RF systems: transmission lines and matching

Welcome to part three, the second unit of exam 1: transmission lines. Last year, four of the ten problems came from here: the input impedance of a line, a VSWR, a coaxial cable from a datasheet, and return loss with cable loss. We build the theory from the line model, then solve every example and problem on the slides, learn to use the Smith chart, and finish with S-parameters and impedance matching.

A transmission line is a physical medium that carries electrical energy, in which the voltage and the current vary along its length. What matters is the length compared with the guided wavelength, the speed of light over f times the square root of the effective permittivity. If the line is much shorter than a wavelength, all points see the same voltage at the same time and ordinary circuit theory works; once it is a sizeable fraction of a wavelength, as a rule of thumb more than a tenth, we must treat it as a line with waves travelling along it. For example, one metre of polyethylene coaxial cable is half a wavelength long at 99 MHz, so it is a transmission line there, but at 50 kHz it is about a four-thousandth of a wavelength and behaves as a simple wire.

A short piece of line is modelled by four parameters per unit length: a series resistance R for the conductor losses and a series inductance L, both in the direction of the current. And a shunt conductance G for the losses in the dielectric, and a shunt capacitance C between the two conductors. Solving the circuit equations for a chain of these pieces, the voltage along the line is the sum of a forward wave, V plus times e to the minus gamma x, and a reflected wave, V minus times e to the plus gamma x; the current has the same two waves with a minus sign between them. Gamma, the propagation constant, is the square root of R plus j omega L times G plus j omega C, with real part alpha, the attenuation, and imaginary part beta, the phase constant; and the characteristic impedance Z zero, the ratio of voltage to current in a single wave, is the square root of R plus j omega L over G plus j omega C.

In most problems the line is lossless, R and G are zero, and everything becomes simpler. The characteristic impedance is the square root of L over C, a real number such as 50 or 75 ohms. The propagation constant is purely imaginary, j beta, with beta equal to omega times the square root of L C, which is also two pi over the guided wavelength. The phase velocity is f times lambda g, omega over beta, which is one over the square root of L C. The waves travel without attenuation, only changing phase. Combining the two formulas gives a result we will use for coaxial cables: Z zero is one over the phase velocity times the capacitance per metre.

When the line ends in a load different from Z zero, part of the wave is reflected. The reflection coefficient is the ratio of the reflected to the forward wave, and at the load it is Z L minus Z zero over Z L plus Z zero. If the load equals Z zero, the coefficient is zero and nothing is reflected: the load is matched. Looking into a line of length l terminated in Z L, the input impedance is Z zero times Z L plus j Z zero tan beta l, over Z zero plus j Z L tan beta l; since the tangent repeats every half wavelength, so does the input impedance. Moving back from the load towards the generator, the reflection coefficient keeps its modulus and its phase decreases by twice beta l, a fact that is the basis of the Smith chart. For power, the reflected power is the modulus of gamma squared times the incident power, and the load receives the rest, one minus gamma squared; maximum power is delivered when gamma is zero.

The forward and reflected waves interfere and create a standing-wave pattern along the line; watch the three cases, with the load on the right. With a matched load there is only the forward wave, and the envelope of the voltage, in orange, is flat. With a partial reflection, here gamma equal to one half, the envelope swings between a maximum of one plus gamma and a minimum of one minus gamma. The voltage standing-wave ratio, SWR, is the maximum over the minimum, one plus the modulus of gamma over one minus the modulus of gamma; here 1.5 over 0.5, which is 3. Inverting it, the modulus of gamma is SWR minus one over SWR plus one, the formula you need in almost every exam question. With a short or an open circuit everything is reflected, the minima are true zeros and the SWR is infinite, while a matched line has an SWR of one; the maxima and minima repeat every half wavelength.

The slides ask you to discuss the special cases of the input impedance formula, and they are worth memorising. At low frequency beta l is almost zero, the tangent vanishes and the input impedance is just the load: the line is invisible. A matched line shows Z zero at its input whatever its length. A short-circuited line gives j Z zero tan beta l, and an open line minus j Z zero cotangent beta l: both are pure reactances, which is how stubs make inductors and capacitors. A half-wavelength line repeats the load exactly. And a quarter-wavelength line inverts it, Z zero squared over Z L, so a short becomes an open and an open becomes a short; we will use this in the quarter-wave transformer.

This question looks hard but is a one-liner once you see the trick. Beta l is two pi over lambda times an eighth of a wavelength plus n halves, which is pi over 4 plus n pi; the n pi does not change the tangent, so tan beta l is one for every n. Substituting, the numerator is j Z zero plus j Z zero, twice j Z zero, and the denominator is Z zero plus j times j Z zero, which is Z zero minus Z zero, zero. So the input impedance is infinite: the line looks like an open circuit, whatever n is. You can confirm it with the reflection coefficient: for Z L equal to j Z zero, gamma at the load is j minus one over j plus one, which is j, and moving an eighth of a wavelength towards the generator rotates it by 90 degrees clockwise, to plus one, the open-circuit point.

This is the worked example on the slides. Z L minus Z zero is minus 35 plus j 20, and Z L plus Z zero is 115 plus j 20, so the modulus of gamma is 40.3 over 116.7, which is 0.345. The SWR is 1.345 over 0.655, 2.06; the slide rounds gamma to 0.35 and gets 2.07, both are fine. The electrical length is 2 pi times 0.3, 0.6 pi, 108 degrees, whose tangent is minus 3.078. Substituting in the input impedance formula, the numerator is 40 minus j 210.83, the denominator 136.55 minus j 123.11, and the result is 69.71 minus j 52.95 ohms: a complex division that you must be able to do on the calculator.

Now the problem left for you on the slides. Gamma at the load is minus 60 minus j 80 over 140 minus j 80, which is minus 0.08 minus j 0.62, a modulus of 0.620 at minus 97.1 degrees. The line is 0.7 wavelengths long, which is the same as 0.2, because everything repeats every half wavelength; twice beta l is 4 pi times 0.7, 504 degrees. Gamma at the input has the same modulus, and its angle decreases by 504 degrees: minus 97.1 minus 504 is minus 601.1, which is the same as plus 118.9 degrees. Converting back, the input impedance is 100 times one plus gamma over one minus gamma, 31.0 plus j 54.8 ohms: the capacitive load looks inductive at the input. And the SWR is 1.62 over 0.38, 4.27, the same everywhere on the lossless line.

Datasheets of microwave components specify three things: the SWR over the operating band, the insertion loss, which is the loss in transmission, and the return loss, which measures the input mismatch; both losses are positive numbers in dB. For an SWR of 1.5, the modulus of gamma is 0.5 over 2.5, which is 0.2, so the reflected power is 0.2 squared, 4 % of the incident power. An insertion loss of 0.5 dB means the output power is ten to the minus 0.05 times the input, 0.891, so 89 % of the power gets through. The return loss is minus 20 log of 0.2, 13.98 dB. So of the input power, 89 % comes out, 4 % is reflected at the input, and the remaining 7 % is lost in the circuit and in the output mismatch.

The slides ask you to fill this table and judge the matching of a coaxial antenna system with a limit of 10 % reflected power. For each SWR, the modulus of gamma is SWR minus one over SWR plus one, the return loss is minus 20 log of it, and the reflected power is its square. An SWR of 1.2 reflects less than one percent, 1.5 reflects 4 %, 2 reflects 11 % and 3 reflects a quarter of the power. The 10 % limit corresponds to a gamma of the square root of 0.1, 0.316, an SWR of 1.92 and a return loss of 10 dB, so an SWR of 1.5 is a good match and an SWR of 2 is just outside the criterion.

Three of last year's problems were quick specification checks. In problem 5, a VSWR of 1.5 gives a gamma of 0.2, and the reflected power is its square, 4 %. In problem 8, a return loss of 21 dB means the modulus of gamma is ten to the minus 21 over 20, 0.089, which meets the requirement of less than 0.1. And 300 feet of cable at 6.3 dB per hundred feet loses 3 times 6.3, 18.9 dB, below the 20 dB limit, so both specifications are met. The other solved copy also has a matched-load question: the load equals the characteristic impedance, so gamma is zero, the SWR is one and no power is reflected, whatever the frequency.

The coaxial cable is the line you will meet most in aircraft, from the radio to the antenna. Its characteristic impedance depends only on the ratio of the outer to the inner diameter and on the dielectric: 138 over the square root of epsilon r, times the logarithm of D over d. The velocity factor is the phase velocity over the speed of light, one over the square root of epsilon r: 66 % for solid polyethylene and 69 % for PTFE. The capacitance per metre is 24.12 epsilon r over the logarithm of D over d picofarads per metre, which can also be written 3326 times the square root of epsilon r over Z zero. Combining them, the characteristic impedance is one over the phase velocity times the capacitance, so a datasheet that gives the velocity factor and the capacitance gives you Z zero. For the LMR-400 cable in the brochure, 84 % and 78.4 picofarads per metre give 50.6 ohms: it is a 50 ohm cable. The same glossary also gives the attenuation in dB per 100 m as a conductor term, 1.43 R over Z zero, plus a dielectric term, 9.15 root epsilon r times the frequency in megahertz times the loss factor F, so loss grows with frequency, which is why datasheets quote dB per 100 feet at each frequency.

Two coaxial problems. For part a, invert the impedance formula: D is d times ten to the power Z zero root epsilon r over 138. In air, 50 ohms needs ten to the 0.362, so D is 2.99 mm, and 75 ohms needs 4.54 mm. With polyethylene the exponent grows by the root of 2.3, and the diameters become 4.61 and 8.67 mm: a dielectric makes the cable fatter for the same impedance. For last year's exam problem, the velocity factor of 85 % means the square root of epsilon r is one over 0.85, 1.176, and the capacitance formula gives 3326 times 1.176 over 78.4, 49.9 ohms. The shortcut, one over the phase velocity times the capacitance, gives 50.0 ohms, the answer of the first solved copy. The other copy had a variant with the inductance instead, 0.20 microhenries per metre: then Z zero is the square root of L over C, 50.5 ohms, and the velocity factor is one over c times the square root of L C, 0.84.

The slides close the line section with a comparison table. Waveguides work from 1 to 300 GHz and handle high power with very low loss, but they are bulky and expensive. Coaxial cables work up to about 50 GHz, are shielded, and are made almost always at 50 or 75 ohms. Planar lines, stripline, microstrip, coplanar waveguide and slotline, are printed on circuit boards: small, cheap and easy to connect to transistors, which is why every RF circuit board uses them. In a microstrip part of the field travels in the air above the board, so its effective permittivity lies between one and that of the substrate, which is why the matching problems give you an epsilon effective.

The Smith chart is a graphical calculator for transmission lines. It is a polar plot of the reflection coefficient, on which a grid shows the normalised impedance, z equal to Z over Z zero, which is one plus gamma over one minus gamma. The circles are lines of constant normalised resistance and the arcs are lines of constant reactance: the upper half is inductive and the lower half capacitive. The centre is the matched point, the left end of the axis is the short circuit and the right end the open circuit. Since moving along the line only changes the phase of gamma, moving from the load towards the generator is a clockwise rotation on a circle centred at the origin, and a full turn is half a wavelength; the outer scale is graduated in wavelengths for this reason. Where that circle crosses the right half of the real axis, the value of r read there is the SWR. And because a quarter wavelength inverts the impedance, the normalised admittance is the point diametrically opposite the impedance, a 180 degree rotation.

Now the example on the slides, step by step on the chart. Normalise first: 100 plus j 50 over 50 is 2 plus j 1, which we place at the intersection of the r equals 2 circle and the x equals 1 arc; analytically gamma is one plus j over three plus j, 0.447 at 26.6 degrees. Draw the circle of constant gamma through the load: it cuts the real axis on the right at r equal to 2.62, so the SWR is 2.62, read as about 2.6 on the chart. To find the input impedance, read the load on the wavelengths-towards-generator scale, 0.213 lambda, which the slide rounds to 0.214, add the line length, 0.2, and stop at 0.413 lambda; the point reads 0.5 minus j 0.5, so Z in is about 25 minus j 25 ohms, against 24.8 minus j 24.6 from the formula. Finally the admittance is the point opposite the load, 0.4 minus j 0.2, which de-normalised, divided by 50 ohms, is 0.008 minus j 0.004 siemens, that is 8 minus j 4 millisiemens.

This is the classic problem from Pozar's book that the slides leave for you. The normalised load is 0.4 plus j 0.7, and gamma at the load is minus 60 plus j 70 over 140 plus j 70, 0.589 at 104 degrees. The SWR is 1.589 over 0.411, 3.87, and the return loss is minus 20 log of 0.589, 4.6 dB. Moving 0.3 wavelengths towards the generator rotates gamma clockwise by twice 360 times 0.3, 216 degrees, so gamma at the input is 0.589 at minus 112 degrees. Converting back, Z in is 36.5 minus j 61.1 ohms; on the chart you read about 0.365 minus j 0.611, Pozar's answer.

At high frequency impedance and admittance matrices are difficult to measure directly, because they need ideal open or short circuits at the ports, which are hard to realise, so we describe networks with scattering parameters, which a vector network analyser measures directly. At each port we define a normalised incident wave a and a reflected wave b, the voltage waves divided by the square root of the port impedance, and the scattering matrix relates them: b equals S times a. Each parameter S i j is b i over a j, measured with all the other ports terminated in matched loads, so that nothing comes back into them. For a two-port, S one one and S two two are the reflection coefficients at the input and the output, S two one is the forward transmission, the gain of an amplifier, and S one two the reverse transmission. That is why last year's exam gave the amplifier gains as S two one equal to 10 dB, and why minus 20 log of S one one is the return loss.

Matching networks transform the load so that the line sees Z zero. That delivers maximum power to the load, improves the signal-to-noise ratio of a receiver when the antenna is matched to the LNA, and in a transmitter minimises wasted power, saving battery and reducing radiation hazards. The simplest network is the L-section, two reactive elements, which is practical with lumped components up to about 1 GHz. If the normalised load lies inside the r equals one circle, that is if R L is larger than Z zero, the element next to the load is in shunt and the second one in series. If it lies outside, R L smaller than Z zero, the element next to the load is in series and the second one in shunt. Each case has two solutions; de-normalise the reactance by multiplying by Z zero and the susceptance by dividing by Z zero, and turn a positive reactance into an inductor and a negative one into a capacitor, while a positive susceptance is a capacitor and a negative one an inductor. For real loads there is also the quarter-wave transformer, which we will see after the L-section examples.

Now the L-section example from the slides, which comes from Pozar. The normalised load is 2 minus j, inside the r equals one circle, so the first element, next to the load, is in shunt, and we work with admittances: y L is 0.4 plus j 0.2. The susceptance comes from Pozar's formula, X L plus or minus root R L over Z zero times root of R L squared plus X L squared minus Z zero R L, all over R L squared plus X L squared, which here is minus 100 plus or minus 244.9 over 50 000, so b is plus 0.29 or minus 0.69; on the chart, adding plus j 0.29 moves the point to 0.4 plus j 0.49, on the circle that maps to r equals one, z equal to one minus j 1.22, and the reactance follows from X equals one over B plus X L Z zero over R L minus Z zero over B R L. Then a series reactance of plus j 1.22 cancels the imaginary part and lands at the centre; de-normalised, the shunt element is a capacitor of 0.92 picofarads and the series element an inductor of 38.8 nanohenries, or 39.0 if you keep more decimals. The second solution uses b equal to minus 0.69 and x equal to minus 1.22: a shunt inductor of 46.1 nanohenries and a series capacitor of 2.61 picofarads, which is the answer printed on the next slide.

The second matching problem on the slides has no printed solution, so let us solve it. The normalised load is 0.5 plus j 0.3, outside the r equals one circle, because R L is smaller than Z zero, so the series element goes next to the load and the shunt element on the line side. Pozar's formulas for this case give X equal to plus or minus the square root of R L times Z zero minus R L, minus X L, that is plus or minus 25 minus 15, and B equal to plus or minus one over Z zero times the square root of Z zero minus R L over R L, which is plus or minus 0.02 siemens. Solution one: X is plus 10 ohms, an inductor of 0.80 nanohenries at 2 GHz, and B is plus 0.02 siemens, a shunt capacitor of 1.59 picofarads. Solution two: X is minus 40 ohms, a series capacitor of 1.99 picofarads, and B is minus 0.02 siemens, a shunt inductor of 3.98 nanohenries; substituting back, both give exactly 50 ohms.

For a real load, a quarter-wavelength section of line does the matching on its own, because it inverts the impedance: the input impedance is Z one squared over Z L, and setting it equal to Z zero gives Z one equal to the square root of Z zero Z L. For a 10 ohm load on a 50 ohm line, Z one is the square root of 500, 22.36 ohms. Its length is a quarter of the guided wavelength, 3 times ten to the 8 over 4 times 3 times ten to the 9 times the square root of 4.6, 1.17 cm. The match is perfect only at f zero; with a limit of an SWR of 1.5, the maximum tolerable gamma is 0.2. Plugging it into the bandwidth formula, the arc cosine of 0.2 over the root of 0.96, times twice the root of 500 over 40, gives a fractional bandwidth of 0.29. That is 29 %, from 2.56 to 3.44 GHz, which the slides read from their plot as 2.55 to 3.42, as in the plot of gamma against frequency, where the curve stays under the dashed limit.

The problem that follows on the slides uses the same three steps. Z one is the square root of 100 times 350, 187.1 ohms. At 4 GHz with an effective permittivity of 4.6, the quarter wavelength is 0.874 cm. An SWR of 2 allows a gamma of one third. The bandwidth formula then gives 0.71, a fractional bandwidth of 71 %: much wider than before, because the load is closer to the line impedance and the tolerance is looser.

Before the summary, the traps in this unit. Beta l uses the guided wavelength, not the free-space one, and everything repeats every half wavelength. Gamma rotates by twice beta l, that is 720 degrees per wavelength, clockwise when you move towards the generator. From the SWR, gamma is SWR minus one over SWR plus one; the reflected power fraction is gamma squared, and the return loss is minus 20 log of gamma, not 10 log. On the Smith chart, normalise before you start and de-normalise at the end, multiplying impedances by Z zero and dividing admittances by it. For an L-section, check where the load is: inside the r equals one circle, the shunt element goes at the load; outside, the series element does. And for a coaxial cable from a datasheet, Z zero equals one over the phase velocity times the capacitance is the quickest route.

To summarise part three: the reflection coefficient at the load and the input impedance formula solve most line problems. The SWR, the return loss and the insertion loss describe how good a match is, and a reflected power of 10 % corresponds to an SWR of about 1.9. A coaxial cable's impedance follows from its diameters and dielectric, or from its velocity factor and capacitance. On the Smith chart, move clockwise towards the generator, half a wavelength per turn, and find the admittance on the opposite side. To match a load, use an L-section of two reactances, or a quarter-wave transformer with Z one equal to the square root of Z zero Z L. To prepare, redo last year's ten problems against the clock, about six minutes each, then the problems on the slides, checking every number against these three videos; good luck on Tuesday.