\[\nabla\times\mathbf E=-\frac{\partial \mathbf B}{\partial t}\]
Faraday's lawA magnetic field that changes in time creates an electric field.
EE 499 · Satellite Technologies
Week 5
We Have a Spacecraft in Orbit. Now How Do We Talk to It?
Four weeks of orbital mechanics told us where the spacecraft is. This week we build the link to it — and every step is an engineering decision that forces the next one.
By the end: you should be able to explain why the link frequency must sit far above the ionospheric limit, why antenna directivity pushes it higher still, why that lands satellite communications in the microwave bands, why microwave power is carried in waveguide rather than wire, and how to size a rectangular waveguide for a given operating frequency.
Primary source: the original lecture notes, Topic 3 — sections 1.1, 1.2, 1.3, 1.3.1 and 1.3.2.
01 — The link problem
The link we have to build
- transmitter
- transmission structure
- antenna
- space channel
- satellite
01 — The link problem
One decision forces the next
This lecture is a single chain of engineering decisions. Nothing in it is chosen for mathematical convenience — each step is forced by the one before it, and we will walk the chain from the left.
- pick a frequency
- cross the ionosphere
- get directivity
- so: microwaves
- feed the antenna
- so: waveguide
- modes & cutoff
Part 1 — the first decision
What frequency should we use?
Before power, before modulation, before antennas: a carrier frequency. It is the one choice that constrains everything else, and the first thing that constrains it is what the signal has to travel through.
02 — Choosing a frequency
What the signal has to get through
03 — The ionosphere
Sunlight makes a conductor out of thin air
High in the atmosphere the gas is extremely thin and completely exposed to the Sun. Ultraviolet and X-ray photons arriving there carry enough energy to knock electrons off the atoms and molecules they strike:
- solar radiation
- electrons stripped from atoms
- free electrons & ions
- a plasma
- radio propagation changes
The result is a region containing free charge. Everywhere else on the path the wave simply travels; here it meets charges that can move in response to its own electric field, and a medium that responds is a medium that changes how the wave behaves.
Because the energy supply is sunlight, the layer is not constant. It is stronger by day than by night and varies with season and with solar activity — which is exactly what the notes record.
03 — The ionosphere
Reflected, or passed through?
One threshold decides it, and it is set by how many free electrons there are. The notes give it as:
\[f_{\text{critical}}=9\sqrt{N_{\max}}\]
\(N_{\max}\) is the maximum electron density in electrons per cubic metre; \(f_{\text{critical}}\) comes out in hertz. This threshold is the layer's plasma frequency.
Below \(f_{\text{critical}}\) the electrons can follow the wave's field, the layer acts as a reflector and the wave is turned back. Above it the field reverses faster than the electrons can respond, and the wave passes through.
03 — The ionosphere
From electron density to a frequency floor
- A known figure for the maximum electron density is \(N_{\max}=10^{6}\ \text{electron/cm}^{3}\). Convert to SI first: \(1\ \text{cm}^{3}=10^{-6}\ \text{m}^{3}\), so \(N_{\max}=10^{12}\ \text{electron/m}^{3}\).
- Substitute: \(f_{\text{critical}}=9\sqrt{10^{12}}=9\times10^{6}\ \text{Hz}=9\ \text{MHz}\).
- That figure is for a wave arriving perpendicular to the layer. Arriving at an angle makes reflection easier, and raises the limit by a factor of about three to four.
- \(9\ \text{MHz}\times(3\ \text{to}\ 4)\approx30\ \text{MHz}\) — the highest frequency that can practically be reflected by the ionosphere.
03 — The ionosphere
The same layer, two completely different outcomes
03 — The ionosphere
Check yourself
1 · A 10 MHz uplink
An engineer proposes a \(10\ \text{MHz}\) uplink to a geostationary satellite. What is the decisive objection?
Part 2 — the second decision
Is getting through the ionosphere enough?
No. A frequency above \(30\ \text{MHz}\) will leave the atmosphere — and then spread its power over the whole sky. The spacecraft intercepts almost none of it.
04 — Directivity
We do not want to radiate in every direction
A geostationary satellite is a small target \(36{,}000\ \text{km}\) away. An antenna that radiates equally in all directions sends almost all of its power somewhere the spacecraft is not, and interferes with everyone else while doing it.
What we actually need is directivity: the ability to concentrate radiated energy into a narrow cone aimed at the spacecraft. That requirement, not the ionosphere, is what pushes the frequency much higher than \(30\ \text{MHz}\).
04 — Directivity
The same dish, at two different frequencies
04 — Directivity
Why shorter waves buy directivity
For an aperture antenna — a dish, a horn, an array — directivity scales with the aperture area measured in square wavelengths:
\[D\;\propto\;\frac{A}{\lambda^{2}},\qquad\quad \lambda=\frac{c}{f}\]
So for a fixed physical aperture \(A\), doubling the frequency halves \(\lambda\) and quadruples the directivity.
That is the whole argument. The dish on the roof is a fixed number of metres across, and we cannot keep making it bigger. The only remaining way to make it electrically larger is to shorten the wavelength — which means raising the frequency.
Antennas are a topic of their own later in the course; here we need only the scaling, not the full aperture theory.
04 — Directivity
Check yourself
2 · From 3 GHz to 12 GHz
A fixed \(1\ \text{m}\) dish is moved from \(3\ \text{GHz}\) to \(12\ \text{GHz}\). What happens to the wavelength, and what does the dish gain?
05 — Into the microwave region
The physics pushes us toward microwaves
Three independent requirements, all pointing the same way:
Get through the ionosphere
A floor of roughly \(30\ \text{MHz}\), and we want to be well clear of it because the threshold moves.
Get directivity
The aperture must be many wavelengths across, so \(\lambda\) must be small.
Keep the antenna buildable
The dish has to fit on a roof, a ship or a spacecraft, so we cannot buy electrical size with metres.
- all three constraints
- short wavelength
- GHz — the microwave region
A fourth, historical reason from the notes: by the end of the 1930s radio and television already occupied the MF, HF, VHF and UHF bands, so satellite systems in the 1950s had to be placed above them.
06 — Satellite bands
Where the satellite bands sit
Now the band chart means something. The window opens where the ionosphere stops reflecting, and closes where the atmosphere and the available hardware make the link impractical.
06 — Satellite bands
The bands by name, range and use
| Band | Frequency | Wavelength | Typical satellite services |
|---|---|---|---|
| L | \(1\)–\(2\ \text{GHz}\) | \(30\)–\(15\ \text{cm}\) | Mobile satellite services and satellite navigation; small, non-directive terminals. |
| S | \(2\)–\(4\ \text{GHz}\) | \(15\)–\(7.5\ \text{cm}\) | Mobile services, and telemetry, tracking and command. |
| C | \(4\)–\(8\ \text{GHz}\) | \(7.5\)–\(3.75\ \text{cm}\) | The classical fixed-satellite service band; robust in rain, but large antennas. |
| X | \(8\)–\(12\ \text{GHz}\) | \(3.75\)–\(2.5\ \text{cm}\) | Government and military fixed services; radar and imaging. |
| Ku | \(12\)–\(18\ \text{GHz}\) | \(2.5\)–\(1.67\ \text{cm}\) | Broadcast satellite services and VSAT networks; small dishes, more rain fade. |
| Ka | \(26.6\)–\(40\ \text{GHz}\) | \(1.13\)–\(0.75\ \text{cm}\) | High-throughput links and very small terminals; strongly rain-affected. |
Ranges follow the IEEE column of Table 3 in the Rohde & Schwarz Satellite Technologies guide held in the course reference library, which also places K band between Ku and Ka, from \(18\ \text{GHz}\) up to the start of Ka. Band boundaries differ between the IEEE, ITU and NATO conventions, so always state which convention you are quoting.
06 — Satellite bands
So why not simply go as high as possible?
If every argument pointed upward there would be one band and no discussion. Several bands exist because the arguments disagree:
| Moving to a higher frequency | |
|---|---|
| Helps | Smaller antennas for the same beam; more absolute bandwidth available; less congested spectrum; further above the ionospheric limit and its variability. |
| Hurts | Stronger interaction with the lower atmosphere, rain in particular; tighter pointing tolerance, because the beam that was easier to form is also easier to miss; harder and more expensive hardware. |
| Constrains | What technology can actually deliver at acceptable power and cost, and what the international allocations permit for that service. |
Frequency selection is a system decision: propagation, antennas, bandwidth, hardware, interference and regulation together. No single band is "the" satellite frequency.
Part 3 — the consequence
We have entered microwave engineering
Choosing gigahertz was a propagation and antenna decision. It is also a hardware decision, and from here the problem changes character: at these frequencies the connection between the source and the antenna stops being a detail.
07 — Microwave engineering
Source → transmission line → antenna → channel
08 — Transmission lines
At microwave frequencies, a wire stops being a wire
In a low-frequency circuit a connecting wire is an ideal node: the same voltage appears at both ends at the same instant. That picture quietly assumes the wire is much shorter than a wavelength.
At \(12\ \text{GHz}\) the free-space wavelength is \(25\ \text{mm}\). A \(50\ \text{mm}\) connection is two whole wavelengths long. The consequences are immediate:
Voltage varies along the line
\(V\) and \(I\) are now functions of position as well as time, not single circuit values.
The structure guides a wave
Energy travels as a field between conductors, so the geometry of the structure is part of the circuit.
Loss and radiation matter
Conductor loss rises with frequency, and an open structure radiates the power we were trying to deliver.
So the connection becomes a designed component in its own right: a transmission line.
08 — Transmission lines
Three ways to guide the energy
08 — Transmission lines
What actually drives the choice
| Conductor loss | Rises with frequency and concentrates in a thin surface layer. The notes record that above about \(1\ \text{GHz}\) two-conductor lines become impractical for long runs because of it. |
|---|---|
| Dielectric loss | A coaxial line needs a dielectric to hold the inner conductor in place, and that dielectric absorbs. An air-filled guide has almost none. |
| Power handling | Hollow air-filled waveguides handle much higher power than coaxial cables or printed lines — which matters for a high-power transmit chain. |
| Dimensions and hardware | Waveguide is bulky and rigid, and its size is set by the operating frequency. Coax and printed lines are small, flexible and cheap to connect. |
This is why the notes conclude that waveguides are very important for satellite communications: in the high-power, low-loss parts of a microwave chain, the trade comes out in their favour.
Part 4 — the structure
Can a wave travel through a hollow metal pipe?
There is no second conductor, no signal wire, no return path. Just air, surrounded by metal. It is worth pausing on how unreasonable this sounds before finding out that it works.
09 — Inside a waveguide
Yes — but not just any field
09 — Inside a waveguide
The geometry, and the pattern it allows
09 — Inside a waveguide
Naming the modes: TEM, TE and TM
Modes are classified by what the field does along the direction of travel, taken here as \(z\):
| TEM | \(E_z=H_z=0\): both fields entirely transverse. This is the free-space plane wave, and it exists on two-conductor transmission lines — but not inside a hollow waveguide. |
|---|---|
| TE | \(E_z=0\): no electric field along the direction of travel. The dominant mode of a rectangular guide is \(TE_{10}\); of a circular guide, \(TE_{11}\). |
| TM | \(H_z=0\): no magnetic field along the direction of travel. These are higher-order modes and are usually not the ones we operate on. |
The two subscripts \(m\) and \(n\) count how the field varies across the cross-section — \(m\) along the broad dimension \(a\), \(n\) along the narrow dimension \(b\). \(TE_{10}\) therefore means one half-cycle of variation across \(a\) and none across \(b\): exactly the arch in the previous figure.
Part 5 — the rules underneath
Why can a wave propagate at all — in space, and inside a pipe?
Both answers come from the same four equations. We have needed them twice now: once for the wave that crosses the ionosphere, and once for the pattern that fits inside the guide.
10 — Maxwell's equations
The four rules the fields must obey
\[\nabla\times\mathbf H=\frac{\partial \mathbf D}{\partial t}\]
Ampère–Maxwell lawAn electric field that changes in time creates a magnetic field.
\[\nabla\cdot\mathbf D=\rho_v\]
Gauss's lawElectric charge is the source of the electric field.
\[\nabla\cdot\mathbf B=0\]
Gauss's law for magnetismThere is no magnetic charge: no isolated north or south pole exists.
Equations (3) and (4) say where fields come from, and neither contains time. Equations (1) and (2) couple the two fields together, and they are the pair that makes propagation possible. \(\mathbf D=\varepsilon\mathbf E\) and \(\mathbf B=\mu\mathbf H\) tie the fields to the medium. Topic 3, p. 8 →
10 — Maxwell's equations
The coupling is what makes a wave
10 — Maxwell's equations
Four coupled equations become one
- Assume a source-free, homogeneous, linear medium: \(\rho_v=0\) so \(\nabla\cdot\mathbf E=0\), with \(\varepsilon\) and \(\mu\) constant. Free space satisfies all of this exactly; so does the air inside a waveguide.
- Equations (1) and (2) each contain both fields. Eliminate one by taking the curl of (1): \(\nabla\times(\nabla\times\mathbf E)=-\dfrac{\partial}{\partial t}(\nabla\times\mathbf B)\).
- Substitute \(\mathbf B=\mu\mathbf H\) and then (2), \(\nabla\times\mathbf H=\varepsilon\,\partial\mathbf E/\partial t\). The right-hand side becomes \(-\mu\varepsilon\,\partial^{2}\mathbf E/\partial t^{2}\), and \(\mathbf H\) has gone.
- On the left use \(\nabla\times(\nabla\times\mathbf E)=\nabla(\nabla\cdot\mathbf E)-\nabla^{2}\mathbf E\); the first term vanishes by step 1, leaving \(-\nabla^{2}\mathbf E\).
Starting from the curl of (2) instead gives the same equation in \(\mathbf H\). The two fields obey the same law, which is another way of saying they are two aspects of one wave.
10 — Maxwell's equations
The result, and the speed it predicts
\[\nabla^{2}\mathbf E=\mu\varepsilon\,\frac{\partial^{2}\mathbf E}{\partial t^{2}}\]
This is the wave equation, the same form a vibrating string obeys: \(\nabla^{2}f=\frac{1}{v^{2}}\partial^{2}f/\partial t^{2}\). Comparing the two fixes the propagation velocity:
\[v=\frac{1}{\sqrt{\mu\varepsilon}},\qquad\quad c=\frac{1}{\sqrt{\mu_0\varepsilon_0}}\approx3\times10^{8}\ \text{m/s}\]
Two constants measured with batteries and magnets predict the speed of light.
10 — Maxwell's equations
The free-space solution, and its parameters
Working one frequency at a time, every quantity varies as \(e^{j\omega t}\) and the wave equation becomes the form the notes write as equation (5), \(\nabla^{2}\mathbf E+k^{2}\mathbf E=0\) with \(k^{2}=\omega^{2}\mu_0\varepsilon_0\). For a wave travelling along \(z\) with its electric field along \(x\), one solution is equation (6):
\[\mathbf E=\hat a_x\,E_0\,e^{j(\omega t-\beta z)},\qquad \beta=k=\omega\sqrt{\mu_0\varepsilon_0}=\frac{\omega}{c}\]
\(\omega=2\pi f\) is radians of phase per second at a fixed position — the cycle counted in time, fixed by the transmitter.
\(\beta=2\pi/\lambda\) is radians of phase per metre at a fixed instant — the cycle counted in space. The notes call it the phase constant.
A crest keeps \(\omega t-\beta z\) constant, so it advances at \(v=\omega/\beta=f\lambda\). In free space that is \(c\), and \(\lambda=c/f\) — the relation the whole antenna argument rested on.
10 — Maxwell's equations
Every symbol is a feature of the picture
Part 6 — the design rules
Maxwell plus metal walls gives modes and cutoff
The same equations, solved in the same way, but now with a boundary condition: the electric field tangential to a conducting wall must vanish. Everything practical about waveguide design comes out of that one extra requirement.
11 — Modes and cutoff
The same wave equation, with walls
Inside the hollow guide the region is still source free, so the field still satisfies \(\nabla^{2}\mathbf E+k^{2}\mathbf E=0\). What is new is that the solution must also vanish on the conducting walls. The notes give the resulting field as equation (8):
\[\mathbf E=\hat a_y\,E_0\sin\!\left(\frac{\pi x}{a}\right)e^{j(\omega t-\beta z)}\]
The \(\sin(\pi x/a)\) factor is the arch in the figure: it is zero at \(x=0\) and at \(x=a\), which is exactly what the side walls demand. Substituting back gives equations (9) and (10):
\[k^{2}=\left(\frac{\pi}{a}\right)^{2}+\beta^{2}=\omega^{2}\mu_0\varepsilon_0\qquad\Longrightarrow\qquad \beta=\sqrt{\omega^{2}\mu_0\varepsilon_0-\left(\frac{\pi}{a}\right)^{2}}\]
11 — Modes and cutoff
What that square root is telling us
11 — Modes and cutoff
Every mode has its own cutoff frequency
Every allowed field pattern has its own cutoff. For an air-filled rectangular guide the notes give equation (14):
\[f_{c,mn}=\frac{c}{2}\sqrt{\left(\frac{m}{a}\right)^{2}+\left(\frac{n}{b}\right)^{2}}\]
\[\text{a mode propagates only if}\quad f>f_{c}\]
Below its cutoff a mode does not travel down the guide at all: the field decays away from the source instead.
Here \(m=0,1,2,\dots\) and \(n=0,1,2,\dots\), not both zero. Topic 3, p. 12 →
11 — Modes and cutoff
Which mode appears first?
The mode with the lowest cutoff frequency is called the dominant mode — it is the first pattern an electromagnetic wave can take inside the guide. Put \(m=1,\ n=0\) into the general formula, with \(a>b\):
\[f_{c,10}=\frac{c}{2a}\]
It depends on the broad dimension \(a\) alone. This is a design equation: choose \(a\), and you have chosen the lowest frequency the guide will carry.
That is the mode whose field we drew earlier: one half-cycle across \(a\), no variation across \(b\), electric field everywhere vertical. Its cutoff wavelength is \(2a\) — the guide must be at least half a wavelength wide before anything can propagate at all.
11 — Modes and cutoff
Check yourself
3 · A quick cutoff
An air-filled rectangular waveguide has \(a=20\ \text{mm}\). Roughly what is its dominant-mode cutoff, and what happens to a \(5\ \text{GHz}\) signal fed into it?
12 — Design: a 12 GHz feed
We need to carry a 12 GHz satellite signal
A ground terminal operates at \(f=12\ \text{GHz}\), a frequency representative of Ku-band satellite downlinks. We have to get that signal from the equipment to the feed of the dish. What waveguide can we use?
\[\lambda_0=\frac{c}{f}=\frac{3\times10^{8}}{12\times10^{9}}=25\ \text{mm}\]
Four questions decide the answer, and we will work through them in order:
Is \(f\) above the dominant-mode cutoff of the guide we are considering?
Is \(f\) below the cutoff of the next mode up?
Is \(f\) comfortably clear of both edges, rather than sitting on one?
12 — Design: a 12 GHz feed
Question 1 — will 12 GHz propagate?
- Take a standard rectangular guide with \(a=22.86\ \text{mm}\) and \(b=10.16\ \text{mm}\). Note \(b<a/2\), which is the industry convention the notes describe.
- Dominant-mode cutoff: \(f_{c,10}=\dfrac{c}{2a}=\dfrac{3\times10^{8}}{2\times0.02286}=6.56\ \text{GHz}\).
- Compare with the operating frequency: \(12\ \text{GHz}>6.56\ \text{GHz}\).
\[f=12\ \text{GHz}\;>\;f_{c,10}=6.56\ \text{GHz}\quad\checkmark\]
The \(TE_{10}\) mode propagates. Question 1 is answered — but on its own, that is not enough to call the design finished.
12 — Design: a 12 GHz feed
Question 2 — what else can propagate?
If the guide is large enough, higher-order modes propagate too. Each one travels at its own rate, so a signal split between modes arrives smeared — what the notes call intermodal distortion. We want exactly one mode in the operating band.
- Next mode along the broad wall: \(f_{c,20}=\dfrac{c}{a}=13.12\ \text{GHz}\).
- First mode across the narrow wall: \(f_{c,01}=\dfrac{c}{2b}=14.76\ \text{GHz}\).
- So at \(12\ \text{GHz}\) only \(TE_{10}\) propagates — but the next mode begins just \(1.1\ \text{GHz}\) higher.
- The notes give the recommended band as \(1.2f_{c,10}<f_{op}<0.8f_{c,20}\), which here is \(7.87\)–\(10.50\ \text{GHz}\).
12 — Design: a 12 GHz feed
Question 3 — so choose a smaller guide
Shrinking \(a\) raises every cutoff frequency together, sliding the whole single-mode window upward until it contains \(12\ \text{GHz}\):
| a = 22.86 mm | a = 19.05 mm | |
|---|---|---|
| \(f_{c,10}=c/2a\) | 6.56 GHz | 7.87 GHz |
| \(f_{c,20}=c/a\) | 13.12 GHz | 15.75 GHz |
| Recommended band | 7.87 – 10.50 GHz | 9.45 – 12.60 GHz |
| Is 12 GHz inside? | No — above the window | Yes |
The second guide, with \(a=19.05\ \text{mm}\) and \(b=9.525\ \text{mm}\), carries \(12\ \text{GHz}\) in a single mode with margin at both edges. That is the answer to the design question.
12 — Design: a 12 GHz feed
The wave is longer inside the guide
Rewriting the notes' \(\beta=\sqrt{\omega^{2}\mu_0\varepsilon_0-(\pi/a)^{2}}\) in terms of the cutoff frequency gives the two quantities a microwave engineer works with:
\[\beta=k_0\sqrt{1-\left(\frac{f_c}{f}\right)^{2}},\qquad \lambda_g=\frac{2\pi}{\beta}=\frac{\lambda_0}{\sqrt{1-\left(f_c/f\right)^{2}}}\]
The square root is always less than one, so \(\lambda_g>\lambda_0\): the wavelength along the guide is always longer than in free space, and it stretches without limit as \(f\) approaches cutoff. For our chosen guide at \(12\ \text{GHz}\), \(\lambda_g=33.1\ \text{mm}\) against \(\lambda_0=25\ \text{mm}\), and \(\beta=2\pi/\lambda_g=190\ \text{rad/m}\).
\(\beta\) is the phase change per metre along the guide — the quantity that sets the electrical length of every section of the feed, and the reason a waveguide run cannot be measured in free-space wavelengths.
12 — Design: a 12 GHz feed
Check yourself
4 · Two things to avoid
Why should a waveguide feed be operated neither too close to cutoff nor in an arbitrarily large guide?
13 — Putting it back together
From a spacecraft in orbit to a waveguide dimension
Every step in this lecture was forced by the one before it. Read the chain backwards and the waveguide on the bench is explained by an orbit.
- pick a frequency
- cross the ionosphere
- get directivity
- so: microwaves
- feed the antenna
- so: waveguide
- modes & cutoff
| The ionosphere | sets a floor: reflection below roughly \(30\ \text{MHz}\), so an Earth–space link must sit far above it. |
|---|---|
| Directivity | raises the floor much further: \(D\propto A/\lambda^{2}\), so a buildable antenna needs a small \(\lambda\), which means GHz. |
| Microwave hardware | then makes the feed a design problem, and low loss with high power favours waveguide. |
| Maxwell plus walls | gives modes, a cutoff for each, and \(f_{c,10}=c/2a\) — which turns the operating frequency into a dimension in millimetres. |
13 — Putting it back together
Carry these forward
Frequency floor
\(f_{\text{critical}}=9\sqrt{N_{\max}}\); with obliquity the practical reflection limit is about \(30\ \text{MHz}\).
Why microwaves
\(\lambda=c/f\) and \(D\propto A/\lambda^{2}\): a fixed dish is electrically larger, and far more directive, at higher \(f\).
Why waveguide
Above about \(1\ \text{GHz}\), low loss and high power handling favour a hollow single-conductor guide.
Cutoff
\(f_{c,mn}=\frac{c}{2}\sqrt{(m/a)^{2}+(n/b)^{2}}\), dominant \(f_{c,10}=c/2a\), and a mode travels only if \(f>f_c\).
EE 499 · Week 5
Week 5 complete
- frequency
- ionosphere
- directivity
- microwaves
- transmission lines
- waveguide modes
We started with a spacecraft in a known orbit and finished holding a waveguide whose width is set by the frequency we chose in the first ten minutes. Nothing in that chain was arbitrary: the ionosphere set a floor, directivity pushed us to microwaves, and microwaves made the feed a design problem with its own equations.
Next: Week 6 turns the waveguide dimension into hardware — circular guides, standard sizes including WR-340, and coaxial cable. Week 7 then looks inside the guide at the field itself, and Week 8 asks what happens at the joints, loads and components along the path.