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

A cavity loses energy through several channels: the walls (conductor loss), the filling (dielectric loss) and the couplings to the outside world, through which the power leaves to the source and the load. Each channel adds its own power to $P_{L}$ , and since $1/ Q = P_{L} / (ω_{0} W)$ , the reciprocal $Q$ values add:

\frac{1}{Q _{L}} Q_{d} = 1/ Q_{0} \frac{1}{Q _{c}} + \frac{1}{Q _{d}} + \frac{1}{Q _{ext}}, = \frac{1}{tan δ} . (7.8.26)

The unloaded $Q_{0}$ describes the cavity itself (walls and filling), $Q_{ext}$ the power that leaves through the couplings, and the loaded $Q_{L}$ the whole. The result $Q_{d} = 1/ tan δ$ holds for a cavity completely filled with a dielectric of loss tangent $tan δ = ε^{''} / ε^{'}$ : the dielectric dissipates $\frac{ω ε ^{''}}{2} \int ∣ E ∣^{2} d v = 2 ω tan δ W_{e} = ω tan δ W$ , and $ω W / P_{d} = 1/ tan δ$ .

The strength of a coupling is measured by the coupling coefficient $β = Q_{0} / Q_{ext}$ of each port, the ratio of the power that leaves through the port to the power lost inside. At $f_{0}$ the cavity looks to the line like a shunt conductance of normalised value $g = 1/ β$ . For a symmetric transmission cavity with two ports (Figure 7.8.9a) the line at port 1 sees $g$ in parallel with the matched port 2, the normalised admittance $1 + g$ , so that $S_{11} (f_{0}) = \frac{1 - ( 1 + g )}{1 + ( 1 + g )} = - \frac{1}{1 + 2 β}$ and the transmitted wave is $S_{21} = 1 + S_{11}$ (the shunt element leaves the voltage continuous).

A symmetric two-port cavity at resonance

S_{21} (f_{0}) Q_{L} S_{11} (f_{0}) = \frac{2 β}{1 + 2 β} = 1 - \frac{Q _{L}}{Q _{0}}, = \frac{Q _{0}}{1 + 2 β}, = - \frac{1}{1 + 2 β} . (7.8.27)

A one-port cavity has $Q_{L} = Q_{0} / (1 + β)$ and $S_{11} (f_{0}) = (β - 1) / (β + 1)$ . It is matched at critical coupling, $β = 1$ , where $Q_{L} = Q_{0} /2$ .

In plain words

Couple strongly and most of the power passes through, but the band is wide (low $Q_{L}$ ). Couple weakly and the band is narrow, but most of the power that enters is lost in the walls. A narrow, low-loss filter therefore needs a resonator with a high $Q_{0}$ .

(b) |S_{21}| for three couplings. — (a) Cavity between two irises.

Common mistake · Reading

Q_{0}

from a measured bandwidth

Taking the measured half-power width as $f_{0} / Q_{0}$ . A measured resonance curve always includes the loading by the ports, so $f_{0} /Δ f$ is the loaded $Q_{L}$ . Only with very weak couplings ( $β \to 0$ ) does it approach $Q_{0}$ . Otherwise $Q_{0}$ must be recovered from $Q_{L}$ and the coupling, for example from the peak transmission with Eq. (7.8.27) (Problem 7.8.4) or from the reflection of a one-port cavity (Exercise 7.8.5).

Quick check

A one-port cavity with $Q_{0} = 8000$ is critically coupled. What bandwidth does one measure at $10 GHz$ , and how much of the incident power is reflected at $f_{0}$ ?

$β = 1$ gives $Q_{L} = Q_{0} /2 = 4000$ and $Δ f = 10 GHz /4000 = 2.5 MHz$ . $S_{11} (f_{0}) = (1 - 1) / (1 + 1) = 0$ : nothing is reflected, all the incident power is absorbed in the walls.

Application 7.8.1 · The microwave oven cavity

A MULTIMODE CAVITY

A domestic microwave oven is a metal box of about $32 \times 20 \times 30 cm$ fed by a $1 kW$ magnetron through a short TE $_{10}$ guide (Section 7.3 and Figure 7.8.10a). The perforated metal sheet in the door is part of the wall: its holes are much smaller than the wavelength and the field does not leak through them. At $2.45 GHz$ ( $λ = 12.2 cm$ ) the box is several wavelengths wide in every direction. Its lowest resonance (TE $_{101}$ ) is at only $0.69 GHz$ , and near $2.45 GHz$ the resonances crowd together, about $9 MHz$ apart. A large cavity has a dense spectrum, and the number of modes in a band $Δ f$ (both polarisations) is estimated by the mode-counting formula

f_{mn p} Δ N = \frac{c}{2} (\frac{m}{a})^{2} + (\frac{n}{b})^{2} + (\frac{p}{d})^{2}, \approx \frac{8 π V f ^{2}}{c ^{3}} Δ f, (7.8.28)

where $V = ab d$ . It counts the integer triples $(m, n, p)$ inside a thin shell of an ellipsoid in index space, two modes per triple.

(b) |E_y|^2 of TE_{304} on the floor. — (a) The oven: magnetron, feed guide, cavity.

Typical numbers (

32 \times 20 \times 30 cm

)

Quantity	Value
Resonances in $2.40$ to $2.50 GHz$	$11$ (estimate $10.7$ )
Mean spacing	$\approx 9 MHz$
Loaded $Q$ with food	$\sim 100$ , so $Δ f \approx 25 MHz$
Hot-spot spacing (TE $_{304}$ )	$a /3 = 10.7 cm$ , $d /4 = 7.5 cm$
Empty oven ( $Q$ of steel walls)	thousands

The food, wet and very lossy, lowers $Q_{L}$ to about $100$ . The bandwidth $f_{0} / Q_{L}$ then spans several resonances, and several modes are driven together, each with its own hot spots (Figure 7.8.10b shows one of them). Their spacing is never less than $λ /2 = 6.1 cm$ , since each of the three components of the wave vector is at most $k$ .
The turntable carries the food through the pattern, and the mode stirrer, a rotating metal fan, shifts the resonances in time so that the pattern itself moves. Both even out the heating.
Never run an oven empty. Its $Q$ is then high, little power is absorbed, and most of it returns to the magnetron (Home exercise 7.8.4 shows how a circulator protects it). Worked example 7.8.1 puts numbers in.

With $Q_{L} = 100$ below, several resonances respond to the $2.45 GHz$ magnetron at once. Count the hot spots of TE $_{304}$ .

InteractiveLoading simulation…

7.8.10 Loaded Q, coupling and bandwidth