NNTP-Posting-Date: Sat, 02 Nov 2002 22:10:16 -0600
Message-ID: <3DC4A09F.7020907@ieee.org>
Date: Sat, 02 Nov 2002 22:05:51 -0600
From: Roy McCammon
Reply-To: rbmccammon@ieee.org
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Newsgroups: sci.electronics.design
Subject: Re: Transmission line fun--re:"who thinks this?"
References: <3DC42479.3040207@BOGUS.earthlink.net>
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Chris Carlen wrote:
> Hmm. Your response got me to go back and fiddle with the equations some
> more, and I've discovered things that I didn't realize before. Somehow
> I had gotten the impression after studying the dispersionless case which
> is mentioned in a problem in my text, that *all* "commercial"
> transmission lines, like RG58 etc., must be designed to be nearly
> dispersionless. Actually, my text refers to the condition
RG58 is pretty good in the MHz+ range
> It didn't occur to me at that time, that this is really quite
> impossible, as it requires a very high G to ensure that when frequency
> -> zero, the characteristic Z remains the same as it was at high f.
The ratio of G/wC where G and C are functions of w
tends to be constant as w -> 0, hence it is series
R that gives you the dispersion and loss at low freqs.
At low freqs, you can take G = 0 as an acceptable
approximation.
Even at high freaks, the losses due to R are usually
larger than the losses due to G, though both are
significant.
Plastic filled coax is typically lossier than open pair
lines (more G. air is better than plastic), but
picks up much less noise.
> I discovered this when I tried to "design" a line, by assuming some
> reasonable values (Zo and R/G), and trying to find the missing ones (L
> and C). For instance, let R/G=10^12, which is reasonable for a typical
> low-loss line. Try to design for Zo=50 ohms, dispersionless. Since
> R/G=L/C=Zo^2, Zo=10^6. Oops. Thus, very nearly lossless lines which
> are practically realizable, must not have dispersionless behavior. One
> must either employ very hefty conductors to get a very low R, or
> increase the G, making it quite lossy.
It is practical to make L larger by putting discrete
series inductors ever so often, or by embedding iron
filings in the dielectric. That was how they "fixed"
the early trans Atlantic telegraph cables.
> Then I took the following assumed values for what might be typical of
> ordinary "commercial" cables. Let R=.001ohm/m, G=10^-15S/m, C=10pF/m,
> and L=25nH/m. Next I looked at what happens to the real and imaginary
> components of the Zo over frequency. Of course, you are right, at very
> low frequency, the Zo->Sqrt(R/G) which in this case is 1 Megohm. Then
> there is a range of frequencies over which Re{Zo} ~= Im{Zo}, and these
> values are quite high, at several hundred ohms each. Above 1MHz, the Zo
> is almost entirely real, with a value of 50ohms.
Here are numbers for telephone wiring (22 gauge at 70F)
freq R(ohms/Kft) L(mH/kft) G(uS/kft) C(nF/Kft)
1Hz 33 .187 0.000 15.72
10Hz 33 .187 0.000 15.72
100Hz 33 .187 0.002 15.72
1KHz 33 .187 0.014 15.72
10kHz 33.2 .186 0.123 15.72
100KHz 40.8 .175 1.066 15.72
1MHz 110.5 .151 9.202 15.72
5Mhz 240 .141 41.51 15.72
> This is very perplexing I thought, that the Zo could have such high
> values and imaginary ones too. What implications does this have to
> power transfer at low frequencies? So I began to calculate the input
> impedance for a 50 ohm load, discovering that for an example of 1kHz,
> where Zo=(708-j706)ohms, the load reflection coefficient would be
> Gamma=-0.9294+j0.0658.
>
> At this point I encountered the remarkable result, since I figured I
> wanted to know the propagation velocity of this line to be able to think
> about real lengths in meters, instead of wavelengths, that:
>
> Propagation velocity of my line = 2*10^9 m/s !!!
I calculate a number smaller by about the square
root of 2, but that would be phase velocity, not
group velocity. No problem.
> Now we see that for my somewhat hypothetical cable, if the length is
> long, ie., some substantial fraction of a wavelength (which *is* very
> long on this cable, for one wavelength is 1000km at 1kHz), then the
> input impedance can have quite high values.
In a sense, all that series R decouples the input
from most of all that C out there.
> What is most bizarre, is
> that at one half wavelength, for which Zin=Zo always for lossless lines,
> this line/load combination exhibits a Zin=(452+j385) ohms.
All that 500 Km of series R also decouples
that puny 50 ohms load.
> That basically means that one cannot terminate line that is either not
> lossless, or not distortionless (satisfying RC=LG), at all frequencies.
Of course one can, all one needs is the proper
frequency dependent load. Actually quite trivial
for narrow band operation.
> Now what are the implications of this for digital wave propagation, the
> problem which in the practical realm I deal with most often (albeit at a
> much simpler level than the real engineers who are designing high speed
> PCBs and all that, I just design simple TTL compatible 50R coax drivers
> and receivers in a day's work.)??? Or for any sort of application where
> one is transmitting a distribution of frequencies, as opposed to a
> single f?
>
> It seems to suggest that the high frequency components of an edge, will
> terminate properly. But the lower frequency components, will exhibit
> reflections.
sure. The reflections are just the load's way
of saying that it is different from the characteristic
impedance of the cable. But, in the PCB world, the trace
length is so short compared to the wavelength at 1KHz,
that the reflections get to bounce back and forth 1000's
of times and get smaller on each reflection. Think
of the reflections as negotiation being carried out
between the source and the load.
Source: "You look like 1Kohm to me, here take this
1000mV and you can have 1mA for your trouble"
Load: "For 1 mA, I'm only going to take 50 mV. Take
950 back."
Source: "I really want you to take 1000mV. How about
if I give you 2mA.
Load: "For that, I take 100mV. Take 900 back."
and so forth
> Depending on the line length relative to the edge
> risetime, this may or may not be much of an issue. Thus, this picture
> is far more complex than what I had originally perceived.
Yeah, but it sounds like you are well on your
way to throwing it to the floor and pinning
it down. Knowledge has to be wrestled with
and conquered. Get it under your thumb. Hold
it down for a count of three, and it will do
your bidding.