EE 499 · 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.

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

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 (pp. 17–19) and the loss and power-handling remarks on p. 7. The impedance and reflection sections go beyond the notes and are marked as such. Presentation View — study slide by slide →

01 — Where we are

The wave reaches a joint. How much gets through?

Week 7 ended with a question. 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 not one long uniform guide. It is a source, a cable, a connector, a transition, a length of guide, perhaps a filter or an amplifier, and a load at the end. At every one of those places the structure changes, and at every change 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

Week 8 builds the language for that question and the instrument that answers it. It ends at a measurement: how effectively does RF power pass through a real component or interface, and how would we know?

02 — Where the power goes

Loss, reflection and transmission are three different things

The notes open their treatment of waveguide technology with an argument about loss. Above about \(1\ \text{GHz}\), two-conductor transmission lines become impractical because of their high conductor loss. A single-conductor hollow guide has relatively low loss by comparison, and hollow air-filled guides can handle much higher power than coaxial cables or printed lines. Therefore waveguides are very important for satellite communications.

That argument uses the word loss in one sense: power turned into heat in the metal and the dielectric. But power reaching a component can meet three different fates, and only one of them is a loss.

TransmittedThe power that gets through to the far side. This is what we want.
ReflectedThe power that is turned back towards the source. It is not destroyed: it still exists, travelling the other way. It is the signature of an interface where the structure changes.
DissipatedThe power converted to heat in conductors and dielectrics, or radiated away. This is genuine loss, and it is gone.
\[P_{\text{incident}}=P_{\text{transmitted}}+P_{\text{reflected}}+P_{\text{dissipated}}\]

For a lossless component the third term is zero, and whatever is not reflected must be transmitted. The notes state exactly this later as a rule for S-parameters.

The vocabulary of loss

Several loss words are in daily use, and they are easy to confuse. Fix them now, because the rest of the week depends on them.

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 well-matched component it is \(-S_{21}\) in decibels.reflection and dissipation together
Return lossHow far below the incident power the reflected power 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)

Decibels

Powers in microwave work span many orders of magnitude, and cascaded components multiply their effects, so engineers work in decibels, where multiplication becomes addition. A power ratio is \(10\log_{10}(P_2/P_1)\) decibels. A field or wave amplitude, whose square is a power, is \(20\log_{10}\) of its ratio.

Want practice with dB, dBm and 10 log versus 20 log? The auxiliary dB Gym is not part of any week.

\(0\ \text{dB}\)the same power: nothing lost
\(-3\ \text{dB}\)about half the power
\(-10\ \text{dB}\)one tenth of the power
\(-20\ \text{dB}\)one hundredth of the power

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

03 — Characteristic impedance

What does a “50 Ω cable” actually mean?

Beyond the notes. The notes use two-conductor lines without defining a characteristic impedance; this section supplies it, because \(S_{11}\) is a reflection measured against it.

Put an ohmmeter across a \(50\ \Omega\) coaxial cable and it will not read \(50\ \Omega\). Along the inner conductor it reads a fraction of an ohm; between inner and shield it reads an open circuit. So the number on the jacket is not a resistance you can measure with a meter.

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 it drives has \(Z_0=50\ \Omega\) and is terminated in \(50\ \Omega\). Everything about reflection, matching and \(S_{11}\) below is measured against that reference.

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.

Three things are worth reading straight off it. Only the ratio \(b/a\) matters, so a thick cable and a thin cable can share an impedance. The dependence is logarithmic, so large changes in geometry make modest changes in \(Z_0\). And a denser dielectric lowers \(Z_0\), because it raises the capacitance per unit length.

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.

This expression is standard transmission-line theory rather than one of the notes' numbered equations; it is included because \(Z_0\) is the reference impedance every S-parameter below is defined against.

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 it became self-reinforcing: once connectors, attenuators, loads, amplifiers and instruments were all built to the same reference, the cost of departing from it outweighed any remaining advantage. Standardisation, not physics, is what fixed the number.

The takeaway is not that \(50\ \Omega\) is optimal — it is not optimal in any single respect. It is that once a microwave system is designed around a reference impedance, its connectors, cables, instruments, loads and S-parameter measurements are all referenced consistently to that same value. That shared reference is what makes a number like \(S_{11}=-20\ \text{dB}\) mean the same thing on two different benches.

04 — Matching and reflection

What happens if the load is not 50 Ω?

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

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.

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 has some loss, so a measured \(|\Gamma|\) is always slightly under one. Keep \(\Gamma\) in view: it reappears below, under the name \(S_{11}\).

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
0.506.0 dB3.0025 %poor
0.3210 dB1.9210 %the notes' “low” threshold
0.1020 dB1.221 %good
0.0330 dB1.070.1 %excellent

Return loss is a positive number that gets bigger as the match gets better. VSWR is a positive number that gets closer to \(1\). Both are just \(|\Gamma|\) rescaled, so quoting one is enough — but you have to be able to move between them. The same reflection measured at the input of a terminated two-port is called \(S_{11}\), so return loss is equally often written \(\mathrm{RL}=-20\log_{10}|S_{11}|\).

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: \(\mathrm{VSWR}=\dfrac{1+0.20}{1-0.20}=1.50\).
  4. Reflected power fraction: \(|\Gamma|^{2}=0.04\), so \(4\%\) returns and \(96\%\) reaches the load.

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.

05 — S-parameters

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

From the notes — section 1.3.4, equations (16) and (17).

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.

Why do we prefer travelling waves at microwave frequencies?

It is worth being precise here, because the usual shorthand overstates the case. Voltage and current do not stop being meaningful at \(12\ \text{GHz}\). On a two-conductor transmission line such as coax they remain perfectly well defined quantities. What changes is how convenient they are to measure, and how naturally they describe a structure that has no second conductor.

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 a single pair of terminal values — the lumped picture no longer applies.

Open and short are hard to realise

\(Z\), \(Y\) and \(ABCD\) parameters are defined with ideal open- and short-circuit terminations. Up here an open circuit radiates and a short has inductance, so those conditions are both difficult to realise accurately and undesirable to impose on a live circuit.

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 in a waveguide than it is on a two-conductor line.

Incident and reflected travelling waves, by contrast, exist in both structures, and a measuring instrument can separate them directly. That is why they became the practical characterisation language at these frequencies — not because the older quantities are wrong, but because the wave description is the one the hardware and the instrument both agree with. The notes put the waveguide half of it in one sentence: it is clear that in waveguides we cannot define potential, as the structure has only one conductor. Instead, we define waves.

Four waves, before any matrix

Draw the component as a two-port network. At port 1 there is a wave going in, \(a_1\), and a wave coming out, \(b_1\); at port 2 there is \(a_2\) going in and \(b_2\) coming out. 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.

A scattering parameter is then the ratio of a wave leaving the network to a wave sent into it, with no wave arriving at any other port:

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

The subscript condition is less forbidding than it looks. It says: to measure one S-parameter, excite one port and terminate every other port in the reference impedance — \(50\ \Omega\) for the systems in this course — so that nothing is reflected back into the network from those ports. That is a realisable microwave measurement, unlike an ideal open or short.

For a two-port that unpacks into exactly four measurements, two with port 2 terminated and two with port 1 terminated:

\[S_{11}=\left.\frac{b_1}{a_1}\right|_{a_2=0},\qquad S_{21}=\left.\frac{b_2}{a_1}\right|_{a_2=0}\]
\[S_{12}=\left.\frac{b_1}{a_2}\right|_{a_1=0},\qquad S_{22}=\left.\frac{b_2}{a_2}\right|_{a_1=0}\]

Collecting the four combinations gives the matrix form the notes write as equation (17):

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

Read a row at a time: what leaves port 1 is whatever bounces off port 1, plus whatever travels back through the network from port 2.

The four entries, in plain language

\(S_{11}\)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 of the previous section, measured at the input.
\(S_{21}\)Forward transmission. How much gets from port 1 through to port 2? Loss for a passive component, gain for an amplifier.
\(S_{12}\)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.
\(S_{22}\)Output reflection. How well is port 2 matched, seen from the other side?
\[S_{11}:\ \text{how much comes back}\qquad S_{21}:\ \text{how much gets through}\]

For most practical work those two answer the question. The other two say what the component looks like from the far end, and whether it passes signal backwards.

\(\Gamma\) and \(S_{11}\) are the same quantity

These are not two unrelated ideas that happen to describe reflection. For a one-port, or at the input of a two-port whose output is properly terminated, they are identical under the chosen reference impedance:

\[S_{11}=\Gamma_{in}\]

So everything already established about \(\Gamma\) carries straight over. In particular the input return loss is

\[\mathrm{RL}=-20\log_{10}|S_{11}|\]

The change of name reflects a change of context rather than of physics: \(\Gamma\) is what we write when thinking about a load on a line, \(S_{11}\) is what the instrument reports when the same reflection is measured at the input of a component. The reference impedance must be stated either way, because both numbers are defined against it.

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

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

From the original lecture notes · Topic 3, pp. 17–18, equations (16) and (17) →

06 — Components and the VNA

What the numbers look like for real hardware

From the notes — equations (18) and (19) and notes (i)–(iii), with worked component examples added.

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

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 most useful diagnostic in this lecture: add up the reflected and transmitted power fractions of a real component, and the shortfall is the power that was dissipated or radiated.

The instrument that measures all four

A vector network analyzer sweeps a known signal across the band and separates what returns from what passes through, recording both magnitude and phase. Drive port 1 with port 2 terminated and it reports \(S_{11}\) and \(S_{21}\); drive port 2 with port 1 terminated and it reports \(S_{22}\) and \(S_{12}\). The notes add the practical footnote that VNAs are usually much more expensive than oscilloscopes and spectrum analyzers.

The reference plane — the point in the setup beyond which the numbers describe the device rather than the cables — is established by calibration. SOLT, TRL and their relatives are the standard procedures, and they are a topic of their own; for now it is enough to know that an uncalibrated VNA measures your cables as enthusiastically as it measures your component.

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

07 — RF connectors

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.

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, air dielectricaround 34 GHzprecision measurement, calibration kits
2.92 mm (K)threaded, mates with SMA and 3.5 mm50 Ω, air dielectricaround 40 GHzKa-band work and VNA test ports
2.4 mmthreaded, does not mate with SMA50 Ω, air dielectricaround 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 between grades of the same family, and a connector's usable range also depends on the cable behind it and on how it was fitted — so read the specific datasheet rather than memorising this column.

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

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

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.

08 — An RF path

One complete RF path, end to end

Every idea in Weeks 5 to 8 appears somewhere in a real transmit chain:

  1. source
  2. SMA
  3. 50 Ω coax
  4. transition
  5. WR-75
  6. circular feed
  7. antenna

Each stage exercises a concept from these weeks, and the middle column names it:

StageConceptWhat it means here
Sourcethe frequency choice\(12\ \text{GHz}\) — forced in Week 5 by the ionosphere and by directivity, and it fixes every dimension downstream.
Coaxial cable\(Z_0\)\(Z_0=\frac{60}{\sqrt{\varepsilon_r}}\ln(b/a)=50\ \Omega\). TEM, so no cutoff; flexible, and the reference impedance everything else on this path is measured against.
SMA connectormaintain geometry, and so impedanceThe joint is a short length of line. Hold the \(b/a\) ratio through it and \(Z_0\) is continuous; disturb it and you have made a discontinuity.
Any impedance step\(\Gamma\)\(\Gamma=\frac{Z_L-Z_0}{Z_L+Z_0}\). Wherever the impedance changes — a connector, a transition, a flange, the feed — part of the wave turns round.
Any component in the chain\(S_{11}\) and \(S_{21}\)\(S_{11}\) is what comes back from it, \(S_{21}\) is what gets through it. Two numbers, measured on a VNA against the same \(50\ \Omega\) reference, decide whether the part is doing its job.
Coax-to-waveguide transitionmode conversionWhere TEM becomes \(TE_{10}\). A designed component with its own \(S_{11}\) and \(S_{21}\), and the usual home of a band-edge reflection.
Rectangular runmode and cutoffWR-75, \(f_{c,10}=c/2a=7.87\ \text{GHz}\). Operating at \(12\ \text{GHz}\) is above that cutoff and below \(f_{c,20}\), so one mode travels. Low loss, high power, rigid.
Circular feed and antennalaunching into free space\(2a=16.8\ \text{mm}\), \(f_{c_1}=10.46\ \text{GHz}\), \(TE_{11}\) with its two orthogonal orientations. This is where the guided wave stops being guided and becomes the beam Week 5 set out to form.

The mode changes twice along this path — TEM in the coax, \(TE_{10}\) in the rectangular run, \(TE_{11}\) in the circular feed — and each change happens inside a component that has to be designed and measured. Every interface is a possible mismatch and so a possible \(\Gamma\); the whole chain answers to one measured \(S_{11}\) at the input and one \(S_{21}\) end to end. That is the conceptual payoff of Week 8: how effectively RF power passes through real components and interfaces is answered, component by component, by two measured numbers.

09 — Week summary

How effectively power passes through real components

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 how we describe and measure the result.

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 the forward wave; 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, with calibration setting the reference plane.

Connectors

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

Study this week as slides On to Week 9

10 — Check yourself

Three quick checks

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?

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?

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?

11 — Week 8 exercises

Practice problems

The lecture ends above. What follows is a separate set of four 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.

These four problems assess impedance, reflection and S-parameters. The waveguide design problems are in Week 6.

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.

Show worked 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.

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.

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

Show worked 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.

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 without using a formula, what an engineer learns from those two numbers.
Show worked 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. In plain engineering language: “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 of the far side, and the missing fifth is lost inside the part. It is a well-matched component with about a decibel of insertion loss.” Notice that neither number on its own would have supported that sentence: \(S_{11}\) speaks only about the match, \(S_{21}\) only about what gets through.

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.

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.

Show worked 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.

1 · SMA connectorThe most-mated and most easily damaged part in the path. Inspect the centre pin and the interface; substitute a known-good cable.
2 · Coax-to-waveguide transitionThe only mode conversion in the path, and the usual home of a band-edge reflection. Check that its band actually covers \(12\ \text{GHz}\).
3 · The cable itselfConfirm that \(2\ \text{m}\) at \(12\ \text{GHz}\) should lose only a fraction of a dB for its type. A crushed or water-ingressed cable will not.
4 · Waveguide section and flangesLook for a misaligned, dirty or partly bolted joint, and confirm the guide is the right size for the band.
5 · The feedLast, because it is hardest to reach: wrong band, wrong polarisation or mechanical damage.

A VNA in time domain, or a swept \(S_{11}\) whose ripple period reveals the distance to the discontinuity, will find the fault faster than disassembly will.