clear all; close all; clc;

%% Question 5

t = -2:0.01:2;
y = rect(t);
plot(t, y);
xlim([-3 3]);
ylim([-2 2]);
xlabel('t');
ylabel('rect(t)');
title('rect(t) vs t');


%% Question 6

T0 = 2;
f0 = 1/T0;
Ts = T0/10;
t = -10*T0:Ts:10*T0;
s = sinc(f0*t);

figure(2);
plot(t, s);
xlabel('t');
ylabel('s(t)');
title('Signal s(t) vs t');
grid on;

% We observe that A sinc pulse passes through zero 
% at all positive and negative integers (i.e., t = ±1, ±2, ±3, ...),
% but at t=0, it reaches its maximum of 1.

%% Question 7

t = -2:0.01:2;

f1 = 10;
x1 = 2*sin(2*pi*f1*t);
figure(3);
plot(t, x1);
xlabel('t');
ylabel('x1(t)');
title('Signal x1(t) vs t');

x2 = 2 + x1;
figure(4);
plot(t, x2);
xlabel('t');
ylabel('x2(t)');
title('Signal x2(t) vs t');

f3 = 10*f1;
x3 = sin(2*pi*f3*t);
figure(5);
plot(t, x3);
xlabel('t');
ylabel('x3(t)');
title('Signal x3(t) vs t');

x4 = x1.*x3;
figure(6);
plot(t, x4);
xlabel('t');
ylabel('x4(t)');
title('Signal x4(t) vs t');

%% Question 8

% 8.a

Ts = 1/50;
t = 0:Ts:10-Ts;                   
x = sin(2*pi*15*t) + sin(2*pi*20*t);

% 8.b

X = fft(x);   
fs = 1/Ts;
f = (0:length(X)-1)*fs/length(X);

% 8.c

figure(7);

subplot(2,1,1);
plot(t,x);
xlabel('Time (seconds)')
ylabel('Amplitude')

subplot(2,1,2);
plot(f,abs(X));
xlabel('Frequency (Hz)')
ylabel('Magnitude')
title('Magnitude')

% 8.d

% When we plot the magnitude of the signal as a 
% function of frequency, the spikes in magnitude 
% correspond to the signal's frequency components 
% of 15 Hz and 20 Hz.

%% 8.e

% Changing the length of the signal 
% keeping it 5 seconds (instead of 10 seconds)

Ts = 1/50;
t = 0:Ts:5-Ts;                   
x = sin(2*pi*15*t) + sin(2*pi*20*t);

X = fft(x);   
fs = 1/Ts;
f = (0:length(X)-1)*fs/length(X);

figure(8);

subplot(2,1,1);
plot(t,x);
xlabel('Time (seconds)')
ylabel('Amplitude')

subplot(2,1,2);
plot(f,abs(X));
xlabel('Frequency (Hz)')
ylabel('Magnitude')
title('Magnitude')

% We find that there is no difference in the solution
% except for the case that A longer FFT will divide up 
% the frequency range into more "bins" between 0 and Fs/2. 
% So the result bin of a particular frequency may be 
% in a different bin number for different lengths of FFT.

%% 8.f

% Changing the sampling period of the signal 
% keeping it 1/20 seconds (instead of 1/50 seconds)

Ts = 1/20;
t = 0:Ts:10-Ts;                   
x = sin(2*pi*15*t) + sin(2*pi*20*t);

X = fft(x);   
fs = 1/Ts;
f = (0:length(X)-1)*fs/length(X);

figure(9);

subplot(2,1,1);
plot(t,x);
xlabel('Time (seconds)')
ylabel('Amplitude')

subplot(2,1,2);
plot(f,abs(X));
xlabel('Frequency (Hz)')
ylabel('Magnitude')
title('Magnitude')

% Since the FFT puts the signal into as many frequency bins 
% as there are signal samples, so more samples we have 
% in the time domain, the better our resolution in the 
% frequency domain will be that is more bins 
% therefore smaller bandwidth of each bin

%% 8.g

Ts = 1/500;
Vth = 1;
Vin = 5; 
t = 0:Ts:1;
fs = 10;
y1 = Vin*sin(2*pi*fs*t);
y = Vin*sign(y1 - Vth);


X = fft(y);   
f = (0:length(X)-1)*fs/length(X);

figure(10);

subplot(2,1,1);
plot(t,y);
xlabel('Time (seconds)')
ylabel('Amplitude')

subplot(2,1,2);
plot(f,abs(X));
xlabel('Frequency (Hz)')
ylabel('Magnitude')
title('Magnitude')

% we get the frequency value in the multiples of 0.2
% like 0.2, 0.4, 0.6, etc using a sampling frequency of 10 Hz and 
% sampling period of 1/500


%% Functions

function [y] = rect(t)
    y = zeros([1 numel(t)]);
    idx = (t<=0.5) & (t>=-0.5);
    y(idx) = 1;
end

function [y] = sinc(t)
    y = sin(pi*t)./(pi*t);
end