Red Wave Radio

Quote:

> It is a common error to assume that digital communications are similar
> to analog RF. One reason is very fast edge times are required to
create
> the most eye margin possible at the decoding end of a data stream so
the
> bandwidth required is much greater. A good rule of thumb is 3.7 times
> the clock rate as a minimum. Usually the engineering shoots for the
> fastest edge times practical.

Well, I thought we started out with harmonics of the BPL carriers.  I
don't see why there would have to be BPL harmonics due to the digital
modulation anymore than a RTTY transmission would have to have
harmonics.

If you're saying the sidebands of carriers will be spaced as far as 3.7
times as far as the data clock rate, sure, why not?  I don't know any of
the specifics.  But if the BPL carriers stop at 80 Mhz, I suppose the
total spectrum won't go much past 80 Mhz..

Quote:

> An one/zero pattern and multiples thereof are square waves but I
should
> not have used that term because it looks like I just threw you off the
> path of understanding.

> --
> Telamon
> Ventura, California

That's the rocky path of understanding, for ya.  Can't even take take my
shoes off when I need to count past 10.

Frank Dresser

In article

No it just that since the data is sent without a clock the data stream
regardless of the encoding need fast and precise (low jitter) edge
times. Faster edges provide more timing margin. Fast edges have most of
the energy in the odd harmonics 1, 3, 5, 7 etc. Most of the energy is in
the lowest odd harmonics 1, 3 and 5 being the most important. This
explanation only makes sense for a single carrier two level scheme. I do
not know what BPL employs but I expect a high frequency scheme be used
to reduce the coupling requirements across transformers in the power
system.

--
Telamon
Ventura, California

Quote:

> No it just that since the data is sent without a clock the data stream
> regardless of the encoding need fast and precise (low jitter) edge
> times. Faster edges provide more timing margin. Fast edges have most
of
> the energy in the odd harmonics 1, 3, 5, 7 etc. Most of the energy is
in
> the lowest odd harmonics 1, 3 and 5 being the most important. This
> explanation only makes sense for a single carrier two level scheme.

Let's say one of the BPL carriers is at 10 Mhz.  Let's say it's
modulated at 10 khz.  If you're saying the modulation is making a
channel which covers something like 9.960 Mhz to 10.040 Mhz, that sounds
OK to me.

If you're saying the modulation creates harmonics at 20, 30, 40 Mhz, I
can't see how.

Quote:> I do
> not know what BPL employs but I expect a high frequency scheme be used
> to reduce the coupling requirements across transformers in the power
> system.

> --
> Telamon
> Ventura, California

Power pole transformers should have a nice grounded copper
electrostactic shield between the primary and secondary windings.  This
reduces capacitive coupling between the windings to almost zero.  The
BPL company will have to bypass the shield with some sort of bandpass
coupling.  I suppose something as simple as a capacitor would do the
job, but they probably have something more elaborate.

Maybe they're using a small ferrite transformer with enough insulation
to withstand the full primary voltage.  Bypassing the power
transformer's internal shield would be a lightning hazard.

Frank Dresser

Unfortunately there's no readily available description of
how BPL works, so none of us can really claim he knows exactly
what he's talking about. My general understanding is that
BPL uses a set of spaced carriers, each modulated at some
reasonably low data rate. These carriers are arrayed, I think,
over a range like 2MHz to 75 Mz or maybe 2 - 40 or something
of that order of magnitude.

The transmitted signal is not a pure digital baseband signal -
anytime any wideband signal has to propagate over any appreciable
distance this can't be used because the medium be it wire or "ether"
has a response which varies in frequency, and is not flat enough
over the passband of the entire signal bandwidth. There are two
broad classes of approach to combatting this (very oversimplified).
A single carrier approach takes the baseband signal and uses
it to modulate a carrier. This moves the spectrum away from DC
which most media don't like (including power lines as you note in the
paragraph I've quoted below). The resulting signal would have a
natural bandwidth of a modest multiple of the symbol rate
as you have noted in another post. Even this is usually too wide
in most applications, so the baseband pulses are shaped in time
to concentrate them (around DC at baseband, but around the carrier
when using the signal to modulate one). This pulse shaping causes the
pulses to have very long time durations compared to the signalling
interval, but a matched filter is used at the receiver, to
re-concentrate the energy temporally.

The other broad approach is to use a bunch of evenly spaced RF
carriers (as Frank has suggested, and is what I think they actually
do). The tones are spaced at frequency intervals of 1/(signalling
interval), and the tradeoff between length and frequency spacing is
a design decision depending on medium - it does not affect the overall
data rate. You can key twice as fast but then carry only half as many
tones in a given bandwidth. This is OFDM. In this case, since the
signalling rate of each tone is only a small fraction of the total
bandwidth the fact that the effective bandwidth of a rectangular pulse
is 3 or 5 or whatever times the signalling frequency doesn't effect the
total bandwidth by much since the total bandwidth is that of hundreds or
thousands of tones. Again, like the case of single carrier, since each
tone modulates an RF carrier, there is no LF energy in the resulting
signal. Now the medium is roughly flat over the effective BW of each
individual tone, and all is basically well.

Lastly there was a confusion in the thread between harmonics and excess
bandwidth. A randomly modulated square wave does not have discrete
harmonics, because the modulation eliminates the periodicity. It does
have excess BW as per your discussion of the need for sharp edges.
Again in practice the pulses used are not rectangular so the spectrum
does not have the ideal (sin(x)/x)^2 roll-off, but something that falls
off much more quickly, but there is still significant energy beyond
the Nyquist frequency.

Oz

Quote:

> Bypassing the power
> transformer's internal shield would be a lightning hazard.

I meant to say that bypassing the internal shield in a simple way could
be a lightning hazard.  I suppose some kind of opto-isolator would work
well.

Frank Dresser