How to calculate average
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Dhawal Beohar
on 10 Jul 2019
Commented: Dhawal Beohar
on 21 Jul 2019
Hi. I have an equation below for which I need to calculate the average value for mu_o.
First, by calculating the sum of it, and secondly by dividing it with no of iterations, but I am unable to do it. Please help it's urgent!
I am doing for mu_o the same as I have calculated the average for EHH but I am not getting correct values.
rt=2;
NoB=1;
Pmax=100;
EHH(i)=0;
mu_o(i)=0[
for j=1:100000
mu_o(i)= mu_o(i)+ min((((exp(rt)-1)*NoB)./r),Pmax);
EHH(i)=EHH(i)+exp(ii*mu_o);
end
EHHH_avg(i)=(EHH(i))/100000;
mufix(i)=(1/ii)*log(EHHH_avg(i));
mu_o_avg(i)=(mu_o(i))/100000;
Epsilon=((mu_o_avg(i))/mufix(i));
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Accepted Answer
Image Analyst
on 10 Jul 2019
Edited: Image Analyst
on 10 Jul 2019
Did you try using sum()?
EHHH_avg = sum(EHH) / 100000;
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Image Analyst
on 20 Jul 2019
See if this works for you:
clc
clear
close all;
fprintf('Working...\n');
NoB=1; % Noise Density
rt=0.5; % Rate in bits/sec
EH=0; % Initialising Energy Harvesting
Bmax= 50; % Threshold level of battery in Joules
Bslot= 3; % Energy threshold in Joules
Pout=[0.001:0.001:0.1];
QoS_Component_u=[0.001:0.001:0.1];
maxj = 100000;
for i=1:length(QoS_Component_u)
u=QoS_Component_u(i);
Pmax=100;
muo = zeros(1, maxj);
EHH = zeros(1, maxj);
for j = 1 : maxj
hx=(1/sqrt(2))*randn(1,1);
hy=(1/sqrt(2))*randn(1,1);
ht_rayleigh = hx.^2 + hy.^2;
muo(i) = min((((exp(rt)-1)*NoB)./ht_rayleigh),Pmax);
EHH(i) = EHH(i)+exp(u*muo(i));
end
EHHH(i)= mean(EHH);
mufix(i)=(1/u)*log(EHHH(i));
mu_o_avg(i)= mean(muo);
Epsilon(i)=((mu_o_avg(i))/mufix(i));
Pout(i)= (Epsilon(i)*(exp(-u*(mufix(i))*(Bmax-Bslot))));
end
plot(QoS_Component_u, Pout, 'LineWidth', 2);
grid on;
hold on;
title('Comparison between Pout and QoS Component u', 'FontSize', 13);
xlabel('QoS Component u', 'FontSize', 13);
ylabel('Pout', 'FontSize', 13);
fprintf('Done!\n');
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