Richard Harrison wrote:

Cecil, W5DXP wrote:
"Net current doesn`t flow."

I`m inclined to agree, but at first current flows with a volts to amps
ratio rqual to the Zo of the line until a reflection returns to the
connection point of the stub. Then the total phase rotation within the
stub has reached 360-degrees. The complete reflection supplies a
reflected voltage wqual to the incident voltage. There is no difference
of potential or phase to evoke current. It`s the equivalent of a very
high impedance. Almost no more current is motivated to flow, once the
steady-state condition is reached.

Richard, some people use that exact same argument to try to prove
that no current is flowing in the middle of a transmission line
where forward power equals reflected power. The strange thing is
that the current 1/4WL away from that zero net current point is
sky high.

If the current 1/4WL away from your above source output terminals
is indeed sky high, it is because the forward and reflected currents
are in phase at that point which means they are 180 degrees out of
phase at the source output which means they are both still there.
If the net current is zero, all it means is that |Ifor|-|Iref| = 0.
It tells us nothing about the magnitudes of Ifor and Iref.

Trying to treat a distributed network as a lumped circuit can lead
to mistakes.

Alexander Wing wrote on page 29 of "Transmission Lines, Antennas, and
Wave Guides":
"Suppression of Even Harmonics.- An application of a short-circuited
quarter-wavelength line is to suppress any unwanted even harmonics in
the output of a radio transmitter. A short-circuited one-quarter
wavelength long at the desired output frequency may be connected across
the output terminals or across the antenna feeder at any point without
placing much load on the transmitter at the fundamental or desired
output frequency, since at this frequency such a section has an
impedance ideally infinite, actually about 400,000 ohms."

How much current flows into 400.000 ohms?

How much current flows in the shorted end of the stub? Let's say we
measure it at 2 amps. Where does that current come from?

The 400,000 ohms is (Vfor+Vref)/(Ifor+Iref) at the mouth of the stub
where V's and I's are phasors. Knowing that (Ifor+Iref) is a small
value doesn't tell us anything about the magnitudes of Ifor and Iref
except that they are nearly equal as is always the case for a
low-loss stub. Knowing (Inet=|Ifor|-|Iref|) is small doesn't tell us
a thing about the magnitude of (|Ifor|+|Iref|) that exists at the
shorted end of the stub. Does that make sense?
--
73, Cecil http://www.qsl.net/w5dxp

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