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What is correct to calculate Rate or shannon capacity?
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Users are uniformly distributed in the circular area and base station is at the center. If the distance is varied then their corresponding rate should also be different. Here is the code, I am getting the approximate same rate for all users. Where is the mistake in my code? any suggestion is very thankful.
n_users=10;%no. of users
N0dbm=-124; %%%noise power
N0 = 10^((N0dbm-30)/10);
P_MBS=20;%Power in watt
MBS_XY = [0 0];
r_MBS=100;
alpha=0.3; %bandwidth allocation
FREQ = 2.15;
angle_users = 2*pi*rand(n_users,1);
r_users = r_MBS*sqrt(rand(n_users,1));
%x and y coordinate of users
X_users = r_users.*cos(angle_users)+ MBS_XY(1);
Y_users = r_users.*sin(angle_users)+ MBS_XY(2);
XY_users=X_users + 1i.*Y_users;
MBS_XY = MBS_XY(1)+1i*MBS_XY(2);
%%Calculate channel gain
dist_users = abs(XY_users-MBS_XY);
path_loss = 10*alpha*log10(dist_users) + 27.3 + 26*log10(FREQ);
shadowing = 4*ones(size(dist_users));
shadowing = shadowing.*randn(size(shadowing));
sub_power_array = 10.^((shadowing-path_loss)./10); % caculate shadowing
size_h = size(sub_power_array); % caculate h
h = sqrt(0.5)*randn(size_h)+sqrt(0.5)*randn(size_h)*1j;
% final power array
channel_gain = (abs(h)).^2.*sub_power_array;
G_MBS = channel_gain;
%%%%
%%Rate
R_MBS = log2(1+(P_MBS.*G_MBS)./N0); %%Rate or shannon capacity
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Answers (1)
KALYAN ACHARJYA
on 18 May 2018
Edited: KALYAN ACHARJYA
on 18 May 2018
The output is not same, where you got the same results if you vary the n_user
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/188599/image.jpeg)
6 Comments
KALYAN ACHARJYA
on 18 May 2018
Edited: KALYAN ACHARJYA
on 20 Nov 2019
Why not plot Shannon's capacity limit graph, you will get the clue whether the results are OK or not?
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