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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
Modulated and UnmodulatedSignal
Amplitude Modulation (AM)The carrier frequency is held constant and its
amplitude is varied by the measurand
Modulated and UnmodulatedSignal
Amplitude Modulation (AM)
Modulated and UnmodulatedSignal
Amplitude Modulation (AM)
Modulated and UnmodulatedSignal
Amplitude Modulation (AM)
Modulated and UnmodulatedSignal
Amplitude Demodulation
Modulated and UnmodulatedSignal
Amplitude Modulation (AM)
Modulated and UnmodulatedSignal
Frequency Modulation (FM)The carrier amplitude is held constant and its
Frequency is varied by the measurand
Modulated and UnmodulatedSignal
• 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
The Simple Current-Sensitive Circuit
• 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
The Simple Current-Sensitive Circuit
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 −−=
The Voltage-Dividing Loading Error
+−−
=
+−−=
)/()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
=
×
+−−
=
The Voltage-Dividing Loading Error
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
=+
=
==
Small Changes in Transducer Resistance
∆+
∆+=
∆+∆+
=∆++
∆+=∆+
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 ∆⇔∆
Small Changes in Transducer Resistance
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.
Small Changes in Transducer Resistance
• 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
+=
+=
===
Null-type d.c. bridge
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
++
+−=
+−=+==
+==
+==
Null-type d.c. bridge
.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
Outputs from ideal filters
0 ∞frequency
Signalamplitude
Pass-band
Stop-bandHigh-pass filter
0 ∞frequency
Signalamplitude
Pass-band
Stop-band
Low-pass filter
fc
fc
Outputs from ideal filters
0 ∞frequency
Signalamplitude
Pass-band
Stop-bandBand-stop filter
0 ∞frequency
Signalamplitude
Pass-band
Stop-band
Band-pass filter
fc2
fc1
fc1
fc2
Outputs from ideal filters
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