[opendtv] Re: Latest S/N test
- From: "Manfredi, Albert E" <albert.e.manfredi@xxxxxxxxxx>
- To: "OpenDTV (E-mail)" <opendtv@xxxxxxxxxxxxx>
- Date: Tue, 25 Jan 2005 18:21:27 -0500
Bob Miller wrote:
> I am waiting for more info. They did say that
> the data rates were as close as possible to
> 19.3 Mbps for all modulations tested.
Okay, so I'll assume either an 8 MHz or a 6 MHz
RF channel. And I'll use the same percentage of
the band in each case, to provide for credible
guard bands.
6 MHz (5.38 MHz used), at 19.3 Mb/s
Shannon limit 10.42 dB S/N
8 MHz (7.17 MHz used), at 19.3 Mb/s
Shannon limit 7.37 dB S/N
There's also the propagation channels to think
about. If the channel is Gaussian, you will
approach the Shannon limit more easily. If the
RF channel is degraded by multipath fading,
you will require progressively more S/N for
solid reception, moving away from the Shannon
limit.
When robust schemes are achieved through pilots
or other modulation tricks, the S/N required for
reception in degraded channels will suffer
least compared with Gaussian channel
performance, but under benign conditions you'll
pay a price in higher S/N required.
If robustness is achieved through clever signal
processing, then you can more easily approach
the Shannon limit in benign conditions, but the
robustness will depend entirely on your cunning
SP routines. So you tend to pay a price in
degraded channels.
This is another example of the no free lunch
hypothesis.
Also, to understand the numbers and to get a
good idea about where you are wrt the Shannon
limit, it helps to use (S+I)/N rather than
S/(N+I), where I is the sum of interfering
echoes of the main signal. Gives much less
impressive numbers unless you know what to
look for.
What's impressive is to see (S+I)/N in
Rayleigh channels that approach or even beat
the S/N in a Gaussian channel. Even more
impressive when these two numbers are close to
the Shannon limit.
In principle, (S+I)/N in a fading channel can
beat S/N of a Gaussian channel because
knowledge of the fading mechanism can give the
receiver information to better untwist the
distorted symbols. Whereas in a Gaussian
channel, the distortion is random, which means
unpredictable.
Bert
----------------------------------------------------------------------
You can UNSUBSCRIBE from the OpenDTV list in two ways:
- Using the UNSUBSCRIBE command in your user configuration settings at
FreeLists.org
- By sending a message to: opendtv-request@xxxxxxxxxxxxx with the word
unsubscribe in the subject line.
Other related posts:
