Consider a CAN system. We use the bit time for time unit, i.e., one time unit is the time needed to transmit one bit. The longest propagation delay is two time units. Two nodes, A, and B, are trying t...
Consider a CAN system. We use the bit time for time unit, i.e., one time unit is the time needed to transmit one bit. The longest propagation delay is two time units. Two nodes, A, and B, are trying to access the bus at time 0. Node C tries to access the bus at time 5. The identifier for node A is: 00011100110; the identifier for node B is: 00010101010; the identifier for node C is: 00010011010. Who will the first to successfully send the data? Justify your answer
Homework Answers
Among node A and node B we can see that 11-bit identifier of node B is lower than 11-bit identifier of node A and as per CAN protocol, lower value of identifier will have higher priority, hence node B will be first start sending the data.
But since size of message is more than 11 bits, hence by the time node B will be in the process of sending data, node C will try to access bus at time unit 5( note that in time unit 5 node B has not completed sending data since each bits transmission takes 1 unit time) and since 11 bit identifier of node C is even less than node B, so node B will halt transmission because C has higher priority and hence node C will first successfully sends the data and then node B will resume transmission.
Hence node C will be first to successfully send the data.
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Please comment for any clarification.
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