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.
- 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.
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.
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.
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:
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 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:
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.
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:
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.
| Mode | Cutoff \(f_c\) (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 |
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:
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:
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.
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):
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.
Worked Examples
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 \(f_c=c/2a\):
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 \(f_c=\tfrac{c}{2}\sqrt{(m/a)^2+(n/b)^2}\):
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 \(\sqrt{1-(f_c/f)^2}\):
Problem. Find \(u_p\) and \(u_g\) for that case, and verify their product.
Solution. \(u_p=c/\sqrt{1-(f_c/f)^2}\), \(u_g=c\sqrt{1-(f_c/f)^2}\):
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 \(\eta'=377\,\Omega\):
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 \(k_c=\pi/a\):
Chapter Summary
A hollow single conductor supports only TE (\(E_z=0\)) and TM (\(H_z=0\)) modes.
\(m,n\) count half-waves across \(a,b\); \(k_c=\sqrt{(m\pi/a)^2+(n\pi/b)^2}\).
\(f_c=\tfrac{u'}{2}\sqrt{(m/a)^2+(n/b)^2}\); below it the mode is evanescent.
\(\text{TE}_{10}\): \(f_c=u'/2a\), \(\lambda_c=2a\); used for single-mode operation.
\(\lambda_g=\lambda'/\sqrt{1-(f_c/f)^2}\); \(u_p u_g=u'^2\).
\(\eta_{\text{TE}}=\eta'/\sqrt{1-(f_c/f)^2}\), \(\eta_{\text{TM}}=\eta'\sqrt{1-(f_c/f)^2}\).
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.
- An air guide has \(a=4\,\text{cm}\). Find the \(\text{TE}_{10}\) cutoff frequency and cutoff wavelength.
- Why is \(\text{TM}_{10}\) not a valid mode, while \(\text{TE}_{10}\) is?
- For \(a=2.5\,\text{cm}\), \(b=1.25\,\text{cm}\) (air), list the cutoffs of \(\text{TE}_{10}\), \(\text{TE}_{20}\), \(\text{TE}_{01}\).
- What is the lowest-order TM mode, and what is its cutoff for the guide in problem 3?
- A WR-90 guide (\(a=2.286\,\text{cm}\)) runs at \(9\,\text{GHz}\). Find \(\beta\) for \(\text{TE}_{10}\).
- For the same case, find \(\lambda_g\) and compare it with the free-space wavelength.
- Find \(u_p\) and \(u_g\) at \(9\,\text{GHz}\) and verify \(u_p u_g=c^2\).
- Find the \(\text{TE}_{10}\) wave impedance at \(9\,\text{GHz}\) and at \(12\,\text{GHz}\).
- A guide is filled with a dielectric of \(\varepsilon_r=2.25\). By what factor does every cutoff frequency change?
- An air guide with \(a=1.5\,\text{cm}\) is fed at \(8\,\text{GHz}\). Determine whether \(\text{TE}_{10}\) propagates.
- Show that at exactly \(f=f_c\), the group velocity is zero and the guide wavelength is infinite.
- Explain why a waveguide acts as a high-pass filter, and why two-conductor lines have no such cutoff.