WDM (Question on replace of M40 by D40)
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Somebody can help: If wants to organize WDM system (point-to-point). For the purpose of multiplexing and demultiplexing in a standard situation it is necessary to apply two boards of M40 and two boards of D40. On one from each side. But available there are only four boards of M40. Whether it is possible to organize a full-fledged linear path with their help, using M40 boards instead of D40? Are interesting both theoretical and practical comments! |
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| Very interesting all comments posted here. I will look forward for more. |
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Dear Ilnur, its a good question. Firstly I guess you can do it. In fact if you connect directly a Line board (for instance 192.10) to M40-Out port you can measure a significant power level in M40 Channel Port (192.10) - around (-5dBm)-, of course that power level decreased by M40 Insertion Loss and patch cord loss. I remenber when we performed that test -it was for curiosity |
- once removed the patch cord between Line Board and M40 Board the power outgoing for M40 Channel Port (192.10) disappeared. I will look forward for more comments or experiences. Reply 2 #
Thanks for the comment  Theoretically it seems to me that so it is possible to make too. As these boards are made on the AWG technology, and it is bidirectional. But practically I never tried to connect so. We will wait, someone can already tried. Whether also there will be any features of operation of such decision? We now had a situation that it is necessary to organize a new path. Financings arent present. But in spare cards there are many M40 which were in large quantities released long ago after M40V installation, and some amplifiers. And now we think to create from this that there is a new path, without D40 use |
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