27  Phasors and alternating waveforms

A rotating complex number traces a sine wave

Two AC voltages arrive at a junction in a circuit. They oscillate at the same frequency but at different phases — one peaks a little before the other. What is the combined voltage?

You could set up a trigonometric identity and expand. It works, but it is laborious. Or you could represent each voltage as a complex number, add them in two lines, and read off the result. That shortcut — replacing a trigonometric addition with a complex addition — is the phasor method, and it is how electrical engineers have analysed circuits since the 1890s.

27.1 What this chapter helps you do

Symbols to keep handy

These are the bits of notation you'll see a lot. If a line of symbols feels like a fence, read it out loud once, then keep going.

  • Z = R + jX: complex impedance: R is resistance, X is reactance

  • A: a phasor with amplitude A and phase angle phi

Definitions to keep handy

These are the words we keep coming back to. If one feels slippery, come back here and steady it before you push on.

  • phasor: A complex number representing a sinusoidal quantity by its amplitude and phase angle.

  • amplitude: The peak value of a sinusoidal wave — the distance from zero to the crest.

  • phase angle: The horizontal shift of a wave, measured in radians or degrees from a reference.

  • impedance: The complex-valued generalisation of resistance for AC circuits, written Z = R + jX.

Here is the main move this chapter is making, in plain terms. You do not need to be fast. You just need to keep the thread.

  • Coming in: You know that complex numbers live in a plane, and that multiplying by i rotates a number by 90 degrees. You also know that sine and cosine describe oscillation. These two ideas are the same idea.

  • Leaving with: A phasor is a complex number that rotates. As it spins, its real part traces a cosine wave and its imaginary part traces a sine wave. This means that adding sinusoidal signals — the core problem of AC circuit analysis — is equivalent to adding complex numbers, which is far simpler. Engineers use j (not i) for the imaginary unit; the notation is different, the mathematics is identical.

27.2 Rotation and oscillation are the same thing

Here is the connection at the heart of this chapter. Take Euler’s formula:

e^{j\theta} = \cos\theta + j\sin\theta

Read it as: the complex number e^{j\theta} — that is, e to the power of j times theta — sits on the unit circle at angle \theta from the positive real axis. Its real part is \cos\theta and its imaginary part is \sin\theta.

Now replace \theta with \omega t — that is, angle increasing at a constant rate \omega (called the angular frequency, read as omega) over time t:

e^{j\omega t} = \cos(\omega t) + j\sin(\omega t)

As t increases, the angle \omega t increases, and the point e^{j\omega t} rotates steadily around the unit circle. Its real part traces a cosine wave. Its imaginary part traces a sine wave. The rotation and the oscillation are the same object viewed two different ways.

This is not a trick or an approximation. It is an exact identity. A phasor is a complex number that rotates in exactly this way. The wave you observe — say, the voltage across a component in a circuit — is the real part of the phasor as it spins.

One more notation to be clear about: engineers write j where mathematicians write i for the imaginary unit (to avoid confusion with current, which is also called i). The mathematics is identical.

Drag the angle slider. Watch the dot on each wave move exactly as the phasor tip moves. They are describing the same thing.

27.3 What the notation is saying

A sinusoidal signal takes the form:

v(t) = A\sin(\omega t + \phi)

Read it as: “v of t equals A times sine of omega t plus phi.” Four pieces:

  • A — the amplitude (ay): the peak value, the distance from zero to the crest
  • \omega — the angular frequency (omega, in radians per second): how fast the phasor rotates; \omega = 2\pi f where f is frequency in hertz
  • t — time in seconds
  • \phi — the phase angle (phi, in radians or degrees): how much the wave is shifted from a reference wave

The period — the time for one complete cycle — is T = 2\pi/\omega.

Now, the phasor. When we write the full rotating complex number Ae^{j(\omega t + \phi)}, the e^{j\omega t} part is the same for every signal in a circuit operating at frequency \omega. It is just the shared rotation. The part that distinguishes one signal from another is Ae^{j\phi} — the amplitude and the phase shift.

So we define the phasor for the signal A\sin(\omega t + \phi) as:

\mathbf{V} = A\angle\phi

Read that as: “A at angle phi” — a complex number with magnitude A and argument \phi. The \angle symbol is shorthand for polar form: A\angle\phi = A(\cos\phi + j\sin\phi).

When we need to add sinusoidal signals, we convert each to a phasor, add the phasors (as complex numbers), and convert the result back to a sinusoid. The e^{j\omega t} rotation is implicit — it is the same for all signals and can be dropped during the calculation, then restored at the end.

Why we can drop e^{j\omega t}

Every signal in the circuit rotates at the same angular frequency \omega. The relative angles between phasors do not change during rotation — only the absolute positions change. So for addition and subtraction, the rotating part is irrelevant. We work with the amplitudes and phase angles alone, then restore the rotation at the end when we write the answer as a waveform.

27.4 Adding sinusoidal waveforms — the key technique

Suppose you need to add these two voltages:

v_1 = 120\sin(100\pi t) \qquad v_2 = 80\sin(100\pi t + 60°)

Both oscillate at the same frequency (\omega = 100\pi rad/s). The question is: what is v_1 + v_2?

The trigonometric route. You can expand \sin(100\pi t + 60°) using the angle-addition formula, collect terms, and simplify. It takes half a page and is error-prone.

The phasor route. Convert each signal to a phasor:

\mathbf{V}_1 = 120\angle 0° \qquad \mathbf{V}_2 = 80\angle 60°

Convert to Cartesian form (\mathbf{V} = A\cos\phi + jA\sin\phi):

\mathbf{V}_1 = 120\cos 0° + j\cdot 120\sin 0° = 120 + 0j

\mathbf{V}_2 = 80\cos 60° + j\cdot 80\sin 60° = 80(0.5) + j\cdot 80(0.866) = 40 + 69.28j

Add them component by component:

\mathbf{V} = (120 + 40) + j(0 + 69.28) = 160 + 69.28j

Convert back to polar form:

|\mathbf{V}| = \sqrt{160^2 + 69.28^2} = \sqrt{25600 + 4799.7} = \sqrt{30399.7} \approx 174.4

\phi = \arctan\!\left(\frac{69.28}{160}\right) = \arctan(0.433) \approx 23.4° = 0.408 \text{ rad}

The combined voltage is:

v = 174.4\sin(100\pi t + 0.408) \text{ V}

Two lines of arithmetic instead of half a page.

Adjust the amplitudes and phases. Notice that the resultant phasor (green arrow) is always the vector sum of the other two, and the green waveform always matches.

27.4.1 The a\sin\omega t + b\cos\omega t form

A common variation: you have one sine and one cosine at the same frequency, and you want to write their sum as a single sinusoid.

a\sin\omega t + b\cos\omega t = R\sin(\omega t + \phi)

Expand the right side using the angle-addition formula:

R\sin(\omega t + \phi) = R\cos\phi\sin\omega t + R\sin\phi\cos\omega t

Match coefficients:

R\cos\phi = a \qquad R\sin\phi = b

Square and add: R^2\cos^2\phi + R^2\sin^2\phi = a^2 + b^2, so:

R = \sqrt{a^2 + b^2}

Divide: \tan\phi = b/a, so \phi = \arctan(b/a).

This is exactly the polar-form conversion of the complex number a + jb. The phasor a + jb has magnitude \sqrt{a^2+b^2} and argument \arctan(b/a) — which confirms that the combination formula is just a restatement of complex arithmetic.

27.5 Worked examples

Example 1: phasor addition (engineering context)

Two AC currents enter a node: i_1 = 15\sin(\omega t) A and i_2 = 10\sin(\omega t + 45°) A. Find the total current.

Write the phasors:

\mathbf{I}_1 = 15\angle 0° \qquad \mathbf{I}_2 = 10\angle 45°

Convert to Cartesian:

\mathbf{I}_1 = 15 + 0j

\mathbf{I}_2 = 10\cos 45° + j\cdot 10\sin 45° = 7.071 + 7.071j

Add:

\mathbf{I} = 22.071 + 7.071j

Convert to polar:

|\mathbf{I}| = \sqrt{22.071^2 + 7.071^2} = \sqrt{487.1 + 50.0} = \sqrt{537.1} \approx 23.18 \text{ A}

\phi = \arctan\!\left(\frac{7.071}{22.071}\right) = \arctan(0.3204) \approx 17.8°

Total current: i = 23.18\sin(\omega t + 17.8°) A.

Check: the magnitude of the sum (23.18) is less than 15 + 10 = 25 because the two currents are not in phase — they partially cancel.

Example 2: the a\sin + b\cos form (signal processing)

Simplify 3\sin\omega t + 4\cos\omega t to the form R\sin(\omega t + \phi).

R = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5

\phi = \arctan\!\left(\frac{4}{3}\right) = \arctan(1.333) \approx 53.1°

So 3\sin\omega t + 4\cos\omega t = 5\sin(\omega t + 53.1°).

Check: at \omega t = 0: left side = 3(0) + 4(1) = 4. Right side = 5\sin(53.1°) = 5(0.8) = 4. Yes.

This is a 3-4-5 right triangle in the phasor plane — the amplitude 5 is the hypotenuse.

Example 3: all four methods for adding two phasors

Add \mathbf{V}_1 = 100\angle 0° and \mathbf{V}_2 = 60\angle 30°.

Method 1 — Graphical. Draw \mathbf{V}_1 as a horizontal arrow of length 100. Draw \mathbf{V}_2 from the tip of \mathbf{V}_1 at 30° above horizontal, length 60. Measure the resultant arrow from origin to the final tip. (Requires accurate drawing; gives approximate answer.)

Method 2 — Sine and cosine rules. The angle between the two phasors is 30°. The phasor triangle has sides 100, 60, and unknown, with included angle 180° - 30° = 150° (exterior). By the cosine rule:

|\mathbf{V}|^2 = 100^2 + 60^2 - 2(100)(60)\cos 150° = 10000 + 3600 - 12000(-0.866) = 13600 + 10392 = 23992

|\mathbf{V}| = \sqrt{23992} \approx 154.9 \text{ V}

For the angle, use the sine rule to find how much above zero the resultant sits.

Method 3 — Resolving components (Cartesian). This is the method we used above:

\mathbf{V}_1 = 100 + 0j \qquad \mathbf{V}_2 = 60\cos 30° + j\cdot 60\sin 30° = 51.96 + 30j

\mathbf{V} = 151.96 + 30j

|\mathbf{V}| = \sqrt{151.96^2 + 30^2} = \sqrt{23092 + 900} = \sqrt{23992} \approx 154.9 \text{ V}

\phi = \arctan(30/151.96) \approx 11.2°

Method 4 — Complex arithmetic directly. Same as Method 3, but written explicitly as complex number addition. This is the preferred method for more than two phasors.

All four methods give |\mathbf{V}| \approx 154.9 V at approximately 11.2°. Methods 3 and 4 are fastest when working with three or more phasors. Methods 1 and 2 help build geometric intuition.

27.6 Complex impedance

In a DC circuit, resistance R (in ohms) relates voltage and current by Ohm’s law: V = IR. Resistance is a single real number — it scales the current but does not shift the timing.

In an AC circuit, components affect both the amplitude and the phase of the current. The complex-valued generalisation of resistance is called impedance, written Z = R + jX — read as “Z equals R plus j times X.” The symbol j is the imaginary unit in engineering notation. R is the resistance (real part) and X is the reactance (ree-AK-tense, imaginary part).

Three types of component:

Resistor. Impedance Z_R = R. Purely real — no phase shift. A resistor opposes current equally at all frequencies.

Inductor. Impedance Z_L = j\omega L where L is inductance in henries. Voltage leads current by 90°. At higher frequencies, an inductor has higher impedance — it resists change in current.

Capacitor. Impedance Z_C = \dfrac{1}{j\omega C} = \dfrac{-j}{\omega C} where C is capacitance in farads. Current leads voltage by 90°. At higher frequencies, a capacitor has lower impedance — it passes high-frequency signals easily.

For series components, total impedance is:

Z = Z_R + Z_L + Z_C = R + j\!\left(\omega L - \frac{1}{\omega C}\right) = R + jX

The magnitude |Z| = \sqrt{R^2 + X^2} tells you the overall opposition to current. The argument \phi = \arctan(X/R) tells you the phase shift between voltage and current. The power factor — how efficiently the circuit uses electrical power — is \cos\phi.

Worked example — RLC series circuit.

Given: R = 100\ \Omega, L = 0.1\ \text{H}, C = 100\ \mu\text{F}, \omega = 100\ \text{rad/s}.

Compute the reactances:

X_L = \omega L = 100 \times 0.1 = 10\ \Omega

X_C = \frac{1}{\omega C} = \frac{1}{100 \times 100 \times 10^{-6}} = \frac{1}{0.01} = 100\ \Omega

Total impedance:

Z = R + j(X_L - X_C) = 100 + j(10 - 100) = 100 - 90j\ \Omega

Magnitude:

|Z| = \sqrt{100^2 + 90^2} = \sqrt{10000 + 8100} = \sqrt{18100} \approx 134.5\ \Omega

Phase angle:

\phi = \arctan\!\left(\frac{-90}{100}\right) = \arctan(-0.9) \approx -42.0°

Negative angle means capacitive dominates at this frequency — current leads voltage. Power factor:

\cos\phi = \cos(-42.0°) \approx 0.743

Try setting the sliders above to these values to confirm the readout.

27.7 Where this goes

This chapter is the gateway to the frequency-domain methods that run through most of Vol 7.

Fourier series (Vol 7, Fourier and PDEs section) extends what you have done here: instead of adding two or three phasors at one frequency, it decomposes an arbitrary periodic signal into a sum of phasors at many frequencies — the fundamental and its harmonics. The coefficient of each phasor tells you how much of that frequency is present. Every concept in Fourier analysis rests on the rotating-phasor picture you have just built.

Laplace transforms (Vol 7) take the phasor method further: instead of assuming sinusoidal steady-state signals (as phasors do), the Laplace transform handles any signal — including switch-on transients, impulses, and arbitrary time functions — by working with complex frequencies s = \sigma + j\omega. The phasor you know (j\omega) is a special case of s with \sigma = 0. Understanding phasors first makes the Laplace transform feel like a natural extension rather than an abstract tool.

Where this shows up

  • Electrical engineering: every AC circuit calculation — power factor correction, filter design, motor drives, transmission lines — uses phasors. This chapter is how ECE 202 begins.
  • Acoustics: sound waves from two speakers interfere constructively or destructively depending on the phase difference. The loudness at any point is the magnitude of the phasor sum.
  • Optics and physics: thin-film interference, diffraction, and quantum mechanical probability amplitudes all use phasor addition. The double-slit pattern is a phasor diagram.
  • Signal processing: digital filters are specified and analysed in the frequency domain using the complex frequency response H(j\omega) — the impedance concept applied to signals rather than circuits.
  • Structural engineering: vibrating structures have resonant frequencies where the impedance analogue (dynamic stiffness) passes through zero. The mathematics is identical to RLC resonance.

27.8 Exercises

These are puzzles. Each one has a clean numerical answer, but the interesting step is always the translation: turning the waveform description into a phasor, doing the arithmetic in the complex plane, and reading the answer back out.

1. Express v = 50\sin(200\pi t + 0.4) in phasor form, then state the amplitude, angular frequency, frequency in Hz, and phase angle.

2. Add the two currents i_1 = 12\sin(\omega t) A and i_2 = 8\sin(\omega t + 30°) A. Express the result as a single sinusoidal current.

3. Express 3\sin\omega t + 4\cos\omega t in the form R\sin(\omega t + \phi).

4. An RLC series circuit has R = 50\ \Omega, L = 0.2\ \text{H}, C = 200\ \mu\text{F} at \omega = 50\ \text{rad/s}. Find the total impedance Z, its magnitude, phase angle, and power factor. Is the circuit inductive or capacitive at this frequency?

5. Two parallel branches of an AC circuit carry currents \mathbf{I}_1 = 5\angle 0° A and \mathbf{I}_2 = 3\angle{-45°} A. Find the total supply current phasor and express it as a sinusoid (leave \omega symbolic).