ex05_p01.m

clc; clear; close all;
%% Problem 1
%% c)
% Plot x(n)
%set(0,'defaultaxescolororder');
%set(0,'defaultaxescolororder',[0 0 0; 0.5 0.5 0.5]) %black and gray
set(0,'defaultaxeslinestyleorder',{'-','.-','--'}) %or whatever you want
f = linspace(-0.5,0.5,1e2);
n = linspace(0,50,200);
l = linspace(-50,50,200);

% Plot x(n), Sxx(f), Rxx(l) for a=0.5
for a = [0.5 0.9 -0.9];
x = a.^n; % signal
Sxx = 1./(1-2*a*cos(2*pi*f)+a^2); % Energy density spectrum
rxx = a.^abs(l)/(1-a^2); % autocorelation

figure(); subplot(3,1,1); hold on;
stem(n,real(x), '.'); title(sprintf('Signal x(n), a = %f',a)); xlabel('Sample,n'); ylabel('x(n)');
subplot(3,1,2); plot(f,Sxx); title(sprintf('Energy Density spectrum of x(n), a = %f',a)); xlabel('Digital frequency,f'); ylabel('Sxx(f)');
subplot(3,1,3); stem(l,real(rxx), '.'); title(sprintf('Autocorrelation of x(n), a = %f',a)); xlabel('Length,l'); ylabel('Rxx(l)');
end;
hold off;

%% Result
% With a = 0.5 the signals rate of change is much higher. This is due to
% more high frequency components than a=0.9, this also makes the enerdy density spectrum wider.
% The correlation of two samples l lenght appart is thus much smaller (less
% similar)
% When a=-0.9 signals change a lot more rappid than a=0.5 and 0.9, for
% samples next to each other, but are similar for 3 and three samples (at
% this number of data points). The autocorrelation between groups is negative, while inside each group the
% the correlation is high. This gives the doubble sided pyramid autocorrelation function. 
% The energy density spectrum show that the
% signal only contains high frequency components. (eg all the energy in the
% signal is at higher frequencies).
%

%% d)
a = 0.5;
x = a.^n;
h = cos(2*pi*f)-a*w*cos(2*pi*f);