Fourier and Laplace 389
ANALYTICAL Solution
Part 1
ft() Fen jnw t
∞
∞
∑^0
FedtTwnjnw t
1
2
12
2
2
0
05
05
() ; 0
.
.
∫ then =
Fedt
jn
njnw t ejnw t
1
2
1
1
2
00
05
05
05
05
()
.
.
.
.
∫ |
F
n
n n
1
2
2
2
sin( )
()
Therefore,
ft
n
n n
()
sin( )
()
1
2
1
2
2
1 2
∞
∑
MATLAB Solution
% Script file: Fourier _ series
n = -5:1:5; % creates harmonics
% add eps to avoid division by zero
Fn = 0.5*sin(n.*pi/2)./((n.*pi/2)+eps); % creates the coefficients Fn
t =linspace (-2,2,50); % limits from -1 to 1,one
period
F1=Fn(5)*exp(j*pi.*t)+Fn(7)*exp (-j*pi.*t); % first component
F2=Fn(4)*exp(j*2*pi.*t)+Fn(8)*exp (-j*2*pi.*t);
% second component
F3=Fn(3)*exp(j*3*pi.*t)+Fn(9)*exp (-j*3*pi.*t);
% third component
F4=Fn(2)*exp(j*4*pi.*t)+Fn(10)*exp (-j*4*pi.*t);
% fourth component
F5=Fn(1)*exp(j*5*pi.*t)+Fn(11)*exp (-j*5*pi.*t);
% fifth component
figure(1); % see Figure 4.40; part (2a)
F0 = .5.*ones(1,50);
subplot (3,2,1);plot (t,F0);
title (‘DC component’); ylabel (‘Amplitude (F0)’)
subplot (3,2,2);plot (t,F1);
FIGURE 4.39
Plot of f(t) of Example 4.4.
f(t)
1.5
−0.5 0.5
0.5
− 1 1
T
t