10.04.04 · em-sr / maxwell-fields

Waveguides and transmission lines: TE, TM, TEM modes and the cutoff frequency

shipped3 tiersLean: none

Anchor (Master): Jackson, Classical Electrodynamics, 3rd ed. (1999), Ch. 8.1-8.5; Collin, Field Theory of Guided Waves, 2nd ed. (1991)

Intuition Beginner

A waveguide is a hollow metal tube that guides electromagnetic waves from one end to the other. Unlike an antenna that radiates in all directions, a waveguide confines the wave to travel along the tube. The most common shape is a rectangular waveguide — a metal pipe with a rectangular cross-section.

Not all frequencies can travel through a waveguide. Each waveguide has a cutoff frequency $f_{c}$ below which the wave is reflected back and cannot propagate. The cutoff depends on the width of the guide: wider guides support lower frequencies. For a rectangular waveguide of width $a$ , the cutoff for the lowest mode is $f_{c} = c / (2 a)$ .

There are three types of waveguide modes, classified by which field components are transverse (perpendicular to the direction of propagation):

TEM (Transverse ElectroMagnetic): both E and B are transverse. Found in coaxial cables and two-wire lines, not in hollow pipes.
TE (Transverse Electric): E is entirely transverse, B has a longitudinal component. The dominant mode in rectangular waveguides.
TM (Transverse Magnetic): B is entirely transverse, E has a longitudinal component.

A coaxial cable carrying TV signals uses the TEM mode — the simplest and most broadband. A microwave oven feed uses the TE mode of a rectangular waveguide. Waveguides are essential for radar, satellite communications, and particle accelerators.

Visual Beginner

Mode	$E_{z}$	$B_{z}$	Found in
TEM	0	0	Coaxial cable, two-wire line
TE	0	Non-zero	Rectangular waveguide
TM	Non-zero	0	Rectangular waveguide (higher order)

Worked example Beginner

A rectangular waveguide has width $a = 2.3$ cm and height $b = 1.0$ cm. The cutoff frequency for the dominant TE $_{10}$ mode is:

$f_{c} = c / (2 a) = 3 \times 1 0^{8} / (2 \times 0.023) = 6.52$ GHz.

Frequencies below 6.52 GHz cannot propagate. The waveguide is used for X-band radar (8-12 GHz), which is above the cutoff.

At $f = 10$ GHz, the guide wavelength is:

$λ_{g} = λ / 1 - (f_{c} / f)^{2} = (3 \times 1 0^{- 2}) / 1 - 0.65 2^{2} = 0.03/0.758 = 3.96$ cm.

The guide wavelength (4.0 cm) is longer than the free-space wavelength (3.0 cm) because the wave bounces off the walls, zigzagging down the guide.

Check your understanding Beginner

Formal definition Intermediate+

Waveguide geometry. A waveguide is a cylindrical metal tube of arbitrary cross-section extending in the $z$ -direction. The walls are perfect electric conductors (PEC). For a rectangular waveguide: cross-section $a \times b$ with $a > b$ .

Mode decomposition. The fields in a waveguide are decomposed into modes with definite $z$ -dependence $e^{i (β z - ω t)}$ where $β$ is the propagation constant. The transverse fields satisfy the two-dimensional Helmholtz equation:

(\nabla_{t}^{2} + k_{c}^{2}) ψ = 0

where $k_{c}$ is the cutoff wave number and $ψ$ is either $E_{z}$ (TM modes) or $B_{z}$ (TE modes). The boundary conditions are:

TM: $ψ = 0$ on the wall (Dirichlet)
TE: $\partial ψ / \partial n = 0$ on the wall (Neumann)

Dispersion relation. The propagation constant is:

β = k^{2} - k_{c}^{2} = k 1 - (f_{c} / f)^{2}

where $k = ω / c$ and $f_{c} = c k_{c} / (2 π)$ is the cutoff frequency. For $f > f_{c}$ : $β$ is real and the mode propagates. For $f < f_{c}$ : $β$ is imaginary and the mode is evanescent (exponentially decaying).

Phase and group velocities.

v_{p} = \frac{ω}{β} = \frac{c}{1 - ( f _{c} / f ) ^{2}} > c

v_{g} = \frac{d ω}{d β} = c 1 - (f_{c} / f)^{2} < c

The product $v_{p} v_{g} = c^{2}$ . The phase velocity exceeds $c$ (the wavefronts move faster than light), but the group velocity (which carries information and energy) is always less than $c$ .

Rectangular waveguide modes. For a rectangular guide of dimensions $a \times b$ :

k_{c, mn} = π (m / a)^{2} + (n / b)^{2}

The TE $_{10}$ mode ( $m = 1$ , $n = 0$ ) has the lowest cutoff: $f_{c} = c / (2 a)$ .

TEM mode. The TEM mode requires $k_{c} = 0$ (zero cutoff frequency) and exists only in structures with two conductors. The fields satisfy Laplace's equation in the transverse plane. The voltage and current on the transmission line satisfy the telegrapher's equations:

\frac{\partial V}{\partial z} = - L^{'} \frac{\partial I}{\partial t}, \frac{\partial I}{\partial z} = - C^{'} \frac{\partial V}{\partial t}

where $L^{'}$ and $C^{'}$ are the inductance and capacitance per unit length. The characteristic impedance is $Z_{0} = L^{'} / C^{'}$ and the wave speed is $v = 1/ L^{'} C^{'}$ .

Key derivation Intermediate+

Derivation (TE $_{10}$ mode of a rectangular waveguide).

Theorem. The dominant TE $_{10}$ mode of a rectangular waveguide of dimensions $a \times b$ has cutoff frequency $f_{c} = c / (2 a)$ , field components $E_{y} = E_{0} sin (π x / a) cos (ω t - β z)$ , and guide wavelength $λ_{g} = λ_{0} / 1 - (f_{c} / f)^{2}$ .

Proof. For TE modes, $E_{z} = 0$ and $B_{z} = B_{0} cos (mπ x / a) cos (nπ y / b) e^{i (β z - ω t)}$ . The TE $_{10}$ mode has $m = 1$ , $n = 0$ : $B_{z} = B_{0} cos (π x / a)$ .

The transverse fields are obtained from $B_{z}$ using the standard TE relations:

E_{x} = 0, E_{y} = \frac{iω}{k _{c}^{2}} \frac{\partial B _{z}}{\partial x} = - \frac{iω π}{k _{c}^{2} a} B_{0} sin \frac{π x}{a}

The cutoff wave number is $k_{c} = π / a$ and $β = k^{2} - π^{2} / a^{2}$ . The cutoff frequency is $ω_{c} = c π / a$ , i.e., $f_{c} = c / (2 a)$ .

The guide wavelength is $λ_{g} = 2 π / β = 2 π / (k 1 - f_{c}^{2} / f^{2}) = λ_{0} / 1 - (f_{c} / f)^{2}$ . $□$

Bridge. The waveguide mode theory builds toward a complete description of guided electromagnetic wave propagation. The foundational insight is that the boundary conditions on the conducting walls quantise the allowed modes, producing discrete cutoff frequencies. The central message is that each mode behaves like a separate transmission channel with its own dispersion relation, phase velocity, and group velocity. This is exactly the same mathematical structure as quantum mechanical wavefunctions in a box. Putting these together, the waveguide modes generalise to cavity resonators 10.04.05 (standing waves in a closed box), the attenuation is determined by the skin depth 10.04.03 on the guide walls, and the TEM mode provides the basis for transmission line theory used in all RF engineering.

Exercises Intermediate+

Exercise 3 (medium, symbolic).

Compute the attenuation constant $α$ of a rectangular waveguide due to wall losses for the TE $_{10}$ mode, given surface resistance $R_{s}$ .

Hint

The power dissipated per unit length is $d P / d z = - 2 α P$ where $P$ is the power flow. The dissipation comes from the surface currents on the four walls.

Answer

The attenuation constant is $α = P_{loss} / (2 P)$ where $P_{loss}$ is the power dissipated per unit length and $P$ is the power flow. For TE $_{10}$ :

α = \frac{R _{s}}{a ^{3} b β k η} (2 b π^{2} + a^{3} k^{2})

where $η = μ_{0} / ϵ_{0} \approx 377 Ω$ . The dominant contribution comes from the broad walls (top and bottom). Near cutoff ( $f \to f_{c}^{+}$ ), $β \to 0$ and $α \to \infty$ — the losses diverge at cutoff because the group velocity vanishes and the wave spends infinite time in the guide.

For copper walls at 10 GHz ( $R_{s} \approx 0.026 Ω$ ) in a WR-90 guide ( $a = 2.286$ cm, $b = 1.016$ cm): $α \approx 0.011$ Np/m $\approx 0.096$ dB/m. A 10-metre run loses about 1 dB.

Exercise 5 (medium, symbolic).

Show that a TEM mode cannot exist in a hollow conducting waveguide (single conductor).

Hint

A TEM mode requires $k_{c} = 0$ , so the transverse fields satisfy Laplace's equation. The boundary condition on the single conductor means the potential is constant on the boundary.

Answer

For a TEM mode, $E_{z} = 0$ and $B_{z} = 0$ . The transverse E-field satisfies $\nabla_{t}^{2} E_{t} = 0$ (Laplace's equation with $k_{c} = 0$ ). The potential $Φ$ satisfies $\nabla_{t}^{2} Φ = 0$ with $Φ =$ constant on the conducting boundary.

By the maximum principle for harmonic functions, $Φ$ is constant everywhere in the cross-section (since it is constant on the boundary and harmonic inside). Therefore $E_{t} = - \nabla_{t} Φ = 0$ everywhere, and there is no field.

A TEM mode requires an inner conductor (a second boundary) to sustain a nonzero potential difference. Coaxial cables have two conductors at different potentials; hollow pipes have only one.

Exercise 6 (hard, symbolic).

Derive the complete set of TE $_{mn}$ and TM $_{mn}$ cutoff frequencies for a rectangular waveguide of dimensions $a \times b$ with $a > b$ . Show that the TE $_{10}$ mode always has the lowest cutoff.

Hint

$f_{c, mn} = (c /2) (m / a)^{2} + (n / b)^{2}$ . For TM modes, $m \geq 1$ and $n \geq 1$ . For TE modes, $m$ and $n$ cannot both be zero.

Answer

The cutoff frequencies are $f_{c, mn} = (c /2) (m / a)^{2} + (n / b)^{2}$ for integers $m, n$ .

TM modes: Both $m \geq 1$ and $n \geq 1$ (because $E_{z} = 0$ if either index vanishes). The lowest TM mode is TM $_{11}$ : $f_{c, 11} = (c /2) 1/ a^{2} + 1/ b^{2}$ .

TE modes: At least one of $m, n$ must be nonzero. TE $_{10}$ : $f_{c, 10} = c / (2 a)$ . TE $_{01}$ : $f_{c, 01} = c / (2 b)$ . Since $a > b$ : $f_{c, 10} < f_{c, 01} < f_{c, 11}$ .

TE $_{10}$ has the lowest cutoff: $f_{c, 10} = c / (2 a)$ . The bandwidth of single-mode operation (TE $_{10}$ only) extends from $c / (2 a)$ to $c / (2 b)$ . For $a = 2 b$ : the single-mode bandwidth is one octave ( $f_{c, 01} / f_{c, 10} = a / b = 2$ ). Most practical waveguides are designed with $a \approx 2 b$ for maximum single-mode bandwidth.

Exercise 7 (hard, symbolic).

The evanescent mode of a waveguide below cutoff has $β = i α$ where $α = k_{c} 1 - f^{2} / f_{c}^{2}$ . Show that a section of below-cutoff waveguide of length $L$ acts as a high-pass filter with attenuation $e^{- α L}$ and that no power is dissipated in the walls (for a lossless guide).

Hint

Below cutoff, the fields decay exponentially but no energy flows. The Poynting vector averaged over a cross-section is zero.

Answer

Below cutoff ( $f < f_{c}$ ): $β = i α$ with $α = (2 π / c) f_{c}^{2} - f^{2}$ . The fields decay as $e^{- α z}$ . After length $L$ : the field is attenuated by $e^{- α L}$ .

The Poynting vector has components proportional to $e^{- 2 α z}$ . But the transverse and longitudinal components of $E$ and $H$ are 90 degrees out of phase for an evanescent mode, so $⟨ S ⟩ = (1/2) Re (E \times H^{*}) = 0$ . No net power flows.

The wave does not transport energy — it is a reactive (stored) field, not a propagating one. The guide acts as an attenuator that absorbs no power: the attenuation is purely reactive. This property is used in evanescent-mode filters, where a below-cutoff section provides frequency-selective attenuation without dissipating power.

Exercise 8 (hard, symbolic).

For a coaxial cable with inner radius $a$ , outer radius $b$ , and dielectric $ϵ$ , derive the power handling capacity $P_{m a x}$ in terms of the dielectric breakdown field $E_{bd}$ .

Hint

The maximum E-field in a coaxial cable occurs at the inner conductor surface: $E_{m a x} = V / (a ln (b / a))$ . The power flow is $P = V^{2} / (2 Z_{0})$ .

Answer

The E-field in a coaxial TEM mode: $E_{r} = V / (r ln (b / a))$ . The maximum occurs at $r = a$ : $E_{m a x} = V / (a ln (b / a))$ . Setting $E_{m a x} = E_{bd}$ : $V_{m a x} = E_{bd} a ln (b / a)$ .

The power flow: $P = V^{2} / (2 Z_{0}) = V^{2} ϵ_{r} / (120 ln (b / a))$ . At the maximum voltage:

P_{m a x} = \frac{E _{bd}^{2} a ^{2} ln ( b / a ) ϵ _{r}}{120}

For an air-filled cable ( $ϵ_{r} = 1$ , $E_{bd} = 3$ MV/m) with $a = 1$ mm, $b = 3.55$ mm: $P_{m a x} = (9 \times 1 0^{12} \times 1 0^{- 6} \times 1.267) /120 = 9.5 \times 1 0^{4}$ W $\approx 95$ kW. The inner conductor limits the power handling because the field is strongest there.

Lean formalization Intermediate+

Mathlib has the Laplacian eigenvalue problem in rectangular domains but does not contain the waveguide mode classification, the cutoff frequency formulas, the guide wavelength, the phase/group velocity relations, or the telegrapher's equations. lean_status: none.

Advanced results Master

Circular waveguides. A circular waveguide of radius $R$ supports modes based on Bessel functions. The TE $_{11}$ mode (cutoff $f_{c} = 1.841 c / (2 π R)$ ) is the dominant mode, followed by TM $_{01}$ ( $f_{c} = 2.405 c / (2 π R)$ ). The TE $_{01}$ mode has the special property that its attenuation decreases with increasing frequency — making it attractive for long-distance millimetre-wave transmission.

Mode coupling. Imperfections in a waveguide (bends, twists, discontinuities) couple energy between modes. The coupled-mode theory describes the power transfer:

\frac{d A _{m}}{d z} = - i n \sum κ_{mn} A_{n} e^{i (β_{m} - β_{n}) z}

where $A_{m}$ is the amplitude of mode $m$ and $κ_{mn}$ is the coupling coefficient. Phase-matched coupling ( $β_{m} \approx β_{n}$ ) transfers power efficiently between modes.

Optical fibers. An optical fiber is a dielectric waveguide: a core of index $n_{1}$ surrounded by cladding of index $n_{2} < n_{1}$ . The modes are hybrid (HE and EH modes, with both $E_{z}$ and $B_{z}$ nonzero). Single-mode fibers have a core diameter small enough that only the HE $_{11}$ mode propagates, eliminating intermodal dispersion and enabling long-distance communication at terabit data rates.

Synthesis. Waveguide theory provides the mathematical framework for understanding all guided electromagnetic wave propagation, from microwave radar to optical fiber communications. The foundational insight is that the conducting (or dielectric) boundary quantises the allowed field patterns into discrete modes, each with its own cutoff frequency and dispersion relation. The central message is that the mode decomposition reduces a complex three-dimensional wave problem to a set of independent one-dimensional propagation problems, one per mode. Putting these together, the waveguide mode theory generalises to cavity resonators 10.04.05, the skin depth 10.04.03 determines the wall losses, and the TEM mode provides the foundation for transmission line theory in all RF and microwave engineering.

Full proof set Master

Proposition (Completeness of waveguide modes). The TE $_{mn}$ and TM $_{mn}$ modes of a rectangular waveguide form a complete set: any field distribution in the waveguide can be expanded as a superposition of modes.

Proof. The transverse field components of each mode satisfy the Helmholtz equation $\nabla_{t}^{2} ψ_{mn} + k_{c, mn}^{2} ψ_{mn} = 0$ with either Dirichlet (TM) or Neumann (TE) boundary conditions on the rectangle $[0, a] \times [0, b]$ . The eigenfunctions $ψ_{mn} (x, y) = cos (mπ x / a) cos (nπ y / b)$ (for TE) or $sin (mπ x / a) sin (nπ y / b)$ (for TM) form orthogonal sets:

\int_{0}^{a} \int_{0}^{b} ψ_{mn} ψ_{m^{'} n^{'}} d x d y = \frac{ab}{4} δ_{m m^{'}} δ_{n n^{'}}

By the spectral theorem for the self-adjoint operator $- \nabla_{t}^{2}$ on the rectangle with PEC boundary conditions, these eigenfunctions form a complete orthogonal basis for $L^{2} ([0, a] \times [0, b])$ . Any square-integrable transverse field can be expanded in this basis. $□$

Connections Master

EM waves 10.04.02 are the building blocks; waveguide modes are constrained superpositions of plane waves.
Plane waves in matter 10.04.03 provide the skin depth that determines waveguide wall losses.
Cavities 10.04.05 are waveguides closed at both ends, forming standing-wave resonators.
Maxwell equations 10.04.01 are the governing equations from which all mode properties are derived.
Green's functions 10.08.01 can be expanded in waveguide modes for boundary-value problems.

Historical & philosophical context Master

Waveguide theory was developed independently by several groups during the 1930s. George C. Southworth at Bell Labs and Wilmer L. Barrow at MIT independently demonstrated waveguide propagation in 1936. The theoretical framework was developed by Chu (1938), Schelkunoff (1943), and others during the rapid development of radar technology in World War II.

The distinction between phase velocity and group velocity in waveguides caused considerable confusion. Sommerfeld and Brillouin (1914) had already resolved the apparent paradox of superluminal phase velocities in dispersive media, showing that the front velocity (not the phase or group velocity) determines the signal speed and always satisfies $v_{front} \leq c$ .

Optical fiber waveguides were proposed by Kao and Hockham (1966) and first demonstrated with losses below 20 dB/km by Corning in 1970. Single-mode fiber (1983) revolutionised telecommunications by eliminating modal dispersion.

Bibliography Master

Southworth, G. C., "Hyper-frequency wave guides," Bell Sys. Tech. J. 15, 284-309 (1936).
Schelkunoff, S. A., "Electromagnetic Waves in Conducting Tubes," Phys. Rev. 63, 232 (1943).
Kao, K. C. and Hockham, G. A., "Dielectric-fibre surface waveguides for optical frequencies," Proc. IEE 113, 1151-1158 (1966).
Jackson, J. D., Classical Electrodynamics, 3rd ed. (Wiley, 1999).
Collin, R. E., Field Theory of Guided Waves, 2nd ed. (IEEE Press, 1991).

Prerequisites

10.04.03
10.04.02

Used in

10.04.05

Tier anchors

beginner: Griffiths, Introduction to Electrodynamics, 4th ed. (2017), Ch. 9.5
intermediate: Griffiths, Introduction to Electrodynamics, 4th ed. (2017), Ch. 9.5; Jackson, Classical Electrodynamics, 3rd ed. (1999), Ch. 8.1-8.4
master: Jackson, Classical Electrodynamics, 3rd ed. (1999), Ch. 8.1-8.5; Collin, Field Theory of Guided Waves, 2nd ed. (1991)

References

Griffiths — Introduction to Electrodynamics, 4th ed. (Cambridge, 2017) · Ch. 9.5 Wave guides; the rectangular waveguide; TE and TM modes; the cutoff frequency
Jackson — Classical Electrodynamics, 3rd ed. (Wiley, 1999) · Ch. 8.1 Waveguides; Ch. 8.2 Rectangular waveguides; Ch. 8.3 Waveguides with circular cross-section; Ch. 8.4 Attenuation in waveguides
Collin — Field Theory of Guided Waves, 2nd ed. (IEEE Press, 1991) · Ch. 2-3 Waveguide modes; Ch. 4 Transmission line theory; the TEM, TE, and TM mode classification

Estimated time

beginner: 15m
intermediate: 40m
master: 60m