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Transcript of 1st Year Lab Instrumentation
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7/31/2019 1st Year Lab Instrumentation
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Department of Electronic & Electrical Engineering
Experiment L
INSTRUMENTATION AND DATA ANALYSIS
1. INTRODUCTION
Aims
This experiment concerns a practical demonstration of the synthesis of anelectronic instrument and the analysis of measurements made with the instrument.
The instrument responds to the strain in a bending beam.The key sub-units of the demonstration instrumentation system are a sensor bridge
circuit and a differential amplifier. The sensing element in the bridge is a strain
gauge, but the amplifier and bridge circuits are very versatile and can readilyaccommodate different types of resistive sensor with little alteration.
The chapter provides a set of sample results for the calibration of the amplifier andmeasurements from the bending beam. The sample results lead to a calculation of
a value of the modulus of elasticity for the steel beam and an estimate of the errorin that calculation.
The objectives of the experiment are:
! To highlight considerations in precision amplifier design and in thenumerical treatment of measurements.
! To provide a practical demonstration of a measuring instrument.! To collect and analyse data from a measuring instrument.
! To outline the steps in the analysis of experimental results.! To conduct an analysis of measurement error.! To determine the modulus of elasticity of a steel beam and an estimate of
the error.
Resistance codes
The BS 1852 resistance codes are used in this laboratory sheet to indicatecomponent values on an electronics diagram. Their purpose is to ensure that the
positions of decimal points are clearly marked. The position of the decimal point
is marked by an upper case symbol R, K or M, where R indicates a value in ", Kindicates a value in K" and M indicates a value in M".
The table below illustrates the use of the code through a series of examples:
0.47" R47 100" 100R1" 1R0 1K" 1K04.7" 4R7 10K" 10K47" 47R 10M" 10M
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Description Specification Values
LM301 Bipolar operational amplifiermilitary/aerospace precision with low drift
and high input resistance.
Input offset voltageDrift
Input resistanceCMRCost
2mV
0.6#V/K
2M"90dB0.82
OPA177GP Precision bipolar operational
amplifier, very low offset voltage and low
drift. High performance and low cost suit itto precision instrumentation amplifiers,
bridge, thermal couple amplifiers.
Input offset voltage
Drift
Input resistanceCMR
Cost
4#V0.1#V/K45M"140dB
1.30
OP27GNB Instrumentation grade
operational amplifier with very low noiseand ultra-low offset voltage.
Input offset voltage
Input resistanceCMR
Input noise
Cost
30#V4M"120dB3.8nV/!Hz2.85
LF356 high precision operation amplifierJFET input with offset null.
Input offset voltageInput resistanceInput noise
CMRCost
3mV
1012"12nV/!Hz100dB0.77
1NA118P Precision instrumentationamplifier using 3 op-amp design with over
voltage protection.Applications for bridge amplifiers,
thermocouple amplifiers and medicalinstrumentation.
Input offset voltageDrift
Input impedanceGain range
Input noiseCost
25#V0.25#V/K1010"||5pF1 to 1000012nV/!Hz5.25
Table 1 Example of precision components for use in instrumentationamplifiers
2.2 Common-mode rejection stage
The Common-mode rejection circuit
Figure 2a shows the right hand side of the amplifier circuit in figure 1 whichprovides the common-mode rejection for the amplifier. This sub-circuit acts as adifferential amplifier with a differential gain of R1/R2, while removing any
common-mode voltage from the output. Its performance depends upon the close
matching of resistor values, Resistors marked R1 must be the same value, andthose marked R2 likewise.
Figure 2b shows a practical realisation of the circuit based on the LM301
operational amplifier. The tuning of the variable resistors gives a means of
ensuring the resistors can be correctly balanced once the circuit is built. Thefunction of the capacitor is for frequency compensation and explained shortly.
(a) (b)
Figure 2, The common-mode rejection sub-circuit and practical realisation ofthe circuit that includes trimming resistors and compensation capacitor.
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Practical Task - The differential gain
Figure 4 shows some sample results from a differential amplifier circuit on a
graph with the r.m.s. value of the input on the horizontal axis and the r.m.s. valuesof the output on the vertical axis. Your graph should look similar. The
uncertainty in the measurements of the output voltages was found to be 0.01V and
was limited by the resolution of the equipment. For the best straight linedetermined using the least-squires calculation the gradient, is m +"m=492.4+0.24.
Calibration of differential amplifier
Figure 4, Example calibration graph for a differential amplifier. The plotshows the r.m.s. values of output voltage and input voltages. The differential
gain is the gradient of the best fit straight line.
Practical Task - The common mode gain
The common mode gain does not enter the calculations for the modulus ofelasticity of the steel beam. An approximate figure suffices to establish that the
common mode rejection exceeds the performance of 80dB. For this task thepotential divider circuit of figure 3b is not used. The oscillator output from
function generator is applied directly to the inputs of the amplifier.
Apply a 200Hz sine wave signal with an amplitude of up to 10V to both inputs ofthe amplifier. Monitor input and output on the oscilloscope. The amplitude of the
common-mode output should be measured, or if it is too small to measure make a
note instead of an upper bound, a value which you are sure is larger than theamplitude of the actual voltage output. The common mode gain is the ratio of
output and input signal amplitudes:
GCM = Vo / Vin
What is the common mode gain for your instrument? Sample results for the
differential amplifier gave an output of less than 15mV for an input of 7V and thecommon-mode gain was therefore less than 0.00214. Your results should be of
similar magnitude. As with the differential gain, the d.c. common mode gain istaken to be the same as the common mode gain at 200Hz.
Calculation of common mode rejection
The common-mode rejection characteristics performance of the differential
amplifier. The quantity to be calculated is:
20 log10 (GDIFF / GCM)
Calculate the CMR for your amplifier. Sample results gave a common mode gain
less than 0.00214, while the differential gain was 4912.45. The common moderejection for the amplifier was at least 107dB. Your own amplifier should have a
CMR of around 80dB.
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Sensor bridge circuit
A strain gauge bridge circuit for use with the apparatuses of figure 5 is shown infigure 7a. It is a basic bridge circuit with a single sensing element, and is most
sensitive to changes in the strain gauge resistance when the fixed resistors in thecircuit have the same value as the unstrained resistance of the strain gauge (120").The set zero resistor provides a means of adjusting the bridge to account forcomponent-to-component variability; the bridge output should be zero when the
sensor is unstrained.The bridge circuit has a 5V d.c. source giving a current of about 20mA in each arm
of the bridge and power dissipation of 50mW in each resistor. Resistors rated at
0.33W are adequate for a 5V bridge circuit, but would need upgraded to 0.5W with a15V bridge supply.
The sensitivity of the bridge transducer circuit is given by the expression below inwhich the increment "R is the change in resistance of the sensor that creates a d.c.voltage "V between the two sides of the bridge:
$ %$ %
4"V/V"R/R '
V: d.c. supply voltage to the bridgeR: The resistance of the resistors in the bridge circuit
Alternative configurations for bridge circuits are shown in figures 7b and 7c. Their
advantages are discussed in chapter 10 of the recommended text.
Figure 7 The sensor bridge circuits. In (a) a bridge with a single sensingelement. (b) and (c) use additional elements to improve sensitivity.
Practical task - The bridge circuit
The circuit you have been given is shown in figure 7a, using the indicated resistorvalues and a 5V supply set up the bridge. Monitor the difference in voltage between
the 2 sides of the bridge using the mV scale on a digital voltmeter. A voltage of zero
must be achieved when the rule is in its unstrained (resting) position by trimming theset-zero resistor.
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Figure 8 Sample results for measurement of the modulus of
elasticity
Practical task - Other measurements
The analysis of the experimental data calls for some additional measurements givingdetails of the geometry the apparatus. The thickness of the steel rule, d, can be
measured using a hand-held micrometer (ask technician for this, then return it). Thedistance, l, from the pivot point of the rule to the centre of the hole that holds the pan
must be measured as must the width, w, In the region where the strain gauge islocated. You should measure these quantities for your own apparatus.
3. CALCULATION OF THE MODULUS OF ELASTICITY
3.1 Review of engineering formulae
Use of engineering formulae
Electronics engineers engaged in instrumentation need to use formulae and theoryfrom other branches of engineering. The theory underlying the calculation of the
modulus of elasticity, for example, is the static beam theory used by mechanical andstructural engineers, and the relevant expressions are published in the handbooks
used by those engineers.
It is good practice to review an expression taken from a handbook. A review ensuresthat all the terms in the expressions are understood and that the units of quantities
required in the calculation are consistent with the units of the measurements. Theintention of this sub-section is to outline the meanings of the relevant terms and to
inspect the validity of the available measurements for use in the engineeringformulae.
Review of static beam theory:
The modulus of elasticity relates the stress at the surface of a strained component to
the strain causing the stress:
e
#E '
E : modulus of elasticity of the material
*: stress
e : strain
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The geometric parameters of the steel beam determine the stress at the pivot point
resulting from the application of a force at a distance ! from the pivot:
FI
b#
!'
b : the half thickness of the beam, b = d/2
! : distance between pivot point and point where force is applied
F : the applied force
I : the second moment of the cross-sectional area
A combination of the two expressions above gives the relationship between the
strain and the force causing the deflection:
FEI
be
!'
Geometrical parameter, I, is a weighted integral of the cross sectional area at the
pivot with the weighting factor being the square of the distance from the centre line.The effect of the weighting is that parts of the beam far away from the centre line
contribute more to the integral than those near the centre and I therefore indicatesthe distribution of material in the beam:
Prove that for the steel rule with a rectangular cross section the expression for I isI = 2w b3 / 3.
Figure 9. A geometrical interpretation of I, the second moment of
the cross-sectional area.
Dimensional analysis
The following analysis itemises the units and dimensions of the terms to be used inthe engineering formula from the static beam theory. The aim is to ensure that all the
measurements are expressed consistently and that when used in the engineering
formula the formula are dimensionally correct.
The units of the quantities are:
F : Force in Newton, given by mass times the acceleration due to gravity, g
*+, Stress is a force per unit area, its unit is Pa, or N .m-2
e : Strain has no units. It is the dimensionless ratio )x/xoE : Modulus of elasticity, also measured in Pa
! , b , w , d : All these quantities are lengths in metres, m
I : The integral definition of I shows its units are m4
Write down the dimensions of the quantities in the static beam expressions in terms
of the fundamental dimensions of mass (M). length [L] and time [T]. The left handside of the expression for the strain, /EIbe !' is dimensionless. Prove that theright hand side is also dimensionless.
The conclusion from the dimensional analysis is that the measured quantities we
propose to substitute into the expression for the strain are correctly dimensioned andconsistent with the engineering formula.
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The random error in the derived quantity f(x, y) taking account of the averaging
effect mentioned above is:
2
2
y
2
2
xf dy
df
#dx
df
## //0
1
223
4
7/0
1
23
4
'
As an example, consider the derived value f(x, y) = xy. An expression for thestandard deviation as a fraction of the derived value is:
22
y
#y
x
#x
xy
#f
y)f(x,
#f//
0
1223
47/
0
123
4''
The above result for the bivariate case with random error is not of the same form as
the deterministic result.
Summary of results for propagation of errors:
Derived
quantity
Propagated error:
deterministic case
Propagated standard
deviation:random case
f(x) = x2
x
"x2
f
"f'
x
#2
f
#f x'
f(x) = x1/2
x
"x
2
1
f
"f'
x
#
2
1
f
#f x'
f(x) = x-1
x
"x
f
"f'
x
#
f
#f x'
f(x) = ln(x)x
"x
ln(x)1
f
"f ' x
#
ln(x)1
f#f
x'
f(x) = ex"x
f
"f' x
f
#f*'
f(x, y) = x . y
//0
1223
47/
0
123
4'y
"y
x
"x
f
"f
2
y
2
x
y
#
x
#
f
#f//
0
1223
47/
0
123
4'
f(x, y) = x / y
//0
1223
47/
0
123
4'y
"y
x
"x
f
"f
2
y
2
x
y
#
x
#
f
#f//
0
1223
47/
0
123
4'
f(x, y) = xp. yq
//01
22347/
012
34'
y
"yqx
"xpf
"f2
y
2
x
y
#q
x
#p
f
#f//
0
1223
47/
0
123
4'
Table Al. Expressions for fractional worst case uncertainties of
derived quantities and standard deviations and derived quantities
when the errors are random.
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APPENDIX B. STATISTICAL ANALYSIS OF
MEASUREMENTS
B.1 The correlation coefficient
Correlation
It is useful to determine a measure of correlation in order to answer the questionhow much of the variation in the y-axis values is caused by the spread in the x-axis
values? The correlation coefficient, r, is +l if the data lie exactly on a straight linewith a positive gradient because then each x-axis value predicts exactly the
corresponding y-axis value. A correlation coefficient of -1 applies to the situation
when the x-y data lie on a perfect straight line with a negative gradient.On the other hand if the y-axis values to bear no relation to the x-axis values the
value of r would be close to zero. That situation applies if the data are distributed
randomly on the x-y plane, intermediate values of r indicate that a part of thevariability in the y-axis values is predictable from the x-axis values, and a partrandom and nothing to do with the x-axis values. The first step in analysing
experimental data is often a correlation analysis because the correlation coefficient
indicates the strength of the relationship between two variables.For more details and deeper insight into the interpretation of the correlation
coefficient please see Chapter 9 of the recommended text.
Calculation of the correlation coefficient
Denoting the mean-centred and standardised variables by the symbols x and y, thecorrelation coefficient is given by the sum other products:
&'
'n
1i
ii 'y'xr
The steps in the calculation of r are outlined below. They can easily be implemented
in a package such as Excel or MATLAB.
Step1 : Find the sample means:
&''
n
li
ix
n
1x
&''
n
li
iy
n
1y
Step2 : Determine the standard deviations:
&'
9'n
li
2
ix )x(xn
1# &
'
9'n
li
2
iy )y(yn
1#
Step3 : Mean centre and standardise the data to unit standard deviation:
x
ii
#
xxx'
9'
y
ii
#
yyy'
9'
Step4 : The correlation coefficient r is calculated from the mean centred and
standardised values:
&'
'n
1i
ii 'y'xr
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B.2 Regression analysis
The concept of a best fit
Figure B1 shows a set of measurementsand two straight lines. The equation ofeach line gives a predicted value of y
denoted as yp and is of the form
yp = mx + c, but the parameters m and care different for the two lines. One of the
straight lines is clearly a much betterrepresentation of the trend in the data than
the other. The question to be addressed inthis section is the selection of the
combination of m and c that is best forthis particular set of data and its aim is to
determine mathematically the
parameters of the best line.
The sum of squares cost function
The least squares fitting procedure finds the values of m and c which minimise the
sum of the squares of the residuals in the fit to the experimental data. A residual isthe deviation between an experimental value and the straight line. An expression for
the ith residual for the measurement pair { x i, yi } is given by the differencebetween y and the prediction of yi from the straight line mx + c:
piii yye 9'
c)(mxy ii 79'
A measure of how well the line fits the measured data for a given m and c is the sumof the squares of the residuals for the n data points. This quantity is known as the
sum of squares cost function, J. It is written as J(m,c) to emphasise that its value
depends upon m and c:
&'
'n
li
2
iec)J(m,
The least squares method
The sum of squares cost function depends on m and c. Substituting the expression
for ei:
$ %$ %&'
79'n
li
2
ii cmxyJ
The above expression shows that J is a quadratic function of m and c. The analyticalprocedure for finding the minimum value for a quadratic polynomial function of two
variables is to set each partial derivative to zero. Thus:
0dm
dJ' 0
dc
dJ'
$ %$ % /0
123
4999'799' & & &&
' ' ''
n
li
n
li
n
li
i
2
iii
n
li
iii xcxmyx2cmxyx2dm
dJ
$ %$ % /0
123
4999'799' & &&
' ''
n
li
n
li
ii
n
li
ii cnxmy2cmxy2dc
dJ
And so:
0xcxmyxn
li
n
li
i
n
li
2iii '/
012
34 99& &&
' ''
0ncxmyn
ln
n
li
ii '/0
123
499& &
' '
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These two equations are known as the normal equations, and are simultaneous
equations for those special values of m and c which give the best straight line fit.The terms under the summation signs are numerical values that can easily be
calculated from the experimental measurements. The special values are indicated by
the notation m"
and c and the equations can be solved for m"
and c to give theresult:
/0
123
49/
0
123
4
/0
123
4/
0
123
49/
0
123
4
'
&&
&&&
''
'''
n
li
i
n
li
2
i
n
li
i
n
li
i
n
li
ii
xxn
yxyxn
m
/0
1
23
4
9/0
1
23
4
/0
123
4/
0
123
49/
0
123
4/
0
123
4
'
&&
&&&&
''
''''
n
lii
n
li
2
i
n
li
ii
n
li
i
n
li
i
n
li
2
i
xxn
yxxyx
c
You are recommended to use MATLAB or EXCEL to carry out the summationsrather than doing them by hand. You may find it useful to visualise a table layout as
follows:
force/N
xi
output/V
yi
xi2
xiyi
0.391
0.79
0.154
0.31
Column sums:xi :yi :xi
2 :xiyiTable B1. Layout of table for measurements of force and voltage
output from the bending beam instrument, together with derived
quantities needed for the least square calculations.
Matrix formulation of the least squares method
An equivalent formulation of the least squares equations uses a matrix-vectormethod. The matrix formulation is easy and convenient to use in MATLAB.
Consider a table in which the experimental measurements are laid out in columns.
The columns of the y-axis and x-axis measurements are regarded as vectors v and x.A vector of predicted y-axis value lying on a straight line y = mx + c would be given
by a matrix-vector equation:
//0
1223
4'
c
mAy
p
Such a matrix equation is a compact way of writing the element-by-element
expressions y1p = mx1 + c, Y2p = mx2 + c, and so on for the n measurements as
shown below:
//0
1223
4'//
0
1223
4
////////
0
1
22222222
3
4
'
////////
0
1
22222222
3
4
7
77
'
////////
0
1
22222222
3
4
'c
mA
c
m
1x
......
......
......
1x1x
cmx
...
...
...
cmxcmx
y
...
...
...
yy
y
n
2
1
n
2
1
np
2p
p1
p
Value for the parameters of the best straight line, m"
and , can be expressed in
terms of the matrix A and the vector of the measurements. The expression does not
need explicit evaluations of the summations over the experimental data becausethose summations occur automatically during the matrix multiplication and
inversion operations. The matrix-vector expression for the best values of m and c is:
c
$ % yAAAc
mT1T 9'//
0
1223
4
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the parameters of the best quadratic fit. The A matrix in this case would be the 3 x n
matrix with the columns of xi2, xi and unity terms:
$ % yAAAc
b
a
T
1
T
9
'///
0
1
222
3
4
Figure B2 shows the best fit quadratic model, which is clearly an improvement overthe best straight line for the data in the graph.
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