Uncertainty in Measurements Joshua Russell January 4, 010 1 Introduction Error analysis is an important part of laboratory work and research in general. We will be using probability density functions PDF) to calculate the uncertainty in our measurements. We will be using 3 types of PDFs, the triangle shaped, box shaped, and bell or gaussian shaped. For a PDF, the area under the function is equal to one, which tells us that the sum of all probabilities is equal to one. Direct Measurement Uncertainties.1 Single Analog Measurements Analog measurements are the basic measurement type you will be taking in the lab. For analog measurements, we will use the triangle PDF. This type of PDF lets us say that the measured value has the highest probability and as we go out from either side of this value the probability goes down. We will look at an example to see how we find single analog measurement uncertainly. We want to measure the length of a metal rod with a ruler, as shown in figure??. We place the metal rod next to the ruler and record the position of one end to be 1.00 units and the position of the second end to be.63 units. Now we need to find the uncertainty in the length of the rod. Figure 1: Measuring the length of a metal rod. When estimating the uncertainty we need to take into account the significant digits and accuracy of the measuring device. We find the two ends of the measurement where we know the probability of finding the measured value there is 0. We can with certainty that the first end of the rod is between 0.975 and 1.05 units and the second end of the rod is between.60 and.65 units. The distance between these two points is the width, a, of the PDF shown in figure??. We find the uncertainty of our measurement by finding the squared width of the triangle shaped PDF, fx) T. We find the squared width by adding up the squared distance from the center of 1
Figure : PDF of a analog measurement. the PDF and multiplying them by the probability density function at that point. We center the triangle shaped PDF at 0 knowing this will not change the width. u = a/ a/ u = a 4 u = a 6 fx) T x dx 1) This result tells us that the uncertainty of an analog measurement is given by the width of our PDF divided by 6. What does this mean? If we find the area underneath the PDF which is bounded by the uncertainly on either side of the center of the PDF, we find the probability that our measured value is within this bound. For the triangle PDF, we have a probability of our measured value being within this bound is 65 percent. Plugging in our values for the rod into equation??, we find the uncertainty of the measured position of the ends of the rod is u = 0.01 units. We would state the position of the first end of the rod to be 1.00 ± 0.01 units and the second end of the rod to be.63 ± 0.01 units. )
. Single Digital Measurements On digital readouts we use a box PDF. We use the box PDF because we can only see a set number of significant figures on the digital display. We do not know what the significant figure is pass the last shown significant figure. The first significant figure pass the last shown significant figure has equal probability to be any number. Let us look at an example to illustrate how to find the uncertainty for the box PDF. Figure 3: Measuring the time it takes a ball to fall. In figure??, we measure the time it takes a ball to fall a known distance h. There are 3 significant digits shown on the digital display of the stop watch. We know that the measured value is some where between 3.604 s and 3.595 s, but all the values between these bounds have equal probability of being the measured value. These two values will make up the two ends on the box PDF. Figure 4: PDF of a digital measurement. Plugging the box shaped PDF, fx) B, into equation?? we find the uncertainty. Again we center 3
the PDF at 0 knowing this will not change the width. u = a/ a/ u = a 1 u = a 3 fx) B x dx 3) For the box PDF, we have a probability of 58 percent that our measured value is within this uncertainty bound. Plugging in our values for the falling ball into equation??, we find that the uncertainty in the time it took the ball to fall is u = 0.003. We would state the time for the ball to fall a distance h to be 3.600 ± 0.003. This may be miss leadingly simple to do as you need to include other sources of uncertainly as well. We will illustrate this with an example. We want to time how long a ball drops for a given distance h, see figure??. The first measurement is 3.600 s. The uncertainty of ±0.01 s will give us 100 percent probability of our measured value be between the uncertainty bound. Now we take a second measurement and now we get 3.9 ± 0.01 s. We can see the two measured values do not overlap. There must be a difference in how the two measurements were taken. Most likely the difference is the reaction time of the person taking the measurement. Make sure to include all sources of uncertainly and not just the uncertainly of the digital measurement..3 Multiple Measurements When taking multiple measurements we will want to use a bell shaped or Gaussian PDF. We will want to use the standard deviation of the mean as the uncertainty. The idea behind the standard deviation is that all the measured values are centered around a single value and the probability of measuring a value falls off from this single value in a bell shaped or Gaussian curve. This holds if we assume the errors are random and not systematic. Here I will define the standard deviation, σ, and the standard deviation of the mean, σ. σ = 1 N x i x) N 1 4) σ = i=1 σ N 5) Where N is the number of measurements, x i is the i th measurement, and x is the average of all N measurements. As an example, we take 10 measurements of the second end of the metal rod in figure??. We find the values of the position of the end of the rod to be,.60,.65,.64,.61,.63,.63,.6,.64,.63,.65 units. First we find the average value, x, of the measured positions with equation??. N i=1 x = x i N This gives use an average value of.63 units. We find the standard deviation by plugging in the 10 measured values into equation??. 6) 4
σ = 1 N x i x) N 1 i=1 1 = 9 {.60.63) +.65.63) +.64.63) + +.61.63) +.63.63) +.63.63) + +.6.63) +.64.63) +.63.63) +.65.63) } =0.016 Now we find the standard deviation of the mean by plugging σ into equation??. σ = σ N = 0.016 10 = 0.005 The uncertainly for the multiple measurement is 0.005 units with a probability of our measured value being with in the uncertainty bound of 68 percent. This is a smaller uncertainty than the single measurement which had an uncertainly of 0.01 units. We would state the measured position of the second end of the rod to be.630 ± 0.005..4 Counting Measurements At times we will need to make a measurement by counting the number of events. The uncertainty associated with this measurement is given by a Poisson PDF. The standard deviation of a Poisson PDF is the square root of the total number counted. σ = N 7) If we count the number of rotations of a wheel to be 00 then the uncertainty of the measurement is 00 or 14. We would state the measured number of rotations of the wheel to be 00 ± 14 rotations. 3 Propagation of Uncertainties When calculating a result from many different measured values, which each has an uncertainly associated with it, we need to know what the uncertainly of the result will be. In general we can expand a function fx, y, z,..) in a power series. ) ) f fx, y) fx, Y ) + f x x X) + y Y ) 8) x=x Y Where X and Y are the values we are expanding the function about. We can ignore the first term since it just shifts the graph without changing its shape. The second and third ) terms are constants ) multiplied by the PDF of x and y. So the width of these terms is f x x=x σ x and f y y=y σ y. 5 y=y
Now looking at the distributions of x and y as gaussian, P robx) = Ae x σx we can see that the probability of getting one result for x and one result for y is P robx) P roby). Here we have A and B as normalization coefficients such that P robx) dx = 1. y σ and P roby) = Be y, P robx, y) = Ae x σx = ABe 1 We can rearrange the term to simplify the exponent. y σ Be y x σx + y σy ) 9) x σx + y σ y = σ yx σx + σy ) ) σxσ y σ x + σy + σ xy σx + ) σxσ y σ x + σy) = σ yσxx + σyx 4 + σxσ yy + σxy 4 + σxσ yxy σxσ yxy σxσ y σ x + σy) = σ xσy x + y) + σyx σxy ) σxσ y σ x + σy) = x + y) σ x + σ y) + σ y x σxy ) σxσ y σ x + σy) = x + y) σ x + σ y) + z Where z is a function of x and y. Now plug this result into equation??. P robx, y) = ABe 1 = ABe 1 ) x+y) σx +σ y) +z ) x+y) σx +σ y) e z Now to make the probability a function of only x and y we need to integrate over all of z. P robx, y) = allz Now we have the PDF in terms of only x+y. Where the width, σ f = σ x + σ y or σ f = terms from equation??. P robx + y, z) dz = πp robx + y) P robx, y) = Ce σ f = ) x+y) σx +σ y) σ x + σ y. Now we plug in the widths of the nd and 3 rd f x σ x This is the general form for error propagation. uncertainly propagation. ) f + y σ y ) The following sections are the special cases for 6