Is There Life After HART?
4800 BPS HART: A Proposal
Bit Error Rate
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Probability of Undetected Message Error
A concern in error analysis is the frequency of undetected message errors (UME). A UME occurs when a message that has errors gets past the error detection scheme and appears to be perfectly good. The frequency of UMEs will be derived here as a function of bit error probability. The required bit error probability for distortionless transmission and for the desired rate of UMEs is also calculated.
The proposed method uses a checksum byte as the last byte of the message, just as in existing HART. But the bytes, including the checksum byte, are all encoded into 11-bit words according to the proposed coding scheme. Each 11-bit word has odd parity. Thus, any word that has an odd number of bit errors is caught and the message is considered in error. There are also many words that could have an even number of bit errors and be caught because they violate another of the coding rules. For example, an even number of bit errors could cause a string of 4 ones or 4 zeros, which would be recognized as an error. To simplify the analysis, it will be assumed that these even-numbered bit error conditions are not immediately caught. That is, it will be assumed that every word with odd parity has an equivalent byte into which it is decoded. This simplification leads to a conservative result because it discards a level of error checking that is actually present.
With the simplification, every received word that has odd parity is decoded into a message byte. If a word is in error it is translated into the wrong byte. A single wrong byte would result in a checksum error. But two wrong bytes could produce a correct checksum -- resulting in a UME. Thus, the minimum conditions for a UME are that:
1. There be at least two incorrect words that each have at least two bit errors.
2.
The two incorrect bytes resulting from decoding the erroneous words
must
produce the same
checksum that would have occurred had the bytes been
correct.
If we make the usual assumption that the probability of a bit error is small, then we can get rid of the "at least" in the first listed condition and say that there must be two incorrect words that each have two bit errors.
The erroneous bytes that result from decoding of erroneous words may be thought of as randomly occurring bytes. Thus, the problem becomes one of determining the probability that two randomly selected bytes will produce the same checksum as the correct bytes. But since the correct bytes could be anything, they can also be considered randomly chosen. Since each bit position is independent of the others, we can initially consider the checksum for just one bit position. Let's take the LSB bit position for example. Let R1 and R2 be the LSB bits in each of the "incorrect" bytes. And let T1 and T2 be the LSB bits in each of the "correct" bytes. That is, if there had not been any errors we would have R1 = T1 and R2 = T2. If the LSB checksum is to appear correct, then we need (R1 EXOR R2) = (T1 EXOR T2), where (A EXOR B) means that bit A is exclusive-ORed with bit B.
R1, R2, T1, and T2 are a set of 4 random variables. Each can have a value of 0 or 1. This results in 16 possible combinations of the four bits. It can be shown that the equality (R1 EXOR R2) = (T1 EXOR T2) holds for half of them. Thus, the probability that a given bit position will expose a message error is 1/2. It can be shown that the probability that at least one of the 8 bit positions will expose a message error is 255/256. And, therefore, the probability that a message error will not be exposed is 1/256. Intuitively, we expect this number to be relatively small because it is unlikely that two random bytes will have the same checksum as two other random bytes.
The overall probability of UME is approximately given by
where P1 is the probability that any word has two bit errors, P2 is the probability that any other word has two bit errors, and P3 is the probability that the two words in error, when decoded, will result in a UME. P1 is given by
where N is the number of message bytes and Pb is the probability of a bit error.
P2 is given by
And P3 was found above to equal 1/256. Thus, the expression for Pume becomes
BER For Integrate-and-Dump Detection
Because the energy per bit of the proposed method is 1/4 what it is in existing HART, the proposed method is far more susceptible to a bit error if nothing is done to prevent it. A very narrow receive filter or its equivalent are needed. Such a filter exists in the form of an integrate-and-dump type of detector. The expression for Pb is then given in [1]:
where Q(x) = area in tail of normal distribution, A = peak amplitude. The received signal is assumed sinusoidal.
The expected FASTHART Channel and FASTHART error requirements are given in a separate document (Click here to view.) The target Pume is 5.2e-8 with noise density of 100 microvolt/root Hz. Using Pume = 5.2e-8 and 40-byte messages we get Pb = 0.00128. For Pb = 0.00128 the required argument for Q is about 3. Then
Plugging in A = 75 mV and T = 1/4800 second, we get No = 98e-9 volt squared/Hz. The voltage noise density is square root(98e-9 volt squared/Hz) = 313 microvolt/root Hz. Since this is far greater than the required 100 microvolt/root Hz, there is a large margin against error.
It turns out that some of the margin will be eroded by distortion caused by various filters. The BER simulations with all filters present and with minimum received signal are presented in the Conclusions section of the main document (Click here to view.) At BER = 0.001, and with all filters and the channel present, the noise level is 180 microvolt/root Hz. This is a margin of about 4 dB. At BER = 0.0001 this drops to 2.9 dB.
References
1. Ziemer & Tranter, Principles of Communications, 4th ed., Wiley, 1995.
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