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

Problem 7.8.2

Interactive: Cavity Resonator

Problem 7.8.2

TE $_{101}$ energy, loss and $Q$

The cavity of Problem 7.8.1 is made of copper ( $σ_{c} = 5.8 \times 1 0^{7} S/m$ ) and oscillates in the TE $_{101}$ mode with the peak field $E_{0} = 10 kV/m$ .

Write the phasors of $E$ and $H$ with numbers.
Find the stored energy.
Find the power lost in the walls and the quality factor.
Find the 3-dB bandwidth and the time in which the stored energy falls to $1/ e$ after the source is switched off.
What peak field does a source sustain that delivers $1 W$ to the cavity at resonance?

Try it yourself first.

Strategy

At $f_{101} = 6.25 GHz$ , $λ = 4.8 cm$ , so $λ /2 a = 0.8$ and $λ /2 d = 0.6$ . These two numbers are the $sin θ$ and $cos θ$ of the TE $_{10}$ zig-zag and give the magnetic amplitudes directly. Then $W$ from Eq. (7.8.17), the wall-loss bracket of Eq. (7.8.23), $Q = ω W / P_{L}$ , and everything else scales from there.

Bookkeeping

$a = 0.03$ , $b = 0.015$ , $d = 0.04 m$ , $ω = 2 π \times 6.25 \times 1 0^{9} = 3.927 \times 1 0^{10} s^{- 1}$ , $E_{0} / η_{0} = 1 0^{4} /377 = 26.53 A/m$ , $π / a = 104.7 rad/m$ , $π / d = 78.54 rad/m$ .

$26.53 \times 0.6 = 15.92 A/m$ and $26.53 \times 0.8 = 21.22 A/m$ , so, with $x$ and $z$ in metres,

E_{y} H_{x} H_{z} = 10 sin (104.7 x) \times sin (78.54 z) kV/m, = - j 15.92 sin (104.7 x) \times cos (78.54 z) A/m, = j 21.22 cos (104.7 x) \times sin (78.54 z) A/m .

W = \frac{ε _{0} E _{0}^{2} ab d}{8} = \frac{8.842 \times 1 0 ^{- 12} \times 1 0 ^{8} \times 1.8 \times 1 0 ^{- 5}}{8} = 1.99 nJ .

R_{s} = \frac{π \times 6.25 \times 1 0 ^{9} \times 4 π \times 1 0 ^{- 7}}{5.8 \times 1 0 ^{7}} = 20.6 mΩ

(skin depth $δ = 1/ (σ_{c} R_{s}) = 0.84 μ m$ ). The bracket of Eq. (7.8.23) is

0.36 \times 4.5 \times 1 0^{- 4} + 0.64 \times 6 \times 1 0^{- 4} + 6 \times 1 0^{- 4} = (1.62 + 3.84 + 6.00) \times 1 0^{- 4} = 1.146 \times 1 0^{- 3} m^{2},

and with $E_{0}^{2} / (2 η_{0}^{2}) = 351.8 A^{2} / m^{2}$ ,

P_{L} Q = 0.0206 \times 351.8 \times 1.146 \times 1 0^{- 3} = 8.32 mW, = \frac{ω W}{P _{L}} = \frac{3.927 \times 1 0 ^{10} \times 1.99 \times 1 0 ^{- 9}}{8.32 \times 1 0 ^{- 3}} = 9395.

$Δ f = f_{0} / Q = 6.25 GHz /9395 = 665 kHz$ . The energy decays as $e^{- ω t / Q}$ , so it reaches $1/ e$ after $τ_{W} = Q / ω = 0.239 μ s$ , about $Q /2 π = 1500$ periods.

$P_{L} \propto E_{0}^{2}$ , so $E_{0} = 10 kV/m \times 1/0.00832 = 110 kV/m$ , with $W = 1.99 nJ \times (110/10)^{2} = 239 nJ$ .

Answer

W = 1.99 nJ, P_{L} = 8.32 mW, Q = 9395 Δ f = 665 kHz, τ_{W} = 0.239 μ s E_{0} (1 W) = 110 kV/m

Read $W = 1.99 nJ$ , $P_{L} = 8.32 mW$ and $Q = 9395$ below, then halve $σ_{c}$ and watch $Δ f$ grow by $2$ , from $665$ to $940 kHz$ .

InteractiveLoading simulation…

Check

The closed formula Eq. (7.8.24) gives

Q_{101} = \frac{π f μ _{0} ab d ( a ^{2} + d ^{2} )}{R _{s} [ 2 b ( a ^{3} + d ^{3} ) + a d ( a ^{2} + d ^{2} )]} = \frac{1.110 \times 1 0 ^{- 3}}{0.02063 \times ( 2.73 + 3.00 ) \times 1 0 ^{- 6}} = 9395,

the same value. Also $W / P_{L} = 1.99 nJ /8.32 mW = 0.239 μ s = τ_{W}$ . Units: $ε_{0} E_{0}^{2}$ is $(F/m) (V/m)^{2} = J/ m^{3}$ , times $m^{3}$ gives J.

(a) Response near f_0=6.25\,\mathrm{GHz}. — (a) Response near $f_{0} = 6.25 GHz$ .

(b) Where the 8.32\,\mathrm{mW} are lost. — (a) Response near $f_{0} = 6.25 GHz$ .

Reading the graphs

A detuning of half the bandwidth, $333 kHz$ or one part in $19000$ , already halves the response power (Figure 7.8.23a): a copper cavity is a very sharp filter. The loss splits $14 %$ (end walls), $34 %$ (side walls) and $52 %$ (top and bottom, which carry both $H_{x}$ and $H_{z}$ ). Making $b$ larger raises the volume, and with it $W$ , faster than the loss, since the loss on the top and bottom walls does not depend on $b$ : $Q$ grows.

Problem 7.8.2: TE_{101} energy, loss and Q