Every circular cutoff is a fixed number divided by the radius. Halve the radius and every cutoff doubles.
EE 499 · Satellite Technologies
Week 6
Last Week We Designed the Path. Now How Do We Build It?
Week 5 finished holding a waveguide dimension in millimetres. Week 6 turns it into hardware — the other shape of guide, the standard sizes you can actually order (including the WR-340 guide used in class), and the cable that does the short runs.
By the end: you should be able to size a circular waveguide as well as a rectangular one, read a standard waveguide designation and work out its cutoff — including the WR-340 used in class — explain why a coaxial cable has no cutoff while a hollow guide does, and choose between rectangular, circular and coaxial structures for a given job.
Primary source: the original lecture notes, Topic 3 — sections 1.3.2 and 1.3.3.
01 — Where we are
We designed the path. We have not built it.
01 — Where we are
How do we select and physically implement the transmission structure?
Week 5 was one chain of decisions that ended at a dimension. Week 6 turns that dimension into hardware, and ends at a choice: which structure, which size, which band.
- rectangular
- single-mode window
- circular
- standard sizes
- WR-340
- coax
- choosing a structure
Part 1 — another geometry
We have seen rectangular waveguide. Is that our only option?
The notes are explicit that it is not: a waveguide can have any cross-sectional shape, as long as that shape is uniform along the length of the guide. So before we order anything, we should look at the other shape that matters in satellite hardware.
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 has its own cutoff:
\[f_{c,mn}=\frac{c}{2}\sqrt{\left(\frac{m}{a}\right)^{2}+\left(\frac{n}{b}\right)^{2}},\qquad f_{c,10}=\frac{c}{2a}\]
| 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. |
02 — Rectangular recap
Why the working band is a window, not a floor
The shape of this diagram, and the \(1.2f_{c,10}\) to \(0.8f_{c,20}\) rule drawn on it, are from Topic 3, p. 13 →
03 — Circular waveguide
What if the cross-section is circular?
Nothing in Week 5 required the cross-section to be rectangular. The walls were conductors and the shape was constant along \(z\); that is all the physics needed. A circular pipe satisfies both conditions, and circular structures appear throughout satellite hardware for reasons that have nothing to do with mathematics:
Rotational symmetry
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.
Feeds and horns
A reflector feed, a conical horn and a circularly symmetric radiation pattern all want a circular aperture behind them.
Rotary joints
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.
03 — Circular waveguide
The geometry, and the one dimension that sets everything
04 — Circular modes
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 | \(TE_{10}\) | \(TE_{11}\) |
04 — Circular modes
The first two patterns a circular guide allows
05 — 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, which are not integers. The notes give the first two cutoffs directly:
\[f_{c_1}=\frac{c}{2a}\left(\frac{1.841}{\pi}\right),\qquad f_{c_2}=\frac{c}{2a}\left(\frac{2.405}{\pi}\right)\]
Here \(a\) is the radius, so the internal diameter is \(2a\). The numbers come from Bessel functions and their derivatives.
The general form used for any circular mode is \(f_c=x'_{mn}c/2\pi a\) for \(TE\) modes and \(f_c=x_{mn}c/2\pi a\) for \(TM\) modes. Substituting \(x'_{11}=1.841\) gives \(TE_{11}\), and \(x_{01}=2.405\) gives \(TM_{01}\) — which reproduces the notes' two equations exactly.
From the original lecture notes · Topic 3, p. 14, equation (15) →
05 — Circular cutoff
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 \(\rho=a\). That is what puts a root into the cutoff formula, and it is the whole reason the constants are irrational rather than whole numbers.
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.
For this course the roots are given numbers, exactly as the notes give them. What matters is what they do to the design, not where they come from.
05 — Circular cutoff
A narrower window than the rectangle offers
Between \(f_{c_1}\) and \(f_{c_2}\) only the dominant mode propagates, and because both cutoffs share the factor \(c/2a\) their ratio is fixed by the roots alone. This is the notes' first closing remark on circular guides: the bandwidth for single-mode operation of circular waveguides is less than that of a rectangular waveguide with \(a\ge 2b\).
\[f_{c,\text{dominant}}\;<\;f_{op}\;<\;f_{c,\text{next mode}}\]
Single-mode operation, in any guide. What differs is how far apart those two edges are: \(f_{c_2}/f_{c_1}=2.405/1.841=1.31\) for a circular guide, against \(f_{c,20}/f_{c,10}=2\) for a rectangular one with \(b\le a/2\).
From the original lecture notes · Topic 3, p. 15, note (i) →
05 — Circular cutoff
Could a circular guide also 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 — the radius.
- 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 \(f_{c_2}=1.31\times10.46=13.67\ \text{GHz}\).
\[10.46\ \text{GHz}\;<\;12\ \text{GHz}\;<\;13.67\ \text{GHz}\quad\checkmark\]
\(12\ \text{GHz}\) sits \(15\%\) above the dominant cutoff and \(12\%\) below the next mode — margins chosen for this example, not a standard.
06 — 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 | \(TE_{10}\) — one orientation, fixed by the broad wall | \(TE_{11}\) — two degenerate orthogonal orientations, 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 |
| Typical use | runs, filters, couplers and most catalogue components | 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.
06 — Rectangular vs circular
Check yourself
1 · Halving the radius
A circular waveguide has its radius halved. What happens to the frequency range over which only the dominant mode propagates?
Part 2 — from equations to a catalogue
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 the calculation into an order.
07 — Standard sizes
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:
\[\text{WR-}90\ \Longrightarrow\ a=0.90\ \text{inch}=22.86\ \text{mm}\]
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 the Week 5 result \(f_{c,10}=c/2a\). 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.
07 — Standard sizes
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.
07 — Standard sizes
The guide on the bench: WR-340
The guide used with this class: \(a=3.40\ \text{inch}=86.36\ \text{mm}\) and \(b=1.70\ \text{inch}=43.18\ \text{mm}\).
\[\begin{gathered}f_{c,10}=\frac{c}{2a}=1.74\ \text{GHz}\\ f_{c,20}=\frac{c}{a}=3.47\ \text{GHz},\qquad f_{c,01}=\frac{c}{2b}=3.47\ \text{GHz}\end{gathered}\]
\(b=a/2\) exactly, so \(TE_{20}\) and \(TE_{01}\) share a cutoff, as in WR-75. Single-mode window: \(1.74\) to \(3.47\ \text{GHz}\).
| The notes' rule | \(1.2f_{c,10}\) to \(0.8f_{c,20}\): \(2.08\) to \(2.78\ \text{GHz}\) |
|---|---|
| Published band | \(2.20\) to \(3.30\ \text{GHz}\): about \(1.27f_{c,10}\) to \(0.95f_{c,20}\) |
WR-340 is WR-90 scaled by \(3.78\): every length \(3.78\times\) larger, every cutoff \(3.78\times\) lower. It cannot carry \(1.5\ \text{GHz}\) at all (below cutoff); a coaxial cable can.
07 — Standard sizes
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. 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 tolerances all can.
07 — Standard sizes
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. So the honest engineering move is to size the guide from the notes' own equations:
| 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 |
Every number in the middle three columns is \(f_{c_1}=1.841c/2\pi a\) and \(f_{c_2}=2.405c/2\pi a\) evaluated for that diameter — 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 on the previous slide.
07 — Standard sizes
Our 12 GHz link, in either geometry
Week 5 chose a guide for \(12\ \text{GHz}\) by shrinking \(a\) until the window contained it. With both geometries now available, the same link can be built either way:
| 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 — Standard sizes
Check yourself
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?
Part 3 — the other transmission structure
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.
08 — Coaxial line
Two conductors, one axis
08 — Coaxial line
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. |
\[\text{coax: TEM possible}\qquad\text{hollow guide: TEM impossible}\]
A TEM field needs a potential difference between two separate conductors. A hollow guide has only one conductor, 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.
08 — Coaxial line
Where each one wins
| 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 |
08 — Coaxial line
Check yourself
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?
Part 4 — putting it together
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.
09 — Choosing a structure
Which structure for which job?
| The job | Usually the answer | Why |
|---|---|---|
| Bench jumpers, test leads, short runs | coaxial cable | Flexible, routable, cheap to terminate; works from DC upward — 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 long, high-power run to the antenna | rectangular waveguide | Low loss, highest power handling, one fixed polarisation; rigid, about an octave of single-mode band. |
| The feed, and any rotating joint | circular waveguide | Symmetry: it can rotate and carry either polarisation of \(TE_{11}\); window only \(1.31\times\) wide. |
09 — Choosing a structure
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.
- Size. For a guide, set the dimension from the cutoff equations so that the 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 \(a\).
09 — Choosing a structure
Carry these forward
Rectangular recap
\(f_{c,mn}=\frac{c}{2}\sqrt{(m/a)^{2}+(n/b)^{2}}\); dominant \(TE_{10}\); operate in \(1.2f_{c,10}
Circular guides
\(f_{c_1}=\frac{c}{2a}\frac{1.841}{\pi}\), \(f_{c_2}=\frac{c}{2a}\frac{2.405}{\pi}\); dominant \(TE_{11}\); window only \(1.31\times\) wide.
Standard sizes
WR-\(n\) means \(a=n/100\) inch, so a designation is a cutoff in disguise. Circular guides have no equivalent universal scheme.
WR-340
\(a=86.36\ \text{mm}\), \(b=43.18\ \text{mm}\): \(f_{c,10}=1.74\ \text{GHz}\), \(f_{c,20}=f_{c,01}=3.47\ \text{GHz}\); WR-90 scaled by \(3.78\).
Coax and TEM
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.
Choosing
Coax for flexible runs and signals below cutoff; rectangular guide for high-power runs; circular guide for feeds and rotating joints.
After the lecture
Week 6 exercises — practice problems
The lecture ends here. What follows is a set of three problems to work through afterwards, with a worked solution after each one. Attempt the problem before advancing to its solution; use \(c=3\times10^{8}\ \text{m/s}\) throughout, as the notes do. Problems on impedance, mismatch and S-parameters follow their own lecture in Week 8.
10 — Exercises
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 that the arithmetic is there to support.
10 — Exercises
Exercise 1 — 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 is unpredictable. A design can be electromagnetically possible and still be a bad design.
This is the same conclusion Week 5 reached: WR-90 works at \(12\ \text{GHz}\) but is a guide intended for a lower band.
10 — Exercises
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 propagation 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.
10 — Exercises
Exercise 2 — 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 \(1\ \text{m}\) of this guide is electrically shorter than \(1\ \text{m}\) of free space.
10 — Exercises
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.
- 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.
10 — Exercises
Exercise 3 — 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 \(f_{c_1}\).
- Single-mode interval: \(17.58\) to \(22.97\ \text{GHz}\) — a width of \(5.39\ \text{GHz}\).
- \(20\ \text{GHz}\) is a sensible choice: \(13.8\%\) above \(f_{c_1}\) and \(12.9\%\) below \(f_{c_2}\), so it is clear of both edges by about the same amount. That is roughly the best either margin can be, because the whole interval is only \(1.31\) wide.
Part 5. No. \(24\ \text{GHz}>f_{c_2}=22.97\ \text{GHz}\), so \(TM_{01}\) propagates alongside \(TE_{11}\). The two modes travel at different speeds, so the signal arrives smeared — the intermodal distortion the notes warn about. A larger operating frequency needs a smaller diameter, not the same tube.
EE 499 · Week 6
Week 6 complete
- rectangular recap
- circular guides
- standard sizes
- WR-340
- coax & TEM
- choosing
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, which job — and used the WR-340 on the bench to check that the equations describe real hardware.
Next: Week 7 looks inside the guide — what is the field actually doing above cutoff, at cutoff and below it?