Part 7 · Chapter 27

Rectangular Waveguides

At microwave frequencies the two-wire line and the coax run out of road: they radiate, they heat up, and they cannot carry much power. The answer is to throw away the centre conductor entirely and send the wave down a hollow metal pipe. But a single conductor cannot support the simple TEM wave we have relied on since Part 5 — so the wave reorganizes itself into modes that bounce diagonally between the walls, each with a cutoff frequency below which it simply will not propagate. The pipe becomes a high-pass filter with structure. This chapter solves the rectangular guide, finds those modes, identifies the dominant one that nearly every microwave system uses, and works out how fast its energy actually travels.

Electromagnetic Field Theory Prof. Mithun Mondal Reading time ≈ 56 min
i What you'll learn
  • Why a single hollow conductor cannot carry a TEM wave, only TE and TM modes.
  • The field solutions, set by two integer mode indices \(m\) and \(n\).
  • The cutoff wavenumber \(k_c\) and the cutoff frequency \(f_c\) below which a mode is evanescent.
  • The dominant mode \(\text{TE}_{10}\), with \(f_c=u'/2a\) and \(\lambda_c=2a\).
  • The guide wavelength and the phase and group velocities, with \(u_p u_g=u'^2\).
  • The wave impedance of TE and TM modes.
Section 27-1

Why a Hollow Pipe?

A transmission line needs two conductors — a signal path and a return — to support the transverse electromagnetic (TEM) wave, in which both \(\vec{E}\) and \(\vec{H}\) lie entirely across the line with nothing along the direction of travel. A waveguide is a single hollow conductor; there is no second conductor to hold the return, so a TEM wave is impossible inside it. Yet a hollow copper pipe guides microwaves superbly, with far lower loss and far higher power handling than coax. How?

The wave survives by tilting. Instead of running straight down the guide, it bounces at an angle between the walls, zig-zagging forward. That tilt gives the field a component along the axis — either an axial electric field or an axial magnetic field — and that is exactly what distinguishes the two allowed families of solutions.

a b z perfectly conducting walls · hollow interior (μ, ε)
A rectangular waveguide: a hollow conducting pipe of internal cross-section a × b, guiding waves along z
Section 27-2

TE and TM Modes

Since the axial direction is no longer field-free, we classify solutions by which axial component is present:

  • Transverse magnetic (TM) modes: \(H_z=0\), \(E_z\neq0\). The magnetic field is purely transverse; the electric field has an axial part.
  • Transverse electric (TE) modes: \(E_z=0\), \(H_z\neq0\). The electric field is purely transverse; the magnetic field has an axial part.

Every field inside the guide can be built from these two families. The trick is that once you know the single axial component (\(E_z\) for TM, \(H_z\) for TE), Maxwell's equations hand you all four transverse components algebraically. So the whole problem reduces to solving one scalar Helmholtz equation for that axial component, subject to the metal-wall boundary conditions.

Section 27-3

Field Solutions and Mode Indices

With a propagation factor \(e^{-j\beta z}\), the axial component obeys the two-dimensional Helmholtz equation \(\nabla_t^2\psi+k_c^2\psi=0\), where \(k_c^2=k^2-\beta^2\) and \(k^2=\omega^2\mu\varepsilon\). Separation of variables on the rectangle, with \(E_z=0\) on the walls (TM) or \(\partial H_z/\partial n=0\) on the walls (TE), gives sinusoidal solutions labelled by two integers:

Axial fields (m, n = mode indices)
\[ \text{TM}_{mn}:\;E_z=E_0\sin\!\frac{m\pi x}{a}\sin\!\frac{n\pi y}{b}\,e^{-j\beta z}, \qquad \text{TE}_{mn}:\;H_z=H_0\cos\!\frac{m\pi x}{a}\cos\!\frac{n\pi y}{b}\,e^{-j\beta z} \]

The indices \(m\) and \(n\) count the half-wave field variations across the width \(a\) and the height \(b\). For TM modes both must be at least one (if either is zero the sine collapses the field to nothing), so the lowest TM mode is \(\text{TM}_{11}\). For TE modes either may be zero but not both, so \(\text{TE}_{10}\) exists. The transverse wavenumber that sets everything is:

Cutoff (transverse) wavenumber
\[ k_c=\sqrt{\left(\frac{m\pi}{a}\right)^2+\left(\frac{n\pi}{b}\right)^2} \]
Section 27-4

Cutoff Frequency

The axial phase constant follows from \(\beta^2=k^2-k_c^2\). For the wave to travel, \(\beta\) must be real, which requires \(k>k_c\) — the frequency must exceed a threshold. Below it, \(\beta\) is imaginary, the field decays exponentially, and the mode is evanescent: it does not propagate at all. That threshold is the cutoff frequency:

Cutoff frequency and wavelength (u' = 1/√(με))
\[ f_{c,mn}=\frac{u'}{2}\sqrt{\left(\frac{m}{a}\right)^2+\left(\frac{n}{b}\right)^2}, \qquad \lambda_{c}=\frac{2}{\sqrt{(m/a)^2+(n/b)^2}} \]
β ω evanescent (ω < ω_c) light line ω=u'β ω_c (cutoff) guided mode
Dispersion: the guided mode exists only above ω_c and approaches the light line at high frequency

Each mode has its own cutoff, so a waveguide passes a band of frequencies with a definite set of modes. The dispersion curve makes the behaviour vivid: at cutoff the wave is purely transverse (\(\beta=0\), it bounces straight across without advancing); well above cutoff it hugs the light line and behaves almost like free space.

Section 27-5

The Dominant Mode TE₁₀

The mode with the lowest cutoff is the dominant mode. For the usual convention \(a>b\), that is \(\text{TE}_{10}\) — one half-wave across the wide dimension, none across the narrow one. Its cutoff depends only on the wide wall:

Dominant mode TE₁₀
\[ f_{c,10}=\frac{u'}{2a}, \qquad \lambda_{c,10}=2a \]
E_y ∝ sin(πx/a) a (width)
TE₁₀ cross-section: the electric field is a half-sine across the width — strongest at centre, zero at the walls

This is the workhorse of microwave engineering. By making \(a\) roughly twice \(b\), the next modes — \(\text{TE}_{20}\) at \(u'/a\) and \(\text{TE}_{01}\) at \(u'/2b\) — sit well above \(\text{TE}_{10}\), leaving a clean single-mode band between \(f_{c,10}\) and the next cutoff. Operating there guarantees the field has just one predictable shape, which is essential for couplers, filters, and antennas. Standard guides like WR-90 are dimensioned precisely to place that band where it is needed.

ModeCutoff \(f_c\) (air, a > b)Note
\(\text{TE}_{10}\)\(c/2a\)dominant — lowest cutoff
\(\text{TE}_{20}\)\(c/a\)twice the TE₁₀ cutoff
\(\text{TE}_{01}\)\(c/2b\)set by the narrow wall
\(\text{TE}_{11},\,\text{TM}_{11}\)\(\tfrac{c}{2}\sqrt{(1/a)^2+(1/b)^2}\)lowest TM mode
Section 27-6

Guide Wavelength and Velocities

Because the wave travels diagonally, the distance between wavefronts measured along the axis is stretched. Writing \(\beta=k\sqrt{1-(f_c/f)^2}\), the axial wavelength — the guide wavelength — is longer than the wavelength the same frequency would have in the unbounded medium:

Guide wavelength (λ' = u'/f is the unbounded wavelength)
\[ \lambda_g=\frac{2\pi}{\beta}=\frac{\lambda'}{\sqrt{1-(f_c/f)^2}}>\lambda' \]

The two velocities split apart. The wavefronts' axial sweep — the phase velocity — exceeds \(u'\) (even exceeding \(c\) in an air guide, with no contradiction, since no energy moves that fast). The energy itself travels at the group velocity, which is correspondingly slower:

Phase and group velocity
\[ u_p=\frac{u'}{\sqrt{1-(f_c/f)^2}}, \qquad u_g=u'\sqrt{1-(f_c/f)^2}, \qquad u_p\,u_g=u'^2 \]

The clean product \(u_p u_g=u'^2\) is a handy check. As \(f\to f_c\), \(u_g\to0\) (the energy stalls — the wave bounces straight across without advancing); as \(f\gg f_c\), both approach \(u'\) and the guide behaves like open space.

Section 27-7

Wave Impedance

The ratio of the transverse electric to transverse magnetic field — the wave impedance — also depends on frequency, and differs between the two mode families. Relative to the medium's intrinsic impedance \(\eta'=\sqrt{\mu/\varepsilon}\) (377 Ω for air):

TE and TM wave impedance
\[ \eta_{\text{TE}}=\frac{\eta'}{\sqrt{1-(f_c/f)^2}}>\eta', \qquad \eta_{\text{TM}}=\eta'\sqrt{1-(f_c/f)^2}<\eta' \]

So a TE mode looks more inductive than free space and a TM mode more capacitive, both converging to \(\eta'\) far above cutoff and diverging from it (\(\eta_{\text{TE}}\to\infty\), \(\eta_{\text{TM}}\to0\)) as cutoff is approached. These impedances matter when matching a guide to an antenna or joining sections of different size — the same matching logic as transmission lines, carried into the microwave pipe.

Section 27-8

Worked Examples

1 Dominant-mode cutoff

Problem. An air-filled X-band guide (WR-90) has \(a=2.286\,\text{cm}\), \(b=1.016\,\text{cm}\). Find the cutoff frequency of the dominant mode.

Solution. The dominant mode is \(\text{TE}_{10}\), with \(f_c=c/2a\):

Working
\[ f_{c,10}=\frac{3\times10^{8}}{2(0.02286)}\approx 6.56\ \text{GHz} \]
2 Mode ordering and single-mode band

Problem. For the same guide, find the cutoffs of \(\text{TE}_{20}\), \(\text{TE}_{01}\), and \(\text{TE}_{11}\), and state the single-mode band.

Solution. Use \(f_c=\tfrac{c}{2}\sqrt{(m/a)^2+(n/b)^2}\):

Working
\[ f_{20}=13.1,\quad f_{01}=14.8,\quad f_{11}\approx16.2\ \text{GHz}\;\Rightarrow\;\text{single-mode } 6.56\text{–}13.1\ \text{GHz} \]
3 Guide wavelength

Problem. Operate the WR-90 guide at \(10\,\text{GHz}\) in \(\text{TE}_{10}\) (\(f_c=6.56\,\text{GHz}\)). Find the guide wavelength.

Solution. \(\lambda'=c/f=3\,\text{cm}\); divide by \(\sqrt{1-(f_c/f)^2}\):

Working
\[ \lambda_g=\frac{3}{\sqrt{1-(0.656)^2}}=\frac{3}{0.755}\approx 3.97\ \text{cm} \]
4 Phase and group velocity

Problem. Find \(u_p\) and \(u_g\) for that case, and verify their product.

Solution. \(u_p=c/\sqrt{1-(f_c/f)^2}\), \(u_g=c\sqrt{1-(f_c/f)^2}\):

Working
\[ u_p=\frac{3\times10^8}{0.755}=3.97\times10^8,\quad u_g=3\times10^8(0.755)=2.26\times10^8,\quad u_p u_g=c^2 \]
5 Wave impedance

Problem. Find the TE₁₀ wave impedance at \(10\,\text{GHz}\) for the air guide above.

Solution. \(\eta_{\text{TE}}=\eta'/\sqrt{1-(f_c/f)^2}\) with \(\eta'=377\,\Omega\):

Working
\[ \eta_{\text{TE}}=\frac{377}{0.755}\approx 499\ \Omega \]
6 Operating below cutoff

Problem. An air guide has \(a=2\,\text{cm}\). A \(5\,\text{GHz}\) signal is applied in \(\text{TE}_{10}\). Does it propagate, and what is its attenuation?

Solution. \(f_c=c/2a=7.5\,\text{GHz}>5\,\text{GHz}\), so it is below cutoff and evanescent. The decay rate is \(\alpha=k_c\sqrt{1-(f/f_c)^2}\) with \(k_c=\pi/a\):

Working
\[ \alpha=\frac{\pi}{0.02}\sqrt{1-\left(\tfrac{5}{7.5}\right)^2}\approx 157(0.745)\approx 117\ \text{Np/m}\;(\text{no propagation}) \]
Review

Chapter Summary

No TEM

A hollow single conductor supports only TE (\(E_z=0\)) and TM (\(H_z=0\)) modes.

Mode indices

\(m,n\) count half-waves across \(a,b\); \(k_c=\sqrt{(m\pi/a)^2+(n\pi/b)^2}\).

Cutoff

\(f_c=\tfrac{u'}{2}\sqrt{(m/a)^2+(n/b)^2}\); below it the mode is evanescent.

Dominant mode

\(\text{TE}_{10}\): \(f_c=u'/2a\), \(\lambda_c=2a\); used for single-mode operation.

Wavelength & speed

\(\lambda_g=\lambda'/\sqrt{1-(f_c/f)^2}\); \(u_p u_g=u'^2\).

Wave impedance

\(\eta_{\text{TE}}=\eta'/\sqrt{1-(f_c/f)^2}\), \(\eta_{\text{TM}}=\eta'\sqrt{1-(f_c/f)^2}\).

Practice

Problems

For each item, identify the mode and compute \(f_c\) first; whether a signal propagates, and how fast, both follow from the ratio \(f_c/f\). Assume air filling unless stated. Difficulty rises down the list.

  1. An air guide has \(a=4\,\text{cm}\). Find the \(\text{TE}_{10}\) cutoff frequency and cutoff wavelength.
  2. Why is \(\text{TM}_{10}\) not a valid mode, while \(\text{TE}_{10}\) is?
  3. For \(a=2.5\,\text{cm}\), \(b=1.25\,\text{cm}\) (air), list the cutoffs of \(\text{TE}_{10}\), \(\text{TE}_{20}\), \(\text{TE}_{01}\).
  4. What is the lowest-order TM mode, and what is its cutoff for the guide in problem 3?
  5. A WR-90 guide (\(a=2.286\,\text{cm}\)) runs at \(9\,\text{GHz}\). Find \(\beta\) for \(\text{TE}_{10}\).
  6. For the same case, find \(\lambda_g\) and compare it with the free-space wavelength.
  7. Find \(u_p\) and \(u_g\) at \(9\,\text{GHz}\) and verify \(u_p u_g=c^2\).
  8. Find the \(\text{TE}_{10}\) wave impedance at \(9\,\text{GHz}\) and at \(12\,\text{GHz}\).
  9. A guide is filled with a dielectric of \(\varepsilon_r=2.25\). By what factor does every cutoff frequency change?
  10. An air guide with \(a=1.5\,\text{cm}\) is fed at \(8\,\text{GHz}\). Determine whether \(\text{TE}_{10}\) propagates.
  11. Show that at exactly \(f=f_c\), the group velocity is zero and the guide wavelength is infinite.
  12. Explain why a waveguide acts as a high-pass filter, and why two-conductor lines have no such cutoff.
Tip: a waveguide is a transmission line that has lost its second conductor, and with it the TEM wave. What remains are TE and TM modes, each a standing pattern across the cross-section set by two integers \(m,n\) and a single number, the cutoff \(f_c=\tfrac{u'}{2}\sqrt{(m/a)^2+(n/b)^2}\). Below \(f_c\) the mode is evanescent and carries no power; above it everything else — the guide wavelength \(\lambda_g\), the velocities (\(u_p u_g=u'^2\)), the wave impedance — follows from the one factor \(\sqrt{1-(f_c/f)^2}\). The dominant \(\text{TE}_{10}\) (\(f_c=u'/2a\)) is the one to know cold, because almost every microwave system lives in its single-mode band. Next, Chapter 28 closes the box at both ends to trap a standing wave — the cavity resonator — and bends the cross-section into a circle for circular waveguides.