01 — Where we are
We designed the path. How do we build it?
Week 5 was a single chain of engineering decisions: the ionosphere set a floor on the carrier frequency, the need for directivity pushed it much higher, that landed satellite communications in the microwave bands, and microwave hardware made the connection between source and antenna a design problem with its own equations. The chain ended at a number — a broad-wall dimension in millimetres.
A number on paper is not yet a waveguide. Between the amplifier and the feed of a real earth station there is a cable, a transition, a length of rectangular guide, a rotary joint and a circular feed. Each of those is a transmission structure with its own rules about which modes it carries, which frequencies it passes and how large it must be, and none of them was settled by choose a value of \(a\).
- rectangular
- single-mode window
- circular
- standard sizes
- WR-340
- coax
- choosing a structure
Week 6 is the engineering continuation of Week 5. It ends not at a dimension but at a choice: which structure, which size, which band. It stops deliberately at the hardware. Two questions are left for the weeks after it — what the field is actually doing inside the guide (Week 7), and what happens to RF power at the joints, loads and components along the path (Week 8).
02 — Rectangular recap
Week 5 in four lines
A hollow rectangular guide of broad dimension \(a\) and narrow dimension \(b\), with \(a>b\), supports discrete \(TE\) and \(TM\) modes, each with its own cutoff frequency:
| Dominant mode | \(TE_{10}\) — the lowest cutoff, and the first pattern a wave can take. |
|---|---|
| Propagation | A mode travels only if \(f>f_c\); below that the field decays instead. |
| Second mode | With the industry proportion \(b\le a/2\), the next mode is \(TE_{20}\) at \(f_{c,20}=c/a\). |
| Working band | \(1.2f_{c,10}<f_{op}<0.8f_{c,20}\) — one mode only, clear of both edges. |
The last line is the one that matters most this week. Waveguide design is a window problem, not a threshold problem: below the dominant cutoff nothing travels, and above the second cutoff several modes travel at different speeds and the signal arrives smeared — what the notes call intermodal distortion. The usable band is the gap between those two, with margin at each end.
03 — Circular waveguide
What if the cross-section is circular?
Nothing in Week 5 required the cross-section to be rectangular. The walls had to be conductors and the shape had to be constant along \(z\); that is all the physics needed. The notes say so directly: the waveguides discussed can be of any cross-sectional shape, as long as that shape is uniform along the length of the guide.
A circular pipe satisfies both conditions, and circular structures appear throughout satellite hardware for reasons that have nothing to do with mathematics:
Nothing in the cross-section singles out a direction, so the guide does not impose an orientation on the wave the way a broad wall does.
A reflector feed, a conical horn and a circularly symmetric radiation pattern all want a circular aperture behind them.
A tracking antenna has to turn while the signal keeps flowing. A joint that rotates about the axis of a circular guide can do that; a keyed rectangular flange cannot.
Circular guides are also used for high-power runs, and for applications where the polarisation of the wave carries information in its own right — dual-polarisation and circular-polarisation feeds in particular.
The geometry, and the one dimension that sets everything
A rectangular guide is described by two independent dimensions, \(a\) and \(b\). A circular guide has only one: the radius \(a\), so the internal diameter is \(2a\). The designer therefore has one degree of freedom instead of two, and every cutoff frequency moves together when that one number changes.
The same rule, applied to a circle
The argument is identical to Week 5. Maxwell's equations are solved inside the guide, the field is required to satisfy the conducting-wall boundary condition, and only a discrete set of patterns survives. What changes is the shape of the boundary — and therefore the functions that describe the solution.
| Rectangular guide | Circular guide | |
|---|---|---|
| Cross-section described by | \(x\) and \(y\) | radius \(\rho\) and angle \(\phi\) |
| Solutions built from | sines and cosines | Bessel functions of \(\rho\); sines and cosines of \(\phi\) |
| First index \(m\) | half-cycles across the broad wall \(a\) | full cycles of variation around the guide, in \(\phi\) |
| Second index \(n\) | half-cycles across the narrow wall \(b\) | which root of the Bessel condition — how many maxima along the radius |
| Dominant mode | \(TE_{10}\) | \(TE_{11}\) |
The notes confirm both halves of that classification: modes inside metallic waveguides made of one conductor are not TEM, they are either TE or TM; and the dominant modes in rectangular waveguides and also in the circular ones are TE modes.
04 — Circular cutoff
Cutoff, with a Bessel root in place of an integer
In a rectangular guide the boundary condition is met at integer multiples of a half wavelength, which is why the integers \(m\) and \(n\) appear in the cutoff formula. In a circular guide it is met at the zeros of a Bessel function, and those zeros are not integers. The notes give the first two cutoffs directly:
Here \(a\) is the radius, so the internal diameter is \(2a\). The numbers \(1.841\) and \(2.405\) come from Bessel functions and their derivatives.
Written in the general form used for any circular mode, those same two lines are
with \(x'_{11}=1.841\) giving \(TE_{11}\) and \(x_{01}=2.405\) giving \(TM_{01}\). Substituting those two roots reproduces the notes' equations exactly.
What that number is, without a lecture on Bessel functions
A Bessel function is the circular relative of a sine: it is the shape a standing field takes along a radius. Where a sine crosses zero at \(\pi,2\pi,3\pi,\dots\), a Bessel function crosses zero at values that have to be looked up — \(2.405\), \(5.520\), and so on — and its derivative crosses zero at a different list, beginning \(1.841\). The conducting wall requires one of those curves to be zero at \(\rho=a\), which is what puts a root into the cutoff formula and the whole reason the constants are irrational rather than whole numbers.
For this course the roots are given numbers, exactly as the notes give them. What matters is what they do to the design:
Every circular cutoff is a fixed number divided by the radius. Halve the radius and every cutoff doubles.
The ordering of the modes is just the ordering of the roots, so the smallest root gives the dominant mode.
With only one dimension available, choosing the diameter chooses the entire usable frequency region at once.
A narrower window than the rectangle offers
Single-mode operation always means the same thing, whatever the geometry: the operating frequency sits above the dominant-mode cutoff and below the cutoff of the next mode up.
What differs between geometries is how far apart those two edges are.
For the circular guide both cutoffs share the same factor \(c/2a\), so their ratio is fixed by the two roots alone and does not depend on the guide at all:
Compare the rectangular guide with \(b\le a/2\), where \(f_{c,20}/f_{c,10}=2\).
This is the first of the notes' three closing remarks on circular guides: the bandwidth for single-mode operation of circular waveguides is less than that of a rectangular waveguide with \(a\ge 2b\). The circular geometry gives up roughly a factor of two in usable span, and that single fact explains most of how the two geometries divide the work in a real terminal.
A second remark from the notes: dielectric filling
The notes' second closing remark applies to both geometries. A waveguide filled with a homogeneous dielectric of relative permittivity \(\varepsilon_r\) has lower cutoff frequencies than its air-filled counterpart, and the correction is a single substitution: replace \(c\) by \(c/\sqrt{\varepsilon_r}\) in the rectangular formula and in the circular ones. A filled guide is therefore electrically larger than its dimensions suggest — useful when a guide must be made physically small, at the cost of the dielectric loss that an air-filled guide does not have.
Could a circular guide carry our 12 GHz signal?
Week 5 sized a rectangular guide for a \(12\ \text{GHz}\) Ku-band downlink. The same question, put to a circular guide, has only one unknown.
- Design choice for this example: place \(12\ \text{GHz}\) near the middle of the window, so target \(f_{c_1}\approx0.87\times12=10.4\ \text{GHz}\).
- Invert the notes' expression: \(a=\dfrac{1.841\,c}{2\pi f_{c_1}}=\dfrac{1.841\times3\times10^{8}}{2\pi\times10.4\times10^{9}}=8.45\ \text{mm}\).
- Round to a buildable radius \(a=8.4\ \text{mm}\), an internal diameter of \(2a=16.8\ \text{mm}\).
- Check both cutoffs: \(f_{c_1}=\dfrac{1.841\times3\times10^{8}}{2\pi\times0.0084}=10.46\ \text{GHz}\), and \(f_{c_2}=1.31\times10.46=13.67\ \text{GHz}\).
- \(10.46<12<13.67\ \text{GHz}\): \(12\ \text{GHz}\) sits \(15\%\) above the dominant cutoff and \(12\%\) below the next mode. Those margins are the choice made in this example, not a standard — they are simply what this narrow window allows.
From the original lecture notes · Topic 3, pp. 14–15, equation (15) and notes (i)–(iii) →
05 — Rectangular vs circular
Which geometry suits the system we are building?
| Rectangular | Circular | |
|---|---|---|
| Free dimensions | two, \(a\) and \(b\) | one, the radius \(a\) |
| Dominant mode | \(TE_{10}\) — a single orientation, fixed by the broad wall | \(TE_{11}\) — two degenerate orthogonal orientations at the same cutoff, because the circle has no preferred direction |
| Cutoff relation | \(f_{c,mn}=\frac{c}{2}\sqrt{(m/a)^{2}+(n/b)^{2}}\) | \(f_c=x'_{mn}c/2\pi a\), from Bessel roots |
| Single-mode span | up to \(2\times\) the dominant cutoff | \(1.31\times\) — markedly narrower |
| Polarisation | the broad wall fixes it, which is usually what is wanted | symmetry allows rotation, so it must be controlled — but also allows dual and circular polarisation |
| Manufacture | drawn or machined rectangular tube; flanges are keyed | plain tube is simple, but roundness and concentricity are critical |
| Interfacing | standard flanges, and most catalogue components are rectangular | transitions are usually needed to join the rectangular world |
| Typical use | runs, filters, couplers and most standard hardware | feeds, horns, rotary joints, polarisers and high-power sections |
Neither geometry is better in general. The rectangle buys operating bandwidth and a fixed polarisation; the circle buys symmetry, rotation and polarisation freedom. Which of those you need is a property of the system, not of the guide — and a real terminal uses both, joined by a transition.
06 — Standard sizes and WR-340
Nobody machines a waveguide from first principles
The cutoff equations give a dimension in millimetres. Real hardware comes in standard sizes, with flanges that mate and bands that manufacturers publish. Knowing how to read those designations is what turns a calculation into an order.
Reading a standard rectangular designation
Rectangular waveguide is supplied against a designation of the form WR-n — waveguide, rectangular. The number is the nominal broad internal dimension \(a\), expressed in hundredths of an inch:
The narrow dimension \(b\) is not in the name. For the standard sizes it follows the industry proportion the notes describe, \(b\le a/2\), which is what keeps \(TE_{20}\) as the next mode up.
Once \(a\) is known, the dominant cutoff follows immediately from \(f_{c,10}=c/2a\). So a designation is really a cutoff frequency written in imperial units, and the published operating band is the single-mode window that cutoff creates. The sizes and flanges were standardised in the 1940s and have never been renamed, because changing the designation would orphan every flange, bend and coupler already in service; the underlying physics is metric, only the label is not.
The sizes a satellite engineer actually meets
| Designation | \(a\) | \(b\) | \(f_{c,10}=c/2a\) | Published band | Where it appears |
|---|---|---|---|---|---|
| WR-340 | 86.36 mm | 43.18 mm | 1.74 GHz | 2.20–3.30 GHz | S band — the physical guide used in this course |
| WR-137 | 34.85 mm | 15.80 mm | 4.30 GHz | 5.85–8.20 GHz | C-band earth-station uplink chains |
| WR-112 | 28.50 mm | 12.62 mm | 5.26 GHz | 7.05–10.0 GHz | upper C and lower X band |
| WR-90 | 22.86 mm | 10.16 mm | 6.56 GHz | 8.20–12.4 GHz | X band — the Week 5 worked example |
| WR-75 | 19.05 mm | 9.53 mm | 7.87 GHz | 10.0–15.0 GHz | the guide Week 5 chose for 12 GHz |
| WR-62 | 15.80 mm | 7.90 mm | 9.49 GHz | 12.4–18.0 GHz | Ku-band feeds and LNB waveguide ports |
| WR-28 | 7.11 mm | 3.56 mm | 21.10 GHz | 26.5–40.0 GHz | Ka-band high-throughput terminals |
Dimensions and bands are the long-established industry values, quoted here to connect the notes' equations to real hardware; the notes themselves define waveguides by \(a\) and \(b\) rather than by designation. The cutoff column is not looked up — it is \(c/2a\) evaluated for each row, and you should be able to reproduce every entry.
The guide on the bench: WR-340
Every row in the table is a drawing until you can hold one. The guide used with this class is a WR-340: broad internal dimension \(a=3.40\ \text{inch}=86.36\ \text{mm}\) and narrow dimension \(b=1.70\ \text{inch}=43.18\ \text{mm}\). It is large enough to look into, which makes it a good place to check that the equations describe a real object.
Here \(b=a/2\) exactly, so \(TE_{20}\) and \(TE_{01}\) share one cutoff — the same degeneracy that Exercise 2 finds in WR-75. The theoretical single-mode window is therefore \(1.74\) to \(3.47\ \text{GHz}\).
| WR-340, \(a=86.36\ \text{mm}\) | |
|---|---|
| Mode limits | \(1.74\) to \(3.47\ \text{GHz}\) — the range in which exactly one mode propagates. |
| The notes' rule | \(1.2f_{c,10}\) to \(0.8f_{c,20}\), i.e. \(2.08\) to \(2.78\ \text{GHz}\). |
| Published band | \(2.20\) to \(3.30\ \text{GHz}\) — about \(1.27f_{c,10}\) to \(0.95f_{c,20}\), the same proportions as WR-90. |
WR-340 is WR-90 scaled up by a factor of \(86.36/22.86=3.78\). Every length is \(3.78\times\) larger, so every cutoff is \(3.78\times\) lower: \(6.56\ \text{GHz}\) becomes \(1.74\ \text{GHz}\), and the X-band guide becomes an S-band one. The physics has not changed, only the size. That scaling is worth remembering, because next week it lets a single picture serve both guides.
A WR-340 cannot carry a \(1.5\ \text{GHz}\) signal at all, however much power is applied: \(1.5\ \text{GHz}\) is below its cutoff. A coaxial cable can. We return to that difference when choosing a structure.
Two different answers to “what band is this guide for?”
Take WR-90. The notes' rule and the manufacturer's catalogue do not give the same answer, and the difference is worth understanding rather than smoothing over.
| WR-90, \(a=22.86\ \text{mm}\) | |
|---|---|
| Mode limits | \(f_{c,10}=6.56\ \text{GHz}\) to \(f_{c,20}=13.12\ \text{GHz}\) — the range in which exactly one mode propagates. |
| The notes' rule | \(1.2f_{c,10}\) to \(0.8f_{c,20}\), i.e. \(7.87\) to \(10.50\ \text{GHz}\). |
| Published band | \(8.20\) to \(12.40\ \text{GHz}\) — about \(1.25f_{c,10}\) to \(0.95f_{c,20}\). |
Both are defensible. The lower edge is essentially agreed: nobody operates close to cutoff, where \(\beta\) collapses and \(\lambda_g\) stretches rapidly with frequency. The upper edge is where they differ. The catalogue runs nearer to \(f_{c,20}\) because a well-made straight run excites almost no \(TE_{20}\); the notes keep a wider margin because bends, steps, flanges and manufacturing tolerances all can.
Circular guides do not work the same way
There is no circular equivalent of WR that is used as universally. The notes define a circular guide by its radius \(a\) alone and give no designation scheme at all; in industry some suppliers use a WC-n label for the internal diameter in hundredths of an inch, others simply quote the bore in millimetres, and a feed assembly is usually specified by its band rather than by any tube name. Do not assume that a circular guide carries a WR-style designation, and do not invent one.
The reliable move is to size the guide from the notes' own equations. Every number in the middle three columns below is \(f_{c_1}=1.841c/2\pi a\) and \(f_{c_2}=2.405c/2\pi a\) evaluated for that diameter:
| Internal diameter \(2a\) | \(f_{c_1}\ (TE_{11})\) | \(f_{c_2}\ (TM_{01})\) | Single-mode window | Satellite use |
|---|---|---|---|---|
| 50.0 mm | 3.52 GHz | 4.59 GHz | 3.5–4.6 GHz | C-band downlink feed, 3.7–4.2 GHz |
| 25.4 mm | 6.92 GHz | 9.04 GHz | 6.9–9.0 GHz | X-band feeds and rotary joints |
| 17.2 mm | 10.22 GHz | 13.35 GHz | 10.2–13.4 GHz | Ku-band downlink feed, 10.7–12.75 GHz |
| 10.4 mm | 16.90 GHz | 22.08 GHz | 16.9–22.1 GHz | Ka-band downlink feed, 17.7–21.2 GHz |
This is a worked table, not a manufacturer's standard, and it is deliberately not called WR-anything. Note how much narrower each window is than the rectangular bands above.
Our 12 GHz link, in either geometry
| Rectangular — WR-75 | Circular — \(2a=16.8\ \text{mm}\) | |
|---|---|---|
| Dominant cutoff | \(f_{c,10}=c/2a=7.87\ \text{GHz}\) | \(f_{c_1}=10.46\ \text{GHz}\) |
| Next mode | \(f_{c,20}=15.75\ \text{GHz}\) | \(f_{c_2}=13.67\ \text{GHz}\) |
| 12 GHz sits | \(1.52\times\) cutoff, well inside | \(1.15\times\) cutoff, comfortably inside a tighter window |
| Margin to next mode | \(24\%\) | \(12\%\) |
| Sensible role | the run from the equipment to the antenna | the feed itself, and any rotating joint |
This is how a real terminal is built: not one geometry or the other, but rectangular guide for the run, a transition, and circular guide into the feed — each used where its properties are the ones needed.
07 — Coaxial line
Do we always need a hollow waveguide?
No. A waveguide is rigid, bulky, band-limited and expensive, and most of a microwave bench is not built from it. For short runs, for test equipment, and for everything below the high-power output stage, the answer is a coaxial cable.
A coaxial line is a solid inner conductor of radius \(a\), a dielectric out to radius \(b\), and an outer conductor — usually a braid or a solid tube — beyond that. The jacket outside it is mechanical protection and plays no part in the electromagnetics. The shield is a conductor, not packaging: it carries the return current, and it is why the field is confined entirely between the two conductors and nothing escapes.
Two conductors change what modes are possible
Week 5 classified modes by what the field does along the direction of travel. The notes' rule is short, and it is the whole difference between these two structures:
| TEM | \(E_z=H_z=0\). Exists on two-conductor transmission lines, but not in waveguides. |
|---|---|
| TE | \(E_z=0\). Dominant mode of a rectangular guide is \(TE_{10}\); of a circular guide, \(TE_{11}\). |
| TM | \(H_z=0\). Higher-order modes, usually not the ones a guide is operated on. |
A TEM field needs a potential difference between two separate conductors. A hollow guide has only one, so there is nothing to hold that difference — which is exactly why a waveguide has a cutoff and a coaxial line does not.
That single structural fact drives the practical differences: coax works from DC upward with no lower cutoff, it can be bent and routed, it is small, and it connects to instruments easily. Its price is a dielectric that absorbs, a thin inner conductor that heats, and conductor loss that climbs with frequency.
| Coaxial cable | Hollow waveguide | |
|---|---|---|
| Conductors | two — inner and shield | one — the wall |
| Mode | TEM, plus higher modes if the cable is too fat for the frequency | TE or TM only |
| Low-frequency limit | none — works down to DC | cutoff; below it nothing propagates |
| Loss | conductor and dielectric loss, both rising with frequency | low — air filled, and the current spreads over a large wall |
| Power handling | limited by the thin inner conductor and the dielectric | much higher, as the notes record |
| Mechanics | flexible, routable, cheap to terminate | rigid, bulky, precise flanges |
What a coaxial line needs in addition to a mode is a way of being joined to things: to loads, to connectors, to the next component. That is where a number such as \(50\ \Omega\) becomes important, and it belongs to Week 8, once we have seen what the field does inside a guide.
08 — Choosing a structure
Which structure for which job?
We now have three transmission structures and a way to size each of them. None is better in general; each is the right answer to a different question.
| The job | Usually the answer | Why |
|---|---|---|
| Bench jumpers, test leads, short runs | coaxial cable | Flexible and routable, cheap to terminate, and it works from DC upward — there is no cutoff. |
| A signal below the guide's cutoff | coaxial cable | A waveguide carries nothing below \(f_{c,10}\). A WR-340 cannot carry \(1.5\ \text{GHz}\); a coaxial cable can. |
| A long, high-power run to the antenna | rectangular waveguide | Low loss and the highest power handling of the three, in one fixed polarisation. The price is rigidity and a single-mode band about an octave wide. |
| The feed, and any rotating joint | circular waveguide | Symmetry: no preferred orientation, so it can rotate and can carry either polarisation of \(TE_{11}\). The price is a window only \(1.31\times\) wide. |
A selection procedure
- Frequency. Fix the operating frequency or band from the link requirements (Week 5).
- Structure. Choose rectangular, circular or coaxial from the job, using the table above.
- Size. For a guide, set the dimension from the cutoff equations so that the operating band sits inside \(1.2f_{c,\text{dominant}}
- Order. Round to the nearest standard size and re-check the window with that size's real \(a\).
A real terminal uses all three: coax from the equipment, a transition, a rectangular run and a circular feed, each where its properties are the ones needed. The mode changes at every change of structure — TEM, then \(TE_{10}\), then \(TE_{11}\) — and that is the first hint of a new question. Every joint between structures is a place where the wave can be partly turned back. How much, and how we measure it, is the subject of Week 8.
09 — Week summary
Which structure, which size
Week 5 asked why microwaves, and how a waveguide can carry the signal. Week 6 asked how to select and physically implement the transmission structure: which geometry, which size, and which job each one suits.
\(f_{c,mn}=\frac{c}{2}\sqrt{(m/a)^{2}+(n/b)^{2}}\); dominant \(TE_{10}\); operate in \(1.2f_{c,10}
\(f_{c_1}=\frac{c}{2a}\frac{1.841}{\pi}\), \(f_{c_2}=\frac{c}{2a}\frac{2.405}{\pi}\); dominant \(TE_{11}\); window only \(1.31\times\) wide.
WR-\(n\) means \(a=n/100\) inch, so a designation is a cutoff in disguise. Circular guides have no equivalent universal scheme.
\(a=86.36\ \text{mm}\), \(b=43.18\ \text{mm}\): \(f_{c,10}=1.74\ \text{GHz}\), \(f_{c,20}=f_{c,01}=3.47\ \text{GHz}\); WR-90 scaled by \(3.78\).
Two conductors allow TEM and remove the cutoff; a hollow guide has one conductor, so only TE and TM modes exist, each with a cutoff.
Coax for flexible runs and signals below cutoff; rectangular guide for high-power runs; circular guide for feeds and rotating joints.
Next: Week 7 looks inside the guide — what is the field actually doing above cutoff, at cutoff and below it? Then Week 8 asks what happens when RF power meets the joints, loads and components along the path.
10 — Check yourself
Three quick checks
1 · Halving the radius
A circular waveguide has its radius halved. What happens to the frequency range over which only the dominant mode propagates?
2 · A signal below cutoff
A WR-340 waveguide (\(a=86.36\ \text{mm}\)) is driven with a \(1.5\ \text{GHz}\) signal. What happens?
3 · Why does only the guide have a cutoff?
A coaxial cable carries a signal from DC upward, but a hollow waveguide has a cutoff. What is the structural reason?
11 — Week 6 exercises
Practice problems
The lecture ends above. What follows is a separate set of three problems to work through afterwards. Each one hides its worked solution behind a disclosure — attempt the problem first, then open it. Use \(c=3\times10^{8}\ \text{m/s}\) throughout, as the lecture notes do.
Exercises 1 and 2 carry forward the practical rectangular-waveguide design work introduced in Week 5; exercise 3 is new to Week 6. Problems on impedance, mismatch and S-parameters follow their own lecture in Week 8.
Exercise 1
WR-90 at 12 GHz
Given. An air-filled WR-90 rectangular waveguide, broad dimension \(a=22.86\ \text{mm}\), with \(b\le a/2\). The intended operating frequency is \(12\ \text{GHz}\).
- Calculate the dominant-mode cutoff frequency \(f_{c,10}\).
- Determine whether \(12\ \text{GHz}\) propagates in this guide.
- Find the conservative operating range using the rule given in the lecture notes.
- Explain, in your own words, why above cutoff does not automatically mean preferred operating region.
Part 4 is the point of the exercise. The first three parts are arithmetic; the fourth is the engineering judgement the arithmetic exists to support.
Show worked solution
- \(f_{c,10}=\dfrac{c}{2a}=\dfrac{3\times10^{8}}{2\times0.02286}=6.56\ \text{GHz}\).
- \(12\ \text{GHz}>6.56\ \text{GHz}\), so the \(TE_{10}\) mode propagates. The next mode is \(f_{c,20}=c/a=13.12\ \text{GHz}\), which is above \(12\ \text{GHz}\) — so propagation is single-mode.
- \(1.2f_{c,10}=7.87\ \text{GHz}\) and \(0.8f_{c,20}=10.50\ \text{GHz}\), so the conservative range runs from \(7.87\) to \(10.50\ \text{GHz}\) — and \(12\ \text{GHz}\) lies outside it.
Part 4. Above cutoff is a statement about the modes: at \(12\ \text{GHz}\) exactly one mode can travel, so a signal does get through. Preferred operating region is a statement about margin: \(12\ \text{GHz}\) sits only \(8\%\) below \(f_{c,20}\), so any bend, step, flange or manufacturing tolerance that disturbs the field can couple energy into \(TE_{20}\), and behaviour near that edge becomes unpredictable. A design can be electromagnetically possible and still be a bad design — which is exactly why a catalogue quotes an operating band rather than just a cutoff.
This is the same conclusion Week 5 reached: WR-90 works at \(12\ \text{GHz}\) but is a guide intended for a lower band.
Exercise 2
WR-75 in full
Given. An air-filled WR-75 rectangular waveguide with \(a=19.05\ \text{mm}\) and \(b=9.525\ \text{mm}\), operated at \(12\ \text{GHz}\).
- Find the dominant-mode cutoff frequency.
- Find the nearest relevant higher-order cutoff, and say which mode or modes it belongs to.
- Determine whether \(12\ \text{GHz}\) lies inside the design range given by the notes' rule.
- Calculate the free-space wavelength \(\lambda_0\) at \(12\ \text{GHz}\).
- Calculate the guide wavelength \(\lambda_g\).
- Calculate the phase constant \(\beta\).
Note that \(b=a/2\) exactly for this guide. That has a consequence for part 2 which is worth spotting before you calculate anything.
Show worked solution
- \(f_{c,10}=\dfrac{c}{2a}=\dfrac{3\times10^{8}}{0.0381}=7.87\ \text{GHz}\).
- \(f_{c,20}=\dfrac{c}{a}=15.75\ \text{GHz}\) and \(f_{c,01}=\dfrac{c}{2b}=15.75\ \text{GHz}\). Because \(b=a/2\) exactly, \(TE_{20}\) and \(TE_{01}\) are degenerate — they share a cutoff. The next after those is \(f_{c,11}=\frac{c}{2}\sqrt{(1/a)^{2}+(1/b)^{2}}=17.61\ \text{GHz}\).
- \(1.2\times7.87=9.45\ \text{GHz}\) to \(0.8\times15.75=12.60\ \text{GHz}\). \(12\ \text{GHz}\) is inside ✓
- \(\lambda_0=\dfrac{c}{f}=\dfrac{3\times10^{8}}{12\times10^{9}}=25.0\ \text{mm}\).
- \(\lambda_g=\dfrac{\lambda_0}{\sqrt{1-(f_c/f)^{2}}}=\dfrac{25.0}{\sqrt{1-(7.87/12)^{2}}}=\dfrac{25.0}{0.755}=33.1\ \text{mm}\).
- \(\beta=\dfrac{2\pi}{\lambda_g}=\dfrac{6.2832}{0.0331}=190\ \text{rad/m}\).
WR-75 carries \(12\ \text{GHz}\) single-mode with margin at both edges, and the wave inside it is \(33.1\ \text{mm}\) long against \(25.0\ \text{mm}\) in free space — so a metre of this guide is electrically shorter than a metre of free space.
Exercise 3
Design a circular waveguide
Given. An air-filled circular waveguide has an inner diameter \(D=10.0\ \text{mm}\), so its radius is \(a=5.0\ \text{mm}\). It is intended for a Ka-band downlink feed.
- Calculate the dominant-mode cutoff \(f_{c_1}\) (the \(TE_{11}\) mode).
- Calculate the next-mode cutoff \(f_{c_2}\) (the \(TM_{01}\) mode).
- State the theoretical single-mode interval this guide offers, and its width in GHz.
- Choose a sensible operating frequency inside that interval, and justify the margin you leave at each edge.
- Would \(24\ \text{GHz}\) be acceptable in this guide? Explain what would happen.
Use \(f_{c_1}=\frac{c}{2a}\left(\frac{1.841}{\pi}\right)\) and \(f_{c_2}=\frac{c}{2a}\left(\frac{2.405}{\pi}\right)\), where \(a\) is the radius, not the diameter. Single-mode operation means \(f_{c,\text{dominant}}<f_{op}<f_{c,\text{next mode}}\); there is no fixed margin rule to apply here, so part 4 asks you to justify your own.
Show worked solution
- The cutoff equations take the radius, so use \(a=D/2=5.0\ \text{mm}\): \(f_{c_1}=\dfrac{1.841\times3\times10^{8}}{2\pi\times0.005}=17.58\ \text{GHz}\).
- \(f_{c_2}=\dfrac{2.405\times3\times10^{8}}{2\pi\times0.005}=22.97\ \text{GHz}\), which is \(\tfrac{2.405}{1.841}=1.31\) times \(f_{c_1}\), as it must be for any circular guide.
- The theoretical single-mode interval is \(17.58\) to \(22.97\ \text{GHz}\) — a width of \(5.39\ \text{GHz}\). Below \(17.58\ \text{GHz}\) nothing propagates; above \(22.97\ \text{GHz}\) a second mode joins in.
- \(20\ \text{GHz}\) is a sensible choice: it sits \(13.8\%\) above \(f_{c_1}\) and \(12.9\%\) below \(f_{c_2}\), so it is clear of both edges by about the same amount. Those are roughly the largest margins available, because the whole interval spans only a factor of \(1.31\). A choice nearer either edge would be more sensitive to frequency, to tolerance and to any discontinuity in the run.
Part 5. No. \(24\ \text{GHz}>f_{c_2}=22.97\ \text{GHz}\), so \(TM_{01}\) propagates alongside \(TE_{11}\). Two modes travelling at different speeds smear the signal — the intermodal distortion the notes warn about. Operating higher needs a smaller diameter, not this tube: every circular cutoff scales as \(1/a\).