1st year lab instrumentation

7/31/2019 1st Year Lab Instrumentation

1/22

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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University College London: Department of Electronic & Electrical Engineering

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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1st year lab instrumentation

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