|
An Interval Timer
(Designing a Printed Circuit Board)
Intro to Eng. II - Module 2
|
Developed by
Web Development by |
Part 2. Component Variations Effects
Nominal Values and Tolerance
Nothing made is exact. If we try to produce
something that is supposed to have some exact value or dimension,
we would fail. We must specify, in addition to the values or dimensions
we are trying to achieve, what variations we can allow in these
things.
For example, consider a process to manufacture
470 ohm resistors. If we were to measure the resistors made by
the process, none of them would be exactly 470 ohms. We
would probably find some with values larger than 470 ohms and
some with values less than 470 ohms. If the process is designed
to make 10% resistors, we would find the resistance values may
be as large as 470+47 ohms (517 ohms) and as small as 470-47 ohms
(423 ohms). If the process is making 1% resistors, then we would
expect the values to be between 465.3 and 474.7 ohms.
The exact value that a component, such as a
resistor, is intended to have is called its nominal value,
and variations about that nominal value are called its tolerance.
The resistors mentioned above have tolerances of 10% and 1%.
Resistors are often made with 5% tolerances, and even 0.1% tolerance
resistors are available.
Generally the smaller the tolerance allowed
in a component, the more expensive it is to make. Consequently,
if cost is an issue in what we are making, it behooves us not
to select components with unreasonably small tolerances if they
are not really necessary.
Suppose we are designing a product made with
a number of components. Our product undoubtedly has certain requirements
or specifications which it must meet. Clearly the tolerances
in the components going into our product will affect its behavior.
It is usually not an easy task to determine what tolerances can
be allowed in the components of our product so that the specifications
can be met.
In this project we are concerned with how the components in the interval timer affect the timer's performance. To understand how component tolerances affect the overall behavior of the circuit, it is necessary to understand probability density functions.
What a Probability Density Function Is
Suppose we have a large number of balls in
a container as shown in Figure 18. Each of the balls has a value
between 0 and 1 printed on it. For example, the value on one
of the balls may be 0.2368; another may have the value 0.9824.
The values seem to be random.
To understand more about the nature of the
values on the balls, we separate the balls into 10 bins as shown,
each bin corresponding to a range of values. Bin 1 corresponds
to the range between 0 and 0.1; bin 2 to the range between 0.1
and 0.2, and so forth. The ball with the value 0.2368 will be
placed in bin number 3 for example, and the ball with the value
0.9824 will be placed in bin number 10. (Of course we will have
to decide in which bin to place a ball if its number is exactly
0.0000, 0.1000, 0.2000, etc.)
Figure 19(a) shows a typical result after we
have sorted all of the balls. In this example, there are more
balls with values between 0.4 and 0.5, than between 0.8 and 0.9
for instance; generally there are more balls in the bins in the
middle ranges than in the end ranges.
 | |
| (a) The Bins after All Values are Sorted | (b) Corresponding Probability Density Function |
The height of the sorted balls in the bins
has a particular shape. We could represent this shape approximately
by a continuous function as shown in Figure 19(b). This curve,
properly normalized, is called the "probability density function".
It can be used to determine what the probability is that the
value on a randomly selected ball is within a particular range
of values. This probability is related to the area under the
probability density function as shown in Figure 20. The probability
that the value on a ball is between 0 and 1 is 1. (It is certain.)
Hence the total area under the curve is 1 as shown in Figure
20(a). The probability that the value is between 0.3 and 0.5
is the area under the curve between 0.3 and 0.5 as shown in Figure
20(b). It is more probable that the value on a selected ball
will be between 0.3 and 0.5 than between 0.1 to 0.3.
 | |
| (a) | (b) |
Determining the Probability Density Function
from Samples
Now suppose we have a huge number of
balls in the container. (So many that it is not possible to sort
them all.) We could get an estimate of the shape we obtained
above by selecting enough balls at random. (That means that any
ball in the container can be selected with equal likelihood.)
We read the value of each ball and put a marker in the particular
bin corresponding to the value as shown in Figure 21. Each ball
is placed back in the container before another ball is selected
at random. We repeat this process until we have enough markers
to properly indicate the shape of the probability density function.
This process, called "sampling", is how the probability description of a manufactured item would be determined. You can see that it is important to use enough samples, and the samples must be randomly selected.
The Uniform Distribution
The uniform distribution is an important probability
density function. For this, any value is equally likely over
a range of values as shown in Figure 22. Each bin would be filled
to approximately the same height as shown in Figure 22(a). The
continuous probability density function is shown in Figure 22(b).
The Probability Density Function for the
Sum of 2 Uniformly Distributed Samples
Suppose we have 2 large sets of balls, each
of which has a uniform distribution between 0 and 1. (Call 1
set of balls 'x', and the other set 'y' as shown in Figure 23.)
We randomly select a ball from 'x' and a ball from 'y', add their
values, and return the balls to their respective sets. We place
a marker in the bin that corresponds to their sum. Note that
the sum is necessarily between 0 and 2. We have chosen each bin
to have a range of 0.1 units.
We continue selecting 2 samples and marking
the sum until we have enough samples to define the shape of the
function.
The shape of the probability density function
of the sum of two uniformly distributed values can be determined
simply as follows:
Suppose, rather than continuous uniform distributions,
the distribution of the values of the balls in the bins have only
10 possible values, namely, 0.05, 0.15, 0.25, ..., 0.95 as shown
in Figure 24. We would say that this probability density function
is "discrete". You can see that this discrete distribution
is an approximation of the continuous distribution.
Using the discrete distribution, we can determine
a discrete distribution for the sum. The trick is to do this
is as follows:
Select a particular value for the sum. (It
can only be either 1, 2, 3, ..., 19.)
For the particular value for the sum, determine
the number of ways this can be obtained by using the discrete
values of the values to be added. The probability of this sum
being produced is proportional to the number of ways it can be
obtained.
This is represented in Figure 25 where the
first three components of the distribution are shown with indications
of how many ways they can be obtained.
The results of calculating all of the possible
sums is shown in Figure 26. Since there are 100 possible sets,
[2(1+2+3+4+5+6+7+8+9)+10 = 100], then the actual probabilities
for each outcome is found by dividing the number of possible ways
to achieve a sum by. Hence the probability that the sum obtained
by adding the values on a random 'x' and 'y' ball is 0.3 is 3/100
= 0.03.
It is easy to see that for the continuous case,
the probability distribution function would be as shown in Figure
28. (Why is the maximum value 1?)
A More General Case
The probability distribution functions above
all had random values distributed between 0 and 1. What would
be the probability distribution if this were not so? Suppose
that the 'x' values were uniformly distributed between 6 and 9,
and the 'y' values were uniformly distributed between 3 and 5
for example. Clearly the sum has to be between 9 (6+3) and 14
(9+5), but what does the probability distribution function really
look like. Using a technique equivalent to enumerating all of
the possible ways a particular sum can be obtained, the distribution
shown in Figure 29 would result.
A Computer Technique for Determining Probability
Density Functions: The Monte Carlo Method
It is possible to obtain probability density
functions in the manner described above (by selecting balls from
a container) but by using a computer to do the selecting. The
general technique, termed the "Monte Carlo method",
is conceptually very simple.
Basically the computer must be able to calculate
outputs of concern for a given set of input parameters. It is
the effect of variations, or tolerances, in these input parameters
that are of concern. The program sets realistic values of these
parameters by generating random numbers using the nominal values
and tolerances or the parameters and makes a single run. The
outputs, generated in this single run, are saved. Another set
of parameter values are set using new random numbers and another
run is made. The outputs again are saved. Enough runs are made
so that the probability characteristics of the saved outputs can
be determined.
The computer is fast enough so that it can
run run enough cases in a reasonable amount of time to obtain
the desired information.
Monte Carlo method requires that a routine
for generating random numbers be available. Most programming
languages have such a function built in, and that is true about
FORTRAN.
APPENDIX B illustrates computer programs for
determining probability density functions using the Monte Carlo
method.
Part 2 Exercise: Probability
Density Functions
The objective of these exercises is for you
to get an appreciation of random numbers and the probability density
function so that it can be applied to the interval timer behavior.
Logon to the VAX
Copy the three files RAND1.FOR, RAND2.FOR and
RAND3.FOR to your account using the following VMS command:
Compile, link and run them using the following
sequence of commands:
(Do the 20 numbers produced by RAND1 look like
they are uniformly distributed between 0 and 1?)
(In RAND2, 10000 uniformly distributed random
numbers with a range from 0 to 1 are sorted into 10 bins. Do
the bins contain roughly the same values? They should. See Figure
22.)
(RAND3 generates a character plot, so it is
necessary to link plot routines with RAND3.OBJ. This is what
the LINK statement above does. Does the plot look like Figure
28? It should.)
Modifying the Programs a Little
a) Change RAND1.FOR (using the EDIT command) so that it uses a different value for the seed.
Are the numbers it produces different from
the original ones?
b) Change RAND2.FOR so that the number of BINS used is 20 rather than 10.
Are the values in the BIN array about half
of the original values?
Probability Density Function for the
Sum of 3 Uniformly Distributed Random Numbers
Modify RAND3.FOR so that it generates a plot
of the probability function for the sum of 3 random numbers rather
than 2.
Use the approach suggested in Figure 19 to
determine the probability density function on paper.
To do this find the number of ways the discrete
values of the following 2 probability density functions can be
added to produce a particular sum of three numbers (The sum must
be between 0 and 3).
IMPORTANT: Don't do this by listing all the
possible sums. Try to find a pattern.
Modify RAND3.FOR so that it generates the probability
density function for the sum of 2 random numbers. The first number
is uniformly distributed between 3 and 5, the second number between
6 and 9. The resulting plot should look like Figure 29.
Use 70 BINs between the value 8 and 15 (so
each bin has a width of 0.1).
When the tabular results of the measured time
interval, resistance value, and capacitance value for the set
of students are available (in the form of a spreadsheet), develop
a probability density function as described below.
Add the average and standard deviation to your
"bins" plot.
Notice that the ratio of the measured time
to the calculated time is nearly constant with a value of about
1.06. This systematic error indicates a deficiency in the model
used to derive the time interval equation.
