abe425 engineering measurement systems abe425 engineering measurement systems measurement systems...
TRANSCRIPT
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ABE425 Engineering Measurement ABE425 Engineering Measurement SystemsSystems
Measurement Systems with Electrical Signals
Dr. Tony E. Grift
Dept. of Agricultural & Biological EngineeringUniversity of Illinois
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Agenda
1. AC and DC signals2. Transducers3. OpAmps4. Active Filters5. Loading error examples
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AC and DC signals
Alternating Current (AC)Direct Current (DC)Most signals have both!
DC component (offset) measurement: Put DMM on DCAC component measurement: Put DMM on AC
Scope gives Amplitude and peak-peak valueTo get the scope trace at 0: Put input on gnd.
How is the scope amplitude related to the AC value on the DMM?
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Let’s figure out what the AC RMS value is of this signal
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
x 10-3
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
1
11,000
2* *
T ms
f HzT
radpi f
s
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RMS value represents Power
2 212
1.44100DC
UP Watt
R
2
2 2
U U
2
2
U
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
x 10-3
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
x 10-3
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1DC: 12 Volt, R = 100 Ω
AC RMS value must give same power as DC of the same value
Mean value
Root Mean Square
2 212
1.4410
RMSAC
UP Watt
R
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Question
If an AC Voltmeter shows 12V RMSWhat is the amplitude of this signal?
Simple AC Voltmeters measure the amplitude and divide by sqrt(2)
This only works for Sinusoidal signals!!
True RMS voltmeters measure real Power, Resistance and take the square root of the ratio
This works for ANY signal since it follows the definition of the power equalization of DC and AC signals
The RMS value is NOT the mean of the AC signal. It is the Root of the Mean of the Squared value!
* 2 12* 2 16.97RMSU U V
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Digital TRUE RMS meters digitize the signal and compute the RMS value from the definition
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
x 10-3
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Digitize at least one cycle of the signalSquare itCompute mean value
Integrate signalDivide by cycle time
Take square root
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A Transducer converts a physical measurand into an electric signal
AntennaCathode Ray Tube. LCD monitorFluorescent light, light bulb. Light Emitting Diode (LEDs)Magnetic stripe cards
Photocells/Light
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Combined model of (a) input source, (b) amplifier and (c) output load
You want to prevent loading errorsChoose Ri high
Choose Ro low
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Non-inverting OpAmp
Ofi
i
i
O
VRR
RV
VV
VVAV
i
fii
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fiO
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iiO
R
RVV
R
RRV
A
V
RR
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A
V
V
RR
RA
AV
AVRR
RAV
VRR
RVAV
1lim
1
1
1
1
i
fiO R
RVV 1
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Non-inverting OpAmp : Virtual ‘ground’ principle
i
fiO R
RVV 1
i
O
VV
VVVVAV
Since no current is flowing intothe OpAmp:
f
iO
i
i
R
VV
R
V
What is the input resistance for the source ?
iV
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OpAmps have a limited band width (741 is about 1 MHz)
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Inverting OpAmp
0
V
RR
RV
RR
RVV
VVAV
if
iO
if
fi
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ii
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fi
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fi
fi
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lim
11
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ii
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Inverting OpAmp: Virtual ground principle
0
V
VVVVAVO
ii
fO V
R
RV
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f
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Since no current is flowing intothe OpAmp:
What is the input resistance for the source ?
iV
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Capacitor equation
Charge across Capacitor (Coulomb)Capacitance value (Fahrad (German for bicycle))Voltage across Capacitor (Volt)Current through Capacitor (Ampere)
dt
dQti
tVCtQ
CC
cC
*
ti
tV
C
tQ
C
c
C
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Integrating Inverting OpAmp : Virtual ground
dttVRC
tV iO
1
0
V
VVVVAVO
dttVRC
tVdt
tdVRCtV
R
tV
dt
tdVC
dt
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tVti
iOO
i
iOCC
OCi
C
1
,
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Differentiating Inverting OpAmp: Virtual ground
0
V
VVVVAVO
dt
tdVRC
dt
tdVRCtV
R
tV
dt
tdVC
dt
tdQti
R
tVti
iCO
OCCC
OC
dt
tdVRCtV i
O
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A buffer gives an near infinite input resistance and a near zero output resistance
O
O IN
V A V V V V
V V
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An instrumentation amplifier has two high impedance (resistance) inputs
OpAmps have a very high input impedance (resistance)This configuration has superb Common Mode Rejection Ratio (CMRR) up to 70 dB
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A simple way to attenuate a signal is by using a voltage divider
2
1 2O i
RV V
R R
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Decibel notation
Addition is much simpler than multiplication
Notation in Bel (after Alexander Graham Bell)
For Power
For Voltages (Power ~ Voltage2)
In deciBel (0.1 Bel)
Belin log10 P
log*2 log 10210 UU
(dB) deciBelin log*20Belin log*2 1010 UU
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Common filters arrangements are low-pass, high-pass, band-pass and band-stop (notch)
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Butterworth filters are smooth, but have a high roll-on roll-off factor.
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Chebyshev filters have sharp roll offs but lots of ripple
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Bessel filters are tame (no ripples) but a gradual roll off
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Active filters combine amplification and filtering in one circuit!
What is the input impedance the source ‘sees’?
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Active Low-Pass filter analysis (1st order)
1 1
22
2 22
2
1*
1\ \
1 1
Z R
RRj C
Z Rj C j CRR
j C
20
1
O
i
V RG
V R 2
1
O
i
V ZG
V Z
2 20
1 1
1 1
1 1
Z RG G
Z R j j
Without C In general with impedances
0
1
1dB dB
dB
G Gj
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Notation differences
Wheeler / Ganji System
frequency in Hz
Corner frequency
Grift System
frequency in rad/s
Corner frequency rad/s
Time constant (s)
2 f
cyclef Hz
s
RC
1
1G j
j
11 C
1
1 2G j
j f RC
1
2Cf RC
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Active High-Pass filter with OpAmp and Capacitor / Resistor pair (1st order)
What is the input impedance the source ‘feels’?
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Active High-Pass filter analysis
1 1
2 2
1Z R
j C
Z R
20
1
O
i
V RG
V R 2
1
O
i
V ZG
V Z
2 2 2 20 0
1 1 11
1 1 1
Z R j R C jG G G
Z j RC jRj C
Without C In general with impedances
0 21
1
1dB dB dBdB
G G jj
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High-pass and low pass section separated by OpAmp
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Bandwidth and Distortion
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ABE425 Engineering Measurement ABE425 Engineering Measurement SystemsSystems
Measurement Systems with Electrical Signals
The End
Dept. of Agricultural & Biological EngineeringUniversity of Illinois