Henry Kolesnik wrote:
I know they work! One of my reasons for asking the question is I've not
found any mention in the literature of where the
"attenuation/rejection/reflection/filter residue" goes.

An open 1/4WL stub detours all the associated frequency
energy into the stub and then reflects it back to the source.
Consider the following:

XMTR---------feedline-------+----load
|
| 1/4WL
| open
| stub

The stub causes complete destructive interference between the
forward voltage and reflected voltage at the mouth of the stub.
Since the net voltage at the mouth of the stub is zero, no
current flows into the load. All the current in the feedline
is flowing in the stub and is reflected at the open end. The
reflected waves head back to the source. Therefore, for lossless
feedlines, all the attenuation/interference occurs in the source.
Of course, for real-world feedlines, I^2*R losses will occur
because of the high SWR at the associated frequency.

The effect of the above stub is much like shorting out the load.
Shorting out the load increases the dissipation in the source.
--
73, Cecil http://www.qsl.net/w5dxp

"Cecil Moore" wrote in message
...
Henry Kolesnik wrote:
I know they work! One of my reasons for asking the question is I've
not
found any mention in the literature of where the
"attenuation/rejection/reflection/filter residue" goes.

An open 1/4WL stub detours all the associated frequency
energy into the stub and then reflects it back to the source.

a new theory! now i'll have to add the traffic theory of waves to the list.
just who posts this detour sign? so does all the energy go down to the end
of the stub and have to make a u-turn to get back? are you sure they aren't
issued a ticket for doing that?? and then again at the junction they have
to be directed back to the source, sounds like a job for robo-cop! got to
be fast to direct all those waves back and forth!

The stub causes complete destructive interference between the
forward voltage and reflected voltage at the mouth of the stub.
Since the net voltage at the mouth of the stub is zero, no
current flows into the load

oh, oh! i know this one!! but if the voltage at the mouth of the stub is
zero, how does any wave flow down into the stub? there is no voltage there
to drive it?? how does the incoming wave know to not flow past the stub
connection and get reflected instead?? doesn't that make it a virtual short
and just reflect back to the source from there?? but wait, how does the
stub get hot then?? where does the current at the end of the stub come
from?

Dave wrote:
"Cecil Moore" wrote:
An open 1/4WL stub detours all the associated frequency
energy into the stub and then reflects it back to the source.

a new theory!

Not a new theory. I learned that at Texas A&M back in the 50's.
I had a pretty smart RF prof. It's the same thing that happens
without a load.

XMTR---------feedline---------------open

so does all the energy go down to the end
of the stub and have to make a u-turn to get back?
Energy is normally completely reflected at a physical open
circuit in a piece of transmission line. Stubs are no exception
to transmission line theory.

oh, oh! i know this one!! but if the voltage at the mouth of the stub is
zero, how does any wave flow down into the stub?

If the standing-wave minimum voltage is zero, why is the current
at that exact point at its maximum value. Maybe you should review
"Plumber's Delight" beam construction rules and standing-waves.

there is no voltage there
to drive it?? how does the incoming wave know to not flow past the stub
connection and get reflected instead??

Because reflections only occur at a physical impedance discontinuity.
There is no physical impedance discontinuity until the open is reached.

XMTR---one wavelength feedline---+---1/4WL---open
|
load

Note the load is connected at a "Plumber's Delight" V=0 point.

The voltage 1/4WL from the open end is zero. What do you think the
current is 1/4WL from the open end? What do you think the voltage
is at the open. Sounds like you believe that there is no current
or voltage on the entire length of the above since it is an odd
number of 1/4WL's long.

doesn't that make it a virtual short
and just reflect back to the source from there??

Nope, in the above example, there is a virtual short at the XMTR output.
Does that mean to you that there's no voltage or current on the entire
feedline? Virtual impedances don't cause reflections.

On a 50 ohm feedline with an SWR of 2:1, an impedance of 100 ohms
repeats every 1/2WL but it doesn't cause any reflections at the
100 ohm points because there is not a physical impedance discontinuity
there.
--
73, Cecil http://www.qsl.net/w5dxp

What if the source doesn't or can't dissipate?

--
73
Hank WD5JFR
"Cecil Moore" wrote in message
...
Henry Kolesnik wrote:
I know they work! One of my reasons for asking the question is I've
not
found any mention in the literature of where the
"attenuation/rejection/reflection/filter residue" goes.

An open 1/4WL stub detours all the associated frequency
energy into the stub and then reflects it back to the source.
Consider the following:

XMTR---------feedline-------+----load
|
| 1/4WL
| open
| stub

The stub causes complete destructive interference between the
forward voltage and reflected voltage at the mouth of the stub.
Since the net voltage at the mouth of the stub is zero, no
current flows into the load. All the current in the feedline
is flowing in the stub and is reflected at the open end. The
reflected waves head back to the source. Therefore, for lossless
feedlines, all the attenuation/interference occurs in the source.
Of course, for real-world feedlines, I^2*R losses will occur
because of the high SWR at the associated frequency.

The effect of the above stub is much like shorting out the load.
Shorting out the load increases the dissipation in the source.
--
73, Cecil http://www.qsl.net/w5dxp

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

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!!!

Henry Kolesnik wrote:
What if the source doesn't or can't dissipate?

Heh, heh, you're on your own for that discussion. I personally
believe it is possible for reflected energy to wind up decreasing
the power consumption from the DC source that is supplying the finals,
which winds up decreasing the load on the 60 Hz power grid, which
decreases the oil required from the Middle East, which tends to
decrease the possibility of world war, which bodes well for the human
race, but that is another thread for another time.
--
73, Cecil http://www.qsl.net/w5dxp



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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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Yuri Blanarovich wrote:
Thanks for freeing me and millions of others from behind the Iron Curtain!!!

.... and from the IRS by lowering the maximum tax rate by 300%? :-)



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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?