University of Waterloo
Faculty of Engineering
Department of Electrical and Computer Engineering
NE 216
Laboratory 1
FALL 2011
Prepared by
Surname/Last Name, Legal Given/First Name(s)
UW Student ID Number: 2NNNNNNN
UW User ID: uwuserid @uwaterloo.ca
Prepared by
Surname/Last Name, Legal Given/First Name(s)
UW Student ID Number: 2NNNNNNN
UW User ID: uwuserid @uwaterloo.ca
2A Nanotechnology Engineering
Today’s Date
1.0 Differentiating your UW User ID and your UW Student ID Number
Your UW User ID is up to eight letters or numbers that allows you to log onto Quest,
Unix, etc. Enter it your:
Your UW User ID
Your UW Student ID number is eight digits starting with 2 that uniquely identifies you as
a student at the University of Waterloo. Enter it your UW Student ID Number here:
Your UW Student ID Number
In general, do not give your UW Student ID Number to anyone except those working in
an official capacity with the University of Waterloo. Especially, do not include it
automatically in your signature; however, you should include both your UW User ID and
your UW Student ID Number in any e-mail correspondence with a university official.
In balance of this laboratory will look at halting conditions.
1.1 Determining the problem of non-convergence
Any numerical algorithm may never actually halt—perhaps because there is no solution,
or there may be a characteristic of the problem that does not allow a solution to be found.
For example, the following is an implementation of the secant method. Given two points,
we find the interpolating line that passes through them and then we find the root of that
line.
Figure 1. Interpolating two points and finding
the root of the interpolating polynomial.
2
function [x2] = secant( f, x0, x1 )
while true
x2 = ...;
if x1 == x2
return;
end
x0 = x1;
x1 = x2;
end
end
As a reminder, File→New→Function and cut-and-paste the above code into the editor
and the name of the file must be the name of the function followed with a ".m"; in this
case, the file name would be "secant.m".
The interpolating polynomial is
y  f  x0 
x  x0
x  x1
 f  x1 
x0  x1
x1  x0
0  f  x0 
x  x0
x  x1
 f  x1 
x0  x1
x1  x0
Solve the equation
for x which is the root of the interpolating polynomial (the blue point in Figure 1) and
assign that value to x2.
You can try your function out:
>> secant( @cos, 1, 1.1 )
ans =
1.570796326794897
Once this works, you can continue.
3
What happens if you execute
>> secant( @log, 4, 2 )
? You should make use of Ctrl-C in order to halt your program. To observe what is
happening, you can either remove a semi-colon from one of the lines in your function
(this will print it to the screen as the function executes) or you can use the debugger by
clicking to the left of the line you want Matlab to stop at, as shown in Figure 2.
Figure 2. Using the debugger.
The next time you run, you be in the debugger as soon as the execution reaches the line
you highlighted. To clear the break point, click on the break point a second time. In the
Matlab command window, you can now determine the value of a variable:
K>> x0
x0 =
1.100000000000000
4
You can now perform one of a number of options, as is shown in Figure 3.
Figure 3. Options available in debugging.
These options perform the following actions:
1. Step: execute the statement we are currently on. If it is a for, while or if
statement or a function call, it will execute the entire loop, condition, or function
call.
2. Step in: execute the statement we are currently on. If it is a for, while or if
statement or a function call, step into the body of that structure and stop at the
first statement in that for, while or if statement, or function call.
3. Step out: whatever body you are in, whether it be a for, while or if statement or a
function call, continue executing it to the end and stop at the next statement
following it.
4. Continue: Continue executing until either we return to the Matlab prompt or we
reach the next break point.
5. Exit debug mode: This should be self evident....
These names are visible if you hover over the icons.
Based on your use of the debugger and the information that was provided during the
presentation, why is secant( @log, 4, 2 ) going into an infinite loop?
Your answer here.
5
In the Matlab command window, enter your UW User IDs and take a screen shot of the
Matlab (in Windows, Ctrl-Alt-PrtScn) and paste that image (Ctrl-V) into Figure 4. You
should be see the offending value of x2 that is causing Matlab to go into an infinite loop.
Figure 4. Your screenshot.
6
1.2 Limiting the number of iterations.
Thus, rather than using a while-true loop, we will allow the user to pass a second
parameter, N_max, the maximum number of iterations.
function [x2] = secant( f, x0, x1, N_max )
for i = 1:N_max
% the body of the while loop here...
end
end
Notice that the parameters that define the mathematical problem come first (we want to
find a root of f(x) using the secant method starting with x0 and x1. The parameter Nmax
determines the behaviour of the numerical algorithm and does not affect the problem, so
we place it at the end.
Thus, if the function iterations more than Nmax times and does not converge, the for loop
will end and the function will exit. At this point, we must indicate that there is a
problem; otherwise, the function will just return the last value of x2—something we did
not want because it is not valid. Instead, we will signal that an exception has occurred;
that is, we will throw an exception:
throw( MException( 'MATLAB:numeric_exception', ...
'..' ) );
In this case, the throw command indicates a message is to be sent. That message is an
MException and the first argument of MException is the message identifier. The
identifier must be a string starting with the seven characters 'Matlab:' and the
remaining characters (letters, numbers, or underscores) should describe what is
happening. In the above, the identifier is that a numeric exception has occurred.
After the for loop, replace the .. add a message that you think would be useful for the
user. In this message, you can use regular any character on the keyboard with a few
exceptions which you can find if you do a search for strings in the Matlab help dialog
shown in Figure 5.
7
Figure 5. The Matlab help dialog.
First, save the function
function [y] = f1a( x )
y = x.^2 + 1;
end
and then test your output by running the two instructions:
secant( @sin, 3, 3.1, 20 )
secant( @f1a, 3, 3.1, 20 )
Copy and paste your Matlab commands and the output here.
8
1.3 When to Stop in the x-direction
Consider the following example:
function [y] = f1b( x )
y = cos(1./x);
end
When the following command
>> secant( @f1b, 0.000722, 0.0007221, 100 )
is executed, at least with the author’s implementation, the algorithm goes into an infinite
loop with the values of x2 being:
7.226104113139403e-004
7.226104113139402e-004
7.226104113139401e-004
7.226104113139403e-004
7.226104113139402e-004
7.226104113139401e-004
7.226104113139403e-004
As you can see, even though the correct answer is 0.00072261041131394023···, the
answer oscillates around the solution. Thus, we really should stop, but what value of x
should we return as the approximation of the root? In addition, is there any engineer in
the world who really needs sixteen digits of precision, or even 10? Would not under
many conditions an approximation of 0.00072261 be good enough; after all, the relative
error of this approximation is 0.000057 %.
What we need is an alternate terminating condition: one that the user can control.
We will say that an approximation close enough if two approximations are
sufficiently close. That is, if x2  x1   step . If two successive approximations are
sufficiently close, we will also assume that x2  x*   step where x* is the actual
solution.
Thus, rather than checking if x1 == x2, we will instead check that
abs( x2 - x1 ) < eps_step
This parameter must also be passed. We pass it as the fourth parameter and move N_max
to the fifth location with the justification that eps_step indicates when to halt
successfully (relatively more important) while N_max indicates when to signal a failure.
9
function [x2] = secant( f, x0, x1, eps_step, N_max )
% Enter the body of your function here
end
Now, test your code here:
secant( @f1b, 0.000722, 0.0007221, 1e-4, 100 )
Copy and paste your Matlab commands and the output here.
10
1.4 When to Stop in the y-direction
Using the same function,
>> x1b = secant( @f1b, 0.000722, 0.0007221, 1e-6, 100 )
x1b =
7.230088870592349e-004
>> f1b( x1b )
ans =
0.690876363485700
>> x1b = secant( @f1b, 0.000722, 0.0007221, 1e-7, 100 )
x1b =
7.226118709557649e-004
>> f1b( x1b )
ans =
0.002795351509144
>> x1b = secant( @f1b, 0.000722, 0.0007221, 1e-8, 100 )
x1b =
7.226104111736519e-004
>> f1b( x1b )
ans =
-2.686662956663953e-007
Thus, in the first case, even though we stopped when successive steps in the x-direction
were sufficiently small, the answer was still not optimal: f1b( 0.000723) is very large—
around 0.69—and not something we would call a root! Thus, we cannot stop just
because step size is sufficiently small between approximations: we must also make sure
that we are getting something that is close enough to a real answer.
Therefore, we will add a second condition: f  x2    abs . Because both conditions must
be met, we must use && and add eps_abs as a fifth parameter, relegating N_max again to
the sixth location.
function [x2] = secant( f, x0, x1, eps_step, eps_abs, N_max )
% Enter the body of your function here
end
11
Now, when you execute:
x1 = secant( @f1b, 0.000722, 0.0007221, 1e-6, 1e-16, 100 )
f1b( x1b )
the answer should converge and the answer should be a reasonably good approximation
of a root.
Copy and paste your Matlab commands and the output here.
12
1.5 Comments
Next, any significant amount of code that you write today, even if you read it again six
months from now, you will likely not remember what you meant to do. It is therefore
best if you start now learning how to comment: get into the habit now and it will become
part of your routine. More importantly, when you are employed as an engineer, others
will be doing the coding for you and if you do not set good commenting standards, your
employees, too, will become lazy costing you your bonus (most significantly,
maintenance costs will increase—the author remembers spending in some cases hours
trying to determine why the original author of a piece of code wrote what he did—it was
wrong, but because there was no description or justification, it cost approximately $200
of developer time to determine the mistake—probably much more than it cost to have the
original author write those few lines of code.
Cut and paste the following into your code and then replace it once you have added
comments. Any red text must be removed.
% Secant method
%
%
%
%
%
%
%
%
%
%
%
%
%
%
%
%
Write a description of what the function is suppose to do here.
paragraph or two.
Parameters
==========
f
x0
x1
eps_step
eps_abs
N_max
It should be a
For each of these, replace the ... with a comment about
the parameter, what it means, what it controls.
...
...
...
...
...
...
Return Values
=============
x2
...
For each return value, indicate the significance.
In the body of the function, any significant block of code, for-loop, or ifstatement should have a comment before it indicating the purpose. This need
only be a few lines or even one line. Do not over comment--we do not need one
comment per line.
function [x2] = secant( f, x0, x1, eps_step, eps_abs, N_max )
% Enter the body of your function here
end
13
These are short-answer questions. Please note, all subsequent laboratories will not have
such a section. This is here to help your understanding of floating-point numbers.
1.6a General knowledge: if you accidently assign to a name that is being used as a
function, you get a very specific error. For example, consider
>> sum = sum( 1:100 );
What happens if you try to call sum again? The exact error message is “Enter Error
Message Here”. Does this differ from the error message when you try the following?
>> M = [1 2 3; 4 5 6; 7 8 9];
>> M( 1:100 )
Yes/No
You can unassign names that are accidently assigned by typing, for example,
>> clear sum
1.6b Split the following numbers into the sign bit, the eleven exponent bits, and the 52
mantissa bits:
format hex
10532.0947265625
ans =
40c4920c20000000
1 sign bit:
11 exponent bits:
52 mantissa bits:
You can use 0...0 to indicate all trailing zero bits. Consider using the Microsoft
Calculator with the View→Scientific option selected. Alternatively, you can also use:
0
1
2
3
4
5
6
7
0000
0001
0010
0011
0100
0101
0110
0111
8
9
a
b
c
d
e
f
14
1000
1001
1010
1011
1100
1101
1110
1111
1.6c Place an x in the appropriate column of Table 1 to indicate whether the doubleprecision floating-point numbers represent positive or negative values.
Table 1. Positive and negative floating-point numbers.
Representation
Positive Negative
3f293ac9d935310c
1ce0f5ff5675cf94
d933f307fea5b029
b5fe6a8a831e7905
28a43e05659feaf3
3744be1ee8f9f90c
x
1.6d For each of the numbers in Table 2, find the exponent as a decimal power of two.
As an example, the first three questions have already been completed:
Table 2 Exponents of double-precision floating-point numbers.
Representation
Exponent
3ff0000000000000
3f193ac9d935310c
4020f5ff5675cf94
4033f307fea5b029
408e6a8a831e7905
3f743e05659feaf3
3f84be1ee8f9f90c
3ef0f5ff5675cf94
0
-14
3
?
?
?
?
?
Note that you can check your answer by comparing the first three hexadecimal digits of
the number. For example, the first three hexadecimal digits of 2-14, 2-13, 22 and 23 are
3f1, 3f1, 401 and 402, respectively.
15
1.6e The variable eps is automatically assigned the distance from 1 to the next largest
floating-point number; that is, 1 + eps is not equal to 1:
>> 1
ans =
3ff0000000000000
>> 1 + eps
ans =
3ff0000000000001
Find the next smaller floating point number before 1 by subtracting off an appropriate
multiple of eps. Hints: it’s not 1 – eps and use format hex.
>> format hex
>> 1 - ???
% Enter your code and the output here
1.6f The variable eps is equal to 2–n for what value of n? What other number does this
value of n represent with respect to the double format?
Your answer here.
1.6g What is the correct mantissa of the following double-precision floating point
numbers shown in Table 3?
Table 3. Mantissas of floating-point numbers.
Representation
Mantissa
3f193a0000000000 1.1001001110100...0
4020f10000000000 1.0000111100010...0
4033f00000000000 1.????
414e600000000000 1.????
3f74300000000000 1.????
1.6h What is the smallest integer power n of 10 such that 10n is not a denormalized
number (that is, the exponent as hexadecimal numbers are not yet 000).
>> 1e-???
% Enter your code and the output here
1.6i What is the smallest integer power n of 10 such that 10n is not zero?
>> 1e-???
% Enter your code and the output here
1.6j What is the largest integer power n of 10 such that 10n is not infinity?
>> 1e???
% Enter your code and the output here
16
1.6k Calculate each of the following to determine the result.
Expression
Inf
Inf
Inf
Inf
+
*
/
Output
0
0
0
0
Inf
?
?
?
?
?
?
?
?
?
?
?
?
0 - Inf
0 / Inf
Inf + Inf
Inf – Inf
Inf * Inf
Inf / Inf
Inf^-Inf
0^0
Inf^0
1.6l Is the representation of NaN and -Nan different in as double-precision floating-point
numbers (use format hex)?
Yes or No?
1.6m You are given these five double-precision floating point numbers.
1ec08f8bf99a36fe
67caccf63d443628
6910b6f1ba816f22
46b6acf2673e096a
183dc6edceef302d
Order them from smallest to largest and place them in Table 4.
Table 4. Order of floating-point numbers.
Smallest 3ff0000000000000
Largest
3ff0000000000000
3ff0000000000000
3ff0000000000000
3ff0000000000000
17
1.6nYou are given these six double-precision floating point numbers.
cd617c809c482961
54d3c359e37c5187
4d41c8436fa8ce8e
bb825a24626f4112
c42c3c937cc00418
2d91ca5f2651c8b8
Order them from smallest (the largest negative number) to largest (the largest positive
number) and place your answer in Table 5.
Table 5. Order of floating-point numbers.
Smallest 3ff0000000000000
Largest
3ff0000000000000
3ff0000000000000
3ff0000000000000
3ff0000000000000
3ff0000000000000
18
Download

here - Electrical and Computer Engineering