Today, I had an epiphany about something as fundamental as phasor analysis. It was actually something so embarrassingly simple that I had to actually consider if I dare to publish it at all. Well, here we are. Anyway, the following might be nothing new for some of you, but I’d wager not for most. Anyway, here goes.

If you have had any contact whatsoever with electrical engineering, you are certainly familiar with phasor analysis. Let’s say you have a sinusoidally-varying voltage

U(t) = U_0 \cos(\omega t + \theta)

over a resistance R, and you want to solve the current I flowing in it. Of course, you do this by defining a voltage phasor

\underline{U} = U_0 e^{j\theta},

and then solve the current phasor as

\underline{I} = \frac{\underline{U}}{R}

Finally, the actual time-varying current is then obtained as

I(t) = \mathrm{real}[\underline{I}e^{j\omega t}]

Some authors also like to include the time-dependent term e^{-j\omega t} also in the phasor definition, but the overall principle remains the same.

Nevertheless, most people (including yours truly until today) interpret phasors as a handy way of compressing both the amplitude and phase angle into a single variable, and nothing else. In the other words, they are not thought to directly correspond to an actual physical instantaneous quantity per se.


However, things get quite different look when viewed from another angle. Let’s begin by writing the voltage U with its complex Fourier-series representation, i.e. the positive e^{j\omega t} and negative components e^{-j\omega t}

U(t) = \frac{1}{2} U [\cos(\theta) - j\sin(\theta)] e^{-j\omega t} + \frac{1}{2}U[\cos(\theta) + j\sin(\theta)] e^{j\omega t}

Note that this representation is exact, and holds at any instant t of time. Similarly, we write the unknown current (again exactly) as

I(t) = \overline{c} e^{-j\omega t} + c e^{j\omega t}

where c is the unknown Fourier coefficient to be solved, and \overline{c} its complex conjugate. Then, we plug both these expressions U(t), I(t) into the Ohm’s law

R I(t) = U(t)

and apply the Galerkin’s approach: we multiply both sides of the equation with \frac{1}{T} e^{-j\omega t}, and integrate over the period T (i.e. the period corresponding to \omega).

Thus, we have to calculate the integrals

\int\limits_{0}^{T} \frac{1}{T} e^{-j\omega t} e^{j\omega t} \mathrm{d}t and
\int\limits_{0}^{T} \frac{1}{T} e^{-j\omega t} e^{-j\omega t} \mathrm{d}t,

the former of which equals one, while the latter is of course zero. This leaves us with the equation

R c = \frac{1}{2} U [\cos(\theta) + j\sin(\theta)],

from which the positive Fourier coefficient c is easy to solve. Then, we just substitute it back to Fourier series representation of I

I(t) = \overline{c} e^{-j\omega t} + c e^{j\omega t},

which, after some simplifications, yields the same expression for the current

I(t) = \frac{U(t)}{R}

that the pure-phasor approach did.


Now, if we take a closer look at the equation

R c = \frac{1}{2} U [\cos(\theta) + j\sin(\theta)],

we see that the right-hand side can we re-written as

\frac{1}{2} U e^{j\theta}

which of course equals \frac{1}{2} \underline{U}. In other words, we can say that we have a voltage phasor even here. On the other hand, it turns out that when we are doing phasor analysis, we are in reality solving for the positive Fourier coefficient of the unknown quantity, multiplied by 2.

Due to the slight differences in post-processing, both methods obviously yield the same current I(t).


So, what’s the point?

Well, it’s always nice to understand things better. Like, that phasors actually do have direct 1-on-1 correspondence to physical time-dependent quantities. That relationship has simply been integrated into invisibility, an operation which also makes half of the terms (the negative components) vanish.

Moreover, it exposes some important limitations in phasor analysis. Namely, if – in our example – the resistance R was time-dependent R(t) rather than constant, the negative components e^{-j\omega t} might not vanish under the integration at all. If standard phasor analysis, with only those positive components, was applied in this case, the results might of course be incorrect. The same thing could probably happen in non-linear circuits as well. This could lead to some nasty consequences, like repeatedly banging your head against the wall for a whole day.

So, it’s not a good idea to try ramming a square peg into a round hole. And to avoid that, you have to be able to realize that a square peg is actually what you have.


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A new angle on phasor analysis

2 thoughts on “A new angle on phasor analysis

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