Electromagnetic Fields and Waves: Notes, Solved Problems and Interactive Lab

Calculating with sines and cosines is tedious: every derivative turns one into the other, and adding two of them requires trigonometric identities. Euler's formula $e^{j θ} = cos θ + j sin θ$ offers a way out, because $cos θ = Re {e^{j θ}}$ . The wave of Eq. (5.6.1) can then be written as

u (r; t) = A (r) cos [ω t - ϕ (r)] = Re {A (r) e^{j [ω t - ϕ (r)]}} = Re {A (r) e^{- j ϕ (r)} e^{j ω t}} .

The factor $e^{j ω t}$ is common to every quantity at the same frequency. Everything that distinguishes one field from another, its amplitude and its phase, sits in the complex function in front of it.

Definition · Phasor

The phasor of the time-harmonic quantity $u (r; t) = A (r) cos [ω t - ϕ (r)]$ is the complex function of position

U (r) = A (r) e^{- j ϕ (r)}, u (r; t) = Re {U (r) e^{j ω t}} . (5.6.7)

It keeps the amplitude, $∣ U ∣ = A$ , and the phase, $ar g U = - ϕ$ . The time factor $e^{j ω t}$ is suppressed and must be put back to return to the time domain.

Now differentiate with respect to time. The phasor $U (r)$ does not depend on $t$ , and the real part of a complex function can be taken before or after the differentiation, since $Re {z} = \frac{1}{2} (z + z^{*})$ is a linear operation:

\frac{\partial u ( r ; t )}{\partial t} = Re {U (r) \frac{\partial}{\partial t} e^{j ω t}} = Re {j ω U (r) e^{j ω t}} . (5.6.8)

The derivative of $u$ is again a time-harmonic quantity at the same frequency, and its phasor is $j ω U$ .

The phasor rule

u (r; t) \frac{\partial u ( r ; t )}{\partial t} phasor U (r), phasor j ω U (r) . (5.6.9)

In plain words

A phasor is a snapshot of a rotating arrow, taken at $t = 0$ . Differentiating in time multiplies the arrow by $j ω$ : it stretches it by $ω$ and turns it forward by $9 0^{\circ}$ , because the derivative of a cosine peaks a quarter period earlier than the cosine itself.

Figure 5.6.6 gives the picture behind the name. The complex number $U e^{j ω t}$ is an arrow of length $A$ that turns anticlockwise at $ω rad/s$ . Its shadow on the real axis is the physical quantity $u (t)$ , which therefore swings between $- A$ and $+ A$ as a cosine. The phasor $U$ is the position of the arrow at $t = 0$ : it fixes where the arrow starts.

(b) Its shadow on the real axis. — (a) The rotating arrow.

Watch the arrow's shadow trace the cosine below, here for the wave $10 cos (6 π \times 1 0^{8} t - 3 π z) V/m$ of Problem 5.6.1. Then drag the observation point along the middle plot and see the phasor turn back by $β = 3 π rad$ per metre.

InteractiveLoading simulation…

Table 5.6.2 collects the rules used in practice. The second line follows from $sin θ = cos (θ - π /2)$ : a sine is a cosine delayed by a quarter period, and its phasor carries the factor $e^{- j π /2} = - j$ . Repeating Eq. (5.6.8) gives the second derivative, and reading it backwards gives the integral, provided the integration constant is chosen zero, the only choice that keeps the integral time-harmonic (Home exercise 5.6.1). A vector field is handled component by component: $E (r; t) = Re {E (r) e^{j ω t}}$ , where the phasor $E (r) = E_{x} a_{x} + E_{y} a_{y} + E_{z} a_{z}$ has three complex components, each with its own amplitude and phase. In this book the phasor and the time-domain field carry the same symbol, and the argument tells them apart: $E (r)$ or $E (z)$ is a phasor, $E (r; t)$ or $E (z; t)$ the real field.

Table 5.6.2. The phasor rules, with $u (r; t) = Re {U (r) e^{j ω t}}$ .

Time domain	Phasor
$A cos (ω t - ϕ)$	$A e^{- j ϕ}$
$A sin (ω t - ϕ)$	$- j A e^{- j ϕ}$
$\partial u / \partial t$	$j ω U$
$\partial^{2} u / \partial t^{2}$	$- ω^{2} U$
$\int u d t$	$U / (j ω)$
$E (r; t) = Re {E (r) e^{j ω t}}$	$E (r)$ , three complex components

Common mistake · Forgetting that the reference is a cosine

The rule $u = Re {U e^{j ω t}}$ uses a cosine reference. A field $0.8 sin (ω t - 4 x)$ does not have the phasor $0.8 e^{- j 4 x}$ : that phasor describes $0.8 cos (ω t - 4 x)$ , a field a quarter period early. Write the sine as a cosine first, $sin θ = cos (θ - π /2)$ , and the phasor is $- 0.8 j e^{- j 4 x}$ (Problem 5.6.14). The factor $- j$ is the sine.

Common mistake · Multiplying phasors

Phasors add, subtract and differentiate exactly like the fields they represent, because these operations are linear. They do not multiply: $Re {a} Re {b} \neq = Re {ab}$ . The product of two time-harmonic fields contains a constant and a term at $2 ω$ , and has no phasor at $ω$ . Power, the product of $E$ and $H$ , needs its own rule, $P_{a v} = \frac{1}{2} Re {E \times H^{*}}$ , derived in Section 6.1.

Quick check

Write the phasor of $H (z; t) = 2 cos (ω t + 3 z + 6 0^{\circ}) a_{y} A/m$ . Which way does the wave travel?

$H (z) = 2 e^{j 6 0^{\circ}} e^{+ j 3 z} a_{y} A/m$ . The term $ω t$ and the term $3 z$ carry the same sign, so a crest $ω t + 3 z =$ const moves towards $- z$ as $t$ grows. A phasor $e^{+ j β z}$ marks a wave towards $- z$ , a phasor $e^{- j β z}$ a wave towards $+ z$ .

5.6.4 Phasors