CHAPTER TWO Measurement Errors

Objectives: Explain the various types of errors that occur in measurements. Explain and apply the following measurement terms: Accuracy, Precision, Resolution. Determine the resultant error for various calculations involving instrument and component error combinations. Use basic statistical methods for analyzing measurement error.

Ways of Expressing Errors The measurement errors are usually expressed in two ways. Absolute Errors Definition: it is the difference between the measured value (Xm) and the true value (X). ΔX = Xm-X

Ways of Expressing Errors Relative Errors Definition: is the ratio between the absolute error (ΔX) and the true value (X) δX = ± ΔX /X Percentage Relative Errors δX% = (± ΔX /X)x100 The percentage relative errors are often called Accuracy or Tolerance especially.

If the errors are still very small {specially, when dealing with Temperature Coefficient of Resistance} the relative percentage errors can be expressed as the part per million (PPM) i.e PPM = (±ΔX /X)x106  δX% =(± ΔX /X)x100 ±ΔX= PPM(X)10-6

Ex: The temperature coefficient (ΔR/ΔT) of a 1MΩ resistor might be stated as 100 ppm/°C which means: A 1oC change in temp. may be cause the 1MΩ resistance to increase or decrease by ΔR: ΔR = (temp. coff.) x ΔT ΔR =(100 ppm/°C) X 1°C ΔR =100x1MΩ x10-6 x 1 =100 Ω

Illustrative Examples 1-The true value of a resistor is 500Ω. It is measured using repeated experiments. The result value is between 490 Ω and 510 Ω. Calculate the absolute error, the relative error, and the percentage relative error of measurements.

Illustrative Examples Solution: ΔR = Rm– R = 490 – 500 = -10Ω ΔR = Rm– R = 510 – 500 = +10Ω ΔR = ±10Ω The relative error is δR = ± ΔR /R = ±10Ω/500Ω= ±0.02 The percentage relative error is: δR% = (± ΔR /R) x100 =( ±10Ω/500Ω)x100 = ± 2%

Illustrative Examples 2- If the temperature coefficient of a 100KΩ resistor is TCR=100PPM/oc, find its value when its temperature rise from 25oc to 40oc. Solution: R40oc=R25oc+TCR . ∆T R40oc = 100KΩ +100 X 100KΩ X 10-6 (40-25) R40oc =100.15KΩ

Accuracy, Precision, Resolution, and Significant Figures Accuracy: It is the closeness of the individual results of repeated measurements to its true value. Precision دقة الثبات It is the closeness of the individual results of repeated measurements to some mean value. مقياس الثبات فى القياس

Take experimental measurements for example of precision and accuracy. If you take the measurements of the mass of a 50.0 gram standard sample and get values of 47.5, 47.6, 47.5, and 47.7 grams, your scale is precise, but not very accurate. If your scale gives you values of 49.8, 50.5, 51.0, 49.6, it is more accurate than the first balance, but not as precise. The more precise scale would be better to use in the lab, providing you made an adjustment for its error.

Illustrative Examples The accuracy of ±1 % defines how close the measured result to the actual value. (Measured value = 99 % to 101 % of the actual value). Precision of the deflection instrument =(1/4)(its minimum scale division). Precision of the digital instrument = its least significant digit.

The term accuracy and tolerance are also used as: A resistor with a possible error of ±10% is said to be accurate to ± 10% Or to have a tolerance of ± 10% .

Conclusion The accuracy of the instrument depends on the accuracy of its internal components. Where as, its precision depends on its scale division or the number of its displayed digits.

Resolution It is defined as the smallest change in the measured quantity that can be observed. Resolution Precision

Resolution

Illustrative Examples Resolution of the potentiometer whose total resistance =100 Ω and its number of turns = 1000 ……is equal to the resistance of one turn which is 100/1000 = 0.1 Ω, Because the resistance of one turn is its minimum adjustable resistance.

Illustrative Examples The resolution of the instruments is the same as its precision. For example the resolution of the voltmeter shown in Fig. 2-1a = ±1 mV and that of the voltmeter shown in Fig. 2-1b = ±50 mV

Significant Figures Ex: The precision of the 4-significant figures voltmeter read out 6.432 V is ±1 mV. But the precision of the 3-significant figures voltmeter read out 6.43 V is ±10 mV.

Significant Figures The number of significant figures of the result of any calculations must be less than or equal to the least number of significant figures of any number used in calculation. If R= V/I = 8.14/2.330 =3.493562 KΩ =3.49KΩ

Significant Figures The numerator has 4-significant figures only, then the resultant significant figures must not exceed 3-figures. { i.e. the precision of the result (R) must not better than precision of the used instruments (V, and I)}.

Classifications of Measurement Errors The Measurement Errors are classified into three main categories: Grossجسيمة errors, Random errors, and Systematic errors

Gross Errors These errors are happened due to fatigue fəˈtēg اجهادand/or carelessness of the operators. They are sometimes called human errors. They are non-determinant errors (can't be calculated or estimated)

Examples of These Errors Incorrect reading of instrument. Incorrect recording of experimental data. Incorrect use of instruments “reading on a wrong scale of the multi-scale instruments” These errors can be minimized by being careful when doing the experiments

Random Errors These errors are happened due to the unexpected changes of the environments such as: Changing of the mains power supply voltage or frequency, Changing of the ambient temperature, aging درجة الحرارة المحيطة، of the instruments,…………etc.

Random Errors These errors are not determinate. It can be minimized by repeating the experiment many times and taking the average value of its results as the final result.

Systematic Errors They are produce due to the used system of measurement. These errors are determinate (Their values can be calculated). They are classified into: 1- Connection Errors 2- Instrumental Errors

1-Connection Errors It happened due to ignoring some essential quantities during measurements. It depends on the chosen connection diagram.

Voltmeter and Ammeter Methods of Resistance Measurements The voltmeter measures the voltage E across the resistor’s terminals. The ammeter indicates the current I +Iv. The calculated R= E /I+Iv

Voltmeter and Ammeter Methods of Resistance Measurements Since Ractually = E/I Then Iv constitutes an error. If Iv << I the error may be negligible.

Voltmeter and Ammeter Methods of Resistance Measurements The ammeter indicates the resistor current I. The voltmeter measures the voltage E + EA .

Voltmeter and Ammeter Methods of Resistance Measurements Calculated resistor R = E+ EA / I Ractually= E/I If R>> ammeter resistance. So EA<< E the error may be negligible.

2- Instrumental Errors This type of errors is due to the limited accuracy of the used instruments. For example: if the accuracies of voltmeter and ammeter in the above figures are δV%, δI% respectively, then the instrumental error in both figures is: δR%= ±(δV%+ δI%)

Specified Accuracy This accuracy is called the device accuracy or its specified accuracy a, (aI for ammeter, av for voltmeter). It is specified of the full scale deflection FSD or the range. The absolute error (Δ) at any pointer deflection on the scale is obtained as follows: Δ=(±a/100) x FSD

Specified Accuracy Ex: An instrument which indicates 100 µA at FSD has specified accuracy of ±1%. Calculate the upper and lower limits of measured current and the percentage error in the measurement if the measured value is: FSD 0.5 FSD 0.1 FSD

SOLUTION The absolute error ΔI at any pointer deflection on the scale is obtained as follows: ΔI=( ±a/100 )x FSD ΔI= (±1/100)X100= ±1 µA At FSD: Minimum and Maximum measured current are: 100 µA± ΔI= 100 µA ± 1 µA =99µA to 101µA. The percentage error of the measurement (measurement accuracy) δI% = (± ΔI/ Reading) x100 = (± 1 µA/100 µA)x100 = ±1 % of measured current.

SOLUTION At 0.5 FSD: Minimum and Maximum measured current are: 50 µA± ΔI= 50 µA ± 1 µA =49µA to 51µA. The percentage error of the measurement (measurement accuracy) δI% = (± ΔI/ Reading)x100 =± (1 µA/50 µA)x100=±2 % of measured current

SOLUTION At 0.1 FSD: Minimum and Maximum measured current are: 10 µA± ΔI= 10 µA ± 1 µA =9µA to 11µA. The percentage error of the measurement (measurement accuracy) δI% = (± ΔI/ Reading) x100 =± (1 µA/10 µA)x100=±10 % of measured current COMMENT ………??????

Measurement Error Combinations Sum of Quantities E=(V1+V2)±(ΔV1+ ΔV2)

Measurement Error Combinations Difference of Quantities

Measurement Error Combinations Product or quotient ˈkwō SHənt of Quantities % error in P = % error in I + % error in E % error in R = % error in I + % error in E

Measurement Error Combinations Quantity Raised to a Power See Reference