RadioBanter - Standing-Wave Current vs Traveling-Wave Current

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Standing-Wave Current vs Traveling-Wave Current

Gene Fuller wrote:
I recall from some previous messages that you may have read Feynmann's
QED. Why don't you add some Feynmann diagrams into this discussion and
edumacate us all?

They would actually be easier to do than some of the ASCII
schematics that I have drawn.

More to the point. If you would stop the silly business about "net"
everything you might begin to understand that standing waves have
effects that move at the speed of light just in the same manner that
traveling waves do. Do you really believe that the fields from the
standing waves don't propagate?

Yes, the standing wave fields do not propagate so that should
be a clue that they are somewhat unrelated to reality. That the
standing wave fields do not propagate can be seen in the phase
of the standing wave current as reported by EZNEC - it doesn't
change by more than a negligible amount over the entire length
of a 1/2WL dipole just as Kraus reported:

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

Perhaps you should review the difference between static and stationary.
The standing wave is stationary. It is not static.

I don't recall uttering the word "static". Perhaps you would like
to quote me and prove me wrong. If you cannot, would you like
to admit that you are just blowing smoke?
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current

Dave wrote:
"Tom Donaly" wrote in message
. net...
Yuri Blanarovich wrote:
"Dave" wrote in message
news:cwPbj.1073$ML6.117@trndny04...
you can do it when it makes physical sense. it does not make sense in
standing waves for all the obvious reasons that i have pointed out. it
does make sense in the individual traveling waves. just accept what
your little swr meter tells you, it shows the forward power and
reflected power, that is all you need and the only powers that make
sense.

Little SWR meter shows forward AND reflected power in one direction, and
reflected power only in reverse direction. Why is the Bird wattmeter
calibrated in Watts, measuring power (forward and reverse) and has chart
to calculate SWR, when there are no standing waves and no power in them?
Laying waves or sitting waves???

Seems to me that the PROBLEM is that some consider standing wave to be
some imaginary, stopped, frozen wave, no good, while some of us consider
standing wave to be the result of superposition of forward and reverse
waves, that can be (their components) measured, current heats when
flowing through resistance, voltage "burns" when poor dielectric.
Like there is standing wave current, but no standing wave, huh????
Or are we forgetting that we are dealing with electromagnetic waves?
Can someone sort out the terminology and definitions?

Yuri, K3BU
Hi, Yuri,
Cecil and Dave aren't taking resistance into account when they
talk about waves. In other words, they're not writing about real
transmission lines, coils, and such. The only lines they care about are
the ones in their minds: the simple ones where the attenuation constants
are always zero, and current and voltage are always either in phase, or
90 degrees out of phase, and there is never any dielectric breakdown, no
matter the voltage. You can only feel sorry for guys like that.
73,
Tom Donaly, KA6RUH

using any decent coax of a reasonable length and typical amateur power
levels the assumptions we have stated are very close to the actual results.
if you want to examine lossy lines in detail then go ahead, the formulas get
much messier and without proper formula rendering on a newsgroup they are
almost impossible to discuss... and for the concepts that have been proposed
the ideal lossless line case is perfectly acceptable.

Not when you put the kind of stress on it that Yuri does. Yuri has a
choice: he can either believe you, or what he can see with his own eyes,
in which case your fantasy line is not at all "perfectly acceptable."
73,
Tom Donaly, KA6RUH

Standing-Wave Current vs Traveling-Wave Current

Gene Fuller wrote:
Cecil Moore wrote:
3. Do standing waves move at the speed of light? _______

3. Yes.

Do I win the prize?

Of course not. Please explain how standing waves can
move at the speed of light yet only change phase by 1/10
of one degree in 90 degrees of transmission line or antenna?
Please be technically specific and include the math.

We cannot wait to hear your answer to that one.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current

Roy Lewallen wrote:
Well, let's look at that problem. Make the line 1/8 wavelength long
instead of 1/4 wavelength. The ratio of V to I at the source can be
calculated directly with a single formula or by separately calculating
the forward and reverse traveling voltage and current waves and summing
them. The result, for my 50 ohm transmission line terminated with 25
ohms, is 40 + j30 ohms.

Whoops, that is not possible for your V & I scalar values.
Please don't tell us that you are using ASCII characters,
normally reserved for scalar values, for phasor values,
without telling anyone what you were doing. Do you think
that is ethical?
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current

Roy, try a real simple case.

50 ohm line, 1/2 wavelength long, open circuit, lossless, sinusoidal steady
state, fed with a 1v p-p voltage source.

Everyone will agree there is a standing wave on this line of course.

now, to make everyone happy... in the middle of the line calculate v(t),
i(t), and p(t), these are the standing wave voltage, current, and power.

now, calculate vf(t), if(t),pf(t), vr(t), ir(t), pr(t) at that point, where
the 'f' terms are the forward wave, the 'r' terms are the reflected wave.

compare and comment on the relation between pf, pr, and p.

extra credit:
repeat calculations at the far end of the line. again compare pf, pr, and
p.

Standing-Wave Current vs Traveling-Wave Current

Roy Lewallen wrote:
Roger wrote:
. . .
You give a good example Keith. It would be correct for measurement at
the load and at every point 1/2 wavelength back to the source from the
load, because the standing wave has the same measurements at these
points. At the 1/4 wavelength point back from the load and every
successive 1/2 wave point back to the source, the equation would also
be correct as demonstrated in Roy's example earlier today.

Excepting for these points, we would also be measuring a reactive
component that could be described as the charging and discharging of
the capacity or inductive component of the transmission line.
(Imagine that we are measuring the mismatched load through a 1/8 wave
length long transmission line, using an Autek RX VECTOR ANALYST
instrument) The inclusion of this reactive component would invalidate
the power reading if we were assuming that the measured power was all
going to the load.
. . .

Well, let's look at that problem. Make the line 1/8 wavelength long
instead of 1/4 wavelength. The ratio of V to I at the source can be
calculated directly with a single formula or by separately calculating
the forward and reverse traveling voltage and current waves and summing
them. The result, for my 50 ohm transmission line terminated with 25
ohms, is 40 + j30 ohms. V at the input is fixed by the source at 100
volts RMS. I = V / Z = 1.6 - j1.2 amps = 2.0 amps magnitude with a phase
angle of -36.87 degrees.

Now I'll translate V and I into time domain quantities. (I could have
calculated I directly in the time domain, but this was simpler.)

Using w for omega, the rotational frequency,

If V(t) = 141.4*sin(wt) [100 volts RMS at zero degrees] then
I(t) = 2.828*sin(wt-36.87 deg.) [2 amps RMS at -36.87 degrees]

Multiplying V * I we get:

V(t) * I(t) = 400 * sin(wt) * sin(wt - 36.87 deg.)

By means of a trig identity, this can be converted to:

= 200 * [cos(36.87 deg) - cos(2wt - 36.87 deg.)]

cos(36.87 deg) = 0.80, so

V(t) * I(t) = 160 - 200 * cos(2wt - 36.87 deg.)

This is a waveform I described in my previous posting. The cosine term
is a sinusoidal waveform at twice the frequency of V and I. The 160 is a
constant ("DC") term which offsets this waveform. The fact that the
waveform is offset means that the power is positive for a larger part of
each cycle than it is negative, so during each cycle, more energy is
moved in one direction than the other. In fact, the offset value of 160
is, as I also explained earlier, the average power. It should be
apparent that the average of the first term, 160, is 160 and that the
average of the second term, the cosine term, is zero.

Let's see how this all squares with the impedance I calculated earlier.

Average power is Irms^2 * R. The R at the line input is 40 ohms and the
magnitude of I is 2.0 amps RMS, so the power from the source and
entering the line is 2^2 * 40 = 160 watts. Whaddaya know. Let's try
Vrms^2 / R. In this case, R is the shunt R. The line input impedance of
40 + j30 ohms can be represented by the parallel combination of 62.50
ohms resistive and a reactance of 83.33 ohms. The 62.50 ohms is what we
use for the R in the V^2 / R calculation. Vrms^2 / R = 100^2 / 62.50 =
160 watts.

We can also calculate the power in the load from its voltage and current
and, with the assumption of a lossless line I've been using, it will
also equal 160 watts.

P = V(t) * I(t) always works. You don't need power factor or reactive
power "corrections", or to have a purely resistive impedance.

This is really awfully basic stuff. Some of the posters here would come
away with a lot more useful knowledge by spending their time reading a
basic electric circuit text rather than making uninformed arguments.

Roy Lewallen, W7EL

Thanks for the example of P = V(t) * I(t).

I think the example illustrates why the instantaneous power equation P =
V * I that Keith was referencing is not appropriate at all points on the
line. If I understood Keith correctly, he would have calculated 200
watts input for your example (100 volts at 2 amps).

73, Roger, W7WKB

Standing-Wave Current vs Traveling-Wave Current

Roger wrote:

Thanks for the example of P = V(t) * I(t).

I think the example illustrates why the instantaneous power equation P =
V * I that Keith was referencing is not appropriate at all points on the
line. If I understood Keith correctly, he would have calculated 200
watts input for your example (100 volts at 2 amps).

Huh? I calculated 160 watts using P = V(t) * I(t). Why do you think
Keith would have calculated 200 from the same equation? I have no doubt
that his math skills exceed mine, and I used nothing more than complex
arithmetic and high school trig.

I calculated power at the input end of the line, but I can calculate it
from the same equation P = V(t) * I(t) at any point on the line, and
will get exactly the same result. At what point(s) do you think it's not
"appropriate" to use?

Roy Lewallen, W7EL

Standing-Wave Current vs Traveling-Wave Current

Dave wrote:
Roy, try a real simple case.

50 ohm line, 1/2 wavelength long, open circuit, lossless, sinusoidal steady
state, fed with a 1v p-p voltage source.

Everyone will agree there is a standing wave on this line of course.

now, to make everyone happy... in the middle of the line calculate v(t),
i(t), and p(t), these are the standing wave voltage, current, and power.

No problem.

Let's use the voltage source as the phase reference, so it will be
(using w to represent the rotational frequency omega):

v(t) at the line input = 0.5 * sin(wt)

At the center of the line, i(t) = 0.02 * sin(wt - 90 deg.) and v(t) = 0.

p(t) = v(t) * i(t) = 0

Note that this isn't the average power (although the average power is
also zero), but the instantaneous power -- meaning that the power at
that point is zero at all times.

These aren't "standing wave" voltage, current, and power, but simply the
total voltage and current, and the power, at that point.

now, calculate vf(t), if(t),pf(t), vr(t), ir(t), pr(t) at that point, where
the 'f' terms are the forward wave, the 'r' terms are the reflected wave.

At the center of the line,

vf(t) = 0.5 * sin(wt - 90 deg.)
if(t) = 0.01 * sin(wt - 90 deg.)
vr(t) = -0.5 * sin(wt - 90 deg.)
ir(t) = 0.01 * sin(wt - 90 deg.)

where I've taken the positive direction of ir to be forward, the same as if.

compare and comment on the relation between pf, pr, and p.

Calculation of P(t) is as follows:

p(t) = v(t) * i(t) = [vf(t) + vr(t)] * [if(t) + ir(t)]

= 0 * 0.02 sin(wt - 90 deg.) = 0 as calculated before.

I haven't seen a definition of pr and pf, but they're not relevant to
the discussion. If you get a different result for power than zero by
using whatever you take them to mean, then the concept is invalid. There
is no average power leaving the source and no average power being
dissipated in the load(*). So there had better be no average power
anywhere in the line. There will be non-zero instantaneous power
everywhere along the line except at the input, far end, and midway, but
its average value will be zero, indicating the movement of energy back
and forth but no net energy flow.

extra credit:
repeat calculations at the far end of the line. again compare pf, pr, and
p.

Sure.

At the far end of the line,

v(t) = sin(wt - 180 deg.)
i(t) = 0

p(t) = v(t) * i(t) = 0 at all times.

vf, if, vr, and ir, are the same everywhere on the line, so see the
previously calculated values. Likewise, the calculation of p(t) from the
forward and reverse traveling waves is the same as before, with the same
result.

I've shown that I can calculate the correct power at two points along
the line by using p(t) = v(t) * i(t) as I can for any point on any line
with any termination. Your question implies that the results I got (zero
at all times at both points) are incorrect. What power (instantaneous
and average) do you calculate for those two points?

(*) In fact, calculation of the power at the source will show that the
power there is also zero at all times. The energy moving back and forth
in the line at all points except the quarter wavelength nulls was put
into the line during the turn-on process. After steady state has been
reached, the source can be turned off (converted to a short circuit)
with no change in what's happening on the line.

Roy Lewallen, W7EL

Standing-Wave Current vs Traveling-Wave Current

Roy Lewallen wrote:
Roger wrote:

Thanks for the example of P = V(t) * I(t).

I think the example illustrates why the instantaneous power equation P
= V * I that Keith was referencing is not appropriate at all points on
the line. If I understood Keith correctly, he would have calculated
200 watts input for your example (100 volts at 2 amps).

Huh? I calculated 160 watts using P = V(t) * I(t). Why do you think
Keith would have calculated 200 from the same equation? I have no doubt
that his math skills exceed mine, and I used nothing more than complex
arithmetic and high school trig.

Why is it the same equation? I understand your P = V(t) * I(t) to be V
and I as functions of time, but Keith to be using what ever he reads
from his voltmeter and ammeter.

I calculated power at the input end of the line, but I can calculate it
from the same equation P = V(t) * I(t) at any point on the line, and
will get exactly the same result. At what point(s) do you think it's not
"appropriate" to use?

Roy Lewallen, W7EL

73, Roger, W7WKB