PRACTICAL MATLAB® FOR ENGINEERS PRACTICAL MATLAB

618 Practical MATLAB® Applications for Engineers

w =-5*a:10000:5*a;

t =linspace(0,2e-3,1000);

f0 = 1000; % input frequencies

f1 = 35000;

input0 =5.*cos(2*pi*f0.*t);input1=12.5*sin(2*pi*f1.*t);

input = input0+input1; % input signal

H0lp = a./(j*2*pi*f0+a); % mag of LPF at f = f0

H1lp = a./(j*2*pi*f1+a); % mag of LPF at f = f1

resp0lp = abs(H0lp).*(cos((2*pi*f0.*t)+angle(H0lp)));

resp1lp = abs(H1lp).*(sin((2*pi*f1.*t)+angle(H0lp)));

outputlp = resp0lp+resp1lp; % output of LPF

H0hp = j*2*pi*f0./(j*2*pi*f0+a); % mag of HPF at f = f0

H1hp = j*2*pi*f1./(j*2*pi*f1+a); % mag of HPF at f = f1

resp0hp = abs(H0hp).*(cos((2*pi*f0.*t)+angle(H0hp)));

resp1hp = abs(H1hp).*(sin((2*pi*f1.*t)+angle(H0hp)));

outputhp = resp0hp+resp1hp; % output of HPF

figure(1)

subplot(3,1,1)

plot(t,input);

axis([0 2e-3 -18 18]);

ylabel(‘Input Amplitude’);

subplot(3,1,2)

plot(t,outputlp);

ylabel(‘Output Amplitude’);

axis([0 2e-3 -1.5 1.5]);

subplot(3,1,3)

plot(t,input0);

axis([0 2e-3 -6.5 6.5]);

ylabel(‘Amplitude’);

xlabel(‘time in sec’)

figure(2)

subplot(3,1,1)

plot(t,input);

axis([0 2e-3 -18 18]);

ylabel(‘Input Amplitude’);

subplot(3,1,2)

plot(t,outputhp);

ylabel(‘Output Amplitude’);

axis([0 .1e-3 -1.2 1.2]);

subplot(3,1,3)

plot(t,input1);

a xis([0 .1e-3 -12.5 12.5]);

ylabel(‘Amplitude’);

xlabel(‘time in sec’)

Back in the command window, the script fi le test_fi lter is executed and the fi lters’ out-

puts are shown in Figures 6.45 and 6.46.

Note that the LPF output consists of a 1000 Hz sinusoidal wave plus a small high-

frequency 35 kHz ripple. The HPF passes the 35 kHz frequency, while completely

removing the low-frequency component of 1000 Hz.