Avoid infinity in the answer
3 Ansichten (letzte 30 Tage)
Ältere Kommentare anzeigen
Chanaka Navarathna
am 9 Feb. 2019
Kommentiert: Chanaka Navarathna
am 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 Kommentare
Star Strider
am 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.
Akzeptierte Antwort
John D'Errico
am 9 Feb. 2019
Bearbeitet: John D'Errico
am 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.
Weitere Antworten (0)
Siehe auch
Kategorien
Mehr zu Loops and Conditional Statements finden Sie in Help Center und File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!