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mixing wireless standards?

ocicat002
Level 1
Level 1

If I have a 802.11g access point which has 802.11a, b, & g clients attached, do all transmissions run at the lowest common denominator (802.11a) for *all* conversations, or only for those conversations specifically involving the 802.11a client(s)? If two 802.11b clients are communicating with each other, does their conversation run at 802.11b?

Thanks for any clarification offered.

1 Reply 1

scottmac
Level 10
Level 10

802.11a is a completely different frequency band than 802.11g/b. An 802.11g/b AP couldn't talk to an 802.11a client (or vice-versa) ... it'd be like trying to listen to AM radio on an FM receiver.

802.11b uses he same frequency band and channel scheme, but the encoding & modulation are different, so an 802.11b client couldn't talk to an 802.1g (only) AP; it'd be like trying to speak Chinese to person that only understands American English (right tone range, wrong encoding).

Now, 802.11g supports a compatibility mode that allows an 802.11g AP to pass 802.11b traffic, but it requires that the AP coordinate with both sides, since they cannot see each other.

The overhead caused by the coordinating protocol(s) reduces the overall throughput of the AP for all traffic.

So, a B-client can talk to other B clients over wireless, or to other wired clients via the AP .... as well as G-clients though the AP due to the compatibility mode (which also acts as a traffic cop to keep the two sides from stepping on each other).

Finally, regarding "lowest common denominator" ... 802.11a is a higher frequency and has more non-overlapping channels than 802.11 g/b. It has fewer users too. Those two factors tend to provide faster, cleaner throughput than anything available on 802.11g/b (which is cluttered, only has three non-overlapping channels, and is interfered with by things like microwave ovens, bluetooth, or a much larger number of cordless phones).

802.11a is hardly the lowest common denominator. 802.11b is the oldest and slowest of the bunch in common utilization.

Good Luck

Scott

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