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

About the further exercises of Section 7.2

Five further exercises on the parallel-plate guide, in the spirit of Problems 7.2.1–7.2.5 and the home exercises, graded from easy to harder. Exercise 7.2.1 is about the TEM line, Exercise 7.2.2 about exciting the modes, Exercises 7.2.3 and 7.2.4 about the TE $_{1}$ and TM $_{1}$ modes at work, and Exercise 7.2.5 about the conductor losses.

Exercise 7.2.5

Why the TE $_{1}$ loss falls with frequency

Derive the conductor attenuation $α_{c}^{T E_{n}}$ of the TE $_{n}$ mode of a parallel-plate guide, quoted in Eq. (7.2.37), from $α = P_{L}^{'} / (2 P^{'})$ .
Above which frequency does TE $_{1}$ lose less in the plates than TEM? Give the exact ratio $f / f_{co, 1}$ .
Copper plates, air, $b = 5 mm$ : find the crossover frequency and $α_{c}$ of TEM, TE $_{1}$ and TM $_{1}$ at $60$ and $120 GHz$ .
Explain the trend with the zig-zag picture.

Try it yourself first.

Strategy

Write the TE $_{n}$ fields with one constant $C$ . The power carried is the integral of $S_{z}$ across the gap, the power lost is $\frac{1}{2} R_{s} ∣ J_{s} ∣^{2}$ on both plates, and on a plate the only tangential magnetic field of TE $_{n}$ is $H_{z}$ .

With $H_{z} = C cos (h_{n} y) e^{- j τ z}$ , Eq. (7.2.12) gives $E_{x} = \frac{j ω μ}{h _{n}} C sin (h_{n} y) e^{- j τ z}$ , so $E_{0} = ω μ ∣ C ∣/ h_{n}$ and

P^{'} = \frac{b E _{0}^{2}}{4 Z _{T E}} = \frac{b ω ^{2} μ ^{2} ∣ C ∣ ^{2}}{4 h _{n}^{2}} \frac{τ}{ω μ} = \frac{b ω μτ ∣ C ∣ ^{2}}{4 h _{n}^{2}} .

On each plate $H_{y} = 0$ (normal) and $∣ H_{z} ∣ = ∣ C ∣$ , hence $∣ J_{s} ∣ = ∣ C ∣$ and $P_{L}^{'} = 2 \times \frac{1}{2} R_{s} ∣ C ∣^{2} = R_{s} ∣ C ∣^{2}$ . Then

α_{c} = \frac{P _{L}^{'}}{2 P ^{'}} = \frac{2 R _{s} h _{n}^{2}}{b ω μτ},

and with $h_{n} = k f_{co, n} / f$ , $τ = k s$ , $ω μ = k η$ :

α_{c}^{T E_{n}} = \frac{2 R _{s} ( f _{co, n} / f ) ^{2}}{η b s} .

$α_{c}^{T E_{1}} / α_{c}^{T E M} = 2 u / 1 - u$ with $u = (f_{co, 1} / f)^{2}$ . Equal losses: $4 u^{2} = 1 - u$ , $u = (17 - 1) /8 = 0.3904$ , so

\frac{f}{f _{co, 1}} = \frac{8}{17 - 1} = 1.6005.

Above this frequency TE $_{1}$ has the lower conductor loss.

$f_{co, 1} = c /2 b = 30 GHz$ , crossover at $48.0 GHz$ . With $R_{s} = 63.9 m Ω$ at $60 GHz$ and $90.4 m Ω$ at $120 GHz$ ( $η_{0} b = 1.885 Ω$ ):

60 GHz (s = 0.866) : α_{c}^{T E M} = 0.294, α_{c}^{T E_{1}} = 0.170, α_{c}^{T M_{1}} = 0.680 dB/m, 120 GHz (s = 0.968) : α_{c}^{T E M} = 0.416, α_{c}^{T E_{1}} = 0.054, α_{c}^{T M_{1}} = 0.860 dB/m .

The two plane waves of TE $_{1}$ meet the plates at the angle $θ$ with $sin θ = f_{co, 1} / f$ , and their magnetic field has the component $H sin θ$ along $z$ at the plates. As $f$ grows, $θ \to 0$ : the waves graze the plates, the wall current $∣ H_{z} ∣$ shrinks and so do the bounces per metre. TEM and TM modes always have their full $H$ tangential to the plates.

Answer

α_{c}^{T E_{n}} = \frac{2 R _{s} ( f _{co, n} / f ) ^{2}}{η b s}, f / f_{co, 1} = 1.6005 (48.0 GHz), TE_{1} : 0.170 \to 0.054 dB/m from 60 to 120 GHz

Check

Far above cutoff $α_{c}^{T E_{1}} \propto f / f^{2} = f^{- 3/2}$ : from $60$ to $120 GHz$ this predicts a factor $2^{3/2} = 2.83$ , and with the change of $s$ ( $0.968/0.866$ ) the factor is $3.16$ , exactly $0.170/0.054$ . At $120 GHz$ $f_{co, 3} = 90 GHz$ , so TE $_{2}$ , TM $_{2}$ , TE $_{3}$ and TM $_{3}$ propagate too: low loss in TE $_{1}$ is bought with a strongly overmoded guide.

Remark

The same mechanism makes the TE $_{01}$ mode of a circular guide (Section 7.4) famous: its loss falls without limit as the frequency rises, and it was once the candidate for long-distance millimetre-wave trunk lines, until optical fibres took over.

cf. Problem 7.2.4

Exercise 7.2.5: Why the TE_1 loss falls with frequency