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# physics114_lecture11 (1).ppt

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# physics114_lecture11 (1).ppt

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### physics114_lecture11 (1).ppt

1. 1. Physics 114: Lecture 11 Error Analysis Dale E. Gary NJIT Physics Department
2. 2. Mar 5, 2010 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%.
3. 3. 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.
4. 4. 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  So the relative error is  In our case of the 100 ohm resistor with ±1 ohm tolerance, there is a relative error of which is why we said the resistor has a 1% tolerance.  Our example ± ½ ohm temperature effect gives an additional 0.5% error. SNR.    Relative Error (parent distribution). Relative Error (sample) s x     1 0.01 1% 100  
5. 5. 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:  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. 100 0.01 1% 10000  
6. 6. 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 how does the error in resistance affect the current?  Say V = 1.5 volts. We can always calculate using the two extremes  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. ? V I R R    1.5 volts 14.85 mA 101 ohm 1.5 volts 15.15 mA 99 ohm V I R R V I R R          
7. 7. 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:  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 V I R R      . V I R 
8. 8. Mar 5, 2010 Method of Propagation of Errors  Start with original relation, in this case:  Use the chain rule:  Here, dI represents the deviations of individual measurements :  We now square both sides to give the square deviations:  Finally, average over many measurements: 1 2 1 . dV dV dV V dI Vd VdR dR R R R R R              i i dI I I x x       2 2 2 2 2 2 2 3 ( ) 2 . i dV V dV V V dI I I dR dR dVdR R R R R R                          . V I R    2 2 2 2 2 3 2 2 2 2 2 4 3 ( ) 2 1 2 i I V R dV V V dI I I dR dVdR R R R V V dVdR R R R                       
9. 9. Mar 5, 2010 A Great Simplifier—Relative Error  Notice that we can take:  And divide through by I2 = V2/R2:  What about this term ? 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:  Test it. 2 2 2 2 2 4 3 1 2 I V R V V dVdR R R R       2 2 2 2 2 1 2 V I R dVdR I V R VR        Square of relative error in I dVdR 2 2 2 2 V I R I V R     
10. 10. Mar 5, 2010 Earlier Example  If there are only errors in resistance (i.e. ) then in our earlier example we saw by calculating using the two extremes  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:  Notice that if, say, the relative error of the voltage was also 1%, then the relative error in current would instead be larger: 1.5 volts 14.85 mA 101 ohm 1.5 volts 15.15 mA 99 ohm V I R R V I R R           0, V V       2 2 2 2 2 0 0.01 0.01 V I R I V R             2 2 2 2 2 2 0.01 0.01 0.014 1.4% V I R I V R         