Engineering 25
Problems
1.[23, 26]
Solutions
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
Engineering-25: Computational Methods
1
Bruce Mayer, PE
ENGR-25_Prob_9_3_Solution.ppt
P1.23 FT approximation
 Consider the Step Function
 1 for
f x   
 1 for
0 x 
0 x 
 The Fourier
Transform
4 k  sin 2k  1x 
f x   
for the
 k 1
2k  1
above Function
 Taking the First FOUR
Terms of the Infinite Sum
4  sin  x  sin 3 x  sin 5 x  sin 7 x 

f x   



 
 1
3
5
7

Engineering-25: Computational Methods
2
Bruce Mayer, PE
ENGR-25_Prob_9_3_Solution.ppt
P.1-23 FT m-file & plot
% Bruce Mayer, PE
% EGNR25 * 23Jun11
% file = P1_23_fft_1106.m
%
delx = 1e-6 % for small increment on either side of ZERO
%
% Make STEP-function plotting vector
x1 = [-pi,0-delx, 0+delx, pi]
f1 = [-1,-1,1,1]
%
% Make FFT plotting vectors
x2 =-pi:0.01:pi; % Row Vector Containing 629 elements
% Calc the value of f at all 629 points using elementby-element addition
f2 = (4/pi)*(sin(x2)/1 + sin(3*x2)/3 + sin(5*x2)/5 +
sin(7*x2)/7);
%
% Plot Both with a grid
plot(x1,f1, x2,f2, 'linewidth',3), grid
1.5
1
0.5
0
-0.5
-1
-1.5
-4
-3
-2
-1
Engineering-25: Computational Methods
3
0
1
2
3
4
Bruce Mayer, PE
ENGR-25_Prob_9_3_Solution.ppt
P1.23 MATLAB m-file
% Bruce Mayer, PE
% EGNR25 * 23Jan11
% file = P1_23_fft_1201.m
%
%
delx = 1e-6 % for small increment on either side of ZERO
%
% Make STEP-function plotting vector
x1 = [-pi,0-delx, 0+delx, pi]
f1 = [-1,-1,1,1]
%
% Make FFT plotting vectors
x2 =-pi:0.01:pi;
f2 = (4/pi)*(sin(x2)/1 + sin(3*x2)/3 + sin(5*x2)/5 + sin(7*x2)/7);
%
% Plot Both with a grid
plot(x1,f1, x2,f2, 'linewidth',3), grid, title('Vectorized Solution')
%
disp('Showing NONLoop plot; hit ANY KEY to continue')
pause
%
%
% try using FOR Loop and element-by-element addition to calc
components of the FT
fTot = zeros(1, length(x2)); % initialize total as array of zeros same
length of x2
for k = 1:2:7
fc = sin(k*x2)/k;
fTot = fTot + fc; % element-by-element addition
end
%
% Now mult fTot Vector by 4/pi to create f3
f3= (4/pi)*fTot;
%
% use same x2 as before to plot f3
plot(x1,f1, x2,f3, 'linewidth',3), grid, title('Loop and zeros
Solution')
Engineering-25: Computational Methods
4
Bruce Mayer, PE
ENGR-25_Prob_9_3_Solution.ppt
P1.23 MATLAB Plots
Vectorized Solution
1.5
1
0.5
0
-0.5
-1
-1.5
-4
-3
-2
-1
0
1
2
3
4
Loop and zeros Solution
1.5
1
0.5
0
-0.5
-1
-1.5
-4
-3
-2
-1
Engineering-25: Computational Methods
5
0
1
2
3
4
Bruce Mayer, PE
ENGR-25_Prob_9_3_Solution.ppt
P1-26 Law of CoSines
 Given Irregular Quadrilateral
 By Law of CoSines
a  b  c  2b1c1 cos A1
2
a  b  c  2b2c2 cos A2
2
2
1
2
2
Engineering-25: Computational Methods
6
2
1
2
2
Bruce Mayer, PE
ENGR-25_Prob_9_3_Solution.ppt
P1-26 Law of CoSines
 Given Quantities 
 Find c2 when
b1 = 180 m
b2 = 165 m
c1 = 115 m
A1 = 120°
A2 = 100°
• Instructor to WhtBd to Build
Quadratic Eqn in c2
Engineering-25: Computational Methods
7
Bruce Mayer, PE
ENGR-25_Prob_9_3_Solution.ppt
P1-26 Law of CoSines
 Build Quadratic Eqn in c2
• Instructor to Write m-file from
Scratch (start w/ blank m-file)
Engineering-25: Computational Methods
8
Bruce Mayer, PE
ENGR-25_Prob_9_3_Solution.ppt
P1-26 mFile & Results
% Bruce Mayer, PE
% EGNR25 * 23Jun11
% file = P1_26_LawOfCos_1106.m
%
%
% Set Known ParaMeters
b1 = 180; b2 = 165; c1 = 115; % all in meters
A1 = 120; A2 = 100; % all in Degrees
%
% Calc the constant a^2
asq = b1^2 + c1^2 -2*b1*c1*cosd(A1) % note use of cosd
%
% Make Quadratic PolyNomial
c2Poly = [1, -2*b2*cosd(A2), (b2^2 - asq)]
%
% Find roots of PolyNomial
c2roots = roots(c2Poly)
%
% NOTE: since c2 is a DISTANCE it MUST be POSITIVE
asq =
66325
c2Poly =
1.0e+004 *
0.0001
0.0057
-3.9100
c2roots =
-228.4542
171.1503
ANSWER
Engineering-25: Computational Methods
9
Bruce Mayer, PE
ENGR-25_Prob_9_3_Solution.ppt
Another Solution
% Bruce Mayer, PE
% 20Aug12 * ENGR25
% file P1_26_LawOfCos_Alternative_1208.m
%
% Side Lengths in meters
b1 = 180, b2 = 165, c1 = 115 % commas do NOT
suppress ReadBack
%
% Angles in Degrees
A1 = 120; A2 = 100; % semicolons DO suppress
ReadBack
%
% calc parameter "asqd"
asqd = b1^2 + c1^2 - 2*b1*c1*cosd(A1)
% note use of cosd; not cos. cos operates on
radians
%
c2QuadCoeff = [1 -2*b2*cosd(A2) b2^2-asqd]
% find Roots of Quadratic
c2 = roots(c2QuadCoeff)
% Note that c2 is a DISTANCE and hence MUST
be POSITIVE
Engineering-25: Computational Methods
10
Bruce Mayer, PE
ENGR-25_Prob_9_3_Solution.ppt