%% ELE 632 Lab 1 Report
% *Author:*  Scott Goddard 500723518
%
%% A.1)
delta = @(n) 1.0.*(n==3);
u = @(n) 1.0.*(n>=-1);
x = @(n) cos((pi*n)/5.0).*u(n);
x1 = @(n) x(n-3);
x2 = @(n) x(-n);
n = (-20:20)';

figure(1);
stem(n,delta(n)); grid on; title('I.'); xlabel('n');

figure(2);
stem(n,u(n)); grid on; title('II.'); xlabel('n');

figure(3);
stem(n,x(n)); grid on; title('III.'); xlabel('n');

figure(4);
stem(n,x1(n)); grid on; title('IV.'); xlabel('n');

figure(5);
stem(n,x2(n)); grid on; title('V.'); xlabel('n');

% x1[n] is translating x[n] 3 units right.
% x2[n] is horizontally mirroring x[n] across the y axis.

%% A.2)
n = [-10:70];
u = @(n) 1.0.*(n>=0);
y = @(n) 5.0.*exp(-1.0*n/8).*(u(n)-u(n-10));
y1 = @(n) y(3.*n);
y2 = @(n) y(n/3);

figure(6);
stem(n,y(n)); grid on; title('I.'); xlabel('n');
axis([-10 70 0 6]);

figure(7);
stem(n,y1(n)); grid on; title('II.'); xlabel('n');
axis([-10 70 0 6]);

figure(8);
stem(n,y2(n)); grid on; title('III.'); xlabel('n');
axis([-10 70 0 6]);

% y1[n] is horizontally compressing x[n] by 3.
% y2[n] is horizontally stretch x[n] by 3.

%% A.3)
t = [-10:0.1:70];
u = @(t) 1.0.*(t>=0);
z = @(t) 5.0.*exp(-1.0*t/8).*(u(t)-u(t-10));
y3 = @(t) z(t./3);

n = [-10:70];

figure(9);
stem(n,y3(n)); grid on; title('I.'); axis([-10 70 0 6]);

% II. y3[n] is not the same as y2[n] because it was sampled after the
% signal was generated, so it has more values.

%% B.2)
n = 1:13;
y = [2000,zeros(1,length(n)-1)];
i = 1.02;
for j = 1:13
    y(j+1) = i*y(j);
end
figure(10);
stem(n,y(n)); grid on;
xlabel('Month'); ylabel('Balance [$]'); title('B.2');
axis([0 14 1000 12000]);

%% B.3)
n = 1:13;
y = [2000,zeros(1,length(n)-1)];
x = @(n) 100.*n;
i = 1.02;
for j = 1:13
    y(j+1) = i*y(j) + x(j);
end
figure(11);
stem(n,y(n)); grid on;
xlabel('Month'); ylabel('Balance [$]'); title('B.3');
axis([0 14 1000 12000]);

%% C.1)
delta = @(n) 1.0.*(n==0);
x = @(n) cos(pi*n./5)+delta(n-20)-delta(n-35);
n = 0:44;

figure(12);
stem(n,x(n)); title('Input signal'); xlabel('n'); grid on;

%  Matlab function given below
%
%  function [y] = maximum(x,N)
%    M = length(x);
%    y = [zeros(1,M)];
%    x2 = [zeros(1,(N-1)),x];
%    for i = 1:M
%        out = max(x2(i:(i+N-1)));
%        y4(i) = out;
%    end

%% C.2)
figure(13);
y = maximum(x(n),4);
stem(n,y); title('N = 4'); xlabel('n'); grid on;

figure(14);
y = maximum(x(n),8);
stem(n,y); title('N = 8'); xlabel('n'); grid on;

figure(15);
y = maximum(x(n),12);
stem(n,y); title('N = 12'); xlabel('n'); grid on;


%% C.3)

% For each value of N the signal is essentially taking the max value for
% the last N points. That means a local maximum will persist for N values.
% We lose a lot of information about the signal by increasing N.

%% D.1)

% Function given below:
%
% function [e,p] = enpow(x)
%    sq = abs(x).^2;
%    en = 0;
%    for i = 1:length(x)
%        en = en + sq(i);
%    end
%    e = en
%
%    N0 = length(x)+1;
%    N = (length(x)-1)/2;
%    pow = 0;
%    for i = 1:N
%        pow = pow + sq(i);
%    end
%   p = N0^(-1)*pow
% end

%% D.2)
u = @(n) 1.0.*(n>=0);
x = @(n) 3.0.*n.*(u(n+3)-u(n-4));
n = -3:3;
figure(16);
stem(n,x(n));title('x[n] original'); xlabel('n'); grid on;
enpow(x(n));



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