mCBand_CNo_Question3.m

% Under conditions of heavy rain, the C-Band 
% path from from the transmitting station in a
% suffers an attenuation of 2-dB. Calculate
% overall C/N at the earth station in a 
% in a bandwidth of 27 MHz under these conditions
% and find the link margin. Reminder: The uplink
% margin is the number of dB of attenuation that
% can occur on the uplink before the receiver
% overall C/N reaches the limit of 9.5 dB

% Parameters
c = 3e8; % Speed of light (m/s)
f_UL = 6.285e9; % UL Frequency (Hz)
f_DL = 4.06e9; % DL Frequency (Hz)
lambda = c / f_DL; 
R = 38500e3; % distance to sat in meters
D = 9; % Diameter of Earth Station antenna in meters


BW_Hz = 27e6; % IF Noise Bandwidth (Hz)
BW_dB = 10*log10(BW_Hz);

K = -228.6; % Boltzmann constant (dBW/K/Hz)
To_k = 290; % Average room temp
T_in = 20; % Antenna noise temperature (K)
T_LNA = 55; % LNA noise temperature (K)
T_sys = T_in + T_LNA; % System noise temperature (K)
T_sys_dB = 10*log10(T_sys);

Pt_sat_W = 40; % Saturated output power of the satellite (W)
Pt_sat_dB = 10*log10(Pt_sat_W);
G_sat_dB = 31; % Antenna gain (dB)
Aeff_VSAT = 0.65; % VSAT antenna aperture efficiency
Aeff_ES = 0.68; % Earth Station antenna aperture efficiency

BO = 1; % Back off (dB)
T_grnd_k = 12; % Assumed ground temperature due to sidelobes in kevlin degrees

L_ptr = 3; % Edge of contour Loss (dB)
L_misc = 0.5; % Clear air atmospheric attenuation and other losses (dB)
Lp_dB = 197.3215; % Path Loss of the CNo
A_rain_dB = 2;    % Rain attenuation
FEC_dB = 5.5;     % FEC Gain        

Pt_dB = 17.7932; % Power of the Earth Station Antenna
A_ca = 0.5; % Clear sky attenuation dB
Nca_dB = -126.0473; % Clear sky Noise Power

% Answer: The earth station C/N ratio will be reduced when the uplink
% is attenuated by 2 dB of rain. If we assume linear operation of the
% transponder (C/N)up is reduced by 2 dB and Pt_sat is reduced by 2 dB
% leading to a similar redcution in (C/N)dn. When both (C/N)up and (C/N)dn
% fall by the same amount, the reduction in over C/N is equal to the uplink
% attenuation. Hence:
%% Part A: (2 dB rain attenuation)

% Required overall (C/N)_rain
minCNo_dB = 9.5; % Minimun required CNo (dB)
CNup_dB = 26; % Inherited value from Question1 in decibels
CNup_lin = 10^(CNup_dB / 10); 
CNdn_dB = 13.2; % Inherited value from Question2 in decibels
CNdn_lin = 10^(CNdn_dB / 10);
CNo_ca_dB = 13;
CNo_rain_dB = CNo_ca_dB - A_rain_dB; % CNo in dB under rain

% The minimum permitted CNo ratio is 9.5 dB. We must find the lowest value
% of CNup_dB which results in (C/N)o_min to determine the uplink margin. 

UL_margin_dB = CNo_ca_dB - minCNo_dB;
fprintf('The UL margin in clear sky conditions is: %.2f dB\n', UL_margin_dB);

%% Check: when the UL is attenuated by 3.5 dB (margin)
CNup_check_dB = CNup_dB - UL_margin_dB;
CNup_check_lin = 10^(CNup_check_dB / 10);
CNdn_check_dB = CNdn_dB - UL_margin_dB;
CNdn_check_lin = 10^(CNdn_check_dB / 10);

% Using the reciprocal formula
invCNo_check_lin = (1 / CNup_check_lin) + (1 / CNdn_check_lin);
CNo_check_lin = 1 / invCNo_check_lin;
CNo_check_dB = 10*log10(CNo_check_lin);

fprintf(' The C/No using the UL Margin is %.2f: dB\n', CNo_check_dB)

%% Part B:
%Answer: When rain affects the downlink attenuation is 1.5 dB and 0.3 dB
%atm loss. The sky noise temperature increase because the DL attenuation
%has increased. For a path attenuation of 1.8 dB, a gain with a ratio
%0.661, the sky noise temperature can be found from equation 4.19 since the
%rain is treated as a lossy medium. 


A_ca_b = 0.3;
A_rain_dB2 = 1.5;
A_b = A_ca_b + A_rain_dB2;
fprintf('Total path sky attenuation A: %.2f dB\n', A_b);

% 2. Determine the sky noise temperature
T_skyrains_k = (To_k)*(1 - 10^(-A_b / 10 ));
T_sca_k = (To_k)*(1 - 10^(-A_ca_b / 10 ));
fprintf('Sky noise temperature under rain T_sky_k: %.2f K\n', T_skyrains_k);
fprintf('Sky noise temperature under clear skies: %.2f dB\n', T_sca_k);
% 3. Find the antenna noise temperature
T_sys_rain_k = T_skyrains_k + T_LNA;
fprintf('Antenna noise temperature T_Ant_k: %.2f K\n', T_sys_rain_k);

% 5. Determine the increase in noise power
deltaN_dB = 10*log10(T_sys_rain_k / T_sys );

fprintf('Increase in noise power deltaN_dB: %.2f dB\n', deltaN_dB);


% Hence, the carrier power has fallen by 1.5 dB because of rain
% attenuation, so the total effect of C/N is a reduction of 1.5 + 3.1 = 4.6
% dB

linkM_dB = A_rain_dB2 + deltaN_dB;
% Hence the DL CN ratio of 1.5 dB of rain attenuation in the DL path is 

% 6. Determine (C/N)rain
CNdn_rain_dB_b = CNdn_dB - linkM_dB;
CNdn_rain_lin_b = 10^(CNdn_rain_dB_b / 10);
CNo_rain_lin = 1 / ((1 / CNup_lin) + (1 / CNdn_rain_lin_b));
CNo_rain_dB_b = 10*log10(CNo_rain_lin);

% 7. Print results:

fprintf('Overall C/N with 2-dB of rain attenuation \n is: %.2f dB\n', CNo_rain_dB);


if (CNdn_rain_dB_b < minCNo_dB )
    fprintf(' The link has faded beyond repair: %.2f is below the required \n overall C/N ratio %.2f\n ', CNdn_rain_dB_b, minCNo_dB);

    CNo_rain_lin = 1 / ((1 / CNup_lin) + (1 / CNdn_rain_lin_b));
    CNo_rain_dB_b = 10*log10(CNo_rain_dB_b);
    CNo_min = 10*log10(1 / ( (1/minCNo_dB) - (1 / CNup_lin )  )); % Calculated C/N minimum for each UL and DL
    CNdn_rain_dB_b_FEC = CNdn_rain_dB_b + FEC_dB;
    fprintf('C/N down in rain in 2.0-dB of rain attenuation \n is: %.2f dB\n which is smaller than %.2f dB\n but %.2f dB with the margin provided \n by FEC \n', CNdn_rain_dB_b, CNo_min, CNdn_rain_dB_b_FEC);
end