Standing Wave Phase
 

50 ohm Shorted line 8.5655 meters long. Frequency = 7 Mhz. 2 volt 50 ohm
generator. Current at input = 12.361 milliamps. Current at short =
40 milliamps. Divide the current at the imput of the line by the current
at the short and take the arc sine (in radian mode) of the result. This
is 1.2566. Now take Beta = .14671 and multiply it by the length 8.5655.
This also equals 1.2566, which is the angular length of the shorted
line. Will someone explain how this works to Cecil?
73,
Tom Donaly, KA6RUH

Standing Wave Phase
 

Tom, KA6RUH wrote:
"Will somebody explain how this works to Cecil?"

Cecil is capable of solving practically any transmission line problem
without help from anyone. Tom could have been more descriptive by saying
if his line were 50-ohm coax with a velocity factor of about 2/3.

Best regards, Richard Harrison, KB5WZI

Standing Wave Phase
 

On Dec 4, 9:06 pm, "Tom Donaly" wrote:
50 ohm Shorted line 8.5655 meters long. Frequency = 7 Mhz. 2 volt 50 ohm
generator. Current at input = 12.361 milliamps. Current at short =
40 milliamps. Divide the current at the imput of the line by the current
at the short and take the arc sine (in radian mode)

arc cosine, perhaps?

of the result. This
is 1.2566. Now take Beta = .14671 and multiply it by the length 8.5655.
This also equals 1.2566, which is the angular length of the shorted
line. Will someone explain how this works to Cecil?

Cecil either knows this, in which case it is unnecessary to explain
it,
or he does not, in which case it will be impossible.

....Keith

Standing Wave Phase
 

Richard Harrison wrote:
Tom, KA6RUH wrote:
"Will somebody explain how this works to Cecil?"

Cecil is capable of solving practically any transmission line problem
without help from anyone. Tom could have been more descriptive by saying
if his line were 50-ohm coax with a velocity factor of about 2/3.

Best regards, Richard Harrison, KB5WZI


No he isn't, Richard, or he wouldn't have made the ignorant statement
that there is no phase information in a standing wave. As to the
velocity factor, the line is one of Cecil's ideal, lossless lines. The
velocity factor is 1. Nice try.
73,
Tom Donaly, KA6RUH

Standing Wave Phase
 

Tom Donaly wrote:
50 ohm Shorted line 8.5655 meters long. Frequency = 7 Mhz. 2 volt 50 ohm
generator. Current at input = 12.361 milliamps. Current at short =
40 milliamps. Divide the current at the imput of the line by the current
at the short and take the arc sine (in radian mode) of the result. This
is 1.2566. Now take Beta = .14671 and multiply it by the length 8.5655.
This also equals 1.2566, which is the angular length of the shorted
line. Will someone explain how this works to Cecil?

You won't understand what I am talking about until you perform
the stub experiments that I previously posted.

---600 ohm line---+---10 deg, 100 ohm line---open-circuit

How many degrees of 600 ohm line does it take to resonate
that stub to an electrical 1/4WL?

--5 deg, 100 ohm line--+--600 ohm line--+--5 deg, 100 ohm line--open

How many degrees of 600 ohm line does it take to resonate
that stub to an electrical 1/4WL?
--
73, Cecil http://www.w5dxp.com

Standing Wave Phase
 

Richard Harrison wrote:
Tom, KA6RUH wrote:
"Will somebody explain how this works to Cecil?"

Cecil is capable of solving practically any transmission line problem
without help from anyone. Tom could have been more descriptive by saying
if his line were 50-ohm coax with a velocity factor of about 2/3.

Tom apparently doesn't realize that he needs to be dealing
with more than one Z0 to observe the difference between
electrical degrees and physical degrees. I invite everyone
to solve the dual-Z0 stub problem that I earlier presented
and repeated tonight.
--
73, Cecil http://www.w5dxp.com

Standing Wave Phase
 

Tom Donaly wrote:
No he isn't, Richard, or he wouldn't have made the ignorant statement
that there is no phase information in a standing wave.

Please get it right, Tom. I said there is no phase
information in the standing-wave phase. The phase
information is certainly there but it is in the
amplitude, not in the phase. The following graph
proves it:

http://www.w5dxp.com/travstnd.gif

The phase of the standing-wave is absolutely flat.
--
73, Cecil http://www.w5dxp.com

Standing Wave Phase
 

Keith Dysart wrote:
On Dec 4, 9:06 pm, "Tom Donaly" wrote:
50 ohm Shorted line 8.5655 meters long. Frequency = 7 Mhz. 2 volt 50 ohm
generator. Current at input = 12.361 milliamps. Current at short =
40 milliamps. Divide the current at the imput of the line by the current
at the short and take the arc sine (in radian mode)

arc cosine, perhaps?

of the result. This
is 1.2566. Now take Beta = .14671 and multiply it by the length 8.5655.
This also equals 1.2566, which is the angular length of the shorted
line. Will someone explain how this works to Cecil?

Cecil either knows this, in which case it is unnecessary to explain
it,
or he does not, in which case it will be impossible.

...Keith
Right. I lied.
73,
Tom Donaly, KA6RUH

Standing Wave Phase
 

Tom Donaly wrote:
Keith Dysart wrote:
On Dec 4, 9:06 pm, "Tom Donaly" wrote:
50 ohm Shorted line 8.5655 meters long. Frequency = 7 Mhz. 2 volt 50 ohm
generator. Current at input = 12.361 milliamps. Current at short =
40 milliamps. Divide the current at the imput of the line by the current
at the short and take the arc sine (in radian mode)

arc cosine, perhaps?

of the result. This
is 1.2566. Now take Beta = .14671 and multiply it by the length 8.5655.
This also equals 1.2566, which is the angular length of the shorted
line. Will someone explain how this works to Cecil?

Cecil either knows this, in which case it is unnecessary to explain
it,
or he does not, in which case it will be impossible.

...Keith
Right. I lied.
73,
Tom Donaly, KA6RUH
To amplify:
Arc Cosine is correct. And the comment on Cecil is right on the mark.
The same thing can be accomplished, above, using an open stub and
measuring the voltage at both ends. All this is just theoretical,
though, because line loss will skew the results. Besides, why try to
measure current or voltage when all you have to do is measure length
and frequency?
73,
Tom Donaly, KA6RUH

Standing Wave Phase
 

On Dec 4, 11:13 pm, Cecil Moore wrote:
Tom Donaly wrote:
50 ohm Shorted line 8.5655 meters long. Frequency = 7 Mhz. 2 volt 50 ohm
generator. Current at input = 12.361 milliamps. Current at short =
40 milliamps. Divide the current at the imput of the line by the current
at the short and take the arc sine (in radian mode) of the result. This
is 1.2566. Now take Beta = .14671 and multiply it by the length 8.5655.
This also equals 1.2566, which is the angular length of the shorted
line. Will someone explain how this works to Cecil?

You won't understand what I am talking about until you perform
the stub experiments that I previously posted.

---600 ohm line---+---10 deg, 100 ohm line---open-circuit

How many degrees of 600 ohm line does it take to resonate
that stub to an electrical 1/4WL?

--5 deg, 100 ohm line--+--600 ohm line--+--5 deg, 100 ohm line--open

How many degrees of 600 ohm line does it take to resonate
that stub to an electrical 1/4WL?
--
73, Cecil http://www.w5dxp.com

You have said multiple times that the electrical length of a
quarter wave stub must be 90 electrical degress, so the
computation is too easy...

1) x + 10 = 90
x = 80 degrees for the 600 Ohm line
2) 5 + x + 5 = 90
x = 80 degrees for the 600 Ohm line

although I suspect others will disagree with your solution.

Standing Wave Phase
 

Tom Donaly wrote:
The same thing can be accomplished, above, using an open stub and
measuring the voltage at both ends. All this is just theoretical,
though, because line loss will skew the results. Besides, why try to
measure current or voltage when all you have to do is measure length
and frequency?

Tom, you are not going to understand what I am saying until
you perform the stub exercise I provided. Just do one at a
time. Assume ideal lossless conditions with VF=1.0.

---600 ohm line---+---10 deg 100 ohm line---open

How many degrees of 600 ohm line does it take to make the
above stub look like 1/4 wavelength, i.e. 90 degrees?

Until you perform the exercise, you are just creating
diversions and avoiding the technical truth.
--
73, Cecil http://www.w5dxp.com

Standing Wave Phase
 

Keith Dysart wrote:
On Dec 4, 11:13 pm, Cecil Moore wrote:
Tom Donaly wrote:
50 ohm Shorted line 8.5655 meters long. Frequency = 7 Mhz. 2 volt 50 ohm
generator. Current at input = 12.361 milliamps. Current at short =
40 milliamps. Divide the current at the imput of the line by the current
at the short and take the arc sine (in radian mode) of the result. This
is 1.2566. Now take Beta = .14671 and multiply it by the length 8.5655.
This also equals 1.2566, which is the angular length of the shorted
line. Will someone explain how this works to Cecil?
You won't understand what I am talking about until you perform
the stub experiments that I previously posted.

---600 ohm line---+---10 deg, 100 ohm line---open-circuit

How many degrees of 600 ohm line does it take to resonate
that stub to an electrical 1/4WL?

--5 deg, 100 ohm line--+--600 ohm line--+--5 deg, 100 ohm line--open

How many degrees of 600 ohm line does it take to resonate
that stub to an electrical 1/4WL?

You have said multiple times that the electrical length of a
quarter wave stub must be 90 electrical degress, so the
computation is too easy...

1) x + 10 = 90
x = 80 degrees for the 600 Ohm line
2) 5 + x + 5 = 90
x = 80 degrees for the 600 Ohm line

although I suspect others will disagree with your solution.

I have not yet provided a solution. Your's is *wrong*.

The 90 degree physical solution is *wrong* because it results
in more than 90 electrical degrees. Please try again.
--
73, Cecil http://www.w5dxp.com

Standing Wave Phase
 

On Dec 5, 9:14 am, Cecil Moore wrote:
Keith Dysart wrote:
On Dec 4, 11:13 pm, Cecil Moore wrote:
You won't understand what I am talking about until you perform
the stub experiments that I previously posted.

---600 ohm line---+---10 deg, 100 ohm line---open-circuit

How many degrees of 600 ohm line does it take to resonate
that stub to an electrical 1/4WL?

--5 deg, 100 ohm line--+--600 ohm line--+--5 deg, 100 ohm line--open

How many degrees of 600 ohm line does it take to resonate
that stub to an electrical 1/4WL?

You have said multiple times that the electrical length of a
quarter wave stub must be 90 electrical degress, so the
computation is too easy...

1) x + 10 = 90
x = 80 degrees for the 600 Ohm line
2) 5 + x + 5 = 90
x = 80 degrees for the 600 Ohm line

although I suspect others will disagree with your solution.

I have not yet provided a solution. Your's is *wrong*.

The 90 degree physical solution is *wrong* because it results
in more than 90 electrical degrees. Please try again.

I thought that when you specified 5 and 10 degrees in your
problem statement, you meant electrical degrees. That is, the
phase shift encountered by the forward travelling wave.

Certainly, the answer was in terms of electrical degrees. That is,
the phase shift encountered by the forward travelling wave.

Or have I misunderstood the meaning of 'electrical degrees'?

Or perhaps 'electrical degrees' do not sum either?

Standing Wave Phase
 

Keith Dysart wrote:
I thought that when you specified 5 and 10 degrees in your
problem statement, you meant electrical degrees. That is, the
phase shift encountered by the forward travelling wave.

That specification is the same for physical and electrical
degrees because we are dealing with a single Z0 piece of
transmission line. The 100 ohm line is indeed 10 degrees
long both physically and electrically.

Certainly, the answer was in terms of electrical degrees. That is,
the phase shift encountered by the forward travelling wave.

You, and others, are going to be surprised to find out the
600 ohm section is only 43 degrees of physical length. How
can 43 degrees of 600 ohm line add to 10 degrees of 100 ohm
line to equal 90 electrical degrees of stub? Hint: Like I
told Roy and Tom years ago, there's a 37 degree phase shift
at the impedance discontinuity between the 600 ohm line and
the 100 ohm line. 43+37+10 = 90 electrical degrees.

Understand that simple stub example and you will understand
loaded mobile antennas. Most of the "experts" here are just
full of you-know-what.
--
73, Cecil http://www.w5dxp.com

Standing Wave Phase
 

On Dec 5, 10:01 am, Cecil Moore wrote:
Keith Dysart wrote:
I thought that when you specified 5 and 10 degrees in your
problem statement, you meant electrical degrees. That is, the
phase shift encountered by the forward travelling wave.

That specification is the same for physical and electrical
degrees because we are dealing with a single Z0 piece of
transmission line. The 100 ohm line is indeed 10 degrees
long both physically and electrically.

Certainly, the answer was in terms of electrical degrees. That is,
the phase shift encountered by the forward travelling wave.

You, and others, are going to be surprised to find out the
600 ohm section is only 43 degrees of physical length.

Hardly surprised. After all, the same can be achieved with an
inductor and/or capacitor which has essentially
0 physical (or electrical) length.

How
can 43 degrees of 600 ohm line add to 10 degrees of 100 ohm
line to equal 90 electrical degrees of stub? Hint: Like I
told Roy and Tom years ago, there's a 37 degree phase shift
at the impedance discontinuity between the 600 ohm line and
the 100 ohm line. 43+37+10 = 90 electrical degrees.

So the important take-away is that the system phase shift is
NOT equal to the sum of the phase shifts of the components.

This then begs the question, is the physical (or electrical) phase
shift in the components of much interest?

Standing Wave Phase
 

Keith Dysart wrote:
Hardly surprised. After all, the same can be achieved with an
inductor and/or capacitor which has essentially
0 physical (or electrical) length.

Only true for a lumped inductor which doesn't exist
in reality. Any large coil, such as the coil tested
by W8JI, has considerable electrical length at 4 MHz.
This electrical length is what some folks have been
denying for years even though they should certainly
know better by now.

This then begs the question, is the physical (or electrical) phase
shift in the components of much interest?

It is - when someone tries to convince the world that
there is a 3 ns delay through a 2" dia, 10 TPI,
100 turn coil. The electrical length (phase shift)
through a coil is necessary and sufficient to kill
the old wives tales being supported by some so-called
"experts" on this newsgroup.
--
73, Cecil http://www.w5dxp.com

Standing Wave Phase
 

On Dec 5, 10:40 am, Cecil Moore wrote:
Keith Dysart wrote:
Hardly surprised. After all, the same can be achieved with an
inductor and/or capacitor which has essentially
0 physical (or electrical) length.

Only true for a lumped inductor which doesn't exist
in reality. Any large coil, such as the coil tested
by W8JI, has considerable electrical length at 4 MHz.
This electrical length is what some folks have been
denying for years even though they should certainly
know better by now.

This then begs the question, is the physical (or electrical) phase
shift in the components of much interest?

It is - when someone tries to convince the world that
there is a 3 ns delay through a 2" dia, 10 TPI,
100 turn coil. The electrical length (phase shift)
through a coil is necessary and sufficient to kill
the old wives tales being supported by some so-called
"experts" on this newsgroup.

But again, given that the key message is "that the system phase shift
is NOT equal to the sum of the phase shifts of the components.",
why is the question of delay through the coil important?

Is it just to have a "debate" with some called experts?

Or does it offer some advancement in the solution of antenna
problems? Having computed (or measured) the delay through
the coil, how would this alter the design of the antenna?

....Keith

Standing Wave Phase
 

"Keith Dysart" wrote in message
...
On Dec 5, 10:40 am, Cecil Moore wrote:
Keith Dysart wrote:
Hardly surprised. After all, the same can be achieved with an
inductor and/or capacitor which has essentially
0 physical (or electrical) length.

Only true for a lumped inductor which doesn't exist
in reality. Any large coil, such as the coil tested
by W8JI, has considerable electrical length at 4 MHz.
This electrical length is what some folks have been
denying for years even though they should certainly
know better by now.

This then begs the question, is the physical (or electrical) phase
shift in the components of much interest?

It is - when someone tries to convince the world that
there is a 3 ns delay through a 2" dia, 10 TPI,
100 turn coil. The electrical length (phase shift)
through a coil is necessary and sufficient to kill
the old wives tales being supported by some so-called
"experts" on this newsgroup.

But again, given that the key message is "that the system phase shift
is NOT equal to the sum of the phase shifts of the components.",
why is the question of delay through the coil important?

Is it just to have a "debate" with some called experts?

Or does it offer some advancement in the solution of antenna
problems? Having computed (or measured) the delay through
the coil, how would this alter the design of the antenna?

...Keith


That train of arguments and nitpicking developed from the main argument
about distribution of (standing wave) current along the loading coil and
antenna.
"Gurus" and some literature claimed that current is the SAME at both end of
the coil (Kirchoff "said so"). That would mean that current remains constant
along the coil and then drops drastically towards (almost) zero at the tip.
This makes loaded whip antenna look better than it is.

The reality is that current drops around 40 - 60 % along the loading coil,
which makes the distribution along the remaining stinger starting with less
and overall efficiency less.

The key to understand the loaded radiator is to trying to maximize the
current in the physical "straight wire" - so the higher the coil, larger
hat, will stretch the high current portion of the radiator and make it more
efficient. Fooling yourself by modeling the loading coil as a lumped
inductance and making it look better in modeling program does not help.
Again, this effect is magnified in multi element loaded arrays, so while
some might consider this not a big deal in a mobile whip, the errors would
magnify in multielement designs.

I hope I can get the main "problem" across, the rest was digging into the
smaller effects like coil radiates, junction impedance discontinuity, bla,
bla....

Now I also understand the small "bump" increase in the current at the bottom
of the coil due to some loses that reflected wave encounter on the way
"there and back" to the tip of the radiator from the bottom of the coil (?)

73 Yuri, www.K3BU.us

Standing Wave Phase
 

Cecil Moore wrote:
Keith Dysart wrote:
Hardly surprised. After all, the same can be achieved with an
inductor and/or capacitor which has essentially
0 physical (or electrical) length.

Only true for a lumped inductor which doesn't exist
in reality. Any large coil, such as the coil tested
by W8JI, has considerable electrical length at 4 MHz.
This electrical length is what some folks have been
denying for years even though they should certainly
know better by now.

This then begs the question, is the physical (or electrical) phase
shift in the components of much interest?

It is - when someone tries to convince the world that
there is a 3 ns delay through a 2" dia, 10 TPI,
100 turn coil. The electrical length (phase shift)
through a coil is necessary and sufficient to kill
the old wives tales being supported by some so-called
"experts" on this newsgroup.

Cecil,

No one has ever said that there is a 3 ns delay *through* the coil. Ask
Richard Harrison if Faraday screens work, even without any conduction
path at all. Radiation is real.

You keep trying to change the topic, but the only debate is the relative
contributions of "round and round the wire" vs. other coupling. The math
is not easy, and the problem is not readily amenable to solution by
intuition and word games.

73,
Gene
W4SZ

Standing Wave Phase
 

On Wed, 5 Dec 2007 07:18:47 -0800 (PST), Keith Dysart
wrote:

This then begs the question, is the physical (or electrical) phase
shift in the components of much interest?

Hi Keith,

Its consideration is listed by the Department of Commerce as an
important occupation for the home bound and invalid during the Winter
months.

73's
Richard Clark, KB7QHC

Standing Wave Phase
 

Cecil Moore wrote:
Tom Donaly wrote:
The same thing can be accomplished, above, using an open stub and
measuring the voltage at both ends. All this is just theoretical,
though, because line loss will skew the results. Besides, why try to
measure current or voltage when all you have to do is measure length
and frequency?

Tom, you are not going to understand what I am saying until
you perform the stub exercise I provided. Just do one at a
time. Assume ideal lossless conditions with VF=1.0.

---600 ohm line---+---10 deg 100 ohm line---open

How many degrees of 600 ohm line does it take to make the
above stub look like 1/4 wavelength, i.e. 90 degrees?

Until you perform the exercise, you are just creating
diversions and avoiding the technical truth.

If you know how to do it, Cecil, don't be coy about it. Just state
your case and be done with it. Since you already stated that
the total electrical length is 90 degrees, you're just asking me
to prove your point. Do it yourself, and then I'll tell you
whether I agree with you or not.
73,
Tom Donaly, KA6RUH

Standing Wave Phase
 

Cecil Moore wrote:
Keith Dysart wrote:
On Dec 4, 11:13 pm, Cecil Moore wrote:
Tom Donaly wrote:
50 ohm Shorted line 8.5655 meters long. Frequency = 7 Mhz. 2 volt 50
ohm
generator. Current at input = 12.361 milliamps. Current at short =
40 milliamps. Divide the current at the imput of the line by the
current
at the short and take the arc sine (in radian mode) of the result. This
is 1.2566. Now take Beta = .14671 and multiply it by the length 8.5655.
This also equals 1.2566, which is the angular length of the shorted
line. Will someone explain how this works to Cecil?
You won't understand what I am talking about until you perform
the stub experiments that I previously posted.

---600 ohm line---+---10 deg, 100 ohm line---open-circuit

How many degrees of 600 ohm line does it take to resonate
that stub to an electrical 1/4WL?

--5 deg, 100 ohm line--+--600 ohm line--+--5 deg, 100 ohm line--open

How many degrees of 600 ohm line does it take to resonate
that stub to an electrical 1/4WL?

You have said multiple times that the electrical length of a
quarter wave stub must be 90 electrical degress, so the
computation is too easy...

1) x + 10 = 90
x = 80 degrees for the 600 Ohm line
2) 5 + x + 5 = 90
x = 80 degrees for the 600 Ohm line

although I suspect others will disagree with your solution.

I have not yet provided a solution. Your's is *wrong*.

The 90 degree physical solution is *wrong* because it results
in more than 90 electrical degrees. Please try again.

No, Cecil, it's your theory. You have to provide the method and
then everyone else will decide whether or not they agree with you.
You're not chicken are you?
73,
Tom Donaly, KA6RUH

Standing Wave Phase
 

Cecil Moore wrote:
Keith Dysart wrote:
I thought that when you specified 5 and 10 degrees in your
problem statement, you meant electrical degrees. That is, the
phase shift encountered by the forward travelling wave.

That specification is the same for physical and electrical
degrees because we are dealing with a single Z0 piece of
transmission line. The 100 ohm line is indeed 10 degrees
long both physically and electrically.

Certainly, the answer was in terms of electrical degrees. That is,
the phase shift encountered by the forward travelling wave.

You, and others, are going to be surprised to find out the
600 ohm section is only 43 degrees of physical length. How
can 43 degrees of 600 ohm line add to 10 degrees of 100 ohm
line to equal 90 electrical degrees of stub? Hint: Like I
told Roy and Tom years ago, there's a 37 degree phase shift
at the impedance discontinuity between the 600 ohm line and
the 100 ohm line. 43+37+10 = 90 electrical degrees.

Understand that simple stub example and you will understand
loaded mobile antennas. Most of the "experts" here are just
full of you-know-what.

Actually, you got it wrong, Cecil. It's -43 degrees, which means
if you wanted to make one it would be about 317 degrees. (Actually,
I got -46.613 degrees.) Maybe you can make a length using negative
degrees, but it's tough for me to do.
73,
Tom Donaly, KA6RUH

Standing Wave Phase
 

Keith Dysart wrote:
But again, given that the key message is "that the system phase shift
is NOT equal to the sum of the phase shifts of the components.",
why is the question of delay through the coil important?

Is it just to have a "debate" with some called experts?

That's as good a reason as any. :-)

Or does it offer some advancement in the solution of antenna
problems? Having computed (or measured) the delay through
the coil, how would this alter the design of the antenna?

I don't know.
--
73, Cecil http://www.w5dxp.com

Standing Wave Phase
 

Yuri Blanarovich wrote:
Now I also understand the small "bump" increase in the current at the bottom
of the coil due to some loses that reflected wave encounter on the way
"there and back" to the tip of the radiator from the bottom of the coil (?)

I suspect that current bump is caused by magnetic linkage
between the central windings while the end windings would
not experience it to such a degree. It is the thing that
increases the velocity factor over the purely helical
path.
--
73, Cecil http://www.w5dxp.com

Standing Wave Phase
 

Gene Fuller wrote:
No one has ever said that there is a 3 ns delay *through* the coil.

That's simply a false statement. Here's W8JI exact words:

"On 80-meters ... time delay is about 3nS. How does the
current travel through the inductor so fast?

Gene, "through the inductor" is through the coil. W8JI
said there is a 3 ns delay *through* the coil. W7EL
seems to agree.
--
73, Cecil http://www.w5dxp.com

Standing Wave Phase
 

Tom Donaly wrote:
If you know how to do it, Cecil, don't be coy about it. Just state
your case and be done with it. Since you already stated that
the total electrical length is 90 degrees, you're just asking me
to prove your point. Do it yourself, and then I'll tell you
whether I agree with you or not.

Does that mean you don't know how to do it?

I already did it on another thread, Tom. Adding 43 degrees
of Z0=600 ohm feedline to the 10 degrees of Z0=100 ohm
feedline will turn the stub into an electrical 1/4
wavelength (90 degree) open stub. And that's exactly
how base-loaded mobile antennas work.
--
73, Cecil http://www.w5dxp.com

Standing Wave Phase
 

Tom Donaly wrote:
No, Cecil, it's your theory. You have to provide the method and
then everyone else will decide whether or not they agree with you.
You're not chicken are you?

Actually, I wanted to see if anyone besides me could
solve the problem - no one else has.
--
73, Cecil http://www.w5dxp.com

Standing Wave Phase
 

Tom Donaly wrote:
Actually, you got it wrong, Cecil. It's -43 degrees, which means
if you wanted to make one it would be about 317 degrees.

Tom, you are just showing your ignorance. 43 degrees
of 600 ohm line attached to 10 degrees of 100 ohm
line makes a dandy 1/4WL stub. Why are you so totally
ignorant of that fact of physics?
--
73, Cecil http://www.w5dxp.com

Standing Wave Phase
 

Cecil Moore wrote:
Gene Fuller wrote:
No one has ever said that there is a 3 ns delay *through* the coil.

That's simply a false statement. Here's W8JI exact words:

"On 80-meters ... time delay is about 3nS. How does the
current travel through the inductor so fast?

Gene, "through the inductor" is through the coil. W8JI
said there is a 3 ns delay *through* the coil. W7EL
seems to agree.

Cecil,

You're losing your touch. That isn't exactly a subtle selective quote.

Here are the rest of the words following the question, "How does the
current travel through the inductor so fast?"

"At first this seems impossible, but the answer is actually quite
obvious. Time-varying current gives rise to time-varying magnetic flux.
This magnetic flux, since conductor spacing is close and the distance
very small, links the starting turn very tightly to the next turn. The
rapidly changing magnetic flux causes charges to move in the next
conductor, and the changing magnetic field couples through all the close
spaced turns with very little time delay. It is this magnetic flux
coupling that provides the primary mechanism for energy transfer through
the inductor, and the path is much shorter than the circuitous and much
longer path along the conductor."


But you already knew that . .

73,
Gene
W4SZ

Standing Wave Phase
 

Gene Fuller wrote:
But you already knew that . .

Of course I did. "Through the coil" does NOT mean "through
the coil wire". It means "through the coil". You still uttered
a falsehood but I doubt that you will ever admit it.
--
73, Cecil http://www.w5dxp.com

Standing Wave Phase
 

Cecil Moore wrote:
Tom Donaly wrote:
No, Cecil, it's your theory. You have to provide the method and
then everyone else will decide whether or not they agree with you.
You're not chicken are you?

Actually, I wanted to see if anyone besides me could
solve the problem - no one else has.

You're chicken.
73,
Tom Donaly, KA6RUH

Standing Wave Phase
 

Cecil Moore wrote:
Tom Donaly wrote:
Actually, you got it wrong, Cecil. It's -43 degrees, which means
if you wanted to make one it would be about 317 degrees.

Tom, you are just showing your ignorance. 43 degrees
of 600 ohm line attached to 10 degrees of 100 ohm
line makes a dandy 1/4WL stub. Why are you so totally
ignorant of that fact of physics?

O.k., now prove it using standard transmission line theory.
You can say it over and over again, but if you can't give anyone
a reason to believe it, you're just blowing hot air.
73,
Tom Donaly, KA6RUH

Standing Wave Phase
 

Cecil Moore wrote:
Tom Donaly wrote:
If you know how to do it, Cecil, don't be coy about it. Just state
your case and be done with it. Since you already stated that
the total electrical length is 90 degrees, you're just asking me
to prove your point. Do it yourself, and then I'll tell you
whether I agree with you or not.

Does that mean you don't know how to do it?

I already did it on another thread, Tom. Adding 43 degrees
of Z0=600 ohm feedline to the 10 degrees of Z0=100 ohm
feedline will turn the stub into an electrical 1/4
wavelength (90 degree) open stub. And that's exactly
how base-loaded mobile antennas work.

It will, will it? I'm waiting for you to prove it. Do you
really expect it to be resonant at the right frequency?
This is getting tiresome, Cecil. I guess you think that
stating things over and over again will make them come
true. Oh, well, it's my fault. I shouldn't waste my time
trying to get blood out of a turnip.
73,
Tom Donaly, KA6RUH

Standing Wave Phase
 

Tom Donaly wrote:

...
It will, will it? I'm waiting for you to prove it. Do you
really expect it to be resonant at the right frequency?
This is getting tiresome, Cecil. I guess you think that
stating things over and over again will make them come
true. Oh, well, it's my fault. I shouldn't waste my time
trying to get blood out of a turnip.
73,
Tom Donaly, KA6RUH

You mean at the frequency where the 600 line length is 53 degrees and
the 100 line 10 degrees length ... well, I guess that already answers
your own question, doesn't it?--but then, you should have already knew
that ...

JS

Standing Wave Phase
 

Cecil Moore wrote:
Tom Donaly wrote:
No, Cecil, it's your theory. You have to provide the method and
then everyone else will decide whether or not they agree with you.
You're not chicken are you?

Actually, I wanted to see if anyone besides me could
solve the problem - no one else has.

O.K., Cecil, I finally figured out what you want to do. You want
a zero ohm input impedance, just like a 1/4 wave open stub. In that
case, you're absolutely right, the 600 ohm line should be 43.387
degrees long. If you call the 100 ohm line, line 1, and the 600
ohm line, line 2, then the criterion for what you want is:
tan(Bl1)*tan(Bl2)= Z01/Z02. This behaves sort of like a backwards,
transmission-line, Helmholtz resonator. I still don't know where you
come up with the 90 degree stuff. It isn't necessary to explain the
phenomenon, but if it waters your lawn, go for it.
73,
Tom Donaly, KA6RUH

(P.S. Check my math. Bl2 = atan((Zo1/Zo2)/tan(10)) =
atan(.166667/.17633) = 43.387.)

Standing Wave Phase
 

John Smith wrote:
Tom Donaly wrote:

...
It will, will it? I'm waiting for you to prove it. Do you
really expect it to be resonant at the right frequency?
This is getting tiresome, Cecil. I guess you think that
stating things over and over again will make them come
true. Oh, well, it's my fault. I shouldn't waste my time
trying to get blood out of a turnip.
73,
Tom Donaly, KA6RUH

You mean at the frequency where the 600 line length is 53 degrees and
the 100 line 10 degrees length ... well, I guess that already answers
your own question, doesn't it?--but then, you should have already knew
that ...

JS

I'm beginning to see why Roy plonked you, Smith, or whatever your
present pseudonym is. It looks as if you've been taking opacity
lessons from Art.
73,
Tom Donaly, KA6RUH

Standing Wave Phase
 

Tom Donaly wrote:
Actually, you got it wrong, Cecil. It's -43 degrees, which means
if you wanted to make one it would be about 317 degrees. (Actually,
I got -46.613 degrees.) Maybe you can make a length using negative
degrees, but it's tough for me to do.

What you apparently don't realize is the negative sign
simply indicates the direction on the Smith Chart of
"toward the source". This is absolutely reasonable since
we are indeed going in the direction away from the open
end of the stub toward the mouth of the stub, i.e. from
right to left if the source side is on the left.

From -j567.128 ohms to zero ohms "toward the source" is about
43.4 degrees. From -j567.128 ohms to zero ohms "toward the load"
is 316.6 degrees. Your signs are correct - you just didn't
understand what they mean.
--
73, Cecil http://www.w5dxp.com

Standing Wave Phase
 

Tom Donaly wrote:
No, Cecil, it's your theory. You have to provide the method and
then everyone else will decide whether or not they agree with you.

I appreciate your crediting me with a transmission
line theory, Tom, but after hundreds of years of
established theory, I seriously doubt that I have
discovered anything new. It is much more likely
that you slept through a few days of Fields and
Waves 301.
--
73, Cecil http://www.w5dxp.com

Standing Wave Phase
 

Tom Donaly wrote:
O.k., now prove it using standard transmission line theory.
You can say it over and over again, but if you can't give anyone
a reason to believe it, you're just blowing hot air.

Does that imply that you are incapable of verifying
it for yourself? - incapable of proving it wrong? I
went through the math in a reply to Dan and I have
verified it using MicroSmith. IMO, these dual-Z0
shortened stubs are best understood using the Smith
Chart.

Let's see how much you have learned. If we make the
600 ohm section equal in length to the 100 ohm
section, how many physical degrees of length will each
section have to be to achieve 1/4 wavelength resonance?
--
73, Cecil http://www.w5dxp.com