Exercise 7.4.5: Peak power and loss of a TM01 rotary joint

About the further exercises of Section 7.4

Five further exercises on the circular guide, in the spirit of Problems 7.4.1–7.4.2, the worked examples and the home exercises, graded from easy to harder. They use only the results listed at the head of this subsection. (Five more, on the coaxial line, close Section 7.5.)

Exercise 7.4.5

Peak power and loss of a TM $_{01}$ rotary joint

An S-band surveillance radar ( $f = 2.9$ GHz) uses an air-filled copper TM $_{01}$ rotary joint of radius $a = 4.5$ cm. Use $\int_{0}^{a} J_{1}^{2} (x_{01} ρ / a) ρ d ρ = \frac{a ^{2}}{2} J_{1}^{2} (x_{01})$ , $J_{1} (2.405) = 0.519$ , and the maximum of $J_{1}$ , $0.582$ at $x = 1.841$ .

Check that only TE $_{11}$ and TM $_{01}$ propagate.
With $E_{z} = E_{0} J_{0} (h ρ) e^{- j τ z}$ , find $E_{ρ}$ , $H_{ϕ}$ and the power $P$ in terms of $E_{0}$ .
Where is $∣ E ∣$ largest? Find the peak power for $E_{br} = 3$ MV/m, and with a safety factor $2$ on the field.
From the wall current find $α_{c}$ , and compare with Eq. (7.4.23).

Try it yourself first.

Strategy

For $m = 0$ the TM formulas leave $E_{z}$ , $E_{ρ}$ and $H_{ϕ}$ , all built from $J_{0}$ and $J_{0}^{'} = - J_{1}$ . The power is the integral of $\frac{1}{2} E_{ρ} H_{ϕ}^{*}$ over the disc, which the given Bessel integral evaluates. The loss is $P_{L}^{'} / (2 P)$ with $P_{L}^{'} = \frac{1}{2} R_{s} ∣ H_{ϕ} (a) ∣^{2} 2 π a$ .

$c / (2 π a) = 1.061$ GHz: TE $_{11}$ $1.95$ GHz, TM $_{01}$ $2.55$ GHz, TE $_{21}$ $3.24$ GHz. At $2.9$ GHz only TE $_{11}$ and TM $_{01}$ propagate.

$h = 2.405/0.045 = 53.44$ rad/m, $k = 60.74$ rad/m, $f / f_{c} = 1.137$ , $s = 0.475$ , $τ = 28.86$ rad/m. From Eq. (7.4.17) with $m = 0$ :

E_{ρ} H_{ϕ} = - \frac{j τ}{h} E_{0} J_{0}^{'} (h ρ) = j \frac{τ}{h} E_{0} J_{1} (h ρ), = - \frac{j ω ε _{0}}{h} E_{0} J_{0}^{'} (h ρ) = j \frac{ω ε _{0}}{h} E_{0} J_{1} (h ρ),

with $τ / h = 0.540$ . Then $S_{z} = \frac{1}{2} Re {E_{ρ} H_{ϕ}^{*}} = \frac{τ ω ε _{0}}{2 h ^{2}} E_{0}^{2} J_{1}^{2} (h ρ)$ and, with $ω ε_{0} = k / η_{0}$ ,

P = \frac{τ ω ε _{0}}{2 h ^{2}} E_{0}^{2} 2 π \frac{a ^{2}}{2} J_{1}^{2} (x_{01}) = \frac{π a ^{2}}{2 η _{0}} s (\frac{f}{f _{c}})^{2} J_{1}^{2} (x_{01}) E_{0}^{2} = 1.40 \times 1 0^{- 6} E_{0}^{2} W .

$E_{z}$ and $E_{ρ}$ are in time quadrature, so at each point the field vector traces an ellipse whose longer half-axis is the larger of $∣ E_{z} ∣$ and $∣ E_{ρ} ∣$ . $∣ E_{z} ∣ \leq E_{0}$ (on the axis), while $∣ E_{ρ} ∣ \leq 0.540 \times 0.582 E_{0} = 0.314 E_{0}$ . The largest field is $E_{0}$ , on the axis:

P_{m a x} = 1.40 \times 1 0^{- 6} \times (3 \times 1 0^{6})^{2} = 12.6 MW, with E_{0} = 1.5 MV/m : 3.1 MW .

On the wall $∣ H_{ϕ} (a) ∣ = \frac{ω ε _{0}}{h} E_{0} J_{1} (x_{01})$ and $P_{L}^{'} = \frac{1}{2} R_{s} ∣ H_{ϕ} (a) ∣^{2} 2 π a$ , so

α_{c} = \frac{P _{L}^{'}}{2 P} = \frac{π a R _{s} ( ω ε _{0} / h ) ^{2} J _{1}^{2}}{( τ ω ε _{0} / h ^{2} ) π a ^{2} J _{1}^{2}} = \frac{R _{s} ω ε _{0}}{a τ} = \frac{R _{s}}{a η _{0}} \frac{k}{τ} = \frac{R _{s}}{a η _{0} s},

the formula Eq. (7.4.23). $R_{s} = 14.0$ m $Ω$ at $2.9$ GHz: $α_{c} = 0.0140/ (0.045 \times 377 \times 0.475) = 1.74 \times 1 0^{- 3}$ Np/m $= 0.015$ dB/m.

Answer

P = 1.40 \times 1 0^{- 6} E_{0}^{2} W ∣ E ∣_{m a x} = E_{0} on the axis P_{m a x} = 12.6 MW (3.1 MW with the factor 2) α_{c} = \frac{R _{s}}{a η _{0} s} = 0.015 dB/m

Check

$E_{ρ} / H_{ϕ} = τ / (ω ε_{0}) = η_{0} s = 179 Ω = Z_{T M}$ . A numerical integration of $S_{z}$ over the disc gives the same $1.396 \times 1 0^{- 6} E_{0}^{2}$ W.

Remark

A megawatt pulse needs only $E_{0} = 0.85$ MV/m on the axis. Unlike the coaxial line, whose field crowds onto the inner conductor (Section 7.5), TM $_{01}$ concentrates its field in the middle of the pipe, away from the metal, which is one reason high-power joints use it. Arcing in practice starts at the edges of the coupling probes and of the choke gap, so real ratings are set lower and the joints are often pressurised.

cf. Worked example 7.4.2

Exercise 7.4.5: Peak power and loss of a TM_{01} rotary joint