Resonant cavities, stored energy and Q
Close the waveguide into a rectangular PEC cavity; derive TM110 and TE101 eigenfields, identify allowed mode triples, calculate resonance, stored energy and conductor/dielectric loss, and distinguish Q0, external Q and loaded Q in VNA measurements.
- Derive the vector cavity eigenproblem from Maxwell with PEC conditions on all six walls.
- Explain the distinct TE-to-z and TM-to-z index restrictions for a closed cavity.
- Derive rectangular cavity eigenfrequencies by separation rather than memorizing a table.
- Reconstruct TM110 and TE101 fields and check their signed Maxwell curls.
- Audit resonant mode ordering and geometric degeneracy.
- Derive peak-phasor electric and magnetic stored-energy integrals.
- Prove equality of average electric/magnetic energy at ideal eigenresonance.
- Derive the perturbative good-conductor and weak-dielectric loss models.
- Distinguish unloaded Q0, external coupling Qe and loaded QL.
- Interpret a VNA resonance bandwidth only within a stated coupling/response model.
A hollow guide becomes a resonator when conducting end walls close its longitudinal direction. This changes a traveling-wave problem into a three-dimensional eigenvalue problem: all six PEC walls must be satisfied by the complete vector field, not just by a scalar frequency formula. We construct allowed standing-wave modes, then add controlled losses, quality factors and the interpretation of a VNA measurement.
0105.1 — Maxwell eigenproblem in a closed cavity
Consider a PEC rectangular box 0<x<a, 0<y<b, 0<z<d with homogeneous, lossless, nondispersive ε and μ. Use peak phasors with exp(+jωt). Without impressed sources, Maxwell gives ∇×E=−jωμH and ∇×H=+jωεE; the divergence constraints then yield the vector Helmholtz equations.
A perfect conductor imposes n×E=0 and n·B=0 on all six walls. These are vector conditions: an allowed mathematical scalar eigenvalue does not automatically produce a nonzero physical mode. The ideal resonator has real eigenfrequencies and no intrinsic power loss.
0205.2 — Three-dimensional quantization and valid indices
Separate the Cartesian Helmholtz equation along x, y and z. Sinusoidal factors have transverse wavenumbers mπ/a, nπ/b and longitudinal wavenumber pπ/d. Combining their squares produces the rectangular-cavity resonance relation, but the full wall conditions select which integers are legal.
For TM-to-z, E_z can be chosen proportional to sin(mπx/a)sin(nπy/b)cos(pπz/d); m,n≥1 and p≥0. For TE-to-z, H_z can be chosen proportional to cos(mπx/a)cos(nπy/b)sin(pπz/d); p≥1 and (m,n)≠(0,0). Thus TM110 and TE101 are valid, while TE100 and TM100 are not with this choice of axis and indexing.
0305.3 — Explicit TM110 and TE101 field reconstruction
Take the TM110 mode with E_z=E₀sin(k_xx)sin(k_yy), k_x=π/a, k_y=π/b and no z dependence. Maxwell gives E_x=E_y=H_z=0, H_x=j(∂_yE_z)/(ωμ), H_y=−j(∂_xE_z)/(ωμ). The electric field is normal at the z end walls and tangentially zero on the four side walls, as required.
For TE101, choose H_z=H₀cos(k_xx)sin(k_zz), k_x=π/a, k_z=π/d and k_y=0. A consistent nonzero set is E_y=−j(ωμ/k_x)H₀sin(k_xx)sin(k_zz), H_x=−(k_z/k_x)H₀sin(k_xx)cos(k_zz), H_z as specified. It satisfies n×E=0 on the four side walls and both endcaps, and n·B=0 on each wall. Signs follow from exp(+jωt).
0405.4 — Mode ordering, geometric tuning and degeneracy
The resonance formula must be evaluated only over allowed mode triples. For vacuum dimensions a=30 mm, b=20 mm, d=40 mm, the selected TE101, TE011 and TM110 frequencies are approximately 6.246 GHz, 8.379 GHz and 9.008 GHz, respectively. These are not a claim that no other cavity mode exists: a global lowest-mode statement requires enumeration of both families.
Changing d moves TE101 and TE011 because they contain p=1, but does not shift TM110 in the ideal nondispersive cavity because its longitudinal index p is zero. Degeneracies occur when different allowed integer triples produce the same weighted sum of squares; at degeneracy any orthogonal basis in the eigenspace can be used.
Figure 5.1 — How cavity length changes modal frequencies
For fixed transverse dimensions a=30 mm and b=20 mm, TE101 changes with length d; TM110 does not, in the ideal PEC model. The two traces describe different eigenmodes.
0505.5 — Stored energy, equal partition and normalization
Under peak phasors in a nondispersive lossless filling, the time-averaged electric energy is W_e=(1/4)∫ε|E|²dV and magnetic energy is W_m=(1/4)∫μ|H|²dV. The total cycle-averaged stored energy is W=W_e+W_m; do not accidentally replace peak phasors by RMS phasors without changing the factors.
At a nonzero ideal PEC eigenfrequency, integrate the identity ∇·(E×H*)=H*·(∇×E)−E·(∇×H*). PEC walls give zero normal real/complex flux, while Maxwell yields equality of the volume electric and magnetic integrals. Consequently W_e=W_m. Field amplitudes must be fixed by an explicit normalization, such as a specified stored energy.
0605.6 — Intrinsic losses and unloaded quality factor
For a nonideal cavity define Q₀=ω₀W/P_internal. An exactly PEC and perfectly lossless cavity has P_internal=0 and formally infinite Q₀. With weak perturbative wall loss in the classical good-conductor regime, surface resistance R_s=√(ωμ_c/(2σ)) leads to P_wall=(R_s/2)∮|H_t|²dS using idealized tangential magnetic fields.
A dielectric with complex ε_c=ε'−jε'' under exp(+jωt), ε''>0 for passivity, dissipates P_d=(ω/2)∫ε''|E|²dV. The reciprocal internal-Q contributions add only when power loss channels can be evaluated consistently in the weak-loss model: 1/Q₀≈1/Q_wall+1/Q_dielectric+.... A surface resistance model can fail for very thin, nonlocal or rough conducting films.
0705.7 — External coupling and loaded quality factor
A cavity accessible through a probe or aperture exchanges energy with the outside world. The corresponding external quality factors Q_ei are separate from internal wall/dielectric losses. Energy balance at weak coupling gives 1/Q_L=1/Q₀+Σ_i1/Q_ei. A strongly coupled resonator can therefore have a low loaded Q_L even if its intrinsic Q₀ is high.
For a single-port convention define the coupling coefficient g=Q₀/Q_e. Then Q_L=Q₀/(1+g). A simple bandwidth estimate Q_L≈f₀/Δf_3dB applies to an isolated resonance with a proper Lorentzian power response; direct 3 dB reading on arbitrary distorted S11 or S21 spectra is not justified.
0805.8 — Numerical design and VNA interpretation
With a=30 mm, b=20 mm and d=40 mm in vacuum, compute allowed modal frequencies before selecting a probe position. A probe located at a node of its intended electric-field component couples weakly even if its frequency matches an eigenvalue; field symmetry and polarization therefore matter as much as resonance frequency.
As an illustrative one-port model, take f₀=6.246 GHz, Q₀=3000 and Q_e=1500. Then g=2, Q_L=1000 and the idealized Lorentzian power width is about 6.246 MHz. This width is not guaranteed by an arbitrary raw VNA trace: correct calibration/reference plane, background coupling, cable loss and possible mode overlap are necessary before interpreting it. Frequencies and Q may drift with wall finish, temperature and dielectric loading.
Worked exercises — cavity modes and quality factors
- Starting with source-free phasor Maxwell equations, derive the vector Helmholtz equation and state every PEC boundary condition for a closed cavity.
- Separate the scalar cavity eigenproblem and derive the general frequency formula, explaining why not all integer triples are allowed.
- Show that TM110 is permitted whereas TM100 is forbidden, and identify which PEC conditions constrain E_z.
- Starting from TM110 E_z, calculate H_x and H_y and verify the z component of Ampère’s law.
- Reconstruct the TE101 mode and verify electric tangential boundary data on both z endcaps.
- For a=30 mm, b=20 mm, d=40 mm in vacuum, calculate f_TE101, f_TE011 and f_TM110 with units.
- Find a condition for geometric degeneracy between TE101 and TE011; explain why identical eigenfrequency does not imply identical fields.
- Using the complex Poynting identity and PEC walls, derive the equality W_e=W_m at an ideal eigenfrequency.
- Given W=1 μJ, f₀=6 GHz and internal loss P=20 mW, calculate Q₀. Explain the PEC limit.
- Given Q₀=3000, Q_e=1500 and f₀=6.246 GHz, find g, Q_L and an idealized resonance linewidth; explain a measurement limitation.
Worked solutions
1. Apply ∇× to ∇×E=−jωμH. Since μ is uniform, ∇×∇×E=−jωμ(jωεE)=ω²μεE. The identity ∇×∇×E=∇(∇·E)−∇²E and ∇·E=0 give (∇²+k²)E=0 with k²=ω²με; similarly for H. On each PEC wall n×E=0 and n·B=0. The six wall normals have different directions and must all be considered.
2. Assume X(x)Y(y)Z(z), divide the scalar Helmholtz equation by XYZ and introduce separation constants k_x²,k_y²,k_z². PEC boundaries quantize them as mπ/a,nπ/b,pπ/d, giving ω²με=π²[(m/a)²+(n/b)²+(p/d)²]. Thus f=(2√(με))⁻¹√[...]. Vector PEC conditions eliminate some index triples, so this scalar formula alone is insufficient.
3. For TM_z, E_z must vanish on x and y walls, giving sin(mπx/a)sin(nπy/b) and m,n≥1. At z endcaps E_z is normal, so it can be nonzero; cos(pπz/d) with p=0 is allowed. TM110 is therefore nontrivial, while TM100 has n=0 and vanishes identically.
4. For TM110, E_z=E₀sin(k_xx)sin(k_yy). Faraday gives (∇×E)_x=∂_yE_z=−jωμH_x, so H_x=j∂_yE_z/(ωμ); (∇×E)_y=−∂_xE_z=−jωμH_y gives H_y=−j∂_xE_z/(ωμ). Then (∇×H)_z=∂_xH_y−∂_yH_x=−j∇_t²E_z/(ωμ)=j(k_x²+k_y²)E_z/(ωμ)=jωεE_z.
5. Take H_z=H₀cos(k_xx)sin(k_zz) with k_x=π/a, k_z=π/d. A Maxwell-consistent pair is E_y=−jωμH₀sin(k_xx)sin(k_zz)/k_x and H_x=−k_zH₀sin(k_xx)cos(k_zz)/k_x; all other E components vanish. E_y is zero on x=0,a and on z=0,d, while on y walls E_y is normal. The normal magnetic H_z also vanishes at the endcaps.
6. Set c=299792458 m/s. For TE101, f=(c/2)√[a⁻²+d⁻²]≈6.246 GHz; TE011 gives (c/2)√[b⁻²+d⁻²]≈8.379 GHz; TM110 gives (c/2)√[a⁻²+b⁻²]≈9.008 GHz. These comparisons concern three allowed modes, not every possible cavity mode.
7. TE101 and TE011 share p=1 and have squared frequencies proportional to a⁻²+d⁻² and b⁻²+d⁻². Equality therefore requires a=b. Their spatial variations lie along different transverse coordinates; at degeneracy the vector eigenfunctions form an eigenspace and can be combined, but their field maps are not identical.
8. Use ∇·(E×H*)=H*·(∇×E)−E·(∇×H*)=−jωμ|H|²+jωε|E|² with the +jωt convention. PEC imposes n×E=0, so the surface integral of n·(E×H*) vanishes. For ω≠0, the volume integrals satisfy ∫μ|H|²dV=∫ε|E|²dV, and therefore W_m=W_e for the peak-phasor factors 1/4.
9. Q₀=ω₀W/P=(2π×6×10⁹×10⁻⁶)/(20×10⁻³)≈1.885×10⁶. Units cancel: angular frequency (s⁻¹) times energy (J) over power (J/s). The strictly lossless PEC cavity has P=0 and Q₀ formally tends to infinity; a finite Q requires an actual loss channel.
10. The coupling coefficient is g=Q₀/Q_e=2. Inverse-Q addition gives Q_L=(1/3000+1/1500)⁻¹=1000 and Δf_3dB≈f₀/Q_L=6.246 MHz for an isolated Lorentzian power resonance. Real VNA measurements may require background correction, coupling-model fitting and calibrated reference planes, so direct width reading is conditional.
With the Maxwell eigenproblem, power normalization and loss channels established, a later RF module can connect resonators to port variables, scattering parameters and calibrated VNA response.