Phasors and the Unit Circle

 

Pretty much every science teacher/professor I’ve ever had told me the same thing about science education:

“Everything I tell my students is partially a lie.  It’s a simple lie, designed to help them understand the next lie, which will be more complex.  And once you add enough levels of complexity, you eventually get to the truth.  But once you see the truth, you’ll probably say, ‘Hey, that looks like what my old teachers taught me!’”

Today we’re going to confront a simple lie you learned in high school, and replace it with something more complicated.  The topic today is the unit circle:

Personally, I don’t see that many uses for the unit circle, aside from teaching students how to deal with trig identities and right triangles.  But the unit circle become much more interesting when you use it to describe imaginary numbers.  I’m getting ahead of myself though—let’s start with some basics:

Euler’s equation (pronounced “oiler’s”) is considered to be one of the most beautiful and famous mathematical formulas ever discovered.  In it, Euler writes,

$$e^{j \theta}=\cos (\theta)+j \sin (\theta)$$

I’m using \(j\) as the imaginary number \(j=\sqrt{-1}\).  It’s common to use \(j\) instead of \(i\) in the engineering world because \(i\) is already reserved for other values, namely current.  Anyway, back to the equation…

Euler’s equation is supposed to be incredible and awe-inspiring, but the first time I saw it I was just confused.  The equation isn’t exactly self-explanatory.  But then one day, I saw the equation visually depicted on a graph:

This new picture tells quite a story.  Obviously, it’s a unit circle, but it’s more than that.  First of all, it shows us that it’s okay to graph things on an imaginary axis.  Note that the y-axis in the picture above is completely imaginary—any vector that lies outside of the real axis contains an imaginary component; we would call it a “complex vector”.  Another thing that the graph reveals is that we can conserve all the implications of the standard unit circle in our imaginary unit circle. Any complex vector \(a+b j\) can be treated as a right triangle with a magnitude and an angle \(\theta\).

Now I know what you’re thinking: “Jonathan, this is all well and good, but how is this related to my ENG17 class?”  Great question; thanks for participating.  The answer is that we can define any component of a circuit as having a complex resistance—we call this value impedance.  Like any complex vector, impedance has a real component, which we call resistance.  It also has an imaginary component, which we call reactance.  Resistors get their name from having only real impedance values; they are made entirely out of resistance.  Meanwhile, the other circuit components you know and love—capacitors and inductors—are considered “reactive” circuit components, because they only contain imaginary impedance.  When you put reactive components in series or in parallel with resistors, you can combine their impedance values to get complex impedances.

That was a whole lot of words.  Let’s draw a picture depicting some of that:

image39.jpg

Before we can do anything with this new knowledge, we have to know one more thing: how much reactance does a capacitor or inductor have?  The reactance of an inductor is \(\omega L\), where \(L\) is the inductance of the inductor, measured in Henrys.  Meanwhile, the reactance of a capacitor is \(\frac{-1}{\omega C}\), where \(C\) is the capacitance of the capacitor, measured in Farads.  In both equations, \(\omega\) is the frequency at which the circuit oscillates.

If the idea of \(\omega\) confuses you, don’t panic.  Realize that most signals in the real world come in the form of electromagnetic waves (like the radio waves in your cell phone), and the frequency of one of these signals is represented by \(\omega\).  Most real-world circuits therefore function on frequencies defined by their voltage or current source.

One more quick side note: I wish I had a good explanation for why reactance values are \(\omega L\) and \(\frac{-1}{\omega C}\), but I don’t.  Apparently, those values are a consequence of the geometry and physics principles that make inductors and capacitors function the way that they do.  I know, I wasn’t satisfied by that explanation either—you’ll just have to memorize those values until someone can explain it better.

Okay, that was a lot of explaining stuff.  Now it’s time for me to prove that memorizing reactance equations and complex impedance stuff is worth all your trouble.  Let’s use what we’ve learned to solve a tough circuit.  Today’s question:

Consider the following circuit.  Find the current, \(i(j \omega)\), in terms of the voltage \(v(j \omega)\), and the values \(\omega\), \(R\), \(C\), and \(L\).

I am intentionally making this problem more difficult than something you would see on a test.  In a testing situation, you would get concrete values for \(\omega\), \(R\), \(C\), and \(L\). But we’re hardcore champions, so I know we can solve this one together.  Time to jump right in…

The first step is always my favorite.  Why would you deal with an ugly circuit like the original one, when you could combine all the components into a black box?  We say that the black box has some complex impedance, \(Z_{e q}\).  The value of \(Z_{e q}\) is equal to the impedance of the inductor added to the impedance of the parallel branch \(C \| R\).  Let’s write that out.

$$Z_{e q}=Z(L)+Z(C \| R)$$

Let’s plug in the values of reactance that we just learned:

$$Z_{e q}=j \omega L+\frac{1}{\frac{1}{R}+\frac{1}{\frac{-j}{\omega C}}}$$

If that went a bit fast for you, feel free to slow down and redo that step more slowly.  Start by finding just the value of \(Z(C \| R)\), the same way you would find the resistance of two resistors in parallel.  Then add \(Z(L)\). Let’s simplify the ugly fraction up there:

$$Z_{e q}=j \omega L+\frac{1}{\frac{1}{R}+j \omega C}$$

Where did the negative sign go?  I multiplied the fraction by \(\frac{j}{j}\). Notice that \(j^{2}=-1\) because it’s the imaginary number.  We continue to simplify…

$$Z_{e q}=j \omega L+\frac{1}{\left(\frac{1+j \omega C R}{R}\right)}$$

$$Z_{e q}=j \omega L+\frac{R}{1+j \omega C R}$$

No need to be afraid of these letters—they’re just constants, after all.  The next step is to remove the imaginary number \(j\) from the fraction’s denominator.  No one likes \(j\) in a denominator because it doesn’t mean anything there.  However, if we can move it to the numerator, our equation will be in the form of a complex number: \(a+b j\).  To facilitate this, we multiply the fraction by the denominator’s complex conjugate.

$$Z_{e q}=j \omega L+\left(\frac{R}{1+j \omega C R} * \frac{1-j \omega C R}{1-j \omega C R}\right)$$

$$Z_{e q}=j \omega L+\left(\frac{R-j \omega C R^{2}}{1+\omega^{2} C^{2} R^{2}}\right)$$

This is the point in the problem at which we would normally plug in all our values for \(\omega\), \(R\), \(C\), and \(L\). Once we do that, our equation will immediately simplify to the form of a complex number.  Then, we can simply invoke Ohm’s law, \(i=\frac{v}{Z}\), to find the current as some complex number, \(a+b j\). Obviously, the values of \(a\) and \(b\) depend on \(\omega\), \(R\), \(C\), and \(L\), so we can’t calculate them today.  But we can write out our final answer to the question we were asked:

$$\boldsymbol{i}(j \omega)=\frac{\boldsymbol{v}(j \omega)}{j \omega L+\left(\frac{R-j \omega C R^{2}}{1+\omega^{2} C^{2} R^{2}}\right)}$$

When you’re given real values for this problem, you’ll be able to plug everything in, remove \(j\) from the denominator again, and get the complex number form of the current: \(a + bj \). Normally, however, no one wants to see an imaginary value for current.  So instead of writing \(i = a + bj\), we write the current as a phasor.  A phasor is basically a glorified use of polar coordinates.  Remember the reactance/resistance plane from earlier?  Any complex vector on that plane can be described by a magnitude: \(r=\sqrt{a^{2}+b^{2}}\), and a direction: \(\theta=\arctan \left(\frac{b}{a}\right)\).  So whatever \(a\) and \(b\) turn out to be, your final answer can be written as: \(i=\sqrt{a^{2}+b^{2}} \angle \arctan \left(\frac{b}{a}\right)\).  The weird right-triangle shape is just a symbol we use to refer to the “phase angle” \(\theta\).  Thus, the name “phasor”.

I’ll admit that I raced through that problem pretty quick, so don’t feel bad if you have to review it again, or try a similar problem with actual numbers in it.  But regardless of how well you followed the example, two major things about circuits should now become clear: Firstly, that complex impedance and phasors are entirely dependent on the use of the imaginary plane, so don’t forget the rules governing trigonometry in such a plane.  And second of all, all the tricks you’ve ever used to solve purely-resistive circuits work with complex impedance values.  All you have to do is write inductors and capacitors as black boxes with imaginary impedance, and then you’re free to use Ohm’s law, Kirchhoff’s laws, and anything else to your heart’s content.  Just don’t forget to replace resistance, R, with impedance, Z, in your calculations.