io

===

+=

kRRe

Riii

m

ti

moo

+

==1

1

max

0 0.2 0.4 0.6 0.8 1.00

0.20

0.40

0.60

0.80

1.00

i o/i m

ax

k

Rt/Rm=0.5

1

2

4

10

The Ballast Circuit

Voltage indicatoror recorderSensing outputvoltage

Rb

kRtResistance-type

transducer

ei

The Ballast Circuit

• Use a voltage-sensitive device placed across the transducer

• It would always indicate full source voltage

• Two different situations may exist– The meter may be of high impedance, as would be

the case if some form of electronic voltmeter were used

– The meter may be of low impedance, so that consideration of such current flow is required

The Ballast Circuit

tb

tito

mt

io

kRRkRekRie

RkRei

ance meterhigh imped

+==

+=

)(

lawsOhm' using

)/(1/

bt

bt

i

o

RkRRkR

ee

+=

outputRkR

inputee

b

t

i

o

⇒

⇒

The Ballast Circuit

3)/()(

bt

btti

b RkRRkRRe

dRd −

=η

2)(y sensitivit the

bt

btio

RkRRRe

dkde

+==η

obtained. isy sensitivit maximumfor which ,kRRfor and y,sensitivit minimumin resultswhich ,Rfor

tb

b

=∞=

The Ballast Circuit

• Input and output relation curves

0 0.2 0.4 0.6 0.8 1.00

0.20

0.40

0.60

0.80

i o/i m

ax

k

Rt/Rb=2.0

0.5

1.0

Not linear

Voltage-Dividing Circuits

• A ubiquitous element of instrumentation circuits

• Uses a pair of resistors to divide

R1

ei

R2

+

_

eoio

i

eRR

RiRe

RRei

21

22

21 )(

law sOhm' using

+==

+=

The Voltage-Dividing potentiometer

• It would be in the ballast circuit, but across the complete resistance element

• A simple pressure-measuring device

i

o

iip

po

eek

keeR

kRe

=

==ei Rp

eo

RL

kRp

The Voltage-Dividing Loading Error

Lp

Lpp RkR

RkRkRR

++−= )1(

Lpp

Lpii

RRkkR

RkReRei

+−

+==

)1(

)(2

2)/()/(1 kRRkRRk

ee

LpLpi

o

−+=

)1( kiRee pio −−=


+−−

=

+−−=

)/()1()1(

1)/)(1(

error

2

pLi

Lpi

RRkkkke

RRkkkke

zero. iserror which the,10

100)/()1(

)1(

errorPercent output scale-full2

or kfor

RRkkkk

pL

=

×

+−−

=


k

RL/Rp=1.0

2

510

0 0.2 0.4 0.6 0.8 1.00

2.0

4.0

6.0

8.0

10.0

Erro

r, %

12.0

14.0

Small Changes in Transducer Resistance

• Some resistance transducers show only very small changes in their resistance– For example, a foil strain gage (0.0001%)

R1 =R0

ei

R2 = R0

→R0 + R

+

_

eo →eo + eo 221

2

021

iio

eeRR

Re

RRR

=+

=

==


∆+

∆+=

∆+∆+

=∆++

∆+=∆+

oo

ii

io

o

o

oi

oo

ooo

RRRRee

eRR

RRR

ReRRR

RRee

2/11

222

2/1/1

2)(

io

oo

iioo

o

eRRe

RReeee

RR

4222

12/

∆+=

∆+≈∆+

<<∆

. with variation shows thechanges, resistance smallfor Thus,

∆Rinestraight-lvoltageouput

Reo ∆⇔∆


output. theof reading final and initial hebetween tslightly drifts if ie

io

iooo e

RReeee

42∆

+∆

+≈∆+

6102

120240:

−=∆

=∆

==

oo

o

o

RR

ee

ΩµΩ;R∆Rexamplefor

An important principle in measurement:Avoid measurements based on a small difference between large numbers.


• Eliminated eo circuit

R0

ei

R0 + R

eout

io

oooobaout eRReeeeeee

4)( ∆

=∆=−∆+=−=

R0

R0

ea eb

Resistance Bridges

• Connecting passive transducers to measuring systems

• The Wheatstone resistance bridge– By S. H. Christie in 1833– Bridge circuits enable high-accuracy

resistance measurements– Application

• Resistance thermometers• Thermistors• Resistance type strain gage

Wheatstone Bridge Circuit

4

2

3

1

4

3

2

1 or RR

RR

RR

RR

==

Wheatstone Bridge Circuit

Wheatstone Bridge Circuit• In order for the Wheatstone resistance bridge to

balance, the ratio of resistance of any two adjacent arms must equal the ratio of resistance of the remaining two arms, taken in the same sense

• Types of electrical bridge circuits– Voltage (current)-sensitive bridge– Null balance (deflection) bridge– Ac (dc) bridge– Constant voltage (current)– Resistance (impedance) bridge

DC Resistance Balance Bridges

Null-type d.c. bridge

• Wheatstone bridge with d.c. excitation• The unknown resistance Ru

• Two equal-value resistors R2 and R3

22

31

4231

:law sOhm'by and

0

RRVI

RRVI

IIIIIfor

v

i

u

i

m

+=

+=

===


32

22

31

:thus

ionsuperposit of principle by the

;

dropvoltage thecalculate

RRRV

RRRVV

VVVVVV

RRRVRIV

RRRVRIV

u

ui

v

vio

ADABADBABDo

v

viuAB

u

uiuAD

++

+−=

+−=+==

+==

+==


.32

2

323

23

23

then if Thus,

or i.e.

:sidesboth inverting

:so ,0point null at the

vu,

vu

vu

v

v

u

u

v

v

u

u

o

RRRRRRRR

RR

RR

RRR

RRR

RRR

RRR

V

==

==

+=

+

+=

+

=

Deflection-type d.c. bridge

• The unknown resistance Ru

+

−+

=

21

1

3 RRR

RRRV

V

u

ui

o

Example: [deflection-type d.c. bridge]

A certain type of pressure transducer, designed to measure pressures in the range 0-10 bar, consists of a diaphragm with a strain gauge cemented to it to detect diaphragm deflections. Thestrain gauge has a normal resistance of 120Ω and forms one arm of a Wheatstone bridge circuit, with the other three arms each having a resistance of 120 Ω. The bridge output is measured by an instrument whose input impedance can be assumed infinite. If,in order to limit heating effect, the maximum permissible gauge current 30mA, calculate the maximum permissible bridge excitation voltage. If the sensitivity of the strain gauge is 338m Ω/bar and the maximum bridge excitation voltage is used, calculate the bridge output voltage when measuring a pressure of10 bar.

Solution:Given data: R1=R2=R3=120ΩIn path ADC of the bridge Vi=I1(Ru+R3)At balance, Vi=0.03(120+120)=7.2 VThus the maximum bridge excitation voltage allowable is 7.2 volts.

For a pressure of 10 bar applied, we can write:

Thus, if the maximum permissible bridge excitation voltage is used, the output voltage is 50mA when a pressure of 10 bar is measured.

mVRR

RRR

RVVu

uio 50)

240120

38.24338.123(2.7)(

21

1

3

=−=+

−+

=

Example: [deflection-type d.c. bridge]

A bridge circuit, as shown in below, is used to measure the value of the unknown resistance Ru of a strain gauge of normal value 500Ω. The output voltage measured across point DB in the bridge is measured by a voltmeter. Calculate the measurement sensitivity in volts per ohm change in Ru. If the resistance Rm of the measuring instrument is neglected.

Solution:Given: Ru=500Ω, Vm=0.Find: To determine sensitivity, calculate Vm for Ru=501ΩSolution:

Applying equation and substituting in values:

Thus, if the resistance of the measuring circuit is neglected, the measurement sensitivity is 5.00mV per ohm change in Ru.

mVRR

RRR

RVVu

uio 00.5

1000500

100150110

21

1

3

=

−=

+

−+

=

The voltage-Sensitive W. Bridge

• Readout instrument does not “load” bridge; that is, it require no current; e.g., electronic voltmeter or CRO

+∆++−∆+

=

+∆++

−∆+=∆+

)]/(1)][/()/(1[()/()/(1

))(())(

342221

321422

43221

14322

RRRRRRRRRRRRe

RRRRRRRRRReee

i

ioo

Resonant Circuits

• Impedance bridge– The inductance offers small opposition to

current flow at low frequencies– The capacitive reactance is low at high

frequencies• Resonance frequency

LCf

π21

=

LC Circuit and Frequency-impedance

Variation in capacitance caused by variation in an input signal would then alter the resonance frequency, which could be used as a measure of input.

Electronic Amplification or Gain

• The ratio of output to input– Gain– Amplification ratio (greater then unity)– Attenuation (less then unity)

• Power gain (decibel)– A decibel(dB) is one-tenth of a bel and is

based on a ratio of power )/(log10(dB) decibel 10 io PP=

Electronic Amplification or Gain

RiReei 22 /power

resistance pure aFor

===

)/(log10)/(log20dB)/(log10)/(log20dB

1010

1010

ioio

ioio

RRiiRRee

+=−=

Frequency sees voltage ratios

=

i

o

ee

10log20dB

Ideal Electronic Amplifiers

• Infinite input impedance• Infinite gain• Zero output impedance• Instant response• Zero output for zero input• Ability to ignore or reject extraneous inputs

Operational Amplifiers (op amp)

• A dc differential voltage amplifier

)( −+ −= eeGeo

Op-Amp Output Response

(e+-e-)

eo

0

≈Vcc

≈Vee

1G

Saturated Saturated

Typical Op-Amp

DIP integrated circuit TO integrated circuit

Op-Amp Satisfies the Ideal Amplifier

• Very high input impedance• Capable of very high gain• Very low output impedance• Very fast response or high slew rate• Quite effective in rejecting common-mode

inputs

=

cmGG

10log20CMRR

ratiorejection mode-common

Shielding

• Purposes– To isolated or retain electrical energy within

an apparatus– To isolated or protect the apparatus from

outside source of energy• Basic type

– Electrostatic– electromagnetic

Shielding Rules• An electrostatic shield enclosure, to be effective,

should be connected to the zero-signal reference potential of any circuitry contained within the shield

• The shield conductor should be connected to the zero-signal reference potential at the signal-to-earth connection

• The number of separate shields required in a system is equal to the number of independent signals being processed plus one for each power entrance

Grounding

• Required reason– To provide an electrical reference for the

various sections of a device– To provide a drainage path for unwanted

currents• Ground reference type

– Earth ground– chassis ground

Earth Chassis

Grounding Rules• An entire system can be grounded and need not

involve earth at all• The word circuit need not imply wires or

components• Shielding can be at any potential and still

provide shielding• Two nearly points are at the same potential is

often invalid• Potential characteristics of an element are not

the same at radio or high frequencies as they are at power or low frequenciesOther rules see text p.304

Filters • The process of attenuating unwanted

components of a measurand while permitting the desired components to pass

• Basic classes– Active : uses powered components, commonly

configurations of op amps– Passive : made up of some form of RLC

arrangement• Classified

– High-pass, low-pass, band-pass– Notch or band-reject

Outputs from ideal filters

0 ∞frequency

Signalamplitude

Raw signal


0 ∞frequency

Signalamplitude

Pass-band

Stop-bandHigh-pass filter

0 ∞frequency

Signalamplitude

Pass-band

Stop-band

Low-pass filter

fc

fc


0 ∞frequency

Signalamplitude

Pass-band

Stop-bandBand-stop filter

0 ∞frequency

Signalamplitude

Pass-band

Stop-band

Band-pass filter

fc2

fc1

fc1

fc2


0 ∞frequency

Signalamplitude

Pass-band

Stop-band

Notch filter

fc1 fc2

Outputs from practical constant-k filters

Low-Pass Filter

Band-Pass Filter

The RC Low-Pass Filters Theory

R

ei

+

eo

+i

C

RCfc π2

1= 2)/(1

1

ci

o

ffVV

+=

−= −

cff1tanφ

Low-Pass Filters Frequency Response

1 2 3

0.5

1.0

2/1

f/fc

Vo/Vi

The RC High-Pass Filters Theory

Rei

+

eo

+i

C

RCfc π2

1=

2)/(1

/

c

c

i

o

ff

ffVV

+=

−= −

cff1tan90oφ

Filter Frequency Response (Bode Plot)

Basic Active Filters

Passive filter networks are linked to an op amp, which provide power and improves impedance characteristics.

Passive filter

network

Passive filter

network_

+

ei eo

LC Filter ArrangementsBand passHigh passLow pass

LC Filter ArrangementsBand passHigh passLow pass

First-Order Active Filters

Low pass High pass

Band pass

Differentiators Op-Amp

_

+ei eo

C

R

dtdeRCe

Re

dtdeC i

ooi −=−= or

Integrators Op-Amp

_

+

ei eo

CR

.1or constdteRC

e)(-edtdC

Re

iooi +−== ∫

Component Coupling Methods

• Coupling problems– Obtaining proper impedance matching– Maintaining circuit requirements such as

damping– Cause by the desire for maximum energy

transfer and optimum fidelity of response• I most cases, driving a high-impedance circuit

component with a low-impedance source presents fewer problems than does the reverse

Simple circuit for component coupling

ZsZLEs EL

+

=sL

LsL RR

REE

222

L is R todeliveredpower the

+

==sL

L

L

s

L

L

RRR

RE

REP

The maximum power is transferred if RL=RsIn general terms, maximum power is transferredwhen ZL=Zs

Power Transfer coupling methods

• Proper coupling may be important in providing adequate dynamic response

• Methods– Matching transformers– Impedance transforming– Coupling networks

Impedance MatchingBy means of a coupling transformer

L

s

L

s

ZZ

NN

=Zs= the source impedance,ZL= the load impedance,Zs/ZL= the turns ratio of the transformer.

Impedance MatchingBy means of a resistance pad

Rd= the output impedance of the driver,RL= the load resistance,Rp= the paralleling resistance,Rs= the series resistance.

++=

Lp

Lpsd RR

RRRR

Documents

Signal Conditioningdragon.ccut.edu.tw/~mejwc1/e-mea/PDF/em_7.pdf · Signal Conditioning ... balance, the ratio of resistance of any two ... strain gauge cemented to it to detect diaphragm