Current across the antenna loading coil - from scratch
 

Dave wrote:
thats the basic problem in this whole discussion. you are all talking
about
the same thing, just using different notation and incomplete statements
so
that none of you understands exactly what the others are trying to talk
about... when really you are all saying the same thing. its kind of like
after i graduated from college with an ee degree and my sister graduated
from an air force basic electronics course, she tried to ask me something
about currents in a transistor and i saw it all backwards... well of
course
she was talking electron flow and i was talking hole flow. we both got
the
same result but the notation was all different.

That's not true at all Dave. Most of us know that current is current.
It really only flows one direction at any instant of time. We can
indeed consider systems as having current that flows two directions at
one instant of time, but the results of that better agree with the
actual real current that flows only in one direction at any instant of
time or they are wrong.

Also, behavior of basic components cannot change. A two terminal device
like a loading coil cannot have differences in the current flowing
through it at each terminal without a third path. (I assume we all
know current is not an across vector and it does not "drop", the person
who started this thread just used poor wording.)

ARGH! maybe it really is more basic than different notations and
terminology. when working with antennas and 'component's that are a
significant fraction of a wavelength in size you must take into account the
'third path'... the 'third path' consists of the distributed capacitance and
resistance that CAN be modeled with lumped components if you want to go
through all the approximations and extra calculations that are required. if
you are ignoring that 'path' when talking about relatively large loading
coils then you will be wrong, how wrong depends on how large of course.

i haven't been following all the different threads and junk in here, but if
you are trying to analyze a significant sized loading coil without taking
into account all the paths then you are going to likely be less accurate
than cecil using a more complete distributed model. OBVIOUSLY if you are
using a strict lumped model the current can't be different from one end to
the other. And just as obviously if you make a really large loading coil,
like a full '1/2 wave' slinky dipole, the current at the feedpoint end will
be MUCH different than at the open end. You can both get the same results,
but to do it with lumped elements requires the same calculations that are
done by finite element simulations that try to do enough small lumped
elements as possible to approximate the distributed equations that would
give nice smooth results. Unfortunately cecil does not do a good job in
relating the distributed model, and his constant references to 'optics' and
the use of terms related to that field do nothing but confuse many of the
people in here to think that he is in a different world. admit it cecil,
while you may be correct, using a different set of terminology than most of
the people in here has done nothing but add to the confusion factor in many
of this long drawn out threads.

I still think that if each of you explained the WHOLE problem in your own
terminology, INCLUDING all the assumptions that are required for the models
you are using, that you would find that each of you is correct. but because
you are starting from different sets of assumptions you will never find a
common ground.

enough of this, back to assembling my new linear loaded 40m beam... why
don't you go analyze that loading system for a while.

Current across the antenna loading coil - from scratch
 

wrote:

I had this example at the bottom of my posting but you seem
to ignore such. So I am moving it to the top of the posting.
If you ignore it now, at least everyone will know you couldn't
possibly have missed it.

You are in a room with a 50 ohm transmission line routed
through a hole in one side of the room, across the room,
and through a hole in the other side of the room. You don't
know which is the source end of the line. A directional
wattmeter reads 200 watts forward power and 200 watts
reflected power but you don't know which direction is
forward. Here's a diagram:

200W-- 2 amps--
hole-------------------50 ohm coax-------------------hole
--200W --2 amps

Which direction is the standing wave current flowing?

If you knew forward current was moving left to right which
direction would the standing wave current be flowing?

That's not true at all Dave. Most of us know that current is current.

Too bad EZNEC disagrees with you as seen in the graphic at:

http://www.qsl.net/w5dxp/travstnd.GIF

The traveling wave current is virtually the opposite of the
standing wave current as can be seen by their different
equations. There is no phase information in the standing
wave current phase. Yet that is exactly the phase W7EL
used to try to measure the delay/phase shift through a
coil.

DC current is different from AC current. That's why the
DC or AC designations are necessary. RF forward current
is different from RF standing wave current. That's why
the different designations are necessary.

It really only flows one direction at any instant of time.

Or not at all at a standing wave current node.
Too bad we are talking RMS values here which is what EZNEC
reports. I'll ask the question of you: If one amp of RF
current is flowing in one direction and one amp of RF
current is flowing in the opposite direction, which direction
does the phasor sum of those two currents flow?

We can
indeed consider systems as having current that flows two directions at
one instant of time, but the results of that better agree with the
actual real current that flows only in one direction at any instant of
time or they are wrong.

The phase of standing wave current is unchanging. It doesn't "flow"
in the commonly accepted sense of the word. As Hecht says in "Optics":

"This is the equation for a STANDING or STATIONARY WAVE, as opposed
to a traveling wave. Its profile does not move through space; it is
clearly not of the form Func(x +/- vt)."

[Standing wave phase] "doesn't rotate at all, and the resultant
wave it represents doesn't progress through space - its a standing
wave."

If standing wave light doesn't move through space, then standing
wave RF also doesn't move through a wire.

(I assume we all
know current ... does not "drop", ...)

EM current does indeed drop exactly like EM voltage drops both
according to the attenuation factor. Just one more proof that
EM waves are not lumped circuit currents. The only difference
in the equation for transmission line voltage and current is
the voltage gets divided by the characteristic impedance which
is usually a resistive constant.

I think the basic problem is Cecil wants to used some definition of
current that does not allow models to be freely exchanged and does not
produce results that match real world systems. It always has to match.
We can't have different results unless someone has an error.

Exactly correct and the reason for the different results is your
error. In any conflict between the distributed network model and
the lumped circuit model, the distributed network model wins
every time since it is a superset of the lumped circuit model.

This has gone on for perhaps three years now. It is really up to Cecil
and Yuri to let it go, since they are the ones who seem to disagree
with measurements and accepted theory.

On the contrary, it is up to you and others to correct your
misconceptions about standing wave current. Your "accepted
theory" has holes in it that I could drive my GMC pickup
through.
--
73, Cecil http://www.qsl.net/w5dxp

Current across the antenna loading coil - from scratch
 

Cecil Moore wrote:
Gene Fuller wrote:

I thought you denounced and denied this "concept" earlier today.


Guess you misunderstood. A coil can replace 30 degrees of
an antenna but it won't use the same amount of wire as
30 degrees of wire. What I said is that an inductor is
more efficient than linear loading.

Cecil,

I am feeling dizzy. I am quite comfortable with my understanding of the
entire problem, but I am seriously confused about your position. Nobody
has ever talked about efficiency or the length of wire needed. The issue
has always been replacing "degrees of antenna". I have captured a few
excerpts from April 7.

73,
Gene
W4SZ


Excerpt follow:

9:03 am -- From Cecil

K7ITM wrote:

Another 'speriment occured to me, for those who think the coil current
MUST be different at the two ends by the amount corresponding to the
antenna section it replaces:


To the best of my knowledge, nobody believes that. The coil
is much more efficient at the loading function than is the
straight wire from which it is made. That's why inductive
loading is more efficient than fractal antennas or other
types of linear loading.
--
73, Cecil http://www.qsl.net/w5dxp


9:24 am -- From Yuri

[excerpt]

Now you move that coil say half way up the must, to higher impedance
point at the antenna, and that coil now, in order to maintain the
"match" has to have higher impedance, more turns and will exhibit MORE
current drop across it, while replacing THE SAME NUMBER OF "missing"
DEGREES AT THE RADIATOR.
Assuming that our goal is to stay with the same physical length of the
whip (which we do) and maintaining 90 degrees of resonant radiator. So
the radiator stays 50 degrees ()+50, 10+40, 20+30, 30+20, 40 + 10) long
and coil replaces the same "missing" 40 degrees.

[emphasis was in the original message]


9:44 am -- From Cecil

Roy Lewallen wrote:

Of course loading coils can be expressed in electrical degrees. But
extrapolating this to mean that a loading coil has the same properties
as an antenna with the same number of "degrees" has no justification.

I haven't heard anybody make that assertion in years. Coils
occupy whatever number of degrees that they occupy.


8:49 pm -- From Cecil

[excerpt]

Example: The phase shift from 30% to 60% in the traveling wave
antenna is taken from the tabular data as 54.2-27.6 = 26.6 degrees.

The phase information is in the *phase* in a traveling wave.

For the standing wave current, the situation is completely
different. The phase measured between any two current probes
will always be zero. The phase of a standing wave current is
useless for measuring phase shift. The way to extract the
phase information is to measure the *amplitude* at two points
and then calculate the phase shift by taking the arc-cos of
the normalized amplitude.

Example: The phase shift from 30% to 60% in the standing wave
antenna is arc-cos(0.8843) - arc-cos(0.5840) = 26.5 degrees.

The phase information is in the *amplitude* in a standing wave.

Thus in both antennas, the phase shift in 30 percent of the
wire is about 27 degrees. (90*.3 = 27) If we had a coil installed
in that 30 degrees of the antenna instead of a wire, the same
concepts would apply.

Current across the antenna loading coil - from scratch
 

Dave wrote:
admit it cecil,
while you may be correct, using a different set of terminology than most of
the people in here has done nothing but add to the confusion factor in many
of this long drawn out threads.

I have a limited technical library. I wish my RF references spelled out
everything as well as "Optics", by Hecht, but mine don't. Light and RF
are the same kind of EM waves, just at different frequencies. Hecht's
material is certainly relevant to RF waves. And I make every effort to
translate the technical jargon from one field to the other as best I
know how.

Hecht presents the best treatment of superposition, interference,
and standing waves that I have ever seen. I wish I had an RF
reference book as well written as "Optics".
--
73, Cecil http://www.qsl.net/w5dxp

Current across the antenna loading coil - from scratch
 

Gene Fuller wrote:
I am feeling dizzy. I am quite comfortable with my understanding of the
entire problem, but I am seriously confused about your position. Nobody
has ever talked about efficiency or the length of wire needed. The issue
has always been replacing "degrees of antenna". I have captured a few
excerpts from April 7.

What you quoted from me is my reporting of what EZNEC says about
standing wave current Vs traveling wave current at:

http://www.qsl.net/w5dxp/travstnd.GIG

The 'x' axis for both conditions is just a piece of 1/4WL wire.
One can calculate the phase shift in any section of wire in two
ways:

1. For traveling waves, the phase shift is given by the graph
of the phase (red line). The magnitude (blue line) contains
no phase information.

2. For standing waves, the phase shift is given by taking the
arc-cosine of the magnitude (blue line). The phase (red line)
contains no phase information.

Before we talk about replacing a piece of wire with a coil,
do you understand the above graphic and concepts? You seemed
to understand when you posted:

Regarding the cos(kz)*cos(wt) term in a standing wave:

Gene Fuller, W4SZ wrote:
In a standing wave antenna problem, such as the one you describe,
there is no remaining phase information. Any specific phase
characteristics of the traveling waves died out when the startup
transients died out.

Phase is gone. Kaput. Vanished. Cannot be recovered. Never to be
seen again.

The only "phase" remaining is the cos (kz) term, which is really
an amplitude description, not a phase.
--
73, Cecil http://www.qsl.net/w5dxp

Current across the antenna loading coil - from scratch
 

Cecil Moore wrote:
(snip)
You are in a room with a 50 ohm transmission line routed
through a hole in one side of the room, across the room,
and through a hole in the other side of the room. You don't
know which is the source end of the line. A directional
wattmeter reads 200 watts forward power and 200 watts
reflected power but you don't know which direction is
forward. Here's a diagram:

200W-- 2 amps--
hole-------------------50 ohm coax-------------------hole
--200W --2 amps

Which direction is the standing wave current flowing?

Lets also say that the wavelength of the waves passing through this
transmission line are as long as the room is wide (one wavelength fits
inside the room.

Then we can say that at any moment when the instantaneous magnitude of
he current is not at zero (and it will pass through zero all along the
line simultaneously, right?), the current will going in one direction
in half of that length *with varying magnitude) and the other way in
the other half (also with varying magnitude). I am not saying that
the direction reversal will necessarily be at the center, but that is
one possibility.

For example, if we took a snapshot of the current, all along the line
at the moment it peaked it might look like this:(length of arrow
represents current magnitude, and head shows direction)(view in fixed
width font)

.....--- --- -- - - -- --- --- -- - - --......
hole-------------------50 ohm coax-------------------hole

This is a snapshot of the current all along the line at an instant.

A quarter cycle later, the current would be zero, everywhere.

A half cycle later (than the first snapshot) it would look like this:

.....--- --- -- - - -- --- --- -- - - --......
hole-------------------50 ohm coax-------------------hole

This is the pattern the standing wave function describes. The current
at every point has one of two phases, which are 180 degrees from each
other.

Is this how you see it?

Current across the antenna loading coil - from scratch
 

Cecil Moore wrote:
Gene Fuller wrote:

I am feeling dizzy. I am quite comfortable with my understanding of
the entire problem, but I am seriously confused about your position.
Nobody has ever talked about efficiency or the length of wire needed.
The issue has always been replacing "degrees of antenna". I have
captured a few excerpts from April 7.


What you quoted from me is my reporting of what EZNEC says about
standing wave current Vs traveling wave current at:

http://www.qsl.net/w5dxp/travstnd.GIG

The 'x' axis for both conditions is just a piece of 1/4WL wire.
One can calculate the phase shift in any section of wire in two
ways:

1. For traveling waves, the phase shift is given by the graph
of the phase (red line). The magnitude (blue line) contains
no phase information.

2. For standing waves, the phase shift is given by taking the
arc-cosine of the magnitude (blue line). The phase (red line)
contains no phase information.

Agreed, with one exception.

There is a phase reversal each time you pass through a node, so you
can tell by phase measurement, if you are on the far end of an odd
numbered node or an even numbered node, once you decide which of the
two possibilities of the phase is at the far end of node zero (or some
other reference point). But between any pair of nodes, yes, you have
to use the phase information obtained from the ARC-COS(magnitude), or
the distance from that point to a node (as a fraction of a wavelength
in the line), to infer where you are within that half wavelength.

Of course, you can find the node either as a point with zero
magnitude, or the point between phase reversals.

Current across the antenna loading coil - from scratch
 

John Popelish wrote:
This is the pattern the standing wave function describes. The current
at every point has one of two phases, which are 180 degrees from each
other.

Is this how you see it?

Yes, now which direction is that current flowing? If the
source were known to be to the left, would that change
your answer?
--
73, Cecil http://www.qsl.net/w5dxp

Current across the antenna loading coil - from scratch
 

John Popelish wrote:

Cecil Moore wrote:
http://www.qsl.net/w5dxp/travstnd.GIF

Agreed, with one exception.
There is a phase reversal each time you pass through a node, ...

You are correct if two sides of a node exist in the system.
But since the context was my above 1/4WL wire, there is no
"passing through a node". I was limiting my statements
in context to a 1/4WL long conductor.
--
73, Cecil http://www.qsl.net/w5dxp

Current across the antenna loading coil - from scratch
 

Cecil Moore wrote:
John Popelish wrote:

This is the pattern the standing wave function describes. The current
at every point has one of two phases, which are 180 degrees from each
other.

Is this how you see it?

Yes, now which direction is that current flowing?

You deleted the arrows which I drew that showed one possible case. Do
you have some argument with what you deleted?
(I'll replace it, so you don't have to go back to look at it)
(begin paste)

For example, if we took a snapshot of the current, all along the line
at the moment it peaked it might look like this:(length of arrow
represents current magnitude, and head shows direction)(view in fixed
width font)

.....--- --- -- - - -- --- --- -- - - --......
hole-------------------50 ohm coax-------------------hole

This is a snapshot of the current all along the line at an instant.

A quarter cycle later, the current would be zero, everywhere.

A half cycle later (than the first snapshot) it would look like this:

.....--- --- -- - - -- --- --- -- - - --......
hole-------------------50 ohm coax-------------------hole
(end paste)

See all those arrows of various length representing current direction
and magnitude? Why do you ask me about something after erasing my answer?

If the source were known to be to the left, would that change
your answer?

For a pure standing wave, there is effectively a source at each end,
so this question is meaningless.