%% Signal in Time Domain
% Use Fourier transforms to find the frequency components of a signal buried in noise.
% Specify the parameters of a signal with a sampling frequency of 1 kHz and a signal duration of 1.5 seconds
Fs = 1000; % Sampling frequency
T = 1/Fs; % Sampling period
L = 1500; % Length of signal
t = (0:L-1)*T; % Time vector
% Form a signal containing a 50 Hz sinusoid of amplitude 0.7 and a 120 Hz sinusoid of amplitude 1.
S = 0.7*sin(2*pi*50*t) + sin(2*pi*120*t);
% Corrupt the signal with zero-mean white noise with a variance of 4.
X = S + 2*randn(size(t));
% Plot the noisy signal in the time domain. It is difficult to identify the frequency components by looking at the signal X(t).
subplot(311);
plot(1000*t(1:200),X(1:200), 'b')
title('Signal Corrupted with Zero-Mean Random Noise')
xlabel('t (milliseconds)')
ylabel('X(t)')
hold on
plot(1000*t(1:200),S(1:200),'r')
hold off
% Signal in Frequency Domain
% Compute the Fourier transform of the signal.
Y = fft(X);
% Compute the two-sided spectrum P2. Then compute the single-sided spectrum P1 based on P2 and the even-valued signal length L.
P2 = abs(Y/L);
P1 = P2(1:L/2+1);
P1(2:end-1) = 2*P1(2:end-1);
A2 = angle(Y);
A1 = A2(1:L/2+1);
% Define the frequency domain f and plot the single-sided amplitude spectrum P1.
% The amplitudes are not exactly at 0.7 and 1, as expected, because of the added noise.
% On average, longer signals produce better frequency approximations.
f = Fs*(0:(L/2))/L;
subplot(3,1,2);
plot(f,P1, 'b')
title('Single-Sided Amplitude Spectrum of X(t)')
xlabel('f (Hz)')
ylabel('|P1(f)|')
hold on
% Now, take the Fourier transform of the original, uncorrupted signal and retrieve the exact amplitudes, 0.7 and 1.0.
Y = fft(S);
P2 = abs(Y/L);
P1 = P2(1:L/2+1);
P1(2:end-1) = 2*P1(2:end-1);
plot(f,P1, 'r')
title('Single-Sided Amplitude Spectrum of S(t)')
xlabel('f (Hz)')
ylabel('|P1(f)|')
hold off
% Angulo / fase
subplot(3,1,3);
plot(f,A1)
%ylim([0 1.05])
title('Single-Sided Phase Spectrum of X(t)')
xlabel('f (Hz)')
ylabel('phase(f)')