1. 1 Seventh Lecture Error Analysis Instrumentation and Product Testing

2. 2 1. Basic Concept Systematic error: Shift in the mean value Random error: Standard deviation Repeat measuring the same physical quality a number of time, we shall obtain the following:

3. 3 2. Random errors They are represented by repeatability described in earlier lectures. Repeatability (R) is numerically equal to the half range random uncertainty (U r ) of the measurement.

4. 4 To estimate the population standard deviation from the sample deviation, say, n = 10 or n = 20, we shall use the Student’s 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 , s is usually referred to as the sample standard deviation.

5. 5 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.

6. 6 You have to use a Student’s 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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8. 8 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.

9. 9 We can use Microsoft Excel to determine repeatability…

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11. 11 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 2:1, 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.

12. 12 Example. Three repeatability tests were carried out on the balance introduced in last example. The results obtained were as follows: R 1 = 22g, R 2 = 24g, and R 3 = 28g Find the repeatability of the balance. Solution: R 3 / R 1 = 28/22 = 1.27 < 2 Rounding up, the repeatability, R r.m.s. = 25

13. 13 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.

14. 14 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.

15. 15 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.

16. 16 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.

17. 17 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 U s is the systematic error. 4. Calculations of accuracy and errors 4.1 Accuracy of a measuring instrument

18. 18 Systematic Error Random Error (  Z  ) Accuracy

19. 19 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. Solution: Repeatability, R = U r = 25g Systematic error, U s = 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.

20. 20 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. 4.2 Estimation of total error

21. 21 Solution: From Student’s t distribution, at 95% confidence level, and sample size n = 10, the value of t is 2.26. U r = R = t  s = 2.26  0.37 = 0.8338  m The half range error (uncertainty) of the setting gauge, U l = 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.

22. 22 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 M=M(x 1, x 2, x 3,…) is a function of a number of individual measurements x 1, x 2, x 3, etc. All of these measurements have individual error of  x 1,  x 2,  x 3, etc. Then, the compound error of the measurement M,  M, can be determined by substituting into the equation of M=M(x 1, x 2, x 3,…) the maximum and minimum values of x 1, x 2, x 3, etc., and thus finding the maximum and minimum values for M.

23. 23 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  M.

24. 24 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

25. 25 By taking the worst possible errors in voltage and current,  E =  1V and  I =  0.1A,

26. 26 Alternatively, P max = (100 + 1) (10 + 0.1) = 1020.1W P min = (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.

27. 27 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%.

28. 28 Example. The resistance of a certain size of copper wire is given as where R 0 = 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

29. 29 To determine the uncertainty, the partial differentiation is performed: The uncertainty of each component is:

30. 30 Thus, the uncertainty in the resistance is: It can be noted that the two components and predominate the value of  R. For accuracy improvement, these two terms must be of main concern.

31. 31 Thank you