Download Introduction to Instructor Tony Grift, PhD

ABE425 Engineering Measurement
Systems
Measurement Systems with
Electrical Signals
Dr. Tony E. Grift
Dept. of Agricultural & Biological Engineering
University of Illinois
Agenda
1.
2.
3.
4.
5.
AC and DC signals
Transducers
OpAmps
Active Filters
Loading error examples
AC and DC signals
Alternating Current (AC)
Direct Current (DC)
Most signals have both!
DC component (offset) measurement: Put DMM on DC
AC component measurement: Put DMM on AC
Scope gives Amplitude and peak-peak value
To get the scope trace at 0: Put input on gnd.
How is the scope amplitude related to the AC value on the
DMM?
Let’s figure out what the AC RMS value is of this
signal
T  1 ms 
1
f   1, 000  Hz 
T
 rad 
  2* pi * f 
 s 
1
0.8
0.6
0.4
0.2
0
-0.2
-0.4
-0.6
-0.8
-1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
-3
x 10
RMS value represents Power
DC: 12 Volt, R = 100 Ω
2
P DC
1
0.8
0.6
2
U
12


 1.44 Watt 
R 100
AC RMS value must give
same power as DC of the same
value
0.4
0.2
0
-0.2
-0.4
-0.6
-0.8
-1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
-3
U2
2
Mean value
2
U
U

2
2
Root Mean Square
x 10
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
2
P AC
2
U
12
 RMS 
 1.44 Watt 
R
10
0.2
0.1
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
-3
x 10
Question
If an AC Voltmeter shows 12V RMS
What is the amplitude of this signal?
U  U RMS * 2  12* 2  16.97V
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!
Digital TRUE RMS meters digitize the signal and
compute the RMS value from the definition
Digitize at least one cycle of the signal
Square it
Compute mean value
Integrate signal
Divide by cycle time
Take square root
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
-3
x 10
A Transducer converts a physical measurand into
an electric signal
Antenna
Cathode Ray Tube. LCD monitor
Fluorescent light, light bulb.
Light Emitting Diode (LEDs)
Magnetic stripe cards
Photocells/Light
Combined model of (a) input source, (b) amplifier
and (c) output load
You want to prevent loading errors
Choose Ri high
Choose Ro low
Non-inverting OpAmp
VO  AV   V  
V   Vi
V 
Ri
VO
Ri  R f
 Rf
VO  Vi 1 
Ri






Ri

VO  A Vi 
VO 


Ri  R f



Ri 

VO 1  A
 AVi


Ri  R f 

A
VO 
Vi


R
i
1  A



R

R
i
f


1
VO 
Vi
1

  Ri 
A R R 
i
f 

Ri  R f
lim
 Rf 

VO 
Vi  Vi 1 
A
Ri
Ri 

Non-inverting OpAmp : Virtual ‘ground’ principle
Since no current is flowing into
the OpAmp:


VO  A V   V   V   V 
Vi VO  Vi 

Ri
Rf
V   Vi
 Rf
VO  Vi 1 
Ri




What is the input resistance for
the source Vi ?
OpAmps have a limited band width
(741 is about 1 MHz)
Inverting OpAmp

VO  A V   V 

Rf
Ri
V   Vi
 VO
R f  Ri
R f  Ri
V 0
VO  
Rf
Ri
Vi


Rf
R
i

VO  AVi
 VO
 Ri  R f
Ri  R f 


Rf
Ri 

VO 1  A
 AVi


Ri  R f 
Ri  R f

Rf
Rf
AVi
Vi
Ri  R f
Ri  R f
VO 


 
1  A Ri   1  Ri

  A Ri  R f
R

R
i
f

 
Rf
lim
VO  
Vi
A
Ri




Inverting OpAmp: Virtual ground principle
Since no current is flowing into
the OpAmp:
i
VO  AV   V    V   V 
V 0
VO  
Rf
Ri
Vi
Vi
V
 O
Ri
Rf
Rf
VO

Vi
Ri
What is the input resistance for
the source Vi ?
Capacitor equation
QC t   C *Vc t 
iC t  
dQC
dt
QC t  Charge across Capacitor (Coulomb)
C
Capacitance value (Fahrad (German for bicycle))
Vc t  Voltage across Capacitor (Volt)
iC t 
Current through Capacitor (Ampere)
Integrating Inverting OpAmp : Virtual ground
VO  AV   V    V   V 
V 0
Vi t 
iC t  
, VC t   VO t 
R
dQC t 
dVO t  Vi t 
iC t   
 C

dt
dt
R
dV t 
1
Vi t    RC O  VO t   
Vi t dt

dt
RC
1
VO t   
Vi t dt

RC
Differentiating Inverting OpAmp: Virtual ground
VO  AV   V    V   V 
V 0
VO t 
iC t   
R
dQC t 
dVC t 
VO t 
iC t  
C

dt
dt
R
dVC t 
dVi t 
VO t    RC
  RC
dt
dt
dVi t 
VO t    RC
dt
A buffer gives an near infinite input resistance
and a near zero output resistance
This method can be used to prevent a loading error
Virtual ground principle:


VO  A V   V   V   V 
VO  VIN
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
A simple way to attenuate a signal is by using a
voltage divider
R2
VO  Vi
R1  R2
Decibel notation
Addition is much simpler than multiplication
Notation in Bel (after Alexander Graham Bell)
For Power
10
log P  in Bel
For Voltages (Power ~ Voltage2)
10
 
log U 2  2*10 log U 
In deciBel (0.1 Bel)
2*10 log U  in Bel  20*10 log U  in deciBel (dB)
Common filters arrangements are low-pass, highpass, band-pass and band-stop (notch)
Butterworth filters are smooth, but have a high
roll-on roll-off factor.
Chebyshev filters have sharp roll offs but lots of
ripple
Bessel filters are tame (no ripples) but a gradual
roll off
Active filters combine amplification and filtering
in one circuit!
What is the input impedance the source ‘sees’?
Active Low-Pass filter analysis (1st order)
VO
R
 G0   2
Vi
R1
Without C
G
VO
Z
 2
Vi
Z1
In general with impedances
Z1  R1
1
R2 *
R2
1
jC
Z 2  R2 \ \


1
jC R 
1  j CR2
2

jC
 1 
Z2
1 R2
G

 G0 

Z1
R1 1  j
1

j



G dB
 1 
 G0 dB  

1

j


 dB
Notation differences
Wheeler / Ganji
System
G  j  
1
1  j 2 f RC

frequency in Hz
Corner frequency
Grift
System
frequency in rad/s
 cycle 
f 
 Hz

 s 
1
fC 
2 RC
G  j  
1
1  j
  2 f
Corner frequency rad/s
  1  C 
Time constant (s)
  RC
1


Active High-Pass filter with OpAmp and Capacitor
/ Resistor pair (1st order)
What is the input impedance the source ‘feels’?
Active High-Pass filter analysis
VO
R2
 G0  
Vi
R1
VO
Z
 2
Vi
Z1
Without C In general with impedances
G
1
Z1  R1 
jC
Z 2  R2
G
Z2
R2
j R2C
j 2

 G0
 G0
1
Z1
1  j R1C
1  j 1
R1 
jC
G dB


1
 G0 dB   j 2  dB  

1

j

1  dB

High-pass and low pass section separated by OpAmp
Bandwidth and Distortion
ABE425 Engineering Measurement
Systems
Measurement Systems with
Electrical Signals
The End
Dept. of Agricultural & Biological Engineering
University of Illinois