% EE4541 Fall 2008
% Assignment01 - Prob 8, Moving average filter
% due Thursday Sept 11

% initialization
clear all; close all; clc;

% load Assignment01.mat
load Assignment01.mat   % eeg_ch_1, eeg_ch_2, and Fs exist

% plot 1
t = (0:length(eeg_ch_1)-1) / Fs;
figure; 
subplot(211); plot_1_1 = plot(t, eeg_ch_1);
set(plot_1_1, 'LineWidth', 1.25);
title('EEG Ch 1');
ylabel('[uV]');

subplot(212); plot_1_2 = plot(t, eeg_ch_2);
set(plot_1_2, 'LineWidth', 1.25);
title('EEG Ch 2');
xlabel('Time [sec]');
ylabel('[uV]');

% plot 2
figure;
plot_2 = plot(t, eeg_ch_1, t, eeg_ch_2);
set(plot_2, 'LineWidth', 1.25);
title('EEG waveforms');
xlabel('Time [sec]');
ylabel('[uV]');
legend('EEG Ch 1', 'EEG Ch 2');

% Hints
% a. refer to 'filter' function
% b. refer to plot 2, especially 'set' and 'legend' functions
% c. moving average filter's characteristics
% d. order = # of input terms - 1

% solution
B = ones(15,1)/15;  % the order 14 filtering vector
filtered_ch_1 = filter(B, 1, eeg_ch_1);
filtered_ch_2 = filter(B, 1, eeg_ch_2);

% plot
figure;
subplot(211); plot_3 = plot(t, eeg_ch_1, t, filtered_ch_1);
set(plot_3, 'LineWidth', 1.25);
title('EEG waveforms, order = 14');
xlabel('Time [sec]');
ylabel('[uV]');
legend('EEG Ch 1', 'filtered EEG Ch 1', 'Location', 'NW');

subplot(212); plot_4 = plot(t, eeg_ch_2, t, filtered_ch_2);
set(plot_4, 'LineWidth', 1.25);
xlabel('Time [sec]');
ylabel('[uV]');
legend('EEG Ch 2', 'filtered EEG Ch 2', 'Location', 'NW');

% solution to (a)
B = ones(15,1)/15;  % the order 14 filtering vector
filtered_ch_1 = filter(B, 1, eeg_ch_1);
filtered_ch_2 = filter(B, 1, eeg_ch_2);

% (b)
figure;
subplot(211); plot_5 = plot(t, eeg_ch_1, t, filtered_ch_1);
set(plot_5, 'LineWidth', 1.25);
title('EEG waveforms, order = 8');
xlabel('Time [sec]');
ylabel('[uV]');
legend('EEG Ch 1', 'filtered EEG Ch 1', 'Location', 'NW');

subplot(212); plot_6 = plot(t, eeg_ch_2, t, filtered_ch_2);
set(plot_6, 'LineWidth', 1.25);
xlabel('Time [sec]');
ylabel('[uV]');
legend('EEG Ch 2', 'filtered EEG Ch 2', 'Location', 'NW');

% (c)
% i.  The MA removes some of the high frequency components, for example the sharp transients are significantly reduced.
% ii.  The output of the filter is a delayed version of the input (approximately 7-sample delay)
% iii.  Transient response during the first 14 samples of the output can be observed
% Here, first 2 answers are sufficient

% (d)
% Repeat with C=ones(9,1)/9;
% Here the order is equal to # of input terms - 1
% Less loss in high frequency somponents, less delay (4 samples), and shorter transient effect (8 samples)