Standing Wave Phase
 

Tom Donaly wrote:
Cecil Moore wrote:
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?

I have proved it in a reply to Dan and verified it
with MicroSmith. I don't know what else you are
asking for. Yes, it will resonate at the design
frequency. Are you incapable of those simple
calculations? Note that everything is rounded
off to the nearest degree.
--
73, Cecil http://www.w5dxp.com

Standing Wave Phase
 

John Smith wrote:
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 ...

Make that 43 degrees instead of 53 degrees.
--
73, Cecil http://www.w5dxp.com

Standing Wave Phase
 

Tom Donaly wrote:
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.

For an open stub to exhibit a zero ohm input impedance,
it must be electrically 90 degrees long (or 270 ...).
That's where the 90 degrees comes from. The example
stub is electrically 90 degrees long while being 53
degrees long physically.

Good for you, Tom, now you have it - "just like a 1/4WL
open stub" from 53 degrees of transmission line. Here's
another tidbit for you.

Using 600 ohm line and 100 ohms line, if you make the two
sections equal length, the dual-Z0 stub will be very close
to 1/2 the physical length of a single-Z0 stub, i.e. physically
45 degrees long for an electrical 1/4WL (90 deg) stub. On 75m,
that cuts the 1/4 stub physical length from ~66 feet to ~33
feet, a much more manageable length.

Here's a useful equation. For a 1/4WL stub with equal length
sections, the physical length in degrees of each section is:

ARCTAN[SQRT(Z0Low/Z0High)]
--
73, Cecil http://www.w5dxp.com

Standing Wave Phase
 

"Cecil Moore" wrote in message
t...
Tom Donaly wrote:
O.K., Cecil, I finally figured out what you want to do. You want
snip

Using 600 ohm line and 100 ohms line, if you make the two
sections equal length, the dual-Z0 stub will be very close
to 1/2 the physical length of a single-Z0 stub, i.e. physically
45 degrees long for an electrical 1/4WL (90 deg) stub. On 75m,
that cuts the 1/4 stub physical length from ~66 feet to ~33
feet, a much more manageable length.

Here's a useful equation. For a 1/4WL stub with equal length
sections, the physical length in degrees of each section is:

ARCTAN[SQRT(Z0Low/Z0High)]
--
73, Cecil http://www.w5dxp.com


lurking off

NEAT

lurking back on

Standing Wave Phase
 

Cecil Moore wrote:

...
Make that 43 degrees instead of 53 degrees.

Sorry, don't know if that is brain atrophy from age or the 20mg
hydrocodone the doc has me on for the fractured bone and arthritis in
the spine--laying down a bike at ~60mph is something better left for the
younger generation.

Can't seem to find my glasses after I lay 'em down--wife claims she has
already given me something, I claim no, then find 'em in my pocket.

By the way, one week in the hospital cost: Hospital $80,000, emergency
room $1,900, ambulance $1

Standing Wave Phase
 

John Smith wrote:
Cecil Moore wrote:

...
Make that 43 degrees instead of 53 degrees.

Sorry, don't know if that is brain atrophy from age or the 20mg
hydrocodone the doc has me on for the fractured bone and arthritis in
the spine--laying down a bike at ~60mph is something better left for the
younger generation.

Can't seem to find my glasses after I lay 'em down--wife claims she has
already given me something, I claim no, then find 'em in my pocket.

By the way, one week in the hospital cost: Hospital $80,000, emergency
room $1,900, ambulance $1,500, etc.

Oh yeah, and then there is constantly hitting the wrong key and sending
email early ...

Sorry guys/gals ... keep your medical PAID UP!

Regards,
JS

Standing Wave Phase
 

Cecil Moore wrote:
Tom Donaly wrote:
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.

For an open stub to exhibit a zero ohm input impedance,
it must be electrically 90 degrees long (or 270 ...).
That's where the 90 degrees comes from. The example
stub is electrically 90 degrees long while being 53
degrees long physically.

(The rest deleted.)
O.k., Cecil, you said it, now prove it. There's no requirement
for a 90 degree phase shift when you do the math. Don't expect me
to do it for you this time. Since I did some math for you, you can
do some for me: Given the above formula, if you know l1, l2, and
Z01, and Z02, what's the formula for B? It should be easy, right?
73,
Tom Donaly, KA6RUH

Standing Wave Phase
 

Cecil Moore wrote:
John Smith wrote:
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 ...

Make that 43 degrees instead of 53 degrees.

Some, like me, might like to review some info like this, easily
digest-able ...:

http://courses.ece.uiuc.edu/ece450/N...sionLines2.pdf

Regards,
JS

Standing Wave Phase
 

John Smith wrote:
By the way, one week in the hospital cost: Hospital $80,000, emergency
room $1,900, ambulance $1

I'm sorry that happened. Get well quick.
--
73, Cecil http://www.w5dxp.com

Standing Wave Phase
 

Tom Donaly wrote:
O.k., Cecil, you said it, now prove it. There's no requirement
for a 90 degree phase shift when you do the math.

Are you using the equation for forward and reflected
current? If not, you need to do so. The phase shift
is not in the standing-wave current. Standing-wave
current phase is fixed with respect to the source.
Absolutely *nothing* happens to the standing-wave
current at the impedance discontinuity.

The reflected current is known to be in phase with
the forward current at the feedpoint. The forward
current is reflected at the tip of the antenna and
undergoes a 180 degree phase shift. Something must
account for the other 180 degrees or else the
feedpoint impedance would not be resistive. I am
working on a graphic that illustrates what happens
at the impedance discontinuity.

Please enlighten us on how the reflected current
gets back in phase with the forward current without
undergoing a phase shift of 180 degrees in its
round-trip path. It is my understanding that the
forward phasor rotates in one direction while the
reflected phasor rotates in the opposite direction.
The key concept there is that a phasor is always
rotating.

I have waded through the math before but I cannot
locate my notes after moving. If you can figure
out a reasonable answer to the above question, I
will certainly consider it.
--
73, Cecil http://www.w5dxp.com

Standing Wave Phase
 

Cecil Moore wrote:
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.

Cecil,

You got me. I omitted the word "wire".

Of course your time delay "impossibility" complaint makes absolutely no
sense at all if you accept that W8JI was not talking about a wave
traveling through 50 feet of wire in 3 ns. He was talking about a wave
traveling 10 inches in 3 ns. I mistakenly assumed that no one in this
discussion even remotely considered superluminal wave propagation.

Please accept my humble apology for an utter lack of fine craftsmanship
in word games.

73,
Gene
W4SZ

Standing Wave Phase
 

Cecil Moore wrote:
Tom Donaly wrote:
O.k., Cecil, you said it, now prove it. There's no requirement
for a 90 degree phase shift when you do the math.

Are you using the equation for forward and reflected
current? If not, you need to do so. The phase shift
is not in the standing-wave current. Standing-wave
current phase is fixed with respect to the source.
Absolutely *nothing* happens to the standing-wave
current at the impedance discontinuity.

The reflected current is known to be in phase with
the forward current at the feedpoint. The forward
current is reflected at the tip of the antenna and
undergoes a 180 degree phase shift. Something must
account for the other 180 degrees or else the
feedpoint impedance would not be resistive. I am
working on a graphic that illustrates what happens
at the impedance discontinuity.

Please enlighten us on how the reflected current
gets back in phase with the forward current without
undergoing a phase shift of 180 degrees in its
round-trip path. It is my understanding that the
forward phasor rotates in one direction while the
reflected phasor rotates in the opposite direction.
The key concept there is that a phasor is always
rotating.

I have waded through the math before but I cannot
locate my notes after moving. If you can figure
out a reasonable answer to the above question, I
will certainly consider it.

How about answering the other part of my post, Cecil.
I didn't use reflection mechanics to reach my conclusion.
I did use two port ABCD parameters (the hard way) which can
be derived from reflection mechanics. Anyway, you're being too
simple. If you're going to use reflection mechanics, you have to
account for all the reflections, and you have to explain yourself
each step of the way. Anyway, destroy a few brain cells
thinking about the other part of my post and get back to me.
73,
Tom Donaly, KA6RUH

Standing Wave Phase
 

Cecil Moore wrote:
John Smith wrote:
By the way, one week in the hospital cost: Hospital $80,000,
emergency room $1,900, ambulance $1

I'm sorry that happened. Get well quick.

Off topic, I know you have a harley though ...

Had putts since I was 16, never had an accident out of the dirt, nothing
ever serious, etc.

Be careful on that monster of yours--wishing ya all the luck--and, oh
yeah, TAKE CARE on that road! They ARE out to get ya ... ;-)

Regards,
JS

Standing Wave Phase
 

Gene Fuller wrote:
Cecil Moore wrote:
"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.

You got me. I omitted the word "wire".

In that case, I apologize for jumping on you. The VF of
the coil is approximately double what it would be if
the current followed the wire. That's because of the
inter-winding interaction. But the magnetic fields
from coil#1 do not magically jump 10 inches to coil#100
as W8JI implies they do. His measurements are off by
a magnitude because he used standing-wave current for
his measurements. I suspect he actually measured less
than 3 ns "delay" but knew he couldn't afford to post
a faster-than-light measurement.
--
73, Cecil http://www.w5dxp.com

Standing Wave Phase
 

Tom Donaly wrote:
If you're going to use reflection mechanics, you have to
account for all the reflections, and you have to explain yourself
each step of the way.

I can do that, Tom, but I am standing by for an emergency
trip to New York which involves my daughter's life.

In the meantime, let's see if we can agree if 43.4 degrees
of 600 ohm line is terminated in -j567 ohms, the forward
current will be in phase with the reflected current, i.e.
the impedance looking into the line will be zero ohms. Have
you used any current reflection coefficients recently? :-)
--
73, Cecil http://www.w5dxp.com

Standing Wave Phase
 

Cecil Moore wrote:
Tom Donaly wrote:
If you're going to use reflection mechanics, you have to
account for all the reflections, and you have to explain yourself
each step of the way.

I can do that, Tom, but I am standing by for an emergency
trip to New York which involves my daughter's life.

In the meantime, let's see if we can agree if 43.4 degrees
of 600 ohm line is terminated in -j567 ohms, the forward
current will be in phase with the reflected current, i.e.
the impedance looking into the line will be zero ohms. Have
you used any current reflection coefficients recently? :-)

Actually, I have. They're built into the formulas for transmission line
voltages and currents on my calculator. My calculator is a Cecil
disciple.
I hope your daughter comes through in good shape. I understand the
worry, believe me.
73,
Tom Donaly, KA6RUH

Standing Wave Phase
 

Cecil Moore wrote:
... I am standing by for an emergency
trip to New York which involves my daughter's life.
...

Prayers and best wishes ... God protect and speed.

Warmest regards,
JS

Standing Wave Phase
 

Tom Donaly wrote:
Cecil Moore wrote:
In the meantime, let's see if we can agree if 43.4 degrees
of 600 ohm line is terminated in -j567 ohms, the forward
current will be in phase with the reflected current, i.e.
the impedance looking into the line will be zero ohms. Have
you used any current reflection coefficients recently? :-)

Actually, I have. They're built into the formulas for transmission line
voltages and currents on my calculator.

So are we agreed that a 43.4 degree stub terminated in
0-j567 ohms impedance is electrically 1/4WL, i.e. 90
degrees long?
--
73, Cecil http://www.w5dxp.com

Standing Wave Phase
 

Cecil Moore wrote:
Tom Donaly wrote:
Cecil Moore wrote:
In the meantime, let's see if we can agree if 43.4 degrees
of 600 ohm line is terminated in -j567 ohms, the forward
current will be in phase with the reflected current, i.e.
the impedance looking into the line will be zero ohms. Have
you used any current reflection coefficients recently? :-)

Actually, I have. They're built into the formulas for transmission line
voltages and currents on my calculator.

So are we agreed that a 43.4 degree stub terminated in
0-j567 ohms impedance is electrically 1/4WL, i.e. 90
degrees long?

No, but I agree with myself that whatever voltage is applied to
the input will be canceled by all the reflections in the circuit
adding up to a voltage at said input that will cancel the input voltage.
It's like finding the zeros of a network equation.
73,
Tom Donaly, KA6RUH

Standing Wave Phase
 

On Dec 6, 9:48 pm, Cecil Moore wrote:
So are we agreed that a 43.4 degree stub terminated in
0-j567 ohms impedance is electrically 1/4WL, i.e. 90
degrees long?

There are many ways to get the 0 input impedance:
- 43.4 degrees of 600 ohm line terminated in a lumped 0-j567 impedance
(assuming I recall the problem corrrectly and you did the math
correctly)
- 43.4 degrees of 600 ohm line followed by 46.6 degrees of 600 ohm
line, open at the end
- 43.4 degrees of 600 ohm line followed by 10 degrees (IIRC) of 100
ohm line, open at the end
- a short
- 180 degrees of any impedance line shorted at the end
- and many, many more

Are you claiming that all of these are electrically 1/4WL ?
Even the 180 degree line? And the short? Seems like a stretch.

And when looked at in detail (think time domain for moment), they each
behave quite differently.

....Keith

Standing Wave Phase
 

Keith Dysart wrote:
On Dec 6, 9:48 pm, Cecil Moore wrote:
So are we agreed that a 43.4 degree stub terminated in
0-j567 ohms impedance is electrically 1/4WL, i.e. 90
degrees long?

There are many ways to get the 0 input impedance:
(1)- 43.4 degrees of 600 ohm line terminated in a lumped 0-j567 impedance
(assuming I recall the problem corrrectly and you did the math
correctly)
(2)- 43.4 degrees of 600 ohm line followed by 46.6 degrees of 600 ohm
line, open at the end
(3)- 43.4 degrees of 600 ohm line followed by 10 degrees (IIRC) of 100
ohm line, open at the end
(4)- a short
(5)- 180 degrees of any impedance line shorted at the end
- and many, many more

If you calculate the complex rho and calculate the phase
shift provided by a -j567 impedance, you will agree with
my statement above.

Are you claiming that all of these are electrically 1/4WL ?

Of course not!!!! I numbered your examples above. Examples
1-3 are electrically 1/4WL long. Example 4 is 0 WL long.
Example 5 is 1/2WL long.

All my remarks apply only to stubs and antennas that are
electrically 1/4WL long. My remarks do NOT apply to any
stub or antenna that is not electrically 1/4WL long. An
ideal open stub that is 3/4WL long has the same impedance
as a 1/4WL stub but is it obviously not 1/4WL long.
--
73, Cecil http://www.w5dxp.com

Standing Wave Phase
 

On Dec 7, 12:37 pm, Cecil Moore wrote:
Keith Dysart wrote:
On Dec 6, 9:48 pm, Cecil Moore wrote:
So are we agreed that a 43.4 degree stub terminated in
0-j567 ohms impedance is electrically 1/4WL, i.e. 90
degrees long?

There are many ways to get the 0 input impedance:
(1)- 43.4 degrees of 600 ohm line terminated in a lumped 0-j567 impedance
(assuming I recall the problem corrrectly and you did the math
correctly)
(2)- 43.4 degrees of 600 ohm line followed by 46.6 degrees of 600 ohm
line, open at the end
(3)- 43.4 degrees of 600 ohm line followed by 10 degrees (IIRC) of 100
ohm line, open at the end
(4)- a short
(5)- 180 degrees of any impedance line shorted at the end
- and many, many more

If you calculate the complex rho and calculate the phase
shift provided by a -j567 impedance, you will agree with
my statement above.

Are you claiming that all of these are electrically 1/4WL ?

Of course not!!!! I numbered your examples above. Examples
1-3 are electrically 1/4WL long. Example 4 is 0 WL long.
Example 5 is 1/2WL long.

All my remarks apply only to stubs and antennas that are
electrically 1/4WL long. My remarks do NOT apply to any
stub or antenna that is not electrically 1/4WL long. An
ideal open stub that is 3/4WL long has the same impedance
as a 1/4WL stub but is it obviously not 1/4WL long.

Your original claim was that 43.4 degrees of 600 ohm
line terminated with 0-j567 was electricall 90 degrees long.
You made no reference to how the 0-j567 was obtained.

You have said that 1, 2 and 3 from above are electically
90 degrees.

How about:
(6) 43.4 degrees of 600 ohm line, 180 degrees of arbitrary
line terminated in a lump of 0-j567. This is just another
way of placing 0-j567 at the end of the 43.4 degrees of
600 ohm line.

And (5), if we use 600 ohm line is also 43.4 degrees of
600 ohm line terminated with 0-j567, this being obtained
with 136.6 degrees of 600 ohm line that is short
circuited.

So while I can accept your statement , "My remarks do
NOT apply to any stub or antenna that is not electrically
1/4WL long.", I am having great difficulty coming up for
a rule so that I will know when your remarks apply.

Can you provide a rule for discerning when a stub or
antenna is electrically 1/4WL long?

....Keith

Standing Wave Phase
 

Keith Dysart wrote:
You have said that 1, 2 and 3 from above are electically
90 degrees.

How about:
(6) 43.4 degrees of 600 ohm line, 180 degrees of arbitrary
line terminated in a lump of 0-j567. This is just another
way of placing 0-j567 at the end of the 43.4 degrees of
600 ohm line.

Of course, if you add 180 degrees you have added 180
degrees to whatever existed before. Do you disagree?

Can you provide a rule for discerning when a stub or
antenna is electrically 1/4WL long?

Of course! When the reflected wave undergoes a phase
shift of 180 degrees in its round trip to the end of
the stub and back, the stub is electrically 1/4WL long.
How can you disagree with that?
--
73, Cecil http://www.w5dxp.com

Standing Wave Phase
 

On Dec 7, 4:21 pm, Cecil Moore wrote:
Keith Dysart wrote:
You have said that 1, 2 and 3 from above are electically
90 degrees.

How about:
(6) 43.4 degrees of 600 ohm line, 180 degrees of arbitrary
line terminated in a lump of 0-j567. This is just another
way of placing 0-j567 at the end of the 43.4 degrees of
600 ohm line.

Of course, if you add 180 degrees you have added 180
degrees to whatever existed before. Do you disagree?

So this meets the criteria you originally proposed and
is an example of 90 degree electical length?

Can you provide a rule for discerning when a stub or
antenna is electrically 1/4WL long?

Of course! When the reflected wave undergoes a phase
shift of 180 degrees in its round trip to the end of
the stub and back, the stub is electrically 1/4WL long.
How can you disagree with that?

So (4), a short, meets this criteria. It did not have to go
far down the stub, but it did arrive back with a 180 degree
phase change. But previously, you did not include (4).

Is it now in the list?

....Keith

Standing Wave Phase
 

Keith Dysart wrote:
Cecil Moore wrote:
Keith Dysart wrote:
Of course, if you add 180 degrees you have added 180
degrees to whatever existed before. Do you disagree?

So this meets the criteria you originally proposed and
is an example of 90 degree electical length?

Don't be silly. 180 degrees plus any positive angle
is more than 180 degrees.

The context was mobile loaded antennas shorter than
a physical 1/4WL.
--
73, Cecil http://www.w5dxp.com

Standing Wave Phase
 

On Dec 8, 12:43 am, Cecil Moore wrote:
Keith Dysart wrote:
Cecil Moore wrote:
Keith Dysart wrote:
Of course, if you add 180 degrees you have added 180
degrees to whatever existed before. Do you disagree?

So this meets the criteria you originally proposed and
is an example of 90 degree electical length?

Don't be silly. 180 degrees plus any positive angle
is more than 180 degrees.

The context was mobile loaded antennas shorter than
a physical 1/4WL.

Hmmmm. So you are no longer in agreement with your
original question:
"So are we agreed that a 43.4 degree stub terminated
in 0-j567 ohms impedance is electrically 1/4WL, i.e.
90 degrees long?"

This is good. You can now understand why some were
not quick to jump to agreement.

The concept of electrical/physical degrees is an
occasionally useful way to think about delay on
a transmission line that is used in a single frequency
environment. It even helps understand stubs where
the reflection arrives back with some phase shift
from the original. But extending the concept to lumped
circuits or expecting to find 90 degrees when different
impedances are involved has little value. It leads
to worthless questions like "where did the missing
degrees go?"

This is much like ascribing excessive reality to
"reflected power" which leads to worthless questions
like "where did the reflected power go?".

Or asking "where is the missing dollar?'. The flawed
underpinnings lead to worthless questions. Well maybe
not worthless, like the hotel puzzle, they test the
ability of the answerer to detect flawed assumptions.

....Keith

Standing Wave Phase
 

Keith Dysart wrote:
On Dec 8, 12:43 am, Cecil Moore wrote:
Keith Dysart wrote:
Cecil Moore wrote:
Keith Dysart wrote:
Of course, if you add 180 degrees you have added 180
degrees to whatever existed before. Do you disagree?
So this meets the criteria you originally proposed and
is an example of 90 degree electical length?
Don't be silly. 180 degrees plus any positive angle
is more than 180 degrees.

The context was mobile loaded antennas shorter than
a physical 1/4WL.

Hmmmm. So you are no longer in agreement with your
original question:
"So are we agreed that a 43.4 degree stub terminated
in 0-j567 ohms impedance is electrically 1/4WL, i.e.
90 degrees long?"

Now I understand your confusion. I was talking about
a -j567 ohm *capacitor*, not a virtual impedance.

I was, of course, using the "impedor" definition of
impedance but since that confused you, let me restate
the question:

"So are we agree that a 43.4 degree stub terminated in
a -j567 ohm impedor is electrically 1/4WL, i.e. 90
degrees long?"

This was the original meaning of the question. I'm sorry
that you took it the wrong way and wasted so many postings
on such a trivial misunderstanding.

It leads
to worthless questions like "where did the missing
degrees go?"

Click on "Load Dat" in the EZNEC model below. There are
*no* missing degrees. All necessary degrees are present
and accounted for. But you will never see them if you
are trying to use standing-wave current to see them.

http://www.w5dxp.com/coil512.ez

This is much like ascribing excessive reality to
"reflected power" which leads to worthless questions
like "where did the reflected power go?".

Since energy must be conserved, the proper question
is: "Where did the reflected wave *energy* go?" Do
you even know the answer? The answer is that there
is exactly the amount of energy existing in a transmission
line to support the forward wave and the reflected wave.
--
73, Cecil http://www.w5dxp.com

Standing Wave Phase
 

On Dec 8, 9:18 am, Cecil Moore wrote:
Keith Dysart wrote:
On Dec 8, 12:43 am, Cecil Moore wrote:
Keith Dysart wrote:
Cecil Moore wrote:
Keith Dysart wrote:
Of course, if you add 180 degrees you have added 180
degrees to whatever existed before. Do you disagree?
So this meets the criteria you originally proposed and
is an example of 90 degree electical length?
Don't be silly. 180 degrees plus any positive angle
is more than 180 degrees.

The context was mobile loaded antennas shorter than
a physical 1/4WL.

Hmmmm. So you are no longer in agreement with your
original question:
"So are we agreed that a 43.4 degree stub terminated
in 0-j567 ohms impedance is electrically 1/4WL, i.e.
90 degrees long?"

Now I understand your confusion. I was talking about
a -j567 ohm *capacitor*, not a virtual impedance.

I was, of course, using the "impedor" definition of
impedance but since that confused you, let me restate
the question:

"So are we agree that a 43.4 degree stub terminated in
a -j567 ohm impedor is electrically 1/4WL, i.e. 90
degrees long?"

This was the original meaning of the question. I'm sorry
that you took it the wrong way and wasted so many postings
on such a trivial misunderstanding.

So does this new question rule out the cases (previously
accepted) where the 0-j567 is obtained with 46.4 degrees
of 600 ohm line or 10 degrees of 100 ohm line? These are
not lumped capacitors.

Some consistency that persists longer than one post would
be valuable.

....Keith

Standing Wave Phase
 

Keith Dysart wrote:
So does this new question rule out the cases (previously
accepted) where the 0-j567 is obtained with 46.4 degrees
of 600 ohm line or 10 degrees of 100 ohm line? These are
not lumped capacitors.

No, but they are an electrical 1/4WL, not any
other length. The electrical length of a stub
is whatever it is. If it is not 1/4WL, it is
some other length. Why is that difficult to
understand?
--
73, Cecil http://www.w5dxp.com

Standing Wave Phase
 

On Dec 8, 5:59 pm, Cecil Moore wrote:
Keith Dysart wrote:
So does this new question rule out the cases (previously
accepted) where the 0-j567 is obtained with 46.4 degrees
of 600 ohm line or 10 degrees of 100 ohm line? These are
not lumped capacitors.

No, but they are an electrical 1/4WL, not any
other length. The electrical length of a stub
is whatever it is. If it is not 1/4WL, it is
some other length. Why is that difficult to
understand?

Well, I know what I mean by 1/4WL and in my
definition there is no way that (46.4 + 10) = 90.

However I am trying to help you articulate your
definition in a way that is sufficiently precise
that I can use it to determine what you would
consider to be 1/4WL. Unfortunately, at the
moment, it is sufficiently fuzzy that the only
way to determine if something is 90 degrees
(according to your definition) is to ask you.
The need of an oracle to answer such questions
is not the basis for sound science.

Tautologies such as "If it is not 1/4WL, it is
some other length." do not further the definition,
but are good if you want to keep the job of
oracle.

....Keith

Standing Wave Phase
 

Keith Dysart wrote:
Well, I know what I mean by 1/4WL and in my
definition there is no way that (46.4 + 10) = 90.

Of course, those are *physical* degrees. We are
talking about *electrical* degrees. It is impossible
to get the reflected wave in phase with the forward
wave unless there is an electrical 90 degree phase
shift.

If you lay the 43.4 degrees out starting at Z=0
toward the load on the Smith Chart and lay the
10 degrees out starting at Z=infinity toward the
source, you will observe the phase shift caused
by the impedance discontinuity.

... the only
way to determine if something is 90 degrees
(according to your definition) is to ask you.

All one has to do is plot it on a Smith Chart
and the number of electrical degrees is obvious.
If you don't know how to use a Smith Chart it
might be time to learn how.
--
73, Cecil http://www.w5dxp.com

Standing Wave Phase
 

On Dec 9, 12:21 am, Cecil Moore wrote:
Keith Dysart wrote:
Well, I know what I mean by 1/4WL and in my
definition there is no way that (46.4 + 10) = 90.

Of course, those are *physical* degrees.

Yes indeed. And they have the benefit of concreteness
and they are easy to account.

We are
talking about *electrical* degrees. It is impossible
to get the reflected wave in phase with the forward
wave unless there is an electrical 90 degree phase
shift.

Except that I have offerred a number of examples
which you, the oracle, have declared are not
90 "electrical degrees".

If you lay the 43.4 degrees out starting at Z=0
toward the load on the Smith Chart and lay the
10 degrees out starting at Z=infinity toward the
source, you will observe the phase shift caused
by the impedance discontinuity.

I, too, can subtract (43.4 + 10) from 90 and get
a number. This does not, by itself, a useful
proposition make.

... the only
way to determine if something is 90 degrees
(according to your definition) is to ask you.

All one has to do is plot it on a Smith Chart
and the number of electrical degrees is obvious.

Please provide your algorithm in sufficient detail
that I can test it against the various examples.

So far, each time you have provided a rule, I
have constructed examples according to the
rule which the oracle has declared are not
90 "electrical degrees". Without a testable
rule that successfully distinguishes those
cases which are 90 "electrical degress"
from those which are not, there is nothing.

Having to ask the oracle does not suffice.

And the Smith chart is insufficient. One of your
examples began with "take the impedance
of 0-j567 and plot it on the chart", which is
okay, but it turned out that how that impedance
was created is important. It had to be a
capacitor (sometimes). No amount of Smith
charting will reveal that detail.

A testable rule, please...

....Keith

Standing Wave Phase
 

Keith Dysart wrote:
Except that I have offerred a number of examples
which you, the oracle, have declared are not
90 "electrical degrees".

If it is 90 electrical degrees then it is 90
electrical degrees. If it is not 90 electrical
degrees, it is not 90 electrical degrees. I don't
know how to make it any clearer than that.

I, too, can subtract (43.4 + 10) from 90 and get
a number. This does not, by itself, a useful
proposition make.

It does if we know the reflected wave undergoes a
180 degree round-trip phase shift or else the
reflected wave would not be in phase with the
forward wave and therefore the feedpoint impedance
would not be purely resistive.

Please provide your algorithm in sufficient detail
that I can test it against the various examples.

It's the same as determining if an antenna is 0.5WL
or 1.5WL or 2.5WL or 3.5WL or ... Do you also have
a problem with that?

If the phase shift end-to-end is 180 degrees, the
device is 90 electrical degrees long.

If the phase shift end-to-end is not 180 degrees,
the device is not 90 electrical degrees long.

So far, each time you have provided a rule, I
have constructed examples according to the
rule which the oracle has declared are not
90 "electrical degrees".

I have provided no rule. Everything is common sense.
If a dipole is 130 feet, it is 1/2WL on ~3.6 MHz.
If the antenna is 403 feet long, it is 1.5WL on
~3.6 MHz. Why do you have a problem telling the
difference between a 130 foot dipole and a 403
foot dipole?
--
73, Cecil http://www.w5dxp.com

Standing Wave Phase
 

Cecil Moore wrote:

...
It does if we know the reflected wave undergoes a
180 degree round-trip phase shift or else the
reflected wave would not be in phase with the
forward wave and therefore the feedpoint impedance
would not be purely resistive.
...


Or, to sum that up in a nut shell, "Does everyone here know we can
reverse the leads on our SWR meter (input goes to output--output goes to
input) and the fwd/ref switch will just work "backwards." (but, readings
should remain the same.)

Indeed, an excellent way to check homebuilt SWR bridges and make sure
they are "balanced."

Regards,
JS

Standing Wave Phase
 

On Dec 9, 3:27 pm, Cecil Moore wrote:
Keith Dysart wrote:
Except that I have offerred a number of examples
which you, the oracle, have declared are not
90 "electrical degrees".

If it is 90 electrical degrees then it is 90
electrical degrees. If it is not 90 electrical
degrees, it is not 90 electrical degrees. I don't
know how to make it any clearer than that.

I suspect you are correct there.

I, too, can subtract (43.4 + 10) from 90 and get
a number. This does not, by itself, a useful
proposition make.

It does if we know the reflected wave undergoes a
180 degree round-trip phase shift or else the
reflected wave would not be in phase with the
forward wave and therefore the feedpoint impedance
would not be purely resistive.

Please provide your algorithm in sufficient detail
that I can test it against the various examples.

It's the same as determining if an antenna is 0.5WL
or 1.5WL or 2.5WL or 3.5WL or ... Do you also have
a problem with that?

I use a measuring tape for that, so there is no
problem. But if I recall correctly, your definition
of 90 degress is not amenable to the use of
measuring tapes.

If the phase shift end-to-end is 180 degrees, the
device is 90 electrical degrees long.

If the phase shift end-to-end is not 180 degrees,
the device is not 90 electrical degrees long.

So far, each time you have provided a rule, I
have constructed examples according to the
rule which the oracle has declared are not
90 "electrical degrees".

I have provided no rule. Everything is common sense.

Everyone thinks they are full of "common sense" and
that few others are. Science is not advanced by claiming
common sense.

If you do not have articulatable rules, then you do
not even have a hypothesis, much less a theory.

If a dipole is 130 feet, it is 1/2WL on ~3.6 MHz.
If the antenna is 403 feet long, it is 1.5WL on
~3.6 MHz. Why do you have a problem telling the
difference between a 130 foot dipole and a 403
foot dipole?

No problem. But I am allowed to use a measuring
tape to answer that question.

And if you wrote a rule using measuring tapes for
this 90 degree stuff, I would have no trouble with it
either.

But if the best you can do is claim "common
sense", you can be sure that my "common
sense" will arrive at different answers than yours.

....Keith

Standing Wave Phase
 

Keith Dysart wrote:
If you do not have articulatable rules, then you do
not even have a hypothesis, much less a theory.

The theory I support is the distributed network
model. It was invented before you or I were born.
I do not have to defend it. If you disagree with
it, you have to prove it false.
--
73, Cecil http://www.w5dxp.com

Standing Wave Phase
 

On Dec 9, 8:50 pm, Cecil Moore wrote:
Keith Dysart wrote:
If you do not have articulatable rules, then you do
not even have a hypothesis, much less a theory.

The theory I support is the distributed network
model. It was invented before you or I were born.
I do not have to defend it. If you disagree with
it, you have to prove it false.

A truly intriguing riposte: completely devoid of
technical content, fails to further the discussion
in any way, and yet deeply revealing about the
thought processes at work.

....Keith

Standing Wave Phase
 

Keith Dysart wrote:
On Dec 9, 8:50 pm, Cecil Moore wrote:
Keith Dysart wrote:
If you do not have articulatable rules, then you do
not even have a hypothesis, much less a theory.

The theory I support is the distributed network
model. It was invented before you or I were born.
I do not have to defend it. If you disagree with
it, you have to prove it false.

A truly intriguing riposte: completely devoid of
technical content, fails to further the discussion
in any way, and yet deeply revealing about the
thought processes at work.

I'm disappointed in you, Keith. You appeared to be
a reasonably intelligent person. Now that you have
apparently performed my suggested experiment and
know that I am right, you try to tuck your tail
and run. I'm disapointed, but not surprised.
--
73, Cecil http://www.w5dxp.com

Standing Wave Phase
 

On Dec 9, 10:14 pm, Cecil Moore wrote:
I'm disappointed in you, Keith. You appeared to be
a reasonably intelligent person.

Thank you. And I continue to be so.

Now that you have apparently performed my suggested
experiment

There is no evidence that I have performed the experiment.
The need for current probes probably means that I will not
be doing so, though you never know. Rest assured that
should I do so, results will be published.

and know that I am right, you try to tuck your tail
and run.

Ahhh, your favourite accusation. You play the game
of last man standing, and when everyone has left
the field (in this case because you had nothing
further technical to add), you convince yourself
that it is because you must be right, and they
must know it. I can't decide if this logic is
more amusing or sad.

Now if you want to recover, make some better
attempts to write the rule. Look back through
the posts for your sentences that begin with
"Can we agree that", or some such. These
were your attempts at a rule. Complete them
to the point that we agree that they are self
consistent and accurately convey your
definition and we will have gotten somewhere.

Your later attempts, which amount to "I
know it when I see it", were not nearly as
good as your earlier ones.

....Keith

Standing Wave Phase
 

Keith Dysart wrote:
Now if you want to recover, make some better
attempts to write the rule. Look back through
the posts for your sentences that begin with
"Can we agree that", or some such. These
were your attempts at a rule. Complete them
to the point that we agree that they are self
consistent and accurately convey your
definition and we will have gotten somewhere.

What is it that you need a rule for? Is it the
electrical length of a stub?

The stub is electrically half as long as the phase
shift undergone by the reflected wave during its
round trip to the open or shorted end and
back to the feedpoint.

If that phase shift in a dual-Z0 stub is the same
as the phase shift in a single-Z0 stub at the same
frequency, the two stubs are the same electrical
length at that frequency.
--
73, Cecil http://www.w5dxp.com