PRACTICAL MATLAB® FOR ENGINEERS PRACTICAL MATLAB

Fourier and Laplace 389

ANALYTICAL Solution

Part 1

ft()
 Fen jnw t







∞

∞

∑^0

FedtTwn


jnw t

1

2

12

2

2

0

05

05

() ; 0

.

.







∫ then =





Fedt

jn

n

jnw 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