
RDM General Implementation Discussion General Discussion and questions relating to implementing RDM in a product. 

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September 20th, 2006  #1 
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Join Date: Jun 2006
Posts: 14

Discovery Checksum Problem Analysis
Hello
David our resident guru , has analysed the probability of getting one good discovery response when in fact 2 devices are replying at the same time (below) The problem seems to appear a lot more often than I have won the lotto ! Maybe there should be some kind of implementation guide that could recommend having a small backoff timer of some sort to reduce the probability of having this problem. (like ethernet PHYs) Considering most lighting rigs are composed of at least multiple devices of a same manuafcturer that were probably purchased at the same time, I't very possible for an end user to have sequential device IDs. Nic www.enttec.com 1. Discovery Collision Analysis 1.1 Uncorrupted msg MID1  0xAA MID1  0x55 MID0  0xAA MID0  0x55 DID3  0xAA DID3  0x55 DID2  0xAA DID2  0x55 DID1  0xAA DID1  0x55 DID0  0xAA DID0  0x55  CSUM1  0xAA CSUM1  0x55 CSUM0  0xAA CSUM0  0x55 CSUM = (MID10xAA + MID00xAA + DID30xAA + DID20xAA + DID10xAA + DID00xAA) + (MID10x55 + MID00x55 + DID30x55 + DID20x55 + DID10x55 + DID00x55) = (MID1 + MID0 + DID3 + DID2 + DID1 + DID0) + 6*0xFF = (MID1 + MID0 + DID3 + DID2 + DID1 + DID0) + 0x5FA 1.2 Single Bit Corruption Suppose there are 2 responders A and B with UIDs which differ only in DIDx. Collided msg would differ from original msgs A and B in the DIDx bytes and the checksum bytes, as follows. CSUMB  CSUMA = (DIDBx0xAA + DIDBx0x55)  (DIDAx0xAA + DIDAx0x55) COLLISION(DIDAx0xAA, DIDBx0xAA) = (DIDAx  0xAA) & (DIDBx  0xAA) = (DIDAx & DIDBx)  0xAA COLLISION(DIDAx0x55, DIDBx0x55) = (DIDAx  0x55) & (DIDBx  0x55) = (DIDAx & DIDBx)  0x55 COLLISION(CSUMA10xAA, CSUMB10xAA) = (CSUMA1  0xAA) & (CSUMB1  0xAA) = (CSUMA1 & CSUMB1)  0xAA COLLISION(CSUMA10x55, CSUMB10x55) = (CSUMA1  0x55) & (CSUMB1  0x55) = (CSUMA1 & CSUMB1)  0x55 COLLISION(CSUMA00xAA, CSUMB00xAA) = (CSUMA0  0xAA) & (CSUMB0  0xAA) = (CSUMA0 & CSUMB0)  0xAA COLLISION(CSUMA00x55, CSUMB00x55) = (CSUMA0  0x55) & (CSUMB0  0x55) = (CSUMA0 & CSUMB0)  0x55 Taking checksum over collided msg gives CSUM(over EUID part of collided msg)  CSUMA = [(DIDAx & DIDBx)  0xAA] + [(DIDAx & DIDBx)  0x55]   [DIDAx  0xAA]  [DIDAx  0x55] = (DIDAx & DIDBx)  DIDAx CSUM1(recovered from CSUM part of collided msg) = (CSUMA1 & CSUMB1) CSUM0(recovered from CSUM part of collided msg) = (CSUMA0 & CSUMB0) => CSUM(recovered from CSUM part of collided msg) = CSUMA & CSUMB = CSUMA & [CSUMA + (DIDBx0xAA) + (DIDBx0x55)  (DIDAx0xAA)  (DIDAx0x55)] = CSUMA & [CSUMA + DIDBx  DIDAx] For collided msg to have good checksum: CSUMA + (DIDAx & DIDBx)  DIDAx = CSUMA & (CSUMA + DIDBx  DIDAx) For the Special Case where DIDBx is obtained from DIDAx by changing a single bit from 0 to 1: bit k of DIDAx = 0 and DIDBx = DIDAx + (1 << k), where 0 <= k <= 7. DIDAx & DIDBx = DIDAx CSUMA = CSUMA & (CSUMA + (1<<k)), which is true whenever bit k of CSUMA is 0. CONCLUSION: For the special case of a single bit difference in the device IDs, there is a 50% chance of the discovery collision having a good checksum when ideally the discovery collision checksum should be bad. Last edited by nic123; September 20th, 2006 at 11:54 PM. 
September 22nd, 2006  #2 
Task Group Member
Join Date: Sep 2006
Posts: 26

A fundamental assumption made above is that the transmission medium acts as a wireand, such that the receiver sees a one if both transmitters transmit a one, and sees a zero otherwise. DMX uses a differential transmission medium, and so a one is transmitted as
Code:
D+ = V+ D = V Code:
D+ = V D = V+ Code:
D+ = V of B D = V of A Code:
D+  D < 0.2 V Cheers, Shaun Jackman Pathway Connectivity 
September 24th, 2006  #3 
Junior Member
Join Date: Jun 2006
Location: London
Posts: 13

You are forgetting two things:
The SN75176 is a very old device, and more modern transceivers usually guarantee that a 0V input will give a high output. The MAX485 datasheet says: "The receiver input has a failsafe feature that guarantees a logichigh output if the input is open circuit." Also, the controller's terminating network is required to put a slight bias on the line. Either of these things would ensure that a collision by drivers of exactly equal strength would result in a one being received. 
September 24th, 2006  #4 
Junior Member
Join Date: Sep 2006
Posts: 1

Thanks for the corrections.
Repeating the analysis (below) assuming the contending bits are random still gives a 25% probability of good checksum on collided message. Is this to be expected? David Nemes 1. Discovery Collision Analysis 1.1 Uncollided msg MID1  0xAA MID1  0x55 MID0  0xAA MID0  0x55 DID3  0xAA DID3  0x55 DID2  0xAA DID2  0x55 DID1  0xAA DID1  0x55 DID0  0xAA DID0  0x55  CSUM1  0xAA CSUM1  0x55 CSUM0  0xAA CSUM0  0x55 CSUM = (MID10xAA + MID00xAA + DID30xAA + DID20xAA + DID10xAA + DID00xAA) + (MID10x55 + MID00x55 + DID30x55 + DID20x55 + DID10x55 + DID00x55) = (MID1 + MID0 + DID3 + DID2 + DID1 + DID0) + 6*0xFF = (MID1 + MID0 + DID3 + DID2 + DID1 + DID0) + 0x5FA 1.2 Collided Msg Suppose there are 2 responders A and B with UIDs which differ only in DIDx. CSUMB  CSUMA = (DIDBx0xAA + DIDBx0x55)  (DIDAx0xAA + DIDAx0x55) = DIDBx  DIDAx In general it is assumed that the collision between 2 bytes would preserve the noncontending bit positions which have the same value in both bytes, and create random bit values in the contending bit positions which have different values in both bytes. COLLISION(X, Y) = (X & Y)  [(X ^ Y) & R], where 0 <= R <= 255 is a uniformly distributed random number. For the special case where DIDBx is obtained from DIDAx by changing a single bit from 0 to 1, and also where bit k of CSUMA is 0 (which would happen 50% of the time): bit k of DIDAx = 0, bit k of CSUMA = 0, DIDBx = DIDAx + (1 << k), where 0 <= k <= 7. DIDAx & DIDBx = DIDAx CSUMB  CSUMA = (1 << k) CSUMA & CSUMB = CSUMA Collided msg would differ from original msgs A and B in the DIDx bytes and the checksum bytes, as follows. COLLISION(DIDAx0xAA, DIDBx0xAA) = [(DIDAx  0xAA) & (DIDBx  0xAA)]  [((DIDAx  0xAA) ^ (DIDBx  0xAA)) & R1] = [(DIDAx & DIDBx)  0xAA]  [(DIDAx ^ DIDBx) & 0x55 & R1] = [DIDAx  0xAA]  [(1 << k) & 0x55 & R1] COLLISION(DIDAx0x55, DIDBx0x55) = [(DIDAx  0x55) & (DIDBx  0x55)]  [((DIDAx  0x55) ^ (DIDBx  0x55)) & R2] = [(DIDAx & DIDBx)  0x55]  [(DIDAx ^ DIDBx) & 0xAA & R2] = [DIDAx  0x55]  [(1 << k) & 0xAA & R2] COLLISION(CSUMA10xAA, CSUMB10xAA) = [(CSUMA1 & CSUMB1)  0xAA]  [(CSUMA1 ^ CSUMB1) & 0x55 & R3] = [CSUMA1  0xAA] COLLISION(CSUMA10x55, CSUMB10x55) = [(CSUMA1 & CSUMB1)  0x55]  [(CSUMA1 ^ CSUMB1) & 0xAA & R4] = [CSUMA1  0x55] COLLISION(CSUMA00xAA, CSUMB00xAA) = [(CSUMA0 & CSUMB0)  0xAA]  [(CSUMA0 ^ CSUMB0) & 0x55 & R5] = [CSUMA0  0xAA]  [(1 << k) & 0x55 & R5] COLLISION(CSUMA00x55, CSUMB00x55) = [(CSUMA0 & CSUMB0)  0x55]  [(CSUMA0 ^ CSUMB0) & 0xAA & R6] = [CSUMA0  0x55]  [(1 << k) & 0xAA & R6] Taking checksum over collided msg gives CSUM(over EUID part of collided msg)  CSUMA = ([DIDAx  0xAA]  [(1 << k) & 0x55 & R1]) + ([DIDAx  0x55]  [(1 << k) & 0xAA & R2])  [DIDAx  0xAA]  [DIDAx  0x55] when k is even: CSUM(over EUID part of collided msg)  CSUMA = ([DIDAx  0xAA]  [(1 << k) & 0x55 & R1]) + ([DIDAx  0x55])  [DIDAx  0xAA]  [DIDAx  0x55] = [(1 << k) & R1] when k is odd: CSUM(over EUID part of collided msg)  CSUMA = ([DIDAx  0xAA]) + ([DIDAx  0x55]  [(1 << k) & 0xAA & R2])  [DIDAx  0xAA]  [DIDAx  0x55] = [(1 << k) & R2] CSUM1(recovered from CSUM part of collided msg) = [CSUMA1  0xAA] & [CSUMA1  0x55] = CSUMA1 CSUM0(recovered from CSUM part of collided msg) = ([CSUMA0  0xAA]  [(1 << k) & 0x55 & R5]) & ([CSUMA0  0x55]  [(1 << k) & 0xAA & R6]) = CSUMA0  [(1 << k) & 0xAA & R6]  [(1 << k) & 0x55 & R5] => CSUM(recovered from CSUM part of collided msg) = CSUMA  [(1 << k) & 0xAA & R6]  [(1 << k) & 0x55 & R5] when k is even: CSUM(recovered from CSUM part of collided msg) = CSUMA  [(1 << k) & R5] when k is odd: CSUM(recovered from CSUM part of collided msg) = CSUMA  [(1 << k) & R6] For collided msg to have good checksum, when k is even: CSUMA + [(1 << k) & R1] = CSUMA  [(1 << k) & R5], which is true whenever bit k of R1 = bit k of R5 (50% chance given that bit k of CSUMA = 0, or 25% chance when bit k of CSUMA has any value). For collided msg to have good checksum, when k is odd: CSUMA + [(1 << k) & R2] = CSUMA  [(1 << k) & R6], which is true whenever bit k of R2 = bit k of R6 (50% chance given that bit k of CSUMA = 0, or 25% chance when bit k of CSUMA has any value). CONCLUSION: For a single bit difference in the device IDs, there is at least a 25% chance of the discovery collision having a good checksum when ideally the discovery collision checksum should be bad. 
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