Basic key sharing principle. To simplify the presentation we consider that a logical ’1’ is recessive while logical ’0’ is dominant and defer the discussion on the actual logical interpretation of recessive and dominant levels for the next section. In a similar fashion with the CAN protocol, the FlexRay physical layer implementation uses a wired-AND mechanism to resolve the bus level when dominant and recessive levels are generated at the same time by distinct nodes. Therefore, the resulting bus level will only be recessive if all nodes generate a recessive output, otherwise the resulting level will be dominant as long as at least one node generates a dominant output. This physical layer property was previously used to build a key agreement scheme for CAN [7]. We make use of the same underlying principle which we describe in what follows. → → high X low low low high Bus signal Idle Data 0 Data 1 RxD RxEN high high low low high low Let Xxxxx and Xxx be two network nodes that have to share → → a secret by taking advantage of the bus wired-AND behavior. → → Each of them independently generates a random bit sequence Fig. 5. FlexRay symbols: Collision Avoidance Symbol (CAS)/Media Access Test Symbol (MTS), Wakeup Symbol (WUS) and Wakeup During Operation Pattern (WUDOP) of predefined length n. They want to exploit the physical layer behavior by simultaneously transmitting their bit sequences in a manner that would allow each of them to identify the value of the bit transmitted by its counterpart. Additionally, it should not be possible for an eavesdropper node Eve to determine the bit sequences sent by Xxxxx and Xxx by observing bus traffic. Clearly, for the information exchange to be successful it is also required that no other node, besides Xxxxx and Xxx, is transmitting during this interval. However, there are several issues in achieving these requirements. First, if both Xxxxx and Xxx transmit a ’1’ bit this will be evident to Xxx since the resulting bus value is ’1’ only in this case. Secondly, if one of the nodes transmits a ’0’ it would be impossible for it to determine the value sent by its counterpart since the resulting bus value will be ’0’ regardless of the value transmitted by the second node. According to the original scheme proposed by Xxxxxxx and Xxxxxxxxxx [7] Xxxxx and Xxx overcome these difficulties by taking two steps. The first step, intended to enable each legit node to determine what its counterpart transmits, is to modify the generated random bit sequences by replacin...
