Richard Clark wrote:
Frankly, I couldn't imagine how you could possibly
continue to disagree - with yourself.

I don't and you don't either. You just haven't realized it yet.
Exactly what is it that you think you and I disagree about?
--
73, Cecil http://www.qsl.net/w5dxp



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W5DXP wrote:

wrote:
This emphasis on NET seems to be the place where the difficulties
begin. An (ideal) voltmeter placed at a voltage minima on the line
indicates a voltage of 0 volts. You appear to be saying that
despite this indication of 0, there is actually voltage present.

There are actually two voltages present of equal amplitude and
opposite phase. Their phasor sum is zero volts.

- how do I determine what these voltages actually are?

|Vfwd| = Sqrt(Pfwd*Z0) |Vref| = Sqrt(Pref*Z0)

If the voltmeter reads zero, these two voltages are equal in magnitude
and opposite in phase.

So when I use my voltmeter to measure a voltage, I actually have
two voltages present which sum to the reading.

When I do this on a D-cell, what are these two voltages?
When I read 0 on a piece of NMD-90, what are these two voltages?
When I read 0 on D-cell, what are these two voltages?
When I read 120 volts on a power outlet, what are these two voltages?
For each of the above, explain how you decided.

It would seem that this belief will lead to some serious decidability
issues.

And don't ever trust your voltmeter again.

....Keith

wrote:
So when I use my voltmeter to measure a voltage, I actually have
two voltages present which sum to the reading.

Yes, that's why you need a directional voltmeter or directional
wattmeter to separate the forward wave from the reflected wave
when reflections exist.

When I do this on a D-cell, what are these two voltages?

Reflected voltage is zero during DC steady-state.

It would seem that this belief will lead to some serious decidability
issues.

If standing waves exist, there is one forward wave component and one
reflected wave component. If standing waves do not exist, there is
no reflected wave component. Shirley, you can tell whether standing
waves exist or not so you will know whether you are dealing with one
or two waves.

And don't ever trust your voltmeter again.

Do you use a DC voltmeter to measure AC? Do you use a 60HZ AC voltmeter
to measure 1 GHz RF voltages? Do you use a 100MHz o'scope to look at
a light wave?

If reflections exist and you are using a non-directional voltmeter, you
will read the NET voltage. USE THE APPROPRIATE MEASURING INSTRUMENTS!
--
73, Cecil http://www.qsl.net/w5dxp



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W5DXP wrote:

wrote:
So when I use my voltmeter to measure a voltage, I actually have
two voltages present which sum to the reading.

Yes, that's why you need a directional voltmeter or directional
wattmeter to separate the forward wave from the reflected wave
when reflections exist.

When I do this on a D-cell, what are these two voltages?

Reflected voltage is zero during DC steady-state.

But how did you know this little cylinder with a bump on the
end produced DC? And why can't DC circuits have net voltages
derived from superposition? Seems that in your model you need
extra information before you know what your voltmeter is telling you.

I observe that you did not provide an answer for the other examples.

What about that power plug on the wall? Too difficult?

Feel free to assume that you own an abstract voltmeter which will
tell you the voltage as a function of time (an oscilloscope maybe?).

It would seem that this belief will lead to some serious decidability
issues.

If standing waves exist, there is one forward wave component and one
reflected wave component. If standing waves do not exist, there is
no reflected wave component. Shirley, you can tell whether standing
waves exist or not so you will know whether you are dealing with one
or two waves.

And don't ever trust your voltmeter again.

Do you use a DC voltmeter to measure AC? Do you use a 60HZ AC voltmeter
to measure 1 GHz RF voltages? Do you use a 100MHz o'scope to look at
a light wave?

Since we are discussing the world of the ideal, we can assume the
existence of ideal voltmeters.

If reflections exist and you are using a non-directional voltmeter, you
will read the NET voltage. USE THE APPROPRIATE MEASURING INSTRUMENTS!

So I will rephrase my question for your benefit:

When we measure a voltage using an ideal voltmeter, how do we know if
this voltage is composed of components (I generalize, since similar
difficulties arise in situations without reflections) so that we can
do extra measurements and computations to learn the 'real' voltages
and 'powers'?

If you wish, the wall outlet provides an excellent opportunity for
explanation by example.

....Keith

wrote:
But how did you know this little cylinder with a bump on the
end produced DC? And why can't DC circuits have net voltages
derived from superposition? Seems that in your model you need
extra information before you know what your voltmeter is telling you.

The extra information is that you are dealing with an RF transmission
line with reflections. The reflections are somewhat equivalent to
a second source. And in a two-source DC circuit, you can indeed have
net voltages derived from superposition of the two source voltages.
Put them in series adding and you get a voltage maximum. Put them in
series subtracting and you get a voltage minimum.

What about that power plug on the wall? Too difficult?

Too many sources, too many feedlines, too many loads. And the
wavelength is really too long to illustrate RF transmission lines.
I get a wavelength just over 3000 miles, which would stretch out
straight from Texas to Alaska.

When we measure a voltage using an ideal voltmeter, how do we know if
this voltage is composed of components ...

For an RF transmission line, use a directional wattmeter to ascertain
if reflected power exists. If your 50 ohm SWR meter is in a 50 ohm Z0
environment, it will indicate the presence or absence of reflected
power.

If you wish, the wall outlet provides an excellent opportunity for
explanation by example.

Nope, it doesn't. That wall outlet doesn't have one source, one feedline,
and one load. If you had one 60 Hz generator hooked up to 3100 miles of
lossless transmission line, you could certainly observe reflections over
that one wavelength of wire.
--
73, Cecil http://www.qsl.net/w5dxp



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On Thu, 28 Aug 2003 05:18:37 -0500, W5DXP
wrote:

Richard Clark wrote:
Frankly, I couldn't imagine how you could possibly
continue to disagree - with yourself.

I don't and you don't either. You just haven't realized it yet.
Exactly what is it that you think you and I disagree about?

Hi Cecil,

Most amusing your disagreeing that you don't disagree. Isn't that
enough? By the way, I will adjourn to my other posting to observe the
quality of your work as this infinite regress (along with the tedium
of other postings to this thread) serves no purpose.

See ya there?

73's
Richard Clark, KB7QHC

W5DXP wrote:

wrote:
And in a two-source DC circuit, you can indeed have
net voltages derived from superposition of the two source voltages.
Put them in series adding and you get a voltage maximum. Put them in
series subtracting and you get a voltage minimum.

But power can only be computed using the resultant voltage and current
(or, if you prefer, NET voltage and current).

What about that power plug on the wall? Too difficult?

Too many sources, too many feedlines, too many loads. And the
wavelength is really too long to illustrate RF transmission lines.
I get a wavelength just over 3000 miles, which would stretch out
straight from Texas to Alaska.

Let me make it simpler then: the lamp cord from the wall to the table
lamp; lamp off and lamp on. Most lamp cord looks like twin-lead; some
have said it is approximately 100 Ohms but for the purposes of
discussion feel free to assume any convenient Z0 for the cord.

When we measure a voltage using an ideal voltmeter, how do we know if
this voltage is composed of components ...

For an RF transmission line, use a directional wattmeter to ascertain
if reflected power exists. If your 50 ohm SWR meter is in a 50 ohm Z0
environment, it will indicate the presence or absence of reflected
power.

And if it is not an RF transmission line?
If it is a DC circuit? Say your phone line when on hook?
If it is a lamp cord?

Do we really need directional wattmeters to understand what is
happening?

If you wish, the wall outlet provides an excellent opportunity for
explanation by example.

Nope, it doesn't. That wall outlet doesn't have one source, one feedline,
and one load. If you had one 60 Hz generator hooked up to 3100 miles of
lossless transmission line, you could certainly observe reflections over
that one wavelength of wire.

Please, replace wall outlet with lamp cord to table lamp.

....Keith

wrote:
But power can only be computed using the resultant voltage and current
(or, if you prefer, NET voltage and current).

That's true for two batteries. That's not true for transmission lines
where Z0 forces the ratio of the traveling waves to be V/I=Z0. If you
could use circuit math on transmission lines, there would be no need
for transmission line math, Smith Charts, etc. You are allowing your
shortcut to try to dictate reality. That won't work. Assume a constant
V/I ratio is forced on each battery in your circuit and see what you get.

Let me make it simpler then: the lamp cord from the wall to the table
lamp; lamp off and lamp on. Most lamp cord looks like twin-lead; some
have said it is approximately 100 Ohms but for the purposes of
discussion feel free to assume any convenient Z0 for the cord.

I don't think you understand the problem. That lamp cord is about
0.0000005 wavelength at 60 Hz. Any reflected wave effects are completely
negligible. It's like expecting reflections due to resistor leads at
HF RF. They are theoretically there but usually too small to measure.

And if it is not an RF transmission line?
If it is a DC circuit? Say your phone line when on hook?
If it is a lamp cord?

The transmission line has to be a non-negligible percentage of a
wavelength for one to have to switch from circuit math to
transmission line math. What's the wavelength of DC? What's the
wavelength of voice audio? What is the wavelength of 60 Hz?

If the phone line is an appreciable percentage of a wavelength,
there will be reflections (echoes) as anyone doing a transcontinental
phone call can attest. Echo cancellation is big business with the
telephone companies. Your theory would make telephone line echoes
impossible for long unterminated lines. It is known that unterminated
telephone lines result in the largest magnitude of echoes.

It just occurred to me that you can revolutionize the telephone
industry if you are right. You could patent your idea of no
reflected energy on a long unterminated telephone line and sell
it to the telephone companies for millions of dollars. Just have
every intercontinental telephone line unterminated by using
high impedance receivers on each end. Since no reflected energy
could cross the voltage minimum point, echoes would be eliminated
without the expensive echo cancellation equipment.

Do we really need directional wattmeters to understand what is
happening?

No, one can learn the principles of EM waves and transmission lines
and alleviate one's ignorance. I am capable of estimating the
forward power, reflected power, and SWR on my 450 ohm window line
without actually measuring it.

Please, replace wall outlet with lamp cord to table lamp.

Are you capable of measuring the 0.0001 degrees of phase shift
between the forward wave and reflected wave? Your measuring
instrument needs to be about plus or minus 0.0000001 degrees
to get an accurate measurement.
--
73, Cecil, W5DXP

W5DXP wrote:

wrote:
Let me make it simpler then: the lamp cord from the wall to the table
lamp; lamp off and lamp on. Most lamp cord looks like twin-lead; some
have said it is approximately 100 Ohms but for the purposes of
discussion feel free to assume any convenient Z0 for the cord.

I don't think you understand the problem. That lamp cord is about
0.0000005 wavelength at 60 Hz. Any reflected wave effects are completely
negligible. It's like expecting reflections due to resistor leads at
HF RF. They are theoretically there but usually too small to measure.

While the standing wave will be difficulty to observe, none of
the equations related to ideal transmission lines require a
minimum length. The reflected voltage and power can be trivially
calculated just as it is for RF. Let me do it for you.

120 Volt RMS line, open circuit, assume 100 Ohms for the lamp cord:
Vf = 60V RMS, Vr = 60 V RMS
Pf = 36 W, Pr = 36 W

And if it is not an RF transmission line?
If it is a DC circuit? Say your phone line when on hook?
If it is a lamp cord?

The transmission line has to be a non-negligible percentage of a
wavelength for one to have to switch from circuit math to
transmission line math.

But you can switch early if you wish and should still get
reasonable results.

What's the wavelength of DC? What's the
wavelength of voice audio? What is the wavelength of 60 Hz?

If the phone line is an appreciable percentage of a wavelength,
there will be reflections (echoes) as anyone doing a transcontinental
phone call can attest. Echo cancellation is big business with the
telephone companies. Your theory would make telephone line echoes
impossible for long unterminated lines. It is known that unterminated
telephone lines result in the largest magnitude of echoes.

Of course, we were discussing sinusoidal steady state for which
there will be no echoes. Echoes only happen when the line energy
state is changing.

It just occurred to me that you can revolutionize the telephone
industry if you are right. You could patent your idea of no
reflected energy on a long unterminated telephone line and sell
it to the telephone companies for millions of dollars. Just have
every intercontinental telephone line unterminated by using
high impedance receivers on each end. Since no reflected energy
could cross the voltage minimum point, echoes would be eliminated
without the expensive echo cancellation equipment.

Unfortunately, such a situation only occurs when the line is in
steady-state which means no information could be sent, which would
render the line economically unviable.

Please, replace wall outlet with lamp cord to table lamp.

Are you capable of measuring the 0.0001 degrees of phase shift
between the forward wave and reflected wave? Your measuring
instrument needs to be about plus or minus 0.0000001 degrees
to get an accurate measurement.

Don't need to measure phase shift to obtain forward and reflected
voltages and powers.

But I think I understand the technique. Whenever the theory would
produce answers with which you might be uncomfortable, rather than
producing those anwsers and attaining a better understanding of the
viability of the theory, you prune the problem space to which the
theory applies. Ergo, not for DC, not for 60 Hz, not for lamp cords,
not for step functions. But the math from the theory works just fine
in all these situations. Try it.

....Keith

wrote:
... assume 100 Ohms for the lamp cord:

Rash assumption. Please prove that six feet of zip cord
exhibits a Z0 of 100 ohms to 60Hz signals.

Of course, we were discussing sinusoidal steady state for which
there will be no echoes. Echoes only happen when the line energy
state is changing.

In reality, there is no such thing as sinusoidal steady-state
and noise in the system cannot be eliminated. The noise can
be used to prove that echoes (reflections) are still happening.
So nothing magic happens at sinusoidal steady-state. The wave
reflection model does NOT magically become dysfunctional at
steady-state.

But the math from the theory works just fine
in all these situations. Try it.

I know the math works *IF* you can define the boundary conditions.
Assuming that 50 ohm coax exhibits a Z0 of 50 ohms at 60 Hz is
a rash assumption.
--
73, Cecil http://www.qsl.net/w5dxp



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