Standing morphing to travelling waves. was r.r.a.a Laugh Riot!!!
 

Cecil Moore wrote:
On Jan 8, 3:54 pm, art wrote:
Yes 50/60 is frequency just like 10 metrrs. Now go homw and cut the
wire in the kitchen to make it an open circuit Now ham r5adio says the
cuyrrent will turn around where you cutt it and it will go back. Well
science has not seen a trace of this new frequency of 100, 110 0r 220
line double its frequency to twice what it was before.

Art, please think about what you are saying.

The wavelength of 20 MHz is 70 ft. The wavelength of 60 Hz is
16,400,000 feet (3100 miles).

There is a power factor correcting capacitor every city block or so on
the 60 Hz system, i.e. approximately every 0.00003 wavelength. Are
there any power factor correcting capacitors hanging on a 20m dipole?
If the standing wave power factor was corrected on a 20m dipole, the
standing waves would disappear just like they do on the power wires.
Correcting the power factor on a standing wave dipole antenna would
turn it into a traveling wave antenna and there would be no
reflections from the end of the dipole. Our 50/60 Hz power
distribution systems are overwhelmingly traveling wave systems. The
power companies simply don't allow reflections to exist because they
don't get paid for reflected energy and they sure don't want it coming
back to the generator.

If you had a 60Hz generator feeding a low-loss 3100 mile open-
circuited transmission line, you would certainly observe a forward
wave, reflected wave, and standing wave all obeying the distributed
network reflection model rules.
--
73, Cecil, w5dxp.com

Cecil,

That is an interesting tale, but mostly baloney.

When you get back to the wilds of Texas go check out some rural power
lines. Count the number of power factor correcting capacitors you see. I
bet it is a lot less than the equivalent of one per city block. Power
factor correcting capacitors are intended to correct for reactive loads,
such as motors, not for reflections or standing waves on open ended
power transmission lines.

But you already know that.

73,
Gene
W4SZ

Standing morphing to travelling waves, and other stupid notions
 

Before Cecil jumps on the typo, see below. My profuse apologies.

Jim Kelley wrote:

Cecil Moore wrote:

On Jan 8, 2:39 pm, Jim Kelley wrote:

But when you write the equation for the superposition of
traveling waves and claim that resultant standing wave is a different
kind of electromagnetic wave, that is a misguided point of view.



That's not true unless you consider Eugene Hecht to be misguided.


Of course it's true, and Dr. Hecht does post here.

and Dr. Hecht does not post here.

He
said standing waves are so different from traveling waves that they
probably shouldn't even be considered to be waves at all since they
are not even moving.


On what page has Dr. Hecht written "a standing wave is a different kind
of electromagnetic wave"?

Those waveforms could hardly be any different yet you asserted that
they are linked by a trig identity.


I asserted that expression for the sum of traveling waves and the
expression for the resulting standing wave pattern are related by trig
identity, as per page 140 of the 28th Edition of the CRC Standard
Mathematical Tables Handbook.

A standing wave is not only different from an EM traveling wave, it
cannot correctly even be called an EM wave because it is not moving at
the speed of light in the medium, a technical requirement for EM
waves.


The 'wave' which stands is merely an amplitude envelope for the waves
which move. It's not a "different kind of electromagnetic wave." If you
were to instead characterize a 'standing wave' as a different kind of
interference pattern, then we would in fact be in agreement.

ac6xg

Standing morphing to travelling waves, and other stupid notions
 

Cecil Moore wrote:

...
P.S. I'm on my daughter's computer posting from Google.
--
73, Cecil, w5dxp.com

Cecil:

Break down and buy a laptop (even a used one.) Make sure it has a
wireless card or, get a USB wireless dongle (make sure the laptop has
USB ports!)

Jiwire(http://www.jiwire.com) can be used to find any free wireless
hotspots in the areas you are in, or even paid ones--for that matter.

Without you, it's too boring ... Sometimes this is better than an
Indiana Jones movie! You know, the type of movies where the natives
want his head on a stick. LOL

Warm regards,
JS

Standing morphing to travelling waves. was r.r.a.a Laugh Riot!!!
 

On Jan 9, 11:03*am, Michael Coslo wrote:
AI4QJwrote:
The NEC program is just a computer model, for discussion purposes only. I
think there are far too many variables in real life for the program to take
into account. It may be valuable but I am not yet convinced it is
infallible.

* * * * Oh darnit! Here I went and built several antennas designed with Eznec,
and they have worked just like the program said they would.

* * * * I guess I'll have to take the remaining ones down, since I was only
supposed to discuss them, not actually make and use them. Thanks for the
correction!

* * * * - 73 de Mike N3LI -

But you may have a problem discussing on the ng exactly "why" they
work. Why would you want to discuss an antenna if you don't care "how"
it works, you only care "that" it works? There are some people that
operate at 27MHz who don't care how their radios work, only that they
can peg your meter at 10 pounds.

Standing morphing to travelling waves, and other stupid notions
 

On Jan 9, 2:42*pm, Gene Fuller wrote:
Cecil Moore wrote:

.



A standing wave is not only different from an EM traveling wave, it
cannot correctly even be called an EM wave because it is not moving at
the speed of light in the medium, a technical requirement for EM
waves.

Cecil,

I showed you exactly how the energy in a standing wave travels at the
speed of light. Did you miss that message, or are you just pretending it
didn't happen?

Neither you nor your buddy Hecht are likely to overturn more than 100
years of successful use of standing waves.

8-)

73,
Gene
W4SZ

Semantics. The standing wave doesn't 'travel' anywhere, much less at
the speed of light. However, it does vibrate. Each vibration cycle
occurs at a frequency that, when multiplied by its wavelength, would
be equal to the speed of light. I find this an interesting concept.

Standing morphing to travelling waves. was r.r.a.a Laugh Riot!!!
 

Roger Sparks wrote:

Using the reflection point as the zero reference seems to correspond with an observation you made about the end of the line controlling the SWR.

The choice of zero reference is entirely arbitrary; any point on the
line, or off the line, for that matter, can be used. I used the input
end of the line as the x = 0 reference, so my equations are correct only
when that reference is used. The choice of a reference has no effect on
the SWR or any other aspect of line operation; it simply modifies the
equations.

For example, my equations for the first forward and first reflected
voltage wave we

vf1(t, x) = sin(wt - x)
vr1(t, x) = Gl * sin(wt + x)

and for the second set:

vf2(t, x) = Gs * sin(wt - x)
vr2(t, x) = Gs * Gl * sin(wt + x)

where here I've explicitly shown the source and load reflection
coefficients as Gs and Gl respectively. They were 0.5 and 1 in my second
analysis (the one with a 150 ohm resistor at the source).

The more general case where the line is some length L, rather than the
integral number of wavelengths in the example,

vf1(t, x) = sin(wt - x)
vr1(t, x) = Gl * sin(wt + x - 2L)

vf2(t, x) = Gs * Gl * sin(wt - x - 2L)
vr2(t, x) = Gs * Gl^2 * sin(wt + x - 4L)

In general,

(1) vfn(t, x) = (Gs * Gl)^n * sin(wt - x - 2nL)
(2) vrn(t, x) = Gl * (Gs * Gl)^n * sin(wt + x - (2n + 1)L)

where L is expressed in the same units as x and wt (degrees or radians).

These equations are correct with x being the distance from the input end
of the line.

You could, as I mentioned, use a different reference, for example x' = L
- x, where L is the line length in radians or degrees (same units as x
and wt). Then you have, simply by substituting L - x' for x:

vf1(t, x') = sin(wt - L + x')
vr1(t, x') = sin(wt + L - x' - 2L) = sin(wt - L - x')

and so forth, and for the general case,

(3) vfn(t, x') = (Gs * Gl)^n * sin(wt + x' - (2n + 1) * L)
(4) vrn(t, x') = (Gs * Gl)^n * sin(wt - x' - 2nL)

or, you can use x for the forward wave and x' for the reverse wave or
vice-versa in order to reference to the point the wave was reflected
from or where it will be reflected from. Any combination of the
equations is equally valid and will give correct results. You can't,
however, simply redefine the reference point without a corresponding
change in the equation. In general, equation 1 and equation 3 will give
different results if you put in the same value for x and x'; likewise
equations 2 and 4. There are some special cases, as you showed, where
you can change the reference without modifying the equations and not
have any impact on the sum of the waves. However, you can see from the
equations that this won't usually work.

The general case with complex reflection coefficients and arbitrary line
length is mathematically a little more difficult than the simple example
I worked earlier. Not only does each reflection have a different
amplitude than the previous one, it also has a different phase angle,
due to the line length and the reflection coefficients. Consequently,
the simple a / (1 - r) formula I used for summing the infinite series of
waves can't be applied to the equations in the form I used. This is
where a change to phasor notation is really beneficial, since the phase
delay simply becomes e raised to an imaginary exponent which can be
treated more conveniently than its constituent sine and cosine
functions. With phasor notation, the summing formula can be used even
for the general case to find the steady state results from the
individual reflected waves. There's a very excellent treatment of this
in Chipman's _Transmission Lines_ (Schaum's Outline Series). He does
just about exactly what I did in my earlier posting, except for the
general case and using phasors rather than time representations. It's an
excellent text and reference, and I highly recommend it for anyone
seriously interested in transmission lines.

Roy Lewallen, W7EL

Standing morphing to travelling waves. was r.r.a.a LaughRiot!!!
 

Art wrote:
"Yes 50/60 is frequency just like 10 meters."

What is wrong with that statement?

The frequency corresponding to 10 meters is 30 million cycles per
second. The wavelength corresponding to 50 Hz is 6 million meters.

A point a few meters from another along a 50 Hz line must have the same
voltage as that of the other point on the line a few meters away because
the 50 Hz wave changes its value so slowly as compared with wave
velocity, about 300 million meters per second.

At the same distance between points on a 30 MHz line, the voltages of
two points are surely different at the same instant because the wave
changes its value rapidly.

Cut a line carying HF of 30 MHz and the voltage doubles at the open
circuit while the current is completely reflected, making a sum of zero
amperes at the cut.

Best regards, Richard Harrison, KB5WZI

Standing morphing to travelling waves, and other stupid notions
 

Jim Kelley, AC6XG wrote:
"On what page has Dr. Hecht written "A standing wave is a different kind
of electromagnetic wave?"

In "Schaum`s College Physics Outline" by Bueche & Hecht on page 214 is
written:
"Standing Waves:....These might better not be called waves at all since
they do not transport energy and momentum."

Best regards, Richard Harrison, KB5WZI

Standing morphing to travelling waves. was r.r.a.a Laugh Riot!!!
 

On 9 Jan, 11:36, Cecil Moore wrote:
On Jan 8, 3:54 pm, art wrote:

Yes 50/60 is frequency just like 10 metrrs. Now go homw and cut the
wire in the kitchen to make it an open circuit Now ham r5adio says the
cuyrrent will turn around where you cutt it and it will go back. Well
science has not seen a trace of this new frequency of 100, 110 0r 220
line double its frequency to twice what it was before.

Art, please think about what you are saying.

The wavelength of 20 MHz is 70 ft. The wavelength of 60 Hz is
16,400,000 feet (3100 miles).

There is a power factor correcting capacitor every city block or so on
the 60 Hz system, i.e. approximately every *0.00003 wavelength. Are
there any power factor correcting capacitors hanging on a 20m dipole?
If the standing wave power factor was corrected on a 20m dipole, the
standing waves would disappear just like they do on the power wires.
Correcting the power factor on a standing wave dipole antenna would
turn it into a traveling wave antenna and there would be no
reflections from the end of the dipole. Our 50/60 Hz power
distribution systems are overwhelmingly traveling wave systems. The
power companies simply don't allow reflections to exist because they
don't get paid for reflected energy and they sure don't want it coming
back to the generator.

If you had a 60Hz generator feeding a low-loss 3100 mile open-
circuited transmission line, you would certainly observe a forward
wave, reflected wave, and standing wave all obeying the distributed
network reflection model rules.
--
73, Cecil, w5dxp.com

Thank you for responding Cecil as well as not being abusive.
Still looking at the first statement as I cannot move on without
closure.
Your point about a antenna having a open circuit at the top is a
troubling point for me.
For a given frequency the wave travells a certain distance forward and
the a certain distance back.What you are saying is if the antenna is
short then the wave or current flow will always about turn.
Thus it is the length of the antenna by your statement is what turns
around the current
regardless of the frequency applied. Thus if I reduced the antenna to
1 inch length the current will turn around multiple times by the time
that a period for say 10 metres has expired.
Since the current supplied travels on the outside of the antenna it
will be radiating all the time ( as I understand what you are saying)
without pulsating . I suppose you can prove this being a electrical
person! For me it is the time that the antenna is actually radiating
for a given frequency that is important so perhaps you will kindly
relate travel distance, time taken
with the period duration of the frequency applied? There will be
efforts to change the subject under discussion which would make this
one small point last a few years but I do think it is worth clarifying
for all so a solid base can be built. I would appreciate it if you
would clarify what sort of antenna we are referring to ie a full
electrical wave or a half electrical wave (resonant or otherwise )so
that I may relate to either a series circuit or a parallel circuit
such that the discussion is neatly enclosed to prevent deviation.
Very Best Regards
Art Unwin

Standing morphing to travelling waves. was r.r.a.a Laugh Riot!!!
 

Corrections:

The equations I gave aren't adequate for complex reflection
coefficients, which I didn't explicitly state. Also, I inadvertently
omitted a multiplying factor in two equations.

Roy Lewallen wrote:
. . .
For example, my equations for the first forward and first reflected
voltage wave we

vf1(t, x) = sin(wt - x)
vr1(t, x) = Gl * sin(wt + x)

and for the second set:

vf2(t, x) = Gs * sin(wt - x)
vr2(t, x) = Gs * Gl * sin(wt + x)

where here I've explicitly shown the source and load reflection
coefficients as Gs and Gl respectively. They were 0.5 and 1 in my second
analysis (the one with a 150 ohm resistor at the source).

The above equations are valid only for purely real reflection
coefficients, as in the analysis I did.

The more general case where the line is some length L, rather than the
integral number of wavelengths in the example,

vf1(t, x) = sin(wt - x)
vr1(t, x) = Gl * sin(wt + x - 2L)

vf2(t, x) = Gs * Gl * sin(wt - x - 2L)
vr2(t, x) = Gs * Gl^2 * sin(wt + x - 4L)

In general,

(1) vfn(t, x) = (Gs * Gl)^n * sin(wt - x - 2nL)
(2) vrn(t, x) = Gl * (Gs * Gl)^n * sin(wt + x - (2n + 1)L)


For complex reflection coefficients,

vf1(t, x) = sin(wt - x)
vr1(t, x) = |Gl| * sin(wt + x - 2L + /_Gl)

vf2(t, x) = |Gs * Gl| * sin(wt - x - 2L + /_(Gs * Gl))
vr2(t, x) = |Gs * Gl^2| * sin(wt + x - 4L + /_(Gs * Gl^2)

In general,

(1) vfn(t, x) = |(Gs * Gl)^n| * sin(wt - x - 2nL + /_((Gs * Gl)^n))
(2) vrn(t, x) = |Gl * (Gs * Gl)^n| * sin(wt + x - (2n + 1)L + /_(Gl *
(Gs * Gl)^n)

where /_(qty) is the phase angle of qty.

where L is expressed in the same units as x and wt (degrees or radians).

These equations are correct with x being the distance from the input end
of the line.

You could, as I mentioned, use a different reference, for example x' = L
- x, where L is the line length in radians or degrees (same units as x
and wt). Then you have, simply by substituting L - x' for x:

vf1(t, x') = sin(wt - L + x')
vr1(t, x') = sin(wt + L - x' - 2L) = sin(wt - L - x')

and so forth, and for the general case,

(3) vfn(t, x') = (Gs * Gl)^n * sin(wt + x' - (2n + 1) * L)
(4) vrn(t, x') = (Gs * Gl)^n * sin(wt - x' - 2nL)

A Gl term was inadvertently omitted from the equations for vr. With that
error corrected, and with complex reflection coefficients,

vf1(t, x') = sin(wt - L + x')
vr1(t, x') = |Gl| * sin(wt + L - x' - 2L + /_Gl) = sin(wt - L - x' +
/_Gl)

and so forth, and for the general case,

(3) vfn(t, x') = |(Gs * Gl)^n| * sin(wt + x' - (2n + 1) * L + /_((Gs
* Gl)^n)
(4) vrn(t, x') = |Gl * (Gs * Gl)^n| * sin(wt - x' - 2nL + /_(Gl + (Gs
+ Gl)^n))

The reflection coefficient phase terms are easily handled in phasor
analysis, in the same manner as the phase term related to the line length.

Roy Lewallen, W7EL