On Dec 23, 11:33*am, Roger wrote:
Keith Dysart wrote:

clip ....



In the setup above used for "standing waves"
it can be seen that there is zero power in
the line every 90 degrees back from the open
end. At a zero power point, no energy is
being transferred. Therefore, the forward
and reverse waves can not be transferring
energy across these points. Conclusion:
forward and reverse waves do not always
transport energy.

....Keith

Hi Keith,

You are basing this conclusion on the observation that Power = V*I, and
because we can not detect V or I at some points in the standing wave,
then V*I is zero at these points. Correct math, but wrong conclusion.

Are you really saying that if I measure the instantaneous
voltage and the instantaneous current then I can NOT multiply
them together to obtain the instantaneous power?

It certainly works some of the time.

If I can not do it all the time, when can I do it?

...Keith

"Keith Dysart" wrote in message
...
On Dec 23, 11:33 am, Roger wrote:

You are basing this conclusion on the observation that Power = V*I, and
because we can not detect V or I at some points in the standing wave,
then V*I is zero at these points. Correct math, but wrong conclusion.

Are you really saying that if I measure the instantaneous
voltage and the instantaneous current then I can NOT multiply
them together to obtain the instantaneous power?

It certainly works some of the time.

If I can not do it all the time, when can I do it?

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.

On Dec 24, 9:39*am, "Dave" wrote:
"Keith Dysart" wrote in message

...
On Dec 23, 11:33 am, Roger wrote:

You are basing this conclusion on the observation that Power = V*I, and
because we can not detect V or I at some points in the standing wave,
then V*I is zero at these points. Correct math, but wrong conclusion.
Are you really saying that if I measure the instantaneous
voltage and the instantaneous current then I can NOT multiply
them together to obtain the instantaneous power?

It certainly works some of the time.

If I can not do it all the time, when can I do it?

you can do it when it makes physical sense. *

Can you kindly articulate the rules you use to know
when it is appropriate to use P = V * I?

At a minimum, the rules should cover DC circuits,
AC circuits, and transmission lines with various
excitations and terminations. Be sure that the
rules cover 60 Hz circuits and transmission lines
since this is a common application of P = V * I.

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.

And for a challenging use case, please consider two
circuits connected together. The circuits are in black
boxes so you do not know their details, but the voltage
on the connection between the circuits is measured as
10 V RMS at 4 MHz. The current is measured as 0.

How much energy is being transferred between the
circuits?

1) P = VI, so 0.
2) P = VI some of the time, so there is insufficient
detail to answer the question.

Do you choose 1), 2) or perhaps some third answer?

...Keith

PS. The connection between the circuits is very small
so that it is possible that it is part of a
transmission line that continues into each box,
but you can not be sure.

"Keith Dysart" wrote in message
...

Can you kindly articulate the rules you use to know
when it is appropriate to use P = V * I?

it is extremely simple. use traveling waves then V*I works everywhere all
the time. use standing waves and it fails. period, end of story.

On Dec 24, 11:18*am, "Dave" wrote:
"Keith Dysart" wrote in message

...

Can you kindly articulate the rules you use to know
when it is appropriate to use P = V * I?

it is extremely simple. *use traveling waves then V*I works everywhere all
the time. *use standing waves and it fails. *period, end of story.

What happens on a line that is terminated in a real
impedance that is not equal to Z0?

There are aspects of both travelling waves and
standing waves present on the line.

Is it appropriate to use P = V * I?

...Keith

"Keith Dysart" wrote in message
...

What happens on a line that is terminated in a real
impedance that is not equal to Z0?

There are aspects of both travelling waves and
standing waves present on the line.

Is it appropriate to use P = V * I?

it is always appropriate to use P=V*I on the forward and reflected traveling
waves. it is never appropriate to use it on the standing wave voltage and
current. period... plonk.

Dave wrote:
"Keith Dysart" wrote:
Is it appropriate to use P = V * I?

it is always appropriate to use P=V*I on the forward and reflected traveling
waves. it is never appropriate to use it on the standing wave voltage and
current. period... plonk.

Arguments like this are usually semantic. Is it possible
that Keith is talking about phasors and you are talking
about scalars?
--
73, Cecil http://www.w5dxp.com

Keith Dysart wrote:
On Dec 24, 11:18 am, "Dave" wrote:
"Keith Dysart" wrote in message

...

Can you kindly articulate the rules you use to know
when it is appropriate to use P = V * I?
it is extremely simple. use traveling waves then V*I works everywhere all
the time. use standing waves and it fails. period, end of story.

What happens on a line that is terminated in a real
impedance that is not equal to Z0?

There are aspects of both travelling waves and
standing waves present on the line.

Is it appropriate to use P = V * I?

...Keith

And from an earlier post, Keith wrote

"Are you really saying that if I measure the instantaneous
voltage and the instantaneous current then I can NOT multiply
them together to obtain the instantaneous power?

It certainly works some of the time.

If I can not do it all the time, when can I do it?"

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.

I would visualize the situation by saying that at the points mentioned,
the peaks of the traveling waves match as they pass each other going in
opposite directions each cycle. At all other points, the matching is
peak of one plus part of the second, so that the resulting measurement
can always be described as containing a quadrature (or reactive) component.

73, Roger, W7WKB

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

Dave wrote:
"Keith Dysart" wrote in message
...

Can you kindly articulate the rules you use to know
when it is appropriate to use P = V * I?

it is extremely simple. use traveling waves then V*I works everywhere all
the time. use standing waves and it fails. period, end of story.

It does not fail. Present any example of a transmission line terminated
with a load. Choose any line and load impedance and any transmission
line length. I'll tell you the V and I at any point on the line,
calculate the instantaneous power from V(t) * I(t) at that point, and
from that the average power. Then I'll show that this average power
equals the power in the load and the power delivered by the source. As
an added bonus, I can tell you the impedance (ratio of V/I) at any point
along the line. I'll provide both equations that always work and
numerical results.

Then you can show where I've made an error and why my calculations are
invalid.

Can I be any more fair than that?

Roy Lewallen, W7EL