# Help for new type of ode45 question: coupled equations but each with large summations on dependent variables

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Jaya il 24 Apr 2021
Modificato: Jaya il 24 Apr 2021
Hello everyone,
I have seen all ode45 answers but no case seems to be like the one below in which each equation has summations comprising of other dependent variables.
I am trying to replicate the following set of equations. i am getting wrong values. Can anyone see the code and please tell me if the method of ode45 implementation is correct? If correct, then it would mean there is something wrong with the input values that is causing wrong results for me. I have commented everything for better understanding.
clc
clear all
% dz= 400; nz=200;
% L= nz*dz;
% zspan = linspace(0,L,nz+1) %was trying to some other way of zspan
ws=[1520:10:1560]*1e-9 ; F=3e8./ws; %wavelength and frequency of the channels
channels=numel(F); num_span=1;
P_dbm= -3.5* ones(channels, num_span); %input power of each channel
p0= 10.^(P_dbm./10)*0.001; %convert input signal power to Watts
pp1=300e-3 ; pp2=310e-3 ; %300 mW and 310 mW initial pump powers
p_in=[p0;pp1;pp2]; %initial condition vector having signal powers for 5 channels+ two pump powers
zspan = [0:40]*1000 %for 40 kms THESE BIG NOS. ARE CAUSING PROBLEMS
[z,p]= ode45(@lorenz,zspan,p_in); %the ode calling
p_in_dBm=(10*log10(p))+30; %back to dBm
%%
function dpdz= lorenz(z,p)
%%
ws=[1520:10:1560]*1e-9; F=3e8./ws;
f_p1= 3e8/1427e-9; f_p2=3e8/1495e-9;
gr=.4/1e3; %Raman gain spectrum
alphaa=[0.1832 0.1932 0.1142 0.0746 0.1781 0.1432 0.1969]./4.343/1e3 ;
%attenuation coefficient encountered by each frequency channel
channels=numel(F);
m=length(p);
for i=1: channels %For each Pi. The first equation
A=0; B=0; f_i=F(i);
for k= i+1:channels %I separately wrote the components of each eqn. and then added them accordingly
A= A-((f_i/F(k))*p(k)*p(i))*gr;
end
for k=1:(i-1)
B= B+(p(k)*p(i)*gr);
end
alpha_i=alphaa(i);
C= p(i)*alpha_i;
D= p(m-1)*p(i)*gr;
E=p(m)*p(i)*gr;
dpdz(i)= A+B-C+D+E; %this gives one equation. As 5 channels, dpdz(1) to dpdz(5), five eqns.
end %end of the every channel P loop
%=======================================================%
%for the pump 1 eqn.
A=0; B=0; C=0;
for k=1:channels
A= A+(((f_p1/F(k))*p(k)*p(m-1))*gr) ;
end
B= -p(m)*p(m-1)*gr;
C=p(m-1)*alpha_i;
dpdz(m-1)= A+B+C;
%=======================================================%
%for the pump 2 eqn.
A=0; B=0; C=0;
for k=1:channels
A= A+ (((f_p2/F(k))*p(k)*p(m))*gr) ;
end
B= (f_p2/f_p1)*p(m)*p(m-1)*gr; %need to add the gr term later
C=p(m)*alpha_i; %later you will have to change the alpha notation
dpdz(m)= A+B+C;
dpdz=dpdz';
%=======================================================%
end
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