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EE 499 · Satellite Technologies

Week 8

The Wave Reaches a Joint. How Much Gets Through?

Week 7 followed the field along a guide. Real paths are not uniform: they have cables, connectors, transitions, loads and components. This week we learn how engineers say how well RF power passes through them — and the instrument that measures it.

By the end: you should be able to explain the three things that can happen to RF power at a component (transmitted, reflected, dissipated), say what a \(50\ \Omega\) cable is and why the number matters, calculate the reflection coefficient, return loss and VSWR of a mismatched load, interpret \(S_{11}\) and \(S_{21}\) for a real component, and use reciprocity and energy conservation to check a measurement.

Primary source: the original lecture notes, Topic 3 — section 1.3.4 on S-parameters and the loss remarks on p. 7. The impedance and reflection slides go beyond the notes and are marked.

01 — Where we are

We understand the field. What about the joints?

We now understand how fields propagate through a transmission structure: above cutoff the pattern travels, at cutoff it stands still, and below cutoff it dies away.

What happens when RF power encounters real interfaces, loads and components?

A real path is a source, a cable, a connector, a transition, a length of guide, perhaps a filter or an amplifier, and a load. At every change of structure some of the power can go somewhere other than forward.

  1. where the power goes
  2. characteristic impedance
  3. reflection
  4. S-parameters
  5. the VNA
  6. connectors
  7. the RF path

Part 1 — the three fates

When power meets a component, where does it go?

Some is transmitted, some is reflected, and some may be dissipated. Keeping those three apart is the whole discipline of this week.

02 — Where the power goes

Loss, reflection and transmission are different things

The notes argue that above about \(1\ \text{GHz}\) two-conductor lines suffer high conductor loss, while hollow air-filled guides have low loss and much higher power handling — “therefore waveguides are very important for satellite communications”. But power reaching a component can meet three different fates.

TransmittedThe power that gets through to the far side. This is what we want.
ReflectedTurned back towards the source. Not destroyed: the signature of an interface where the structure changes.
DissipatedConverted to heat in conductors and dielectrics, or radiated. Genuine loss.
\[P_{\text{incident}}=P_{\text{transmitted}}+P_{\text{reflected}}+P_{\text{dissipated}}\]

02 — Where the power goes

Four loss words, and the decibel

TermWhat it describesWhere it comes from
AttenuationPower dissipated per unit length of a line or guide.conductor and dielectric loss
Insertion lossHow much less power gets through a component than went in; for a matched part, \(-S_{21}\) in dB.reflection and dissipation together
Return lossHow far below the incident power the reflection is. A large return loss is a good match.an impedance change
Evanescent decayThe fall of the field with distance below cutoff. No power is absorbed.the mode cannot propagate (Week 7)

Powers span many orders of magnitude and cascaded parts multiply, so we use decibels: a power ratio is \(10\log_{10}(P_2/P_1)\), and an amplitude ratio is \(20\log_{10}\).

\(0\ \text{dB}\)same power\(-3\ \text{dB}\)about half
\(-10\ \text{dB}\)one tenth\(-20\ \text{dB}\)one hundredth

From the original lecture notes · Topic 3, p. 7 →

02 — Where the power goes

Check yourself

1 · Where did the power go?

A passive component is measured as \(S_{11}=-20\ \text{dB}\) and \(S_{21}=-1\ \text{dB}\). Roughly what fraction of the incident power is dissipated?

Part 2 — the number on the cable (beyond the notes)

What does a “50 Ω cable” actually mean?

Interfaces turn waves back when the structure changes. For a two-conductor line the thing that changes is a number: the characteristic impedance. We need it to say how much turns back, and because \(S_{11}\) is exactly that reflection.

03 — Characteristic impedance

What does a “50 Ω cable” actually mean?

Put an ohmmeter across a \(50\ \Omega\) coaxial cable and it will not read \(50\ \Omega\): a fraction of an ohm along the inner conductor, an open circuit between inner and shield. The number on the jacket is not a measurable resistance.

It is a property of the wave the cable carries. Send a signal into a long cable and, before any reflection has had time to return, the line draws a definite current for the voltage applied. The ratio of those two, for the wave travelling forwards, is fixed by the geometry:

\[Z_0=\frac{V^{+}}{I^{+}}\]

\(V^{+}\) and \(I^{+}\) are the voltage and current of the forward-travelling wave alone. \(Z_0\) is real for a low-loss line, and it does not depend on how long the cable is.

The practical meaning follows directly: a source expecting \(50\ \Omega\) sees a matched load if the line has \(Z_0=50\ \Omega\) and is terminated in \(50\ \Omega\). Everything about reflection and matching below is measured against that reference.

03 — Characteristic impedance

From geometry to a number

For a coaxial line with a homogeneous dielectric of relative permittivity \(\varepsilon_r\), inner radius \(a\) and outer inner-radius \(b\):

\[Z_0=\frac{60}{\sqrt{\varepsilon_r}}\ln\!\left(\frac{b}{a}\right)=\frac{138}{\sqrt{\varepsilon_r}}\log_{10}\!\left(\frac{b}{a}\right)\]

The two forms are the same expression; \(60\ln x=138\log_{10}x\) to three figures.

Only the ratio \(b/a\) matters, so a thick and a thin cable can share an impedance; the dependence is logarithmic, so large changes in geometry make small changes in \(Z_0\); and a denser dielectric lowers \(Z_0\).

Air, \(\varepsilon_r=1\)\(Z_0=50\ \Omega\) needs \(b/a=2.30\); \(Z_0=75\ \Omega\) needs \(b/a=3.49\).
PTFE, \(\varepsilon_r\approx2.1\)\(Z_0=50\ \Omega\) needs \(b/a=3.35\) — the ratio found inside most semi-rigid cable.

Standard transmission-line theory, not one of the notes' numbered equations — included because \(Z_0\) is the reference every S-parameter is defined against.

03 — Characteristic impedance

Why 50 Ω, of all the numbers?

For an air-filled coaxial line the two things an engineer wants do not peak at the same place:

Lowest attenuationoccurs near \(b/a=3.6\), which corresponds to about \(77\ \Omega\).
Highest power handlingoccurs near \(b/a=1.65\), which corresponds to about \(30\ \Omega\).
The compromise\(50\ \Omega\) sits between the two — not optimal for either, acceptable for both.

That compromise was adopted in the 1930s and 1940s for high-power radio and radar hardware, and then became self-reinforcing. Standardisation, not physics, is what fixed the number.

The takeaway. Once a system is designed around a reference impedance, its connectors, cables, instruments, loads and S-parameter measurements all refer to that same value. Consistency, not optimality, is what \(50\ \Omega\) buys.

03 — Characteristic impedance

Check yourself

2 · Cutting the cable

A \(10\ \text{m}\) length of \(50\ \Omega\) cable is cut down to \(2\ \text{m}\). What happens to its characteristic impedance?

Part 3 — what the load does (beyond the notes)

What happens if the load is not 50 Ω?

A transmission line carries a wave towards a load. If the load does not absorb everything the line offers, the remainder has nowhere to go but back. That reflected wave is the single most common fault in a microwave path — and the easiest one to measure.

Beyond the notes. The reflection coefficient is the one-port version of \(S_{11}\), which the notes define next.

04 — Matching and reflection

The load decides how much comes back

Three loads on a fifty ohm line and the reflection each produces Three transmission lines drawn one above the other. The first ends in a matched fifty ohm load with a forward arrow and no returning arrow. The second ends in a mismatched seventy-five ohm load with a forward arrow and a small returning arrow. The third ends in a short circuit with a forward arrow and a full-size returning arrow. 50 Ω 75 Ω short 50 Ω line 50 Ω line 50 Ω line matched partly reflected fully reflected nothing returns one fifth returns all of it returns incident reflected
The line is identical in all three cases. Only the termination differs — and the termination alone decides how much of the incident wave the source gets back.

04 — Matching and reflection

Putting a number on the reflection

The reflected wave is proportional to the incident one, and the constant of proportionality at the load is the reflection coefficient:

\[\Gamma=\frac{Z_L-Z_0}{Z_L+Z_0}\]

\(\Gamma\) is generally complex: its magnitude says how much comes back, its phase says with what delay. Only the magnitude appears in the quantities below.

\(Z_L=Z_0\)\(\Gamma=0\) — matched. The load absorbs everything; the source sees the line as if it went on forever.
\(Z_L=0\) (short)\(\Gamma=-1\) — everything returns, inverted.
\(Z_L\to\infty\) (open)\(\Gamma=+1\) — everything returns, in phase.
Anything else\(0<|\Gamma|<1\) — part is absorbed, part comes back.

\(|\Gamma|=1\) is the ideal limiting case of a lossless, purely reactive termination; a real short or open measures slightly under one. Keep \(\Gamma\) in view — it returns shortly as \(S_{11}\).

04 — Matching and reflection

Two other ways to say the same thing

\(|\Gamma|\) is rarely quoted directly. Datasheets use one of two derived numbers, both carrying exactly the same information:

\[\mathrm{RL}=-20\log_{10}|\Gamma|\ \ \text{(dB)},\qquad \mathrm{VSWR}=\frac{1+|\Gamma|}{1-|\Gamma|}\]
\(|\Gamma|\)Return lossVSWRPower reflectedVerdict
1.000 dB∞100 %total reflection
0.3210 dB1.9210 %the notes' “low” threshold
0.1020 dB1.221 %good
0.0330 dB1.070.1 %excellent

Return loss grows as the match improves; VSWR approaches \(1\). At the input of a terminated two-port the same reflection is \(S_{11}\), so \(\mathrm{RL}=-20\log_{10}|S_{11}|\).

04 — Matching and reflection

A 75 Ω antenna on a 50 Ω feed

An installer terminates a \(50\ \Omega\) feed in a \(75\ \Omega\) load. How bad is that?

  1. Reflection coefficient: \(\Gamma=\dfrac{75-50}{75+50}=\dfrac{25}{125}=0.20\).
  2. Return loss: \(\mathrm{RL}=-20\log_{10}(0.20)=13.98\approx14\ \text{dB}\).
  3. Standing-wave ratio: \(\mathrm{VSWR}=\dfrac{1+0.20}{1-0.20}=\dfrac{1.20}{0.80}=1.50\).
  4. Reflected power fraction: \(|\Gamma|^{2}=0.04\), so \(4\%\) of the incident power returns and \(96\%\) reaches the load.
\[\Gamma=0.20,\quad \mathrm{RL}=14\ \text{dB},\quad \mathrm{VSWR}=1.5\]

Tolerable for a receive path, and a common accident when video-grade \(75\ \Omega\) hardware meets an RF \(50\ \Omega\) system. On a high-power transmit path the \(4\%\) coming back is heat arriving somewhere it was not designed for.

Part 4 — the language of microwave components (from the notes)

At 12 GHz, how do we characterise a component at all?

We now have a way to describe one reflection at one load. A real component has an input and an output, and we need a description that covers both, in both directions, at every frequency of interest.

From the notes — section 1.3.4: the notes call S-parameters the main tool used to analyse microwave devices.

05 — Microwave measurement

Why do we prefer travelling waves at microwave frequencies?

Voltage and current do not stop being meaningful at \(12\ \text{GHz}\). On a two-conductor line such as coax they remain perfectly well defined. What changes is how convenient they are to measure, and how naturally they describe a waveguide.

The structure is distributed

At \(12\ \text{GHz}\), \(\lambda=25\ \text{mm}\). \(V\) and \(I\) vary along the component, so they are functions of position rather than one pair of terminal values.

Open and short are hard to realise

\(Z\), \(Y\) and \(ABCD\) parameters need ideal open- and short-circuit terminations. Up here an open radiates and a short has inductance, so those conditions are both difficult and undesirable.

A waveguide has no terminal pair

With a single conductor there is no potential difference to define, so a terminal \(V\)–\(I\) description is far less natural there than on a two-conductor line.

Incident and reflected travelling waves, by contrast, exist in both structures and can be separated directly by an instrument. That is why they became the practical characterisation language — the notes put it as: in waveguides we cannot define potential, as the structure has only one conductor. Instead, we define waves.

From the original lecture notes · Topic 3, p. 17 →

06 — S-parameters

Four waves, before any matrix

A two-port network with its incident and reflected waves A central box labelled two-port network. At port one on the left, an arrow labelled a one points into the box and an arrow labelled b one points away from it. At port two on the right, an arrow labelled a two points into the box and an arrow labelled b two points away from it. two-port network cable, guide, filter, amplifier, transition… a1 b1 a2 b2 incident on port 1 leaving port 1 incident on port 2 leaving port 2 PORT 1 PORT 2
The notes define \(a_i\) and \(b_i\) as the square roots of the incident and reflected wave powers, so \(|a|^{2}\) and \(|b|^{2}\) are powers and every ratio built from them is dimensionless.

From the original lecture notes · Topic 3, p. 17 →

06 — S-parameters

One definition, four entries

A scattering parameter is the ratio of a wave leaving the network to one sent into it, with nothing arriving at any other port:

\[S_{ij}=\left.\frac{b_i}{a_j}\right|_{a_k=0,\;k\neq j}\]

In practice: excite one port, terminate the others in \(Z_0\) — \(50\ \Omega\) here. A realisable measurement, unlike an ideal open or short. Topic 3, pp. 17–18 →

The four combinations give the notes' equation (17): what leaves port 1 is whatever bounced off it, plus whatever came back from port 2.

\[\begin{bmatrix}b_1\\b_2\end{bmatrix}=\begin{bmatrix}S_{11}&S_{12}\\S_{21}&S_{22}\end{bmatrix}\begin{bmatrix}a_1\\a_2\end{bmatrix}\]

06 — S-parameters

The four entries, in plain language

\(S_{11}=\left.\frac{b_1}{a_1}\right|_{a_2=0}\)Input reflection. How much of what we sent into port 1 comes straight back? With port 2 terminated in \(Z_0\) this is the reflection coefficient at the input: \(S_{11}=\Gamma_{in}\).
\(S_{21}=\left.\frac{b_2}{a_1}\right|_{a_2=0}\)Forward transmission. How much gets from port 1 through to port 2? Loss for a passive component, gain for an amplifier.
\(S_{12}=\left.\frac{b_1}{a_2}\right|_{a_1=0}\)Reverse transmission. How much travels backwards, from port 2 to port 1? Ideally small for an amplifier or isolator; equal to \(S_{21}\) for anything passive.
\(S_{22}=\left.\frac{b_2}{a_2}\right|_{a_1=0}\)Output reflection. How well is port 2 matched, seen from the other side?
\[S_{11}=\Gamma_{in}\quad\text{(port 2 terminated in }Z_0\text{)}\]

Reflection coefficient and S-parameters are not separate ideas: drive port 1 and terminate port 2, and \(S_{11}\) is exactly the \(\Gamma\) of the previous section. Hence \(\mathrm{RL}=-20\log_{10}|S_{11}|\) for the input return loss.

06 — S-parameters

Why they are almost always quoted in decibels

The notes record that S-parameters are usually measured in dB. The magnitudes span many orders, and cascaded components then add instead of multiplying:

\[S_{ij,\mathrm{dB}}=20\log_{10}|S_{ij}|\]
Value\(|S|\)Power fractionMeaning as \(S_{11}\)Meaning as \(S_{21}\)
\(0\ \text{dB}\)1.00100 %everything reflectedeverything through, lossless
\(-1\ \text{dB}\)0.8979 %a bad match1 dB insertion loss
\(-10\ \text{dB}\)0.3210 %the notes' “low” valuea 10 dB attenuator
\(-20\ \text{dB}\)0.101 %a good matchheavy attenuation
\(+15\ \text{dB}\)5.623160 %not physical for a passive partan amplifier with 15 dB gain

\(S_{11}=-20\ \text{dB}\) and “return loss \(20\ \text{dB}\)” are the same statement with opposite sign conventions: \(\mathrm{RL}=-S_{11,\mathrm{dB}}\). The notes' remark that a value of \(-10\ \text{dB}\) is considered low, however this depends on the application refers to the reflection.

07 — Components and the VNA

What the numbers look like for real hardware

ComponentExpected \(S_{11}\)Expected \(S_{21}\)What you are checking
Coaxial cable, 1 mbelow \(-20\ \text{dB}\)\(-0.3\) to \(-1\ \text{dB}\)that the loss is as specified and the connectors are sound
Waveguide sectionbelow \(-25\ \text{dB}\) in bandclose to \(0\ \text{dB}\)the notes' ideal is exactly this: no reflection, full transmission
Attenuator, 10 dBbelow \(-20\ \text{dB}\)\(-10\ \text{dB}\), flatthat the attenuation is right and both ports stay matched
Filterlow in band, near \(0\ \text{dB}\) out of bandflat in band, falling steeply outsidepassband position, insertion loss, rejection
Amplifierbelow \(-10\ \text{dB}\)positive, e.g. \(+20\ \text{dB}\)gain, input match, and that \(S_{12}\) stays small
Coax-to-waveguide transitionbelow \(-20\ \text{dB}\) in band\(-0.2\ \text{dB}\) or betterthat the mode conversion is clean across the band

07 — Components and the VNA

The ideal section, and the rules every S-matrix obeys

The notes work the ideal case explicitly: a lossless length \(\ell\) of waveguide reflects nothing and transmits everything, delayed by its electrical length.

\[S=\begin{bmatrix}0&1e^{-j\beta \ell}\\1e^{j\beta \ell}&0\end{bmatrix},\qquad S_{\mathrm{dB}}=\begin{bmatrix}-\infty&0\\0&-\infty\end{bmatrix}\]

\(\beta\) is the same phase constant Weeks 5 and 7 obtained from the dispersion relation, so the guide's own physics sits inside its S-matrix.

Reciprocity\(S_{ij}=S_{ji}\): the matrix is symmetric. True for every passive device, including the waveguide — so a measured \(S_{12}\ne S_{21}\) on a passive part means a measurement problem, not a discovery.
Energy conservationFor a lossless network, \(|S_{21}|^{2}+|S_{11}|^{2}=1\) and \(|S_{12}|^{2}+|S_{22}|^{2}=1\). Whatever is not reflected must be transmitted.

The second rule is the best diagnostic here: add a component's reflected and transmitted power fractions, and the shortfall is what was dissipated.

From the original lecture notes · Topic 3, pp. 18–19, equations (18) and (19) →

07 — Components and the VNA

The instrument that measures all four

A vector network analyzer connected to a device under test An instrument with a trace display and two ports on its right edge. A cable from port one runs to the left side of a device under test; a cable from port two loops beneath the bench and up to the right side of the same device. Notes beside the cables list which scattering parameters each drive direction produces. vector network analyzer Port 1Port 2 device under test cable, guide, filter, amplifier drive port 1, terminate port 2 gives S11 and S21 drive port 2, terminate port 1 gives S22 and S12 magnitude AND phase, swept across the band the reference plane is set by calibration — the numbers describe whatever lies beyond it
A VNA sweeps a known signal across the band and separates what returns from what passes through, recording both magnitude and phase. The notes add the practical footnote that VNAs are usually much more expensive than oscilloscopes and spectrum analyzers.

Calibration — SOLT, TRL and their relatives — is what moves the reference plane from the instrument to the ends of the cables. It is a topic of its own, not covered here. Topic 3, p. 19, note (iii) →

07 — Components and the VNA

Check yourself

3 · A suspicious measurement

A passive waveguide section is measured as \(S_{11}=-25\ \text{dB}\) and \(S_{21}=-0.1\ \text{dB}\), but \(S_{12}=-6\ \text{dB}\). What should you conclude?

Part 5 — the joints in the path

How do we physically connect microwave hardware?

Every component in this lecture has to be joined to the next one. At low frequency that is a mechanical question. At \(12\ \text{GHz}\) it is an electromagnetic one, because the joint is a length of transmission line in its own right.

08 — RF connectors

The connectors you will meet on a microwave bench

ConnectorCouplingInterfaceTypical upper frequencyWhere it turns up
BNCbayonet, quarter turn50 Ω (a 75 Ω version exists)a few GHzlab instruments, IF and baseband, trigger lines
N-typethreaded, robust50 Ω (75 Ω version exists)around 11–18 GHzantenna feeds, outdoor runs, higher power
SMAthreaded, small50 Ω, PTFE-supportedaround 18–26 GHzthe general-purpose microwave connector
3.5 mmthreaded, mates with SMA50 Ω, airaround 34 GHzprecision measurement, calibration kits
2.92 mm (K)threaded, mates SMA and 3.5 mm50 Ω, airaround 40 GHzKa-band work and VNA test ports
2.4 mmthreaded, no SMA mating50 Ω, airaround 50 GHzmillimetre-wave measurement
Waveguide flangebolted, keyed to the guidenot an impedance — a mode interfacethe guide's own bandeverything after the transition

Upper frequencies are indicative: ratings differ between manufacturers and grades, and depend on the cable and the fitting. Read the datasheet, not this column.

08 — RF connectors

At microwave frequencies, a connector is part of the transmission line

At \(1\ \text{MHz}\) a connector joins two cables. At \(12\ \text{GHz}\) it is \(10\ \text{mm}\) of transmission line with its own geometry, dielectric and discontinuities, and the wave cannot tell it from the cable.

Geometry is electrical

The \(b/a\) ratio has to be held through the connector. Where it changes, \(Z_0\) changes, and every change of \(Z_0\) makes a reflection.

The upper limit is physical

A connector stops working cleanly when its internal dimensions become comparable to a wavelength and higher modes can appear inside it. That is why the precision families get physically smaller as the frequency rises.

Damage is an RF fault

A bent centre pin, a flattened outer, an over-torqued nut or swarf in the interface all change the geometry — so they show up as a degraded \(S_{11}\) long before anything looks broken.

Part 6 — putting it together

One complete RF path, end to end

Every idea in Weeks 5 to 8 appears somewhere in a real transmit chain. Here it is as one picture, with the relevant quantity written against each stage.

09 — An RF path

Source to feed, with the numbers that matter

A complete microwave transmit path from source to antenna feed Six stages in a row: a microwave source, an SMA connector, a length of fifty ohm coaxial cable, a coax-to-waveguide transition, a rectangular waveguide run, and a circular feed into an antenna. Beneath each stage a caption gives the quantity that characterises it. source transition coax to guide WR-75 SMA 50 Ω coax flange circular feed to the dish 12 GHz hold Z0 Z0 = 50 Ω S11, S21 fc,10 = 7.87 GHz TE(1,1) Week 5 geometry sets Z0 TEM, no cutoff TEM becomes TE(1,0) above cutoff into free space every interface is a possible mismatch, and so a possible Γ; the whole chain answers to one S11 at the input and one S21 end to end
Every idea from these weeks has a place on this line: \(Z_0\) in the cable, geometry holding that impedance through the connector, \(\Gamma\) wherever the impedance changes, \(S_{11}\) and \(S_{21}\) for any component in it, mode and cutoff inside the guide, and at the feed the point where the guided wave is launched into free space. The mode itself changes twice — TEM, then \(TE_{10}\), then \(TE_{11}\).

09 — An RF path

Carry these forward

Three fates

\(P_{\text{in}}=P_{\text{transmitted}}+P_{\text{reflected}}+P_{\text{dissipated}}\). Reflection is not loss, and neither is evanescent decay.

Characteristic impedance

\(Z_0=V^{+}/I^{+}\); for coax \(Z_0=\frac{60}{\sqrt{\varepsilon_r}}\ln(b/a)\). \(50\ \Omega\) is a compromise, not an optimum.

Reflection

\(\Gamma=\frac{Z_L-Z_0}{Z_L+Z_0}\), \(\mathrm{RL}=-20\log_{10}|\Gamma|\), \(\mathrm{VSWR}=\frac{1+|\Gamma|}{1-|\Gamma|}\).

S-parameters

\(S_{ij}=b_i/a_j\) with the far port terminated; \(S_{11}\) is what comes back, \(S_{21}\) what gets through.

Two rules

Passive means \(S_{ij}=S_{ji}\); lossless means \(|S_{11}|^{2}+|S_{21}|^{2}=1\). The shortfall is what was dissipated.

In decibels

\(S_{ij,\mathrm{dB}}=20\log_{10}|S_{ij}|\). A value of \(-10\ \text{dB}\) for reflection is low, but what counts as good depends on the application.

The VNA

Measures magnitude and phase of all four S-parameters; calibration sets the reference plane.

Connectors

A joint is a length of line. Geometry, alignment and damage all appear in \(S_{11}\) and \(S_{21}\).

After the lecture

Week 8 exercises — practice problems

The lecture ends here. What follows is a set of four 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.

10 — Exercises

Exercise 1 — coaxial characteristic impedance

Given. A coaxial cable with inner conductor radius \(a=0.50\ \text{mm}\), outer conductor inner radius \(b=1.65\ \text{mm}\), and a solid PTFE dielectric with \(\varepsilon_r=2.1\).

  1. Calculate \(Z_0\).
  2. The outer conductor is now moved out to \(b=2.50\ \text{mm}\), with \(a\) and the dielectric unchanged. Before calculating anything, predict whether \(Z_0\) will rise or fall, and say why.
  3. Now calculate the new \(Z_0\) and check your prediction.
  4. If instead the dielectric were replaced by air with the original dimensions, what would \(Z_0\) become?

Use \(Z_0=\dfrac{60}{\sqrt{\varepsilon_r}}\ln\!\left(\dfrac{b}{a}\right)\). Part 2 is not optional — committing to a prediction first is how you find out whether you understand the formula or are only evaluating it.

10 — Exercises

Exercise 1 — solution

  1. \(b/a=3.30\), \(\ln 3.30=1.194\), \(\sqrt{2.1}=1.449\). So \(Z_0=\dfrac{60}{1.449}\times1.194=41.40\times1.194=49.4\ \Omega\) — a \(50\ \Omega\) cable.
  2. Prediction: \(Z_0\) rises. Moving the outer conductor away increases \(b/a\), and \(Z_0\) increases with \(\ln(b/a)\). Physically, separating the conductors lowers the capacitance per unit length and raises the inductance, both of which raise \(\sqrt{L/C}\).
  3. \(b/a=5.00\), \(\ln 5.00=1.609\), so \(Z_0=41.40\times1.609=66.6\ \Omega\). The prediction holds — and note that a \(52\%\) increase in \(b\) produced only a \(35\%\) increase in \(Z_0\), because the dependence is logarithmic.
  4. With air, \(\varepsilon_r=1\) and \(Z_0=60\times1.194=71.6\ \Omega\). Removing the dielectric raises \(Z_0\) by the factor \(\sqrt{2.1}=1.449\).

Two independent handles on the same number: the geometry through \(\ln(b/a)\), and the dielectric through \(1/\sqrt{\varepsilon_r}\). A cable designer uses both — which is why a \(50\ \Omega\) PTFE cable and a \(50\ \Omega\) air line have completely different proportions.

10 — Exercises

Exercise 2 — a mismatched load

Given. A \(50\ \Omega\) coaxial feed is terminated in a load of \(Z_L=30\ \Omega\) at the operating frequency.

  1. Calculate the reflection coefficient \(\Gamma\), and comment on the sign.
  2. Calculate the return loss in dB.
  3. Calculate the VSWR.
  4. What fraction of the incident power reaches the load?
  5. Would you accept this match on a receive path? On a high-power transmit path? Justify both answers.

\(\Gamma=\dfrac{Z_L-Z_0}{Z_L+Z_0}\), \(\mathrm{RL}=-20\log_{10}|\Gamma|\), \(\mathrm{VSWR}=\dfrac{1+|\Gamma|}{1-|\Gamma|}\).

10 — Exercises

Exercise 2 — solution

  1. \(\Gamma=\dfrac{30-50}{30+50}=\dfrac{-20}{80}=-0.25\). The sign is negative because \(Z_L<Z_0\): the reflected wave returns inverted, as it would from a short. Only \(|\Gamma|=0.25\) enters the formulas below.
  2. \(\mathrm{RL}=-20\log_{10}(0.25)=12.0\ \text{dB}\).
  3. \(\mathrm{VSWR}=\dfrac{1.25}{0.75}=1.67\).
  4. \(|\Gamma|^{2}=0.0625\), so \(6.25\%\) is reflected and \(93.75\%\) reaches the load.

Part 5. On a receive path, yes: losing \(6\%\) of an already small signal costs about \(0.3\ \text{dB}\), which is usually negligible next to the other losses on the path. On a high-power transmit path, be careful: \(6.25\%\) of a \(200\ \text{W}\) output is \(12.5\ \text{W}\) travelling back towards the amplifier. That is a thermal and a stability problem for the amplifier, not just a lost fraction of a decibel, and it is why transmit chains carry isolators and are specified to much tighter VSWR than receive chains.

10 — Exercises

Exercise 3 — read a two-port result

Given. A passive two-port component is measured on a VNA at its operating frequency:

\[S_{11}=-18\ \text{dB},\qquad S_{21}=-0.8\ \text{dB}\]
  1. What are \(|S_{11}|\) and \(|S_{21}|\) as linear magnitudes?
  2. What percentage of the incident power is reflected, and what percentage reaches port 2?
  3. What VSWR does the input present?
  4. Is this component well matched? Is it low loss? Answer each separately.
  5. What would you expect \(S_{12}\) to be, and why?
  6. A second component from the same family measures \(S_{11}=-20\ \text{dB}\) and \(S_{21}=-1\ \text{dB}\). Describe, in one or two plain sentences and with no formulas, what an engineer learns from those two numbers.

10 — Exercises

Exercise 3 — solution

  1. \(|S_{11}|=10^{-18/20}=0.126\); \(|S_{21}|=10^{-0.8/20}=0.912\).
  2. Reflected: \(|S_{11}|^{2}=0.0158\), so \(1.6\%\). Transmitted: \(|S_{21}|^{2}=0.832\), so \(83.2\%\).
  3. \(\mathrm{VSWR}=\dfrac{1+0.126}{1-0.126}=1.29\).
  4. Well matched: yes — an \(18\ \text{dB}\) return loss means only \(1.6\%\) comes back. Low loss: yes — \(0.8\ \text{dB}\) insertion loss is modest for a passive microwave part. The two answers are independent, which is the point of asking them separately.
  5. \(S_{12}=S_{21}=-0.8\ \text{dB}\). The notes state that \(S_{ij}=S_{ji}\) for a reciprocal system, and that this holds for all passive devices.
  6. “Almost nothing bounces back off the input — about one part in a hundred — so the component is well matched to the \(50\ \Omega\) system around it. Of the power that goes in, roughly four fifths comes out the far side; the missing fifth is lost inside the part. It is a good, slightly lossy component.”

Check the energy: \(1.6\%+83.2\%=84.8\%\), so about \(15\%\) of the incident power is neither reflected nor transmitted. The notes' lossless condition \(|S_{11}|^{2}+|S_{21}|^{2}=1\) is therefore not satisfied — and it should not be, because this component is not lossless. That shortfall is the dissipation.

10 — Exercises

Exercise 4 — diagnose a microwave path

Situation. A Ku-band earth-station feed line is measured end to end at \(12\ \text{GHz}\). The run consists of an SMA connector, \(2\ \text{m}\) of coaxial cable, a coax-to-waveguide transition, a short rectangular waveguide section and the antenna feed. The VNA reports:

\[S_{11}=-7\ \text{dB},\qquad S_{21}=-3.5\ \text{dB}\]
  1. Is the input well matched? Support the answer with \(|\Gamma|\), VSWR and reflected power.
  2. Is there significant transmission loss? How much power actually reaches the feed?
  3. Account for the energy. How much is unaccounted for by reflection and transmission together, and where has it gone?
  4. Which physical parts would you inspect, in what order, and what would you measure to isolate the fault?

This is the exercise where the microwave theory becomes troubleshooting. Part 4 carries most of the marks.

10 — Exercises

Exercise 4 — solution

  1. No. \(|\Gamma|=10^{-7/20}=0.447\), \(\mathrm{VSWR}=\frac{1.447}{0.553}=2.6\), and \(|\Gamma|^{2}=20\%\) of the incident power returns. A feed line should be well below \(-15\ \text{dB}\); the notes call \(-10\ \text{dB}\) low, and this is worse than that.
  2. Yes. \(|S_{21}|^{2}=10^{-3.5/10}=0.447\), so only \(44.7\%\) of the power reaches the feed. About \(1.0\ \text{dB}\) of the \(3.5\ \text{dB}\) is explained by the mismatch alone; the remaining \(2.5\ \text{dB}\) is genuine dissipation or radiation.
  3. \(20\%+44.7\%=64.7\%\), so \(35\%\) of the incident power is neither reflected nor delivered. It is being dissipated in the cable and connectors, or radiated from a discontinuity.

Part 4 — what to inspect. \(S_{11}\) at the input reports the whole chain, so the first job is to localise. Disconnect at each interface and measure the sections separately, working from the source outwards. Order of suspicion: the SMA connector first, since it is the most-mated and most easily damaged part; then the coax-to-waveguide transition, which is the only mode conversion in the path and the usual home of a band-edge reflection; then the cable, checking that \(2\ \text{m}\) at \(12\ \text{GHz}\) really should lose only a fraction of a dB for its type; then the waveguide section and its flanges, for a misaligned or dirty joint; and finally the feed itself, which may simply be the wrong band or mispointed. A VNA in time domain, or a swept \(S_{11}\) whose ripple period reveals the distance to the discontinuity, will find it faster than disassembly.

EE 499 · Week 8

Week 8 complete

  1. three fates
  2. impedance
  3. reflection
  4. S-parameters
  5. VNA
  6. connectors

Week 7 asked what the field does inside a guide. Week 8 asked what happens when RF power meets real interfaces, loads and components — and the answer, for every piece of hardware down to the connector, turned out to be two measured numbers, \(S_{11}\) and \(S_{21}\).

Next: We can now characterize individual RF components and interfaces. Next we combine antenna gains, propagation losses and system losses to determine whether the complete satellite link closes.

Read the Week 8 article