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Chapter III – Measurement systems with electrical signals
- Signal attenuation
It happens that the output voltage of one component in the system is very high compared
to the voltage limit of the next component. In this case, attenuation of the signal is
required. For this purpose, a voltage-dividing network can be used.
Figure 3.10. Attenuation using dividing network.
Vo  Vi
R2
R1  R2
However, such network can lead to input and/or output loading problems.
Example (textbook p. 49)
The voltage used to power a heater in an experiment is nominally 120 V. To record this
voltage, it must first be attenuated using a voltage divider. The attenuator reduces the
voltage by a factor of 15 and the sum of the resistors R1 and R2 is 1000 .
a- Find R1 and R2 and the ideal voltage output (neglecting loading effects).
b- If the source resistance Rs is 1 , find the actual divider output Vo and the
resulting loading error in Vo.
c- If the divider output is connected to a recorder that has an input impedance of
5000 , what will be the voltage output (input to the recorder) and the resulting
loading error.
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Chapter III – Measurement systems with electrical signals
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Chapter III – Measurement systems with electrical signals
- Signal filtering
A real signal includes several frequencies. However, some of them are simply noise and
others are non-physical frequencies (i.e. frequencies that normally do not exist in the
physical phenomena; Exp: under normal conditions, the cardiovascular system does not
include frequencies higher than 12 Hz).
It is important, therefore, to try to remove undesirable frequencies: this is performed
through filtering.
Filtering
Spurious noise
suppression
Aliasing
suppression
- Categories of filters
Figure 3.11. Different categories of filters (a) lowpass; (b) highpass;(c) bandpass; (d) bandstop.
- Class of filters
- Butterworth
- Chebyshev
- Elliptic
- Bessel
- Butterworth filters
The main characteristic of Butterworth filters is that they maintain the gain maximally
constant in the passband.
For low passband Butterworth filters, the gain as a function of frequency can be written
as:
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Chapter III – Measurement systems with electrical signals
1
G
 f 
1   
 fc 
2n
fc is the corner frequency
n is the order of the filter.
Note:
f
 1
if
fc
For n=1
n
 f 
G   
 fc 
 f double  G halve
In general the decrease in the stopband is 6n dB/octave.
Figure 3.12. Butterworth filter.
Higher order Butterworth filters do not have a large change in slope after fc. To solve this
problem, a Chebychev filter can be used.
- Chebychev filter
Chebychev filters have a large change in slope after fc, however, the price to pay to
achieve this characteristic is a less constant gain in the passband region.
G
1
 f 
1   2Tn2  
 fc 
 is the ripple factor.
Tn is a Chebyshev polynomial of order n.
The transition between pass and stop bands can be improved even better with an elliptic
filter. However, this filter induces ripples in both pass and stop band regions.
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Chapter III – Measurement systems with electrical signals
Figure 3.13. Chebychev filter.
Introducing a filter in the system does not only affect the gain, it also affects the phase,
introducing sometimes a very significant phase-shift. This phase shift depends also on the
order of the filter.
A very efficient class of filters in limiting phase shift is Bessel filters (linear relationship
between phase angle and f/fc). However, the change in slope after fc is less steep.
Figure 3.14. Bessel filter. (left) phase angle; (right) gain.
- Selecting a filter
In order to get the appropriate filter, you will have to specify:
-
The type: lowpass, highpass, bandpass, …
The class: Butterworth, Chebychev, …
The order: n
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Chapter III – Measurement systems with electrical signals
-
The corner frequency (ies)
For Chebychev and elliptic filters other parameters describing the ripple must be given.
- Building a Butterworth filter using op-amp
Op-amp can be used to construct active filters. The use of op-amp avoids using inductors
(not ideal components).
Figure 3.15. Lowpass Butterworth using op-amp. (left) sketch; (right) frequency response.
- Determination of the frequency response of a lowpass Butterworth filter using opamp:
We will feed the circuit with a single frequency sine wave signal ( Vmi e j 2ft ). Where Vmi is
the amplitude of the signal and f is its frequency.
Since we are in the frequency domain, the gain will be defined using impedance instead
of resistance:
Z
G 2
Z1
Z1
consist of R1
Z2
consist of R2 and C in //:
R2
1  j 2fCR2
1
R2
R1 1  j 2fCR2
R
Let us put: G0   2 , representing the gain at low frequency for the simple inverting opR1
amp (remove C), then:
Then, G  
G  G0
And
1
V
 o
1  j 2fCR2 Vi
G
1
1


2
G0 1  j 2fCR2
1  2fCR2 
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Chapter III – Measurement systems with electrical signals
To get the corner frequency of this filter, we need to remember that fc is defined as the
frequency where the gain decreased by 3 dB (or a fraction of 0.707).
G 0.707G0
1


2
G0
G0
1  2f cCR2 
1
Giving: f c 
2CR2
Knowing the cutoff frequency, we can determine the roll-off rate for frequencies higher
then fc.
At higher frequencies 2f cCR2 >> 1; then:
Then;
G
1

G0 2f cCR2
Now if the frequency is doubled (1 octave) the ratio
G
is divided by 2 which
G0
correspond to -6 dB.
- Phase shift
To get phase shift induced by the lowpass Butterworth filter using an op-amp, we have to
write the gain in terms of fc:
1
G  G0
 f 
1  j  
 fc 
 f 

 fc 
   tan 1 
In the same way;
1st order highpass filter:
fc 
1
2R1C
Figure 3.16. Highpass Butterworth using op-amp. (left) sketch; (right) frequency response.
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Chapter III – Measurement systems with electrical signals
1st order bandpass filter:
f c1 
1
2R1C1
fc2 
1
2R2C2
Figure 3.17. Bandpass Butterworth using op-amp. (left) sketch; (right) frequency response.
NOTE: low order filters suffer from insufficient attenuation. To construct higher order
filters, a cascade of lower order filters can be used.
Example
A transducer measures a sinusoidal signal with an amplitude of 5 V and a frequency up
to 10 Hz. Superimposed on this signal is 60 Hz noise with an amplitude of 0.1 V. It is
desired to attenuate the 60 Hz signal to less than 10% of its value using a Butterworth
filter. Select a filter order to perform this task if the corner frequency is 10 Hz.
- Circuits for integration differentiation and comparison
t
1
Vi (t )dt  V0 (0)
RC 0
dV t 
Vo (t )   RC i
dt
Vo (t )  
Figure 3.18. (left) integrator; (right) differentiator.
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