Poissonian extended dead time

Hello, lets consider a detector with extended dead time. I mean the there is the dead time, but during it other particles are arriving then the dead time is extended.
Is there a way to calculate it by ROOT knowing the expected (by simulation) and measured (by the detectors) poissonian dead time?

I’have to do this

I simulated particles by Geant4 hitting a scintillator, then I know that the estimated number is N. Measuring the counting rate by an ArduSiPM I measured M events, then I’ve to estimate the extended dead time of the ArduSiPM (i.e. taking into account that during the dead time the coming of other particles extends the dead time). Then, I don’t mean the simple static dead time N=M/(1-Mtau).

Maybe @moneta do you know it?


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Look at page at the bottom of page 3 of this document.
There is the rate for paralyzable(extendedI dead time.
Pay attention to the fact that for a given output rate, in most cases, there are 2 possible solutions for the true input rate, where the correct one is the smaller.

On the book Radiation Detection and Measurement by Knoll, Glenn F. there is a good explenation

Stefano

Hello @Dilicus I read your document:

By geant 4 I get N=9.510^5 particles
ArduSiPM measured lambda_in=4.4
10^5 particles

The build of ArduiSiPM said me to take into account (For the estimated counting rate) a dead time 3*10^-7 s

By your document I read (Formula 1.6)

Lambda_out=lambda_ine^(-lambda_intau)
so
lambda_out=4.410^5e^(-4.410^-5310^-7)=5.810^5
is it right?

Hi,
the same calculation gives me 3.48e-5. The lambda_out should be smaller than the physical(real) lambda_in.

See also Detector Dead Time Determination and Optimal Counting Rate for a Detector Near a Spallation Source or a Subcritical Multiplying System

Hello @Dilicus yes…it must be lower…I made a mistake…

Anyway…I also think that my lambda_in isn’t 4.410^5 (i.e. the measurement of ArduSiPM) but lambda_in=9.510^5 (i.e. the value estimated by the simulation)

Therefore, the right calculation should be

lambda_out= 9.510^5e^(-9.510^5310^-7)=9.510^5e^(-0.285)=7.110^5

i.e. the lambda_out =7.110^5 is the number that must be compared with the counting rate measured by ArduSiPM (that is 4.410^5)…
To compare this number…I must still calculate the efficicency of the ArduSiPM… is it right now?

Thank you @ferhue I check it too!

Hello @faca87,
what I think is that λ_in is your particle rate estimated by simulation weighted with the efficiency ( λ_in= N x ε ).
Such λ_in will give the λ_out you should compare with the measured rate of 4.4*10^5

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Thank you @Dilicus ! yes…I must correct the lambda_in by the efficiency, but I must still calculate the efficiency!

Thank you

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