Mike Monett wrote:
Roy Lewallen wrote:
[... very nice explanation]

Sine waves are another problem -- there, we can easily have
overlapping waves traveling in the same direction, so we'll run into
trouble if we're not careful. I haven't worked the problem yet, but
when I do, the energy will all be accounted for. Either the energy
ends up spread out beyond the overlap region, or the energy lost
during reflections will account for the apparent energy difference
between the sum of the energies and the energy of the sum. You can
count on it!

As always, I appreciate any corrections to either the methodology or
the calculations.

Roy Lewallen, W7EL

How about analyzing a vibrating string? If you play guitar, there's a very
nice note you can make by plucking a high string, then putting your finger
at exactly the correct spot and removing it quickly. The note will jump to
a much higher frequency and give a much purer sound. Clearly, the
mechanical energy has split into two waves that cancel at the node.

In principle, you could show the node is stationary, thus contains no
energy. But there is energy travelling on both sides of the null point -
you can hear it.

You can also create other notes by touching different spots on the
vibrating string. These create standing waves with energy travelling in
both directions, but cancelling at the null points. Very similar to
transmission lines.

Regards,

Mike Monett


Most undergraduate physics texts have, or should have, discussions of
vibrating strings. There's a good treatment of the subject in
William C. Elmore's and Mark A. Heald's book _Physics of Waves_
published by Dover. If you wanted to get in an argument you could
say that the energy on both sides of the node isn't traveling, but is
merely alternating between potential and kinetic. Such strings have loss
(or you wouldn't be able to hear them). Loss is a taboo subject on this
newsgroup because it makes wave behavior too hard to understand for the
savants posting here.
73,
Tom Donaly, KA6RUH

Roy Lewallen wrote:

Mike Monett wrote:

[...]

Sounds like a great idea. I'll look forward to seeing your analysis.

Roy Lewallen, W7EL

LOL! I stopped playing guitar years ago!

Regards,

Mike Monett

"Tom Donaly" wrote:

[...]

Most undergraduate physics texts have, or should have, discussions
of vibrating strings. There's a good treatment of the subject in
William C. Elmore's and Mark A. Heald's book _Physics of Waves_
published by Dover.

If you wanted to get in an argument you could say that the energy
on both sides of the node isn't traveling, but is merely
alternating between potential and kinetic.

Yes, I thought about that a bit before posting. It seems logical a
plucked string sends a wave in both directions, where it is
reflected and returns to create a standing wave.

When it forms a standing wave, it seems reasonable to say the energy
is alternating between potential and kinetic. But isn't that similar
to what happens on a transmission line that is exactly some multiple
of a quarter wavelength long?

Such strings have loss (or you wouldn't be able to hear them).

Loss is a taboo subject on this newsgroup because it makes wave
behavior too hard to understand for the savants posting here.

73,
Tom Donaly, KA6RUH

Regards,

Mike Monett

Mike Monett wrote:
"Tom Donaly" wrote:

[...]

Most undergraduate physics texts have, or should have, discussions
of vibrating strings. There's a good treatment of the subject in
William C. Elmore's and Mark A. Heald's book _Physics of Waves_
published by Dover.

If you wanted to get in an argument you could say that the energy
on both sides of the node isn't traveling, but is merely
alternating between potential and kinetic.

Yes, I thought about that a bit before posting. It seems logical a
plucked string sends a wave in both directions, where it is
reflected and returns to create a standing wave.

When it forms a standing wave, it seems reasonable to say the energy
is alternating between potential and kinetic. But isn't that similar
to what happens on a transmission line that is exactly some multiple
of a quarter wavelength long?


Demo 4 of the TLVis1 program I posted reference to, shows that in a
transmission line with a pure standing wave (load reflection coefficient
magnitude of 1), the energy between nodes alternates between the
electric field (line capacitance) and magnetic field (line inductance).
This is true regardless of the line length or the source termination.

Roy Lewallen, W7EL

Roy Lewallen wrote:

Demo 4 of the TLVis1 program I posted reference to, shows that in
a transmission line with a pure standing wave (load reflection
coefficient magnitude of 1), the energy between nodes alternates
between the electric field (line capacitance) and magnetic field
(line inductance).

This is true regardless of the line length or the source
termination.

Roy Lewallen, W7EL

Yes, this is a very nice demo. Thank you for posting it.

I have a question. In demo 4, the bottom window shows the Ee field
in green, Eh in red, and ETot in black.

When the demo starts, you can only see a green and a black trace.

If you pause it just as the wave hits the end, you can now see the
red trace, Eh. (This is an actual statement and has nothing to do
with the fact I am Canadian.)

What happened to the Eh trace as the wave is initally moving to the
right? Is it overlaid by the Ee trace in green? Or is it just not
plotted?

Then, when the wave hits the end and starts reflecting, the red
trace remains attached to ground, and the green trace moves up and
connects with the black trace. (Sorry for the confusing description
- you have to try it yourself to see.)

Now, as you single step, the green trace and the red trace appear to
be 180 degrees out of phase.

My problem here is someone wrote a web page that claims the electric
and magnetic fields are orthogonal:

http://www.play-hookey.com/optics/tr...etic_wave.html

I tried sending him an email to show if the fields were orthogonal
as he claims, it would look like a pure reactance, and no energy
would be transmitted. But he is stuck on his idea and won't budge.

Now my problem is figuring out exactly what happens at the
reflection, and why the Eh field behaves the way shown in your demo.

Regards,

Mike Monett

Cecil Moore wrote:

The bright interference rings
are four times the intensity of one of the two equal
waves. The dark interference rings are, of course,
zero intensity.

If the intensity of one wave is P, the intensity of the
bright rings will be 4P and the intensity of the dark
rings will be zero.

That's right. And we know that intensity is proportional to the
square of the EM field, so if P=9 then field=3. When there are two
such EM fields superposed, then we have 3+3 squared which is four
times greater than 3 squared. And owing to this supposed
'inequality', we have the sophomoric (literally) notion that there is
"extra" energy which must come from somewhere else.

ac6xg

Mike Monett wrote:

Yes, this is a very nice demo. Thank you for posting it.

I have a question. In demo 4, the bottom window shows the Ee field
in green, Eh in red, and ETot in black.

When the demo starts, you can only see a green and a black trace.

If you pause it just as the wave hits the end, you can now see the
red trace, Eh. (This is an actual statement and has nothing to do
with the fact I am Canadian.)

What happened to the Eh trace as the wave is initally moving to the
right? Is it overlaid by the Ee trace in green? Or is it just not
plotted?

The traces are drawn in the order Eh, Ee, and total. During the initial
forward wave, Eh and Ee are equal, so the Ee overwrites the Eh trace.

Then, when the wave hits the end and starts reflecting, the red
trace remains attached to ground, and the green trace moves up and
connects with the black trace. (Sorry for the confusing description
- you have to try it yourself to see.)

Hopefully it'll all make sense once you think about how one trace will
always win when more than one have the same value.

Now, as you single step, the green trace and the red trace appear to
be 180 degrees out of phase.

My problem here is someone wrote a web page that claims the electric
and magnetic fields are orthogonal:

http://www.play-hookey.com/optics/tr...etic_wave.html

You're making the same error that Cecil often does, confusing time phase
with directional vector orientation. The orthogonality of E and H fields
refers to the field orientations of traveling plane TEM waves in
lossless 3D space or a lossless transmission line, at the same point and
time. The E and H fields of these traveling waves are always in time
phase, not in quadrature. The graphs show the magnitudes of the waves at
various points along the line. These represent neither the time phase
nor the spatial orientation of the E and H fields.

I tried sending him an email to show if the fields were orthogonal
as he claims, it would look like a pure reactance, and no energy
would be transmitted. But he is stuck on his idea and won't budge.

Good for him -- he's absolutely correct. If the E and H fields were in
time quadrature, you'd have a power problem. But they're not. They're in
phase in any medium or transmission line having a purely real Z0 (since
Z0 is the ratio of E to H of a traveling wave in that medium). This
includes all lossless media. But they're always physically oriented at
right angles to each other -- i.e., orthogonally, according to the right
hand rule.

Now my problem is figuring out exactly what happens at the
reflection, and why the Eh field behaves the way shown in your demo.

Go for it!

Roy Lewallen, W7EL

Roy Lewallen wrote:

[...]

The traces are drawn in the order Eh, Ee, and total. During the
initial forward wave, Eh and Ee are equal, so the Ee overwrites
the Eh trace.

Good - thanks.

[...]

My problem here is someone wrote a web page that claims the
electric and magnetic fields are orthogonal:

http://www.play-hookey.com/optics/tr...etic_wave.html

You're making the same error that Cecil often does, confusing time
phase with directional vector orientation. The orthogonality of E
and H fields refers to the field orientations of traveling plane
TEM waves in lossless 3D space or a lossless transmission line, at
the same point and time.

Now you are confusing me with Cecil. I have no difficulty with the E
and H field orientation.

The E and H fields of these traveling waves are always in time
phase, not in quadrature.

Yes, that's what I tried to explain to him also.

The graphs show the magnitudes of the waves at various points
along the line. These represent neither the time phase nor the
spatial orientation of the E and H fields.

I tried sending him an email to show if the fields were
orthogonal as he claims, it would look like a pure reactance, and
no energy would be transmitted. But he is stuck on his idea and
won't budge.

Good for him - he's absolutely correct.

There is a bad mixup here. He claims:

"Note especially that the electric and magnetic fields are not in
phase with each other, but are rather 90 degrees out of phase. Most
books portray these two components of the total wave as being in
phase with each other, but I find myself disagreeing with that
interpretation, based on three fundamental laws of physics"

He claims the E and H fields are in quadrature. I claim he is wrong.

If the E and H fields were in time quadrature, you'd have a power
problem.

I believe that is what I tried to tell him. He bases his argument on
the following:

1. "The total energy in the waveform must remain constant at all
times."

Not true. It obviously goes to zero twice each cycle.

2. "A moving electric field creates a magnetic field. As an electric
field moves through space, it gives up its energy to a companion
magnetic field. The electric field loses energy as the magnetic
field gains energy."

Only if the environment is purely reactive. Not true with a pure
resistance.

3. "A moving magnetic field creates an electric field. This is
Faraday's Law, and is exactly similar to the Ampere-Maxwell law
listed above. A changing magnetic field will create and transfer its
energy gradually to a companion electric field."

Again, not true in a resistive environment.

But they're not. They're in phase in any medium or transmission
line having a purely real Z0 (since Z0 is the ratio of E to H of a
traveling wave in that medium). This includes all lossless media.

But they're always physically oriented at right angles to each
other - i.e., orthogonally, according to the right hand rule.

Yes, there is no confusion about this whatsoever.

[...]

Roy Lewallen, W7EL

Regards,

Mike Monett

Gene Fuller wrote:
You appear to use a very precise, quantitative definition of
"interference." I do not recall ever seeing such a quantitative
definition. Could you please give us a reference or an exact quote from
some reasonably reputable source that defines "interference" in a
quantitative and unambiguous manner?

I've already posted what Eugene Hecht said about interference.

In the irradiance (power density) equation,
Ptot = P1 + P2 + 2*SQRT(P1*P2)cos(A)
the last term is known as the "interference term", page 388 of
"Optics" by Hecht. Here's another reference:

http://en.wikipedia.org/wiki/Interference

A Google search for "electromagnetic wave interference" yielded
1,650,000 hits.

You imply that some interactions lead to "interference" and some do not.
How can the unwashed among us know when the magic occurs and when it
does not?

If the interference term in the above irradiance (power
density) equation is not zero, then interference is present.

In the s-parameter equation, b1 = s11*a1 + s12*a2, if b1
equals zero while s11, a1, s12, and a2 are not zero, then
total destructive interference is present.

Assume we superpose two coherent, collinear voltages, V1 and V2:

If (V1+V2)^2 V1^2+V2^2, then constructive interference is
present.

If (V1+V2)^2 V1^2+V2^2, then destructive interference is
present.
--
73, Cecil http://www.w5dxp.com

Jim Kelley wrote:
Cecil Moore wrote:
If the intensity of one wave is P, the intensity of the
bright rings will be 4P and the intensity of the dark
rings will be zero.

That's right. And we know that intensity is proportional to the square
of the EM field, so if P=9 then field=3. When there are two such EM
fields superposed, then we have 3+3 squared which is four times greater
than 3 squared. And owing to this supposed 'inequality', we have the
sophomoric (literally) notion that there is "extra" energy which must
come from somewhere else.

The intensity is watts/unit-area, i.e. real energy.
If the intensity of the bright rings is 4P there is
indeed greater than average energy which requires a
zero P dark ring somewhere else in order to
average out to 2P. The "extra" energy in the bright
rings comes from the dark rings. The conservation of
energy principle allows nothing else. It is not a
sophomoric notion. It is the laws of physics in action.
--
73, Cecil http://www.w5dxp.com