question about Shannon capacity

Hello, can Shannon capacity be used to estimate the rate for all wireless technologies or just certain ones? I've been calculating channel capacities for 802.16 and 802.11 fine, but 802.15.4 is really confusing me. My data rates are an order of 10 times bigger. How do you calculate the capacity of an 802.15.4 channel?

Also, 802.15.4 specifies fixed data rates. But to me, logically, varying an amount of a physical layer resource, such as power or bandwidth, should vary your rate. How are 802.15.4 rates fixed? Doesn't the Shannon capacity prove that you can only fix a data rate by adjusting other physical layer properties to compensate?

Any guidance on this would be appreciated. To me using the shannon limit C = Wlog2(1 + P*G/(w*N + I)) should allow me to estimate the channel capacities for any technology considering their physical layer resource restrictions.

Thanks, Omar

Reply to
Ginu
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Shannon doesn?t lie but that doesn?t mean that real channels can use=20 their theoretical maximum data rate.

I don?t know the particular technology in question but if you beat=20 Shannon, write a paper! ;-)

?Fixed? as in ?repaired? or ?fixed? as in ?constant?? =20

Shannon gives the theoretical limit of the data rate of a=20 communications channel.

The upper bound of the channel capacity, yes.

--=20 Keith

Reply to
krw

Hi Keith, thanks for the clarification. It helped a lot. In reference to your questions:

As in constant. My problem is occurring because the 802.15.4 specification, for example, states that the "UWB PHY supports an over- the-air mandatory data rate of 851 kb/s..." The selection of the term "mandatory" tells me that the rate is fixed or constant. In fact, the Zigbee standard states a "fixed data rate of 250 kbps". The Shannon rate that I am calculating gives me a much higher rate for Zigbee, but your statement that "that doesn't mean that real channels can use their theoretical maximum data rate" makes a lot of sense and would put a limit on the rate that I can achieve.

And again, the reason my upper bound is so high is that real channels can't always use their theoretical maximum data rates. So all I would do is put a cap on the maximum data rate that I can achieve using Shannon capacity in my simulation.

Thanks again Keith.

Reply to
Ginu

(don't know what happened to my '"' marks.)

Yes, understand that Shannon states the maximum theoretical data throughput assuming perfect components in a perfect world.

The real world isn't perfect so you can't have perfection. It's rather like approaching the woman on the park bench by dividing the distance in half in each iteration. The scientist gives up because she's unreachable. The engineer knows he can get "close enough". ;-)

Right. The real world sucks, but keeps us in a job. ;-)

Keep up the hard work. We need more good engineers.

Reply to
krw

anks for the clarification. It helped a lot. In reference

Will do. Thanks for the help.

Reply to
Ginu

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