Exercise 7.3.10: Mixing the degenerate TE11 and TM11 modes

About the further exercises of Section 7.3

Ten further exercises in the spirit of Problems 7.3.1–7.3.5, the home exercises and the worked examples, graded from easy to harder. Exercises 7.3.1–7.3.2 are about cutoffs and dimensions, Exercises 7.3.3–7.3.5 about power, wall currents and loss, Exercises 7.3.6–7.3.8 about dispersion and pairs of modes, and Exercises 7.3.9–7.3.10 about many modes and degenerate modes. They use only the results listed at the head of this subsection, together with the dielectric attenuation $α_{d} = k tan δ / (2 s)$ .

Exercise 7.3.10

Mixing the degenerate TE $_{11}$ and TM $_{11}$ modes

In an air-filled guide with $a = 2 cm$ and $b = 1.5 cm$ at $19 GHz$ , a TM $_{11}$ mode with $E_{z} = C sin \frac{π x}{a} sin \frac{π y}{b} e^{- j τ z}$ , $C = 1 kV/m$ , and a TE $_{11}$ mode with $H_{z} = D cos \frac{π x}{a} cos \frac{π y}{b} e^{- j τ z}$ travel together. (a) Find $f_{co, 11}$ , $τ$ , $Z_{T M}$ and $Z_{T E}$ , and explain why the two modes keep a fixed phase relation along the guide. (b) Show that $E_{x}$ of both modes has the same $(x, y)$ dependence, and find $D$ such that $E_{x} \equiv 0$ everywhere. (c) Show that the combined $E_{y}$ is then $- j (τ b C / π) sin \frac{π x}{a} cos \frac{π y}{b} e^{- j τ z}$ and evaluate its amplitude. (d) Show that $P_{T E} / P_{T M} = s^{2} (b / a)^{2}$ and find the total power.

Try it yourself first.

Strategy

Degenerate modes share $h_{11}$ and hence $τ$ : any combination of them travels unchanged. The transverse fields come from the series (7.3.17) and (7.3.24) with one term each, and the powers of a TE and a TM mode simply add.

$f_{co, 11} = 1.5 \times 1 0^{8} 5 0^{2} + 66. 7^{2} = 12.5 GHz$ , $s = 1 - (12.5/19)^{2} = 0.753$ and $k = 397.9 rad/m$ :

τ Z_{T M} Z_{T E} = k s = 299.7 rad/m, = 377 \times 0.753 = 283.9 Ω, = \frac{377}{0.753} = 500.6 Ω.

Both factors $e^{- j τ z}$ are the same, so the ratio of the modes does not change with $z$ .

From Eqs. (7.3.17) and (7.3.24),

E_{x}^{T M} E_{x}^{T E} = - j \frac{π τ}{a h ^{2}} C cos \frac{π x}{a} sin \frac{π y}{b}, = j \frac{ω μ _{0} π}{b h ^{2}} D cos \frac{π x}{a} sin \frac{π y}{b}

(times $e^{- j τ z}$ ), the same $cos \cdot sin$ dependence. The sum vanishes everywhere when $τ C / a = ω μ_{0} D / b$ :

D = \frac{τ b}{ω μ _{0} a} C = \frac{299.7 \times 0.015}{3000.4} \times 1000 = 1.498 A/m .

$E_{y}^{T M} = - j \frac{π τ}{b h ^{2}} C sin \frac{π x}{a} cos \frac{π y}{b}$ and $E_{y}^{T E} = - j \frac{ω μ _{0} π}{a h ^{2}} D sin \frac{π x}{a} cos \frac{π y}{b} = - j \frac{π τ b}{a ^{2} h ^{2}} C sin \frac{π x}{a} cos \frac{π y}{b}$ . Their sum has the factor

\frac{π τ C}{h ^{2}} (\frac{1}{b} + \frac{b}{a ^{2}}) = \frac{π τ C}{h ^{2}} \frac{a ^{2} + b ^{2}}{a ^{2} b} = \frac{τ b C}{π}, since h^{2} = π^{2} \frac{a ^{2} + b ^{2}}{a ^{2} b ^{2}} .

The amplitude is $299.7 \times 0.015 \times 1000/ π = 1431 V/m$ (TM part $916 V/m$ , TE part $515 V/m$ ).

$P_{T M} = \frac{ab}{8} \frac{ω ε _{0} τ}{h ^{2}} C^{2}$ and $P_{T E} = \frac{ab}{8} \frac{ω μ _{0} τ}{h ^{2}} D^{2}$ , so

\frac{P _{T E}}{P _{T M}} = \frac{μ _{0} D ^{2}}{ε _{0} C ^{2}} = \frac{μ _{0}}{ε _{0}} \frac{τ ^{2} b ^{2}}{ω ^{2} μ _{0}^{2} a ^{2}} = \frac{τ ^{2}}{k ^{2}} \frac{b ^{2}}{a ^{2}} = s^{2} (\frac{b}{a})^{2} = 0.319.

With $h^{2} = 68539 m^{- 2}$ and $ω ε_{0} = 1.056 S/m$ : $P_{T M} = \frac{0.0003}{8} \times \frac{1.056 \times 299.7}{68 539} \times 1 0^{6} = 0.173 W$ , $P_{T E} = 0.055 W$ , total $0.228 W$ .

Answer

f_{co, 11} = 12.5 GHz, τ = 299.7 rad/m Z_{T M} = 283.9 Ω, Z_{T E} = 500.6 Ω D = \frac{τ b}{ω μ _{0} a} C = 1.498 A/m ∣ E_{y} ∣_{m a x} = \frac{τ b C}{π} = 1431 V/m P = 0.173 + 0.055 = 0.228 W

Check

A numerical integration of $\frac{1}{2} Re {E_{x} H_{y}^{*} - E_{y} H_{x}^{*}}$ over the cross-section gives $0.2283 W$ . The cross terms vanish for any TE and TM pair: $\int \nabla_{T} E_{z} \times \nabla_{T} H_{z}^{*} \cdot a_{z} d A$ reduces to a line integral of $E_{z} \partial H_{z}^{*} / \partial l$ round the walls, where $E_{z} = 0$ .

Remark

The combination has $E_{x} \equiv 0$ but both $E_{z}$ and $H_{z}$ : it is neither TE nor TM to $z$ , but “TE to $x$ ”, a longitudinal-section electric (LSE) mode. Such modes are the natural ones of a guide loaded with a dielectric slab whose faces are planes $x =$ const. In the empty guide any small asymmetry converts power between TE $_{11}$ and TM $_{11}$ without phase mismatch.

cf. Problem 7.3.5

Exercise 7.3.10: Mixing the degenerate TE_{11} and TM_{11} modes