"Roy Lewallen" wrote in message
...
I have a few nits to pick with this analysis. Comments interspersed.

David Robbins wrote:

indeed... and directly to the crux of the whole problem. there are a
few
too many powers being tossed around here without proper definition. i
was
just looking at section 7.1 in 'Basic Circuit Theory' by Desoer and Kuh
that
kind of sums up all the problems this discussion has been having in its
title "Instantaneous, Average, and Complex Power". for some reason this
section seems to have many more highlighted formulas than much of the
rest
of the book.

I think many of our problems come from mis-applying power formulas to
the
wrong cases. for example, the well known P=VI=V^2/R=I^2R takes a bit of
modification to work for general complex impedances.

lets look at the simple one first:
Instantaneous power: p(t)=v(t)i(t)
aha! you say, there it is, nice and simple.... but not so fast. this is
in
the time domain. the Vfwd and Vrev we throw around so easily in phasor
notation when talking about transmission lines aren't the same thing and
can't be so easily converted... in fact, in going to the phasor notation
you
intentionally throw away time information....

Here's my first point of disagreement. Phasor notation doesn't throw
away time information. It's still there, in implied form (as an implied
term exp(jwt)). And that's an important distinction, which I'll note
again shortly. One thing it means is that, given the phasor
representation of a waveform, it's possible to determine exactly what
the time waveform is -- you can convert back and forth at will. You do,
of course, have to know the implied radian frequency. If the time
information were thrown away, you wouldn't be able to do that.

so in this case when you write
v(t) and i(t) in their full form they look like:
v(t)=|V| cos(wt+/_V)
i(t)=|I| cos(wt+/_I)
where w=omega, the angular rate, and /_ is used to denote the relative
angle
at t=0....
thus expanding out this power formula you get and ugly thing:
p(t)=.5 |V| |I| cos(/_V-/_I) + .5 |V| |I| cos(2wt+/_V+/_I)
which when averaged becomes:
Pav=.5 |V| |I| cos(/_V - /_I)
which hopefully looks familiar to people out there who deal with power
factors and such.


Fine so far.

Now, the odd one...
Complex Power:
P=.5 V I*
where V and I are the phasor representations used in sinusoidal steady
state
analysis, and I* is the conjugate of I.


Problem here. As I mentioned above, a phasor contains an implied
frequency term. Phasors represent pure sine waves, and the frequency
*must be the same* for all phasors with which calculations are being
done. You can't use phasor analysis where the voltage, for example, is a
different frequency than the current, or to mix two voltages of
different frequencies. And that's the problem with the equation above.
It is *not* an equation for complex power. P is not, and cannot, be a
phasor, because its waveform has a different frequency than V and I. It
appears that you might be confusing this with an equation for *average*
power, which is:

i didn't say Pcomplex was a phasor. in fact it's 'frequency' would be twice
the V and I frequency. as the derivation continues you can calculate the
average power from it though as shown below...


P = Re{V I*}

with a .5 factor if V and I are peak values, and no .5 factor if they're
RMS. The result of this calculation is a purely real number, and the
same as you get with |V||I|cos(phiV - phiI).

RMS is a whole different problem that simplifies the equation even further.


Power can't be represented in phasor form. At least not in the same
analysis as the V and I it's composed of -- some clever PhD candidate
might have devised some use for a phasor analysis involving only power,
which could be done if you can find a consistent way to deal with
power's DC (average) component. But I've never seen such a method,
unless that's what Cecil is doing. So "complex power" doesn't have any
real meaning, although "instantaneous power", or time-domain power,
certainly does.

ah, but it does and when used as defined.... it is an odd concept and rarely
used but it is a way to represent power which keeps intact the phase
relationship it came from.


now here is where it gets weird...substituted in the exponential forms
of
the phasors:
P=.5 |V| |I| e^(j(/_V-/_I))
oh for a good way to represent equations in plain text... but anyway,
here
is a similar notation to the instantaneous power above, but there is no
time
in it... only the magnitudes of the voltage and current multiplied by a
complex exponential from the difference in their phase angles. this can
be
expanded into a sin/cos expression to separate the real and imaginary
parts
like this:
P=.5 |V| |I| cos(/_V-/_I) + j.5 |V| |I| sin(/_V-/_I)
and from this you can show that the real part of this is the average
power,
so:
Pav=.5 |V| |I| cos(/_V-/_I)
but since we are in phasor notation we can transform this one more time
using
V=ZI and I=YV (Y=1/Z, the complex admittance)
to get:
Pav= .5 |I|^2 Re[Z] = .5 |V|^2 Re[Y]


Because of what I said above, I don't believe that this is valid. You
simply can't calculate a phasor power from phasor V and I.


as i said, its not phasor power, it is complex power. there is no attempt
made to show that it represents a phasor.

there are also a couple of important notes to go with this. in a
passive
network Re[Z] and Re[Y] are both =0 which also constrains cos(/_V-/_I)

=0... essentially, no negative resistances in passive networks... which
of

course then results in Pav always being positive(or zero).

now wait a minute you may say.. we have two different Pav equations...
what
happens if we equate them??? (well, actually we have 4 different
equations,
so lets play around a bit)

from time domain:
Pav=.5 |V| |I| cos(/_V - /_I)

from sinusoidal steady state:
Pav=.5 |V| |I| cos(/_V-/_I)
Pav= .5 |I|^2 Re[Z]
Pav= .5 |V|^2 Re[Y]

well, what do you know, the two methods give the same result (the first
formula of the sss method is the same as the time domain formula). but
what
do the other two with the impedance and admitance do for us.... they
show
that power in phasor calculations is not quite as simple as the
P=VI=V^2/R=I^2R we are used to...

I hope we're not used to that. It's an attempt to calculate phasor
power, which is doomed from the start.

thats what i said, you can't use the simple VI formula with phasors to
directly get power.


in fact, what are those equations good
for? basically, just for resistive circuits where I and V are in
phase...
see section 7.3 of that reference for derivation of the I^2R power
formula
for resistive loads... and also for how the rms value gets rid of that
pesky
factor of .5 in all those equations to make I^2R work for sinusoidal
voltages... so I^2R and V^2/R are REALLY only good for rms voltages and
resistive loads.

That's true. And the reason is that when V and I are in phase, then V =
|V| and I = |I|, and Z = |Z|, so you're simply calculating the average
power.

so you can't just use P=V^2/Z0 in the general case!

Absolutely not. Again, for the reason that you can't calculate a phasor
value of P from phasor V and I.

but
didn't we all know that? after all, why is there a power factor added
to
the calculation when working with reactive loads??

Don't confuse power factor or its existence with the fundamental problem.

and why can you have
LOTS more reactive power in a circuit than real power???

Why not? It's very often the case.

why not??? i didn't say your couldn't, i said you do often have lots of
reactive power.... i think you have lost track of what i was trying to show.


don't believe in
reactive power? wait till the report about the blackout comes out, it
was
basically run away loss of control of reactive power that probably
resulted
in circulating currents that brought the grid to its knees.

The idea of reactive power follows easily from observation of the power
time waveform you described. The power becomes negative for part of the
cycle if V and I aren't exactly in phase. This represents energy flowing
back out of the circuit it flows into during the positive portion of the
cycle. If the power waveform is centered around zero, which it will be
if V and I are in quadrature, then the same amount of energy flows out
as flows in, so no work is done -- the power is entirely imaginary. And
the average value -- the waveform DC offset -- is zero. On the other
hand, if the waveform is entirely positive, which it is if V and I are
in phase, then energy flowing in never flows out, so the power is
entirely real, and the average is half the peak-to-peak waveform value.
Real power represents the power that flows in without a corresponding
outward flow; reactive power is that which goes in and comes back out.


so you are right:


Pfwd is not equal to (Vfwd^2/Z0)

and

Prev is not equal to (Vrev^2/Z0)


but by definition:


Vrev = rho * Vfwd


now, who really read and understood that???

I hope I did. If not, please correct me.

Roy Lewallen, W7EL

"Dr. Slick" wrote in message
om...
"David Robbins" wrote in message
...
"Dr. Slick" wrote in message
m...
Cecil Moore wrote in message
...
Dr. Slick wrote:
If you agree that the Pref/Pfwd ratio cannot be greater than
1
for a passive network, then neither can the [Vref/Vfwd]= rho be
greater
than 1 either.

Sqrt(Pref/Pfwd) cannot be greater than one. (Z2-Z1)/(Z2+Z1) can be
greater than one. Both are defined as 'rho' but they are not
always equal. (Z2-Z1)/(Z2+Z1) is a physical reflection coefficient.
Sqrt(Pref/Pfwd) is an image reflection coefficient.


I agree that Sqrt(Pref/Pfwd) cannot be greater than one for a
passive network.

(Z2-Z1)/(Z2+Z1) can be greater than one, for passive networks
and certain combinations of complex Z1 and Z2. I feel this is
incorrect
usage of this formula, which should be limited to purely real Zo.
A [rho] that is greater than one gives meaningless negative SWR
data, and is limited to active devices.

it only gives negative swr values if you incorrectly use the lossless
line
approximation to calculate vswr from rho. that is the incorrectly
applied
formula in this case. that formula is not valid for a lossy line, you
must
go back to the original definition of VSWR=|Vmax|/|Vmin|. which as we
have
also noted is not meaningful on a lossy line as Vmax and Vmin are
different
at each max and min point because of the losses in the line affecting
both
the forward and reverse waves.


What if Z1 represents the impedance at the end of the line? Then
it doesn't matter what the losses are. Then you are trying to find
the
RC right at the meeting of the line and the load, so you don;t care
what the
losses of the line are.

Slick

but Vmax and Vmin don't occur at the same location on the line. when
measuring Vmax and Vmin you have to look 1/4 wave apart on the line... so by
definition it is a distributed measurement so losses do matter.

Dr. Slick wrote:
I'd like to see the derivation too, as Kurokawa seems to skip it
or just copied it from another paper! haha...

The point that you may have missed is that Kurokawa invents a different
kind of wave called a "Power Wave". He dismisses traveling waves as not
being very useful. So, of course, with the invention of a different kind
of wave, he is free to introduce a different kind of reflection coefficient.
The mistake being made here is trying to assume that his different kind of
reflection coefficient applies to normal forward and reflected traveling
waves. It doesn't.
--
73, Cecil, W5DXP

Dr. Slick wrote:
Show me an antenna-coax network that reflects more power than
incident! Impossible!

I've pretty much proven that it can't happen. Simply insert one
wavelength of lossless feedline between the lossy feedline and
the reactive load. Everything becomes clear.
--
73, Cecil, W5DXP

David Robbins wrote:
but Vmax and Vmin don't occur at the same location on the line. when
measuring Vmax and Vmin you have to look 1/4 wave apart on the line... so by
definition it is a distributed measurement so losses do matter.

The measured 'rho' at any *point* can be determined by Sqrt(Pref/Pfwd).
That means, for lossy lines, measured SWR increases toward the load and
decreases toward the source. An open-ended 400 ft. piece of RG-58 will
show a near perfect SWR at the source end on 440 MHz.
--
73, Cecil, W5DXP

Reference Dr. Best's QEX article.

===============================

Who is Dr. Best? Would it be Dr Who with his Daleks?

And what is QEX.

99.999 percents of the world's population has never heard of either.

Reg Edwards wrote:

Reference Dr. Best's QEX article.

Who is Dr. Best? Would it be Dr Who with his Daleks?

Dr. Best is the ham who started all this mess when he posted parts
of his QEX article to this newsgroup in May of 2001.

And what is QEX.

QEX is a technical magazine publication of the ARRL.

99.999 percents of the world's population has never heard of either.

What percentage of the world's population are US hams? I'll bet at
least 20% of them have heard of QEX magazine.
--
73, Cecil, W5DXP

Who is Dr. Best? Would it be Dr Who with his Daleks?

And what is QEX.

99.999 percents of the world's population has never heard of either.


Yea, but those of us who matter, know both, and have been to Llangollen!

W4ZCB

Slick:

[snip]
Not as limited as yours, it would seems! ;^)

Show me an antenna-coax network that reflects more power than
incident! Impossible!


Slick
[snip]

Ever operate your antenna coax in the near field of a commerical broadcast
antenna?

Guess what the reflected power reads?

RF applications, Ham antennas and transmission lines are ho-hum
technology...

Such simple applications do not present any great analysis difficulty or
operating
challenges, they are approximately lossless and distortionless and always
operated narrow band with a purely resistive Zo =50 Ohms.

The question of complex Zo never arises in ham applications or most other
RF applications for that matter...

The most difficult transmission line problems for design and analysis are
those
operating in what is known as DSL [digital subscriber loop] technology and
similar applications.

In the DSL application the Zo of the line, up to 18,000 feet of twisted pair
with at leat 1500 Ohms of DC resistance, is extremely complex and varies
all over the map over 5 - 6 decades of operating frequency range from DC
to tens of MegaHz, supported by full duplex transmitters transmitting
simulaneously
at full power on both ends with the receivers hooked directly to the same
ends.

If you have succesfully designed transceivers to operate on those lines,
maintained
by span powering from one end and shipped in the millions world wide as I
have, why
then my friend you may claim to know something about complex Zo and
reflection
coefficients.

--
Peter K1PO
Indialantic By-the-Sea, FL

W5DXP wrote in message ...
David Robbins wrote:
but Vmax and Vmin don't occur at the same location on the line. when
measuring Vmax and Vmin you have to look 1/4 wave apart on the line... so by
definition it is a distributed measurement so losses do matter.

The measured 'rho' at any *point* can be determined by Sqrt(Pref/Pfwd).
That means, for lossy lines, measured SWR increases toward the load and
decreases toward the source. An open-ended 400 ft. piece of RG-58 will
show a near perfect SWR at the source end on 440 MHz.


I agree with both of you on this one.

This is why people will tell you that your SWR meter should really
be
at the antenna, instead of at the end of 100 ft. of RG-58, if you
wanna measure the SWR at the attenna!

Absolutely agrees with everything i know up to this point.

But, you SWR meter is at one point only! You don't have to
measure at two places 1/4 wave apart! This is because the SWR is
based on the forward and reflected powers.

But i agree, if you move 200 feet down RG-58 from the SWR meter at
the antenna, the SWR will most certainly improve!

My point is that the RC equation calculated the RC at one point,
and so the SWR equation will also apply at that ONE point.

But if you want to get the SWR 200 feet down the line, you will
have to
move the SWR meter and measure at the new location.


Slick