Download MECH 373 Instrumentation and Measurements Lecture 4

MECH 373
Instrumentation and Measurements
Lecture 4
(Course Website: Access from your “My Concordia” portal)
Measurement Systems with Electrical Signals
(Chapter 3)
• Electrical signal measurement systems
• Signal conditioners
Amplification
Attenuation
Filtering
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Lecture Notes on MECH 373 – Instrumentation and Measurements
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Components of Measurement Systems
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Lecture Notes on MECH 373 – Instrumentation and Measurements
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Electrical Signal Measurement Systems
• Measuring systems that use electrical signals to transmit information
between components have substantial advantages over completely
mechanical systems.
• Almost all modern engineering measurements can be made using sensing
devices that have an electrical output.
• In such devices, the measurand causes a change in an electrical property of
the device (e.g. resistance, capacitance or voltage), either directly or
indirectly.
• Electrical output sensing devices have several significant advantages over
mechanical devices:
1. Ease of transmitting the signal from measurement point to the data
collection point
2. Ease of amplifying, filtering, or otherwise modifying the signal
3. Ease of recording the signal
• However, completely mechanical devices are sometimes still the most
appropriate measuring systems.
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Signal Conditioning
There are many possible functions in the signal-conditioning
stage. Some of the common functions are:
• Amplification
• Attenuation
• Filtering
• Differentiation
• Integration
• Linearization
• Combining a measured signal with a reference signal
• Converting a resistance to a voltage signal
• Converting a current signal to a voltage
• Converting a voltage signal to a current signal
• Converting a frequency signal to a voltage signal
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Why Need Signal Conditioning?
• Large amplification for
small signals
• Good transient
response (i.e. small time
constants)
These are difficult to do
with purely mechanical
elements - due to friction
and inertia!
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General Characteristics of Signal Amplification
• Signals in the millivolt range are common, and in some cases,
signals are in microvolt range.
• It is difficult to transmit such signals over wires of great length,
and many processing systems require input voltage on the order of
1 to 10 V.
• The amplification of such signals can be increased using a device
called an amplifier.
• The low-voltage signal, Vi, is amplified to a higher voltage, Vo.
• The degree of amplification is specified by a parameter called the
gain, G.
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General Characteristics of Signal Amplification
• Common instrumentation amplifiers usually have values of gain in the
range 1 to 1000; however, higher gains can readily be achieved.
• The term gain is often used even for devices that attenuate a voltage
(i.e. Vo < Vi).
• Hence, values of gain can be less than unity.
• Gain is more commonly stated using a logarithmic scale, and the result
is expressed in decibels (dB). For voltage gain, it is expressed as:
GdB  20 log 10 G  20 log 10
Vo
Vi
• For example, an amplifier with a gain (G) of 10 would have a decibel
gain (GdB) of 20 dB, and an amplifier with a gain of 1000 would have a
decibel gain of 60 dB.
• If a signal is attenuated, that is, Vo is less than Vi, the decibel gain will
have a negative value.
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General Characteristics of Signal Amplification
• Although increase in signal amplitude is the primary purpose of an
amplifier, an amplifier can affect the signal. For example, frequency
distortion, phase distortion, etc.
• Typically a signal contains a range of frequencies. However, most
amplifiers do not have the same value of gain for all frequencies.
• For example, an amplifier might have a gain of 20 dB at 10 kHz and a
gain of only 5 dB at 100 kHz.
• Frequency response of a typical amplifier is shown in the following
figure. In the figure, the decibel (dB) gain is plotted versus the logarithm
of the frequency.
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General Characteristics of Signal Amplification
• Typically, the gain has a relatively constant value over a wide range of
frequencies.
• However, at extreme frequencies, the gain is reduced (attenuated).
• The range of frequencies over which the gain is almost constant is
called the bandwidth.
• The upper and lower frequencies defining the bandwidth are called
corner or cutoff frequencies. The cutoff frequencies are defined as
frequencies where the gain is reduced by 3 dB.
An amplifier with a narrow bandwidth changes the shape of an input
time-varying signal by an effect known as the frequency distortion.
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General Characteristics of Signal Amplification
• Although the gain of an amplifier is relatively constant over the
bandwidth, another characteristic of the output signal called the phase
angle may change significantly.
• If the voltage input signal to the amplifier is in the form of a sine wave
and expressed as Vi(t)= Vmi sin (2πft)
where, f is the frequency and Vmi is the amplitude of the input sine wave.
• The output signal will be
Vo(t) = GVmi( 2πft +φ )
where, φ is called the phase angle.
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General Characteristics of Signal Amplification
• The figures of amplitude response and phase response
together are called the Bode diagram or Bode plot.
• For pure sinusoidal waveforms, the phase shift is usually
not a problem. However, for complicated periodic
waveforms, it may result in a problem called phase
distortion.
• If the phase angle varies with frequency, the amplifier can
distort the shape of the waveform.
• However, if the phase angle varies linearly with frequency,
the shape of the waveform will not be distorted and the
waveform will only be delayed or advanced in time.
• But if the phase angle varies nonlinearly with frequency,
the shape of the waveform gets distorted.
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General Characteristics of Signal Amplification
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Input Loading and Output Loading
• Input loading and output loading are potential problems
that can occur when using an amplifier (and when using
many other signal-conditioning devices).
• The input voltage to an amplifier is generated by an
input or source device such as a sensor or another signal
conditioning device.
• If the output voltage of the source device is altered
when it is connected to the amplifier, there exists a
loading problem.
• A similar problem occurs when the output of the
amplifier is connected to another device, i.e. the amplifier
output voltage is changed.
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Input Loading and Output Loading
• Consider a source and an amplifier separately without being connected
• Now consider the combined system where the input source,
amplifier and the output load are connected together
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Input Loading and Output Loading
• If the source is not connected to the amplifier, the voltage at the
source output terminals will be Vs. This is because there is no
current flowing through Rs and consequently, there will be no voltage
drop across Rs.
• When the source is connected to the amplifier, the voltage at the
source output terminals will no longer be Vs. As shown in Figure 3.9,
Vs, Rs and Ri form a complete circuit. Consequently, there will be a
current flowing through Rs and a resulting voltage drop across Rs.
That is, the amplifier has placed a load on the source device.
• Similar behavior is observed when the output of the amplifier is
connected to a device.
• To minimize the loading effects at the input and output, an ideal
amplifier (or other signal conditioner) should have a very high value
of input resistance (Ri) and a very low value of the output resistance
(Ro). This can be seen from next slide.
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Input Loading and Output Loading
• To analyze this circuit, we will first solve the amplifier input voltage Vi in
terms of the source voltage VS . The current through the input loop is
Vs /( Rs  Ri ) and hence, V is given by:
i
RiVs
Vi 
Rs  Ri
• Similarly, the voltage of the output loop,
VL 
VL is given by:
RL GVi
Ro  RL
• Substituting first equation into the second equation, we get
Ri
RL
VL 
G
Vs
Ro  RL Ri  RS
• If RL  Ro and Ri  RS, the above equation will be approximated as
VL  GVs
• This is the equation of an ideal amplifier, that is, no loading effects.
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Example – output loading
• Consider Voltage Divide Circuit … if we stick a “VERY
HIGH” impedance meter between terminals … no current
flows from a to b and we can write ….
Rb
a
Vab  I  Ra
I
I
+
Vs
Ra
Vab
-
Vs
I
(Ra  Rb )
Ra
Vab  Vs 
(Ra  Rb )
b
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Example (contd)
• What happens if meter impedance is not “Very High”?
• Current Flows from a to b … circuit is “loaded”
Rb
a
• Voltage drop from a to b
Ib
Vab  I a Ra  Vab  RM I M
Ia
+
Vs
-
Ra
VRabM
IM = 0
b
Lecture 4
Ra
I a Ra  RM I M  I M  I a
RM
• Current splits thru Ra and RM

Ra 
Ib  I a  I M  I a  1 
RM 

Lecture Notes on MECH 373 – Instrumentation and Measurements
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Example (contd)
• Voltage drop through Ra and Rb

R 
a Vs  Rb I b  Ra I a  I b  I a  I m  I a 1  a 

RM 
Rb
Ib
 


Ra 
Ra 
Vs  Rb I a  1 
 Ra I a  I a  Rb  1 
  Ra 
RM 
R


M


Ia
+
Vs
-
Ra
VRabM
IM = 0
b
Lecture 4
• Solve for Ia
Ia 
Vs
 

Ra 
 Ra 
 Rb  1 

RM 
 

• Solve for current thru meter
Ra
Ra
Vs
Ra
I M  Ia

 Vs
RM RM  
Rb RM  Ra  Ra RM

Ra 
 Ra 
 Rb  1 

RM 
 

19
Lecture Notes on MECH 373 – Instrumentation and Measurements
Example (contd)
• Example Calculation …
Rb
a
Ib
10M
Ia
+
Vs
Ra
VRabM
-
=
Ra = 1000 
Rb = 1500 
Vs = 15 V
IM 
Ra
Vs
Rb RM  Ra  Ra RM
15  1000 
6
6
 1500   10 10 + 1000  + 10 10 1000
b
= 5.9996 10-7 amps
OK … almost no
current
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Example (contd)
• Example Calculation …
Rb
Ra = 1000 
Rb = 1500 
Vs = 15 V
a
Ib
10M
Ia
+
Vs
Ra
• Calculate Vab
VRabM
Vab  RM I M
-
=
15  1000 
6
6
6
 1500   10 10 + 1000  + 10 10 1000
10  10 
b
= 5.9996 volts
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Lecture Notes on MECH 373 – Instrumentation and Measurements
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Example (contd)
• Example Calculation …
Rb
Ra = 1000 
Rb = 1500 
Vs = 15 V
a
Ib
Ia
+
Vs
Ra
VRabM

-
• What happens if
Meter has “infinite
impedance” IM=0
Ra
Vab  Vs 
(Ra  Rb )
b
 15  
1000
1500 + 1000
=
 = 6 volts

• meter loading “insignificant”
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Lecture Notes on MECH 373 – Instrumentation and Measurements
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Example (contd)
• Example Calculation …
Rb
Ra = 1000 
Rb = 1500 
Vs = 15 V
a
Ib
Vab  RM I M
Ia
+
Vs
Ra
VRabM
-
How about
If …
RM = 10 k
=
15  1000 
3
3
 1500   10 10 + 1000  + 10 10 1000
b
10 10
3
= 5.66 volts
• meter loading causes significant
Error in voltage reading!
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