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

About the further exercises of Section 7.8

Ten further exercises in the spirit of Problems 7.8.1–7.8.5 and the home exercises, graded from easy to harder. Exercises 7.8.1–7.8.6 are about cavity resonators, Exercises 7.8.7–7.8.10 about microwave devices and scattering parameters. They use only the results of this section, in particular $f_{mn p}^{2} = f_{co, mn}^{2} + (p v /2 d)^{2}$ , $Q_{101}$ of Eq. (7.8.24) with $R_{s} = π f μ_{0} / σ_{c} = 1/ (σ_{c} δ)$ ( $Q = a /3 δ$ for a cube), $Q_{010}$ of Eq. (7.8.33), $1/ Q_{L} = 1/ Q_{c} + 1/ Q_{d} + 1/ Q_{ext}$ , $Q_{d} = 1/ tan δ$ , $β = Q_{0} / Q_{ext}$ , the response $H (f) = 1/ (1 + j 2 Q δ_{f})$ with $δ_{f} = (f - f_{0}) / f_{0}$ , and for S-parameters $b_{i} = \sum_{j} S_{ij} a_{j}$ , reciprocity $S_{ij} = S_{j i}$ , losslessness $[S]^{H} [S] = [I]$ , $R L = - 20 lo g_{10} ∣ S_{11} ∣$ and $VSWR = (1 + ∣Γ∣) / (1 - ∣Γ∣)$ .

Exercise 7.8.9

How wide is a coupler's directivity?

A coupler between two identical WR-90 guides ( $a = 22.86 mm$ , TE $_{10}$ ) is designed for $f_{0} = 10 GHz$ with holes spaced $ℓ = λ_{g 0} /4$ . (a) Find $ℓ$ . (b) Two equal holes, each coupling the amplitude $0.05$ ( $- 26 dB$ ): find the coupling and the directivity at $8.2$ , $9$ , $11$ and $12.4 GHz$ (neglect the frequency dependence of the holes). (c) Three holes with amplitudes $0.025 : 0.05 : 0.025$ : show that $∣ b_{4} / b_{3} ∣ = cos^{2} τ ℓ$ and repeat. (d) Over which band is $D \geq 20 dB$ for each design?

Try it yourself first.

Strategy

Hole $n$ (at $z = n ℓ$ ) sends a wave forwards and a wave backwards in the upper guide. All forward contributions travel the same total path and add in phase, $∣ b_{3} ∣ = \sum c_{n}$ . The backward one from hole $n$ has travelled $2 n ℓ$ further: $∣ b_{4} ∣ = ∣ \sum c_{n} e^{- j 2 n τ ℓ} ∣$ . With $τ = 2 π / λ_{g}$ , $D = 20 lo g_{10} ∣ b_{3} / b_{4} ∣$ and $C = - 20 lo g_{10} ∣ b_{3} ∣$ .

$f_{co, 10} = 6.562 GHz$ , $λ_{g 0} = \frac{3 cm}{1 - 0.656 2 ^{2}} = 3.976 cm$ and $ℓ = 0.994 cm$ .

$∣ b_{3} ∣ = 0.1$ , so $C = 20 dB$ at all frequencies. $∣ b_{4} / b_{3} ∣ = ∣ cos τ ℓ ∣$ , with $τ ℓ = \frac{π}{2} λ_{g 0} / λ_{g}$ :

$f$ [GHz]	$τ ℓ$ [deg]	$∣ cos τ ℓ ∣$	$D$ [dB], two / three holes
$8.2$	$58.7$	$0.520$	$5.7$ / $11.4$
$9$	$73.5$	$0.285$	$10.9$ / $21.8$
$10$	$90$	$0$	$\infty$ / $\infty$
$11$	$105.3$	$0.264$	$11.6$ / $23.1$
$12.4$	$125.5$	$0.581$	$4.7$ / $9.4$

With $u = e^{- j 2 τ ℓ}$ , $b_{4} \propto c (1 + 2 u + u^{2}) = c (1 + u)^{2}$ and $∣1 + u ∣ = 2∣ cos τ ℓ ∣$ , so $∣ b_{4} ∣ = 4 c cos^{2} τ ℓ$ while $∣ b_{3} ∣ = 4 c = 0.1$ ( $C = 20 dB$ again). Hence $∣ b_{4} / b_{3} ∣ = cos^{2} τ ℓ$ , and $D$ doubles in dB (last column of the table).

Two holes: $∣ cos τ ℓ ∣ \leq 0.1$ for $84. 3^{\circ} \leq τ ℓ \leq 95. 7^{\circ}$ , that is $9.64$ to $10.37 GHz$ ( $7 %$ ). Three holes: $∣ cos τ ℓ ∣ \leq 0.316$ for $71. 6^{\circ} \leq τ ℓ \leq 108. 4^{\circ}$ , that is $8.89$ to $11.21 GHz$ ( $23 %$ ).

Answer

ℓ = 0.994 cm, C = 20 dB D \geq 20 dB : 9.64 to 10.37 GHz (2 holes) 8.89 to 11.21 GHz (3 holes)

Check

The rule of thumb of Section 7.8.16, $λ_{g}$ off by $10 %$ gives $D \approx 16 dB$ : at $λ_{g} = 1.1 λ_{g 0}$ , $∣ cos (9 0^{\circ} /1.1) ∣ = 0.142$ and $D = 16.9 dB$ , and at $0.9 λ_{g 0}$ , $D = 15.2 dB$ .

Remark

Binomial weights $1 : N : \dots : 1$ put an $N$ -fold zero of the backward sum at the design frequency, which flattens the directivity around it. Commercial broadband couplers use ten or more holes, often with Chebyshev weights, and reach $D > 30 dB$ over a whole waveguide band.

cf. Problem 7.8.5

Exercise 7.8.9: How wide is a coupler's directivity?