回复:200PLC通讯采用MODBUS方式遇到难题------求解

Laohuai

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普通 普通 如何晋级?

发布于 2012-10-06 23:38:59

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[2012:10:06:23:22:56]03 04 00 00 00 0B B0 2F
[2012:10:06:23:22:57]03 04 00 00 00 0B B0 2F
[2012:10:06:23:22:58]03 04 00 00 00 0B B0 2F
[2012:10:06:23:22:59]04 03 00 04 00 01 C5 9E
[2012:10:06:23:23:00]04 03 00 04 00 01 C5 9E
[2012:10:06:23:23:01]04 03 00 04 00 01 C5 9E
[2012:10:06:23:23:02]05 03 00 04 00 01 C4 4F
[2012:10:06:23:23:03]05 03 00 04 00 01 C4 4F
[2012:10:06:23:23:04]05 03 00 04 00 01 C4 4F
[2012:10:06:23:23:05]06 03 00 04 00 01 C4 7C
[2012:10:06:23:23:06]06 03 00 04 00 01 C4 7C
[2012:10:06:23:23:07]06 03 00 04 00 01 C4 7C
[2012:10:06:23:23:08]07 03 00 04 00 01 C5 AD
[2012:10:06:23:23:09]07 03 00 04 00 01 C5 AD
[2012:10:06:23:23:10]07 03 00 04 00 01 C5 AD
[2012:10:06:23:23:11]08 03 00 04 00 01 C5 52
[2012:10:06:23:23:12]08 03 00 04 00 01 C5 52
[2012:10:06:23:23:13]08 03 00 04 00 01 C5 52
[2012:10:06:23:23:14]09 03 00 04 00 01 C4 83
[2012:10:06:23:23:15]09 03 00 04 00 01 C4 83
[2012:10:06:23:23:16]09 03 00 04 00 01 C4 83
[2012:10:06:23:23:17]0A 03 00 04 00 01 C4 B0
[2012:10:06:23:23:18]0A 03 00 04 00 01 C4 B0
[2012:10:06:23:23:19]0A 03 00 04 00 01 C4 B0
[2012:10:06:23:23:20]03 10 00 00 00 1E 3C B0 00 00 00 17 47 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 4B E7
[2012:10:06:23:23:22]03 10 00 00 00 1E 3C B0 00 00 00 17 47 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 4B E7
[2012:10:06:23:23:23]03 10 00 00 00 1E 3C B0 00 00 00 17 47 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 4B E7
[2012:10:06:23:23:24]04 10 00 01 00 02 04 00 00 00 00 23 AF
[2012:10:06:23:23:25]04 10 00 01 00 02 04 00 00 00 00 23 AF
[2012:10:06:23:23:26]04 10 00 01 00 02 04 00 00 00 00 23 AF
[2012:10:06:23:23:27]05 10 00 01 00 02 04 00 00 00 00 27 53
[2012:10:06:23:23:28]05 10 00 01 00 02 04 00 00 00 00 27 53
[2012:10:06:23:23:29]05 10 00 01 00 02 04 00 00 00 00 27 53
[2012:10:06:23:23:30]06 10 00 01 00 02 04 00 00 00 00 28 17
[2012:10:06:23:23:31]06 10 00 01 00 02 04 00 00 00 00 28 17
[2012:10:06:23:23:32]06 10 00 01 00 02 04 00 00 00 00 28 17
[2012:10:06:23:23:33]07 10 00 01 00 02 04 00 00 00 00 2C EB
[2012:10:06:23:23:34]07 10 00 01 00 02 04 00 00 00 00 2C EB
[2012:10:06:23:23:35]07 10 00 01 00 02 04 00 00 00 00 2C EB
[2012:10:06:23:23:36]08 10 00 01 00 02 04 00 00 00 00 1C FF
[2012:10:06:23:23:37]08 10 00 01 00 02 04 00 00 00 00 1C FF
[2012:10:06:23:23:38]08 10 00 01 00 02 04 00 00 00 00 1C FF
[2012:10:06:23:23:39]09 10 00 01 00 02 04 00 00 00 00 18 03
[2012:10:06:23:23:40]09 10 00 01 00 02 04 00 00 00 00 18 03
[2012:10:06:23:23:41]09 10 00 01 00 02 04 00 00 00 00 18 03
[2012:10:06:23:23:42]0A 10 00 01 00 02 04 00 00 00 00 17 47
[2012:10:06:23:23:43]0A 10 00 01 00 02 04 00 00 00 00 17 47
[2012:10:06:23:23:44]0A 10 00 01 00 02 04 00 00 00 00 17 47
[2012:10:06:23:23:45]03 04 00 00 00 0B B0 2F
[2012:10:06:23:23:46]03 04 00 00 00 0B B0 2F
[2012:10:06:23:23:47]03 04 00 00 00 0B B0 2F
我测试了一下,在只有主站的时候,一个轮训的时间是48秒,一个从站一次读或写占用3秒,如果从站都存在,那么时间不会大于16秒,因为从站不存在的时候一次读或写会发送3次。
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