Two equal-amplitude isotropic radiators are fed in phase and spaced one full wavelength apart along an axis, with the polar angle measured from that axis. Besides the broadside direction at 90 degrees, where does the array factor reach a maximum?

At 0 degrees and 180 degrees, that is straight along the array axis

Nowhere else; broadside is the only maximum the pattern has

An air-filled rectangular waveguide has internal dimensions of 3.2 cm by 1.4 cm and is driven at 6 GHz, with c taken as 3 x 10^8 m/s and the intrinsic impedance of free space as 376.73 ohm. Which of the following statements about it are correct?

The guide wavelength is 80.1 mm, which is longer than the 50 mm free-space wavelength.

Only the mode of lowest cutoff propagates at 6 GHz, the next cutoff lying at 9.375 GHz.

The guide wavelength is 80.1 mm, which is longer than the 50 mm free-space wavelength.

Only the mode of lowest cutoff propagates at 6 GHz, the next cutoff lying at 9.375 GHz.

The group velocity in the guide is 4.81 x 10^8 m/s.

The wave impedance of that mode is 235 ohm, being the free-space impedance multiplied by the square root of one minus the squared frequency ratio.

Transmission Lines, Waveguides, Optical Fibre and Antennas

This chapter is the guided half of Section 8 of the Electronics and Communication Engineering paper: everything that carries a wave along a structure rather than letting it spread. It is also the half where one idea does most of the work. A transmission line is described by four distributed parameters, and from them come Z₀ = √(L/C) and a velocity, and from those come the reflection coefficient and everything a line does. The single most useful fact in the whole section is that travelling down a line does not change the magnitude of the reflection coefficient, only its phase — Γ rotates on a circle of constant radius at 2β radians per metre. That statement is the Smith chart. This chapter teaches the chart that way, as arithmetic rather than as a graphical construction, because the four operations a chart performs — normalise, convert z to Γ, rotate the phase, convert back — are each one line of algebra, and doing them explicitly is both faster on a calculator and immune to reading the wrong circle. S-parameters are then the same reflection coefficient given a name that works for multi-port devices, with S₁₁ = Γ at the input port. A waveguide is what you use when a two-conductor line will no longer do, and its behaviour is set by one number per mode, the cutoff frequency, below which nothing propagates and above which the phase velocity exceeds c while the group velocity falls short of it, with v_p v_g = c² exactly — the same relation as the ionospheric plasma of the previous chapter, for the same mathematical reason. An optical fibre is a dielectric waveguide operated by total internal reflection, and its two governing numbers are the numerical aperture and the V parameter. And an antenna is the reverse of all of this: a deliberately unguided structure, whose radiation resistance, directivity and pattern follow from how its current is distributed — and whose array behaviour follows from adding the fields of several elements with the right phases, which turns out to be arithmetic you can do without drawing anything.

Line Equations, Characteristic Impedance and Velocity

A transmission line is modelled as series R and L and shunt G and C, all per unit length, and the telegrapher's equations dV/dz = −(R + jωL)I and dI/dz = −(G + jωC)V follow from applying Kirchhoff's laws to one infinitesimal section. Eliminating one variable gives a wave equation with propagation constant γ = √[(R + jωL)(G + jωC)] = α + jβ, and the ratio of the forward voltage to the forward current is the characteristic impedance Z₀ = √[(R + jωL)/(G + jωC)]. Almost every examinable result comes from the lossless case R = G = 0, where those collapse to Z₀ = √(L/C) — real, frequency-independent — and β = ω√(LC), so v_p = 1/√(LC), also frequency-independent. Note what Z₀ is and is not: it is the impedance a wave sees while travelling, not a resistance you could measure with a meter on a short piece, and it dissipates nothing on a lossless line. A separate case worth naming is the distortionless line, which has loss but satisfies R/L = G/C; then α = √(RG) is constant with frequency and β is proportional to ω, so every frequency is attenuated equally and delayed equally, and a pulse arrives smaller but not smeared.

🧠 A 50 Ω coax from its L and C, and three answers that follow at once

Take a coaxial cable with L = 250 nH/m and C = 100 pF/m. Then Z₀ = √(250 × 10⁻⁹/100 × 10⁻¹²) = √2500 = 50 Ω exactly, and

v_p = 1/√(LC) = 1/√(250 × 10⁻⁹ × 100 × 10⁻¹²) = 1/√(2.5 × 10⁻¹⁷) = 1/(5 × 10⁻⁹) = 2 × 10⁸ m/s

. Three consequences fall out of those two lines with no further work. The dielectric constant is determined, because v_p = c/√ε_r gives √ε_r = 3/2 = 1.5 and ε_r = 2.25 — the velocity factor is 0.667, which is the figure a cable datasheet quotes. The delay per metre is 1/v_p = 5 ns/m, and that is the number that matters for a digital board: a 150 mm track on such a line is 0.75 ns of delay, comparable with a fast logic edge, which is why transmission-line thinking becomes compulsory above a few hundred megahertz. And the geometry is fixed too, since for coax Z₀ = (60/√ε_r) ln(b/a); with 60/1.5 = 40, we need ln(b/a) = 50/40 = 1.25, so b/a = e^(1.25) = 3.49. A 50 Ω cable in this dielectric therefore has an outer conductor about three and a half times the inner diameter, and that ratio — not the absolute size — is what sets the impedance. Note that Z₀ and v_p are the only two independent quantities here: L and C can be recovered from them as L = Z₀/v_p and C = 1/(Z₀v_p), so a question that gives you any two of the four has given you all four.

Reflection, VSWR, and How a Line Transforms an Impedance

Terminate a line of impedance Z₀ in a load Z_L and the mismatch launches a reflected wave, with reflection coefficient Γ_L = (Z_L − Z₀)/(Z_L + Z₀). Three special cases anchor the range: Z_L = Z₀ gives Γ = 0, a matched line with no reflection at all; an open circuit gives Γ = +1; a short circuit gives Γ = −1. The interference between forward and reflected waves produces a standing pattern whose voltage standing wave ratio is VSWR = (1 + |Γ|)/(1 − |Γ|), running from 1 for a perfect match to infinity for a total reflection. Two derived quantities are how the mismatch is quoted in practice. Return loss = −20 log₁₀|Γ| dB says how far down the reflected wave is, and mismatch loss = −10 log₁₀(1 − |Γ|²) dB says how much of the incident power failed to reach the load. For Z_L = 100 Ω on a 50 Ω line, Γ = 50/150 = 1/3, so VSWR = 2, return loss = 9.54 dB and mismatch loss = 0.51 dB, meaning 88.9 per cent of the power gets in. Note that Z_L = 25 Ω gives Γ = −1/3 and the identical VSWR = 2: VSWR discards the phase, so it cannot tell you whether the load is too high or too low, which is exactly why a network analyser reports S₁₁ as a complex number and not as a VSWR.

Move back along the line and the impedance you see changes, because the two travelling waves re-add with a different relative phase. The impedance transformation formula is Z_in = Z₀ (Z_L + jZ₀ tan βl)/(Z₀ + jZ_L tan βl), and three of its consequences are worth memorising as facts rather than re-derived each time. At l = λ/2, tan βl = 0, so Z_in = Z_L: a half-wavelength of line is invisible. At l = λ/4, the formula inverts the load about Z₀: Z_in = Z₀²/Z_L, so a short looks like an open and an open looks like a short — which is how a quarter-wave stub becomes a component. And at any length, a shorted line gives Z_in = jZ₀ tan βl, which is inductive for l < λ/4 and capacitive between λ/4 and λ/2, so a length of line is a reactance you can dial in by cutting it. Reactive by design and lossless in principle, these stubs are how matching is done above a gigahertz, where a lumped inductor has stopped behaving like one.

The Smith Chart, Taught as Four Lines of Arithmetic

The Smith chart is normally taught as a drawing, and that presentation hides how little it does. Underneath, it is the bilinear map between normalised impedance and reflection coefficient, Γ = (z − 1)/(z + 1) and back z = (1 + Γ)/(1 − Γ), plotted on the complex Γ plane. Every operation the chart performs is one of exactly four steps, and each is a single line of algebra. Normalise: divide by Z₀ to get z = Z/Z₀, which is why the chart works for any Z₀. Convert: Γ = (z − 1)/(z + 1). Rotate: moving a distance l towards the generator multiplies Γ by e^(−j2βl), that is, it subtracts 2βl from the phase and leaves the magnitude untouched. Convert back: z = (1 + Γ)/(1 − Γ). That is the whole chart. The circles printed on it are just the loci of constant resistance and constant reactance under the map, and the concentric circles about the centre are the loci of constant |Γ| — which is the same thing as constant VSWR, since VSWR is a function of |Γ| alone. Reading the chart as arithmetic has a further advantage in an exam: it gives exact numbers instead of the two significant figures a printed chart can be read to.

🧠 One eighth of a wavelength, done four ways, all giving 40 − j30

A 50 Ω line is terminated in Z_L = 100 Ω. Find the input impedance λ/8 back from the load, using the chart's four steps and nothing drawn. Normalise: z_L = 100/50 = 2 + j0. Convert: Γ_L = (2 − 1)/(2 + 1) = 1/3, at zero phase since z_L is real and greater than 1. Rotate: βl = (2π/λ)(λ/8) = π/4 = 45°, so the phase changes by −2βl = −90°, giving Γ_in = (1/3)∠−90° = −j0.3333, with the magnitude still exactly 1/3. Convert back: z_in = (1 − j0.3333)/(1 + j0.3333). Multiply top and bottom by the conjugate of the denominator: the numerator becomes (1 − j0.3333)² = 0.8889 − j0.6667 and the denominator 1 + 0.1111 = 1.1111, so z_in = 0.8 − j0.6 and Z_in = 50(0.8 − j0.6) = 40 − j30 Ω. Now check it two other ways. The transformation formula with tan 45° = 1 gives Z_in = 50(100 + j50)/(50 + j100); multiplying through by the conjugate (50 − j100) gives numerator 10000 − j7500 over 12500, that is 0.8 − j0.6, so Z_in = 40 − j30 Ω ✓. And compute Γ afresh from that answer: (40 − j30 − 50)/(40 − j30 + 50) = (−10 − j30)/(90 − j30), and multiplying by (90 + j30) gives −j3000/9000 = −j0.3333 ✓ — magnitude 1/3 preserved, phase rotated exactly −90°. Four routes, one answer, and the check that matters is that |Γ| never moved. That invariance is the single most useful thing on the chart: because |Γ| = 1/3 everywhere along a lossless line, the VSWR is 2 at every point, and the impedance you measure depends only on where you measure it. Two immediate corollaries: a λ/2 shift is a full 360° rotation and returns you to Z_L, and a λ/4 shift is 180°, which sends z = 2 to z = 0.5, that is Z₀²/Z_L = 2500/100 = 25 Ω ✓.

🎯 Two ways to match 100 Ω to 50 Ω, and why the transformer is the narrow one

Quarter-wave transformer. Insert a λ/4 section whose impedance is the geometric mean: Z₀' = √(Z₀ Z_L) = √(50 × 100) = 70.71 Ω. At the design frequency the match is perfect, because the quarter-wave section transforms 100 Ω to Z₀'²/Z_L = 5000/100 = 50 Ω exactly. Now move off frequency and watch it fail. At half the design frequency the section is λ/8, so θ = βl = 45°, and the standard result |Γ| = 1/√(1 + [4Z₀Z_L/(Z_L − Z₀)²] sec²θ) gives 4(50)(100)/(50)² = 8 and sec²45° = 2, so |Γ| = 1/√17 = 0.2425, that is VSWR = 1.64 and a return loss of only 12.3 dB. Setting a design target of |Γ| ≤ 0.1 — that is VSWR ≤ 1.22 — and solving the same expression gives sec²θ = 12.375, so cos θ = 0.2843 and θ = 73.48°; the fractional bandwidth is 2(1 − θ/90°) = 36.7 per cent. A single quarter-wave transformer matching a 2:1 impedance ratio is good over about a third of an octave, and no more — which is why broadband matching uses several cascaded sections with a tapered impedance profile rather than one section. Single-stub match. The alternative places a shorted shunt stub at a chosen distance from the load. Work it in admittance: y = (1 − Γ)/(1 + Γ), and setting Re{y} = 1 reduces, after multiplying out, to cos φ = −|Γ_L| = −1/3, where φ = −2βl is the rotated phase. So φ = −109.47°, giving βl = 54.74° and a distance from the load of d = 0.1520λ. At that point y = 1 + j0.7071, so the stub must contribute −j0.7071; a shorted shunt stub has y_s = −j cot βl_s, so cot βl_s = 0.7071, tan βl_s = 1.4142, βl_s = 54.74° and l_s = 0.1520λ — the same length, which is a coincidence of this particular load and a pleasant check that the two calculations were done independently. Adding them: y_total = 1 + j0.7071 − j0.7071 = 1, matched ✓. At 1 GHz on an air line, λ = 300 mm, so d = 45.6 mm and l_s = 45.6 mm. There is a second solution at φ = +109.47°, giving d = l_s = 0.3480λ, and the shorter pair is normally preferred because a shorter line is a narrower-band line in reverse — less phase change per unit frequency shift.

The 50 Ω line into 100 Ω, read at four positions — |Γ| never changes
Distance back from the load	Phase rotation `−2βl`	`Γ`	`Z_in` and VSWR
0 (at the load)	`0°`	`+0.3333`	`100 Ω`, VSWR 2
`λ/8`	`−90°`	`−j0.3333`	`40 − j30 Ω`, VSWR 2
`λ/4`	`−180°`	`−0.3333`	`25 Ω` = `Z₀²/Z_L`, VSWR 2
`λ/2`	`−360°`	`+0.3333` — back where it started	`100 Ω`, VSWR 2 — the line is invisible

S-Parameters: the Reflection Coefficient Given a Matrix

At microwave frequencies you cannot measure a voltage or a current at a port with any accuracy, but you can measure incident and reflected waves. So the network description is recast in terms of wave amplitudes: b_i = Σ_j S_ij a_j, where a_j is the wave going into port j and b_i the wave coming out of port i, both normalised so that |a|² is power. Each parameter then has a direct meaning obtained by driving one port with the others terminated in Z₀: S₁₁ = b₁/a₁ with a₂ = 0 is the input reflection coefficient, which is exactly the Γ of the previous section, so S₁₁ = (Z_in − Z₀)/(Z_in + Z₀) and the Smith chart is a plot of S₁₁. S₂₁ is the forward transmission, whose magnitude in dB is the insertion gain or loss. S₁₂ is reverse transmission, and S₂₂ output reflection. Two structural properties are examined more often than any calculation. A reciprocal network has S_ij = S_ji, so its matrix is symmetric — true of any passive structure without ferrites or active devices. A lossless network has a unitary matrix, whose most used consequence for a two-port is |S₁₁|² + |S₂₁|² = 1: whatever is not reflected is transmitted, because nothing is absorbed.

⚠️ 0.41 dB of loss in a device that dissipates nothing at all

Take a two-port that is passive, reciprocal and perfectly lossless, measured to have |S₁₁| = 0.3. Unitarity then forces |S₂₁| = √(1 − 0.09) = √0.91 = 0.9539, so the measured insertion loss is −20 log₁₀(0.9539) = 0.41 dB. A device that dissipates exactly zero power shows 0.41 dB of loss on a network analyser, and a candidate who reasons that loss implies dissipation has the physics backwards. What the 0.41 dB measures is power that never entered the device: it was reflected at the input by the mismatch, and −10 log₁₀(1 − |Γ|²) = −10 log₁₀(0.91) = 0.41 dB gives exactly the same number by the mismatch-loss route, which is the check that the two descriptions are one. The corresponding return loss is −20 log₁₀(0.3) = 10.46 dB. Now contrast a genuinely lossy device with the same transmission: an ideal 3.01 dB attenuator, properly matched, has S₁₁ = S₂₂ = 0 and S₂₁ = S₁₂ = 10^(−3.01/20) = 0.7071, so |S₁₁|² + |S₂₁|² = 0 + 0.5 = 0.5, not 1 — and the deficit of 0.5 is exactly the fraction of the power turned into heat. So the unitarity sum is a dissipation meter: if |S₁₁|² + |S₂₁|² = 1 the device absorbs nothing, and however far short of 1 it falls is how much it absorbs. That single check catches the commonest error in an S-parameter question, which is to assume that a small S₂₁ means a lossy part when it may mean a badly matched one, and the two are cured by completely different means — a matching network in the first case and a better material in the second.

Rectangular and Circular Waveguides: Cutoff Decides Everything

A hollow metal pipe has only one conductor, and a TEM wave requires two, so a waveguide cannot carry the mode a coaxial line carries. What it carries instead are TE modes, with no longitudinal electric field, and TM modes, with no longitudinal magnetic field, each existing only above its own cutoff frequency. For a rectangular guide of internal dimensions a × b with a > b, f_c(mn) = (c/2)√[(m/a)² + (n/b)²], and the dominant mode — the one with the lowest cutoff — is TE₁₀ at f_c = c/(2a), set by the broad dimension alone. Above cutoff the guide propagates with β = (ω/c)√[1 − (f_c/f)²], and three quantities follow from that square root: the guide wavelength λ_g = λ₀/√[1 − (f_c/f)²] is always longer than free space, the phase velocity v_p = c/√[1 − (f_c/f)²] always exceeds c, and the group velocity v_g = c√[1 − (f_c/f)²] always falls short of it, with v_p v_g = c² exactly — the same identity as the plasma of the previous chapter, and for the same reason, since the dispersion has the same form ω² = ω_c² + c²β².

🎯 WR-90 at 10 GHz, and why b is deliberately less than half of a

Take the X-band standard guide WR-90, a = 22.86 mm and b = 10.16 mm, and compute the first four cutoffs with c = 3 × 10⁸ m/s. TE₁₀: c/2a = 6.562 GHz. TE₂₀: c/a = 13.123 GHz. TE₀₁: c/2b = 14.764 GHz. TE₁₁: (c/2)√[(1/0.02286)² + (1/0.01016)²] = 1.5 × 10⁸ × √(1913.6 + 9687.5) = 1.5 × 10⁸ × 107.71 = 16.156 GHz. So the single-mode band runs from 6.562 to 13.123 GHz, a full octave, and it is bounded above by TE₂₀ and not by TE₀₁. That is a design choice, and it is why b < a/2: making b smaller than half of a pushes TE₀₁ above TE₂₀ and lets the single-mode band be as wide as a 2:1 ratio allows, which is the widest it can ever be. WR-90's rated band is 8.2 to 12.4 GHz, comfortably inside the theoretical 6.56–13.12, with margin at the bottom because v_g → 0 near cutoff and margin at the top to keep away from TE₂₀. Now operate it at f = 10 GHz. The square-root factor is √[1 − (6.562/10)²] = √(1 − 0.4306) = √0.5694 = 0.7546. Then k₀ = 2πf/c = 209.44 rad/m, so β = 158.05 rad/m and λ_g = 2π/β = 39.76 mm against a free-space λ₀ = 30.0 mm — the guide wavelength is a third longer, which is the number that sets the physical length of a λ/4 matching section inside the guide and is the most common slip in a waveguide-component question. Velocities: v_p = ω/β = 3.976 × 10⁸ m/s = 1.325c and v_g = 0.7546c = 2.264 × 10⁸ m/s, with the product 9.00 × 10¹⁶ = c² ✓. Wave impedance: Z_TE = η₀/0.7546 = 499.2 Ω, well above free space's 376.7 Ω — and note that Z_TE → ∞ at cutoff, which is another way of saying that at cutoff the mode stores energy and transports none.

A circular waveguide works the same way with Bessel functions in place of sinusoids, so the cutoff of each mode is set by a root of a Bessel function rather than by an integer. For a guide of radius A, f_c = p c/(2πA) where p is the relevant root: the dominant mode is TE₁₁ with p'₁₁ = 1.8412, and the next mode up is TM₀₁ with p₀₁ = 2.4049. Take A = 10 mm, so c/(2πA) = 4.7746 GHz. Then f_c(TE₁₁) = 1.8412 × 4.7746 = 8.791 GHz and f_c(TM₀₁) = 2.4049 × 4.7746 = 11.484 GHz. The single-mode bandwidth is therefore a ratio of only 2.4049/1.8412 = 1.306, against the rectangular guide's 2.0 — and that comparison is the whole reason rectangular guide is the industrial default while circular guide is used only where its own advantages dominate, such as rotary joints, where the axial symmetry lets one section turn inside another, and long low-loss runs, where the very low attenuation of the TE₀₁ mode is worth the trouble of suppressing everything else.

Light Propagation in Optical Fibres: Numerical Aperture and the V Number

An optical fibre is a dielectric waveguide: a core of index n₁ surrounded by a cladding of slightly lower index n₂, so a ray striking the interface beyond the critical angle θ_c = arcsin(n₂/n₁) is totally internally reflected and stays in the core. Two numbers govern it. The numerical aperture NA = √(n₁² − n₂²) fixes how steeply a ray may enter and still be guided; the half-angle of the acceptance cone in air is θ_a = arcsin(NA). And the normalised frequency or V number, V = (2πa/λ)NA for core radius a, fixes how many modes the fibre supports: the fibre is single-mode when V < 2.405, the first zero of the Bessel function J₀, and for a large V a step-index fibre carries roughly V²/2 modes. Those two numbers pull in opposite directions, and that tension is the whole design problem: a large NA collects light easily from a cheap source but admits many modes, while a small NA gives single-mode operation at the cost of a core so small that alignment becomes a precision task.

🎯 The same fibre design carries 1000 modes at 50 µm and 1 mode at 8 µm

Multimode fibre. Take n₁ = 1.48 and n₂ = 1.46. Then NA = √(1.48² − 1.46²) = √(2.1904 − 2.1316) = √0.0588 = 0.2425, so the acceptance half-angle is arcsin(0.2425) = 14.03° — a full acceptance cone of 28.07°, which is wide enough to couple to an LED. The critical angle inside is arcsin(1.46/1.48) = 80.57°, so guided rays graze the interface at within 9.4° of parallel. The relative index difference is Δ = (n₁ − n₂)/n₁ = 0.0135, and the standard approximation NA ≈ n₁√(2Δ) = 1.48 × 0.1644 = 0.2433 agrees with the exact 0.2425 to better than half a per cent, which is the check on both. With a 50 µm core (a = 25 µm) at λ = 850 nm, V = (2π × 25/0.85) × 0.2425 = 184.8 × 0.2425 = 44.8, so the fibre carries about V²/2 = 1004 modes. And that is the problem with it. Different modes take different path lengths, so a pulse spreads by the difference between the fastest and slowest ray: Δτ/L ≈ (n₁ − n₂)/c = 0.02/(3 × 10⁸) = 66.7 ns per km. Taking the usual criterion B ≈ 1/(2Δτ), a one-kilometre link is limited to about 7.5 Mbit/s — regardless of how good the source and detector are, because this is a geometric limit and not a noise limit. Single-mode fibre. Now use a much smaller index step, n₁ = 1.4500 and n₂ = 1.4462, giving NA = √(2.10250 − 2.09149) = √0.011006 = 0.1049. With a = 4.1 µm at λ = 1310 nm, V = (2π × 4.1/1.31) × 0.1049 = 19.66 × 0.1049 = 2.06 < 2.405 ✓ — single mode. The cutoff wavelength below which a second mode appears is λ_c = 2πa·NA/2.405 = 1124 nm, so this fibre is single-mode anywhere above about 1.12 µm and multimode below it, which is why a fibre is specified with a wavelength and never without one. Intermodal dispersion is now identically zero because there is only one mode, and the remaining limits — chromatic dispersion and attenuation — are orders of magnitude weaker. That is the whole reason long-haul telecommunications is single-mode and the 7.5 Mbit/s above became gigabits: not a better laser, but the elimination of an entire dispersion mechanism by making the waveguide too small to support it.

Dipoles, Monopoles and Linear Antenna Arrays

An antenna converts a guided wave into a radiated one, and the parameter that says how well is the radiation resistance R_r, defined so that P_rad = ½ I₀² R_r for a feed current amplitude I₀. For an infinitesimal or Hertzian dipole of length dl, R_r = 80π²(dl/λ)², and the (dl/λ)² is the whole story of why small antennas are hard: at dl = λ/50, R_r = 80π²/2500 = 0.316 Ω, which is comparable with or smaller than the ohmic resistance of the wire itself, so most of the input power heats the conductor instead of leaving. Grow the element to a half-wave dipole and R_r = 73 Ω with a reactance of about +j42.5 Ω at exactly λ/2, so the standard practice is to shorten it slightly to cancel that reactance and land near a real 70 Ω — which is why a 75 Ω feeder exists. Its directivity is 1.64, that is 2.15 dBi. A quarter-wave monopole over a large conducting ground plane is the same antenna seen in a mirror: the image supplies the missing half, the pattern above ground is identical, but only half the power is radiated for the same current, so R_r halves to 36.5 Ω and the directivity doubles to 3.28, that is 5.15 dBi. Both changes come from the same fact and neither is independent of the other, which is worth seeing once, because a candidate who memorises the four numbers separately has four chances to misremember and one who understands the mirror has none.

Two more per-antenna quantities close the single-element story. The effective length of a receiving antenna is the ratio of open-circuit voltage to incident field, and for a half-wave dipole it is λ/π, that is 0.6366 of the physical length — at 100 MHz where λ = 3 m, a 1.5 m dipole has an effective length of 0.955 m, and the shortfall is because the current tapers to zero at the ends rather than being uniform. The effective aperture is A_e = λ²D/(4π), so the same dipole at 100 MHz presents 9 × 1.64/(4π) = 1.17 m² of capture area — far larger than the wire's physical footprint, which is the fact that makes a wire antenna work at all and is why aperture, not size, is the right way to think about receiving.

🧠 A two-element array, computed as coordinates and angles with nothing drawn

For a linear array the total field is the single-element pattern multiplied by the array factor, and for two equal-amplitude elements AF = 2 cos(ψ/2) with ψ = βd cos θ + α, where d is the spacing, α the feed phase difference, and θ is measured from the array axis. Everything below is exact arithmetic on that one expression. Broadside, d = λ/2, α = 0. Then βd = π and ψ = π cos θ. The maximum is where ψ = 0, that is θ = 90° — perpendicular to the axis, hence the name. Nulls are where cos(ψ/2) = 0, that is ψ/2 = ±π/2, so cos θ = ±1 and θ = 0° and 180°: the nulls point straight along the array. Half-power points satisfy cos(ψ/2) = 1/√2, so ψ/2 = ±π/4, giving cos θ = ±0.5 and θ = 60° and 120°. The half-power beamwidth is therefore exactly 60°. End-fire, d = λ/2, α = −βd = −π. Now ψ = π cos θ − π, and the maximum is at ψ = 0, that is cos θ = 1 and θ = 0° — along the axis. Nulls need π cos θ − π = ±π, so cos θ = 2 (impossible) or cos θ = 0, giving θ = 90°. Half-power needs π cos θ − π = ±π/2, so cos θ = 1.5 (impossible) or cos θ = 0.5, giving θ = 60°; the pattern is rotationally symmetric about the axis, so the beam runs from −60° to +60° and the HPBW is 120°. So steering the same two elements from broadside to end-fire has doubled the beamwidth without touching the hardware — a result that surprises people who expect end-fire to be the sharper mode, and one worth knowing because questions test exactly that intuition. More elements. For N equal elements at d = λ/2 broadside, nulls occur at cos θ = 2n/N, so for N = 8 the first null is at arccos(0.25) = 75.52°, a first-null beamwidth of 2(90 − 75.52) = 28.96°, and the standard approximation HPBW ≈ 0.886λ/(Nd) gives 0.886/4 = 0.2215 rad = 12.69°, satisfyingly a little under half the FNBW. Directivity is about 2Nd/λ = 8, that is 9.03 dBi. One design limit to carry: at d = λ the two-element broadside array has ψ = 2π cos θ, so maxima appear at ψ = 0 and ±2π, that is at θ = 90° and at θ = 0° and 180° — unwanted grating lobes along the axis. Keeping d ≤ λ/2 is what prevents them, and it is the reason array spacings cluster at half a wavelength rather than being chosen for convenience.

Four antennas and two arrays, with the numbers a question asks for
Structure	Radiation resistance	Directivity	Pattern feature
Hertzian dipole, `dl = λ/50`	`80π²(dl/λ)²` = 0.316 Ω	1.5 = 1.76 dBi	`sin θ` shape; efficiency dominated by wire loss
Half-wave dipole	73 Ω (with `+j42.5 Ω` at exactly `λ/2`)	1.64 = 2.15 dBi	Effective length `λ/π` = 0.955 m at 100 MHz
Quarter-wave monopole over a ground plane	36.5 Ω — exactly half the dipole's	3.28 = 5.15 dBi — exactly double	Upper half-space only; the image supplies the rest
Two elements, `d = λ/2`, broadside	Per element as above	About 2× a single element	Max at `90°`, nulls at `0°`/`180°`, HPBW 60°
Two elements, `d = λ/2`, end-fire	Per element as above	Similar total, differently distributed	Max at `0°`, null at `90°`, HPBW 120° — broader
Eight elements, `d = λ/2`, broadside	Set by the feed network, not by one element	About `2Nd/λ` = 8 = 9.03 dBi	First null `75.52°`, FNBW `28.96°`, HPBW `12.69°`

Exam Craft: One Working Order for Every Guided-Wave Question

The EC paper is 65 questions for 100 marks in 180 minutes, with General Aptitude accounting for 10 questions and 15 marks. A wrong MCQ costs one third of the question's value, so narrow before you answer; MSQ and NAT carry no negative marking, so a transmission-line or waveguide numerical is always worth submitting. The material in this chapter rewards a fixed working order more than any other in the paper, because nearly every question is a substitution chain and the intermediate quantities are shared across question types. Establish the wavelength or the cutoff first, and most of the rest is one formula away.

On a line, get Z₀ and v_p before anything else. Z₀ = √(L/C) and v_p = 1/√(LC); from those, λ = v_p/f, and every length in the question becomes a fraction of λ. Any two of L, C, Z₀, v_p determine all four.
Do the Smith chart as four lines, not as a drawing. Normalise, Γ = (z−1)/(z+1), rotate the phase by −2βl, then z = (1+Γ)/(1−Γ). The check is that |Γ| is unchanged on a lossless line, so the VSWR you compute at one point must hold at every point.
Remember λ/2 returns you to the load and λ/4 inverts about Z₀. These two turn many multi-step problems into one line, and they also catch errors: a λ/2 answer that differs from Z_L is arithmetic gone wrong, not a subtlety.
In a waveguide, compute f_c and then the factor √[1 − (f_c/f)²] once. Every other quantity is that factor multiplying or dividing something: λ_g = λ₀/factor, v_g = c × factor, v_p = c/factor, Z_TE = η₀/factor. Use λ_g, never λ₀, for physical lengths inside the guide.
In a fibre, NA first and then V. NA = √(n₁² − n₂²) gives the acceptance angle directly; V = (2πa/λ)NA decides single-mode against multimode against 2.405, and only then is a mode count or a dispersion figure meaningful. A fibre question with no wavelength stated is incomplete.
In an array, write ψ = βd cos θ + α and set it to 0, then ±π, then ±π/2. Those three give the maximum, the nulls and the half-power points respectively, and they are the only three things an array-pattern question ever asks for.

ℹ️ The conventions and constants in this chapter that change an answer

Four places in this chapter a convention or a constant silently decides the number. c is taken as 3 × 10⁸ m/s throughout, so WR-90's TE₁₀ cutoff comes out 6.562 GHz; using 2.998 × 10⁸ gives 6.557 GHz, and a question demanding three decimals will expect one or the other, so read what it gives you. Rotation direction on the Smith chart: moving towards the generator subtracts from the phase of Γ, which is clockwise on the standard chart, and moving towards the load is anticlockwise; getting this backwards gives an answer with the right magnitude and the wrong sign of reactance. Antenna gain units: 2.15 dBi for a half-wave dipole is referred to an isotropic radiator, whereas gain quoted in dBd is referred to the dipole itself and is therefore 2.15 dB lower for the same antenna. Array angle reference: θ here is measured from the array axis, so broadside is θ = 90°; some texts measure from broadside instead, which swaps every sin and cos in the array factor and turns a correct method into a wrong answer. In each case the fix is the same and takes one line: state the convention next to the formula before you substitute anything into it.

Key takeaways

A lossless line is two numbers, and any two of four give all four. L = 250 nH/m with C = 100 pF/m gives Z₀ = 50 Ω, v_p = 2 × 10⁸ m/s, hence ε_r = 2.25, a delay of 5 ns/m, and for coax b/a = e^(1.25) = 3.49.
The Smith chart is four lines of algebra, and |Γ| never moves. Z_L = 100 Ω on 50 Ω gives Γ = 1/3; a λ/8 walk rotates the phase by −90° to −j0.333 and gives Z_in = 40 − j30 Ω, with VSWR still exactly 2. λ/4 gives Z₀²/Z_L = 25 Ω and λ/2 returns 100 Ω.
Matching is easy at one frequency and hard over a band. A 70.71 Ω quarter-wave transformer matches 100 Ω to 50 Ω perfectly at design, but at half that frequency |Γ| = 1/√17 = 0.243 and VSWR is 1.64; the |Γ| ≤ 0.1 bandwidth is only 36.7 per cent. A shorted shunt stub does it with d = l_s = 0.152λ, that is 45.6 mm each at 1 GHz in air.
S₁₁ is the same Γ, and the unitarity sum is a dissipation meter. A lossless two-port with |S₁₁| = 0.3 must have |S₂₁| = 0.954, showing 0.41 dB of insertion loss while dissipating nothing — the same figure the mismatch-loss formula gives. An ideal 3.01 dB attenuator has |S₁₁|² + |S₂₁|² = 0.5, and the missing half is heat.
A waveguide is one cutoff and one square root. WR-90 (22.86 × 10.16 mm) has TE₁₀ at 6.562 GHz and TE₂₀ at 13.123 GHz, a full-octave single-mode band because b < a/2. At 10 GHz the factor is 0.7546, so λ_g = 39.76 mm against λ₀ = 30 mm, v_p = 1.325c, v_g = 0.755c, v_p v_g = c², and Z_TE = 499 Ω.
Circular guide has a much narrower single-mode band than rectangular. TE₁₁ at 1.8412c/2πA and TM₀₁ at 2.4049c/2πA give a ratio of only 1.306 against rectangular's 2.0 — for A = 10 mm, 8.791 to 11.484 GHz. That is why rectangular is the default and circular is reserved for rotary joints and low-loss runs.
A fibre is NA and V, and V < 2.405 is the whole difference. n₁ = 1.48, n₂ = 1.46 gives NA = 0.2425, acceptance half-angle 14.03°, and at a = 25 µm, λ = 850 nm, V = 44.8 with about 1004 modes and 66.7 ns/km of modal spread — about 7.5 Mbit/s over 1 km. Drop to NA = 0.105 and a = 4.1 µm and V = 2.06: single mode, and that dispersion mechanism is gone entirely.
The monopole's two changes come from one mirror. A half-wave dipole has R_r = 73 Ω and D = 1.64 = 2.15 dBi; a quarter-wave monopole over ground halves the resistance to 36.5 Ω and doubles the directivity to 3.28 = 5.15 dBi, both because only half the space is radiated into. A Hertzian element at λ/50 has only 0.316 Ω, which is why small antennas are inefficient.
An array is one expression, ψ = βd cos θ + α, set to three values. Two elements at d = λ/2 broadside give maxima at 90°, nulls at 0°/180° and HPBW 60°; the same pair in end-fire gives a maximum at 0° and HPBW 120° — broader, not sharper. Eight elements broadside give HPBW 12.69° and about 9.03 dBi, and d > λ/2 invites grating lobes.

Practice questions (8)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

A lossless transmission line of characteristic impedance 75 ohm is 0.1 wavelength long at the operating frequency and its far end is left open circuit. What input impedance does it present?
1. -j103.2 ohm
2. -j54.5 ohm
3. -j127.6 ohm
4. -j75.0 ohm
Show answer
Answer: A — -j103.2 ohm
Put the open circuit into the impedance transformation formula and the load impedance dominates both numerator and denominator, leaving the input impedance as minus j times the characteristic impedance times the cotangent of the electrical length. Here the electrical length is 2 pi x 0.1 radians, that is 36 degrees, so the cotangent is 1/0.7265 = 1.3764 and the answer is -j75 x 1.3764 = -j103.2 ohm, capacitive as any open line shorter than a quarter wavelength must be. The tempting -j54.5 ohm is 75 times the tangent of 36 degrees, which is the magnitude a short-circuited line of the same length would show, with the sign flipped to look capacitive; tangent and cotangent belong to opposite terminations and swapping them is the standard slip. The value -j127.6 ohm divides by the sine instead of the tangent, and -j75.0 ohm simply quotes the characteristic impedance.
A mismatched load on a lossless feeder produces a voltage standing wave ratio of 2.5. What fraction of the incident power is actually delivered to the load?
1. 57.1 %
2. 81.6 %
3. 18.4 %
4. 42.9 %
Show answer
Answer: B — 81.6 %
Invert the standing wave ratio first: the magnitude of the reflection coefficient is (VSWR - 1)/(VSWR + 1) = 1.5/3.5 = 0.4286. Power goes as the square of that amplitude ratio, so the reflected fraction is 0.4286 squared = 0.1837 and the delivered fraction is 1 - 0.1837 = 0.816, that is 81.6 per cent, equivalent to a mismatch loss of 0.88 dB. The tempting 57.1 per cent subtracts the reflection coefficient itself rather than its square, confusing an amplitude ratio with a power ratio, and it understates the delivered power badly because squaring a number below one makes it smaller. The value 18.4 per cent is the reflected fraction, which is what the load does not receive, and 42.9 per cent is the reflection coefficient written as a percentage.
An air-filled rectangular waveguide has a broad internal wall of 2.5 cm and is operated at 9 GHz in the mode of lowest cutoff, with c taken as 3 x 10^8 m/s and the intrinsic impedance of free space as 376.73 ohm. What is the wave impedance of that mode inside the guide?
1. 251 ohm
2. 281 ohm
3. 505 ohm
4. 377 ohm
Show answer
Answer: C — 505 ohm
The mode of lowest cutoff in a rectangular guide is the transverse electric mode set by the broad wall alone, with cutoff c/(2a) = 3 x 10^8/0.05 = 6 GHz. Form the factor that every guide quantity depends on, the square root of one minus the square of the cutoff over the operating frequency: 1 - (6/9) squared = 1 - 0.4444 = 0.5556, whose square root is 0.7454. For a transverse electric mode the wave impedance is the free-space impedance divided by that factor, 376.73/0.7454 = 505 ohm. The tempting 281 ohm multiplies by the factor instead of dividing, which is the rule for a transverse magnetic mode; the two modes move the impedance in opposite directions from 377 ohm and the direction is the whole content of the question. The value 251 ohm multiplies by the ratio 6/9 without the square root, and 377 ohm is the free-space value unchanged.
A step-index optical fibre has a core radius of 3.5 micrometres and a numerical aperture of 0.12, and is single-mode when its normalised frequency is below 2.405. Above what wavelength does it carry only one mode?
1. 549 nm
2. 1310 nm
3. 2195 nm
4. 1097 nm
Show answer
Answer: D — 1097 nm
The normalised frequency is 2 pi a times the numerical aperture divided by the wavelength, and it falls as the wavelength rises, so the single-mode condition sets a lower bound on wavelength. Putting the normalised frequency equal to 2.405 gives the cutoff wavelength 2 pi x 3.5 x 10^(-6) x 0.12/2.405 = 2.639 x 10^(-6)/2.405 = 1.097 x 10^(-6) m, that is 1097 nm; above that the fibre carries one mode and below it a second appears. The tempting 2195 nm uses the core diameter where the formula asks for the radius, doubling the answer, and the fibre would then be wrongly declared single-mode at wavelengths where it is not. The value 549 nm halves the answer by writing pi a instead of 2 pi a, and 1310 nm is a standard telecommunications wavelength quoted from memory rather than computed from the numbers given.
Two equal-amplitude isotropic radiators are fed in phase and spaced one full wavelength apart along an axis, with the polar angle measured from that axis. Besides the broadside direction at 90 degrees, where does the array factor reach a maximum?
1. At 0 degrees and 180 degrees, that is straight along the array axis
2. At 45 degrees and 135 degrees
3. At 60 degrees and 120 degrees
4. Nowhere else; broadside is the only maximum the pattern has
Show answer
Answer: A — At 0 degrees and 180 degrees, that is straight along the array axis
For two equal elements the array factor is twice the cosine of half of psi, where psi is the spacing in radians times the cosine of the polar angle plus the feed phase difference. With in-phase feeding and a spacing of one wavelength, psi = 2 pi cos(theta), and the factor is a maximum whenever psi is zero or plus or minus 2 pi. Zero gives cos(theta) = 0 and hence broadside at 90 degrees, while plus and minus 2 pi give cos(theta) = plus or minus 1 and hence maxima at 0 and 180 degrees, straight along the axis. Those extra maxima are grating lobes, and they radiate as strongly as the wanted beam. The tempting answer that broadside is the only maximum is correct only while the spacing stays at or below half a wavelength, which is exactly why array spacings cluster there rather than being chosen for mechanical convenience.
On a network analyser a reciprocal microwave component that dissipates no power at all shows a forward transmission magnitude of 0.8, its two ports being terminated in the reference impedance. What return loss appears at its input?
1. 1.94 dB
2. 4.44 dB
3. 8.87 dB
4. 2.22 dB
Show answer
Answer: B — 4.44 dB
For a lossless two-port the scattering matrix is unitary, so the squared magnitudes in each column sum to one: the squared input reflection magnitude plus the squared forward transmission magnitude equals one. With the transmission at 0.8 the reflection magnitude is the square root of 1 - 0.64 = 0.6. Return loss is minus twenty times the logarithm of that reflection magnitude, which is 4.44 dB. The tempting 1.94 dB is minus twenty log of 0.8, which is the insertion loss read from the transmission term, and it answers a different question: insertion loss says how much of the wave gets through, return loss says how much comes back, and on a lossless device the two are two views of the same mismatch. The value 2.22 dB uses ten log where twenty is required, and 8.87 dB doubles the correct figure.
An air-filled rectangular waveguide has internal dimensions of 3.2 cm by 1.4 cm and is driven at 6 GHz, with c taken as 3 x 10^8 m/s and the intrinsic impedance of free space as 376.73 ohm. Which of the following statements about it are correct?
1. The guide wavelength is 80.1 mm, which is longer than the 50 mm free-space wavelength.
2. Only the mode of lowest cutoff propagates at 6 GHz, the next cutoff lying at 9.375 GHz.
3. The group velocity in the guide is 4.81 x 10^8 m/s.
4. The wave impedance of that mode is 235 ohm, being the free-space impedance multiplied by the square root of one minus the squared frequency ratio.
Show answer
Answer: A — The guide wavelength is 80.1 mm, which is longer than the 50 mm free-space wavelength.; B — Only the mode of lowest cutoff propagates at 6 GHz, the next cutoff lying at 9.375 GHz.
The lowest cutoff is set by the broad wall alone, c/(2a) = 3 x 10^8/0.064 = 4.6875 GHz, and the factor that governs every other quantity is the square root of 1 - (4.6875/6) squared = 0.6242. The free-space wavelength at 6 GHz is 50 mm, so the guide wavelength is 50/0.6242 = 80.1 mm, always longer than free space because the wave zigzags. The next cutoffs are twice the broad-wall one at 9.375 GHz and c/(2b) = 10.71 GHz, so at 6 GHz only one mode propagates. The figure 4.81 x 10^8 m/s is the phase velocity, c divided by the factor, and it exceeds c legitimately; the group velocity is c multiplied by the factor, 1.87 x 10^8 m/s, and their product is exactly c squared. For a transverse electric mode the wave impedance is the free-space value divided by the factor, 603 ohm, so multiplying instead of dividing gives 235 ohm and the wrong side of 377 ohm entirely.
A lossless line of characteristic impedance 60 ohm is exactly half a wavelength long at the operating frequency and is terminated in a load of 30 + j40 ohm. What impedance appears at its input?
1. 43.2 - j57.6 ohm
2. 0.5 + j0.667 ohm
3. 60 ohm
4. 30 + j40 ohm
Show answer
Answer: D — 30 + j40 ohm
At half a wavelength the electrical length is pi radians, so its tangent is zero and the impedance transformation formula collapses to the load impedance itself: a half-wavelength of lossless line is invisible, whatever the load and whatever the characteristic impedance. Seen on the reflection coefficient the reason is that half a wavelength rotates the phase by a full 360 degrees while leaving the magnitude untouched, returning the point to where it started. The tempting 43.2 - j57.6 ohm is the quarter-wavelength answer, the characteristic impedance squared divided by the load, and it is right for the wrong length; the quarter-wave section inverts the load about 60 ohm and the half-wave section undoes that inversion a second time. The value 0.5 + j0.667 ohm is the load normalised to the characteristic impedance and never converted back, and 60 ohm would require the load to be matched in the first place.