Oh well, back to my rubber room....:/ MK


and stay there :-)

Some people can see further than others. Stand by for more. More thorough work
and decent article will take a bit more time. I intend to do that, have
promised cooperation of others in our camp. We should have more precise
figures, arguments and 'splanations. I mentioned some things in my previous
post, but if you can't see the benefits for improvement then be happy with what
you have.


Yuri

Yuri Blanarovich wrote:
Some people can see further than others. Stand by for more. More thorough work
and decent article will take a bit more time. I intend to do that, have
promised cooperation of others in our camp. We should have more precise
figures, arguments and 'splanations. I mentioned some things in my previous
post, but if you can't see the benefits for improvement then be happy with what
you have.

One thing I was originally confused about until I sat down and drew some
phasor diagrams. The phase of the *net* current doesn't change much over
the length of the mobile antenna. It's the phase of the forward current
and reflected current that is changing. But those phases are changing in
opposite directions so the sum of those two phasors results in very little
phase change in the net current.

For a 1/4WL vertical, the net current is approximately equal to 90 degrees
of a cosine wave, 1.0 at zero degrees and zero at 90 degrees. The magnitude
of the net current is an indication of where between zero and 90 degrees
the current is being measured. Where = arc-cos(|net current|)
--
73, Cecil http://www.qsl.net/w5dxp



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Cecil Moore wrote:
One thing I was originally confused about until I sat down and drew some
phasor diagrams. The phase of the *net* current doesn't change much over
the length of the mobile antenna. It's the phase of the forward current
and reflected current that is changing. But those phases are changing in
opposite directions so the sum of those two phasors results in very little
phase change in the net current.

It might be helpful to indicate your point of reference when talking
about phase, i.e. phase is changing with respect to voltage, or phase is
changing with respect to the source, phase is changing with respect to
position, etc.

Take the sentence "it's the phase of the forward and reflected current
that is changing" for example. For a traveling wave it is certainly
true that at any given point along a waveguide, the phase of the wave is
constantly changing with time. But it isn't necessarily true that at
that point, the phase of the current relative the voltage is changing
with time.

In the problem we've been discussing both points of reference are
relevant. In some case we're addressing the phase shift between voltage
and current, and in other cases we're discussing the change in phase
along the length of a conductor. I think it's helpful to be specific
about this.


For a 1/4WL vertical, the net current is approximately equal to 90 degrees
of a cosine wave, 1.0 at zero degrees and zero at 90 degrees. The magnitude
of the net current is an indication of where between zero and 90 degrees
the current is being measured. Where = arc-cos(|net current|)

That angle is the 'phase' of the current standing wave as a function of
position, not to be confused with the phase of the current with respect
to voltage. Roger?

73, Jim AC6XG

Jim Kelley wrote:
That angle is the 'phase' of the current standing wave as a function of
position, not to be confused with the phase of the current with respect
to voltage. Roger?

Since any reference to voltage on an antenna seems to be verboten, I
have avoided any such reference. Otherwise, here is a quote from Kraus:

"It is generally assumed that the current distribution of an infinitesimally
thin antenna is sinusoidal, and that the phase is *constant* over a 1/2WL
interval, changing abruptly by 180 degrees between intervals."

Don't you just love the phrase, "It is generally assumed ..."? He doesn't
say the current is sinusoidal. He doesn't say the phase is constant over
a 1/2WL interval.

For that general assumption to be true, the reflected current would have
to equal the forward current on a standing-wave antenna. But we know it
doesn't. However, this implies that the reflected current arriving back
at the feedpoint is not extremely/severely attenuated. Here's a cute
ballpark analysis sure to drive the gurus crazy.

Assume the Z0 of a traveling-wave dipole is 600 ohms. The ratio of forward
voltage to forward current is 600 ohms. The ratio of reflected voltage to
reflected current is 600 ohms. Assume the feedpoint current is one amp and
the feedpoint voltage is 50 volts. Assume the forward current and the reflected
current are in phase at the feedpoint. Assume the forward voltage and reflected
voltage are 180 degrees out of phase at the feedpoint. This is enough information
to solve for the ratio of forward current to reflected current at the feedpoint.
Assuming the net current is one amp at the feedpoint, I get 0.542 amps for
the forward current and 0.458 amps for the reflected current, i.e. the reflected
current is 85% of the value of the forward current. Remember, that is a ballpark
estimate.

That means the current only decreases by 15% in its round trip to the
end of the antenna and back. Same for the voltage. Oops, I mentioned
voltage - sorry. But please note that the power loss is a lot higher
than 15% since both the current and voltage are reduced by the same 15%.

The argument seems to occur due to the ignoring of the component waves
on a standing wave antenna. Such is the steady-state model seduction attended
by its sacred cows. Who wants to join me in a beer bust and barbecue?
--
73, Cecil http://www.qsl.net/w5dxp



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Cecil Moore wrote:

Jim Kelley wrote:
That angle is the 'phase' of the current standing wave as a function of
position, not to be confused with the phase of the current with respect
to voltage. Roger?

Since any reference to voltage on an antenna seems to be verboten, I
have avoided any such reference.

I can understand how at DC, references to current in a dipole might be
verboten. :-)

I wonder if voltage on a dipole could be roughly likened to transverse
velocity at various points along a whip.

Don't you just love the phrase, "It is generally assumed ..."?

I guess it allows: 'I'm not responsible should it turn out not to be a
good assumption'. :-)

For that general assumption to be true, the reflected current would have
to equal the forward current on a standing-wave antenna. But we know it
doesn't. However, this implies that the reflected current arriving back
at the feedpoint is not extremely/severely attenuated.

Seems to me it just implies that current at the end of a dipole isn't
really zero.

Assume the Z0 of a traveling-wave dipole is 600 ohms. The ratio of forward
voltage to forward current is 600 ohms. The ratio of reflected voltage to
reflected current is 600 ohms. Assume the feedpoint current is one amp and
the feedpoint voltage is 50 volts. Assume the forward current and the reflected
current are in phase at the feedpoint. Assume the forward voltage and reflected
voltage are 180 degrees out of phase at the feedpoint. This is enough information
to solve for the ratio of forward current to reflected current at the feedpoint.
Assuming the net current is one amp at the feedpoint, I get 0.542 amps for
the forward current and 0.458 amps for the reflected current, i.e. the reflected
current is 85% of the value of the forward current. Remember, that is a ballpark
estimate.

50 volts difference across a 600 ohm impedance, over 2, plus and minus
half an amp.

That means the current only decreases by 15% in its round trip to the
end of the antenna and back.

I think it may actually make many round trips. There may be multiple
reflections.

The argument seems to occur due to the ignoring of the component waves
on a standing wave antenna. Such is the steady-state model seduction attended
by its sacred cows.

Well, you'll either have to write out everything as a series, which is a
lot of busy work, or use the steady state equivalent.

So what does all this say about reflectivity at the end of the dipole?
;-)

73, Jim AC6XG

Jim Kelley wrote:
Seems to me it just implies that current at the end of a dipole isn't
really zero.

The net current is very close to zero because the forward and reflected
currents are very nearly equal and 180 degrees out of phase.

I think it may actually make many round trips. There may be multiple
reflections.

Of course, but like a transmission line, there is only one forward wave
and one reflected wave. All the multiple reflections are contained in
those two waves.
--
73, Cecil http://www.qsl.net/w5dxp



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Cecil Moore wrote:

Jim Kelley wrote:
Seems to me it just implies that current at the end of a dipole isn't
really zero.

The net current is very close to zero because the forward and reflected
currents are very nearly equal and 180 degrees out of phase.

Is that what you meant by:

"For that general assumption to be true, the reflected current would
have
to equal the forward current on a standing-wave antenna. But we know it
doesn't"?

It's not clear whether you're making this point in recognition of the
fact that wires are not lossless, or whether you're claiming it's
somehow fundamental to the performance of a radiator.

I think it may actually make many round trips. There may be multiple
reflections.

Of course, but like a transmission line, there is only one forward wave
and one reflected wave. All the multiple reflections are contained in
those two waves.

Then apparently you've decided not to completely eschew making at least
some steady state assumptions. Seductive indeed. ;-)

73, Jim AC6XG

Jim Kelley wrote:
Cecil Moore wrote:
The net current is very close to zero because the forward and reflected
currents are very nearly equal and 180 degrees out of phase.

This is at the tip end of a dipole.

Is that what you meant by:

"For that general assumption to be true, the reflected current would
have
to equal the forward current on a standing-wave antenna. But we know it
doesn't"?

This is apparently not at the tip end of a dipole.

It's not clear whether you're making this point in recognition of the
fact that wires are not lossless, or whether you're claiming it's
somehow fundamental to the performance of a radiator.

Both, radiation is a "loss".

Of course, but like a transmission line, there is only one forward wave
and one reflected wave. All the multiple reflections are contained in
those two waves.

Then apparently you've decided not to completely eschew making at least
some steady state assumptions. Seductive indeed. ;-)

There are only two possible directions, forward and reverse, in which
energy can flow. Multiple reflections do not create any more directions.
I am not opposed to the steady-state solution and use it all the time. I am
opposed to people being seduced by the steady-state solution into believing
there is not such thing as forward and reflected waves even though standing
waves require forward and reflected waves.
--
73, Cecil http://www.qsl.net/w5dxp



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Cecil Moore wrote:
I am
opposed to people being seduced by the steady-state solution into believing
there is not such thing as forward and reflected waves even though standing
waves require forward and reflected waves.

I also apologize if this particular dose of reality brought this thread to
a screeching halt.
--
73, Cecil http://www.qsl.net/w5dxp



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Cecil Moore wrote:
I am
opposed to people being seduced by the steady-state solution into believing
there is not such thing as forward and reflected waves even though standing
waves require forward and reflected waves.

Say that in Sacramento, and some knucklehead legislator will pass a law
against it.

73, Jim AC6XG