Phasors

       Introduction

                Phasor Properties

                                Phasors of Time Derivatives

Introduction

        By this time you should know a great deal about DC circuits.  In particular, you should be familiar with the following.

        When you learned these topics you might have assumed that what you were learning was something that only applied to DC circuits.  In a way you were right, but in a way you were wrong.  While the methods you learned only apply to DC circuits, the methods you learned can also be generalized to AC circuits.  It will take a little work, but every one of the analysis tools listed above has a related - and very similar - AC version.

        If you want to generalize your tool set (all of the tools listed above) you will first need to learn about phasors.  Here is a short lesson on phasors and an introduction to using phasors.

l A sinusoidal signal - voltage or current or
    any other physical variable - can be
    represented by a phasor which encodes
    the amplitude and phase information for
    the signal.  Call that phasor V.

l If the sinusoidal signal is given by:

v(t) = Vmax cos(wt + f)

    Then, the signal can be represented by
    a phasor:

v(t) <=> Vmax/f
    and we would write an expression for the
    phasor V:
V = Vmax/f
l Phasors in linear circuits are related if all
    of the signals in the circuit are at the same
    frequency.
Important Properties of Phasors

        In this section we will examine some simple properties of phasors.  We start with one that may or may not be obvious.

l Adding phasors is equivalent to adding
    the corresponding time function for
    each phasor.
        To see how this works out, consider the sum of two voltage phasors

Vsum = V1 + V2 = V1 /f1 + V2 /f2

This sum corresponds to the sum of the two voltages (as might occur when you write KVL).  The actual sum would be written:

vsum(t) = Vsum cos(wt + fsum)

= V1 cos(wt + f1) + V2 cos(wt + f2)

The problem is to be able to use the phasor method to do the addition, then interpret that in terms of the time functions.  You and interpret the problem in at least two different ways.

l One way to interpret this is in terms of complex numbers.

        Vsum = V1 + V2 = V1 /f1 + V2 /f2

    Each phasor can be represented by a complex number:

        V1 = V1cos(f1) + jV1sin(f1)

        V2 = V2cos(f2) + jV2sin(f2)

So, the sum of the two phasors can be computed by adding the real and the imaginary parts separately, giving:

= V1cos(f1) + V2 cos(wt + f2) + j[V1sin(f1)+ V2 sin(wt + f2)]

Then, we can note the the real part and the imaginary part are the real and imaginary parts of the sum.

= Vsumcos(fsum) + jVsumsin(fsum)

l A second interpretation is geometrical.

        There is, of course, also a geometrical interpretation whenever you deal with complex numbers or variables.  If we have two phasors that we are adding, we visualize the situation as shown below.

                        Vsum = V1 + V2
 
 

        We can add the two phasors any way possible.  That includes doing it graphically by hand, breaking the phasors into components and summing the real and imaginary components - as we did above - or any other way you can imagine to sum two vector-like quantities.

        A second operation that we need to perform often is differentiation of a time function represented by a phasor.  Consider a sinusoidal time function, v(t):

v(t) = Vmax cos(wt + f)

with a phasor V:

V = Vmax/f

The derivative of the time function is given by:

vd(t) = dv(t)/dt = - wVmax sin(wt + f)

The phasor for the derivative signal is:

Vd = - wVmax/f-90o= wVmax/f+90o

(Note the minus sign on the first represention where ninety degrees is subtracted!)

l Differentiating a sinusoidal signal is equivalent
    to multiplication of the signal's phasor by w and
    rotation of the phasor by 90o.
        There is also an interpretation in terms of complex numbers.  Represent the phasor V in terms of complex numbers:

V = Vcos(f) + jVsin(f)

Then, consider the phasor for the derivative:

Vd = Vdcos(f) + jVdsin(f)

But, we can also write the phasor for the derivative from the time function for the derivative.

Vd = wVcos(f+90o) + jwVsin(f+90o)

Now, work with this expression.  Take the ninety degree terms out of the arguments since cos(f+90o) = -sin(f), and sin(f+90o) = cos(f).

Vd = - wVsin(f) + jwVcos(f)

and with some insight, we can note that this is the same as (since j2 = -1):

Vd = j2 wVsin(f) + jwVcos(f)

Vd = j w( jVsin(f) + Vcos(f)) = j wV

l Differentiating a sinusoidal signal is equivalent
    to multiplication of the signal's phasor by jw.
Example
        In a capacitor, the voltage and current are related by:

ic(t) = C dvc(t)/dt

so the voltage phasor, Vc, and the current phasor, Ic, are related by:

Ic = jwCVc

        We leave this as an exercise for the interested student.
l The time integral of a sinusoidal signal is equivalent
    to multiplication of the signal's phasor by 1/jw.
A Quick Summary of What You Need to Know

        Phasors are used extensively in AC circuit analysis.  Operations you need to be familiar with include addition and differentiation in the time domain, and the corresponding results for the phasors representing the time domain signals.