Filter löschen
Filter löschen

How to calculate average

15 Ansichten (letzte 30 Tage)
Dhawal Beohar
Dhawal Beohar am 10 Jul. 2019
Kommentiert: Dhawal Beohar am 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));

Akzeptierte Antwort

Image Analyst
Image Analyst am 10 Jul. 2019
Bearbeitet: Image Analyst am 10 Jul. 2019
Did you try using sum()?
EHHH_avg = sum(EHH) / 100000;
  5 Kommentare
Image Analyst
Image Analyst am 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');
Dhawal Beohar
Dhawal Beohar am 21 Jul. 2019
Thank you ! it's being great help...

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (0)

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by