# How to calculate Bit Error Rate of an electronic equipment with several parts?

I am currently studying integrity of data into electronic systems facing Single Event Effects.

I am focusing on data integrity inside an electronic equipment. I've been through a lot of papers providing methods on how to evaluate Bit Error Rate / Soft Error Rate of EEE parts, and BER over transmission lines.

However I couldn't found methodology on how to calculate the resulted BER for data coming from an equipment, taking into account the different BERs of the different parts and how to ponderate them with the use case of the equipment.

BER is

the number of erroneous bits transmitted divided by the total amount of transmitted bytes, over a defined period of time

So I have the radiation analysis of each component of my system. My guess would be to sum all the BER of each part that could introduce error in the data, ponderate it with the data rate, retention time of the data in some parts (e-g buffers), the ECC used in some parts, and the duty cycle of utilization of each part.

However I have no idea if it is pertinent or not and I would like to follow some existing standard if any.

Hints?

• I do not know the answer, but my guess is that the BERs are not added ... imagine two stages of a device, each having 50% BER ... the total BER could not possibly be 100% ... most likely, the BER of one stage is applied to the "good" data percentage of the previous state ... the two 50% stages would have 75% BER overall Commented Feb 5, 2023 at 17:26

You can apply Bayes' Theorem to calculate the total BER or probability of a bit flip. I assume the different parts you refer to in your question are receiving the bits in a sequence. So, if the first part transmits the bit with error probability $$\P(E_1)\$$ and the BER of the second part is $$\P(E_2)\$$. Then, the total BER is, $$P(E_{2T}) = P(E_1)(1-P(E_2))+(1-P(E_1))P(E_2) = P(E_1)+P(E_2)-2P(E_1)P(E_2)$$ That is, we receive a wrong bit after the second part of the system if its first part sends the bit wrong and the second part does not flip it or if the first part sends it right and the second part flips it. Similarly, from third part $$P(E_{3T}) = P(E_1)+P(E_2)+P(E_3)-2(P(E_1)P(E_2)+P(E_1)P(E_3)+P(E_2)P(E_3))+4P(E_1)P(E_2)P(E_3)$$ And so on. So, for instance, if a two-part system has a component BER of 50% each, then the net BER will be 0.50+0.50-2*0.25=50%.