Cecil, W5DXP wrote:
"---energy being always equally divided between current and potential
forms---."

Except at opens and shorts, real and virtual, where amps or volts for an
instant however brief are brought down to zero, which is a transfer of
all energy to either the electric or magnetic field.

Best regards, Richard Harrison, KB5WZI

Richard Harrison wrote:

Cecil, W5DXP wrote:
"---energy being always equally divided between current and potential
forms---."

Except at opens and shorts, real and virtual, where amps or volts for an
instant however brief are brought down to zero, which is a transfer of
all energy to either the electric or magnetic field.

That was the definition of a traveling wave in a Z0 environment, Richard,
and opens and shorts are not a Z0 environment. Can you give an example of
a virtual short and explain how it develops?
--
73, Cecil, W5DXP

Cecil, W5DXP wrote:
"That was the definition of a traveling wave in a Zo enviroment,---."

All this discussion over the insipid subject of flat lines?

A virtual short appears in a line with 100% reflection. The virtual
short appears where the sum of incident and reflected waves produces a
concurrence of zero volts and maximum current, just as in a true short.
The line has zero loss to enable good repetitions of an actual short or
open.

Best regards, Richard Harrison, KB5WZI

Richard Harrison wrote:
A virtual short appears in a line with 100% reflection. The virtual
short appears where the sum of incident and reflected waves produces a
concurrence of zero volts and maximum current, just as in a true short.

A physical short causes the voltage to go to zero. If a virtual
short causes the voltage to go to zero, what causes the virtual
short?

The virtual short cannot cause itself so ...

Either the virtual short causes the voltage to go to zero in
which case: What causes the virtual short?

Or the voltage going to zero causes the virtual short in which
case: What caused the voltage to go to zero?
--
73, Cecil, W5DXP

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

Equal and opposite voltages. Reaction to connecting wires together
generates an opposite voltage which adds to zero with the incident
voltage. Current doubles at the short.

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

1/4-wave short-circuit stubs are used as metallic insulators. They have
the characteristics of resonant circuits constructed of a
parallel-connected capacitor and coil, a very high impedance at
resonance.

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.

All my radar texts say resonant transmission line sections have the same
characteristics as resonant lumped circuits and I trust them because the
radar circuits using tuned transmission lines to route the signal, work.

Best regards, Richard Harrison, KB5WZI

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 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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