New Technology Could Kill WiMax?

GolygyddMax writes "Techworld reports that a Florida-based start-up, xG, has developed a technology that's a 1000 times more efficient than WiMax and which could, in theory, lead to wireless LANs being powered by watch batteries. It is still in early development, but this technology could allow anyone to set up as an ISP. This could kill WiMax before it even gets off the ground." From the article: "At the demonstration with other reporters, we were able to verify that the signals were being sent wirelessly, and checked the distance by GPS, but had to take the 50mW base station - and its omnidirectional antenna - on trust, since it was at the top of an 850ft mast. The demonstration will be repeated for the US press next week. The system carried 7.4 Mbit/s per MHz per Watt, said Professor Schwartz. By comparison, GSM would have around 0.0058, and CDMA/EV-DO about 0.0085 Mbit/s per MHz per Watt. "

Comments

More details here

[Anonymous Coward]

Lots more details in this article, and photo's here. Looks very interesting.

Nope, WiMax will come first

[EriktheGreen]

The poster of this article assumes that the technically superior solution will rise to the top. In fact, the administratively superior solution will... this means that if companies spend millions of dollars preparing a standard and products for market, they won't switch to something else automatically even if it's obviously better.

The reality of the situation is that if the new solution is exactly what it's sold to be (unlikely) then it probably will eventually break into the market, but even if it's made into a useable product immediately its use will be overshadowed by the well advertised and enthusiastically sold solution that the vendors are pushing instead. Vendors really don't care what's superior unless they're picking technologies from a menu and they have no interest in any of them (positive or negative). Vendors care about money, and if they've already spent some on one technology, they won't switch unless it's obvious that another technology will immediately dominate the market (VERY, VERY rarely does this happen).

Take off the rose colored glasses, people. Technically superior solutions MAY eventually win out over poorer ones if all else is equal, but all else NEVER is equal.

Plus, it's unlikely that this "breakthrough" is anything but some ambitious people trying to sell something inferior as if it's the solution to All Our Problems (tm).

Erik

IAARE

[elgatozorbas]

I am a radio engineer...
well, not professionally but I know what it is about.

Digital transmission works as follows: you select a certain waveform out of a set and transmit it. At the receiver you try to figure out which one it was. Unfortunately the reception is distorted because of noise you pick up, such that the distinction is not perfect (e.g. in case you can reliably tell 8 possible waveforms apart three bits will be conveyed each time you do this). Using more power will lead to a better distinction and therefore higher bit rate. Using a larger (RF) band width allows you to send more waveforms per second hence also increasing the number of bits transferred (this is simplified somewhat).

Shannon left us a nice formula to calculate the capacity aka maximum possible throughput EVER, but first you need to calculate the signal and noise power you receive.

1) If we assume the waves travel in free space, the received signal power will be dependent on
- transmit power
- transmit antenna gain (dish is more focused than dipole etc.)
- free space loss (FSL, i.e. field strength getting weaker far from the source because the energy is spread out in all directions)
- receive antenna gain
This is an optimistic assumption because their setup takes place in suburban territory!

We can assume both the antenna gains are 0dB, being small and probably not perfectly matched.
The FSL is equal to: R^2*4pi^2/lambda^2 (R=distance, lambda=wavelength)
At 900 MHz lambda=0.33m, R=18 miles=29e3 m.
FSL= 3e11(in 'power') or 115dB.
The transmit power was 50mW, i.e.17dBm, the total received power will be 17-115=-98 dBm. The thermic background noise is equal to -173dBm/Hz (best case, due to ambient temparature - this is a bit optimistic too because other wireless devices are transmitting there too).

2) The channel capacity is given by Shannon as C=B*log2(1+S/N), where C=capacity (bits/sec), B=bandwidth (physical, in Hz), S=signal power (-98dBm), N=noise power (-173dBm/Hz*B).
You can now play with the bandwidth to influence the capacity. To a certain extent an increased bandwidth will increase the capacity but after a while you are just catching more noise while the signal will be spread out in frequency, so this saturates.
For these numbers the (theoretical) maximum capacity would be about 4.5e7 bits/sec or 45MB/sec. But even to achieve the 3.7Mb mentioned you already need a bandwidth of 700kHz (rough estimate, I made a plot in matlab).
At that point you transmit 3.7Mb/(50mW)/(0.7Mhz)=100Mb/s/W/MHz, so their figure of 7.4 MB/2/W/MHz is not impossible. However it will be difficult to achieve. We have made some assumptions (especially about the loss in the urban envorinment), and their bit rate only has a 'margin' of a factor 12 (45 to 3.7). There you have it.