To calculate illuminace for four LEDs
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Hello,
This is the code to evaluate the illuminance using four LEDs. But the output graph iam getting only one LED output. Can anyone pls help me with the changes that i have to make in the code.
close all;
clear all;
%paper: Fundamental Analysis for Visible Light Communication Systems using
%LED lights
%Authors: Toshihiko Komine, Student
%Implemented by: Mrinmoyee Mukherjee
%This code is for four LEDs
%*****************************************
%This part defines the transmitter properties
%Semi angle at half power
hpa=60; %[in degrees]
%hpa defines the directivity of the source. Smaller the hpa, more is the
%directed power
%Centre luminous intensity
I0=0.73; %[defined in cadela]
%No. of LEDs in the box
N=60*60;
%******************************************
%Derived quantities of transmitter
%Total luminous intensity
I0_total=I0*60*60;
%mode number
%Note that log10 and log are different command sin matlab.
%The first one will give base 10 and second is natural log
%We use the second one
%a=log10(2);
m=-log(2)/log(cosd(hpa));
disp(m)
%Define room diamensions
lx=5; ly=5; lz=3; % room dimension in meter
%the distance between source and receiver plane
h=2.16170;
%No of grid in the receiver plane
Nx=lx*50; Ny=ly*50;% number of grid in the receiver plane
x=-lx/2:lx/Nx:lx/2;
y=-ly/2:ly/Ny:ly/2;
[XR,YR]=meshgrid(x,y); % receiver plane grid
%Define the position of LED
% position of LED1
XTrans1=-0.86; YTrans1=-0.86;
% position of LED2
XTrans2=-0.86; YTrans2=0.86;
% position of LED3
XTrans3=0.86; YTrans3=-0.86;
% position of LED4
XTrans4=0.86; YTrans4=0.86;
% distance vector from source 1
D1=sqrt((XR-XTrans1(1,1)).^2+(YR-YTrans1(1,1)).^2+h^2);
% distance vector from source 2
D2=sqrt((XR-XTrans2(1,1)).^2+(YR-YTrans2(1,1)).^2+h^2);
% distance vector from source 3
D3=sqrt((XR-XTrans3(1,1)).^2+(YR-YTrans3(1,1)).^2+h^2);
% distance vector from source 4
D4=sqrt((XR-XTrans4(1,1)).^2+(YR-YTrans4(1,1)).^2+h^2);
%Now we need to find the irradiation angles
coseno_phi1=h./D1;
coseno_phi2=h./D2;
coseno_phi3=h./D3;
coseno_phi4=h./D4;
%Now we can find the horizontal illuminance at points (lx,ly).
%Note that we are finding the illuminance hence the receiver position is
%not important
E_lux1=(I0_total*coseno_phi1.^m)./(D1.^2);
E_lux2=(I0_total*coseno_phi2.^m)./(D2.^2);
E_lux3=(I0_total*coseno_phi3.^m)./(D3.^2);
E_lux4=(I0_total*coseno_phi4.^m)./(D4.^2);
%Total illuminace at any point is the addition of all illuminace by all
%LEDs
E_lux=E_lux1+E_lux2+E_lux3+E_lux4;
meshc(x,y,E_lux);
xlabel('X (m)');
ylabel('Y (m)');
zlabel('Illuminance(lx)');
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Respuestas (2)
Walter Roberson
el 8 de En. de 2023
%Total illuminace at any point is the addition of all illuminace by all
%LEDs
E_lux=E_lux1+E_lux2+E_lux3+E_lux4;
You are totaling the four components and using meshc() on the total. You can use subplot() or tiledlayout to create multiple axes in which you meshc() each component by itself.
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Image Analyst
el 8 de En. de 2023
You're plotting only the sum of all 4. You need to plot each one (untested code because my MATLAB is busy right now)
subplot(2, 3, 1);
meshc(x,y,E_lux);
subplot(2, 3, 2);
meshc(x,y,E_lux1);
subplot(2, 3, 3);
meshc(x,y,E_lux2);
subplot(2, 3, 4);
meshc(x,y,E_lux3);
subplot(2, 3, 5);
meshc(x,y,E_lux4);
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