Avoid infinity in the answer
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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 个评论
John D'Errico
2019-2-9
It is impossible to answer this. What are the undefined variables? CN, for example? What is N? How about E? C? l?
We cannot help you if you keep it all a secret.
Star Strider
2019-2-9
What are the variable values (‘M’,‘CN’,‘E’), etc.?
Star Strider
2019-2-9
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.
Chanaka Navarathna
2019-2-9
采纳的回答
John D'Errico
2019-2-9
编辑：John D'Errico
2019-2-9
1 个投票
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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