core30 min studyLesson 288 in Physics

Microwave waveguide cutoff

Derive rectangular-guide cutoff from transverse boundary conditions and distinguish propagating TE/TM modes, evanescent fields, and phase/group velocities.

Before you begin

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The question for today

What you’ll understand

Calculate TE and TM cutoff frequencies for a rectangular hollow guide.
Identify a single-mode operating interval and valid index restrictions.
Distinguish below-cutoff decay from dissipative attenuation and signal speed.

A hollow metallic pipe guides electromagnetic fields differently from a two-conductor TEM line. Its boundary conditions force spatial variation across the cross-section, consuming part of the available wavevector before any longitudinal phase variation is possible. A mode propagates only above its cutoff frequency. Below cutoff the field can penetrate a finite distance as an evanescent solution, even in perfectly conducting walls. The resulting decay is geometric rather than automatically a conversion of electromagnetic energy into heat.

Transverse geometry selects the modes

Consider an empty rectangular guide with side lengths a and b, perfect conducting walls, and propagation along z. TE means the electric field has no longitudinal component, while TM means the magnetic field has no longitudinal component. Neither label means that every field component is transverse. The longitudinal field component that remains solves a transverse eigenvalue problem, with boundary conditions inherited from vanishing tangential electric field on the walls.

For rectangular TE modes, the transverse integer indices are nonnegative and cannot both vanish. For TM modes, both indices must be positive. Thus TE10 exists but TM10 does not. A simply connected hollow conducting guide has no nontrivial TEM mode in this homogeneous setting: a transverse electrostatic potential constant on its single conducting boundary has no internal voltage difference. A coaxial line avoids that argument by having two separate conductors and supports a TEM mode with no analogous low-frequency cutoff.

k_{c,mn}^2=\left(\frac{m\pi}{a}\right)^2+\left(\frac{n\pi}{b}\right)^2,\qquad f_{c,mn}=\frac c2\sqrt{\left(\frac ma\right)^2+\left(\frac nb\right)^2}

a and b are in metres, kc is in inverse metres, and cutoff frequency fc is in hertz for vacuum speed c. TE indices satisfy m,n ≥ 0 with (m,n) ≠ (0,0); TM requires m,n ≥ 1. Perfect walls and vacuum filling are assumed.↔ Scroll wide equations horizontally.

Cutoff follows from the longitudinal wavevector

The vacuum wavevector magnitude is fixed by frequency, while the transverse contribution is fixed by the guide and mode indices. Subtracting their squares leaves the longitudinal propagation constant squared. Above cutoff it is positive, allowing a traveling phase. At cutoff the longitudinal phase constant vanishes and guide wavelength diverges. Below cutoff the longitudinal solution becomes exponential rather than oscillatory. A boundary or source can excite that field even though it is not an independently propagating mode of an infinite guide.

In a semi-infinite empty guide below cutoff, choose the decaying branch to avoid growth at infinity. A single such ideal evanescent field carries no net average longitudinal power. In a finite below-cutoff section, both exponential solutions can be needed to satisfy the two boundaries, and power can couple through it. Therefore saying that fields are identically zero below cutoff is wrong, and saying that exponential decay necessarily proves absorption is also wrong. Wall resistance would introduce an additional, genuinely dissipative loss mechanism.

\beta^2=\left(\frac\omega c\right)^2-k_c^2,\qquad \alpha_{\mathrm{ev}}=\sqrt{k_c^2-\left(\frac\omega c\right)^2}\quad(\omega<ck_c)

β is the above-cutoff phase constant in rad/m, and αev is the below-cutoff spatial decay rate in inverse metres. With the e^{jωt} convention a decaying below-cutoff field varies as e^{−αev z}. This is not a conductor-loss coefficient.↔ Scroll wide equations horizontally.

Dispersion separates phase from energy transport

For an isolated propagating mode in a lossless empty guide, differentiating the dispersion relation gives group velocity smaller than c, while angular frequency divided by phase constant gives phase velocity larger than c. Their product is c squared. The superluminal phase velocity moves a phase marker rather than new information. Narrowband pulse envelopes follow group velocity when the dispersion approximation is valid; causal signal fronts remain consistent with Maxwell propagation and do not outrun vacuum light.

If a exceeds b, TE10 has the lowest cutoff, but the next accessible mode depends on both dimensions. Compare TE20 and TE01 rather than automatically doubling the first cutoff. Operating between the lowest and next cutoffs permits only the fundamental propagating mode in the uniform guide. Discontinuities can still generate locally evanescent higher modes, and operation extremely near cutoff brings strong dispersion and practical loss sensitivity. Single-mode design is therefore more than choosing any frequency barely above the first threshold.

v_p=\frac c{\sqrt{1-(f_c/f)^2}},\qquad v_g=c\sqrt{1-(f_c/f)^2},\qquad v_pv_g=c^2

These formulas apply above cutoff to one lossless vacuum-filled guide mode. vp and vg are in m/s. Group-velocity interpretation assumes a sufficiently narrowband wavepacket; it does not describe a below-cutoff decaying field.↔ Scroll wide equations horizontally.

Work it through

A single-mode rectangular guide

An air-filled ideal rectangular guide has a = 3.0 cm and b = 1.0 cm. Using c = 3.0 × 10⁸ m/s, find the lowest cutoff, the next cutoff, and the group velocity at 6.25 GHz.

The fundamental is TE10 with $f_{c,10}=c/(2a)=5.0\,\mathrm{GHz}$ . TE20 cuts on at 10 GHz, and TE01 at 15 GHz.
TM11 and other mixed-index modes have still larger cutoffs than TE20 here. The ideal single-propagating-mode interval is therefore 5.0 < f < 10 GHz.
At 6.25 GHz, fc/f = 0.8. Thus $v_g=c\sqrt{1-0.8^2}=0.6c=1.8\times10^8\,\mathrm{m/s}$ , while phase velocity is c/0.6.

The result

The first two cutoffs are 5.0 and 10 GHz. At 6.25 GHz the guide supports only TE10 propagation and its group velocity is 1.8 × 10⁸ m/s.

The living notebook

Explore the model

Interactive lab

Interactive lab · spectral

Compare transverse scales without inventing guide dynamics

Reset and set BOTH Mode count N to 1 and Coefficient decay p to 1; predict one spatial sine scale. Then Reset and set BOTH N to 4 and p to 1; predict finer spatial features. The lab displays Σ sin(nx − t)/nᵖ with common temporal frequency. It illustrates spatial mode content only: it has no TE/TM boundary fields, cutoff control, physical dispersion, or waveguide group-velocity readout.

Finite sumHighest included mode

Sum at x = π/2: 0.867
Smallest wavelength: 2π / 5
Highest-mode coefficient: 0.2

Mode count N5Coefficient decay p1

t = 0 s

A finite sum of sin(nx − t)/nᵖ. Playback changes t; increasing the cutoff includes finer spatial scales. This is not a proof about a continuum limit or renormalization. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forTransverse spatial structure matters, but the frequency assigned to each physical mode must come from its own dispersion relation.

Pause & reason

Check your understanding

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Read the checkpoint explanation

Rectangular TM modes require both transverse indices to be positive. TE modes can have one zero index, but not both. TM10 would give a trivial longitudinal electric field under the conducting boundary conditions.

Your turn

Try it yourself

An ideal rectangular guide has a = 4.0 cm and b = 2.0 cm. Find the TE10 cutoff and the group and phase velocities at f = 1.25 times that cutoff.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

For this mode only a sets cutoff. The ratio fc/f equals 0.8, making the velocity square root simple.

Compare with the solution

The cutoff is $3.0\times10^8/(2\times0.040)=3.75\,\mathrm{GHz}$ . The operating frequency is 4.6875 GHz, above this cutoff. Since $\sqrt{1-0.8^2}=0.6$ , group velocity is $1.8\times10^8\,\mathrm{m/s}$ and phase velocity is $5.0\times10^8\,\mathrm{m/s}$ . Their product is $9.0\times10^{16}\,\mathrm{m^2/s^2}=c^2$ . The phase speed exceeding c does not imply superluminal information transmission.

Keep the idea

Take it with you

Rectangular TE and TM modes have different allowed transverse indices.
Below-cutoff decay can occur without dissipative material loss.
Guide phase and group velocities differ because transverse confinement creates dispersion.

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