Exercise 6.1.17: Two crossing waves: a waveguide in disguise

About the further exercises of Section 6.1

Seventeen further exercises in the spirit of the problems, the home exercises and the worked examples, graded from easy to harder within each group. Exercises 6.1.1–6.1.4 are about uniform plane waves, $k$ , $λ$ , $η$ and the triad, Exercises 6.1.5–6.1.6 about power, energy and loss, Exercises 6.1.7–6.1.10 about polarisation, Exercises 6.1.11–6.1.12 about the Doppler effect, and Exercises 6.1.13–6.1.17 mix power, exposure and polarisation. Besides the relations listed at the head of this subsection they use the lossy forms $k = ω μ ε_{c} = β - j α$ , $ε_{c} = ε (1 - j σ / ω ε)$ , $η = μ / ε_{c} = ∣ η ∣ e^{j θ_{η}}$ and $P_{a v} = \frac{∣ E _{0} ∣ ^{2}}{2∣ η ∣} e^{- 2 α a_{k} \cdot r} cos θ_{η} a_{k}$ (Exercise 6.1.6 only), the average energy density $w_{a v} = \frac{1}{4} (ε ∣ E ∣^{2} + μ ∣ H ∣^{2})$ , Faraday's law $E = - j ω Φ$ , and the polarisation match $∣ p_{w} \cdot p_{a}^{*} ∣^{2}$ . A field factor $e^{- α z}$ is $8.686 α z dB$ .

Exercise 6.1.17

Two crossing waves: a waveguide in disguise

Two $10 GHz$ plane waves of amplitude $E_{0} = 10 V/m$ travel in air, both with $E$ along $a_{z}$ , one along $a_{1} = cos θ a_{x} + sin θ a_{y}$ , the other along $a_{2} = cos θ a_{x} - sin θ a_{y}$ , $θ = 3 0^{\circ}$ , in opposite phase at the origin:

E = E_{0} a_{z} (e^{- j k_{0} a_{1} \cdot r} - e^{- j k_{0} a_{2} \cdot r}) .

(a) Write $E$ as a product of a function of $y$ and a wave along $x$ , and find $β_{x}$ , $λ_{x}$ and $v_{p}$ along $x$ . (b) On which planes is $E = 0$ ? (c) Find $H$ . (d) Find $P_{a v}$ and the power per metre of width (along $z$ ) between $y = 0$ and the first zero plane $y = b$ . (e) Find the energy per unit area of the $x z$ -cross-section of that strip and the speed $v_{E}$ of the energy along $x$ , and compare with $v_{p}$ . (f) Metal plates are put at $y = 0$ and $y = b$ . Why does nothing change? Below which frequency can this field not exist between them, and what are $θ$ , $v_{p}$ and $v_{E}$ at $6 GHz$ ?

Try it yourself first.

Strategy

Split $k \cdot r$ into its $x$ and $y$ parts. Both waves share $β_{x} = k_{0} cos θ$ , and their $y$ parts $e^{\mp j k_{y} y}$ , $k_{y} = k_{0} sin θ$ , combine into a standing pattern. $H$ follows wave by wave from $H = \frac{1}{η _{0}} a_{k} \times E$ .

$k_{0} = 2 π f / c = 209.4 rad/m$ , $k_{y} = k_{0} sin 3 0^{\circ} = 104.7 rad/m$ , $β_{x} = k_{0} cos 3 0^{\circ} = 181.4 rad/m$ , and

E = E_{0} a_{z} e^{- j β_{x} x} (e^{- j k_{y} y} - e^{j k_{y} y}) = - j 2 E_{0} sin (k_{y} y) e^{- j β_{x} x} a_{z} .

A wave along $x$ with $λ_{x} = 2 π / β_{x} = 3.46 cm$ and $v_{p} = ω / β_{x} = c / cos θ = 3.46 \times 1 0^{8} m/s$ , faster than light, and a standing pattern $sin k_{y} y$ across.

$sin k_{y} y = 0$ : $y = mπ / k_{y} = m λ_{0} / (2 sin θ) = m \times 3 cm$ . So $b = 3 cm$ .

$a_{1} \times a_{z} = sin θ a_{x} - cos θ a_{y}$ and $a_{2} \times (- a_{z}) = sin θ a_{x} + cos θ a_{y}$ . Adding the two waves,

H = \frac{2 E _{0}}{η _{0}} [sin θ cos (k_{y} y) a_{x} + j cos θ sin (k_{y} y) a_{y}] e^{- j β_{x} x},

with $2 E_{0} sin θ / η_{0} = 26.5 mA/m$ and $2 E_{0} cos θ / η_{0} = 45.9 mA/m$ . $H$ has a component along the direction of travel $x$ : this is not a uniform plane wave.

$E \times H^{*} = E_{z} H_{x}^{*} a_{y} - E_{z} H_{y}^{*} a_{x}$ . $E_{z} H_{x}^{*}$ is purely imaginary, so there is no average flow across, and $- E_{z} H_{y}^{*} = 4 E_{0}^{2} cos θ sin^{2} (k_{y} y) / η_{0}$ . Hence

P_{a v} P^{'} = \frac{2 E _{0}^{2} cos θ}{η _{0}} sin^{2} (k_{y} y) a_{x} = 0.459 sin^{2} (k_{y} y) a_{x} W/ m^{2}, = \int_{0}^{b} P_{a v, x} d y = \frac{E _{0}^{2} cos θ b}{η _{0}} = 6.89 mW/m .

$w_{a v} = \frac{1}{4} (ε_{0} ∣ E_{z} ∣^{2} + μ_{0} ∣ H ∣^{2})$ with $μ_{0} / η_{0}^{2} = ε_{0}$ :

w_{a v} W^{'} = ε_{0} E_{0}^{2} [sin^{2} k_{y} y + sin^{2} θ cos^{2} k_{y} y + cos^{2} θ sin^{2} k_{y} y], = \int_{0}^{b} w_{a v} d y = ε_{0} E_{0}^{2} \frac{b}{2} (1 + sin^{2} θ + cos^{2} θ) = ε_{0} E_{0}^{2} b = 26.5 pJ/ m^{2},

v_{E} = \frac{P ^{'}}{W ^{'}} = \frac{E _{0}^{2} cos θ b / η _{0}}{ε _{0} E _{0}^{2} b} = \frac{cos θ}{η _{0} ε _{0}} = c cos θ = 2.60 \times 1 0^{8} m/s, v_{p} v_{E} = c^{2} .

On the planes $y = 0$ and $y = b$ the field has no tangential $E$ ( $E_{z} = 0$ ) and no normal $H$ ( $H_{y} \propto sin k_{y} y = 0$ ): exactly the boundary conditions of a perfect conductor, so plates there change nothing. With $b$ fixed, $sin θ = λ_{0} /2 b = c / (2 b f)$ must stay below $1$ : $f > c /2 b = 5 GHz$ . At $6 GHz$ , $sin θ = 0.833$ , $θ = 56. 4^{\circ}$ , $v_{p} = c / cos θ = 5.43 \times 1 0^{8} m/s$ and $v_{E} = c cos θ = 1.66 \times 1 0^{8} m/s$ .

Answer

E = - j 20 sin (k_{y} y) e^{- j 181.4 x} a_{z} V/m, b = 3 cm, v_{p} = 1.155 c, v_{E} = 0.866 c, P^{'} = 6.89 mW/m, f_{c} = 5 GHz

Check

$β_{x}^{2} + k_{y}^{2} = 181. 4^{2} + 104. 7^{2} = 209. 4^{2} = k_{0}^{2}$ , as Eq. (6.1.15) requires. $v_{E}$ is also the group velocity: with $k_{y} = π / b$ fixed, $β_{x} = ω^{2} / c^{2} - k_{y}^{2}$ and $d ω / d β_{x} = c^{2} β_{x} / ω = c cos θ$ .

Remark

This is the TE $_{1}$ mode of a parallel-plate waveguide (Section 7.2): two plane waves zig-zagging between the plates. The phase pattern runs along the guide faster than light, while the energy, carried along the zig-zag at $c$ , advances only at $c cos θ$ . At the cutoff frequency the waves bounce straight up and down ( $θ = 9 0^{\circ}$ ) and nothing moves along the guide.

cf. Problem 6.1.2