Avoid infinity in the answer
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Chanaka Navarathna
on 9 Feb 2019
Commented: Chanaka Navarathna
on 11 Feb 2019
Hi,
I am trying to run this code and, I always get infinity for my first iteration. Is there a way to avoid that? or at least filter that answers which gives infinity.
for i=1:M %for loop for M repetitions
Nx=N+randn(1,M)*CN; %excitation photon fluctuation
Nt=Nx*10.^-(E*C*l); %Nt = transmitted photons
SD2=sqrt(Nt); %This relationship is only valid for a Poisson distribution.
Nt1=Nt+randn(1,1)*SD2; %Nt = transmitted photon fluctuation
Nt2=mean(Nt1)+sqrt(Nt1)*randn(1,1); %detected photon fluctuation
A(i)=-log(Nt2/Nx); %calculation of absorbance
meanA=mean(A); %mean absorbance signal
standdeviationA=std(A); %noise
SNRUV(i)=meanA/standdeviationA %1/RSD=Signal to noise ratio for absorbance
end
4 Comments
Star Strider
on 9 Feb 2019
Reiterating:
What are the variable values (‘M’,‘CN’,‘E’), etc.?
Knowing the relevant function arguments is important. We could make wild guesses for them and never come close to the expected values for them.
Accepted Answer
John D'Errico
on 9 Feb 2019
Edited: John D'Errico
on 9 Feb 2019
Sorry. I had a typo in what I entered before.
The first time through the loop. what do you expect?
A is a scalar on the first time throguh. What is the standard deviationm of a scalar variable? ZERO.
Then you divide by that standard deviation, so you get inf. Only you know what it is that you SHOULD do in that case.
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