Worked example 7.4.1: The circular feed of a Ku-band satellite receiver

Two short calculations tied to the applications and to the polarisations of TE $_{11}$ . They use only the cutoff conditions of the circular guide and the universal formulas Eq. (7.4.3). Constants: $c = 3 \times 1 0^{8}$ m/s, $η_{0} = 120 π Ω$ , copper $σ_{c} = 5.8 \times 1 0^{7}$ S/m. A good way to use them: read the given quantities, pick the formula yourself, and only then put the numbers in.

Worked example 7.4.1

The circular feed of a Ku-band satellite receiver

Given. A satellite-TV receiver (LNB) takes both linear polarisations of the $10.7$ – $12.75$ GHz band from the dish through a short air-filled circular waveguide, where a vertical and a horizontal probe pick up the two signals.

Find. (a) The radii for which TE $_{11}$ alone propagates over the whole band. (b) For $2 a = 17.5$ mm: the cutoffs of TE $_{11}$ , TM $_{01}$ and the margins. (c) $λ_{g}$ , $Z_{T E}$ , $v_{g}$ at $10.7$ , $11.7$ and $12.75$ GHz. (d) Why do the two probes not disturb each other?

Try it yourself first.

Strategy

TE $_{11}$ must be above its cutoff at the lowest frequency of the band, and TM $_{01}$ , the next mode, below its cutoff at the highest:

f_{c, TE_{11}} = \frac{1.841 c}{2 π a} < f < f_{c, TM_{01}} = \frac{2.405 c}{2 π a} .

TE $_{11}$ on at $10.7$ GHz: $a > \frac{1.841 \times 3 \times 1 0 ^{8}}{2 π \times 10.7 \times 1 0 ^{9}} = 8.22$ mm. TM $_{01}$ off at $12.75$ GHz: $a < \frac{2.405 \times 3 \times 1 0 ^{8}}{2 π \times 12.75 \times 1 0 ^{9}} = 9.01$ mm. So $16.4 mm < 2 a < 18.0$ mm. The window is narrow because the band ratio $12.75/10.7 = 1.19$ uses most of the single-mode ratio $1.31$ (Figure 7.4.15).

$a = 8.75$ mm: $f_{c, TE_{11}} = 10.05$ GHz and $f_{c, TM_{01}} = 13.12$ GHz. The band edges lie at $10.7/10.05 = 1.065 f_{c, TE_{11}}$ and $12.75/13.12 = 0.972 f_{c, TM_{01}}$ : margins of $6.5 %$ and $2.8 %$ .

With $s = 1 - (10.05/ f)^{2}$ , $λ_{g} = c / (f s)$ , $Z_{T E} = 377/ s$ , $v_{g} = sc$ :

$f$ [GHz]	$10.7$	$11.7$	$12.75$
$s = v_{g} / c$	$0.344$	$0.512$	$0.616$
$λ_{g}$ [mm]	$81.5$	$50.0$	$38.2$
$Z_{T E}$ [ $Ω$ ]	$1096$	$736$	$612$

The $cos ϕ$ and $sin ϕ$ versions of TE $_{11}$ are orthogonal (Figure 7.4.6): the vertical probe couples only to the mode whose central field is vertical, the horizontal probe only to the other. The two signals share the pipe without interfering.

Answer

16.4 mm < 2 a < 18.0 mm 2 a = 17.5 mm : f_{c, TE_{11}} = 10.05 GHz f_{c, TM_{01}} = 13.12 GHz λ_{g} = 81.5, 50.0, 38.2 mm Z_{T E} = 1096, 736, 612 Ω v_{g} / c = 0.344, 0.512, 0.616

Check

At $12.75$ GHz, $λ = 23.5$ mm and $λ_{g} = 23.5/0.616 = 38.2$ mm. $Z_{T E} s = 377 Ω$ in every column.

Worked example 7.4.1. Cutoffs of TE_{11} and TM_{01} against the diameter. The whole band 10.7–12.75 GHz must lie above the orange curve and below the blue one: the shaded window 16.4\lt 2a\lt 18.0 mm. The chosen 2a=17.5 mm sits closer to the TM_{01} limit, with a margin of 2.8\,\% at the top of the band and 6.5\,\% at the bottom. — **Figure 7.4.15.** Worked example 7.4.1. Cutoffs of TE $_{11}$ and TM $_{01}$ against the diameter. The whole band $10.7$ – $12.75$ GHz must lie above the orange curve and below the blue one: the shaded window $16.4 < 2 a < 18.0$ mm. The chosen $2 a = 17.5$ mm sits closer to the TM $_{01}$ limit, with a margin of $2.8 %$ at the top of the band and $6.5 %$ at the bottom.

What the numbers say

At the low band edge the guide works at only $1.065 f_{c}$ : $Z_{T E}$ is almost three times $η_{0}$ and changes fast with frequency, so the transitions to the probes need careful matching. A slightly elliptical pipe would give the two polarisations different cutoffs and let them exchange power along the guide (Exercise 7.4.3), which is why the feed is kept short and round.

cf. the TE $_{11}$ polarisations