Signals and Systems - Electrical Engineering

288 C H A P T E R 4: Frequency Analysis: The Fourier Series

FIGURE 4.18

Two periods of the approximate triangular signal

xN(t)using 100 harmonics.

0 0.5 1 1.5 2

0

0.2

0.4

0.6

0.8

1

xN

(t

)

t(sec)

and the average term is

Z 0 =−

∑

m6= 0

Gm

1

j 2 mπ

=

∑

m6= 0

(− 1 )m

cos(πm)− 1

(πm)^2

=0.5

∑∞

m=−∞,m6= 0

(− 1 )m+^1

[

sin(πm/ 2 )

(πm/ 2 )

] 2

where we used 1−cos(πm)=2 sin^2 (πm/ 2 ). We used the following script to obtain the average,

and to approximate the triangular signal using 100 harmonics (see Figure 4.18). The mean is

obtained as 0.498.

%%%%%%%%%%%%%%%%%

% Example 4.18

%%%%%%%%%%%%%%%%

clf; clear all

w0 = 2 * pi; N = 100; % parameters of periodic signal

% computation of mean value

DC = 0;

for m = 1:N,

DC = DC + 2 * (-1) ˆ (m) * (cos(pi * m) -1)/(pi * m) ˆ 2;

end

% computation of Fourier series coefficients

Ts = 0.001; t = 0:Ts:2 - Ts;

for k = 1:N,

X(k) = (-1)ˆ(k + 1)*(cos(pi * k) - 1)/((pi * k)ˆ2);

end

X = [DC X]; % Fourier series coefficients

xa = X(1)*ones(1,length(t));

figure(1)