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