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蓝孝增

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如何为PLC实现Copress中继器算法

大家好,
我正在考虑为PLC实现Copress中继器算法(AN62497)。我的疑问是,如果我们有15个奴隶,而中继站不只是它旁边的1个节点,而是2个或更多的节点,如果我们没有100%个到达能力节点到节点(更少的说它可以在一个节点正确读取一个消息之前接受1-3个消息)。
根据对第2-3页中的AN62488文档的解释,如果中继消息在其前面达到2个节点,则2节点将停止重复消息,因为两个节点都接收到来自彼此的确认,并且加上这2个节点前面的节点尚未接收到消息。事实上我们没有达到100%的能力。在这里,我试图通过一个方案来解释场景:A -B-C--D -E-F-G-H-I--L A想与L A通信,向L发送了直接消息,不能看到A,发送广播消息。B和C从A接收广播消息,在直接消息到L之后,都发送广播消息。两者都收到对方的确认,所以他们停止重复消息。现在,如果没有其他节点(D,E,…)接收到来自B和C的广播消息(坏运气),则消息丢失并且永远达不到L。我的问题是:如果另一个节点没有以100%确定的消息接收到消息,该算法是否有效。谢谢你的支持。-J

以上来自于百度翻译


     以下为原文
  Hi everyone,
     I am considering implementing Cypress repeater Algorithm for PLC (AN62487). My doubt is if we have 15 slaves and the repeater reaches not just 1 node next to it but 2 or more nodes and if we don't have 100% reach ability node-to-node (less say it could take 1-3 messages before one node reads correctly a message).
        According to the explanation of AN62487 document on page 2-3 it seems that if a repeater message reaches 2 nodes ahead of it, the 2 node will stop repeating the message because both receive a confirmation from each other and plus the nodes ahead of these 2 have not received the message yet due to the fact we don't have 100% reach ability.   
             
        Here I try to explain the scenario trough a scheme:   
             
        A--B--C--D--E--F--G--H--I--L   
             
        A wants to communicate with L.   
             
        A has sent direct message to L, which can't see A, A sends broadcast message.   
        B and C receive the broadcasted message from A.   
        After direct message to L, both send broadcast message. Both receive a confirmation from each other so they stop repeating the message.   
        Now, if none of the further nodes ( D,E, ...) has received the broadcasted message from B and C (bad luck), the message is lost and never reach L.   
             
        My question is: Does the algorithm works if a node next to another does't received with 100% certainty a message.      
        Thank you in advance for the support.     
             
        - J   

回帖(3)

杨军

2019-7-29 15:31:56
在我看来,AlgTrthHIM每个节点都作为一个中继器工作。
需要知道,并且每个节点重复它的邻居直到它
实现认知。因此,数据包传播到每个最近的位置。
邻居直到完成。
HTTP://www. CyPress?COM/?DOCID=44756
问候,Dana。

以上来自于百度翻译


     以下为原文
  Seems to me the algortithim has each node operating as a repeater
    requiring acknowlegement, and each node repeats to its neighbor until it
    achieves acknowlegement. So the packet propagates to each nearest
    neighbor until complete.
     
        
              http://www.cypress.com/?docID=44756
     
    Regards, Dana.
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蓝孝增

2019-7-29 15:42:22
我认为在分组跳数消除了我对消息达到1以上的邻居节点的疑问。
现在我关注的案例2个节点同时寄一包。包会碰撞并被破坏吗?
再次感谢您的帮助!!
-J

以上来自于百度翻译


     以下为原文
  I think that with the number of hops value in the packet eliminates my doubt about the message reaching more than 1 neighbour node.
    Now I am concern about the case 2 nodes send at the same time a packet. Will the packets collide and be destroyed?
    Thank you again, your help is appreciated !!
     
    - J
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杨军

2019-7-29 15:57:40
该协议使用CSMA,就像以太网,以避免碰撞。
HTTP://www. CyPress?COM/?DOCID=46702
问候,Dana。

以上来自于百度翻译


     以下为原文
  Use CSMA in the protocol, much like Ethernet, to avoid collisions.
     
        
              http://www.cypress.com/?docID=46702
     
    Regards, Dana.
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