Exercise 7.5.4: A rigid line for a UHF television transmitter

About the further exercises of Section 7.5

Five further exercises on the coaxial line, in the spirit of Problems 7.5.1–7.5.3, the worked example and the home exercise, graded from easy to harder. They use only the results listed at the head of this subsection. (Five more, on the circular guide, close Section 7.4.)

Exercise 7.5.4

A rigid line for a UHF television transmitter

A UHF TV transmitter delivers $30$ kW (average) at $700$ MHz into a $150$ m run of rigid $6 \frac{1}{8}$ -inch air line: a copper inner tube of outside diameter $66.0$ mm and an aluminium outer tube ( $σ_{Al} = 3.5 \times 1 0^{7}$ S/m) of inside diameter $151.9$ mm. Neglect the PTFE support insulators. (a) Find $Z_{0}$ . (b) Estimate the TE $_{11}$ and TM $_{01}$ cutoffs. Can the line serve the whole UHF band up to $862$ MHz? (c) Find $α_{c}$ in dB/100 m and the share of the inner conductor. (d) Find the power reaching the antenna and the heat dissipated per metre at the transmitter end. (e) Compare $30$ kW with the breakdown limit ( $3$ MV/m). What limits the rating?

Try it yourself first.

Strategy

With different metals the two terms of $α_{c}$ get different surface resistances: $α_{c} = \frac{1}{2 η _{0} ln ( b / a )} (\frac{R _{s, Cu}}{a} + \frac{R _{s, Al}}{b})$ . The heat per metre at the input is $P_{L}^{'} = 2 α_{c} P$ .

$a = 33.0$ mm, $b = 75.95$ mm, $b / a = 2.302$ , $Z_{0} = 60 ln 2.302 = 50.0 Ω$ .

f_{c, TE_{11}} f_{c, TM_{01}} \approx \frac{3 \times 1 0 ^{8}}{π \times 0.1090} = 0.876 GHz, \approx \frac{3 \times 1 0 ^{8}}{2 \times 0.0430} = 3.49 GHz .

$862$ MHz lies just below the TE $_{11}$ estimate, with a margin of under $2 %$ (the exact root, $0.895$ GHz, gives $4 %$ ). The line covers the band, but only just: a larger line would let TE $_{11}$ in at the top channels.

At $700$ MHz: $R_{s, Cu} = 6.90$ m $Ω$ , $R_{s, Al} = π f μ_{0} /3.5 \times 1 0^{7} = 8.89$ m $Ω$ ,

α_{c} = \frac{1}{2 \times 377 \times 0.8336} (\frac{0.00690}{0.0330} + \frac{0.00889}{0.07595}) = \frac{0.2092 + 0.1170}{628.6} = 5.19 \times 1 0^{- 4} Np/m = 0.451 dB/100 m .

The inner conductor contributes $0.2092/0.3262 = 64 %$ (with copper outside as well the loss would be $0.415$ dB/100 m).

Over $150$ m: $0.68$ dB, $P_{out} = 30 kW \times e^{- 2 \times 5.19 \times 1 0^{- 4} \times 150} = 25.7$ kW. At the transmitter end $P_{L}^{'} = 2 α_{c} P = 2 \times 5.19 \times 1 0^{- 4} \times 30000 = 31$ W per metre, $64 %$ of it in the inner tube.

$V_{m a x} = E_{br} a ln (b / a) = 3 \times 1 0^{6} \times 0.0330 \times 0.8336 = 82.5$ kV and $P_{m a x} = V_{m a x}^{2} /2 Z_{0} = 68$ MW, more than $2000$ times $30$ kW. At $30$ kW, $V_{0} = 2 Z_{0} P = 1.73$ kV and the inner-conductor field is only $V_{0} / [a ln (b / a)] = 63$ kV/m.

Answer

Z_{0} = 50.0 Ω f_{c, TE_{11}} \approx 0.876 GHz, f_{c, TM_{01}} \approx 3.49 GHz α_{c} = 0.451 dB/100 m (64 % inner) P_{out} = 25.7 kW, 31 W/m of heat P_{m a x} = 68 MW

Check

$P_{in} - P_{out} = 4.3$ kW over $150$ m is an average of $29$ W/m, a little below the $31$ W/m at the input, as it must be for a power that decays along the line.

Remark

Breakdown is irrelevant at television powers: the rating of rigid lines is thermal. $31$ W/m, two thirds of it in an inner tube that can shed its heat only through the gas and the insulators, warms the line, and the published average-power ratings of rigid lines come from that heating, derated further for standing waves. The size of the line is fixed by the highest channel through the TE $_{11}$ cutoff, not by the power.

cf. Problem 7.5.3