Roy Lewallen wrote:
I'm very glad to hear that our postings are being read and considered.

Bill Ogden wrote:

Speaking as a lurker, I find Roy's and Tom's postings very educational
and I
appreciate the time they take to do it.

I am a little dense, but I think I have learned four key points (at
least,
key for me) from this material:

1. One can discuss transmission lines and antennas using pulse
analysis or
steady-state analysis. When these two are mixed together the results
can be
a mess.


True. You can actually translate from one to the other, but it requires
an FFT or its inverse. Attempts to mix the two nearly always leads to
invalid conclusions.

2. When discussing "phase difference" we need to specify the two
components
that have the difference. (I.e., phase difference between the current
into
and out of an inductor is a different animal than the phase difference
between current and voltage at a specific point.)


Yes, although we can use an arbitrary reference as long as it's the same
for all components. For example, if one current has a phase angle of 50
degrees relative to some arbitrary reference and the other has a phase
angle of 30 degrees relative to that same reference, we know that the
phase of the first relative to the second is 20 degrees.

3. Superposition ("adding together") of power computations is not
valid in
reactive circuits.


It's never valid. Let me give you an example. Consider two AC or DC
voltage sources, each of 10 volts amplitude, with their negative
terminals connected together. (If they're AC, have them be of the same
frequency and in phase.) Connect a 10 ohm resistor between their
positive terminals. Superposition says that we can analyze the circuit
with each source individually and the other one turned off (short
circuited in the case of a voltage source), and add the results. What we
get should be the same answer as a full analysis with both the sources
on at the same time. So let's do it. Turn off source #2. The current
from source #1 through the resistor is 1 amp. The voltage across the
resistor is 10 volts. Now turn source #1 off and #2 on. The current
through the resistor is 1 amp going the other way than before, or -1
amp. The voltage across the resistor is 10 volts, but in the opposite
direction as before, or -10 volts. Adding the results gives a total of 0
amps through and 0 volts across the resistor. That's the right answer --
it's what we have when both sources are on. But now look at the power
dissipated by the resistor. With only source #1 on, it's I^2 * R = 1^2 *
10 = 10 watts. With only source #2 on, it's (-1)^2 * 10 = 10 watts. The
sum of the two is 20 watts, which is not the dissipation with both
sources on. Superposition does not apply to power, period. If it ever
seems to, it's only because of coincidence.

Don't be confused by the "forward" and "reverse" power concept. This is
not superposition and the concept must be used with great care to avoid
reaching invalid conclusions.

4. Displacement current is as real as any other current when dealing with
antennas and their components. (I cannot remember "displacement current"
ever being mentioned back in the dark ages when I was in EE school.
Perhaps
the school should remain nameless.)


It's a useful concept, but also has to be used with care because it
isn't a real current consisting of movement of electrons. Current in one
conductor creates a field which induces current in another conductor,
making the current appear to have "flowed" from one conductor to the
other. The classic example is of course current flow "through" a
capacitor. "Displacement current" is a widely used term; it's in the
index of the first four EM texts I grabbed from the bookshelf. Of an
example of a parallel RC circuit in Kraus' _Electromagnetics_, he says,
"The current through the resistor is a *conduction current*, while the
current 'through' the capacitor may be called a *displacement current*.
Although the current does not flow through the capacitor, the external
effect is as though it did, since as much current flows out of one plate
as flows into the opposite one."

Displacement current appears in Ampere's law, one of the four Maxwell
equations. In one formulation it has the quantity i + d(phi)e/dt on one
side. The i is conduction current, and the derivative quantity is known
as the displacement current.

Roy Lewallen, W7EL

Not everyone is happy with the term "displacement current." Albert
Shadowitz, in his book _The Electromagnetic Field_, has a chapter
entitled "The So-called Displacement Current." The term isn't in
the index to Feynman's _Lectures on Physics_. (At least I couldn't
find it.) All that is academic to the fact that AC current seems to
be able to make its way through a capacitor with no more opposition
than the capacitive reactance. Fortunately, no one on this
newsgroup has any objection to the way the term is commonly used.
73,
Tom Donaly, KA6RUH