Integration output is NaN

Question

Amit Patel 2019-9-20

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/481277-integration-output-is-nan

评论： Walter Roberson 2019-9-20

I don't understand why my double integration is resulting in NaN.

%Author: Amit Patel
%Prob. of non eavesdropping event
clc; 
clear all; 
%close all;
%-----------------------------------
neta=0.75;     %Energy harvesting efficiency                          
rho=0.5;    %PSR parameter
T=1e-3;
%-----------------------------------
sigma_sq=1;                        
%-----------------------------------
Eb=(1:10:1000)*1e-3;                              
Eb_end=size(Eb);
in=Eb_end(1,2);
P=10.^(15./10);
d0=2;                              
m=3;                              
var_hsd=(d0^(-m));  
d1=3;                           
m=3;                              
var_hse=(d1^(-m));  
d2=3;                          
m=3;                             
var_hed=(d2^(-m));                
                                                                      
Rs=2;                             
SNRth=2^Rs; 
count=0;
g3=10^(-5);  %-30dB
%g3=0;
%------------------------------------
lambda_0=1/(var_hsd);
lambda_1=1/(var_hse);
lambda_2=1/(var_hed);
gammath=2^Rs;
Ravg_ana=[];
R_avg1=0;
for i=1:in
     
a=(1-rho).*neta.*rho.*P.*g3./(1-neta.*rho.*g3);
b=(1-rho).*Eb(i).*g3./((1-neta.*rho.*g3).*T) + sigma_sq;
   fun=@(x,z)  log2(1+x).*(lambda_1.*lambda_0.*(1-neta.*rho.*g3)./(P.*neta.*rho.*P.*z)).*exp(-lambda_1.*b.*x.*sigma_sq./(P.*(1-rho) -a.*x) ).*exp(-lambda_0.*x.*Eb(i).*z./((1-neta.*rho.*g3).*T.*P)).*exp(-lambda_0.*x.*sigma_sq./P).*( Eb(i).*z./((1-neta.*rho.*g3).*T)+sigma_sq + 1./( (lambda_0.*x./P) +lambda_1.*(1-neta.*rho.*g3)./(neta.*rho.*P.*z) ) )./((lambda_0.*x./P) +lambda_1.*(1-neta.*rho.*g3)./(neta.*rho.*P.*z)).*lambda_2.*exp(-lambda_2.*z);
   R_avg= integral2(fun,0,Inf,0,Inf);
    
   Ravg_ana=[Ravg_ana R_avg]
   count=count+1
   
end;
plot(Eb,Ravg_ana,'b-'); 
hold on;
xlabel('Eb');
ylabel('Average Eavesdropping rate');
grid on;

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Answer 1

Walter Roberson 2019-9-20

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/481277-integration-output-is-nan#answer_392726

在 MATLAB Online 中打开

You have a sub-expression

./(neta.*rho.*P.*z)

When z is 0, that is a division by 0. When you are working numerically and z is exactly 0 (as it is because of the initial bounds) then this ends up leading to NaN.

The expression has a limit when z goes to 0, but in the form written involves a numeric division by 0.

If you have the symbolic toolbox, then

syms x z
F = simplify(expand(fun(x,z)));
Fun = matlabFunction(F, 'vars', [x, z]);

Now you should be able to use Fun with integral2: the process of expand() and then simplify() smooths out the discontinuity -- in this particular case.