01 — Recap from Week 6
We know whether a mode can travel. What does it do?
Week 6 ended with a structure chosen and sized. Four facts carry straight into this week:
| The dominant mode | \(TE_{10}\): the lowest-cutoff mode of a rectangular guide with \(a>b\), and the first pattern a wave can take. |
|---|---|
| Cutoff | \(f_{c,10}=\dfrac{c}{2a}\). A mode travels only if \(f>f_c\). |
| The single-mode window | Exactly one mode propagates between \(f_{c,10}\) and the next cutoff; the notes' conservative design range is \(1.2f_{c,10} |
| The guide on the bench | WR-340: \(a=86.36\ \text{mm}\), \(b=43.18\ \text{mm}\), so \(f_{c,10}=1.74\ \text{GHz}\) and \(f_{c,20}=f_{c,01}=3.47\ \text{GHz}\). |
Those facts answer whether a signal at a given frequency can get through. They do not say what the field is doing when it does, what it is doing when it cannot, or what happens in between. This week answers those questions, and the Waveguide Field Explorer lets us watch the answer instead of taking it on trust.
02 — Objectives
What you should understand by the end
- Propagation above cutoff: the pattern travels along the guide, with a guided wavelength longer than the free-space wavelength.
- Behaviour exactly at cutoff: nothing advances; the same pattern fills every cross-section.
- Evanescent decay below cutoff: the field falls off exponentially with distance and nothing travels.
- Guided wavelength near cutoff: how it stretches, and why that matters for design margin.
- The TE10 field pattern: strongest at the centre, zero at the side walls.
- Phase and group behaviour: why the pattern can move faster than \(c\) while energy cannot.
- Decay is not dissipation: why a field that dies away below cutoff has not been absorbed as heat.
03 — The field in three regimes
One field, three behaviours
Week 5 wrote the \(TE_{10}\) field as a phasor, equation (8) in the notes:
The physical field is the real part, \(E_y=E_0\sin(\pi x/a)\cos(\omega t-\beta z)\). It has two ingredients. The factor \(\sin(\pi x/a)\) is a pattern across the width that is fixed by the walls: zero at both side walls, strongest at the centre. The factor \(\cos(\omega t-\beta z)\) is what happens along the guide, and that is where the three regimes come from.
The dependence on \(z\) can be written for all three regimes at once. The field varies as \(e^{-\gamma z}\), where \(\gamma=\alpha+j\beta\) is the propagation constant. It satisfies \(\gamma^{2}=k_c^{2}-k_0^{2}\), with \(k_c=\pi/a\) set by the walls and \(k_0=\omega/c\) set by the frequency.
| Condition | \(\gamma\) | What the field does | |
|---|---|---|---|
| Above cutoff | \(f>f_c\) | \(\gamma=j\beta\), \(\ \beta=k_0\sqrt{1-(f_c/f)^{2}}\) | The pattern travels along \(+z\); the amplitude does not change. |
| At cutoff | \(f=f_c\) | \(\gamma=0\) | Nothing advances; every cross-section is identical. |
| Below cutoff | \(f| \(\gamma=\alpha\), \(\ \alpha=k_c\sqrt{1-(f/f_c)^{2}}\) | The amplitude falls as \(e^{-\alpha z}\); the phase does not advance. | |
04 — Above cutoff
The pattern travels
With \(\gamma=j\beta\), a crest of \(\cos(\omega t-\beta z)\) stays at constant phase when \(\omega t-\beta z\) is constant, so it moves along the guide. The speed of that crest is the phase velocity, and the distance between successive crests is the guided wavelength:
Both are larger than their free-space values. The phase velocity exceeds \(c\), and the guided wavelength exceeds \(\lambda_0\).
A phase velocity above \(c\) does not break any physical law, because a crest carries no information. Energy and information travel at the group velocity, which for a hollow guide is
so \(v_p\,v_g=c^{2}\). Far above cutoff both approach \(c\); close to cutoff \(v_p\) grows without limit while \(v_g\) falls towards zero. Above cutoff the electric and magnetic fields rise and fall together in time, and that in-phase combination is what carries power along the guide.
The explorer's Advanced panel reports \(\beta\), \(\lambda_g\), \(v_p\) and \(v_g\); the engineering view also draws the magnetic field.
05 — At cutoff
The pattern stops
At exactly \(f=f_c\) we have \(\beta=0\) and \(\gamma=0\). The field no longer depends on \(z\): every cross-section along the guide carries the same pattern, and the whole pattern simply pulses up and down together, \(E_y\propto\sin(\pi x/a)\cos\omega t\). The wave is bouncing between the two side walls without making any progress along the guide, which is why cutoff is also called a transverse resonance.
| Guided wavelength | \(\lambda_g\to\infty\) |
|---|---|
| Phase velocity | \(v_p\to\infty\) |
| Group velocity | \(v_g=0\): no energy is carried along the guide. |
| Power | No net power flows; the electric and magnetic fields are a quarter of a cycle apart in time. |
This is the dividing line: just above it the pattern creeps forward, just below it the field decays. In the explorer you reach it with the cutoff marker on the frequency slider, which selects exactly \(f_c\). The slider itself never snaps, so a frequency a fraction of a percent above cutoff is shown as propagating, with a very long guided wavelength.
06 — Below cutoff
The field dies away, and nothing travels
With \(\gamma=\alpha\) real, the field is \(E_y=E_0\sin(\pi x/a)\,e^{-\alpha z}\cos\omega t\). The shape across the width is unchanged. Along the guide the amplitude falls by a factor of \(e\) every \(1/\alpha\) metres, and the phase does not advance with \(z\): the pattern pulses in place and never travels. The decay rate is
Far below cutoff, \(\alpha\) approaches \(\pi/a\), so the field falls by a factor of \(e\) in about a third of the guide width. Just below cutoff \(\alpha\) is small and the decay is slow, which is what makes the transition at cutoff continuous.
Dissipative loss is a different mechanism. It comes from the finite conductivity of the walls and any dielectric, it happens above cutoff as well, and in a good guide it is small, a fraction of a decibel per metre. The model in the explorer is lossless, so it shows none of it. The engineering view quotes the decay rate in dB/m because that is a convenient unit for a rate of decay; that number is still a decay, not heat. Week 8 returns to this: where power goes at a component or interface is exactly the question of how much is transmitted, how much is reflected and how much is dissipated.
07 — Guided wavelength
Near cutoff the wave stretches without limit
The guided wavelength \(\lambda_g=\lambda_0/\sqrt{1-(f_c/f)^{2}}\) is longer than \(\lambda_0\) at every frequency, and it grows rapidly as \(f\) approaches \(f_c\):
| \(f/f_c\) | 1.01 | 1.05 | 1.2 | 1.5 | 2 | 3 |
|---|---|---|---|---|---|---|
| \(\lambda_g/\lambda_0\) | 7.1 | 3.3 | 1.8 | 1.34 | 1.15 | 1.06 |
| \(v_g/c\) | 0.14 | 0.31 | 0.55 | 0.75 | 0.87 | 0.94 |
The practical consequence is the design margin of Week 6. Close to cutoff the guided wavelength, the phase velocity and the group velocity all change quickly with frequency, so a small change in frequency changes the electrical length of a run of guide and the speed of the signal a great deal. Operating at \(1.2f_c\) or above keeps those changes modest. It also means a length of guide is a different number of wavelengths from the same distance in free space.
08 — Multimode
Above the next cutoff, more than one pattern can travel
The explorer draws the \(TE_{10}\) mode alone, but a guide carries every mode whose cutoff is below the operating frequency. For WR-340 the next cutoff is \(f_{c,20}=f_{c,01}=3.47\ \text{GHz}\), so above it \(TE_{20}\) and \(TE_{01}\) can travel alongside \(TE_{10}\). The next pair, \(TE_{11}\) and \(TM_{11}\), appears at \(3.88\ \text{GHz}\).
The danger is that modes with different patterns have different phase constants at the same frequency, so they travel at different speeds. A signal split across several modes arrives smeared in time, which the notes call intermodal distortion and the reason a guide is designed to carry only one mode in its band of operation. The risk is potential rather than certain: a straight, symmetric run excites almost no \(TE_{20}\), but bends, steps, flanges and tolerances all can. That is why the conservative design range stops at \(0.8f_{c,20}\) and not at \(f_{c,20}\).
09 — Guided activity
The WR-340 on the bench
This activity takes the physical WR-340 guide from a pair of dimensions to a verified prediction. Work the calculations by hand with \(c=3\times10^{8}\ \text{m/s}\), as in the lectures. The explorer uses the exact value of \(c\), so its numbers differ from yours in the third significant figure; close to cutoff, where small changes in \(f/f_c\) matter most, the difference can reach a percent.
| WR-340 frequency | 1.50 GHz | 2.50 GHz | 3.20 GHz | 3.60 GHz |
|---|---|---|---|---|
| Set in the explorer (\(\times3.778\)) | 5.67 GHz | 9.44 GHz | 12.09 GHz | 13.60 GHz: beyond the slider, use the engineering view |
Step 1
Examine WR-340
Look at the guide. Measure the inside of the opening and confirm \(a=86.36\ \text{mm}\) and \(b=43.18\ \text{mm}\). What is the ratio \(a/b\), and what does it tell you about the next mode up? Where is the flange, and which way does a wave enter?
Show expected observation
\(a/b=2\) exactly, which is the industry proportion \(b\le a/2\) from the notes, so \(TE_{20}\) and \(TE_{01}\) will share a cutoff. The guide is a hollow rectangle with a flange at each end and no inner conductor: a single conductor, so only TE and TM modes can exist.
Step 2
Calculate the cutoffs
Calculate \(f_{c,10}\), \(f_{c,20}\) and \(f_{c,01}\) for WR-340.
Show worked solution
\(f_{c,10}=\dfrac{c}{2a}=\dfrac{3\times10^{8}}{2\times0.08636}=1.74\ \text{GHz}\); \(\ f_{c,20}=\dfrac{c}{a}=3.47\ \text{GHz}\); \(\ f_{c,01}=\dfrac{c}{2b}=\dfrac{3\times10^{8}}{2\times0.04318}=3.47\ \text{GHz}\).
\(TE_{20}\) and \(TE_{01}\) are degenerate because \(b=a/2\).
Step 3
The single-mode window
State the theoretical single-mode interval, and the conservative design range given by the notes' rule \(1.2f_{c,10}
Show worked solution
Theoretical: \(1.74\) to \(3.47\ \text{GHz}\), one octave. Conservative: \(1.2\times1.74=2.08\ \text{GHz}\) to \(0.8\times3.47=2.78\ \text{GHz}\). The published band, \(2.20\) to \(3.30\ \text{GHz}\), sits between the two rules.
Step 4
Predict before you look
For each frequency, write down your predictions before using the explorer: does the field travel, decay or stand still? Which modes can propagate? What is \(f/f_c\)? What is \(\lambda_g\), or for the case below cutoff, the decay constant \(\alpha\)? Is the frequency inside the conservative window?
| Frequency | Region | Modes | \(f/f_{c,10}\) | \(\lambda_g\) or \(\alpha\) |
|---|---|---|---|---|
| 1.50 GHz | ||||
| 2.50 GHz | ||||
| 3.20 GHz | ||||
| 3.60 GHz |
Show worked solution
| Frequency | Region | Modes | \(f/f_{c,10}\) | \(\lambda_g\) or \(\alpha\) |
|---|---|---|---|---|
| 1.50 GHz | below cutoff: evanescent | none | 0.864 | \(\alpha=18.3\ \text{Np/m}\) |
| 2.50 GHz | dominant-mode region, inside the conservative window | \(TE_{10}\) only | 1.44 | \(\lambda_g=167\ \text{mm}\) (\(\lambda_0=120\ \text{mm}\)) |
| 3.20 GHz | dominant-mode region, but above the conservative limit of \(2.78\ \text{GHz}\) and just inside the published band | \(TE_{10}\) only | 1.84 | \(\lambda_g=112\ \text{mm}\) (\(\lambda_0=93.8\ \text{mm}\)) |
| 3.60 GHz | above \(f_{c,20}\): higher-order propagation possible | \(TE_{10}\), \(TE_{20}\), \(TE_{01}\) | 2.07 | \(\lambda_g=95.1\ \text{mm}\) for \(TE_{10}\) |
For 1.50 GHz: \(\alpha=\dfrac{\pi}{a}\sqrt{1-(1.50/1.74)^{2}}=36.4\times0.504=18.3\ \text{Np/m}\), so the field falls by a factor of \(e\) every \(1/\alpha=54.5\ \text{mm}\).
Step 5
Verify in the explorer
Set the explorer to \(5.67\), \(9.44\) and \(12.09\ \text{GHz}\). For each, read the status, and open the Advanced panel to read \(f/f_c\). Does it match your table? Then open the engineering view, choose the WR-340 preset, enter \(3.60\ \text{GHz}\) and read the list of modes.
Show expected observation
The explorer reports below cutoff at 5.67 GHz with \(f/f_c=0.86\), and above cutoff at the other two with \(f/f_c=1.44\) and \(1.84\): the same ratios as for WR-340. In the engineering view with WR-340 at 3.60 GHz the mode list marks \(TE_{10}\), \(TE_{01}\) and \(TE_{20}\) as propagating, \(TE_{11}\) and \(TM_{11}\) as not yet, and a note says that other modes can propagate.
Step 6
Below-cutoff decay
Set the explorer to \(5.67\ \text{GHz}\) (WR-340 at 1.50 GHz) and switch to Cross-section. Move the slice position into the guide. By what factor has the field fallen after one broad dimension, \(z=a\)? Predict it first: it is \(e^{-\alpha a}\).
Show worked solution
\(\alpha a=18.3\times0.08636=1.58\), so \(e^{-1.58}=0.205\): after one broad dimension the field is about a fifth of its entrance value. In the explorer the guide is \(3.778\) times smaller, but \(\alpha a\) is the same, so at a slice of \(a=22.86\ \text{mm}\) the arrows are again about a fifth of their length at the entrance. The arrows pulse together in place; they do not move along the guide.
The field has not been absorbed. It has been reflected at the entrance, and the time-averaged power flowing into the guide is zero.
Step 7
Guided wavelength near cutoff
Predict \(\lambda_g\) for WR-340 at \(1.85\ \text{GHz}\) (\(f/f_c=1.07\)). Then set the explorer to \(6.99\ \text{GHz}\), read \(\lambda_g\) from the Advanced panel and scale it by \(3.778\). Now step the frequency down towards cutoff and use the cutoff marker. What happens to the picture along the guide?
Show worked solution
\(\lambda_0=162\ \text{mm}\) and \(\lambda_g=\dfrac{162}{\sqrt{1-(1.74/1.85)^{2}}}=471\ \text{mm}\), about \(5.5\) broad dimensions. The explorer reads about \(124\ \text{mm}\) for its smaller guide, and \(124\times3.778\approx468\ \text{mm}\). Near cutoff the answer is very sensitive: the \(0.07\%\) difference between \(c=3\times10^{8}\ \text{m/s}\) and the exact value changes \(\lambda_g\) here by about \(0.6\%\), so expect agreement to within a percent rather than to three figures. As the frequency falls the crests spread further apart until, exactly at cutoff, the pattern is the same along the whole guide.
Step 8
Why 3.60 GHz is different
At \(3.20\ \text{GHz}\) a single mode propagates, and at \(3.60\ \text{GHz}\) three can. Explain why operation above \(f_{c,20}\) becomes potentially multimode, and why the explorer cannot show it.
Show expected answer
Above \(3.47\ \text{GHz}\) the \(TE_{20}\) and \(TE_{01}\) patterns have a real \(\beta\), so they can propagate alongside \(TE_{10}\). Each mode has a different \(\beta\), and so a different phase and group velocity, at the same frequency; a signal shared among them arrives smeared in time (intermodal distortion). Whether they are excited depends on the source and on discontinuities. The visual explorer draws \(TE_{10}\) only, so it cannot show this; the engineering view's mode list is what reveals it.
Step 9
Engineering interpretation
Which of the four frequencies would you choose as the operating point for this guide, and why? What would you tell a colleague who proposes using the guide at \(1.50\ \text{GHz}\)? At \(3.60\ \text{GHz}\)? Be careful with the words you use for the \(1.50\ \text{GHz}\) case.
Show worked answer
Choose \(2.50\ \text{GHz}\): it is inside the conservative window \(2.08\) to \(2.78\ \text{GHz}\), comfortably above cutoff, and single-mode. \(3.20\ \text{GHz}\) also carries one mode and lies within the published band, but it leaves little margin below \(f_{c,20}\).
At \(1.50\ \text{GHz}\) nothing propagates: the signal is reflected, not heated, and the field decays within a few centimetres. A coaxial cable, which has no cutoff, is the right structure for that signal. At \(3.60\ \text{GHz}\) the guide is potentially multimode, with risk of intermodal distortion.
10 — Check yourself
Three quick checks
1 · Where does the power go?
A lossless waveguide is driven below its cutoff frequency. The field falls off exponentially along the guide. Where has the power gone?
2 · Approaching cutoff
The frequency in a guide is lowered towards the cutoff frequency from above. What happens to the guided wavelength \(\lambda_g\)?
3 · A phase velocity above c
In a waveguide above cutoff the phase velocity \(v_p=\omega/\beta\) is greater than \(c\). Does this violate relativity?
11 — Week summary
From a calculation to a picture
Week 5 asked why microwaves and how a waveguide can carry the signal. Week 6 asked which structure and which size. Week 7 asked what the field is actually doing inside the guide.
\(\gamma=\alpha+j\beta\): above cutoff \(\gamma=j\beta\); at cutoff \(\gamma=0\); below cutoff \(\gamma=\alpha\).
The pattern travels. \(\lambda_g=\lambda_0/\sqrt{1-(f_c/f)^{2}}\); \(v_p>c\) is a crest, \(v_g=c^{2}/v_p
\(\beta=0\), \(\lambda_g\to\infty\), \(v_g=0\): the same pattern in every cross-section, pulsing in time.
Amplitude \(\propto e^{-\alpha z}\), \(\alpha=\frac{\pi}{a}\sqrt{1-(f/f_c)^{2}}\); the phase does not advance and no net power flows.
The power is reflected, not absorbed. Dissipation is a separate, small effect measured in dB/m.
Above the next cutoff more than one pattern can travel; different speeds smear the signal.
Where this leads. We now understand how fields propagate through a transmission structure. What happens when RF power encounters real interfaces, loads and components?
Next: Week 8 builds the language for that question — how much of the power is transmitted, reflected or dissipated at a component, how an impedance change turns a wave back, and how S-parameters and the vector network analyzer describe and measure it.