Section 7.4 · Worked examples · 2 of 2
Worked example 7.4.2: A TM_{01} rotary joint for an X-band radar
About the worked examples of Section 7.4
Two short calculations tied to the applications and to the polarisations of TE. They use only the cutoff conditions of the circular guide and the universal formulas Eq. (7.4.3). Constants: m/s, , copper S/m. A good way to use them: read the given quantities, pick the formula yourself, and only then put the numbers in.
A TM rotary joint for an X-band radar
#Given. A GHz radar uses an air-filled copper circular TM section, cm long, between the fixed and the rotating rectangular guides (Application 7.4.1).
Find. (a) The radii for which TM propagates and TE does not. (b) For mm: all propagating modes. (c) , , of TM and of TE. (d) The conductor loss of the section. (e) Why must TE be suppressed?
TM needs , and TE, the next mode above it, must stay cut off: . The loss follows from , Eq. (7.4.23).
rad/m, mm. Copper at GHz: m.
mm and mm.
mm: GHz, so TE GHz, TM GHz, TE GHz. Propagating: TE and TM.
TM: ,
TE: and mm.
so the cm section loses dB.
TE has a direction across the pipe. Power converted into it reaches the output with an amplitude and a phase that depend on the rotation angle, and with another guide wavelength, so the radar signal would be modulated at the rotation rate. Mode filters (rings, absorbing vanes) remove it.
, as for every TM mode above cutoff, and because TM is closer to its cutoff. The window – mm is around its centre.
TM runs at only , and the loss is negligible anyway: the design problem of a rotary joint is mode purity, not attenuation. The rotating gap is bridged by a quarter-wave choke because TM has an axial wall current that a plain gap would cut (Exercise 7.4.4).
cf. Application 7.4.1