Re: frequency band

On Sat, 13 May 2006 17:16:51 GMT, "DAB sounds worse than FM"
<dab.is@xxxxxxxxxxxxxx> wrote:

Don Pearce wrote:
On Sat, 13 May 2006 14:27:57 GMT, "DAB sounds worse than FM"
<dab.is@xxxxxxxxxxxxxx> wrote:

Don Pearce wrote:
On Sat, 13 May 2006 13:45:04 GMT, "DAB sounds worse than FM"
<dab.is@xxxxxxxxxxxxxx> wrote:


No, the noise drops.


How can the noise drop? We're talking about Gaussian noise.


No we aren't.


http://en.wikipedia.org/wiki/Shannon%E2%80%93Hartley_theorem

"In information theory, the Shannon-Hartley theorem is an
application of the noisy channel coding theorem to the archetypal
case of a continuous-time analog communications channel subject to
Gaussian noise."


Sure - that is the only way to define the theory. The other spacial
channels tend to be sufficiently incoherent that the noise has
Gaussian characteristics, so the theory still holds.


It definitely does not hold. It's either Gaussian or it's not, and it's not.
Take for example an interfering single-frequency tone. How would that be
made Gaussian?

We don't have single tones. The interfering channels are random noise
as far as the wanted receiver is concerned. It looks Gaussian.


It isn't thermal
in origin, though, so it is amenable to reduction by the same
mechanism that boosts the wanted spacial channel.


Noise isn't reduced. Interference can be and is reduced, but not noise.

I'm including interference in the noise, as it is noise-like as far as
the receiver is concerned.

The noise includes interference from other channels,
cells, whatever. That is what sets the noise floor in a MIMO system.
It reduces when you can direct the power where it is needed.


You improve the signal level and reduce the
interference level simultaneously - that is where the huge
improvement comes from.


No, because Shannon's Capacity Theorem assumes there is no
interference -- just signal received with no interference, add
noise, demodulate. So MIMO beats the case where there is no
interference.

The reason why MIMO beats Shannon's Capacity is because you send
*different* data on the different streams, and MIMO allows you to
recover the data. Each individual data stream must obey Shannon's
Capacity Theorem, but as a whole MIMO beats the Capacity Theorem.

No. Certainly in the iBurst system, you send a different signal, but
not different data. Each antenna sounds its channel on the uplink
training burst and builds a matrix of all the multipaths it can
find. It then inverts that matrix for the downlink. This happens on
each antenna. Although the signals from each antenna are thus
different, by the time they reach the wanted remote receiver, they
have cohered into a single, in phase signal. The same thing is
happening simultaneously for the other remotes being served on that
channel. The result of all this is that what the remote receives is
a single, particularly clean channel, while the "interference" from
the other spatial channels is randomly averaged and thus reduced.

So as far as the receiver is concerned it does not receive different
data on each MIMO channel, it simply sums to the same data.


I'm not going to comment on how this iBurst system works, but all the
space-time codes I can remember reading about transmit different
data from different antennas.

Also, just because this iBurst system uses MIMO, i.e. multiple
transmit and receive antennas, does not mean that it's anywhere near
optimal. The equation I quoted before is the optimal case, and which
shows that it is possible to beat Shannon's Capacity by using MIMO.

It is very close to optimal.


If it is very close to optimal then it beats Shannon's Capacity Theorem.

No. Perfectly optimal would match Shannon. Near optimum falls short by
a small amount.


And of course although the signal on the
various antennas is different, the data can not be.


Of course it can be different. Take Alamouti's method: Take two symbols, x0
and x1 and transmit as follows:

First:
Tx1 = x0, Tx2 = -x1*
then
Tx1 = x1, Tx2 = x0*

Different data (x0 and x1) on both antennas in both instants.

What do you consider to be data here? Are you talking about actual
payload?


The system only
works by cohering the various multipaths into a single data stream.


You say that they find the downstream channel by transmitting upstream. If
that is the way they do it then it is far from being optimal, because the
upstream channel is a poor estimate of the downstream channel. It is far
better to transmit training symbols to find the downstream channel than
trying to estimate the downstream channel from the upstream channel.

No - it is a very good estimate. This is TDD, so the channel is the
same frequency and the direction is switched at 500Hz, so the time
interval is very brief. This allows for mobility in the very worst
multipath at over 70mph. It is only by sounding the upstream channel
that you can generate the Eigenvector table for optimizing the
downstream channel (which carries much faster data). There is also a
training burst on the downstream channel so that the remote receiver
can do a final channel estimation tweak to mop up any small changes
since it last received. It really is a very robust system

d

--
Pearce Consulting
http://www.pearce.uk.com
.

Relevant Pages

  • Re: frequency band
    ... We're talking about Gaussian noise. ... noisy channel coding theorem to the archetypal case of a continuous-time ... That is what sets the noise floor in a MIMO system. ... Each antenna sounds its channel on the uplink ...
    (alt.radio.digital)
  • Re: Generalized Channel Capacity Theorem?
    ... >uncorrelated noise. ... but the channel response is frequency ... MIMO systems of which I'm aware don't use any spreading. ... >capacity possible out of a channel, and MIMO design afforded a higher ...
    (comp.dsp)
  • Re: Channel
    ... i m not clear about AWGN channel. ... so you add the noise instead of dividing or multiplying on something. ... It's white, so it has zero mean. ... Finally, the noise is Gaussian, so it's not uniform or a Poisson process or bi-value or anything else. ...
    (comp.dsp)
  • Re: frequency band
    ... We're talking about Gaussian noise. ... case of a continuous-time analog communications channel subject to ... That is what sets the noise floor in a MIMO system. ...
    (alt.radio.digital)
  • Re: frequency band
    ... We're talking about Gaussian noise. ... noisy channel coding theorem to the archetypal case of a continuous-time ... That is what sets the noise floor in a MIMO system. ... Each antenna sounds its channel on the uplink ...
    (alt.radio.digital)