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

Walls of conductivity $σ_{c}$ and permeability $μ_{c}$ have the surface resistance

R_{s} = \frac{π f μ _{c}}{σ _{c}} = \frac{1}{σ _{c} δ} [Ω], (7.8.22)

where $δ = 1/ π f μ_{c} σ_{c}$ is the skin depth (Sections 6.2 and 7.3). A tangential magnetic field $H_{tan}$ at the wall drives the surface current $∣ J_{s} ∣ = ∣ H_{tan} ∣$ , which dissipates $\frac{R _{s}}{2} ∣ H_{tan} ∣^{2}$ per unit area. As for the attenuation of a guide, we compute this loss with the fields of the lossless cavity, which barely change when $δ$ is a micrometre.

For the TE $_{101}$ mode Eq. (7.8.12) the six walls come in three pairs:

the end walls $z = 0, d$ (each $a \times b$ ) carry $H_{x}$ , with $∣ H_{x} ∣^{2} = \frac{∣ E _{0} ∣ ^{2}}{η ^{2}} (\frac{λ}{2 d})^{2} sin^{2} \frac{π x}{a}$ , and contribute $(\frac{λ}{2 d})^{2} ab$ to the bracket below,
the side walls $x = 0, a$ (each $b \times d$ ) carry $H_{z}$ and contribute $(\frac{λ}{2 a})^{2} b d$ ,
the top and bottom walls $y = 0, b$ (each $a \times d$ ) carry both $H_{x}$ and $H_{z}$ , and with Eq. (7.8.13) they contribute $\frac{a d}{4} [(\frac{λ}{2 d})^{2} + (\frac{λ}{2 a})^{2}] \times 2 = \frac{a d}{2}$ .

Adding the three pairs,

P_{L} = \frac{R _{s} ∣ E _{0} ∣ ^{2}}{2 η ^{2}} [(\frac{λ}{2 d})^{2} ab + (\frac{λ}{2 a})^{2} b d + \frac{a d}{2}] . (7.8.23)

With $W$ from Eq. (7.8.17), $Q = ω_{0} W / P_{L}$ , $ε η^{2} = μ$ , and $(λ /2 d)^{2} = a^{2} / (a^{2} + d^{2})$ , $(λ /2 a)^{2} = d^{2} / (a^{2} + d^{2})$ from Eq. (7.8.13), the dimensions collect into one formula.

Q of the TE₁₀₁ mode

Q_{101} = \frac{π f _{101} μ ab d ( a ^{2} + d ^{2} )}{R _{s} [ 2 b ( a ^{3} + d ^{3} ) + a d ( a ^{2} + d ^{2} ) ]} . (7.8.24)

For a cube ( $a = b = d$ ) it reduces to

Q = \frac{π f μ a}{3 R _{s}} = \frac{a}{3 δ} (μ = μ_{c}), (7.8.25)

the size of the cavity over the skin depth, times one third.

In plain words

$Q$ compares the volume where the energy is stored with the thin skin where it is lost. A cavity a few centimetres across with a skin depth below a micrometre has $Q$ of order ten thousand.

A copper cube with $a = 2 cm$ resonates at $f_{101} = 7.5 2 = 10.6 GHz$ , where $δ = 0.64 μ m$ , and $Q = a /3 δ = 10400$ . Figure 7.8.8 shows how this changes with the size. Scaling the cube down raises $f \propto 1/ a$ and lowers $δ \propto f^{- 1/2}$ less quickly, so $Q \propto a^{1/2} \propto f^{- 1/2}$ : small, high-frequency cavities have lower $Q$ .

Q=a/3\delta of an air-filled cubic TE_{101} cavity against its resonant frequency f=c/(\sqrt2\,a), for walls of copper (5.8\times10^7\,\mathrm{S/m}), aluminium (3.5\times10^7\,\mathrm{S/m}) and brass (1.57\times10^7\,\mathrm{S/m}). The dot is the 2\,\mathrm{cm} copper cube at 10.6\,\mathrm{GHz}. — **Figure 7.8.8.** $Q = a /3 δ$ of an air-filled cubic TE $_{101}$ cavity against its resonant frequency $f = c / (2 a)$ , for walls of copper ( $5.8 \times 1 0^{7} S/m$ ), aluminium ( $3.5 \times 1 0^{7} S/m$ ) and brass ( $1.57 \times 1 0^{7} S/m$ ). The dot is the $2 cm$ copper cube at $10.6 GHz$ .

Recipe · Computing the

Q

of a cavity mode

Fields. Write the lossless fields of the mode with one amplitude, $E_{0}$ or $H_{0}$ .
Energy. $W = 2 W_{e} = \frac{ε}{2} \int_{V} ∣ E ∣^{2} d v$ (at resonance $W_{e} = W_{m}$ , so either may be doubled).
Loss. On each wall take the tangential $H$ and integrate $\frac{R _{s}}{2} ∣ H_{tan} ∣^{2}$ over the wall. Add all the walls, then the dielectric loss if the filling absorbs.
Ratio. $Q = ω_{0} W / P_{L}$ . The amplitude cancels, and the result depends only on the geometry, the frequency and $R_{s}$ .

Quick check

All dimensions of a copper cubic cavity are doubled. How do its TE $_{101}$ resonant frequency and its $Q$ change?

The frequency halves. The skin depth grows by $2$ , and $Q = a /3 δ$ grows by $2/ 2 = 2$ . Exercise 7.8.3 uses the same scaling.

7.8.9 Q of the TE₁₀₁ mode with conducting walls