Seventh Lecture

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Seventh Lecture
Instrumentation and Product Testing

• Error Analysis

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1. Basic Concept
Repeat measuring the same physical quality a
number of time, we shall obtain the following
Systematic error Shift in the mean value Random
error Standard deviation
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2. Random errors   They are represented by
repeatability described in earlier lectures.
  Repeatability (R) is numerically equal to the
half range random uncertainty (Ur) of the
measurement.
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To estimate the population standard deviation
from the sample deviation, say, n 10 or n 20,
we shall use the Students t distribution. The
repeatability may be expressed as
Repeatability, R Z? ? t s
where the value of t can be found from the t
distribution table based on the sample size n and
the confidence level a,     s is usually
referred to as the sample standard deviation.
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Example. A balance is calibrated against a
standard mass of 250.0g. The difference in grams
from the 250.0g scale reading were determined
as 10.5, -8.5, 9.5, 9.0, -10.0, 9.0, -9.5,
10.0, 10.0 and -8.0 Find the repeatability R
of the balance at 95 confidence level.
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You have to use a Students t distribution table
to determine the repeatability. Probability
(1 - ?)/2 0.025, Degree of freedom (n - 1)
9. t 2.262 Apply the formula,
s 9.66
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The repeatability, R t s 21.85g. By
calculating the mean of the differences (2.2g),
it is noted that the mean value lies at 250.0
2.2 252.2g A positive bias (systematic error)
of 2.2g, which should be adjusted to zero.
Alternatively, the calculation can be conducted
by using MS Excel.
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We can use MicroSoft Excel to determine
repeatability
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Quite often, the repeatability of an instrument
varies from time to time by a considerable
amount. This does not necessarily indicate that
the instrument is faulty but rather that
repeatability is a somewhat variable
quantity. Some authorities advocate that three
repeatability tests be carried out on three
similar but not identical specimens in quick
succession. If the ratio between the highest and
lowest value is not greater than 21, then the
root mean square value of the three results
should be regarded as the repeatability of the
instrument. If the ratio obtained is greater
than 2, then the instrument should be examined
for faults, and on rectification further tests
should be made.
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Example. Three repeatability tests were carried
out on the balance introduced in last example.
The results obtained were as follows R1 22g,
R2 24g, and R3 28g Find the repeatability of
the balance. Solution R3/ R1 28/22 1.27 lt 2
Rounding up, the repeatability, Rr.m.s. 25
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Random errors can largely be eliminated by
calculating the mean of the measurements, since
in statistical analysis of data   where
is the standard deviation of the sample mean, ?
is the standard deviation of the population, and
n is the sample size (i.e. number of repeated
measurements for the same measurand). The
standard error of the mean is usually expressed
by .
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For example, a set of length measurements with n
20, 18.18mm, and ? 0.318mm,     The
length can be expressed as   18.18 ? 0.07 mm (for
68 confident limit) 18.18 ? 0.14 mm (for 95
confident limit)   It should be noted that the
sample standard deviation s is used as an
estimator of the standard deviation of the
population ? in some textbooks for the
calculation of .
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3. Error reduction using intelligent
instruments   It also includes       a
microcomputer, and     one or more transducers
(secondary transducers)   The secondary
transducers monitor the environmental conditions
(modifying inputs). By reading the outputs of
the primary transducer and the secondary
transducers, the microcomputer processes the
signals based on a pre-loaded programme.   It
can be programmed to take a succession of
measurements of a quantity within a short period
of time (sampling frequency)-1 and perform
statistical calculations on the readings before
displaying an output measurement. This is valid
for reducing random errors.
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The ability of intelligent instruments to reduce
systematic errors requires the following
pre-conditions be satisfied

• The physical mechanism by which a measurement
  transducer is affected by ambient condition
  changes must be fully understood and all physical
  quantities which affect the transducer output
  must be identified.
• The effect of each ambient variable on the output
  characteristic of the measurement transducer must
  be quantified.
• Suitable secondary transducers for monitoring the
  value of all relevant ambient variables must be
  available for input to the intelligent
  instrument.

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4. Calculations of accuracy and errors   4.1 
Accuracy of a measuring instrument
The accuracy of a measurement system is a
function of its ability to indicate the true
value of the measured quantity under specific
conditions of use and at a defined level of
confidence.   The accuracy A is expressed
by     where R is the instrument repeatability
and Us is the systematic error.
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Accuracy
Random Error (? Z?)
Systematic Error
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Example. The systematic error of a balance is
estimated to be ?5g and the random error of its
measurements is ?25g. State the repeatability and
calculate the accuracy of the instrument.   Soluti
on
Repeatability, R Ur 25g Systematic error, Us
5g  
As the systematic error and the repeatability
are both stated in grams, the accuracy of the
instrument is 26g. It is usual to express the
accuracy in terms of its full scale deflection
(f.s.d.). Accuracy f.s.d.
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4.2 Estimation of total error
The sum of the systematic and random errors of a
typical measurement, under conditions of use and
at a defined level of confidence.   Error
assessment must also include the error of the
calibrator itself and take account of the
confidence level upon which it is
founded.   Example. The diameter of the setting
gauge used to a sensitive comparator was stated
on its calibration certificate to be 60.0072mm
and to have an accuracy of determination equal to
?0.0008mm. Although not stated on the
certificate, the level of confidence was known to
be better than 95. The sample standard
deviation of 10 instrument readings yielded a
value s 0.37?m. Estimate the error of the
measurement at better than 95 confidence level.
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Solution From Students t distribution, at 95
confidence level, and sample size n 10, the
value of t is 2.26.   Ur R t?s 2.26 ? 0.37
0.8338?m  
The half range error (uncertainty) of the
setting gauge, Ul 0.8?m. Therefore, the half
range error of the measurement     This result
must be rounded up to the first decimal place as
this is the order of the observations. The error
of measurement is ?1.2?m. This estimate is given
with better than 95 confidence level owing to
the effect of rounding up.
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4.3 Compound error/uncertainty   In many
instances, the ultimate error (uncertainty) of a
measurement is dependent upon the errors of a
number of contributory measurements which are
combined to determine the final quantity. The
measurement output MM(x1, x2, x3,)   is a
function of a number of individual measurements
x1, x2, x3, etc. All of these measurements have
individual error of Dx1, Dx2, Dx3, etc. Then,
the compound error of the measurement M, DM, can
be determined by substituting into the equation
of MM(x1, x2, x3,) the maximum and minimum
values of x1, x2, x3, etc., and thus finding the
maximum and minimum values for M.
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This would obviously be a laborious process, and
the problem is better solved using partial
differentiation       Attentions must be
paid to those high value terms that
give dominant contributions to DM.  
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Example. Consider the measurement of electric
power from   P EI   where E is the e.m.f. and I
is the current. They are measured as   E 100V
? 1V I 10A ? 0.1A  
The nominal value of the power is   P 100 ? 10
1000W  
By differentiating the equation P EI, it can
be obtained that the error (compound) of
calculating the power
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  By taking the worst possible errors in voltage
and current, DE ?1V and DI ?0.1A,    
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Alternatively,   Pmax (100 1) (10 0.1)
1020.1W   Pmin (100 1) (10 0.1)
980.1W     In the above example, it is quite
unlikely that the power would be in error by
these amounts because the chance of obtaining the
largest errors for both voltage and current
measurements at the same is very slim. It is
especially true when ??x is used to specify the
tolerance of measurement.
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A more accurate method of estimating error has
been derived as follows   If this relation is
applied to the electric power relation of the
previous example, the expected error
is       The expected error (or uncertainty)
is ?1.4 instead of ?2.0.
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Example. The resistance of a certain size of
copper wire is given as where R0 6? ? 0.3
is the resistance at 20?C ? 0.004?C-1 ? 1 is
the temperature coefficient of resistance T 30
? 1?C is the temperature of the wire   Calculate
the resistance of the wire and its
uncertainty.   Solution   The nominal resistance
is  
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To determine the uncertainty, the partial
differentiation is performed    The
uncertainty of each component is  
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Thus, the uncertainty in the resistance
is     It can be noted that the two
components and predominate the
value of DR. For accuracy improvement, these two
terms must be of main concern.
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Thank you