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Modes and cutoffs in rectangular waveguides, $f_c = c/2a$ for TE$_{10}$, group and phase velocities, resonant cavities, coaxial TEM modes and optical fibers.
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By the end of this lesson you will be able to find waveguide cutoffs and mode spectra, compute group and phase velocities of guided waves, and size guides and cavities for a given frequency.
You know plane waves, the boundary conditions at conductors, skin depth and the plasma dispersion relation. You know standing waves on strings and in pipes from mechanics. This lesson confines electromagnetic waves between metal walls, where the requirement that the fields fit sets cutoffs, modes and resonances.
| Term | What it means |
|---|---|
| Waveguide | A hollow conductor, often rectangular, that carries electromagnetic waves along its length. |
| Mode | A particular field pattern that fits the guide's boundary conditions, labeled by integers. |
| TE mode | Transverse electric: no electric field along the guide. |
| Cutoff frequency | The lowest frequency at which a mode propagates; for TE$_{10}$, $c/2a$. |
| Group velocity | $c\sqrt{1 - (f_c/f)^2}$, the speed of energy along the guide. |
| Resonant cavity | A closed conducting box with discrete resonant frequencies. |
| Quality factor | $Q$, the number of radians of oscillation over which a cavity's energy falls by $e$. |
Inside a hollow conductor, the fields must satisfy the wave equation and vanish appropriately at the walls: tangential $\vec{E}$ and normal $\vec{B}$ must be zero at a perfect conductor. For a rectangular guide of width $a$ and height $b$, the solutions are standing waves across the guide and traveling waves along it, labeled by the number of half-wavelengths across each direction, $m$ and $n$. Their dispersion relation is
$$\omega^2 = \omega_{mn}^2 + c^2k^2, \qquad f_{mn} = \frac{c}{2}\sqrt{\left(\frac{m}{a}\right)^2 + \left(\frac{n}{b}\right)^2}.$$
Below the cutoff frequency $f_{mn}$, $k$ is imaginary and the mode decays. The lowest mode of a guide with $a > b$ is TE$_{10}$, with $f_c = c/2a$: half a wavelength fits across the width. Above cutoff the wave travels with
$$v_g = c\sqrt{1 - \left(\frac{f_c}{f}\right)^2}, \qquad v_p = \frac{c}{\sqrt{1 - (f_c/f)^2}}, \qquad v_gv_p = c^2.$$
Closing both ends of a guide of length $d$ makes a resonant cavity, which supports only discrete frequencies $f = \tfrac{c}{2}\sqrt{(m/a)^2 + (n/b)^2 + (p/d)^2}$. A coaxial cable, with a center conductor, is different: its TEM mode has no cutoff and travels at the speed of light in the insulator for any frequency.
Another way: picture
Picture a plane wave zigzagging down the guide, bouncing off the side walls. The zigzag's angle is set so that the reflected waves reinforce and the field vanishes at the walls. At high frequency the zigzag is shallow and the wave rushes down the guide; near cutoff the wave bounces almost straight across, making little forward progress. At cutoff it bounces straight across and goes nowhere.
Another way: steps
Checks. Cutoffs must scale inversely with size: halving the guide doubles them. The group velocity must approach $c$ far above cutoff and zero at cutoff. The phase velocity exceeds $c$ but carries no signal. And a guide must be at least half a wavelength wide to carry anything.
The TE$_{10}$ field can be written as two plane waves bouncing between the side walls at angles $\pm\theta$ to the guide's axis. For the field to vanish at both walls, the transverse wave number must be $\pi/a$: exactly half a transverse wavelength across. Since the total wave number is $\omega/c$, the forward component is $k = \sqrt{(\omega/c)^2 - (\pi/a)^2}$.
When $\omega/c < \pi/a$, no real forward wave number exists; the wavelength is too long to fit half of it across the guide. The same mathematics gave the plasma cutoff in the last lesson, and it echoes the quantum particle in a box: confining a wave in one direction costs a minimum frequency, just as confining a particle costs a minimum energy.
The zigzag picture explains both speeds. Each bouncing plane wave travels at $c$ along its own direction, but the energy advances along the guide only at $c\cos\theta$, the group velocity. The crests, where the two waves cross the guide's axis, slide along faster, at $c/\cos\theta$, like the point where a wave meets a beach at an angle.
Because the group velocity depends on frequency, a guide is dispersive: a pulse spreads as it travels, more so near cutoff. Radar and communication engineers operate well above cutoff, typically at $1.25$ to $1.9$ times $f_c$, to keep dispersion and attenuation low while staying below the next mode's cutoff.
Waveguides come in standard sizes, each covering a frequency band. The WR-90 guide, $22.86$ by $10.16$ mm, has TE$_{10}$ cutoff $6.56$ GHz and serves the X band, $8.2$ to $12.4$ GHz, used by weather radar, police radar and satellite links. WR-340, $86$ mm wide, carries $2.45$ GHz, the frequency of microwave ovens and industrial heating.
Waveguides carry high power with low loss, because the fields fill a large volume and the currents spread over wide walls. That is why radar transmitters, particle accelerators and satellite ground stations use them instead of coaxial cable, whose thin center conductor limits power and adds loss at high frequency.
Close a waveguide at both ends and waves bounce back and forth, forming standing waves. Only frequencies that fit a whole number of half-wavelengths in each direction survive. A microwave oven is such a cavity; its many modes have hot spots and cold spots, which is why ovens have turntables or mode stirrers.
A cavity stores energy that decays slowly, over $Q$ radians of oscillation. Copper cavities reach $Q$ of ten thousand; superconducting niobium cavities, used in particle accelerators at Fermilab, Jefferson Lab and SLAC, reach ten billion. Superconducting microwave cavities are also the heart of some quantum computers, storing single photons for milliseconds as long-lived quantum memories.
A coaxial cable has two conductors, and that changes everything. It supports a transverse electromagnetic (TEM) mode, with both fields perpendicular to the cable, radial $\vec{E}$ and circling $\vec{B}$, which has no cutoff: any frequency, down to DC, travels at the speed of light in the insulator. That makes coax the choice for broadband signals.
At high enough frequencies, though, coax also supports waveguide-like modes, which begin when the wavelength approaches the circumference between the conductors. That sets an upper frequency for each cable size, tens of gigahertz for small laboratory coax, and above that, engineers switch to waveguides or to optical fibers, which are dielectric waveguides for light.
An optical fiber is a dielectric waveguide: light is confined by total internal reflection rather than metal walls, but the same ideas apply. A fiber supports modes with cutoffs set by its core size and the index difference. A single-mode fiber, with a core about $9$ μm across, is small enough that only one mode propagates at telecommunication wavelengths, eliminating the dispersion between modes that blurs pulses in larger fibers.
Single-mode fibers carry signals thousands of kilometers, limited only by the slow spreading from the glass's own dispersion and by attenuation. Their design follows exactly the mode-counting logic of metal waveguides, with a cutoff condition that determines when the second mode appears.
A guide operated below cutoff is a powerful filter. Fields decay exponentially along it, so a short section of narrow pipe blocks low frequencies almost completely. Microwave oven doors use this: the metal mesh's holes are far smaller than half the $12$ cm wavelength, so microwaves cannot pass while visible light, whose wavelength is far smaller than the holes, passes freely.
Shielded rooms for MRI scanners and electronic testing let cables and air ducts pass through the walls in metal tubes sized as waveguides below cutoff, admitting air or fiber-optic lines while blocking radio interference. The cutoff formula tells engineers how narrow and long each tube must be: a tube whose length is several times its width attenuates frequencies well below cutoff by many orders of magnitude.
Weather radars operated by the National Weather Service, the NEXRAD network of about $160$ stations, transmit pulses at about $2.8$ GHz with peak powers near a megawatt. Such power cannot pass through coaxial cable; it travels from the klystron transmitter to the dish in rectangular waveguide sized so that the frequency sits well above the TE$_{10}$ cutoff and below the next mode's. At $2.8$ GHz the guide must be wider than $c/2f = 5.4$ cm, and the standard size used is about $7$ cm.
Automotive radars at $77$ GHz, now standard for adaptive cruise control and emergency braking, need guides only about $2$ mm wide, and at those sizes the waveguides are often formed directly in the circuit board or molded plastic with metal plating. The cutoff formula scales the same plumbing from megawatt weather radar down to a car's bumper.
A microwave oven's cavity is a metal box, and its door is a metal mesh you can see through. The oven works at $2.45$ GHz, a free-space wavelength of $12.2$ cm. Each hole in the mesh is a tiny waveguide a millimeter or two across, far narrower than the $6$ cm half wavelength needed to reach cutoff, so microwaves decay by many orders of magnitude within the thickness of the mesh.
Visible light, with wavelengths near half a micrometer, is thousands of times smaller than the holes and passes freely, which is why you can watch your food. Federal regulations from the Food and Drug Administration limit microwave leakage to $5$ mW/cm² at $5$ cm from the oven's surface over its lifetime, a limit the mesh meets easily as long as the door seals properly.
It is natural to think a hollow pipe can carry any wave, like a garden hose carrying any water. But an electromagnetic wave must satisfy the boundary conditions at the walls, and that is possible only if at least half a wavelength fits across. A guide narrower than that carries nothing: the field decays exponentially along it, which is why a microwave oven's door mesh, with holes far smaller than the microwaves' half-wavelength, keeps them in.
A second misconception is that a phase velocity above $c$ lets signals outrun light. The phase velocity is the speed of the crest pattern along the guide; energy and signals travel at the group velocity, which is always less than $c$.
Take WR-90 guide, $a = 22.86$ mm. Write the TE$_{10}$ cutoff.
$f_c = \dfrac{c}{2a}$
Half a wavelength across the width.
Evaluate the cutoff.
$f_c = \dfrac{3.00 \times 10^8}{2 \times 0.02286} = 6.56\ \text{GHz}$
Below the X band.
Find the next cutoff.
$f_{20} = 2f_c = 13.1\ \text{GHz}$
TE$_{20}$ appears here; TE$_{01}$ at $14.8$ GHz with $b = 10.16$ mm.
Identify the single-mode band.
$6.56\ \text{GHz} < f < 13.1\ \text{GHz}$
Only TE$_{10}$ propagates.
Compare with the rated band.
$8.2\text{–}12.4\ \text{GHz}$
Comfortably inside, away from both edges.
A $10$ GHz signal travels in WR-90, $f_c = 6.56$ GHz. Find the ratio.
$\dfrac{f_c}{f} = 0.656$
Well above cutoff.
Find the square root factor.
$\sqrt{1 - 0.656^2} = 0.755$
The zigzag's forward fraction.
Find the group velocity.
$v_g = 0.755c = 2.26 \times 10^{8}\ \text{m/s}$
Energy moves slower than light.
Find the phase velocity.
$v_p = \dfrac{c}{0.755} = 3.97 \times 10^{8}\ \text{m/s}$
Crests move faster than light.
Find the free-space wavelength.
$\lambda_0 = \dfrac{c}{f} = 30\ \text{mm}$
At $10$ GHz.
Find the guide wavelength.
$\lambda_g = \dfrac{\lambda_0}{0.755} = 39.7\ \text{mm}$
Longer than in free space: the spacing of crests along the guide.
A cavity is $a = 30$ cm wide, $b = 30$ cm deep and $d = 20$ cm tall. Write the resonance condition.
$f = \dfrac{c}{2}\sqrt{\left(\dfrac{m}{a}\right)^2 + \left(\dfrac{n}{b}\right)^2 + \left(\dfrac{p}{d}\right)^2}$
Half-waves in all three directions.
Find the free-space wavelength at $2.45$ GHz.
$\lambda = \dfrac{3.00 \times 10^8}{2.45 \times 10^9} = 12.2\ \text{cm}$
The oven's operating frequency.
Express the condition in wavelengths.
$\left(\dfrac{m}{a}\right)^2 + \left(\dfrac{n}{b}\right)^2 + \left(\dfrac{p}{d}\right)^2 = \left(\dfrac{2}{\lambda}\right)^2 = 0.0267\ \text{cm}^{-2}$
With $\lambda = 12.2$ cm.
Try the mode $(3, 3, 2)$.
$\left(\dfrac{3}{30}\right)^2 + \left(\dfrac{3}{30}\right)^2 + \left(\dfrac{2}{20}\right)^2 = 0.030 \quad\Rightarrow\quad f = 2.60\ \text{GHz}$
Slightly above the operating frequency.
Try the mode $(3, 2, 2)$.
$0.0100 + 0.0044 + 0.0100 = 0.0244 \quad\Rightarrow\quad f = 2.34\ \text{GHz}$
Slightly below it.
Try the mode $(4, 1, 2)$.
$0.0178 + 0.0011 + 0.0100 = 0.0289 \quad\Rightarrow\quad f = 2.55\ \text{GHz}$
Another nearby mode.
Count the nearby modes.
$\text{several modes lie within } 10\% \text{ of } 2.45\ \text{GHz}$
A cavity many half-wavelengths across has many modes near any frequency.
Explain the turntable.
$\text{hot spots at antinodes, spaced about } \lambda/2 = 6\ \text{cm}$
Rotating the food averages over them.
Write the cutoff.
$f_c = \dfrac{c}{2a}$
Half a wavelength across.
Substitute the values.
$f_c = \dfrac{3.00 \times 10^8}{2 \times 0.050}$
SI units.
Evaluate the cutoff.
A waveguide's lowest cutoff frequency is $17$ GHz. What happens to a $8.5$ GHz signal fed into it?
Complete the worked solution: a guide is $25$ mm wide. Find its TE$_{10}$ cutoff frequency in GHz, the cutoff wavelength in mm, and the TE$_{20}$ cutoff in GHz.
Find the TE$_{10}$ cutoff.
$f_{10} = \dfrac{c}{2a} =$ f
Half a wavelength across.
Double the width.
$\lambda_c = 2a =$ l
The longest wavelength that propagates.
Find the TE$_{20}$ cutoff.
$f_{20} = 2f_{10} =$ g
Two half-waves across.
Match each quantity to its expression.
| $c/2a$ | $c\sqrt{1 - (f_c/f)^2}$ | $c/\sqrt{1 - (f_c/f)^2}$ | $\tfrac{c}{2}\sqrt{(m/a)^2 + (n/b)^2 + (p/d)^2}$ | |
|---|---|---|---|---|
| the TE$_{10}$ cutoff | ||||
| the group velocity | ||||
| the phase velocity | ||||
| cavity resonances |
A rectangular guide has width $a$ twice its height $b$, and its TE$_{10}$ cutoff is $17$ GHz. Fill in the cutoffs, in GHz, of TE$_{10}$, TE$_{20}$ and TE$_{01}$.
| cutoff (GHz) | |
|---|---|
| TE$_{10}$ | |
| TE$_{20}$ | |
| TE$_{01}$ |
A rectangular waveguide is $30$ mm wide. What is the cutoff frequency of its TE$_{10}$ mode, in GHz? Use $c = 3.00 \times 10^8$ m/s.
Answer: GHz
A rectangular waveguide $15$ mm wide carries a $12.5$ GHz signal in its TE$_{10}$ mode. At what speed does the signal's energy travel down the guide, in units of $10^8$ m/s? Use $c = 2.998 \times 10^8$ m/s.
Answer: × 10⁸ m/s
A rectangular waveguide will feed a $77$ GHz automotive radar. What is the smallest width, in mm, for which the $77$ GHz signal can travel in the TE$_{10}$ mode? Use $c = 3.00 \times 10^8$ m/s.
Answer: mm
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A rectangular guide has width $a$ twice its height $b$, and its TE$_{10}$ cutoff is $17$ GHz. Fill in the cutoffs, in GHz, of TE$_{10}$, TE$_{20}$ and TE$_{01}$.
| cutoff (GHz) | |
|---|---|
| TE$_{10}$ | |
| TE$_{20}$ | |
| TE$_{01}$ |
You can analyze guided waves. Explain to someone why a microwave oven's see-through door keeps the microwaves inside.
19. Your turn: what is the TE$_{10}$ cutoff of a guide $50$ mm wide?, step 3
$f_c = 3.0\ \text{GHz}$
Signals below $3$ GHz cannot pass.