wrote in message ...
Cecil Moore wrote:

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.

On the other hand...

if
Vrev = rho * Vfwd
then
rho^2 = (Vrev^2/Z0) / (Vfwd^2/Z0)

So
(Vrev^2/Z0) / (Vfwd^2/Z0)
can be greater than one.

If Sqrt(Pref/Pfwd) can not be greater than 1
then either
Pfwd is not equal to (Vfwd^2/Z0)
or
Prev is not equal to (Vrev^2/Z0)

Intriguing result, is it not?

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.... 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.

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.
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]

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... 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. so you can't just use P=V^2/Z0 in the general case! but
didn't we all know that? after all, why is there a power factor added to
the calculation when working with reactive loads?? and why can you have
LOTS more reactive power in a circuit than real power??? 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.

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???

Slick:

[snip]
Are you gonna re-write some books?


Slick
[snip]

Did it ever occur to you that most textbook authors actually plagiarize each
other.

Since most textbooks have numerous competitors covering essentially the same
material, it is apparent that most of their content is not original, rather
lots of it is "copied" or re-written using a slightly different [improved]
notation and set of symbols.

Have you never noticed the same error in several textbooks by different
authors?

Why does that happen?

With how many textbook authors do you have a personal relationship?

Have you ever reviewed pre-publication drafts of textbooks for publishers...

Worked with authors on corrections to drafts before publication... believe
me if you have why then you would be a lot more cautious in taking textbooks
as gospel.

Textbooks that have become very popular and which have appeared in many
editions/printings are usually pretty good... but even then still often have
significant errata.

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

Cecil:

[snip]
"Cecil Moore" wrote in message
...
Dr. Slick wrote:
I can't wait to hook it up to see more reflected power than
incident on my DAIWA meter, that would be very interesting.

If that's really what you want to observe, connect the meter backwards.
:-)
--
73, Cecil http://www.qsl.net/w5dxp
[snip]

Or hook a transmitter to the remote end!

Heh, heh... what is the read out of reflected power on a full duplex DSL
line, with transmitters running full bore on both ends simultaneously.

Slick's exposure to real world transmission problems is very limited!

;-)

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

Cecil:

[snip]
rho = (50-50)/(50+50) = 0 i.e. no reflections in either direction
--
73, Cecil http://www.qsl.net/w5dxp
[snip]

Hey... that's my line.

That's Mother Nature's reflection coefficient!

At every infinitesimal length along a transmission line rho = (Zo - Zo)/(Zo
+ Zo) = 0.

It's not nice to fool Mother Nature!

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

Slick:

[snip]
I can't wait to hook it up to see more reflected power than
incident on my DAIWA meter, that would be very interesting.


Slick
[snip]

It sure will be an interesting event for your Daiwa when the transmitter
hooked to the other end is keyed!

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

Slick:

[snip]
Show us a passive circuit that reflects more power than
you feed it (incident), on my Daiwa meter,
and i will be VERY impressed.

I'm be waiting a long time for that schematic....


Slick
[snip]

That's an easy one... simply connect a negative resistance to the end of the
line.

And don't tell me that negative resistance is not passive!

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

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:

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).

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.

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.

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.

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.

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

"Peter O. Brackett" wrote in message hlink.net...
Slick:

[snip]
Show us a passive circuit that reflects more power than
you feed it (incident), on my Daiwa meter,
and i will be VERY impressed.

I'm be waiting a long time for that schematic....



That's an easy one... simply connect a negative resistance to the end of the
line.

And don't tell me that negative resistance is not passive!


I most certainly WILL tell you that! A negative resistance can be found
in a tunnel diode, but you still need to apply a supply voltage to make
it an oscillator.

Negative resistance implies positive feedback, so i would certainly be
an active device.

Ok, show me your schematic, then.........waiting a long time......


Slick

"David Robbins" wrote in message ...


yeah, verily...

note some interesting things about the case you present...

first, to get it out of the way. as i have stated elsewhere in this thread
today the VSWR calculation from rho is really only applicable to lossless
lines because of the simplifications needed to calculate it from
|Vmax|/|Vmin| which is its real definition. and since Vmax and Vmin are not
the same in any two points on a lossy line it really doesn't have much
physical significance either.



Zo and Zl can be the impedances at a connector, with NO
TRANSMISSION LINE.

Zo can be the impedance at the END of a transmission line.

VSWR = ([rho]+1)/([rho]-1) only works for 0=[rho]=1

or situations of return LOSS, not return GAIN (active networks)!



the actual value of rho for this case is -.9372-1.60477i (which mathcad
gives me as 1.8585/_120.283 but i started from R,G,L,C values and only got
it down to Zo=301.5-250.3i which is close enough for this discussion i
think)

now, how can this be real... the important case is to look at the voltages
at the line/load junction. at this point Vf+Vr=Vl according to the
derrivation of rho. so what do we get with a rho like this???

for Vf=1.0v
Vr=rho*Vf = -.9372+1.60477i
and then
Vl=.06279+1.60477i

so there is a small real voltage across the 10+250i load, and a large
reactive voltage... the reactive voltage is equal to a reactive voltage on
the transmission line side of the junction.... can i believe this? I am
still working on that, in some ways it makes sense because you no longer
have a purely resistive cable characteristic, though i haven't come to grips
with the physical meaning of it yet. i do believe that it has a
relationship to circulating currents and reactive power in power
distribution circuits where you can get very odd looking voltages and
currents when you have a reactive load. in looking at this you have a very
capacitive looking line feeding a very inductive load, with a bit of
resistance thrown in on each side... essentially it looks like a current
pumping a resonant circuit which can result in very high voltages and
currents.



But you will never get a Reflected voltage that is greater than
the
incident in a passive network. Sure, you may have an inductor
charging up a capacitor somewhere in a resonant circuit (like parallel
resonance, with
a "fly-wheel" effect). But this voltage will not be reflected.



on the other side, the rho calculated with the conjugate in the numerator
gives: -.9358-.001688i or .9358/_-179.9


Which makes WAY more sense than the "normal" equation result.

Consider: Zo=300-j250 and Zl=10+j250

Essentially, the two reactances should cancel, and it will be
identical to Zo=300 and Zl=10.

Now, this should be fairly close to a short (Zl=0), which it
really is, in the sense that almost all the voltage is reflected
and the phase shift is almost -180 (-179.9), as it should be for
a near-short.

The "Normal" equation's results of: RC = 1.8585/_120.283 is
absolutely
incorrect. You can't use the normal equation for complex Zo!

You folks can believe what you want, but you are convincing me
more that Besser and Kurokawa and the ARRL are all correct on this
one.

I won't hesistate to admit that i'm wrong, but nothing has been
presented to convince me of that.




btw, for whom ever has it... i am still waiting to see the derivation of the

conjugate rho formula. i published one on here for the 'classical' version,
where is the other one???


I'll send you the paper...

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

Anyways, it's the correct formula for complex Zo, you've
convinced me of that with this post.

Thank you for your time and effort David.


Slick (Garvin)

"Peter O. Brackett" wrote in message thlink.net...
Cecil:

[snip]
"Cecil Moore" wrote in message
...
Dr. Slick wrote:
I can't wait to hook it up to see more reflected power than
incident on my DAIWA meter, that would be very interesting.

If that's really what you want to observe, connect the meter backwards.
:-)
--
73, Cecil http://www.qsl.net/w5dxp


Ha! that would be the ONLY way, if you are going into a passive
network like coax and antenna.




Or hook a transmitter to the remote end!

Heh, heh... what is the read out of reflected power on a full duplex DSL
line, with transmitters running full bore on both ends simultaneously.

Slick's exposure to real world transmission problems is very limited!


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

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


Slick