[MLB-WIRELESS] Theoretical limits of 2.4 GHz band

Donovan Baarda abo at minkirri.apana.org.au
Mon Nov 18 08:46:18 EST 2002


On Mon, Nov 18, 2002 at 12:20:45AM +1100, Robert Graham Merkel wrote:
[...]
> <discussion of Shannon's Theorem, skip if you already understand it>
> 
> Channel Capacity = (Hardware Bandwidth) * LOG2(1 + Signal/Noise ratio).
> 
> Now, the hardware bandwidth is simply the chunk of spectrum allocated
> to the communication.  If we used the entire 2.4 Ghz band that's 100
> Mhz, so hardware bandwidth = 10^8 for this calculation.
> 
> The signal/noise ratio, for this formula, has to be the actual ratio, 
> not the decibel ratio.  The decibel ratio is 10 log10(S/N), so the actual
> signal to noise ratio is 10^((signal-to noise in db/10))
> <end discussion>
> 
> Anyway, the point of all this, is given the signal-noise ratio and the 
> amount of the 2.4 Ghz band you want to use, you can determine how much
> data could ultimately be carried on the band.  

I'm not 100% sure if shannon's law can be translated directly into kbps
though... modems take advantage of encoding multiple bit's per baud using
something called "phase shift keying" (PSK) where a single phase change
encodes multiple bits in the degree of phase change. These phase changes are
usualy combinded with a symultanious amplitude change to give you even more
bits/baud.

I have a feeling though that these tricks to encode multiple bits into one
baud actually increase the bandwidth requirements... so shannons law probably
still applies.

> However, I'm not sure what values to use for the signal-noise ratio.  
> Has anybody got some measured or calculated data on S/N ratios for 
> interesting configurations
> that we could plug into Shannon's Theorem and see what technology might
> ultimately give us?

The interesting thing about spread-spectrum is it distributes the signal
evenly over the whole range used. This means it looks like background noise
to everyone else. This stuff was invented by the military who wanted radio
comms that was hard to detect and jam; it looks like white noise, and to jam
it you've goto hammer the whole spectrum.

However, because of the way the recieved data is "reconstituded" from the
signal over the whole spectrum, anything else in the used spectrum range
contributes to the noise. You can overlay multiple wireless LAN's using
different "channels", but they each contribute to the noise. 

Sorry I can't give you any formula's, but I think it will mainly reduce to
"how many different wireless LAN's can be supported at once". I think
the way spread-spectrum ovelays "noise" on "noise", the spectrum usage is
pretty efficient; two 50MHz channels use as much spectrum as one 100MH
channel.

If you assume there is no other noise, that shannons law does translate
directly into Kbps, and that multiple channels are 100% efficient, you get;

signal to noise ratio = 1
total channel capacity = 100MHz * log2( 1 + 1)
                       = 100MHz * 1
                       = 100MHz                 (funny that :-)
                       
So this can fit just under 10 11MHz channels... Throw in any other noise,
and it just goes downhill from there.

-- 
----------------------------------------------------------------------
ABO: finger abo at minkirri.apana.org.au for more info, including pgp key
----------------------------------------------------------------------

To unsubscribe: send mail to majordomo at wireless.org.au
with "unsubscribe melbwireless" in the body of the message



More information about the Melbwireless mailing list