i am getting an error "Index exceeds the number of array elements (2).", please can someone correct it. thanks in advance

K=2;
r=0.75;
tau=1-(K/T);
L=1;
N=K;
t=100;
L=1;
N=K;
t=100;
Z=1;
T=100;
Pt=0.199;
Pb=0.6;
NoAntenna=[5 10 15 20 25 30 35 40 45 50];
Slul=0;
phi22=0;
a=zeros(L,L,N);
b=zeros(L,L,N);
c1=zeros(1,T);
c2=zeros(1,T);
c3=zeros(1,T);
c4=zeros(1,T);
c5=zeros(1,T);
c6=zeros(1,T);
c7=zeros(1,T);
c8=zeros(1,T);
c9=zeros(1,T);
c10=zeros(1,T);
for S=1:T
phi1=zeros(N,L);
phi2=zeros(N,L);
phi3=zeros(N,L);
phi4=zeros(N,L);
d=(poissrnd(r,L,N,L))/1000;
B=10.^(-13.9 - 3.5*log(d) + 0.8*randn(L,N,L));
G=10.^(-13.9 - 3.5*log(d) + 0.8*randn(L,N,L));
for k=1:N
for g=1:L
for z=1:L
Slul=Slul+Pt*t*B(g,k,z);
end
b(:,g,k)=B(:,k,g)/sqrt((1+Slul));
g=g+1;
end
Slul=0;
k=k+1;
end
for k=1:N
a(:,:,k)=inv(b(:,:,k));
k=k+1;
end
M=NoAntenna(1);
c1(S)=tau*log(2)*(1+Pb*b(M-2));
M=NoAntenna(2);
c2(S)=tau*log(2)*(1+Pt*b(M-2));
M=NoAntenna(3);
c3(S)=tau*log(2)*(1+Pt*b(M-2));
M=NoAntenna(4);
c4(S)=tau*log(2)*(1+Pt*b(M-2));
M=NoAntenna(5);
c5(S)=tau*log(2)*(1+Pt*b(M-2));
M=NoAntenna(6);
c6(S)=tau*log(2)*(1+Pt*b(M-2));
M=NoAntenna(7);
c7(S)=tau*log(2)*(1+Pt*b(M-2));
M=NoAntenna(8);
c8(S)=tau*log(2)*(1+Pt*b(M-2));
M=NoAntenna(9);
c9(S)=tau*log(2)*(1+Pt*b(M-2));
M=NoAntenna(10);
c10(S)=tau*log(2)*(1+Pt*b(M-2));
end
c=[sum(c1) sum(c2) sum(c3) sum(c4) sum(c5)...
sum(c6) sum(c7) sum(c8) sum(c9) sum(c10)]./T
 
 

Answer 1

[Walter Roberson]

d=(poissrnd(r,L,N,L))/1000;

d is fairly likely to have entries that are 0.

B=10.^(-13.9 - 3.5*log(d) + 0.8*randn(L,N,L));

Those 0 entries get log(0) which is infinite, so you start getting nan,and the rest of your calculation gets ruined.

b(M-2)

b only has 2 elements but M is 5. Is it possible you want to multiply b by M-2 instead of indexing b and M-2 ?