Richard Harrison wrote:

Cecil, W5DXP wrote:
"What caused the voltage to go to zero?"

Equal and opposite voltages.

What caused the rearward-traveling current to go to zero at the
same time?

The problem is one of cause and effect. You cannot say the virtual
short causes the voltage and current wave conditions and then say
the voltage and current wave conditions causes the virtual short.

1/4-wave back from the short, a virtual open circuit appears. Cecil
claims this open circuit does not impede current.

If the virtual short causes reflections, why doesn't the virtual open
cause reflections?

1/4-wave short-circuit stubs are used as metallic insulators.

That's nice, but we are not discussing physical shorts. We are
discussing virtual shorts.
--
73, Cecil http://www.qsl.net/w5dxp



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It's actually easy to predict the (resistive) impedance seen looking
into a shorted quarter wavelength line, or any odd multiple. The
impedance is simply the Z0 of the line divided by the loss in nepers.
One neper is about 8.7 dB, so the impedance is about 8.7 * Z0 / dB loss.

All else being equal, the impedance gets higher as frequency increases.
That's because the length of a quarter wave stub decreases in inverse
proportion to frequency, while loss (up to 1 - 10 GHz or so, where
conductor loss dominates) increases only as the square root of
frequency. So the impedance of a stub increases as the square root of
frequency. For example, a quarter wave stub, made from solid
polyethylene dielectric coax (VF = 0.66) at 3.5 MHz is about 46 ft. That
length of RG-58 has a loss of about 0.3 dB, so the impedance looking
into a quarter wave stub of RG-58 at 3.5 MHz is about 1450 ohms. Quite a
far cry from the textbook's example of 400 k ohms or Richard's
extrapolation to 33 k ohms! An RG-58 stub at 350 MHz, or 100 times the
frequency, would have an input impedance of about 14,500 ohms.

A more typical VHF example would be a quarter wave of RG-8 at two
meters. It would be about 13.4 inches long and a loss of about 0.03 dB,
for an input Z of about 14,500 ohms.

Incidentally, the formula I'm using is actually on the same page of King
et al's text as the 400 k ohm value Richard quotes. They say the 400 k
value is for "a reasonably low-loss line" -- to get 400 k ohms with a
600 ohm line, the loss would have to be about 0.013 dB.

The input impedance of an open circuited quarter wavelength line or
shorted half wavelength line is Z0 times the loss in nepers, or about
Z0 * dB loss / 8.7.

I actually ran into a case where the finite resistance of an open stub
became a problem, and it illustrates the hazard of blindly following a
"rule of thumb" without checking to see under what conditions it's
valid. The "Field Day Special" antenna, similar to a ZL special, can be
fed at the center of either element. I connected a one wavelength
transmission line to the center of each element, and fed one or the
other to switch directions, leaving the other line open circuited. When
RG-58 was used, the current diverted into the finite resistance of the
open stub disturbed the element current enough to very significantly
degrade the front/back ratio. The lines were one wavelength at 14 MHz,
or about 46 feet. Loss was a seemingly trivial 0.8 dB, but that means
that the input impedance was only about 540 ohms! 400,000 or even 33,000
would be an awfully poor estimate! Changing to 300 ohm twinlead solved
the problem. (Although 300 ohm twinlead can easily be as lossy as RG-58
when wet, the higher Z0 resulted in an adequately high stub impedance
even when it was wet.)

Roy Lewallen, W7EL

Richard Harrison wrote:

. . .
From King, Mimno, and Wing, "Transmission Lines, Antennas, and Wave
Guides" page 29:

"A short-circuited line, one-quarter wavelength long at the desired
output frequency may be connected across the output terminals of a
transmitter or across the antenna feeder at any point without placing
much load on the transmitter at this fundamental or desired output
frequency, since at this frequency such a section has an impedance
ideally infinite, actually about 400,000 ohms."

Since I = E/Z, how much current do you think will flow into 400,000
ohms?

King, Mimno, and Wing`s impedance might scale down to only 33,333 ohms
on a 50-ohm line, still high, as they may have been considering a
600-ohm line.
. . .

Roy Lewallen wrote:
All else being equal, the impedance gets higher as frequency increases.

Double the frequency and you have a shorted 1/2WL stub. Isn't the
impedance of a shorted 1/2WL stub lower than the impedance of a
shorted 1/4WL stub?
--
73, Cecil http://www.qsl.net/w5dxp



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Richard Harrison wrote:

Cecil, W5DXP wrote:
"What caused the rearward traveling current to go to zero?"

The rearward-traveling current did not change phase on reflection from a
load of too-few ohms.

True, there is a re-reflection event exactly as you describe. But the
reflection coefficient for that re-reflection event is not 1.0 so not
all of the reflected power gets re-reflected. What happens to the rest
of it? How do you explain the discrepancy between the physical reflection
coefficient less than 1.0 and the apparent 100% re-reflection of reflected
energy at a Z0-match point?
--
73, Cecil http://www.qsl.net/w5dxp
"One thing I have learned in a long life: that all our science, measured
against reality, is primitive and childlike ..." Guess who said that.



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I'm not sure if that's meant to be humorous, or if you really did
misinterpret what I meant. In case it's the latter, I'll amplify.

If a quarter wavelength shorted stub at frequency f1 has impedance R1,
then a quarter wavelength shorted stub at frequency f2 will have an
impedance of R1 * sqrt(f2 / f1), if it's made of the same type of
transmission line, and the frequency is in the range where conductor
loss dominates (below about 1 - 10 GHz for typical coax).

Roy Lewallen, W7EL

W5DXP wrote:
Roy Lewallen wrote:

All else being equal, the impedance gets higher as frequency increases.


Double the frequency and you have a shorted 1/2WL stub. Isn't the
impedance of a shorted 1/2WL stub lower than the impedance of a
shorted 1/4WL stub?

Roy Lewallen wrote:
I'm not sure if that's meant to be humorous, or if you really did
misinterpret what I meant. In case it's the latter, I'll amplify.

Whoops, I really did misinterpret what you meant. When you said, "All
else being equal," I inferred that you were including the physical
length of the stub. Changing the length of the stub didn't seem to
match the condition, "All else being equal,". Sorry.
--
73, Cecil http://www.qsl.net/w5dxp



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I apologize for the lower-case "C" used to initiate the last "Cecil" in
my most recent posting. My finger must have been tired and weak when I
hit the shift-key. No offense was intended.

It`s fixed phase and high transmitter volts which thwart energy return
to the transmitter according to my profs over 50 years ago in
circumstances posed in the problem I was considering. It seems
reasonable when the stand-off happens at an impedance discontinuity.

Best regards, Richard Harrison, KB5WZI

Richard Harrison wrote:

Cecil, W5DXP wrote:
"But the reflection coefficient for that re-reflection event is not 1.0
so not all of the reflected power gets re-reflected."

100% re-reflection was a given. It is my assumption and I`m sticking
with it..

And my question is, in the absence of a physical reflection coefficient
magnitude of 1.0, what causes 100% re-reflection?

I`m tired of explaining something cecil refuses to accept.

I accept everything you have explained, Richard, but you have not answered
my question. What causes 100% re-reflection? Please be specific. A voltage
to current ratio is a result and not the cause of anything.
--
73, Cecil http://www.qsl.net/w5dxp
"One thing I have learned in a long life: that all our science, measured
against reality, is primitive and childlike ..." Albert Einstein



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Richard Harrison wrote:
It`s fixed phase and high transmitter volts which thwart energy return
to the transmitter according to my profs over 50 years ago in
circumstances posed in the problem I was considering. It seems
reasonable when the stand-off happens at an impedance discontinuity.

Why doesn't the same thing happen 1/2WL back up the transmission line
where the impedance is exactly the same?
--
73, Cecil http://www.qsl.net/w5dxp
"One thing I have learned in a long life: that all our science, measured
against reality, is primitive and childlike ..." Albert Einstein



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Cecil, W5DXP wroyte:
"Why doesn`t the same thing happen 1/2WL back up the line where the
impedance is exactly the same?"

Uniform distribution of inductance and capacitance 1/2WL back up the
line.

John E. Cunningham in "The Complete Broadcast Antenna Handbook" wrote as
I recall, an attempt to explain the reflection operation on a
transmission line as caused by impedance discontinuities.

If I recall, he said a short on a line vitiates capacitance at that
point. It shorts it out. Increased current in the final segment of the
line inductance sets off the reflection. The extraordinary volts
generated, reverse the wave direction and reverse the phase of the
volts. What happens to the current is a turnaround sooner than later,
but it still comes and goes in the same phase.

The discontinuity is essential to the turnaround. Lacking a reduction
point or growth point in inductance to capacitance ratio, you lack an
essential to generate the reversed signal. That`s why the virtual short
does not turn the signal around. It has no discontinuity.

Best regards, Richard Harrison, KB5WZI