On 7/7/2015 1:47 PM, Dave Platt wrote:
In article , rickman wrote:

Lol. You are a trip. I'm not going to spend $100 on a book just to see
if you are right. I was intrigued by the idea that a wire could carry a
signal without the resistance dissipating power according to P = I^2 R.
I guess there is some communication failure.

Yah.

It's a question of terminology. Unfortunately, one term has come to
be used for two (related but different) concepts.

There is "resistance", as in the E=I^2*R sort. If I recall correctly,
Maxwell refers to this as "dissipative" impedance. If you put current
through a dissipative resistance, a voltage drop develops across the
resistance, and power is dissipated.

There are plenty of examples of this, with which I'm sure you're
familiar.

There is also "resistance", as in "the 'real', non-reactive component
of a complex impedance, in which current is in phase with voltage."
This type of "resistance" is fundamentally non-dissipative - that is,
you can run power through it without dissipating the power as heat.

There are also good examples of this. One "textbook" example would be
a perfectly-lossless transmission line... say, one made out of a wire
and tube of a superconductor, cooled to below the superconducting
temperature.

I have analyzed this previously. Unless your transmission line is
infinitely long, eventually the wave reaches the other end and is either
dissipatively absorbed or is reflected back to the source where it
interacts. An infinitely long transmission line is not very
interesting. Dealing with the reflection from a finite transmission
line is what we are trying to analyze, so that model is not very useful.

So, no, I am not familiar with a non-dissipative load with current in
phase with the voltage.


You can (in principle) build such a superconducting coax to have
almost any convenient impedance... 50 or 75 ohms, for example. Since
we're theorizing, let's assume we can built one a few trillion miles
long... so long that the far end is light-years away.

If you hook a transmitter to one end of this and start transmitting,
it will "look" to the transmitter like a 50-ohm dummy load. The
transmitter itself won't be able to tell the difference. The
transmitter puts out an RF voltage, and the line "takes current"
exactly in phase with the voltage, in a ratio of one RF ampere per 50
RF volts.

But, there's a fundamental difference between this "resistance" and
that of a dummy load. A 50-ohm dummy load's resistance is
dissipative... all of the power going into it turns into heat, and is
dissipated in accordance with the fundamental laws of thermodynamics.

*None* of the power being fed into the superconducting coax, is
dissipated as heat in the coax. All of the power still exists, in its
original RF form. It's being stored/propagated down the coax without
loss.

When it hits a load at the other end, it may be dissipated as heat
there.

Or, perhaps not. What if what's at the other end of the
superconducting coax is a superconducting antenna, tweaked to present
an impedance of exactly 50 ohms? The RF will be radiated into space.

And, "free space" is another great example of a medium that has a
well-defined "resistance" (in the non-dissipated sense).

I believe this radiation *is* dissipative in the sense that the power is
removed from the system being analyzed. That is *exactly* why it is
considered to be due to a radiation "resistance". Note they do not
refer to it as a radiation "impedance".


https://en.wikipedia.org/wiki/Impedance_of_free_space

One of the fundamental jobs of an antenna, is to match the impedance
of its feedline to the impedance of free space.

Now, any coax you can buy at the store has *both* types of
"resistance", of course. It has a dissipative component, and a
non-dissipative component. Typically, the more you spend and the more
you have to strain your back carrying it around, the lower the amount
of dissipative resistance (which is only good for keeping the pigeons'
feet warm) and the more predictable and precisely-defined the
non-dissipative part.

What you are calling "non-dissipative resistance" is only a way to
characterize the AC behavior. It has nothing to do with what we are
discussing and is in no way similar to resistance. Trying to analyze a
transmission line without considering the reflections from the other end
is only a transient solution which ignores the behavior of the
transmission line.

How about we construct an example circuit with a conjugate matching
network rather than deal with abstractions that have nothing to do with
the discussion?

If you want to continue to discuss the transmission line, then we need
to consider the reflection and find a steady state solution, not a
transient one, right?

--

Rick