Phasors vs. vectors and power calculations

Discussion in 'Electronic Basics' started by Chris Carlen, Mar 10, 2005.

1. Chris CarlenGuest

Greetings:

This is just a reflection on some mathematical semantics and an interest
in having a neat notation for calculating the average power dissipated
in an AC circuit from rectangular forms of the voltage and current
phasors. It arose out of my deciding to review in detail my
understanding of AC power, so might be of some interest to students of
EE. There is nothing here which is not found in any basic circuits
text, but just my particular description of it.

To get started, my definition of a phasor would go like this:

"A phasor is a complex number that carries the information about the
magnitude and phase of a sinusoidal time varying function of fixed
frequency."

Thus, any old complex number is not a phasor, unless it is a complex
number arising from the phasor transform applied to a sinusoidal time
varying function of fixed frequency.

"Furthermore, a phasor is not a vector, though a phasor may be
graphically represented as a vector in the complex plane."

"Nor do phasors or the vectors representing them "rotate" in the complex
plane. There is nothing in the phasor transform that leads to the real
and imaginary components of the phasor being time dependent functions.
Indeed it's very point is to remove the time variation aspect from the
voltage and current quantities. Thus, the components of the phasors are
simply constants. It is only a demonstrative tool to illustrate how
IF the vector is rotated in time, that for the case of starting with a
cosine time domain function (and applying the respective phasor
transform), then the real component of the phasor traces out the
instantaneous magnitude of the time varying quantity. Likewise for the
case of starting with a sine time domain function (and applying the
respective phasor transform), then the imaginary component of the phasor
traces out the instantaneous magnitude of the time varying quantity."

Average power dissipated in a linear AC circuit driven by a fixed
frequency sinusoidal source can be determined by the dot product of the
vectors representing the phasors for

V(t) = Vm cos(wt+phi_V)
I(t) = Im cos(wt+phi_I)

Let's use the notation ~V and ~I to represent the phasor transforms of
V(t) and I(t) respectively.

But since ~V and ~I are phasors we can't notate or even speak of their
dot product as we would with vectors. Ie., the notation

~V . ~I

doesn't mean the dot product of the vectors, since the quantities ~V and
~I are phasors, not vectors.

Instead we would have to write out something ugly like:

Pave = [ Re{~V} Re{~I} + Im{~V} Im{~I} ]/2

which works, but isn't a concise mathematical notation such as the
simple dot between two vectors.

Obviously, if we have the phasors in polar form, then the calculation is
a more straightforward:

Pave = 0.5 Vm Im cos( phi_V - phi_I )

which of course is the definition of the dot product of the vectors
representing the phasors.

Additionally, there is the defnition of complex power as

S = P + jQ , where

P = 0.5 Vm Im cos( phi_V - phi_I ) = Pave
Q = 0.5 Vm Im sin( phi_V - phi_I ) = reactive power

In which case average power may be neatly expressed as:

Pave = Re{S}

And since S can be shown to be:

S = 1/2 ~V ~I* then

Pave = Re{ 1/2 ~V ~I* }

So perhaps that is it, huh? The real part of the product of a phasor
with the conjugate of another gives the same effect as the dot product
of the vectors representing those phasors.

Good day!

--
_______________________________________________________________________
Christopher R. Carlen
Principal Laser/Optical Technologist
Sandia National Laboratories CA USA

NOTE, delete texts: "RemoveThis" and "BOGUS" from email address to reply.

2. Don PearceGuest

I think that I would have to say a phasor is a rotating vector.

d

Pearce Consulting
http://www.pearce.uk.com

3. Larry BrasfieldGuest

Then I would have to say: A phasor is a
stationary vector used to conveniently
refer to a rotating vector.

4. krwGuest

Either this, or a weapon of mass destruction.

5. John PopelishGuest

And the coordinate plane it is drawn on and the viewer of that plane
are rotating at the same frequency. ;-)

6. Chris CarlenGuest

The phasor transform is found by taking only the non time-dependent
factors of the argument to the real operator in:

V(t)=Vm cos(wt+phi_V)

=Re{Vm [cos(wt+phi_V) + j sin(wt+phi_V)]}

=Re{Vm exp[j(wt+phi_V)]}

=Re{Vm exp(jwt) exp(j phi_V)}

Thus:

~V = Vm exp(j phi_V)

~V = Vm cos phi_V + j sin phi_V

There is nothing time dependent here to allow for any conception of
rotation. The phasor is a constant complex number. How can a constant
rotate? The only difference between it and an ordinary constant complex
number, such as an impedance, is mathematically no difference at all,
but contextually that it is derived from the phasor transform of a time
varying function.

That's why I said it is a semantics issue. Rotating the vector that
graphically represents the phasor in the complex plane is a
demonstrative tool. But it remains that there is nothing about the
mathematical representation of the phasor that allows for rotation.

One last thing. A vector represents a magnitude and a direction in a
coordinate system. It is based upon orthonormal basis vectors such as
what we might consider to be r_hat and j_hat in the complex plane,
analagous to x_hat, y_hat in the XY plane.

It is the definition of the basis vectors that gives mathematical
consistency to a statement such as:

A_vec . B_vec = (A_x x_hat + A_y y_hat).(B_x x_hat + B_y y_hat)
= A_x B_x x_hat.x_hat + 2 A_x B_y x_hat.y_hat
+ A_y B_y y_hat.y_hat
= A_x B_x + A_y B_y

since x_hat.x_hat=1, y_hat.y_hat=1, and x_hat.y_hat=0

But there are no basis vectors in a complex number or phasor. Thus, the
phasor doesn't contain information about "direction" in an orthonormal
coordinate system. That is why I carefully say that a phasor may be
*graphically represented* as a vector in a complex coordinate plane
characterized by orthonormal basis vectors r_hat and j_hat such that

r_hat.r_hat=1 and
j_hat.j_hat=1 and
r_hat.j_hat=0

So that we can represent the phasors as a vector by letting:

V_vec = Re{~V} r_hat + Im{~V} j_hat
I_vec = Re{~I} r_hat + Im{~I} j_hat

Then the real average power:

Pave = 0.5 V_vec.I_vec works.

But the expression:

~V.~I is meaningless.

Good day!

--
_______________________________________________________________________
Christopher R. Carlen
Principal Laser/Optical Technologist
Sandia National Laboratories CA USA

NOTE, delete texts: "RemoveThis" and "BOGUS" from email address to reply.

7. Don PearceGuest

OK - I'll try again. A phasor is a representation of the instantaneous
angular difference between subject and reference vectors, with the
absolute magnitude of the subject vector.

d

Pearce Consulting
http://www.pearce.uk.com

8. Luhan MonatGuest

Something is rotating here; I seem to be getting dizzy.

9. Terry GivenGuest

what about exp(jwt) that you carefully dropped off? that is the rotating
bit.

10. Chris CarlenGuest

Terry, dropping the exp(jwt) is the *definition* of the phasor transform.

For references, try here:

Ulaby, Fawwaz. "Fundamentals of Applied Electromagnetics 2001 media
ed." pg. 25

And here:

Nilsson, James W. and Susan Riedel. "Electric Circuits 6th ed." pg. 418

And there must be a pile of other texts which explain this definition.

Good day!

--
_______________________________________________________________________
Christopher R. Carlen
Principal Laser/Optical Technologist
Sandia National Laboratories CA USA

NOTE, delete texts: "RemoveThis" and "BOGUS" from email address to reply.

11. Terry GivenGuest

Indeed. If machines are your thing, Retter, Krause, Kron, Vas etc. have
written plenty on the subject.

I think this is a semantic issue. Phasors are, by definition, rotating
vectors. The trick with phasor transforms is to rotate synchronously
with some reference phasor, so all synchronous phasors appear
stationary, but with angular displacement.

Mathematically this is achieved by dropping the exp(jwt) term.

If you build a synchronously rotating reference frame controller for
electrical machinery, this "phasor transform" is normally implemented as
follows:

1) Choose a phasor on which to orient your rotating reference frame - eg
AC line voltage. By definition this transformed phasor is purely real
(or purely complex).

2) Calculate the transformed real and imaginary parts by taking the
original phasor and multiplying by exp(j*theta) [in a 3-phase system we
typically assume some symmetry and perform a 3-phase to 2-phase
transformation, resulting in a single complex phasor.

3) use the imaginary (or real) part of the transformed phasor as the
feedback signal for a PI controller to calculate the actual w. The
setpoint for this PI controller will be zero, again by definition

4) the output of the PI controller is angular speed, w. Usually the PI
controller has the desired w (eg 2pi*50Hz) added to its output, so it
only corrects the phase error, but its not strictly necessary - the PI
controller will eventually get the right w.

5) integrate the estimated w, giving theta = wt

6) use this theta in the phasor transform of step 2.

Ive implemented quite a few of these, and it works very well, for a wide
range of PI gains. The magnitude of the error (ie |Eq|) can be used to
detect synchronisation. Once synchronised, all them pesky rotating
things suddenly look DC. If your are controlling an induction motor, the
rotor vectors all spin at the slip speed.

And of course its quite feasible to transform the synchronous PI
controller into a stationary reference frame, it just becomes a resonant
controller (DG Holmes, P Mattiavelli etc have covered these in detail)
but thats boring power electronics control stuff.

Interestingly enough, the electrical model for a machine in the
stationary reference frame implicitly has time-varying inductances
(hardly surprising, flux linkage being a function of rotor angle),
making the analysis a right bastard. But through a phasor
transformation, the time-varying nature of the inductance "disappears",
making the resultant analysis straightforward. Its been around in power
systems engineering since the 1920s, often referred to as Clarke (or is
it Park, one is the 3:2 phase transform, the other is the
stationary-rotating) transforms.

This is another form of feedback linearisation based nonlinear control BTW

Conversely whenever I analyse a 50Hz AC circuit, I implicitly use phasor
transforms by simply ignoring the rotating aspect, and pretending the
relevant vectors are all stationary.

Cheers
Terry

12. Fred AbseGuest

Are we allowed to Park here?

<Snort> LOL

Cheers
Terry