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Physics 114: Lecture 11
Error Analysis
Dale E. Gary
NJIT Physics Department
Instrumental Uncertainties
When evaluating the effect of uncertainties on a result, one has to be
aware of the underlying principles of the experiment.
 As an important example, consider an electrical experiment involving a
resistor with a 1% tolerance. Say you connect such a resistor to a
battery and measure the current through it, I = V/R.
 The resistor’s value can be within 1% of the stated value, so say instead
of 100 ohm it is 99 ohm. But repeated experiments with the same
resistor will all be with a 99 ohm resistor. The error is a systematic
error.
 However, let’s say an entire class is doing the experiment, each with
their own 1%-tolerance resistor. Some will be 99 ohm, some will be 100
ohm, some will be 101 ohm. The collection of results with that 1%tolerance resistor now has a random error of order 1%.

Mar 5, 2010
Statistical Uncertainties





Sticking with the resistor experiment, all resistors are affected by
temperature. Generally, a hotter resistor has greater resistance than a
colder resistor.
This is yet another source of error. If you measure the resistance to be
99 ohm, and then make repeated measurements at different
temperatures (remember that the very act of passing a current through
the resistor affects its temperature), there is another source of
(essentially) random errors.
Let’s say, as an example, that the temperature affect causes the
resistance to fluctuate with a standard deviation of about ½ ohm.
How do we combine the effects of these different types of error, in that
case of a current measurement: I = V/R.
That is the topic for today.
Mar 5, 2010
Relative Error (Percent Error)

I want to introduce the very important concept of relative error, which is
just the inverse of the signal to noise ratio (SNR), which we saw earlier is

 SNR.


So the relative error is

 Relative Error (parent distribution).

s
 Relative Error (sample)
x
In our case of the 100 ohm resistor with ±1 ohm tolerance, there is a
relative error of
1
 0.01  1%
100
which is why we said the resistor has a 1% tolerance.
 Our example ± ½ ohm temperature effect gives an additional 0.5% error.

Mar 5, 2010
Relative Error (Percent Error)


What is we have a 1%-tolerance 10 k-ohm resistor? Then the error in
resistance is larger—it is 100 ohm—but the relative error is still the same:
100
 0.01  1%
10000
Likewise, the random temperature effect causes larger fluctuations in
resistance, ± 50 ohm, but the relative error is still 0.5%.
It is obviously very convenient to talk about the relative error, rather than
the absolute error, when dealing with uncertainty in a measurement, since
many (most) errors affect the results in this multiplicative sense.
 In addition, relative error comes up naturally when we discuss propagation
of errors, which we will do next.

Mar 5, 2010
Propagation of Errors
What is propagation of error? It is the effect on a measurement of errors
in one or more parameters affecting the measurement. In our current
measurement
V
I
?
R  R
how does the error in resistance affect the current?
 Say V = 1.5 volts. We can always calculate using the two extremes
V
1.5 volts
I

 14.85 mA
R   R 101 ohm
V
1.5 volts
I

 15.15 mA
R   R 99 ohm


So we see that the error is about ± 0.15 mA. What is the relative error?

This is the case for combining measurements from a lot of different 1%tolerance resistors, but if we make repeated measurements with a single
resistor, again we will systematically measure either a high or low current.
Mar 5, 2010
Propagation of Errors, cont’d
What we just did is to examine a simple case of propagation of error, but
the actual case can be more complicated. For example, what if different
measurements are done by students in a lab, using different resistors and
different batteries.
 When the measurements of current are combined, there are errors
contributed by both the different voltages of the batteries, and different
resistances of the resistors:
V  V
I
?
R  R

We need a general way to treat such cases, so that we do not have to
resort to trying different extremes.
 We can use differential calculus to help us. Let’s start with the original
equation, which tells us how the measurements of voltage and resistance
combine to give current:
V
I .
R

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Method of Propagation of Errors
V
Start with original relation, in this case: I  .
R
 Use the chain rule:
dV
dV V
 1  dV
dI 
 Vd   
 VdR 1 
 2 dR.
R
R
R
R
R
 

Here, dI represents the deviations of individual measurements :
dI  I i  I  xi  x
 We now square both sides to give the square deviations:

2

2
2
dV V
dV
V
V
 dI   ( I i  I )    2 dR       2 dR   2 3 dVdR.
R
 R R
  R  R

Finally, average over many measurements:
2
2
dV
V
V
2




 dI   ( Ii  I )2      2 dR   2 3 dVdR
R
 R 
R

2
2
1 2 V2 2
V
  2  V  4  R  2 3 dVdR
R
R
R
2
I
Mar 5, 2010
A Great Simplifier—Relative Error

Notice that we can take:
1 2 V2 2
V
  2  V  4  R  2 3 dVdR
R
R
R
2
I

And divide through by I2 = V2/R2:



2
I


2
V
2
1
dVdR
I
V
R
VR
What about this term dVdR ? This is the product of random fluctuations
in voltage and resistance. When we multiply these random fluctuations
and then average them—they should average to zero!
Thus, we have the final result:

I
I


Square of relative error in I
2
R
2


2
 V2
V2

 R2
R2
Test it.
Mar 5, 2010
Earlier Example

If there are only errors in resistance (i.e.  V   V  0, ) then in our earlier
example we saw by calculating using the two extremes
V
1.5 volts
I

 14.85 mA
R   R 101 ohm
V
1.5 volts
I

 15.15 mA
R   R 99 ohm
while the nominal value for I is 15 mA, so the relative error in I is 1%, just
the same as the relative error in R.
 Our result on propagation of error for this case agrees:

I


 V2
 R2


 R2
 0   0.01  0.01
2
I
V 2 R2
Notice that if, say, the relative error of the voltage was also 1%, then the
relative error in current would instead be larger:
I
I

 V2
V2

R2
 0.01   0.01
2
2
 0.014  1.4%
Mar 5, 2010