% 21.11.2019
% 165678 Tomasz Kisielewski
% Zadanie 5 Modelowanie odpowiedzi impulsowej kanalu dla srodowiska miejskiego
% Komputer

Ts = 6 * 10^(-8);
TU50 = [0,    -5.7;
    21.7e-8,  -7.6;
    51.2e-8,  -10.1;
    51.4e-8,  -10.2;
    51.7e-8,  -10.2;
    67.4e-8,  -11.5;
    88.2e-8,  -13.4;
    123.0e-8, -16.3;
    128.7e-8, -16.9;
    131.1e-8, -17.1;
    134.9e-8, -17.4
    153.3e-8, -19;
    153.5e-8, -19;
    162.2e-8, -19.8;
    181.8e-8, -21.5;
    183.6e-8, -21.6;
    188.4e-8, -22.1;
    194.3e-8, -22.6;
    204.8e-8, -23.5;
    214.0e-8, -24.3];
j = 1;

samples = zeros(length(TU50), 2);
for i=0:Ts:TU50(length(TU50), 1)+Ts
    samples(j, 1) = i;
    samples(j, 2) = 0;
    j = j + 1;
end

for i=1:1:length(TU50)
    TU50(i,2) = 10^(TU50(i,2)/20);
end

figure
stem(TU50(:,1),TU50(:,2),'filled');
title('Przed przeprobkowaniem');

samples(1,2) = TU50(1,2);
for n=2:1:length(TU50)
    for m=2:1:length(samples)-1
        if(samples(m,1) >= (TU50(n,1) - Ts))
            delta = TU50(n,1) - samples(m,1);
            samples(m,2) = samples(m,2) + TU50(n,2)*(delta/Ts);
            samples(m+1,2) = samples(m+1,2) + TU50(n,2)*(Ts-delta)/Ts;
            break;
        end
    end
end

figure
stem(samples(:,1),samples(:,2),'filled');
title('Po przeprobkowaniu');

rayleigh = zeros(1,2*length(TU50));
response = zeros(1,length(TU50));
normal_dist = randn(1,4*length(TU50));
j = 1;
for i=1:2:80
    rayleigh(1,j) = sqrt(normal_dist(1, i)^2 + normal_dist(1,i+1)^2);
    j = j+1;
end

for i=1:1:length(samples)
    response(1,i) = rayleigh(1,i) * samples(i,2);
end

figure
stem(samples(:,1),response,'filled');
title('Przemnonenie');

before = 0;
for i=1:1:length(TU50)
    before = before + TU50(i,2);
end

after = 0;
for i=1:1:length(samples)
    after = after + samples(i,2);
end