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

About the worked examples of Section 7.8

Three short calculations tied to the applications, with resonant frequencies, $Q$ , bandwidth, stored energy and wall loss. They use only the results of this section. A good way to use them: read the given quantities, produce the formula from the corresponding application yourself, and only then put the numbers in. Constants: $c = 3 \times 1 0^{8} m/s$ , $ε_{0} = 1 0^{- 9} /36 π F/m$ , $η_{0} = 120 π Ω$ , copper $σ_{c} = 5.8 \times 1 0^{7} S/m$ , $c /2 = 15 GHz cm$ , $c /2 π = 4.775 GHz cm$ .

Worked example 7.8.2

An X-band TE $_{011}$ wavemeter

cf. Application 7.8.2

Given. A copper cylinder, $a = 2.5 cm$ , air. The length $d$ is set by a non-contacting plunger driven by a micrometer with $0.01 mm$ divisions. Weak coupling to WR-90.

Find. (a) the TE $_{01}$ cutoff and the plunger setting for $10 GHz$ , (b) $Q$ and the width of the dip, (c) the tuning slope and the frequency step per division, (d) the range for $d = 2.0$ to $2.6 cm$ , (e) why the degenerate TM $_{111}$ does no harm.

Try it yourself first.

Strategy

TE $_{011}$ is TE $_{01}$ of the circular guide ( $h = x_{01}^{'} / a$ ) with one half guide wavelength between the end plates: $d = λ_{g} /2 = π / k^{2} - h^{2}$ . Its fields, Eq. (7.8.34), $H_{z} = H_{0} J_{0} (h ρ) sin \frac{π z}{d}$ , $H_{ρ} = - \frac{π / d}{h} H_{0} J_{1} (h ρ) cos \frac{π z}{d}$ and $E_{ϕ} = - j \frac{ω μ}{h} H_{0} J_{1} (h ρ) sin \frac{π z}{d}$ , give with $\int_{0}^{a} J_{1}^{2} (h ρ) ρ d ρ = \frac{a ^{2}}{2} J_{0}^{2} (x_{01}^{'})$

W P_{L} = \frac{ε ( ω μ ) ^{2} π a ^{2} d}{4 h ^{2}} H_{0}^{2} J_{0}^{2}, = \frac{π R _{s}}{2} H_{0}^{2} J_{0}^{2} [a d + 2 a^{2} (\frac{π / d}{h})^{2}],

where $J_{0} = J_{0} (x_{01}^{'})$ . The side wall carries $H_{z}$ , the end plates $H_{ρ}$ . With $ω μ = k η_{0}$ ,

Q D = \frac{( k a ) ^{3} η _{0} a d}{4 x _{01}^{'2} R _{s} D}, = \frac{a d}{2} + (\frac{π a ^{2}}{x _{01}^{'} d})^{2} .

$f_{c, TE_{01}} = 4.775 \times 3.832/2.5 = 7.32 GHz$ . At $10 GHz$ $k = 209.4 rad/m$ and $h = 3.832/0.025 = 153.3 rad/m$ :

d = \frac{π}{209. 4 ^{2} - 153. 3 ^{2}} = 2.201 cm .

$R_{s} = 26.1 mΩ$ , $k a = 5.236$ , $a d /2 = 2.751 \times 1 0^{- 4} m^{2}$ and $π a^{2} / (x_{01}^{'} d) = 0.02328 m$ :

Q Δ f = \frac{143.5 \times 377 \times 5.502 \times 1 0 ^{- 4}}{4 \times 14.68 \times 0.0261} \times \frac{1}{( 2.751 + 5.421 ) \times 1 0 ^{- 4}} = 23800, = \frac{10 GHz}{23 800} = 0.42 MHz .

From $f = \frac{c}{2 π} h^{2} + (π / d)^{2}$ ,

\frac{df}{dd} = - \frac{c}{2 π} \frac{π ^{2} / d ^{3}}{k} = - 4.775 \times 1 0^{7} \times \frac{9.870/ ( 0.02201 ) ^{3}}{209.4} \frac{Hz}{m} = - 211 MHz/mm .

One division ( $0.01 mm$ ) is $2.1 MHz$ , five times the dip width.

$d = 2.0 cm$ : $4.775 (3.832/2.5)^{2} + (π /2)^{2} = 10.48 GHz$ . $d = 2.6 cm$ : $9.32 GHz$ . The range is $9.32$ to $10.48 GHz$ (Figure 7.8.28).

TM $_{111}$ has $x_{11} = x_{01}^{'}$ and resonates at the same frequency, but its wall currents flow along $z$ and must cross the gap around the plunger. They are interrupted, its $Q$ collapses and it produces no visible dip. TE $_{011}$ has only $ϕ$ -directed currents, which never cross that gap.

Answer

d = 2.201 cm, Q = 23800, Δ f = 0.42 MHz - 211 MHz/mm (2.1 MHz per division) 9.32 to 10.48 GHz

Worked example 7.8.2. Tuning curve of the TE_{011} wavemeter (a=2.5\,\mathrm{cm}): the plunger range d=2.0 to 2.6\,\mathrm{cm} covers 10.48 to 9.32\,\mathrm{GHz}, and 10\,\mathrm{GHz} is read at d=2.201\,\mathrm{cm}. — **Figure 7.8.28.** Worked example 7.8.2. Tuning curve of the TE $_{011}$ wavemeter ( $a = 2.5 cm$ ): the plunger range $d = 2.0$ to $2.6 cm$ covers $10.48$ to $9.32 GHz$ , and $10 GHz$ is read at $d = 2.201 cm$ .

Check

$f (2.201 cm) = \frac{c}{2 π} 153. 3^{2} + 142. 7^{2} = 4.775 \times 1 0^{7} \times 209.5 Hz = 10.0 GHz$ . The finite difference $(10.48 - 9.32) /6 = 0.19 GHz/mm$ over the whole range is close to the slope at the centre.

Remark

The micrometer, not $Q$ , sets the resolution here: about $2 \times 1 0^{- 4}$ , ample for checking an X-band source. A finer scale would only help down to the dip width, $4 \times 1 0^{- 5}$ .

Worked example 7.8.2: An X-band TE_{011} wavemeter