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

Problem 6.1.2: A plane wave in an arbitrary direction

Problem 6.1.2

A plane wave in an arbitrary direction

A uniform plane wave in free space has the electric field phasor

E (r) = [(8 a_{x} - 6 a_{z}) + j 10 a_{y}] e^{- j π (2.4 x + 3.2 z)}

in V/m, with $x$ and $z$ in m.

Find the direction of propagation, the wavelength and the frequency.
Check that $E$ is transverse. Find the wavelengths $λ_{x}$ , $λ_{z}$ measured along the axes.
Find $H (r)$ .
Determine the polarisation and its handedness.
Find the time-average Poynting vector and the power crossing $1 m^{2}$ of the plane $z = 0$ .

Try it yourself first.

Strategy

Everything is in the exponent and in the vector $E_{0}$ . The exponent is $- j k \cdot r$ , so its coefficients are the components of $k$ . Split $E_{0} = E_{r} + j E_{i}$ for the polarisation.

a) Wave vector

$k \cdot r = 2.4 π x + 3.2 π z$ , hence

k k a_{k} λ f = π (2.4 a_{x} + 3.2 a_{z}), = π 2. 4^{2} + 3. 2^{2} = 4 π rad/m, = \frac{k}{k} = 0.6 a_{x} + 0.8 a_{z}, = \frac{2 π}{k} = 0.5 m, = \frac{c}{λ} = \frac{3 \times 1 0 ^{8}}{0.5} = 600 MHz .

The direction makes $arctan (0.6/0.8) = 36. 9^{\circ}$ with the $z$ -axis.

b) Transversality

$a_{k} \cdot E_{0} = 0.6 (8) + 0.8 (- 6) + 0 \cdot (j 10) = 0$ . Along the axes the phase repeats every

λ_{x} λ_{z} = \frac{2 π}{k _{x}} = \frac{2 π}{2.4 π} = 0.833 m, = \frac{2 π}{k _{z}} = \frac{2 π}{3.2 π} = 0.625 m,

both longer than $λ = 0.5 m$ , since $λ_{x} = λ /0.6$ and $λ_{z} = λ /0.8$ .

c) Magnetic field

$H = \frac{1}{η _{0}} a_{k} \times E$ . The two cross products needed are

a_{k} \times (8 a_{x} - 6 a_{z}) a_{k} \times a_{y} = 10 a_{y}, = 0.6 a_{z} - 0.8 a_{x},

Answer

H (r) = \frac{10}{120 π} [a_{y} - j (0.8 a_{x} - 0.6 a_{z})] \times e^{- j π (2.4 x + 3.2 z)} A/m

Each of the two parts has the amplitude $10/ η_{0} = 26.5 mA/m$ .

d) Polarisation

$E_{r} = 8 a_{x} - 6 a_{z}$ and $E_{i} = 10 a_{y}$ both have the length $10 V/m$ , they are perpendicular, and they are $9 0^{\circ}$ apart in time: the polarisation is circular. With the handedness test,

E (0) \times E (T /4) = (8 a_{x} - 6 a_{z}) \times (- 10 a_{y}) = - 80 a_{z} - 60 a_{x} = - 100 a_{k},

negative: the field turns against the right-hand sense about $a_{k}$ .

Answer

(d) left-hand circular polarisation (LHCP)

e) Power

$∣ E_{0} ∣^{2} = 8^{2} + 6^{2} + 1 0^{2} = 200 V^{2} / m^{2}$ , so

P_{a v} = \frac{∣ E _{0} ∣ ^{2}}{2 η _{0}} a_{k} = \frac{200}{240 π} a_{k} = 0.265 a_{k} W/ m^{2} .

Only the normal component crosses the plane $z = 0$ :

P = P_{a v} \cdot a_{z} \times (1 m^{2}) = 0.265 \times 0.8 = 0.212 W .

Answer

f = 600 MHz, a_{k} = 0.6 a_{x} + 0.8 a_{z}, LHCP, P_{a v} = 0.265 W/ m^{2}, P = 0.212 W

Check

$P_{a v}$ directly from the phasors: with $a_{1} = 0.8 a_{x} - 0.6 a_{z}$ , $E_{0} = 10 (a_{1} + j a_{y})$ and $H_{0}^{*} = \frac{10}{η _{0}} (a_{y} + j a_{1})$ , so

\frac{1}{2} Re {E \times H^{*}} = \frac{100}{2 η _{0}} Re {a_{1} \times a_{y} + j^{2} a_{y} \times a_{1}} = \frac{100}{η _{0}} a_{1} \times a_{y} = \frac{100}{η _{0}} a_{k},

the same $0.265 W/ m^{2}$ , since $a_{1} \times a_{y} = a_{k}$ .

**Figure 6.1.25.** Problem 6.1.2. Left: the $x z$ -plane at $t = 0$ , with wavefronts $0.5 m$ apart and normal to $a_{k}$ . $E (0) = 8 a_{x} - 6 a_{z}$ lies in the plane and $H (0)$ points out of it. Along $z$ the phase repeats only every $λ_{z} = 0.625 m$ , and only $cos 36. 9^{\circ} = 0.8$ of the power density crosses a plane $z =$ const. Right: the transverse plane seen with the wave coming towards the viewer ( $a_{1} \times a_{y} = a_{k}$ , $a_{1} = 0.8 a_{x} - 0.6 a_{z}$ ). The field turns from $a_{1}$ to $- a_{y}$ in a quarter period, clockwise: left-handed.