Uncertainty Examples
11.1
Example Calculations of Uncertainty
Instrument Bias Uncertainty Example #1
A DC voltage of -7.6541 volts is measured with an Agilent #34405A Multimeter. According
to the data sheet (p.51.1) the “DC accuracy” of this meter in the 0-10.0000V range is
±0.025% of the reading ±0.005% of full scale. Therefore, the instrument or bias uncertainty
of this DC voltage is
 0.025 
 0.005 
U V  
 7.6541V   
10.0000V   0.002364V  U V  0.0024V
 100 
 100 
This value was rounded to the ten-thousandths position  0.0024V  to match the nominal
value’s least significant digit in the ten-thousandths position  7.6541V  .
Instrument Bias Uncertainty Example #2
A DC voltage of -7.654 volts is measured with a LG Precision #DM-441B True RMS Digital
Multimeter. According to the data sheet (p.51.2) the “accuracy” of this meter in the 0-20V
range is  0.1%  4 digits . The ±0.1% value applies to the reading (-7.654 volts) and the
±4 digits apply to the resolution, which is 1 mV = 0.001 V at this setting. Therefore, the
instrument or bias uncertainty of this DC voltage is
 0.1 
U V  
 7.654V   40.001V   0.011654V  U V  0.012V
 100 
This value was rounded to the thousandths position  0.012 V  to match the nominal value’s
least significant digit in the thousandths position  7.654V  .
Instrument Bias Uncertainty Example #3
An AC voltage of 21.67 volts is measured with a Fluke 75 Digital Multimeter. According to
the data sheet (p.51.3) the “accuracy” of this meter in the 3.3-32 V range is  2%  2 digits .
The ±2% value applies to the reading (21.64 volts) and the ±2 digits apply to the resolution,
which is 0.01 V at this setting. Therefore, the instrument or bias uncertainty of this AC
voltage is
 2 
U V  
21.67V   20.01V   0.4534V  U V  0.45V
 100 
This value was rounded to the hundredths position  0.45V  to match the nominal value’s
least significant digit in the hundredths position 21.67 V  .
Instrument Bias Uncertainty Example #4
The same AC voltage of 21.67 volts (at 200 Hz) is measured with a LG Precision #DM-441B
True RMS Digital Multimeter. According to the data sheet (p.51.2) the “accuracy” of this
meter in the 0-200 V range is  0.5% 10 digits , and the resolution is 10 mv = 0.01 V. The
instrument or bias uncertainty of this AC voltage is
Uncertainty Examples
11.2
 0.5 
U V  
21.67V   100.01V   0.20835V  U V  0.21V
 100 
This value was rounded to the hundredths position  0.21V  to match the nominal value’s
least significant digit in the hundredths position 21.67 V  .
Instrument Bias Uncertainty Example #5
A resistance of 14.732 k is measured with an Agilent #34405A Multimeter. According to
the data sheet (p.51.1) the “accuracy” of this meter in the 0-100.000k range is ±0.05% of
the reading ±0.007% of full scale. The instrument or bias uncertainty of this resistance
measurement is therefore
 0.05 
 0.007 
U R  
14.732k   
100.000k   0.014366k  U R  0.014k
 100 
 100 
This value was rounded to the thousandths position  0.014 k  to match the nominal
value’s least significant digit in the thousandths position 14.732 k  .
Instrument Bias Uncertainty Example #6
A resistance of 1473 ohms is measured with a Calibeur DT830D Digital Multimeter.
According to the data sheet (p.51.5) the “accuracy” of this meter in the 0-2000 range is
 0.8%  2 digits , and the resolution is 1. The instrument or bias uncertainty of this
resistance measurement is
 0.8 
U R  
1473   21   13.784   U R  14
 100 
This value was rounded to the ones position  14   to match the nominal value’s least
significant digit in the ones position 1473   .
Instrument Bias Uncertainty Example #7
A capacitance of 82.7 nF is measured with an Agilent #34405A Multimeter. According to the
data sheet (p.51.1) the “accuracy” of this meter in the 0-100.0 nF range is ±1% of the reading
±0.5% of full scale. The instrument or bias uncertainty of this capacitance measurement is
therefore
 1 
 0.5 
U C  
82.7nF   
100.0nF   1.327nF  U C  1.3nF
 100 
 100 
This value was rounded to the tenths position  1.3 nF  to match the nominal value’s least
significant digit in the tenths position 82.7 nF  .
Uncertainty Examples
11.3
Uncertainty Example #1
The hoop stress, , in a thin-walled cylinder is found by

Pd 1 1 1 1
 Pd t
2t 2
where P is the interior pressure, d is the cylinder diameter, and t is the wall thickness. If we
assume that each of the three terms on the right contributes to the overall uncertainty in
stress, then
2
2
  U P    U d    U t 
 
 

 

 P    d    t  
U
2
Nominal values for this problem are P = 30 2 psi, d = 2.4500.03 inch, t = 0.0050
0.0002 inch. The nominal value for hoop stress is
 lbf 
 30 2 2.450 in 
lbf
in 
 
 7350
2 0.0050 in 
in 2
Since the equation for stress is a simple polynomial in each of the three variables, the
simplified form works for this problem,
2
2
 U   U   U 
 1 P   1 d     1 t 

t 
 P   d  
U
2
The numerical value of the uncertainty in hoop stress is
U


0.06667 2  0.01225 2   0.04000 2
U   0.0787 *   0.0787 * 7350
lbf
in 2
 0.0787 
7.9
 7.9%
100
 U    580 psi or  7.9%
Note that the first term in the equation above (  0.06667 ) was due to the uncertainty in the
pressure measurement, and that this term contributed the most to the overall uncertainty in
the hoop stress. Any attempt to reduce the overall uncertainty in the measurement of stress
in this experiment should begin with the pressure measurement.
Uncertainty Examples
11.4
Uncertainty Example #2
An aluminum rod of diameter 0.500 ( 0.001) inch is 25.7 ( 0.02) inches long. A
concentrated mass of 4.70 ( 0.01) lbm is attached to the end of the rod. Estimate the natural
frequency of the system, and the uncertainty in the estimate. Assume that the modulus of
elasticity for aluminum is 10.7E6 ( 1%) psi.
The natural frequency of a massless beam with a concentrated mass Mc at the end is given by
the following formula
EI
c  3
M c L3
where
E =
modulus of elasticity of the beam material (lbf/in2),
I =
area moment of inertia of the beam (in4),
L =
cantilever length (in),
Mc =
concentrated mass (lbm, kg, slugs, or lbf-sec2/ft),
c =
natural frequency (rad/sec or sec-1).
The factor of 3 is developed during the solution to a 4th order partial differential equation
and is a dimensionless quantity. The remaining terms must therefore have units of
radians/sec, since this is the units of c.
Assume the beam has a uniform circular cross-section. The area moment of inertia I is
therefore
I

64
d4
Substituting this value for I into the equation for c we have
c  3
E d 4
1/ 2
 
 3 
3
 64 
64 M c L
E 1 / 2 d  4 / 2 M c1 / 2 L 3 / 2
If we assume that each of the 4 parameters (the "" and "64" terms are exact constants) on
the right-hand side of the equation contributed to the overall uncertainty in c, we would
have the following equation
U c
c
2
2
  U    U    c U M c
  c E    c d   
 E  c   d  c   M c  c
2
   c U L 
 

  L  
c 
 
2
Since the equation for stress is a simple polynomial in each of the three variables, the
simplified form works for this problem,
Uncertainty Examples
U c
c
11.5
2
2
  1 U E    4 U d    1 U M c
 
 
 
 2 E   2 d   2 M c
2
   3 U L 2
 
  2 L 

Calculating the Nominal Value, Incorrect Method:
c  3
E d 4
rad
(10.7 E 6) (0.500) 4

 1.96
3
3
3
sec
64 M c L
64 ( 4.70)( 25.7)
Calculating the Nominal Value, Correct Method:
lbf 
ft  lbm  12 in 

4


10.7 E 6 2  0.500 in   32.2
in 
lbf  sec 2  1 ft 
Ed
rad


c  3
 37.8
3
3
3
sec
64 M c L
64 (4.70 lbm)(25.7 in)
4
The uncertainty is found from
U c
c
U c
c
U c
2
2
 1 U E   4 U d    1 U M c
 
 
 
 2 E   2 d   2 M c
2
2
2
   3 U L 2
 
  2 L 

2
  1 1%    4 0.001in    1 0.01lbm    3 0.02 in 

  
  
 
  
 2 100%   2 0.500 in   2 4.70 lbm   2 25.7 in 

2
0.0052  0.0042   0.001062   0.00117 2
c
U c
0.66
 0.0065949 
 0.66%
c
100
The nominal value for frequency was 37.8 rad/sec, therefore the uncertainty in frequency is
U c   0.0066 (37.8 rad / sec)   0.25 rad / sec
Uncertainty Examples
11.6
Uncertainty Example #3
A first order system consisting of two
resistors in series and two capacitors in
parallel is shown in the figure on the right.
The time constant, , is given by
+
R1
Eo(t)
Ei(t)
  R1  R2 C3  C 4 
+
R2
C3
-
C4
Resistor values are R1 = 10 k  0.5 k and R2 = 22 k  1 k. Capacitor values are
C3 = 0.10 F  0.005 F and C4 = 0.22 F  0.01 F. The nominal time constant is therefore
 volt  amp  sec 


  10 x103 ohm  22 x10 3 ohm 0.10 x10  6 farad  0.22 x10  6 farad 
 amp  ohm  volt  farad 



  0.0102 sec
The uncertainty in the time constant is given by
2
2
2
  U R1    U R 2    U C 3    U C 4 
  
  
  

 

 R1    R2    C3    C 4  
U
2
This is an example of a problem that cannot be worked the “easy” way, due to the additions
in the formula. There is a lot of symmetry in the problem, which makes it relatively easy to
find the four partial derivatives.




 C3  C 4 ,
 C3  C 4 ,
 R1  R2 ,
 R1  R2 
R1
R2
C3
C 4
Dividing each term by the time constant  gives
C3  C4 
1
 1
,


R1  R1  R2 C3  C 4  R1  R2 
C3  C4 
1
 1

,

R2  R1  R2 C3  C 4  R1  R2 
R1  R2 
 1
1


,
C3  R1  R2 C3  C 4  C3  C 4 
R1  R2 
 1
1


C3  R1  R2 C3  C 4  C3  C 4 
Substituting into the equation for uncertainty,
2
2
2
 U R1   U R 2   U C 3   U C 4 
  

  
  
 
R

R
R

R
C

C
C
C


2
2
4
4
 1
 1
 3
 3
U
2
Substituting numerical values into the equation,
2
2
2
2
 0.5 k   1k   0.005F   0.01F 
4.9
  
  
  0.0494 
  
 4.9%
 
100

 32 k   32 k   0.32 F   0.32 F 
U
U  0.04940.0102 sec  0.0005 sec
-
Uncertainty Examples
11.7
Uncertainty Example #4
There is another way to work the previous problem with the first order system consisting of
two resistors in series and two capacitors in parallel.
  R1  R2 C3  C 4     Reff Ceff
Reff  R1  R2 , Ceff  C3  C 4 
Resistor values are R1 = 10 k  0.5 k, R2 = 22 k  1 kand Reff = 32 k. Capacitor
values are C3 = 0.10 F  0.005 F, C4 = 0.22 F  0.01 F, and Ceff = 0.32 F.
The uncertainty in the time constant is given by
2
  U Reff
 
 Reff 


   U C eff
 
  Ceff 
 
U
2
U


    1 Reff


Reff


2
U
 
    1 C eff
 
Ceff
 




2
Note the that “easy” form does apply when we define the effective resistance and
capacitance. Calculations for the uncertainty in the effective resistance and capacitance are
relatively easy,
U Reff
Reff
U Reff
Reff
U C eff
Ceff
U C eff
Ceff
2
2
 Reff U R1   Reff U R 2 
Reff
Reff
 ,
 
 1
 1,
 
 R1 Reff   R2 Reff 
R2
R1


 
2
2
2
2

 





U
U


0
.
5
k
1
k
1
2
R
R
    1
  
   1
 32 k    32 k   0.03494

 

R
R

 

eff
eff

 

2
2
 Ceff U C 3   Ceff U C 4 
Ceff
Ceff
 ,
 
 1,
 1
 
 C3 Ceff   C 4 Ceff 


C
C
3
4

 

2
2
2
2

 0.005 F   0.01F 
U C 3  
U C 4 

  0.03494
  
 
  1
  1

Ceff  
Ceff 
 0.32 F   0.32 F 

The calculation for the uncertainty in the time constant is now quite simple,
UR

   1 eff


Reff

U
2
U
 
    1 C eff
 
Ceff
 
2

 


0.034942  0.034942
U  0.04940.0102 sec  0.0005 sec
 0.0494 
4.9
 4.9%
100
Uncertainty Examples
11.8
Uncertainty Example #5
Series resistance for seven resistors can be calculated from:
RS  R1  R2  R3  ...  R7
The simplified or “easy” method does not apply to calculating uncertainty for series
resistance due to the additions.
Calculations for the uncertainty in the series resistance is relatively easy though using the
partial derivative method:
U RS
RS
2
2
2
 R U   R U   R U 
 R U
  S R1    S R 2    S R 3   ...   S R 7
 R1 RS   R2 RS   R3 RS 
 R7 RS



2
The partial derivatives are easily determined from the series resistance formula above,
RS
RS
RS
RS
 1,
 1,
 1, ....,
 1
R1
R2
R3
R7
Substituting the partial derivatives into the partial derivative form of the computed
uncertainty gives:
2
2
2
 U   U   U 
 U 
 1  R1   1 R 2   1  R 3   ...  1  R 7 
RS
 RS   RS   R S 
 RS 
U RS
2