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

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

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

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

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

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

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

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

On Dec 24, 8:15*pm, Roger wrote:
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.

My apologies for being not being completely clear.

My voltmeter is always one that is appropriate for the situation.
A d'Arsonval movement would work at sufficiently low frequencies
but for the ones under discussion an oscilloscope would probably
be more appropriate. It would be a voltmeter that would be able
to measure the actual voltage at a point on the line at a particular
time. I.e. V(x,t)

Again, apologies.

...Keith