# Why can't we get 1 b/s/Hz?January 17, 2007 4:47 AM   Subscribe

Why, in wireless transmissions of data, can we not achieve 1.0 bits per second per Hertz?

(Or have we already done it?) I've been reading that the effectiveness of radio transmission is measured in bits/s/Hz and I realised that we can only get 2.25% effectiveness (54mbit over 2.4GHz) on consumer-level equipment. I hope these calculations are correct and I haven't horribly misunderstood the nomenclature. If I have, please let me know! If I haven't, can someone explain why we can't get closer to 100% efficiency?
posted by PuGZ to Technology (14 answers total) 1 user marked this as a favorite

I think this is what you are looking for:

http://en.wikipedia.org/wiki/Spectral_efficiency#System_spectral_efficiency
posted by stovenator at 5:04 AM on January 17, 2007

posted by stovenator at 5:04 AM on January 17, 2007

To be more clear, you aren't using the entire 2.4 GHz spectrum, but only a small of 20MHz slice of that spectrum. The spectrum for 2.4 runs from 2.400 ~ 2.473 for 802.11b/g and is divided into 11 channels (for North America). so each 20 Mhz channel is carrying 54Mbps of data. that 270% efficiency.

MIMO allows for even greater throughput, but that relies on dual antenna sending mutiple signals, and some tricky algorithms to reassemble the data stream (which may not be feasible in all environments, nor work at large distances).
posted by stovenator at 5:11 AM on January 17, 2007

Response by poster: To be more clear, you aren't using the entire 2.4 GHz spectrum, but only a small of 20MHz slice of that spectrum.

How could I possibly forget this - d'oh! Thanks for the links and brief explantion, they'll go a long way to clearing this up for me. (I'm only 17, but hope to be an EEng one day ;-))
posted by PuGZ at 5:18 AM on January 17, 2007

Response by poster: Forgive another silly question, but here it is. AFAIK, bits/s/Hz simplifies to bits/cycle. How is it possible to transmit more than one bit in a cycle?

(I am sure this is embarrassingly easy, but we're never taught anything of the sort in school and I'm missing some of the steps between the fundamentals and the fun stuff!)
posted by PuGZ at 5:26 AM on January 17, 2007

How is it possible to transmit more than one bit in a cycle?

One answer is compression, ie coding a bitstream to take advantage of redundancies, repetitions and patterns.
posted by unSane at 5:30 AM on January 17, 2007

And another is the various levels of QAM and other phase / vector modulation techniques. Simplified, you can transfer more than 1 bit per cycle.
posted by Pinback at 6:27 AM on January 17, 2007

Imagine amplitude modulation, where that one cycle can have a peak to peak magnitude of +/- 1, 0.75, 0.5, or 0.25. There are 4 states that one cycle could have, so there are 2 bits of information there, in just one cycle.

You could play the same tricks with phase, frequency, or amplitude (or a combination). There are many different protocols out there, many of which are very clever. However, amplitude is often the easiest to understand.
posted by PJensen at 6:29 AM on January 17, 2007

BTW, my example of 4 states in a cycle is just for example, you can actually have many more than 4 states.

As Pinback pointed out, QAM is a class of protocols with both amplitude and phase modulation, and QAM-16 is a typically used instance of that protocol with 16 possible states.
posted by PJensen at 6:32 AM on January 17, 2007

Shannon's Theorem
posted by leapfrog at 7:42 AM on January 17, 2007

As it happens, we have been transmitting more than one bit per cycle for a long time. For those of you old enough to remember modems, we often conflated "bits per second" with "baud." The two are not the same. "Baud" refers to modulations per second. A 300-bps modem was in fact 300 baud, but a 1200-bps modem was also 300 baud.

With modern wireless protocols, I'm not sure if they count error-correcting overhead in the bit rate or not, but there's a lot of that involved as well. Both dealing with bad packets and the built-in redundancy and mechanisms that make it possible to recognize and deal with those bad packets. So that's another way that theoretical and actual bandwidth diverge.
posted by adamrice at 8:25 AM on January 17, 2007

those of you old enough to remember modems

Ouch. I have never felt so old in my entire life.
posted by scalefree at 8:50 AM on January 17, 2007

Scalefree, you have never been so old in your life. But it beats the alternative.
posted by adamrice at 9:33 AM on January 17, 2007

As leapfrog says, you need to read up on Shannon. One critical point is that reliable communications requires redundancy, and redundancy is parasitic on efficiency. A transmission channel which utilized 100% of its bandwidth can't detect or recover from communications errors.
posted by Steven C. Den Beste at 11:20 AM on January 17, 2007

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