Signals and Systems - Electrical Engineering

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

Because the magnitude response of the low-pass filter changes very little in the range of frequencies

of the input, the output signal is very much like the input (see Figure 4.15). The following script is

used to find the response.

% low-pass filtering

% FS coefficients of input

X(1) = 0; % mean value

for k = 2:M - 1,

X(k) = sin((k−1)∗pi/2)/((k−1)∗pi/2);

end

% periodic signal

Ts = 0.001; t1 = 0:Ts:1 - Ts;L = length(t1);

x1 = [ones(1, L /4) zeros(1, L /2) ones(1, L /4)]; x1 = x1−0.5; x = [x1 x1];

% output of filter

t = 0:Ts:2−Ts;

y = X(1)∗ones(1, length(t))∗Ha(1);

plot(t, y); axis([0 max(t)−.6 .6])

for k = 2:M - 1,

y = y + X(k)∗Hm(k)∗cos(w0∗(k−1).∗t + Ha(k));

plot(t, y); axis([0 max(t)−.6 .6]); hold on

plot(t, x, ’r’); axis([0 max(t)−0.6 0.6]); grid

ylabel(’x(t), y(t)’); xlabel(’t (sec)’) ; hold off

pause(0.1)

end

(a) (b)

0 20 40 60 80 100

0

0.5

1

|H(

jΩ

)|

0 20 40 60 80 100

− 1

−0.5

0

<H(

jΩ

)

Ω (rad/sec)

0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8

−0.5

−0.4

−0.3

−0.2

−0.1

0

0.1

0.2

0.3

0.4

0.5

x(

t),

y(

t)

t(sec)

FIGURE 4.15

(a) Magnitude and phase response of the low-pass RC filterH(s)at harmonic frequencies, and (b) response due

to a train of pulses. n