| Author |
Message |
William P. N. Smith
Guest
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 Posted:
Tue Sep 13, 2005 6:08 am Post subject:
Re: Why is there a minimum spacing? |
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DHP <me@privacy.net> wrote:
| Quote: | William P. N. Smith <> said
DHP <me@privacy.net> wrote:
on why you went for multiples of 2.5m rather than have it as a simple
minimum.
It's not a minimum or maximum, it's the spacing that minimizes
reflection-based bit errors. 2M and 3M (for instance) are both worse
than 2.5M.
That is the thing I find odd - could you explain why, please?
|
Well, Rich said:
/*
I did extensive simulations of the resulting reflections from
transceivers at various spacings, and empirically determined that 2.5
m was "good enough."
[...]
The idea is not just a *minimum* 2.5 m spacing; it is that
transceivers are only placed at the 2.5 m markings.
*/
I'd have to imagine that the 2.5m is somewhere between two distances
that will cause problems in the worst case, perhaps 2m and 3m. |
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James Knott
Guest
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| Posted:
Tue Sep 13, 2005 6:50 am Post subject:
Re: Why is there a minimum spacing? |
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DHP wrote:
| Quote: | Incidentally, I read somewhere, that it's not the "DC" level on the
line that's measured, but the DC current taken from the power supply,
but I dare say that's as accurate as all the other bits of lore!
|
A bit of Ohms law and Thevenin's equivalent, will show those to be measuring
the same thing. |
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glen herrmannsfeldt
Guest
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| Posted:
Tue Sep 13, 2005 8:21 am Post subject:
Re: Why is there a minimum spacing? |
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glen herrmannsfeldt wrote:
(snip)
| Quote: | Say the cable has taps spaced 11.75m the whole length, and a signal
starts at one end transmitting all zeros. Some signal will
reflect off the first tap back to the transmitter. A similar
signal will reflect off the next tap and arrive at the transmitter
100ns later. Off the third tap will arrive 100ns later, etc.
Since the original is periodic with period 100ns all the reflections
will be of similar shape, though slightly decreasing amplitude.
|
After doing a few calculations I realized that this determines
the back reflection which may be different than the accumulated
forward reflections. That is, ones that reflect an even number
of times. Still, it should be that 11.75m is bad.
| Quote: | Now, say you take a cable and mark off 11.75m regions, numbered from
the end. If you place a tap at all the 2.5m marks that are in odd
numbered regions they will tend to add more than subtract. That will
be about the limit of 100 taps on a 500m cable, maybe the worst case.
|
I will try a few more calculations assuming that taps are only
placed where they add constructively to the first harmonic within
the specified spacing.
| Quote: | The third harmonic should be about 1/3 the amplitude, maybe less if
the square wave isn't perfect. Still worth worrying about but much
less likely to cause problems.
|
-- glen |
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DHP
Guest
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| Posted:
Tue Sep 13, 2005 8:21 am Post subject:
Re: Why is there a minimum spacing? |
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glen herrmannsfeldt <gah@ugcs.caltech.edu> said
| Quote: | DHP wrote:
glen herrmannsfeldt <gah@ugcs.caltech.edu> said
DHP wrote:
(snip)
I do not understand how the concept of constructive interference -
which applies to a narrow-band signal, a sinewave - can be applied to
a Manchester-encoded bit stream.
I don't believe that a scrambler is used, and repetitive bit streams
are fairly common. One could easily imagine the entire cable filled
with all zero bits. Otherwise, yes, for each combination of bits
there should be an appropriate combination of taps where they will
add constructively.
With Manchester encoding, even a stream of zeros is
actually a square wave.
With the Fourier series containing odd harmonics
proportional to 1/n.
|
See below :)
| Quote: | In any case, the reflection from a capacitive tap is (approximately)
the time-derivative of the origina. With a fast rise-time on the
transmitter, you'll just get a series of short pulses. With random
polarity of data, how can you ensure they cancel rather than add?
Whatever it looks like it will add in phase to one delayed by
one cycle. For a stream of zero bits that is one half a bit time
down the cable.
However, the reflections from a capacitive tap are not the same shape
as the original signal. I can't see how thay can be said to add
constructively or otherwise to the original signal. So we'd be looking
at two or more reflections a whole bit apart, which is 23.4m.
Say the cable has taps spaced 11.75m the whole length, and a signal
starts at one end transmitting all zeros. Some signal will
reflect off the first tap back to the transmitter. A similar
signal will reflect off the next tap and arrive at the transmitter
100ns later. Off the third tap will arrive 100ns later, etc.
Since the original is periodic with period 100ns all the reflections
will be of similar shape, though slightly decreasing amplitude.
|
Oh yea, sorry, I was forgetting the two-way trip... or rather I'd
thought about it and got it wrong :/ Too early in the morning.
| Quote: | Now, say you take a cable and mark off 11.75m regions, numbered from
the end. If you place a tap at all the 2.5m marks that are in odd
numbered regions they will tend to add more than subtract. That will
be about the limit of 100 taps on a 500m cable, maybe the worst case.
The third harmonic should be about 1/3 the amplitude, maybe less if
the square wave isn't perfect. Still worth worrying about but much
less likely to cause problems.
|
30MHz has a wavelength of 7.8m doesn't it? Placing taps with an
accuracy of much better than 1m doesn't make much sense unless you're
talking about much higher frequency components to make things happen
over << 1m.
With capacitive taps the 1/f signal spectrum gives a flat one in the
reflections - the spectrum of a series of spikes. So Nature strikes
back.
| Quote: | -- glen
Which is avoided by using multiples of 2.5m, I suppose. Maybe my
arithmetic is not exact and the figures are tweaked so that the
nearest multiples are 1.25m on each side. Even so, 1.25m is only
5.3ns, so this can't be an issue unless the "dangerous" parts of the
reflection are very narrow.
Maybe if the signal had a rise time of <3ns it would be an issue.
You've got me thinking, now. Not always a good thing :) |
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glen herrmannsfeldt
Guest
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| Posted:
Tue Sep 13, 2005 8:21 am Post subject:
Re: Why is there a minimum spacing? |
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DHP wrote:
| Quote: | glen herrmannsfeldt <gah@ugcs.caltech.edu> said
DHP wrote:
|
(snip)
| Quote: | I do not understand how the concept of constructive interference -
which applies to a narrow-band signal, a sinewave - can be applied to
a Manchester-encoded bit stream.
I don't believe that a scrambler is used, and repetitive bit streams
are fairly common. One could easily imagine the entire cable filled
with all zero bits. Otherwise, yes, for each combination of bits
there should be an appropriate combination of taps where they will
add constructively.
With Manchester encoding, even a stream of zeros is
actually a square wave.
|
With the Fourier series containing odd harmonics
proportional to 1/n.
| Quote: | In any case, the reflection from a capacitive tap is (approximately)
the time-derivative of the origina. With a fast rise-time on the
transmitter, you'll just get a series of short pulses. With random
polarity of data, how can you ensure they cancel rather than add?
Whatever it looks like it will add in phase to one delayed by
one cycle. For a stream of zero bits that is one half a bit time
down the cable.
However, the reflections from a capacitive tap are not the same shape
as the original signal. I can't see how thay can be said to add
constructively or otherwise to the original signal. So we'd be looking
at two or more reflections a whole bit apart, which is 23.4m.
|
Say the cable has taps spaced 11.75m the whole length, and a signal
starts at one end transmitting all zeros. Some signal will
reflect off the first tap back to the transmitter. A similar
signal will reflect off the next tap and arrive at the transmitter
100ns later. Off the third tap will arrive 100ns later, etc.
Since the original is periodic with period 100ns all the reflections
will be of similar shape, though slightly decreasing amplitude.
Now, say you take a cable and mark off 11.75m regions, numbered from
the end. If you place a tap at all the 2.5m marks that are in odd
numbered regions they will tend to add more than subtract. That will
be about the limit of 100 taps on a 500m cable, maybe the worst case.
The third harmonic should be about 1/3 the amplitude, maybe less if
the square wave isn't perfect. Still worth worrying about but much
less likely to cause problems.
-- glen
| Quote: | Which is avoided by using multiples of 2.5m, I suppose. Maybe my
arithmetic is not exact and the figures are tweaked so that the
nearest multiples are 1.25m on each side. Even so, 1.25m is only
5.3ns, so this can't be an issue unless the "dangerous" parts of the
reflection are very narrow.
Maybe if the signal had a rise time of <3ns it would be an issue.
You've got me thinking, now. Not always a good thing :) |
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glen herrmannsfeldt
Guest
|
| Posted:
Tue Sep 13, 2005 8:21 am Post subject:
Re: Why is there a minimum spacing? |
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William P. N. Smith wrote:
(snip on transceiver spacing calculation)
| Quote: | I'd have to imagine that the 2.5m is somewhere between two distances
that will cause problems in the worst case, perhaps 2m and 3m.
|
That could be, but there is also a desire to minimize the spacing
to make it easier for network engineers installing taps.
So, 3m is worse from that point of view, if not from the signal
point of view.
-- glen |
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DHP
Guest
|
| Posted:
Tue Sep 13, 2005 8:21 am Post subject:
Re: Why is there a minimum spacing? |
|
|
glen herrmannsfeldt <gah@ugcs.caltech.edu> said
| Quote: | DHP wrote:
(snip)
(I wrote)
Constructive interference would result from a half wavelength spacing,
I do not understand how the concept of constructive interference -
which applies to a narrow-band signal, a sinewave - can be applied to
a Manchester-encoded bit stream.
I don't believe that a scrambler is used, and repetitive bit streams
are fairly common. One could easily imagine the entire cable filled
with all zero bits. Otherwise, yes, for each combination of bits
there should be an appropriate combination of taps where they will
add constructively.
|
With Manchester encoding, even a stream of zeros is actually a square
wave.
| Quote: | In any case, the reflection from a capacitive tap is (approximately)
the time-derivative of the origina. With a fast rise-time on the
transmitter, you'll just get a series of short pulses. With random
polarity of data, how can you ensure they cancel rather than add?
Whatever it looks like it will add in phase to one delayed by
one cycle. For a stream of zero bits that is one half a bit time
down the cable.
|
However, the reflections from a capacitive tap are not the same shape
as the original signal. I can't see how thay can be said to add
constructively or otherwise to the original signal. So we'd be looking
at two or more reflections a whole bit apart, which is 23.4m.
Which is avoided by using multiples of 2.5m, I suppose. Maybe my
arithmetic is not exact and the figures are tweaked so that the
nearest multiples are 1.25m on each side. Even so, 1.25m is only
5.3ns, so this can't be an issue unless the "dangerous" parts of the
reflection are very narrow.
Maybe if the signal had a rise time of <3ns it would be an issue.
You've got me thinking, now. Not always a good thing :) |
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DHP
Guest
|
| Posted:
Tue Sep 13, 2005 1:39 pm Post subject:
Re: Why is there a minimum spacing? |
|
|
glen herrmannsfeldt <gah@ugcs.caltech.edu> said
....
| Quote: | After doing a few calculations I realized that this determines
the back reflection which may be different than the accumulated
forward reflections. That is, ones that reflect an even number
of times. Still, it should be that 11.75m is bad.
Now, say you take a cable and mark off 11.75m regions, numbered from
the end. If you place a tap at all the 2.5m marks that are in odd
numbered regions they will tend to add more than subtract. That will
be about the limit of 100 taps on a 500m cable, maybe the worst case.
I will try a few more calculations assuming that taps are only
placed where they add constructively to the first harmonic within
the specified spacing.
|
If you're talking about frequencies ~10MHz, moving a tap by 1.25m is
only going to change the phaseof one reflection by 20 degrees - (and
you can't even move them all in the same direction or you're just
moving the whole cluster!)
If you consider a wide bandwidth system though, the reflection is
actually a forest of spikes. So you can interleave them for
(presumably) minimum impact or superimpose them for worst case.
It seems to me that there's no advantage in moving taps around within
the cluster, but it would be useful to ensure that "forests" arriving
from different clusters always interleave. Insisting on regular
tapping points will acheive this as it creates a predicable set of
time slots for reflections even if not all of them are filled. The
next thing is to ensure the time-slot patterns interleave - by setting
the spacing appropriately. I can see it works, sort of, over a few 10s
of m but whether it can be maintained over a 500m LAN I don't know. |
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glen herrmannsfeldt
Guest
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| Posted:
Tue Sep 13, 2005 1:49 pm Post subject:
Re: Why is there a minimum spacing? |
|
|
DHP wrote:
(snip)
| Quote: | Which is avoided by using multiples of 2.5m, I suppose. Maybe my
arithmetic is not exact and the figures are tweaked so that the
nearest multiples are 1.25m on each side. Even so, 1.25m is only
5.3ns, so this can't be an issue unless the "dangerous" parts of the
reflection are very narrow.
|
Using some simple assumptions it does seem that 2.5m is much better.
First, I assume that for whatever spacing is used taps are only
placed where a period 11.75m cosine is positive.
I then compute the phase shift of the signal reflected off all
pairs of such taps and add them up. If you have awk or gawk
(there is a windows version of gawk around) you can run it and
see. The sum for 2.5m is about one fourth that for 2.4m or 2.6m.
There is also a minimum near 2.0m about twice that for 2.5m.
I didn't put any attenuation into the sum, though.
I don't know if this is at all related to what Rich did, but
it is nice to see 2.5m come out low.
# compare the effect of ethernet taps at different spacing
BEGIN {
# s=11.75;
for(s=1;s<30;s=s+0.1) {
n=int(500/s);
m=x=0;
delete z;
for(i=1;i<=n*4;i++) {
z[i]=cos(i*s*2*3.14159/11.75);
}
for(i=1;i<=n;i++) {
if(z[i]<0) continue;
for(j=1;j<i;j++) {
if(z[j]<0) continue;
x += z[i+(i-j)+n-j];
}
}
printf "%3d %5.2f %5.2f\n", m,s,x;
}
} |
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DHP
Guest
|
| Posted:
Tue Sep 13, 2005 2:24 pm Post subject:
Re: Why is there a minimum spacing? |
|
|
glen herrmannsfeldt <gah@ugcs.caltech.edu> said
....
| Quote: | Using some simple assumptions it does seem that 2.5m is much better.
First, I assume that for whatever spacing is used taps are only
placed where a period 11.75m cosine is positive.
I then compute the phase shift of the signal reflected off all
pairs of such taps and add them up. If you have awk or gawk
(there is a windows version of gawk around) you can run it and
see. The sum for 2.5m is about one fourth that for 2.4m or 2.6m.
There is also a minimum near 2.0m about twice that for 2.5m.
I didn't put any attenuation into the sum, though.
I don't know if this is at all related to what Rich did, but
it is nice to see 2.5m come out low.
|
Ethernet signals have a wide spectrum. What happens with a different
wavelength with the taps in the same place?
| Quote: | # compare the effect of ethernet taps at different spacing
BEGIN {
# s=11.75;
for(s=1;s<30;s=s+0.1) {
n=int(500/s);
m=x=0;
delete z;
for(i=1;i<=n*4;i++) {
z[i]=cos(i*s*2*3.14159/11.75);
}
for(i=1;i<=n;i++) {
if(z[i]<0) continue;
for(j=1;j<i;j++) {
if(z[j]<0) continue;
x += z[i+(i-j)+n-j];
}
}
printf "%3d %5.2f %5.2f\n", m,s,x;
}
}
|
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DHP
Guest
|
| Posted:
Tue Sep 13, 2005 3:47 pm Post subject:
Re: Why is there a minimum spacing? |
|
|
glen herrmannsfeldt <gah@ugcs.caltech.edu> said
| Quote: | DHP wrote:
(snip)
Which is avoided by using multiples of 2.5m, I suppose. Maybe my
arithmetic is not exact and the figures are tweaked so that the
nearest multiples are 1.25m on each side. Even so, 1.25m is only
5.3ns, so this can't be an issue unless the "dangerous" parts of the
reflection are very narrow.
Using some simple assumptions it does seem that 2.5m is much better.
First, I assume that for whatever spacing is used taps are only
placed where a period 11.75m cosine is positive.
I then compute the phase shift of the signal reflected off all
pairs of such taps and add them up. If you have awk or gawk
(there is a windows version of gawk around) you can run it and
see. The sum for 2.5m is about one fourth that for 2.4m or 2.6m.
There is also a minimum near 2.0m about twice that for 2.5m.
I didn't put any attenuation into the sum, though.
I don't know if this is at all related to what Rich did, but
it is nice to see 2.5m come out low.
# compare the effect of ethernet taps at different spacing
BEGIN {
# s=11.75;
for(s=1;s<30;s=s+0.1) {
n=int(500/s);
m=x=0;
delete z;
for(i=1;i<=n*4;i++) {
z[i]=cos(i*s*2*3.14159/11.75);
}
for(i=1;i<=n;i++) {
if(z[i]<0) continue;
for(j=1;j<i;j++) {
if(z[j]<0) continue;
x += z[i+(i-j)+n-j]; <------------------
}
}
printf "%3d %5.2f %5.2f\n", m,s,x;
}
}
|
I am not a programmer so I can't read this very easily. I'm not sure
what you're doing. I can guess most of it but I'm lost with the index
manipulation which I've marked. What's all that about? |
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glen herrmannsfeldt
Guest
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| Posted:
Tue Sep 13, 2005 10:00 pm Post subject:
Re: Why is there a minimum spacing? |
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|
DHP wrote:
| Quote: | glen herrmannsfeldt said
Using some simple assumptions it does seem that 2.5m is much better.
First, I assume that for whatever spacing is used taps are only
placed where a period 11.75m cosine is positive.
I then compute the phase shift of the signal reflected off all
pairs of such taps and add them up. If you have awk or gawk
(there is a windows version of gawk around) you can run it and
see. The sum for 2.5m is about one fourth that for 2.4m or 2.6m.
There is also a minimum near 2.0m about twice that for 2.5m.
I didn't put any attenuation into the sum, though.
|
OK, in words there are n possible taps. For taps i and j, with i>j, and
where an 11.75m period cosine is positive, sum over the phase shift from
the beginning to tap i, back to tap j, and then to the end.
So it is i (to tap i), i-j (back to tap j) and n-j (to the end).
| Quote: | I don't know if this is at all related to what Rich did, but
it is nice to see 2.5m come out low.
|
I did some more tests and it might be that the 2.5m dip is a
rounding artifact. It is there with 199 possible taps over
a 500m cable, but gone with 200 possible taps. The 2m dip
seems to stay, though.
| Quote: | Ethernet signals have a wide spectrum. What happens with a different
wavelength with the taps in the same place?
|
Yes, so far I am only considering the fundamental. The third harmonic
is probably also important, but I have to start somewhere.
-- glen |
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DHP
Guest
|
| Posted:
Tue Sep 13, 2005 10:26 pm Post subject:
Re: Why is there a minimum spacing? |
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|
glen herrmannsfeldt <gah@ugcs.caltech.edu> said
| Quote: | DHP wrote:
glen herrmannsfeldt said
Using some simple assumptions it does seem that 2.5m is much better.
First, I assume that for whatever spacing is used taps are only
placed where a period 11.75m cosine is positive.
I then compute the phase shift of the signal reflected off all
pairs of such taps and add them up. If you have awk or gawk
(there is a windows version of gawk around) you can run it and
see. The sum for 2.5m is about one fourth that for 2.4m or 2.6m.
There is also a minimum near 2.0m about twice that for 2.5m.
I didn't put any attenuation into the sum, though.
OK, in words there are n possible taps. For taps i and j, with i>j, and
where an 11.75m period cosine is positive, sum over the phase shift from
the beginning to tap i, back to tap j, and then to the end.
So it is i (to tap i), i-j (back to tap j) and n-j (to the end).
|
Sorry, I must be missing something. From the sound of it, you're
looking at a double reflection - a signal travelling the whole length
of the cable and arriving with noise coming in the same direction. The
model Rich is talking about is where tap i sends to tap j but tap k a
bit further down the line provides an unwanted echo. It's travelling
in the opposite direction of course, but k doesn't know that, k just
sees it as noise.
I'm also not sure about your summation. Adding cosines of phase just
gives you the componet which is in phase with the original signal. You
should also do a summation of sines to get the quadrature term. The
poor old receiver doesn't know which is which, it just sees the total
amplitude which, of course is (c^2+s^2)^.5.
| Quote: | I don't know if this is at all related to what Rich did, but
it is nice to see 2.5m come out low.
I did some more tests and it might be that the 2.5m dip is a
rounding artifact. It is there with 199 possible taps over
a 500m cable, but gone with 200 possible taps. The 2m dip
seems to stay, though.
Ethernet signals have a wide spectrum. What happens with a different
wavelength with the taps in the same place?
Yes, so far I am only considering the fundamental. The third harmonic
is probably also important, but I have to start somewhere.
|
I'm talking about sidebands, not harmonics. These will be frequencies
close to the 10MHz carrier. Well, close in the sense of being a broad
spectrum covering at least 5 MHz and almost certainly a lot more.
So you need two cosine functions, one to find the quadrants for the
10MHz cosine to get the clusters in the right place, and another for
the actual phase. Well, lots of them actually, an integral would do
nicely, but I'll settle for spot frequencies :) |
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DHP
Guest
|
| Posted:
Tue Sep 13, 2005 10:56 pm Post subject:
Re: Why is there a minimum spacing? |
|
|
DHP <me@privacy.net> said
| Quote: | It's travelling
in the opposite direction of course, but k doesn't know that, k just
sees it as noise.
|
I meant j of course. |
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glen herrmannsfeldt
Guest
|
| Posted:
Tue Sep 13, 2005 11:25 pm Post subject:
Re: Why is there a minimum spacing? |
|
|
DHP wrote:
| Quote: | glen herrmannsfeldt <gah@ugcs.caltech.edu> said
|
(snip)
| Quote: | Sorry, I must be missing something. From the sound of it, you're
looking at a double reflection - a signal travelling the whole length
of the cable and arriving with noise coming in the same direction. The
model Rich is talking about is where tap i sends to tap j but tap k a
bit further down the line provides an unwanted echo. It's travelling
in the opposite direction of course, but k doesn't know that, k just
sees it as noise.
|
The one I thought of first was for a near end tap with the reflections
off all the rest of the taps. A far tap, though, won't see single
reflections off many taps. In my case there are 4851 or so pairs of
taps contributing.
| Quote: | I'm also not sure about your summation. Adding cosines of phase just
gives you the componet which is in phase with the original signal. You
should also do a summation of sines to get the quadrature term. The
poor old receiver doesn't know which is which, it just sees the total
amplitude which, of course is (c^2+s^2)^.5.
|
I thought I was doing pretty well to get as far as I did with
only a small amount of work, but yes, it needs that, too.
(snip)
| Quote: | Ethernet signals have a wide spectrum. What happens with a different
wavelength with the taps in the same place?
Yes, so far I am only considering the fundamental. The third harmonic
is probably also important, but I have to start somewhere.
I'm talking about sidebands, not harmonics. These will be frequencies
close to the 10MHz carrier. Well, close in the sense of being a broad
spectrum covering at least 5 MHz and almost certainly a lot more.
|
I believe those will fall off pretty fast. There will be a 5MHz
fundamental for alternating 1 and 0, especially from the preamble,
with the same amplitude as the 10MHz from all 0's or all 1's.
Three bit repeats should have 3.33MHz and 6.66MHz at maybe one third
or so the amplitude. The ones near 10MHz need longer bit patterns,
and will have much smaller amplitudes.
| Quote: | So you need two cosine functions, one to find the quadrants for the
10MHz cosine to get the clusters in the right place, and another for
the actual phase. Well, lots of them actually, an integral would do
nicely, but I'll settle for spot frequencies :)
|
Maybe Rich will say more, but for now I believe that the 5MHz and
10MHz will be much higher amplitude. If we have the tap pattern
that is worst case for that, I don't believe that any other will
be any worse. I just had the thought of someone running a cable
down a hall of rooms spaced 11.75m apart with one or two taps
to each room.
-- glen |
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