Chapter 2: Measurement
Errors

Gross Errors or Human Errors
– Resulting from carelessness, e.g.
misreading, incorrectly recording
1

Systematic Errors
– Instrumental Errors
• Friction
• Zero positioning
– Environment Errors
• Temperature
• Humidity
• Pressure
– Observational Error

Random Errors
2

Absolute Errors and Relative
Errors
Absolute Error e   X t  X m
where X t : True Value
X m : Measured Value
Relative Error %Error  
Xt  Xm
 100%
Xt
3

Accuracy, Precision,
Resolution, and Significant
Figures
– Accuracy (A) and Precision
• The measurement accuracy of 1%
defines how close the measurement
is to the actual measured quality.
• The precision is not the same as the
accuracy of measurement, but they
are related.
Accuracy  1  %Error
x x
Precision  1  n n
xn
xn
x


n
4
a)
b)
If the measured quantity increases or
decreases by 1 mV, the reading becomes
8.936 V or 8.934 V respectively.
Therefore, the voltage is measured with
a precision of 1 mV.
The pointer position can be read to
within one-fourth of the smallest scale
division. Since the smallest scale
division represents 0.2 V, one-fourth of
the scale division is 50 mV.
–
Resolution
•
–
The measurement precision of an
instrument defines the smallest
change in measured quantity that
can be observed. This smallest
observable change is the
resolution of the instrument.
Significant Figures
•
The number of significant figures
indicate the precision of
measurement.
5
Example 2.1: An analog voltmeter is used to
measure voltage of 50V across a resistor.
The reading value is 49 V. Find
a) Absolute Error
b) Relative Error
c) Accuracy
d) Percent Accuracy
Solution
a) e  X t  X m  50V  49V  1V
b) % Error 
Xt  Xm
100%
Xt
50V  49V
100%  2%
50V
c) A  1  % Error  1  2%  0.98

d) % Acc  100%  2%  98%
6
Example 2.2: An experiment conducted to
measure 10 values of voltages and the
result is shown in the table below.
Calculate the accuracy of the 4th
experiment.
No.
(V)
No.
(V)
1
98
6
103
2
102
7
98
3
101
8
106
4
97
9
107
5
100
10
99
Solution
xn
x x



 x2  ...  x10
n
10
98  102  101  97  100  103  98  106  107  99

10
 101.1
1
Precision  1 
xn  xn
97  101.1
 1
 0.959  96%
xn
101.1
7

Class of Instrument
– Class of instrument is the number
that indicates relative error.
– Absolute Error
e(range) 
Class
 range
100
– Relative Error
%Error 
erange
%Error 
erange
Xt
Xm
100% , xt true value
100% , xm measured value
8
Example 2.3 A class 1.0 Voltmeter with range
of 100V, 250V, and 1,000V is used to
measure voltage source with 90V.
Calculate range of voltage and its
relative errors
Solution
1
 100V   1V
100
 100V  1V, 99V  101V
a) e 100V 
 1V
 100%  1.11%
90V
1
b) e 250V 
 250V   2.5V
100
 250V  2.5V, 247.5V  252.5V
%Error 
 2.5V
 100%  2.77%
90V
1
c) e 1,000V 
 1,000V   10V
100
 1,000V  10V, 990V  1,010V
%Error 
%Error 
 10V
 100%  11.11%
90V
9

Measurement Error
Combinations
– When a quantity is calculated
from measurements made on two
(or more) instruments, it must be
assumed that the errors due to
instrument inaccuracy combine is
the worst possible way.
– Sum of Quantities
• Where a quantity is determined as
the sum of two measurements, the
total error is the sum of the absolute
errors in each measurement.
E  V1  ΔV1   V2  ΔV2 
giving
E  V1  V2    ΔV1  ΔV2 
10
– Difference of Quantities
• The error of the difference of two
measurements are again additive
E  V1  ΔV 1   V2  ΔV 2 
 V1 V2   ΔV 1  ΔV 2 
– Product of Quantities
• When a calculated quantity is the
product of two or more quantities,
the percentage error is the sum of
the percentage errors in each
quantity
P  EI
 E  ΔE I  ΔI

 EI  EΔ I  I Δ E  ΔEΔI
since
ΔEΔI is very small ,
P  EI  EΔ I  I Δ E

11
EI  IE
100%
EI
 I   E 
  
 100%
 I   E 
percentage error 
% error in P  % error in I   % error in E 
 Quotient of Quantities
% error in
E
 % error in E   % error in I
I

 Quantity Raised to a Power
% error in AB  B % error in A
Example 2.4 An 820Ω resistance with an
accuracy of  10% carries a current of
10 mA. The current was measured by
an analog ammeter on a 25mA range
with an accuracy of  2% of full scale.
Calculate the power dissipated in the
resistor, and determine the accuracy of
12
the result.
Solution
P  I 2 R  10 mA  820
2
 82 mW
error in R   10%
error in I   2% of 25 mA
  0.5 mA

 0.5 mA
 100%   5%
10 mA
%error in I 2  2 5%   10%


%error in P  %error in I 2  %error in R 
  10%  10%   20%

Basics in Statistical Analysis
 Arithmetic Mean Value
x1  x2  x3  ...  xn
x
n
• Minimizing the effects of random
errors
13
14
– Deviation
• Difference between any one
measured value and the arithmetic
mean of a series of measurements
• May be positive or negative, and the
algebraic sum of the deviations is
always zero
dn  x n  x
• The average deviation (D) may be
calculated as the average of the
absolute values of the deviations.
D
d 1  d 2  d 3  ...  d n
n
15
– Standard Deviation and Probable
of Error
• Variance: the mean-squared value of
the deviations
d 12  d 22  ...  d n2
 
n
2
• Standard deviation or root mean
squared (rms)
d 12  d 22  ...  d n2
SD or σ 
n
• For the case of a large number of
measurements in which only random
errors are present, it can be shown
that the probable error in any one
measurement is 0.6745 times the
standard deviation:
Probable Error  0 . 6745 
16
Example 2.5 The accuracy of five digital
voltmeters are checked by using each of them
to measure a standard 1.0000V from a
calibration instrument. The voltmeter
readings are as follows: V1 = 1.001 V, V2 =
1.002, V3 = 0.999, V4 = 0.998, and V5 =
1.000. Calculate the average measured
voltage and the average deviation.
Solution
V1  V2  V3  V4  V5
5
1.001  1.002  0.999  0.998  1.000

 1.000V
5
d1  V1  Vav  1.001  1.000  0.001V
Vav 
d 2  V2  Vav  1.002  1.000  0.002V
d 3  0.999  1.000   0.001V
d 4  0.998  1.000   0.002V
d 5  1.000  1.000  0V
d1  d 2  ...  d 5
5
0.001  0.002  0.001  0.002  0

 0.0012V
5
D 
17