Yuri Blanarovich wrote:
Very effective way for monoband antennas. The improved version of this is
the
"Bazooka Balun" which besides above mentioned benefits acts as a balun. I
used
it on all my monoband antennas. Provides DC path to ground, harmonic
supression, RF choke function and balanced to unbalanced feed point
conversion.

A 4:1 toroidal balun will also do the same thing -
provide a DC path between conductors.
--
73, Cecil http://www.qsl.net/w5dxp



Not the SAME (thing), but SOME. That is, minus harmonic suppresion.
Jus' to be preeeesize :-)

Yuri, K3BU.us

God Bless President Reagan! RIP
Thanks for freeing me and millions of others from behind the Iron Curtain!!!

Dave wrote:

"Henry Kolesnik" wrote in message
...

I know that a shorted 1/4 wave stub exhibits a very high impedance. But
for the 2nd harmonic it's a 1/2 wave stub and exhibits a very low

impedance

or a short. There are claims that this can be used to filter the even
harmonics. Shorts can't diisipate power and must reflect, so how does a
stub work?


stubs work very nicely. you can get practical stub information at my web
site, including how to build a 40m to 15m 3rd harmonic stub filter:
http://www.k1ttt.net/technote/techref.html#filters

as you may have noticed by now you have kicked the proverbial hornets nest.
reflections are a touchy word in this group, usually attracting the endless
argument that travels from thread to thread. in time this will deteriorate
into name calling and endless argument over reflections, interference,
virtual impedances, and a few other topics.




We bounce between the Physics of Operation and the Practicalities of
Engineering. We have experts on all sides of the issue grin. And,
typically, engineers and physicists both use English but can't
communicate well grin, again grin

But your basic question is how does one answer your question: "Shorts
can't dissipate power and must reflect, so how does a stub work?"

In practical terms, you have answered your own question! Reflections.

In Physics terms, you have to deal with the wave inside the stub and the
[V^2(theta) + I^2(theta)] energy storage as a function of position
within the stub and the corresponding sources of loss also a function of
V^2 and I^2 inside the stub and finally the reflections that exist
within the stub caused by the intentional physical short circuit [or
open circuit].

So, are you pursuing Physics or Applications type knowledge?

Dave Shrader wrote:
So, are you pursuing Physics or Applications type knowledge?

And apparently, never the twain shall meet. :-)



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By how much does the length of the short-circuiting bar at the end of the
1/4-wave stub affect its resonant length?

What open-circuit resonant frequency does the stub change to when the short
circuiting bar is removed?
----
Reg, G4FGQ

Reg Edwards wrote:

By how much does the length of the short-circuiting bar at the end of the
1/4-wave stub affect its resonant length?

I just twist the ends together. I don't use a bar.

What open-circuit resonant frequency does the stub change to when the short
circuiting bar is removed?

Same resonant frequency - virtual impedance at the input is reversed.
--
73, Cecil http://www.qsl.net/w5dxp



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What happens if the twisted short is grounded as in lightning protection?
Is there any succinct difference?

--
73
Hank WD5JFR

"Cecil Moore" wrote in message
...
Reg Edwards wrote:

By how much does the length of the short-circuiting bar at the end of
the
1/4-wave stub affect its resonant length?

I just twist the ends together. I don't use a bar.

What open-circuit resonant frequency does the stub change to when the
short
circuiting bar is removed?

Same resonant frequency - virtual impedance at the input is reversed.
--
73, Cecil http://www.qsl.net/w5dxp



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Henry, WD5JFR wrote:
"Shorts can`t dissipate power, so how does a stub work?"

Henry also wrote:
"I know that a shorted 1/4 wave stub exhibits a very high impedance. But
for the 2nd harmonic it`s a 1/2 wave stub and exhibits a very low
impedance or a short."

Henry is correct.

Connect a resistance directly to a transmitter. How much energy is
absorbed by the resistance? Ohm`s law is the first approximation.
Current is directly proportional to the applied voltage if the
transmitter`s internal reistance is negligible.

The shorted 1/4-wave stub exhibits an open circuit at its mouth and
accepts only enough current to supply its losss which are none in the
perfect stub. So, it takes no power from the transmitter afer its
circulating current is etablished.
At 2X the 1/4-wave freqency, the stub is 1/2 wavelengh and does not
transform the short at one end to an open circuit at its other end.
Instead, the 1/2-wave directly presents the short circuit at its far
end.

How much curent can the transmitter supply to a short circuit? It
depends on the internal impedance of the transmitter.

"Transmission Lines, Antennas, and Wave Guides" by King, Mimno, and Wing
explains how "Even Harmonics" are suppressed by the 1/4-wave
short-circuited stub on page 29. The gist is that the stub is imperfect
and its resistance saps harmonic energy which is allowed into the stub
by its low impedance.

Best regards, Richard Harrison, KB5WZI

Richard Harrison wrote:
The shorted 1/4-wave stub exhibits an open circuit at its mouth and
accepts only enough current to supply its losss which are none in the
perfect stub.

Richard, I think you would be surprised if you measured the RF current
through the short at the shorted end. It will be the in-phase sum of
the forward current and reflected current and is quite high. I once
melted the insulation at the end of a shorted 1/4WL piece of RG8X.
The heat came from high I^2*R losses at the short.

The SWR inside a perfect stub is infinite.
--
73, Cecil http://www.qsl.net/w5dxp



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Cecil, W5DXP wrote:
"Richard, I think you would be surprised if ypu measured RF current
through the short at the shorted end."

I expect high circulating current but without loss not much current is
required of the power source.

In his 1955 edition on page 106 Terman says:
"Thus, if the line is short-circuited at the load, then at frequencies
in the vicinity of a frequency for which the length is an odd number of
quarter wavelengths long, the impedance will be high and will vary with
frequency in the vicinity of resonance (i.e., frequency corresponding to
quarter wavelength) in exactly the same manner as does the impedance of
an ordinary parallel resonant circuit. It is therefore possible to
describe resonance on a transmission line in terms of impedance at
resonance and the equivalent Q of the resonance curve.

On page 107, Terman gives a 200 MHz example. 2-inch air-dielectric coax
is used for 1/4-wave short-circiuited stubs about 15 inches long. The
resonant impedance is more than 250,000 ohms with a Q of 3000.

How much current flows into an impedance of more than 1/4-million ohms?

Best regards, Richard Harrison, KB5WZI

Richard Harrison wrote:
How much current flows into an impedance of more than 1/4-million ohms?

As you know, transmission lines transform impedances. That stub likely
has an SWR around 5000:1. Assuming a driving voltage of 250V, the
current at the 1/4-million ohm point is about 0.001 amps. However,
1/4WL away, at the short in the 1/4WL stub, the current will be
~5 amps with a voltage of ~0.05V. That's an impedance of ~0.01 ohm. The
shorted stub has transformed the impedance from 1/4-million ohms at
the mouth to ~0.01 ohm at the short. That's what transmission lines do.

The question is not, "How much current flows into an impedance of more
than 1/4-million ohms?" The question is: How much current flows 1/4WL
away from that point at the shorted end of the stub? The answer is the
sum of the forward current and reflected current. The voltage at the
shorted end of a 1/4WL stub is the difference between the forward voltage
and the reflected voltage.

The voltage at the mouth of the stub is the sum of the forward voltage
and reflected voltage. The current at the mouth of the stub is the
difference between the forward current and reflected current. In the
example above, that difference is ~0.001 amp. The forward current flowing
inside the stub is ~2.505 amps and the reflected current flowing inside
the stub is ~2.495 amps.

If you don't believe the above, simply measure the RF current at the
shorted end of the stub. Someone modeled it the other day and even
the modeling program indicated that the current was sky high at
the shorted end of a 1/4WL stub.
--
73, Cecil http://www.qsl.net/w5dxp



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