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

Question

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?

Answer 1

In most practical systems, the BER rate is << 1. In this case, to a first order, the total BER is the sum of all BER on the comms path.

Put another way, with low BER, the chance of a single bit being hit twice is very very low.

Answer 2

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%.
The product terms can be neglected in the case of low BER, and the net BER then is just the sum of individual BERs.

Answer 3

I bit more on sarthak's answer which, BTW I think is right.

From a practical standpoint, however, with typical BERs for a link of 1e-8 or 1e-9, the 2P(E1)P(E2) term is many orders of magnitude smaller than P(E1)+P(E2) and so can be ignored. So to a very good approximations, the total BER is just the sum of the individual BERs of each section.