Question

Sampling 2 periods of each signal in (below) with 100 samples and plot the signal with respect to the same frequency reference signal cos(ω₀t):

a.x_a(t)=50sin(100t+0.2)

b.x_b(t)=25cos(5000t−π/4)

c.x_c(t)=50sin(1000t+ π/4)−20cos(1000t−0.5)

d.x_d(t)=30sin(3000(t−5×10⁻⁵))

e.x_e(t)=25cos(5000t−2.25π)

f.x_f(t)=−25cos(5000t+3π/4)

Answer 1

% Plot for Xa(t)

w = 100;
T = 2*pi/w; % Fundamental Time period for one cycle
Ts = T/100; % Ts = 1/Fs i.e. 100 samples per period
t = 0:Ts:(2*T); % For 2 periods 2 * T
x = 50*sin(100*t + 0.2);
plot(t,x);



% Plot for Xb(t)

w = 5000;
T = 2*pi/w; % Fundamental Time period for one cycle
Ts = T/100; % Ts = 1/Fs i.e. 100 samples per period
t = 0:Ts:(2*T); % For 2 periods 2 * T
x = 25*cos(100*t - pi/4);
plot(t,x);



% Plot for Xc(t)

w = 1000;
T = 2*pi/w; % Fundamental Time period for one cycle
Ts = T/100; % Ts = 1/Fs i.e. 100 samples per period
t = 0:Ts:(2*T); % For 2 periods 2 * T
x = 50*sin(1000*t + pi/4) - 20*cos(1000*t - 0.5);
plot(t,x);

% Plot for xd(t)

w = 3000;
T = 2*pi/w; % Fundamental Time period for one cycle
Ts = T/100; % Ts = 1/Fs i.e. 100 samples per period
t = 0:Ts:(2*T); % For 2 periods 2 * T
x = 30*sin(3000*(t - 5*10 - 5));
plot(t,x);


%%%%%%%%%%%%%%%%%%%%%%%%
% plot for xe(t)
w = 5000;
T = 2*pi/w; % Fundamental Time period for one cycle
Ts = T/100; % Ts = 1/Fs i.e. 100 samples per period
t = 0:Ts:(2*T); % For 2 periods 2 * T
x = 25*cos(5000*t - 2.25*pi);
plot(t,x);



%%%%%%%%%%%%%%%%%%%%%%%%%%
% Plot for xf(t)

w = 5000;
T = 2*pi/w; % Fundamental Time period for one cycle
Ts = T/100; % Ts = 1/Fs i.e. 100 samples per period
t = 0:Ts:(2*T); % For 2 periods 2 * T
x = -25*cos(5000*t + 3*pi/4);
plot(t,x);