Transcript Lecture 03 Matrices - NYU Polytechnic School of Engineering

Lecture 3
Creating M-files
Programming tools:
Input/output
(assign/graph-&-display)
Repetition (for)
Decision (if)
© 2007 Daniel Valentine. All rights reserved. Published by
Elsevier.
Review

Arrays
– List of numbers in brackets
 A comma or space separates numbers (columns)
 A semicolon separates row
– Zeros and ones Matrices:
 zeros()
 ones()
– Indexing
 (row,column)
– Colon Operator:
 Range of Data first:last or first:increment:last
– Manipulating Arrays & Matrices
 Transpose
Input

Examples of input to arrays:
– Single number array & scalar:
1×1
>> a = 2
– Row array & row vector:
1×n
>> b = [1 3 2 5]
– Column array & column vector:
nx1
>> b = b’ % This an application of the transpose.
– Array of n rows x m columns & Matrix: n × m
>> c = [1 2 3; 4 5 6; 7 6 9] % This example is 3 x 3.
Basic elements of a program

Input
– Initialize, define or assign numerical values to
variables.

Set of command expressions
– Operations applied to input variables that lead
to the desired result.

Output
– Display (graphically or numerically) result.
An example of technical computing
Let us consider using the hyperbolic tangent
to model a down-hill section of a snowboard
or snow ski facility.
 Let us first examine the hyperbolic tangent
function by executing the command:

>> ezplot( ‘tanh(x)’ )
We get the following graph:
Ezplot(‘tanh(x)’)
Hyperbolic tangent: tanh(X)

Properties (as illustrated in the figure):
– As X approaches –Inf, tanh(X) approaches -1.
– As X approaches Inf, tanh(X) approaches +1.
– X = 0, tanh(X) = 0.
– Transition from -1 to 1 is smooth and most
rapid (roughly speaking) in the range
-2<X<2.
– REMARK: With this familiarity with tanh(X), let
us apply it to solve the following problem.
Problem background
Let us consider the design of a slope that
is to be modeled by tanh(X). Consider the
range -3<X<3, and assume the slope
changes negligibly for |X|>3.
 Thus, for –3 < X < 3, –1 < tanh(X) < 1.
 We want to design a downhill slope such
that in 10 meters the hill drops 2 meters.
 Thus, as shown next, the following
formula is needed: y = 1 – tanh(3*X/5).

Formula for snow hill shape
Problem statement
Find the altitude of the hill at 0.5 meter
intervals from -5 meters to 5 meters using
the shape described and illustrated in the
previous two slides.
 Tabulate the results.

Structure plan

Structure plan
– Initialize the range of X in 0.5 meter intervals.
– Compute y, the altitude, with the formula:
y = 1 – tanh(3*X/5).
– Display the results in a table.

Implementation of the structure plan
– Open the editor by typing edit in the
command window.
– Translate plan into the M-file language.
SOLUTION TO IN-CLASS EXERCISE
Command window OUTPUT
This is from editor: It is lec3.m
%
%
%
%
%
%
%
%
Ski slope offsets for the
HYPERBOLIC TANGENT DESIGN
by D.T. Valentine...January 2007
Points at which the function
is to be evaluated:
Step 1: Input x
x = -5:0.5:5;
% Step 2: Compute the y offset,
% that is, the altitude:
y = 1 - tanh(3.*x./5);
%
% Step 3: Display results in a table
%
disp(' ') % Skips a line
disp('
X
Y')
disp([x' y'])
X
-5.0000
-4.5000
-4.0000
-3.5000
-3.0000
-2.5000
-2.0000
-1.5000
-1.0000
-0.5000
0
0.5000
1.0000
1.5000
2.0000
2.5000
3.0000
3.5000
4.0000
4.5000
5.0000
Y
1.9951
1.9910
1.9837
1.9705
1.9468
1.9051
1.8337
1.7163
1.5370
1.2913
1.0000
0.7087
0.4630
0.2837
0.1663
0.0949
0.0532
0.0295
0.0163
0.0090
0.0049
Use of repetition (‘for’)

Repetition: In the previous example, the
fastest way to compute y was used. An
alternative way is as follows:
Replace:
y = 1 - tanh(3.*x./5); % This is “vectorized” approach.
With:
for n=1:21
y(n) = 1 - tanh(3*x(n)/5);
end
Remark: Of course, the output is the same.
SOLUTION TO IN-CLASS EXERCISE
Command window OUTPUT
This is from editor: It is lec3_2.m
%
% Ski slope offsets for the
% HYPERBOLIC TANGENT DESIGN
% by D.T. Valentine...January 2007
%
% Points at which the function
% is to be evaluated:
% Step 1: Input x
x = -5:0.5:5;
% Step 2: Compute the y offset
for n=1:21
y(n) = 1 - tanh(3*x(n)/5);
end
%
% Step 3: Display results in a table
disp(' ') % Skips a line
disp('
X
Y')
disp([x' y'])
X
-5.0000
-4.5000
-4.0000
-3.5000
-3.0000
-2.5000
-2.0000
-1.5000
-1.0000
-0.5000
0
0.5000
1.0000
1.5000
2.0000
2.5000
3.0000
3.5000
4.0000
4.5000
5.0000
Y
1.9951
1.9910
1.9837
1.9705
1.9468
1.9051
1.8337
1.7163
1.5370
1.2913
1.0000
0.7087
0.4630
0.2837
0.1663
0.0949
0.0532
0.0295
0.0163
0.0090
0.0049
Decision: Application of ‘if’

Temperature conversion problem:
– Convert C to F or F to C.
Decision: Application of ‘if’

Temperature conversion problem:
– Convert C to F or F to C.
– SOLUTION:
%
% Temperature conversion from C to F
% or F to C as requested by the user
%
Dec = input(' Which way?: 1 => C to F? 0 => F to C: ');
Temp = input(' What is the temperature you want to convert? ');
%
% Note the logical equals sgn (==)
if Dec == 1
TF = (9/5)*Temp + 32;
disp(' Temperature in F: ')
disp(TF)
else
TC = (5/9)*(Temp-32);
disp(' Temperature in C: ')
disp(TC)
end
Summary
Introduced, by an example, the structure
plan approach to design approaches to
solve technical problems.
 Input: Assignment with and without ‘input’
utility.
 Output: Graphical & tabular were shown.
 Illustrated array dot-operation, repetition
(for) and decision (if) programming tools.
