In article ,
"Frank Dresser" wrote:

"Telamon" wrote in message
...


Data communications occupy wider bandwidths than the stated clock
rate.
It is not unreasonable to expect harmonics 3 to 5 times the clock rate
because the signaling uses square waves and there is significant power
in the odd harmonics.

--
Telamon
Ventura, California

A square wave, itself, won't convey much information. It needs to be
modulated, and the modulation would have to effect the symmetry and
result in both odd and even harmonics.

I don't know what sort of modulation BPL is using. I can imagine
hundreds of low amplitude sine wave carriers from 2 to 60 Mhz, all of
them phase modulated. In that case, I don't think there would be much
harmonic output. This would certainly still be a big problem for the
radio hobbyist, but not so much for the FM/TV user. There have been
several BPL tests in various communities, and it doesn't seem to have
wiped out normal broadcast use.

If BPL caused enough bothersome interference to keep people in the test
communities from their TVs and radios, the National Association of
Broadcasters would have squashed it like a bug.


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.

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

"Telamon" wrote in message
...

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..



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
,
"Frank Dresser" wrote:

"Telamon" wrote in message
...

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..



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.

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

"Telamon" wrote in message
...

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.


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

"Frank Dresser" wrote in message
...

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