[Phys-l] phasor power and other nonlinearities -- or not

[John Denker <jsd@av8n.com>]

At 5:20 PM -0700 7/24/11, Bernard Cleyet wrote:
> > "Volt-amperes measures all power passed through a distribution 
> > network, including reactive and actual. This is equal to the product 
> > of root-mean-square volts and amperes."
> >
> > Electricity meter - Wikipedia, the free encyclopedia

On 07/25/2011 04:06 AM, chuck britton wrote:
> True only if volts & amps were calculated as complex values??

Not true.

> Phasors?

Same thing.  Still not true.

There are some things you can do with phasors, but multiplying
complex volts times complex amps to get the power is not one
of them.



Let's go back to basics:

Suppose the actual voltage is 
      V = A cos(ωt + φ)                     [1]

for some real amplitude A, angular frequency ω, and phase φ.
This is the most general expression for this frequency component.
For a linear circuit, we can always resolve it into components
like this.

Trig identities allow us to write this as
      V = v1 cos(ωt) + v2 sin(ωt)           [2]

This is rigorously identical to the previous equation.

If we define the phasor
      v^^ := v1 + i v2                      [3]

then we can write
      V = Rp[ v^^ exp(-iωt) ]               [4]

which is again rigorously identical to equation [1] and
equation [2], as you can verify by directly expanding the 
RHS.  Here Rp means "real part".

You may find it convenient to use the expression
     Rp[ X ] = ½( X + X† )

for all X.

Now Rp is a linear operator.  That means it commutes with any
other linear operator.  If you have a linear circuit and you
want to calculate something (such as the input/output relation)
represented by a linear operator F, then

     F(V) = F(Rp[ v^^ exp(-iωt) ])
          = Rp[ F(v^^) exp(-iωt) ]          [5]

which means that you can do the entire calculation in terms of
the phasor v^^ and then find the actual voltage at the very last
step.

Sometimes people get a little bit sloppy and leave off the last
step.  They forget to multiply by exp(-iωt) and take the real
part.  This is not a big deal, except insofar as it leads to 
the following:

Sometimes people forget the restriction that F must be linear.
This is a Bad Thing.  You will get spectacularly wrong answers
if you incautiously plug a phasor into a nonlinear formula and 
turn the crank.  For today's example, namely the dissipated power
        P = V^2 / R                         [6]

if you take the phase φ in equation [1] to be 90 degrees, then
you get a negative power, which is about as wrong as it possibly
could be.

You *can* use phasors to help calculate V, provided you plug 
in the actual value for V, as given by equation [4], before
doing any nonlinear operations ... such as taking the square 
in equation [6].

Furthermore: If the signal is not a pure sinusoid, then in general 
V will be a sum of terms, i.e. a Fourier series ... and if the 
circuit is nonlinear you cannot treat these components separately.  
There will be utterly nontrivial intermodulation terms.


To summarize:

  *) For LINEAR circuits, the behavior of the actual voltage V
    is isomorphic to the behavior of the phasor v^^.

  *) However, isomorphism is not the same as equality.  You
    cannot equate V = v^^.

  *) The correct relationship can be written in various ways
   including:
      V = ½ ∑ v^^ exp(-iωt)  +  cc            [7]

   where "cc" is an often-used shorthand for "complex conjugate",
   and the sum runs over the aforementioned frequency components.
   This is a lot more complicated than V = v^^ but has the
   advantage of being not wrong.