Part 7 · Chapter 28

Cavity Resonators and Circular Waveguides

Two variations on the rectangular guide complete the story of guided microwaves. Round the cross-section into a circle and the same mode physics reappears — but the sine and cosine give way to Bessel functions, and the cutoffs are set by their zeros rather than by simple integers. Then cap both ends of a guide with metal, and the travelling wave is trapped into a standing one that rings only at discrete resonant frequencies: a cavity resonator, the microwave cousin of the LC tank, with a quality factor a thousand times sharper. This chapter handles both, and with them closes the book on waves confined by conductors.

Electromagnetic Field Theory Prof. Mithun Mondal Reading time ≈ 54 min
i What you'll learn
  • Why a circular guide's modes are Bessel functions, with cutoffs set by the zeros of \(J_n\) and \(J_n'\).
  • The dominant TE₁₁ mode (\(p_{11}'=1.841\)) and the uniquely low-loss TE₀₁ mode.
  • How closing both ends of a guide creates a standing wave at discrete frequencies.
  • The rectangular cavity resonant frequency and its dominant TE₁₀₁ mode.
  • The quality factor \(Q=\omega_0 W/P_{\text{loss}}=f_r/\Delta f\) and why cavities reach \(Q\sim10^4\).
  • Where resonators appear — oscillators, filters, accelerators, and ovens.
Section 28-1

Two New Geometries

The rectangular guide of Chapter 27 is the most common, but two relatives are worth knowing. The circular waveguide — a round metal pipe — is the natural choice for rotary joints (a rectangular guide cannot rotate about its axis while preserving its field orientation) and for certain low-loss long runs. The cavity resonator takes any guide and seals both ends, trapping the wave so it bounces back and forth and reinforces itself only at special frequencies.

Both obey the same Maxwell physics as before — no TEM, only TE and TM modes, each with a cutoff. What changes is the geometry of the boundary, and so the mathematics of the solution.

Section 28-2

Circular Waveguide Modes

For a pipe of radius \(a\) we switch to cylindrical coordinates \((\rho,\phi,z)\). The Helmholtz equation for the axial field separates into an angular part — sinusoidal in \(\phi\) with integer order \(n\) — and a radial part that is no longer a sine but a Bessel function \(J_n(k_c\rho)\). The second Bessel solution \(Y_n\) is discarded because it blows up on the axis. So:

Axial fields in a circular guide
\[ \text{TM}_{nm}:\;E_z=E_0\,J_n(k_c\rho)\cos n\phi\,e^{-j\beta z}, \qquad \text{TE}_{nm}:\;H_z=H_0\,J_n(k_c\rho)\cos n\phi\,e^{-j\beta z} \]
a E lines · TE₁₁ (dominant)
Circular guide cross-section (radius a); the dominant TE₁₁ field crosses the interior like a bent dipole field

The wall at \(\rho=a\) supplies the boundary condition. For TM modes the axial electric field must vanish there, so \(J_n(k_c a)=0\); for TE modes the radial derivative must vanish, so \(J_n'(k_c a)=0\). Each condition is satisfied only at the discrete zeros of the relevant Bessel function, which therefore quantize the cutoff wavenumber.

Section 28-3

Cutoff and the Dominant Mode

Writing \(p_{nm}\) for the \(m\)-th zero of \(J_n\) and \(p_{nm}'\) for the \(m\)-th zero of \(J_n'\), the cutoff wavenumber and frequency are:

Circular-guide cutoff
\[ k_c=\frac{p_{nm}}{a}\ (\text{TM}),\quad k_c=\frac{p_{nm}'}{a}\ (\text{TE}), \qquad f_{c}=\frac{u'\,k_c}{2\pi} \]
x J J₀ J₁ p′₁₁=1.841 (TE₁₁) p₀₁=2.405 (TM₀₁)
Bessel zeros set the cutoffs: TE₁₁ from the first peak of J₁ (1.841), TM₀₁ from the first zero of J₀ (2.405)

The smallest zero of all is \(p_{11}'=1.841\), so the dominant mode of a circular guide is TE₁₁, with the lowest cutoff. The next is TM₀₁ at \(p_{01}=2.405\). One mode deserves special mention: TE₀₁ is the only common mode whose wall-loss attenuation decreases as frequency rises, which once made it the candidate for long-haul millimetre-wave waveguide links. The key Bessel zeros are worth tabulating:

ModeBessel zeroValueNote
TE₁₁\(p_{11}'\)1.841dominant — lowest cutoff
TM₀₁\(p_{01}\)2.405lowest TM mode
TE₂₁\(p_{21}'\)3.054
TE₀₁ / TM₁₁\(p_{01}'=p_{11}\)3.832TE₀₁ has uniquely falling loss

Everything downstream — guide wavelength, phase and group velocity, wave impedance — carries over unchanged from Chapter 27, because all of it depends only on \(f_c\) through the single factor \(\sqrt{1-(f_c/f)^2}\). The geometry sets \(f_c\); the rest is identical.

Section 28-4

From Guide to Cavity

Now take a length of rectangular guide and short-circuit both ends with conducting plates a distance \(d\) apart. A wave travelling in \(+z\) reflects off the far wall, travels back, reflects off the near wall, and so on. The forward and backward waves superpose into a standing wave, and — exactly as for a wave between two mirrors — only those frequencies survive for which the length holds a whole number of half-wavelengths:

Standing-wave condition along z
\[ \beta=\frac{p\pi}{d}, \qquad p=1,2,3,\dots \]
a b d all six walls conducting
A cavity resonator: sealing both ends of a guide traps a standing wave in a closed box a × b × d

The box now confines the field in all three directions, so a third index \(p\) joins \(m\) and \(n\). The structure rings only at the discrete frequencies that satisfy all three conditions at once — it is a resonator.

Section 28-5

Resonant Frequencies

Combining the transverse cutoff wavenumber with the axial standing-wave condition in \(k^2=k_c^2+\beta^2=\omega^2\mu\varepsilon\) gives the resonant frequency of a rectangular cavity — symmetric in all three dimensions:

Rectangular cavity resonant frequency
\[ f_{r}=\frac{u'}{2}\sqrt{\left(\frac{m}{a}\right)^2+\left(\frac{n}{b}\right)^2+\left(\frac{p}{d}\right)^2} \]

The dominant (lowest-frequency) mode depends on which dimension is smallest. With \(b\) the smallest, the dominant mode is TE₁₀₁, whose frequency uses only the two larger dimensions \(a\) and \(d\):

Dominant mode TE₁₀₁ (b smallest)
\[ f_{r,101}=\frac{u'}{2}\sqrt{\left(\frac{1}{a}\right)^2+\left(\frac{1}{d}\right)^2} \]

Note the index restrictions carry over: TM cavity modes need \(m,n\ge1\), and TE cavity modes need \(p\ge1\) (a half-wave must fit along \(z\) to satisfy the end walls). Because every mode resonates at a different frequency, a single cavity offers a whole comb of sharply defined resonances.

Section 28-6

The Quality Factor

What makes a cavity valuable is not just that it resonates, but that it resonates sharply. The quality factor \(Q\) measures that sharpness — the ratio of energy stored to energy lost each cycle:

Quality factor — two equivalent definitions
\[ Q=\omega_0\,\frac{W_{\text{stored}}}{P_{\text{loss}}}=\frac{f_r}{\Delta f} \]
f |H| −3 dB f_r Δf → Q = f_r/Δf
The resonance curve: a high Q means a narrow half-power bandwidth Δf around f_r

Loss occurs almost entirely in the cavity walls, confined to a thin skin depth \(\delta\). Roughly, \(Q\) scales as the cavity's volume-to-surface ratio divided by \(\delta\) — bigger boxes and shallower skin depths store more and waste less. The result is spectacular: where a lumped LC circuit struggles to reach \(Q\sim100\), a copper cavity easily reaches \(10^3\) to \(10^4\), and superconducting cavities go far higher. That extreme selectivity is precisely why cavities define frequency so well.

Section 28-7

Where Resonators Are Used

The high \(Q\) makes cavity resonators the frequency-defining element of countless microwave systems. A cavity stabilizes the frequency of klystron and Gunn oscillators; coupled cavities form sharp bandpass and band-reject filters; tuned cavities are the resonant structure inside magnetrons (and therefore microwave ovens, whose cooking chamber is itself a large multimode cavity). In particle accelerators, RF cavities deliver precisely timed kicks of energy to the beam; in metrology, cavities underpin frequency standards and the measurement of material permittivity. Wherever a microwave system needs a clean, stable frequency, a resonator is usually doing the defining.

Section 28-8

Worked Examples

1 Circular-guide dominant cutoff

Problem. An air-filled circular waveguide has radius \(a=2\,\text{cm}\). Find the cutoff frequency of the dominant mode.

Solution. The dominant mode is TE₁₁ with \(k_c=p_{11}'/a=1.841/a\):

Working
\[ f_{c}=\frac{u'p_{11}'}{2\pi a}=\frac{(3\times10^8)(1.841)}{2\pi(0.02)}\approx 4.40\ \text{GHz} \]
2 Next mode and single-mode band

Problem. For the same guide, find the TM₀₁ cutoff and state the single-mode band.

Solution. TM₀₁ uses \(p_{01}=2.405\):

Working
\[ f_{c,01}=\frac{(3\times10^8)(2.405)}{2\pi(0.02)}\approx 5.74\ \text{GHz}\;\Rightarrow\;\text{single-mode } 4.40\text{–}5.74\ \text{GHz} \]
3 Cutoff wavelength

Problem. Find the cutoff wavelength of the TE₁₁ mode in that guide.

Solution. \(\lambda_c=2\pi a/p_{11}'\):

Working
\[ \lambda_c=\frac{2\pi(0.02)}{1.841}\approx 0.0683\ \text{m}=6.83\ \text{cm}\;(=3.41\,a) \]
4 Cavity resonant frequency

Problem. An air cavity has \(a=5\,\text{cm}\), \(b=2\,\text{cm}\), \(d=4\,\text{cm}\). Find the dominant-mode resonant frequency.

Solution. Since \(b\) is smallest, the dominant mode is TE₁₀₁:

Working
\[ f_{r,101}=\frac{3\times10^8}{2}\sqrt{\left(\tfrac{1}{0.05}\right)^2+\left(\tfrac{1}{0.04}\right)^2}=1.5\times10^8\sqrt{1025}\approx 4.80\ \text{GHz} \]
5 Confirming the dominant mode

Problem. For the same cavity, find the TM₁₁₀ frequency and confirm TE₁₀₁ is lower.

Solution. TM₁₁₀ uses \(a\) and \(b\):

Working
\[ f_{r,110}=1.5\times10^8\sqrt{\left(\tfrac{1}{0.05}\right)^2+\left(\tfrac{1}{0.02}\right)^2}=1.5\times10^8\sqrt{2900}\approx 8.08\ \text{GHz}>4.80 \]
6 Q from bandwidth

Problem. A cavity resonates at \(f_r=10\,\text{GHz}\) with a measured half-power bandwidth \(\Delta f=1\,\text{MHz}\). Find its quality factor.

Solution. Use \(Q=f_r/\Delta f\):

Working
\[ Q=\frac{10\times10^{9}}{1\times10^{6}}=10{,}000 \]
Review

Chapter Summary

Circular modes

Fields are Bessel functions \(J_n(k_c\rho)\); cutoffs are the zeros of \(J_n\) (TM) and \(J_n'\) (TE).

Dominant mode

TE₁₁ (\(p_{11}'=1.841\)); next TM₀₁ (2.405); TE₀₁ has uniquely falling loss.

Carried over

\(\lambda_g\), \(u_p\), \(u_g\), \(\eta\) all follow from \(f_c\) as in Chapter 27.

Cavity

Seal both ends: \(\beta=p\pi/d\); a third index \(p\) and discrete resonances.

Resonance

\(f_r=\tfrac{u'}{2}\sqrt{(m/a)^2+(n/b)^2+(p/d)^2}\); dominant TE₁₀₁.

Quality factor

\(Q=\omega_0 W/P_{\text{loss}}=f_r/\Delta f\); cavities reach \(10^3\)–\(10^4\).

Practice

Problems

For circular guides, pick the right Bessel zero for the mode; for cavities, identify the smallest dimension to find the dominant mode. Assume air filling unless stated. Difficulty rises down the list.

  1. An air circular guide has radius \(a=1.5\,\text{cm}\). Find the TE₁₁ cutoff frequency.
  2. For the same guide, find the TM₀₁ cutoff and the single-mode bandwidth.
  3. Why is the dominant mode of a circular guide TE₁₁ and not TM₀₁?
  4. Find the TE₁₁ cutoff wavelength for a guide of radius \(2.5\,\text{cm}\).
  5. A circular guide must have its TE₁₁ cutoff at \(6\,\text{GHz}\). Find the required radius.
  6. State which common circular-guide mode has attenuation that falls with frequency, and why that is useful.
  7. An air cavity is \(a=4\,\text{cm}\), \(b=2\,\text{cm}\), \(d=4\,\text{cm}\). Find the TE₁₀₁ resonant frequency.
  8. For the same cavity, find the TM₁₁₀ frequency and identify the dominant mode.
  9. A cubic cavity has all sides \(3\,\text{cm}\). Find the lowest resonant frequency.
  10. A cavity resonates at \(8\,\text{GHz}\) with \(Q=5000\). Find the half-power bandwidth.
  11. Explain why a cavity reaches a far higher \(Q\) than a lumped LC circuit at the same frequency.
  12. Show that filling a cavity with a dielectric of \(\varepsilon_r\) lowers every resonant frequency by the factor \(1/\sqrt{\varepsilon_r}\).
Tip: both structures are the rectangular guide's physics in new clothes. Round the cross-section and the modes become Bessel functions, with cutoff set not by integers but by Bessel zeros — TE from the zeros of \(J_n'\), TM from the zeros of \(J_n\); the dominant TE₁₁ comes from the smallest of all, \(p_{11}'=1.841\). Everything past cutoff (\(\lambda_g\), the velocities, the wave impedance) is identical to Chapter 27 because it depends only on \(f_c\). Cap both ends and the guide becomes a cavity: a third condition \(\beta=p\pi/d\) quantizes the axial direction too, giving discrete resonances \(f_r=\tfrac{u'}{2}\sqrt{(m/a)^2+(n/b)^2+(p/d)^2}\), sharpened by a quality factor \(Q=f_r/\Delta f\) that dwarfs any lumped circuit. That closes the chapters on guided waves. From here the field breaks free of its conductors entirely — Chapter 29 opens the antenna sections with how an accelerating current launches radiation into open space.