Standing-Wave Current vs Traveling-Wave Current
 

Roger wrote:

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.

If you'll look back through Keith's postings you'll see that he was
referring to the functions of time. It looks like he left off the
explicit (t) at some places which would lead to confusion.

But I hope you realize that you can also find the average power just
fine by calculating Pavg = Vrms * Irms * cos(theta) at any point along
the line, where Vrms and Irms are the total voltage and current at the
point, and theta is the angle between the two. This is true regardless
of the SWR. Also, you can calculate power as Irms^2 * Rser or Vrms^2 /
Rpar where Rser and Rpar are the series and parallel equivalent
resistive parts of the impedance at any point. Like v(t) and i(t), Vrms
and Irms can be found if desired by summing the forward and reflected
waves to find the total value at the point of interest; superposition
applies. It doesn't apply to power, so always do the summation of
voltages and currents before calculating power.

One property of the P(t) = V(t) * I(t) equation is that it also applies
to non-sinusoidal and even non-periodic waveforms -- it can *always* be
used. And you can always find the average power by integrating it then
dividing by the integration period. The average value calculation
reduces to Vrms * Irms * cos(theta) for pure sine waves but not other
waveforms.

Roy Lewallen, W7EL

Standing-Wave Current vs Traveling-Wave Current
 

Correction:

Roy Lewallen wrote:
. . .
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.

This isn't true. The phase angles of these terms are different at
different places along the line, so the p(t) results are different. At
the load end of the line:

vf(t) = 0.5 * sin(wt - 180 deg.)
if(t) = 0.01 * sin(wt - 180 deg.)
vr(t) = 0.5 * sin(wt - 180 deg.)
ir(t) = -0.01 * sin(wt - 180 deg.)

At this point, the result for p(t) is still zero at all times because of
zero ir + if rather than vr + vf as at the center of the line. At other
points along the line, as I mentioned in the original posting, p(t) will
end up being non-zero -- it'll be a sinusoidal function with rotational
frequency 2wt. But it will have an average value of zero, indicating
movement of energy back and forth but no net energy flow over an
integral number of periods.

I apologize for the error.

May everyone have an enjoyable and peaceful holiday.

Roy Lewallen, W7EL

Standing-Wave Current vs Traveling-Wave Current
 

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

Good Grief, Roy, no wonder you are clueless about standing
waves. In a lossless open-circuit line, the total voltage and
current *ARE* the standing-wave voltage and current because
traveling wave current and voltage doesn't exist. Good Grief!
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Roger wrote:
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.

There is a conspiracy to confuse you by using the same
symbol for completely different values - don't fall for
that Tar Baby.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Roy Lewallen wrote:
Roger wrote:

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.

If you'll look back through Keith's postings you'll see that he was
referring to the functions of time. It looks like he left off the
explicit (t) at some places which would lead to confusion.

But I hope you realize that you can also find the average power just
fine by calculating Pavg = Vrms * Irms * cos(theta) at any point along
the line, where Vrms and Irms are the total voltage and current at the
point, and theta is the angle between the two. This is true regardless
of the SWR. Also, you can calculate power as Irms^2 * Rser or Vrms^2 /
Rpar where Rser and Rpar are the series and parallel equivalent
resistive parts of the impedance at any point. Like v(t) and i(t), Vrms
and Irms can be found if desired by summing the forward and reflected
waves to find the total value at the point of interest; superposition
applies. It doesn't apply to power, so always do the summation of
voltages and currents before calculating power.

One property of the P(t) = V(t) * I(t) equation is that it also applies
to non-sinusoidal and even non-periodic waveforms -- it can *always* be
used. And you can always find the average power by integrating it then
dividing by the integration period. The average value calculation
reduces to Vrms * Irms * cos(theta) for pure sine waves but not other
waveforms.

Roy Lewallen, W7EL

Thanks again Roy. Your last two postings have been very helpful. While
I am aware of these relationships, I do not use them often and each use
seems like a first time. So I try to be very careful, but still make
bad errors as you saw yesterday.

I try not to be too critical of any postings, but (as Cecil warns) the
standards of knowledge and explanation are very high on this news group,
and misuse (even accidental) of terms or equations are noted and
frequently flamed or worse. Participants need a thick skin and
persistence, but a great deal can be learned here. I find it to be
quite a learning experience and proving ground.

Merry Christmas!

73, Roger, W7WKB

Standing-Wave Current vs Traveling-Wave Current
 

On Dec 24, 1:28*pm, Cecil Moore wrote:
Keith Dysart wrote:
Except that V(x,t) and I(x,t) are not, in general, related by Z0.

*From "Fields and Waves ..." by Ramo & Whinnery, 2nd edition:

V(x,t) = V*e^j(wt-kx) + V'*e^j(wt+kx)

I(x,t) = [V*e^j(wt-kx) - V'*e^j(wt+kx)]/Z0

You are quite unfair to Ramo & Whnnery when you quote them
out of context. It makes them look like they do not have a
clue.

Just because you find the same string ["V(x,t)"] in their
text does not mean that they are talking about the same
thing. (It better not in this case, or you should throw
their book away.)

You, yourself, have made the point in other posts that
the current and voltage on an open circuited line are
in quadrature, so they can not be in the ratio of Z0.

No. The two wave view is merely an alternate set of expressions
which, when summed (i.e. using superposition), provide the actual
voltage and current on the line. These alternate expressions are
obtained by algebraic maniupulation of the more fundamental
descriptive equations.

Methinks you are confusing cause and effect. The standing
wave is not the cause of the two traveling waves.

Of course not. "Standing wave" is just a short hand description
of the distribution of voltage and current on the line.

In my analysis, P(x,t) = V(x,t) * I(x,t) is the equation that means
the power at any point and time can by obtained by measuring the
actual voltage and current on the line at the point and time of interest..

Make that the *NET* power and you will have it nailed.

Are you sure you want to throw away this ability? Are you sure you
want to claim that instantaneous power can NOT be obtained by
multiplying the instaneous measured voltage by the instanteous
measured current?

When the instantaneous voltage is the sum of two more elementary
voltages (same for current) then you are reporting the *NET*
results, not the underlying component results. The *NET* results
do not dictate reality.

Sorry Cecil. The observed voltage and current on the line
are the reality.

...Keith

Standing-Wave Current vs Traveling-Wave Current
 

so roy has correctly calculated the standing wave 'power' to be zero at two
points on the line. i am sure that yuri will take great exception to this
result showing there is no power in the standing wave.

but he missed the definition of pr and pf...

dave in problem statement:
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.

so he conveniently skipped that step and instead writes this cop out:

roy:
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.

pr and pf are, as i stated, the power in the forward and reflected waves.
There is of course power in these two waves and it is indeed 'sloshing' back
and forth in the line. These are the waves that can be measured by any of
the simple devices such as neon bulbs or bird watt meters that clearly show
equal and opposite powers in the waves. so you can indeed have power in the
traveling waves, but no power in the standing waves... which will always be
the case.

i will give him this point as being correct for a lossless open (or shorted)
line:

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.

but then he loses it again:

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,

the traveling waves will have power EVERYWHERE on the line, the special
cases are are just the ones where the standing wave is most easily shown as
having no power. obviously if there is power in a wave at one point on a
line it is not going to stop and bypass the quarter wave points, the forward
and reflected waves continue end to end and their power goes with them... it
is at those 'special' points where the voltage or currents in the forward
and reflected waves always cancel each other so if you measure with a simple
tool you will see the voltage or current nulls at those points. that does
not mean there is no power passing those points, only that the voltage or
current in the traveling waves has conveniently canceled each other out at
those points.

and then he has to end up with an obvious contradiction:

indicating the movement of energy back
and forth but no net energy flow.

how does energy not flow if it is moving back and forth???

Standing-Wave Current vs Traveling-Wave Current
 

Cecil Moore wrote:
Roy Lewallen wrote:
These aren't "standing wave" voltage, current, and power, but simply
the total voltage and current, and the power, at that point.

Good Grief, Roy, no wonder you are clueless about standing
waves. In a lossless open-circuit line, the total voltage and
current *ARE* the standing-wave voltage and current because
traveling wave current and voltage doesn't exist. Good Grief!

Hmmm, Like Roy, I thought there was still a traveling wave in this
situation. The voltage at the far end of the line must reverse polarity
as time passes, so the waves must continue to travel, or so I would think.

Maybe it could be said better, but I thought Roy was trying to say that
although power could not be detected at the center or ends, it was
flowing as a result of the initial impetus charged into the system. I
would understand that this power would be the power needed to charge the
impedance and capacity of the line as it continually reversed polarity.
This would be real power from energy stored (but constantly moving) on
the 1/2 wavelength line so long as the system is active. We know we
have power present because we find energy distributed as V and I on the
time plot (viewed as a "standing wave" on the time plot).

Is this a "tongue-in-cheek" comment?

73, Roger, W7WKB

Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:
Cecil Moore wrote:

Keith Dysart wrote:
Except that V(x,t) and I(x,t) are not, in general, related by Z0.

From "Fields and Waves ..." by Ramo & Whinnery, 2nd edition:

V(x,t) = V*e^j(wt-kx) + V'*e^j(wt+kx)

I(x,t) = [V*e^j(wt-kx) - V'*e^j(wt+kx)]/Z0

You are quite unfair to Ramo & Whnnery when you quote them
out of context. It makes them look like they do not have a
clue.

It's not out of context. Those are their equations for
standing wave voltage and standing wave current. It is
net voltage and net current because each equation is
the sum of two component values.

Just because you find the same string ["V(x,t)"]

Actually, it wasn't the same string. R&W used 'z'
instead of 'x' for the length of the wire as was
common a half-century ago when I had their textbook
for both undergraduate and graduate level courses.

in their text does not mean that they are talking about
the same thing. (It better not in this case, or you should
throw their book away.)

But they *are* talking about the same thing. The first equation
above is the total (standing wave) voltage. The second equation
is the total (standing wave) current. The only term difference
between the two equations is the Z0 term. There is a sign difference
in the current equation that shifts the reflected current by 180
degrees putting the net voltage and net current in quadrature.

Methinks suggesting that Ramo & Whinnery should be thrown away
is "delusions of grandeur".

You, yourself, have made the point in other posts that
the current and voltage on an open circuited line are
in quadrature, so they can not be in the ratio of Z0.

Nobody said they are "in the ratio of Z0" and that's not
what you said before either. You said they are "not related
by Z0" and they are related by Z0 just as the above equations
demonstrate. You didn't say "ratio", you said "related".

The ratio of standing wave voltage to standing wave current
above is:

V(x,t) V*e^j(wt-kx) + V'*e^j(wt+kx)
------ = --------------------------------- = Z
I(x,t) [V*e^j(wt-kx) - V'*e^j(wt+kx)]/Z0

This is the old familiar SWR circle on the Smith Chart.
Smith Charts are normalized to Z0 = 1.0. The impedance
on the Smith Chart must be multiplied by Z0 to get the
actual impedance. Those impedances are indeed
"related to Z0". The s-parameter signals are normalized
to the square root of Z0. Vfor/Ifor = Vref/Iref = Z0.
Virtually everything about a transmission line is "related"
to Z0 including the standing wave voltage and current.

Methinks you are confusing cause and effect. The standing
wave is not the cause of the two traveling waves.

Of course not. "Standing wave" is just a short hand description
of the distribution of voltage and current on the line.

Yes, a "short hand description" that exists only in the human
mind and gets some folks into trouble - like trying to use
the illusion of moving standing wave current to "measure"
the delay through a coil. There is no standing wave current
movement through a coil or a wire. Standing wave current
doesn't move - it just stands there, oscillating in place.
EM current that doesn't move is obviously an illusion (and
a violation of the laws of physics).

Here's what Eugene Hecht said about standing waves in "Optics".

"It [the standing wave phasor] doesn't rotate at all, and
the resultant wave it represents doesn't progress through
space - it's a standing wave."

Applied to a wire or a loading coil, we can paraphrase -
standing wave current "doesn't progress through" a wire or
a loading coil - "it's a standing wave". EZNEC supports this
concept. Take a look at the current phase for a 1/2WL wire
dipole. From one end of the dipole to the other, the total
current phase varies by only a couple of degrees. Kraus agrees:

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

Sorry Cecil. The observed voltage and current on the line
are the reality.

The net voltage and net current are real but their independent
existence apart from the underlying traveling waves is just
an illusion. They are not independent. They are the *result*
of superposition of the forward and reverse waves. They are
merely an *effect* and not the cause of anything.

There are lots of real illusions. That beautiful Texas
"sunrise" this morning was just an illusion, an effect of
the rotation of the earth.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Dave wrote:
so roy has correctly calculated the standing wave 'power' to be zero at two
points on the line. i am sure that yuri will take great exception to this
result showing there is no power in the standing wave.

Power is the measure of the energy flow past a measurement
point. There is no net energy flow in pure standing waves.
Since the standing wave current phasor is always 90 degrees
different from the standing wave voltage phasor,

Power = V*I*cos(90) = 0 watts Note that V and I do not
have to be zero. There is simply no real power in a pure
standing wave. However, there are plenty of joules in
standing waves that can be converted to real power at
the expense of the standing waves after key-up.

All of the standing wave energy is contained in

Reactive Power = V*I*sin(90) in units of VARS

From the IEEE Dictionary:

"reactive power - For sinusoidal quantities in a two-wire
circuit, reactive power is the product of the voltage,
the current, and the sine of the phase angle between them."
--
73, Cecil http://www.w5dxp.com