Jason: Part 15 contains no requirement that minimum power be used. It will after the final rules for the white spaces are codified — and that will be ONLY for the white spaces. However, there is an inherent problem in it: to determine the minimum power required to communicate, you have to turn the power down regularly so as to cause communications to fail. Broadband providers’ equipment cannot do this; it needs to work perfectly, 24×7. Another reason why the release of the white spaces as unlicensed is a horrible idea.

On a shared channel, only one frame can be successfully
transmitted at a time, so the length of time my neighbors occupy the
channel directly affects my ability to use the channel.

Besides being not true (c.f. any non-TDMA system like CDMA or FDMA)
the implicit model being discussed is non-cooperating users. That is,
other users of the channel are external interferers.

You don’t
need Shannon’s Law to understand that, it’s simple arithmetic.

Like I said in the previous post this has nothing to do with Shannon’s
law (the premise used is the rate-robustness tradeoff which I do not
dispute).

All else being equal the rate at which other users transmit has no impact on the performance on a given user’s communication.

On a shared channel, only one frame can be successfully transmitted at a time, so the length of time my neighbors occupy the channel directly affects my ability to use the channel. You don’t need Shannon’s Law to understand that, it’s simple arithmetic.

The corollary of Shannon’s Law is that lower data rates require less bandwidth, Jason. Work the problem of low signaling rates in a shared channel and I think you’ll see Brett’s point. If it’s not immediately obvious, consider that many of our popular wireless systems use redundant coding to increase robustness, simply adding or subtracting redundancy at constant modulation.

I understand perfectly well that you can trade robustness for rate. I
was disputing that this is a consequence of the channel capacity
theorem. My opinion is that working with capacity and interference is
not informative: rate and BER are meaningful. But this unimportant
in order to make the following point:

All else being equal the rate at which other users transmit has no
impact on the performance on a given user’s communication. The
rate-race-to-the-bottom assertion is not valid.

Jason, you should read the Part 15 regulations. They absolutely do not say that a device should transmit with no more power than necessary

The whitespaces order says:

All devices must include adaptable power control so
that they use the minimum power necessary to accomplish communications.

As for CDMA vs. GSM: CDMA is an entirely different scheme which differentiates between simultaneous transmitters by encoding rather than requiring them to transmit one at a time. But their bandwidth is constrained by the coding, so that the effective throughput is the same in either case. Also, CDMA requires EXTREMELY careful regulation of power levels. And it cannot be used except on licensed spectrum, because it is very sensitive to in-band interference.

Finally, as for the applicability of Shannon’s Law: I give up. You don’t seem to be capable of getting it, or maybe you just do not want to because you have some preconceived notion of how things “ought” to be.

Again, Jason, you’re missing several important points. Firstly, with
GSM, there is one licensee per channel, and therefore there is no
random interference from millions of babbling consumer
devices.

The point was that multiple access communication can be done with many
users. There is no reason multiple access cannot function with loosely
coordinated or even uncoordinated devices.

Secondly, GSM uses time division multiplexing, so the devices
speak in turn — not all at once — and do not interfere with one
another.

This is not fundamental. Take CDMA instead.

Thirdly, GSM devices only transmit with as much power as
necessary; unlicensed consumer devices do not do this.

I believe the FCC order mandates that they must.

As for why this follows from Shannon’s Law: it’s an
obvious corollary.

I hate to keep harping on this because it is not really germane to the
discussion but the channel capacity is not the rate at which practical
systems transmit only an upper bound and rate upper bounds would not
be terribly useful in comparing two systems. More importantly,
Shannon’s theorem only talks about zero (in the limit) probability of
error communication and therefore effectively says nothing about
non-arbitrarily small error rates like real communication systems
have. Normally what would be meant by a rate-robustness tradeoff is
that, for a fixed level of interference, lower data rates result in
lower bit-error rates and vice-versa. *This* does not follow from the channel capacity theorem.

Since the maximum data rate of a channel depends
upon the signal to noise ratio, equipment that can get away with
transmitting at a much slower data rate (e.g. that railroad telemetry
equipment, which can work at 9600 baud instead of megabits per second)
can “win” the battle when there is interference. But in doing so, it
acts as a spoiler. An ISP who needs to do broadband can’t expect its
users to tolerate slower-than-dialup data rates.

How does it act as a spoiler? The only thing that would help the WISP
is if the railroad telemetry produced less interference (or didn’t
transmit at all) not at what rate it transmits. You may be saying that
railroad telemetry is not important and shouldn’t be transmitted in
order to improve performance of more “important” systems but that is a
different discussion.

Richard points out a minor sort of “race to the bottom,” in which a group of similar rate-adaptive devices cause one another to drop to the lowest speed they all support. But this isn’t common, especially if they can avoid one another via CDMA. What’s much worse — and, alas, a much bigger problem — is when a user of unlicensed spectrum chooses a piece of equipment that ignores whether other equipment is transmitting and just tries to “conquer” all of the others. This occurs, for example, on the 900 MHz band, where railroads use very slow (but very robust) frequency hopping radios for down-the-track signaling. Wireless ISPs literally cannot use the 900 MHz band for broadband within a mile of a railroad track as a result of these noisy radios.

As for why this follows from Shannon’s Law: it’s an obvious corollary. Since the maximum data rate of a channel depends upon the signal to noise ratio, equipment that can get away with transmitting at a much slower data rate (e.g. that railroad telemetry equipment, which can work at 9600 baud instead of megabits per second) can “win” the battle when there is interference. But in doing so, it acts as a spoiler. An ISP who needs to do broadband can’t expect its users to tolerate slower-than-dialup data rates.

The point is that, like the 900 MHz band (which
penetrates walls a bit less), the unlicensed “white spaces” will
become completely congested with signals that leak from buildings.

The same logic would imply that GSM band would be “completely
congested” with signals that leak through buildings also. Or any other
long-range multiple-access wireless communication medium. The fact is
that people want to communicate through walls and buildings and
specfically want to use this spectrum to do so. Power control can
mitigate interference transmission for those do not intend to transmit
over long distances.

That’s not what I said. What I said was that, as a basic
corollary of Shannon’s Law, transmissions at slower data rates are
more robust in the face of interference than those at higher data
rates. There is thus a “race to the bottom” in which the slowest
(and, hence, least efficient) equipment “wins” the interference
battle. So, your baby monitor or wireless speaker system (which could
just as well have been on another unlicensed frequency) will work OK,
but it will render the spectrum useless for broadband.

This is obviously false. It does not matter to User A at what rate
User B transmits at, only the interference User B produces. If there is
much interference then users will be forced to transmit at
correspondingly lower rates to achieve the same robustness but this
has nothing to do with the rates that other users choose; the same
problem would exist if other users decided to select high-rate
low-robustness transmissions.

Also, although it is not relevant, I would like to know how the
rate-robustness trade-off follows specifically from Shannon’s channel
capacity theorem. I have access to several communications theory
textbooks from none of which this consequence is apparent. I’m always
happy to learn something new.