Types of Transmission Lines Compared

Fig 1 — The transmission line family in cross-section. Blue lines show the electric field of the dominant mode

Introduction to Transmission-Line Classification

Part 1 showed why any conductor longer than a small fraction of a wavelength must be treated as a transmission line. Engineers have invented many physical forms of line, each a different compromise between loss, bandwidth, power handling, size, cost and ease of manufacture. This article tours the family, derives the key formula for each member, and ends with a practical guide to choosing between them. The four questions below sort every line in Fig 1.

How many conductors?Two or more conductors can carry a TEM wave down to DC. A single hollow conductor — a waveguide — cannot, and has a cutoff frequency.

Homogeneous dielectric?If the field sees one uniform medium (coax, stripline) the wave is pure TEM; if it sees air and substrate (microstrip, CPW) it is quasi-TEM and slightly dispersive.

Shielded or open?Closed lines (coax, stripline, waveguide) neither radiate nor pick up interference. Open lines (two-wire, microstrip) do both, more so at higher frequency.

Balanced or unbalanced?Two-wire and twisted pair carry equal and opposite currents on identical conductors (balanced); coax and microstrip use a ground as the return (unbalanced).

Why a hollow waveguide cannot carry a TEM wave

For a TEM wave, E_z = H_z = 0, and Maxwell's equations reduce in the cross-section to those of electrostatics: the transverse field is the gradient of a potential Φ(x, y) that satisfies Laplace's equation, ∇²Φ = 0 (the wave-equation article shows the reduction). In a hollow pipe there is only one conductor, so Φ is the same constant everywhere on the boundary. The only solution of Laplace's equation with a constant boundary value is Φ = constant inside, which gives E = −∇Φ = 0. A TEM wave needs at least two separate conductors held at different potentials.

So lines divide naturally into TEM and quasi-TEM lines (two or more conductors, usable from DC up to the frequency where a higher-order mode appears) and waveguides (TE and TM modes only, usable only between a lower cutoff and the onset of the next mode). Fig 2 shows the typical useful frequency range of each type.

Fig 2 — Approximate useful frequency range of each line type (logarithmic axis). Waveguide ranges span the whole series of standard sizes; any one size covers less than an octave

Coaxial Transmission Lines

A coaxial line is a centre conductor of radius a inside a cylindrical outer conductor of inner radius b, separated by a dielectric of permittivity εr — air, solid or foamed polyethylene (PE), or PTFE. The outer conductor completely encloses the field, so a well-made coax neither radiates nor picks up interference, and it can be bent and routed anywhere. It is the universal connection between RF instruments, radios and antennas, from audio to above 100 GHz.

Derivation — Z₀ of coax from Gauss's and Ampère's laws

Put charge q per metre on the centre conductor. By Gauss's law the field between the conductors is radial, E = q/(2πεr), and the voltage between them is the integral of E from a to b. The current I on the centre conductor produces a circular magnetic field H = I/(2πr), and the flux per metre between the conductors is the integral of μH:

V = \int_{a}^{b} \frac{q}{2 π ε r} d r = \frac{q}{2 π ε} \ln \frac{b}{a} \Rightarrow C^{'} = \frac{q}{V} = \frac{2 π ε}{\ln (b / a)}

Φ = \int_{a}^{b} \frac{μ I}{2 π r} d r = \frac{μ I}{2 π} \ln \frac{b}{a} \Rightarrow L^{'} = \frac{Φ}{I} = \frac{μ}{2 π} \ln \frac{b}{a}

The characteristic impedance and the phase velocity then follow from Part 1:

Coaxial line

Z_{0} = \sqrt{\frac{L^{'}}{C^{'}}} = \frac{η_{0}}{2 π \sqrt{ε_{r}}} \ln \frac{b}{a} \approx \frac{60}{\sqrt{ε_{r}}} \ln \frac{b}{a} v_{p} = \frac{1}{\sqrt{L^{'} C^{'}}} = \frac{c}{\sqrt{ε_{r}}}

η₀ = √(μ₀/ε₀) ≈ 376.7 Ω, so η₀/2π ≈ 60 Ω. The wave travels at the speed of light in the dielectric, independent of the dimensions.

Why 50 Ω? The optimum coax

Z₀ depends only on the ratio b/a, so for a given outer size the designer chooses b/a. Three different "best" ratios exist:

Derivation — minimum attenuation, maximum power and maximum voltage

Minimum conductor loss. The conductor loss is α_c = R′/(2Z₀) with R′ = (R_s/2π)(1/a + 1/b). For a fixed outer radius b and x = b/a:

α_{c} \propto \frac{1 + x}{\ln x} \frac{d}{d x} (\frac{1 + x}{\ln x}) = 0 \Rightarrow \ln x = \frac{1 + x}{x} \Rightarrow x = 3.591

which gives Z₀ = 60 ln 3.591 = 76.7 Ω in air — the origin of the 75 Ω standard for cable TV, where loss over long runs is what matters.

Maximum power. For a fixed breakdown field E_max at the centre conductor, V_max = E_maxa ln x and P = V²/(2Z₀) ∝ a² ln x ∝ (ln x)/x². Its maximum is at ln x = 1/2, x = 1.649, Z₀ = 30 Ω in air.

Maximum voltage. V_max ∝ (ln x)/x is largest at x = e = 2.718, Z₀ = 60 Ω in air.

Fig 3 — Conductor loss and power capacity of air-filled coax against b/a, each normalised to its best value. 50 Ω (b/a = 2.30) sits between the 30 Ω power optimum and the 77 Ω loss optimum, close to the best for both

relative 1/loss (minimum loss at b/a = 3.59, 77 Ω)relative power capacity (maximum at b/a = 1.65, 30 Ω)

50 Ω is the compromise: at b/a = 2.30 the air line keeps 86% of the best power capacity and has only 11% more loss than the 77 Ω optimum (Fig 3). It became the standard for RF and microwave equipment; 75 Ω survives in video and cable TV. With solid PE (εr = 2.25) the same b/a = 2.30 gives only 33 Ω, so 50 Ω PE cables use b/a ≈ 3.5, close to their own loss optimum.

Higher-order modes in coax

Coax carries its TEM wave down to DC, but above a certain frequency the TE₁₁ waveguide mode can also propagate inside it. Its cutoff occurs approximately when the mean circumference equals one wavelength in the dielectric:

Coax TE₁₁ cutoff (approximate)

λ_{c} \approx π (a + b) \sqrt{ε_{r}} f_{c} \approx \frac{c}{π (a + b) \sqrt{ε_{r}}}

This sets the maximum frequency of every coaxial cable and connector: the smaller the coax, the higher it can go. SMA (4.1 mm outer) is rated to 18–26.5 GHz, 3.5 mm to 34 GHz, 2.4 mm to 50 GHz and 1.0 mm to 110 GHz.

Common coaxial cable families

Real cables trade loss against size, flexibility and cost. Loss rises roughly as √f from the conductors plus a term proportional to f from the dielectric (Part 8), and a larger cable has lower conductor loss because R′ ∝ 1/a + 1/b. The table gives typical attenuation for widely used 50 Ω cables; always check the manufacturer's data for a specific product.

Cable	Outer diameter	Dielectric	Loss at 100 MHz	Loss at 1 GHz	Typical use
RG-174	2.8 mm	Solid PE	≈ 29 dB/100 m	≈ 100 dB/100 m	Short internal jumpers, GPS pigtails
RG-58	5.0 mm	Solid PE	≈ 16 dB/100 m	≈ 65 dB/100 m	Lab cables, low-power HF/VHF
RG-213	10.3 mm	Solid PE	≈ 7 dB/100 m	≈ 26 dB/100 m	HF/VHF transmitter feeders
LMR-400 type	10.3 mm	Foam PE	≈ 4 dB/100 m	≈ 13.5 dB/100 m	Cellular, Wi-Fi and VHF/UHF antenna runs
½″ corrugated hardline	16 mm	Foam PE	≈ 2.2 dB/100 m	≈ 7.5 dB/100 m	Base-station and broadcast feeders
0.141″ semi-rigid	3.6 mm	Solid PTFE	≈ 12 dB/100 m	≈ 37 dB/100 m	Microwave assemblies to 18–26 GHz

Foam dielectrics (εr ≈ 1.3–1.5) cut the dielectric loss and raise the velocity factor to 0.8–0.88; for the same impedance they also allow a thicker centre conductor, which cuts the conductor loss. That is why a foam LMR-400-type cable has half the loss of solid-PE RG-213 of the same diameter. Flexible cables add a braid, whose small gaps leak a little energy and add loss; semi-rigid and hardline use solid or corrugated outer conductors for the best shielding and phase stability.

Parallel-Wire Transmission Lines

The parallel-wire (two-wire, twin-lead or ladder) line is two identical round conductors of radius a with centre spacing D. It is the oldest form of transmission line, the one usually drawn in textbooks, and still the best low-loss feeder for HF antennas: its loss is very low because most of the field is in air, and its high impedance (300–600 Ω) means low current for a given power.

Derivation — Z₀ of a two-wire line

The exact capacitance of two parallel cylinders, found by replacing each cylinder with an equivalent line charge (the method of images), is C′ = πε/cosh⁻¹(D/2a). Since the line is TEM in a uniform medium, L′C′ = με, so L′ = (μ/π) cosh⁻¹(D/2a) and

Parallel-wire line

Z_{0} = \frac{η_{0}}{π \sqrt{ε_{r}}} \cosh^{- 1} (\frac{D}{2 a}) \approx \frac{120}{\sqrt{ε_{r}}} \ln \frac{D}{a}

The approximation cosh⁻¹x ≈ ln 2x holds for D ≫ a; it gives the familiar Z₀ ≈ (120/√εr) ln(D/a).

A two-wire line is balanced: the two conductors are identical and carry equal and opposite currents, so their far fields cancel and the line radiates little — provided the currents stay balanced. Any asymmetry (a nearby metal object, unequal loading, or a direct connection to an unbalanced coax) creates a common-mode current that radiates and picks up noise. Connecting a balanced line to an unbalanced one requires a balun, introduced in Part 12. The other limitations are that the field extends well beyond the line, so it must be kept away from metal, and that it cannot be used on a circuit board.

Twisted-Pair Transmission Lines

A twisted pair is a two-wire line made of insulated wires twisted around each other. The insulation sets the spacing (D is roughly the insulated diameter) and partly fills the field with dielectric, so the same two-wire formula applies with an effective permittivity of about 1.5–2.5 that accounts for the mix of insulation and air. Typical data cables are designed for Z₀ = 100 Ω ± 15%; the older telephone pairs are 100–150 Ω.

The twist is what makes twisted pair useful at all. External magnetic fields induce voltages in each half-twist that are equal and opposite in the next one, so interference cancels, and the pair's own field reverses every half-twist, so it couples very little to neighbouring pairs (crosstalk). In a four-pair Ethernet cable each pair has a different twist rate to keep their coupling low. Because the line is balanced, it is driven differentially — which also makes it immune to common-mode noise picked up along the run.

Twisted pair is the transmission line of wired data: Ethernet (Cat 5e to 1 Gb/s, Cat 6A to 10 Gb/s with 500 MHz bandwidth, Cat 8 to 40 Gb/s with 2 GHz), telephone and DSL, USB, CAN and RS-485. Its loss rises steeply above a few hundred megahertz, so it is not used for RF signal paths.

Microstrip Transmission Lines

Microstrip is a strip conductor on top of a dielectric substrate with a ground plane underneath. It is the dominant line on RF circuit boards because it is made by ordinary PCB etching and components can be soldered directly onto it. Part of its field is in the air above the board, so the wave is quasi-TEM: it travels at c/√εeff with (εr + 1)/2 < εeff < εr, and εeff rises slightly with frequency. Its impedance is set mainly by the ratio of strip width to substrate height, w/h; a 50 Ω line on 1.6 mm FR-4 is about 3.1 mm wide.

Its weaknesses follow from being open on one side: it radiates at discontinuities, couples to nearby traces, and is limited at high frequency by surface waves and higher-order modes. Part 9: Microstrip Transmission Lines covers microstrip in full — fields, εeff, the design equations, dispersion and losses.

Stripline Transmission Lines

Stripline buries the strip in the middle of the substrate, between two ground planes a distance b apart. The field is completely inside one homogeneous dielectric, so stripline carries a pure TEM wave: v_p = c/√εr exactly, with no dispersion, and no radiation. It is used for filters and couplers that need precise, frequency-independent electrical lengths, and for routing RF and high-speed digital signals on the inner layers of multilayer boards.

Stripline impedance (Pozar, zero thickness, accuracy about 1%)

Z_{0} = \frac{30 π}{\sqrt{ε_{r}}} \frac{b}{W_{e} + 0.441 b}

W_e = W for W/b > 0.35, and W_e = W − (0.35 − W/b)²b for narrower strips. For design, W/b = 30π/(√εr Z₀) − 0.441 when √εr Z₀ < 120 Ω.

The price is that the strip is inaccessible: components must be connected through vias, the two ground planes must be tied together by rows of vias to suppress a parasitic parallel-plate mode, and for the same impedance the strip is narrower than a microstrip (more conductor loss). Part 11: Stripline and Grounded Coplanar Waveguide develops stripline in detail, including asymmetric stripline.

Coplanar Waveguides (CPW)

In a coplanar waveguide the signal strip and two ground planes all lie on the same surface of the substrate, separated by gaps of width s. Despite its name it is not a waveguide in the hollow-pipe sense: it has three conductors and carries a quasi-TEM mode. Because the grounds are right next to the strip, shunt components can be connected to ground without vias, and the line is easy to probe — so CPW is the standard for MMICs and on-wafer measurements, and is widely used at millimetre-wave frequencies.

The CPW impedance from conformal mapping

A conformal transformation maps the CPW cross-section onto a parallel-plate capacitor whose width-to-spacing ratio is K(k)/K(k′), where K is the complete elliptic integral of the first kind. For a substrate much thicker than the slot width, half the field is in the substrate and half in air, so εeff ≈ (εr + 1)/2 and

CPW on a thick substrate

Z_{0} = \frac{30 π}{\sqrt{ε_{eff}}} \frac{K (k^{'})}{K (k)} k = \frac{w}{w + 2 s}, k^{'} = \sqrt{1 - k^{2}}, ε_{eff} \approx \frac{ε_{r} + 1}{2}

k is set by the strip width w and slot width s; the impedance depends on their ratio, not on the substrate thickness (if h ≫ w + 2s).

CPW has two design dimensions (w and s), so the same impedance can be achieved with a narrow line and narrow gaps or a wide line and wide gaps — a freedom microstrip does not have. Its weaknesses are an unwanted slot-line mode, excited if the two ground planes are at different potentials (prevented by air bridges or bond wires across the strip), and higher conductor loss for narrow gaps. Part 10: Coplanar Waveguide (CPW) treats CPW in full.

Grounded Coplanar Waveguides (GCPW)

Grounded CPW (also called conductor-backed CPW or CPW-G) adds a ground plane on the back of the substrate and stitches the top ground planes to it with rows of vias. It is effectively a hybrid of microstrip and CPW: the field couples both sideways to the coplanar grounds and down to the bottom ground. The bottom ground gives mechanical and thermal robustness and a solid reference, the top grounds give easy component grounding and excellent isolation between adjacent lines, and the via fences suppress parallel-plate and surface-wave modes.

GCPW is the line of choice for RF boards above about 10 GHz, for 5G and radar modules, and for the transitions to connectors and to antennas. Its impedance depends on w, s and h together, and the via spacing must be well below a quarter wavelength (typically below λ_g/10). Part 11: Stripline and Grounded Coplanar Waveguide covers GCPW design, via fences and the trade-offs against microstrip and stripline.

Rectangular Waveguides

A rectangular waveguide is a hollow metal pipe of inner dimensions a × b (a > b, usually a ≈ 2b). With only one conductor it cannot carry TEM, but it can carry TE (transverse electric, E_z = 0) and TM (transverse magnetic, H_z = 0) modes, each of which propagates only above its own cutoff frequency.

Derivation — cutoff frequency and guide wavelength

Inside the guide each field component satisfies the wave equation, and the boundary condition that tangential E vanishes on the walls forces the transverse variation to be a whole number of half-sinusoids across each dimension, with wavenumbers k_x = mπ/a and k_y = nπ/b. The propagation constant along the guide is what remains of the free-space wavenumber k:

β_{m n} = \sqrt{k^{2} - {(\frac{m π}{a})}^{2} - {(\frac{n π}{b})}^{2}}

β is real (a propagating wave) only when k exceeds the transverse wavenumber. Setting β = 0 gives the cutoff frequency, and above it the wave travels with a guide wavelength longer than in free space:

Rectangular waveguide

f_{c, m n} = \frac{c}{2 \sqrt{ε_{r}}} \sqrt{{(\frac{m}{a})}^{2} + {(\frac{n}{b})}^{2}} λ_{g} = \frac{λ}{\sqrt{1 - (f_{c} / f)^{2}}}

TE_mn exists for m, n ≥ 0 (not both zero); TM_mn needs m, n ≥ 1. The dominant mode is TE₁₀ with f_c = c/(2a√εr), independent of b.

The TE₁₀ mode has one half-sine variation of E across the broad wall and none across the narrow wall (Fig 1). Single-mode operation runs from f_c,10 to the next mode, TE₂₀ at twice f_c,10 (for a ≥ 2b). Standard guides are used over roughly 1.25–1.9 f_c, avoiding the high loss and strong dispersion near cutoff. The wave impedance of a TE mode, Z_TE = η/√(1 − (f_c/f)²), is higher than η and frequency dependent, so waveguide impedance needs more careful definition than for TEM lines.

Fig 4 — Dispersion of WR-90 (TE₁₀, f_c = 6.56 GHz): normalised phase constant β/k and group velocity v_g/c against frequency, compared with a TEM line. The recommended band is 8.2–12.4 GHz

WR-90 TE₁₀: β/k = v_g/c = √(1 − (f_c/f)²)TEM line (air)

Waveguide's great advantages are very low loss (no centre conductor and no solid dielectric: WR-90 loses about 0.11 dB/m at 10 GHz, compared with about 1 dB/m for good flexible coax) and very high power handling (hundreds of kilowatts peak in WR-90). Its disadvantages are size and weight at low frequency, the narrow bandwidth of each size, and the cost of precision machining. It dominates in radar transmitters, satellite ground stations, high-power microwave heating and at millimetre-wave frequencies, where standard sizes go down to WR-1 (0.254 mm wide) for 750–1100 GHz.

Circular Waveguides

A circular waveguide is a round pipe of radius a. Its modes have Bessel-function patterns, and their cutoff frequencies are set by the roots of the Bessel functions or their derivatives:

Circular waveguide cutoffs

f_{c, {TE}_{n m}} = \frac{p_{n m}^{'} c}{2 π a \sqrt{ε_{r}}} f_{c, {TM}_{n m}} = \frac{p_{n m} c}{2 π a \sqrt{ε_{r}}}

p′ₙₘ and pₙₘ are roots of J′ₙ and Jₙ. The dominant mode is TE₁₁ (p′ = 1.841), followed by TM₀₁ (2.405), TE₂₁ (3.054) and TE₀₁ / TM₁₁ (3.832).

The single-mode band of a circular guide (TE₁₁ to TM₀₁, a ratio of only 1.31) is narrower than that of a rectangular guide, and the TE₁₁ mode has no preferred polarisation, so it can rotate along an imperfect guide. Those properties are useful in two places. Rotary joints in radar use the rotationally symmetric TM₀₁ mode to pass a signal through a rotating interface, and feeds for circularly or dual-polarised antennas use the circular TE₁₁ mode to carry two orthogonal polarisations at once. The TE₀₁ mode has the remarkable property that its conductor loss falls as frequency increases, which made it a candidate for long-distance communication before optical fibre.

Substrate Integrated Waveguides (SIW)

A substrate integrated waveguide builds a rectangular waveguide inside a printed circuit board: the top and bottom copper layers form the broad walls, and two rows of plated vias form the narrow walls (Fig 1). The result has the low loss, high Q and complete shielding of a waveguide, but is made by standard PCB processes and integrates directly with microstrip or GCPW circuits through simple tapered transitions. SIW is widely used for millimetre-wave filters, antenna feeds and slot arrays at 24–100 GHz.

Because the via walls are not continuous, an SIW of via-row spacing a (centre to centre) behaves like a solid-walled guide of slightly smaller width a_eff. For vias of diameter d and pitch s, a widely used empirical rule is

SIW equivalent width and cutoff

a_{eff} = a - \frac{d^{2}}{0.95 s} f_{c, {TE}_{10}} = \frac{c}{2 a_{eff} \sqrt{ε_{r}}}

Design rules to keep leakage between the vias negligible: s ≤ 2d and d ≤ λ_g/5 (stricter: s/d < 2.5 and s < λ_c/4). Only TE_m0 modes exist, because the surface current of TM and TE_mn (n ≥ 1) modes would have to cross the gaps between vias.

The substrate dielectric fills the guide, so SIW is εr times narrower than an air waveguide but suffers dielectric loss, and its small height (the board thickness) increases conductor loss. A low-loss substrate is essential.

Comparison of Transmission-Line Structures

Line	Mode	Typical Z₀	Frequency range	Loss	Power	Shielding	Typical use
Coax (flexible)	TEM	50, 75 Ω	DC – 18…110 GHz (size dependent)	Medium	Medium–high	Excellent	Equipment, antennas, test
Parallel-wire	TEM, balanced	300–600 Ω	DC – ≈300 MHz	Very low	High	None	HF antenna feeders
Twisted pair	TEM, balanced	100–150 Ω	DC – ≈2 GHz	High above 100 MHz	Low	Twist cancels	Ethernet, data, telephone
Microstrip	Quasi-TEM	20–130 Ω	DC – ≈100 GHz	Medium	Low–medium	Partial	RF PCBs, patch antennas
Stripline	TEM	25–100 Ω	DC – ≈60 GHz	Medium	Low–medium	Excellent	Multilayer boards, filters, couplers
CPW	Quasi-TEM	30–120 Ω	DC – >200 GHz	Medium–high	Low	Partial	MMICs, on-wafer probing
GCPW	Quasi-TEM	30–100 Ω	DC – ≈110 GHz	Medium	Low–medium	Very good	mm-wave PCBs, connectors, 5G
Rectangular waveguide	TE₁₀	— (≈400–600 Ω wave impedance)	One band per size, 0.32–1100 GHz	Very low	Very high	Excellent	Radar, satellite, high power
Circular waveguide	TE₁₁, TM₀₁, TE₀₁	—	One band per size	Very low	Very high	Excellent	Rotary joints, dual-pol feeds
SIW	TE₁₀	—	≈5–300 GHz, one band per width	Low–medium	Low–medium	Very good	mm-wave filters, antenna arrays

Selecting Transmission Lines for RF Applications

The choice is usually made by working down a short list of questions:

Is it a connection between boxes or a line on a board? Between equipment, use coax (or waveguide at high power and millimetre-wave frequencies). On a board, choose among microstrip, GCPW, stripline and SIW.
What is the highest frequency? Below a few GHz any planar line works and microstrip is simplest. From about 10 GHz, GCPW gives better isolation and cleaner transitions; above about 30 GHz, thin substrates, GCPW and SIW dominate. For coax, choose the connector series whose rating exceeds the highest frequency.
How much loss can be tolerated? The substrate loss tangent dominates on boards (Part 8). For long or very low-loss runs use air-dielectric lines: waveguide, rigid air coax, or parallel-wire at HF.
How much power? Kilowatts at microwave frequencies call for waveguide or large air coax; planar lines handle watts to tens of watts, limited by heating.
Is isolation important? For dense layouts and sensitive receivers, use fully shielded lines — stripline, GCPW with via fences, or SIW.
Balanced or unbalanced? Dipoles and differential circuits want balanced lines (or a balun); most RF circuits and coax-fed antennas are unbalanced.
Cost and manufacturing. Microstrip on the outer layer is the cheapest RF line; stripline needs a multilayer board; waveguide needs machining.

📌 A typical modern RF product

A 28 GHz 5G module uses GCPW on a low-loss laminate for its RF traces, stripline on inner layers for the beamformer feed network, an SIW or patch antenna array, and a short coax or board-to-board connector to the rest of the radio. Each line is chosen for the job it does.

Worked Examples

The examples use c = 299.79 mm/ns (= mm·GHz), η₀ = 376.7 Ω and ε₀ = 8.854 pF/m.

Coax Parameters from Geometry: RG-58

Problem: RG-58 has a 0.81 mm diameter centre conductor and a 2.95 mm diameter solid polyethylene dielectric (εr = 2.25). Find Z₀, C′, L′ and the velocity factor.

Step 1 — Geometry ratio

a = 0.405 mm, b = 1.475 mm, \ln \frac{b}{a} = \ln 3.642 = 1.2925

Step 2 — Impedance and per-unit-length parameters

Z_{0} = \frac{60}{\sqrt{2.25}} \times 1.2925 = 51.7 Ω

C^{'} = \frac{2 π \times 8.854 \times 10^{- 12} \times 2.25}{1.2925} = 96.8 pF/m

L^{'} = 2 \times 10^{- 7} \times 1.2925 = 258.5 nH/m

Step 3 — Velocity factor

VF = \frac{1}{\sqrt{ε_{r}}} = \frac{1}{1.5} = 0.667

Answer: Z₀ ≈ 51.7 Ω, C′ ≈ 97 pF/m, L′ ≈ 259 nH/m, VF = 0.667

The calculation reproduces the datasheet values of RG-58 (nominal 50 Ω, 101 pF/m, VF 0.66). The small difference in Z₀ comes from the stranded centre conductor, whose effective radius is slightly larger than the bare-wire figure, and from the dielectric tolerance.

Highest Usable Frequency of a Coax

Problem: Estimate the TE₁₁ cutoff frequency of RG-58 (from Example 1) and of an air-dielectric 3.5 mm coax (a = 0.76 mm, b = 1.75 mm).

RG-58 (εr = 2.25)

f_{c} \approx \frac{299.79}{π (0.405 + 1.475) \times 1.5} = \frac{299.79}{8.859} = 33.8 GHz

3.5 mm air line

f_{c} \approx \frac{299.79}{π (0.76 + 1.75)} = \frac{299.79}{7.885} = 38.0 GHz

Answer: about 34 GHz for RG-58 and 38 GHz for 3.5 mm air line

The 3.5 mm connector is specified to 34 GHz, below the ideal cutoff, because its dielectric support bead lowers the cutoff locally. RG-58 will never be used near 34 GHz — its loss above a few GHz is prohibitive — so loss, not moding, limits it. For precision coax the moding limit is the real ceiling, which is why each step up in frequency needs a smaller connector.

The Three Optimum Coax Impedances

Problem: For an air-filled coax with fixed outer radius, find b/a and Z₀ for minimum conductor loss, maximum power and maximum voltage. What b/a gives 50 Ω, and how much worse is its loss than the optimum?

Minimum loss — solve ln x = (1 + x)/x by iteration, x ← e^((1+x)/x)

x = 3.0 \to 3.79 \to 3.54 \to 3.61 \to \dots \to 3.591

Z_{0} = 60 \ln 3.591 = 76.7 Ω

Maximum power and maximum voltage

P_{max} : x = e^{1 / 2} = 1.649, Z_{0} = 30 Ω

V_{max} : x = e = 2.718, Z_{0} = 60 Ω

The 50 Ω line

x = e^{50 / 60} = 2.301

\frac{α_{c} (2.301)}{α_{c} (3.591)} = \frac{(1 + 2.301) / \ln 2.301}{(1 + 3.591) / \ln 3.591} = \frac{3.966}{3.585} = 1.106

Answer: 77 Ω (b/a 3.59), 30 Ω (1.65) and 60 Ω (2.72); 50 Ω has b/a = 2.30 and about 11% more loss than the optimum

The loss curve is flat near its minimum, so moving from 77 Ω to 50 Ω costs only 11% in loss while raising the power capacity by about 60% compared with 77 Ω — a good compromise.

Designing a 450 Ω Ladder Line

Problem: An HF antenna feeder is to be built from 1.3 mm diameter copper wire with thin plastic spacers (treat as air). What centre spacing D gives 450 Ω?

Invert the exact two-wire formula

\cosh^{- 1} (\frac{D}{2 a}) = \frac{π Z_{0}}{η_{0}} = \frac{π \times 450}{376.7} = 3.753

\frac{D}{2 a} = \cosh 3.753 = 21.33 D = 21.33 \times 1.3 = 27.7 mm

Answer: D ≈ 28 mm

Commercial "450 Ω window line" uses a similar wire size with about 25 mm spacing; its plastic web raises εeff slightly and brings the impedance down to around 400 Ω, which is why measured values of window line are often quoted as 370–400 Ω.

Impedance of a Data-Cable Twisted Pair

Problem: A Cat 6 pair uses 0.573 mm diameter (23 AWG) conductors whose insulated diameter, and hence centre spacing, is about 0.96 mm. Taking εeff ≈ 2.0 for the mix of insulation and air, estimate Z₀.

Two-wire formula with εeff

Z_{0} = \frac{376.7}{π \sqrt{2.0}} \cosh^{- 1} (\frac{0.96}{0.573})

= 84.8 \times \cosh^{- 1} (1.675) = 84.8 \times 1.105 = 93.7 Ω

Answer: Z₀ ≈ 94 Ω — within the 100 Ω ± 15% specification

Note that for closely spaced wires the exact cosh⁻¹ form matters: the approximation (120/√εeff) ln(D/a) would give 103 Ω. A slightly thicker insulation (D = 0.98 mm) or a lower εeff brings the pair to 98–100 Ω, which is how cable makers trim the impedance.

50 Ω Stripline Width

Problem: Find the strip width for a 50 Ω stripline with ground-plane spacing b = 1.0 mm in RO4350B (εr = 3.66). Neglect strip thickness.

Synthesis formula (√εr Z₀ = 95.7 Ω < 120 Ω)

\frac{W}{b} = \frac{30 π}{\sqrt{3.66} \times 50} - 0.441 = \frac{94.25}{95.66} - 0.441 = 0.544

Check with the analysis formula (W/b > 0.35, so Wₑ = W)

Z_{0} = \frac{30 π}{1.913} \times \frac{1}{0.544 + 0.441} = 49.27 \times 1.015 = 50.0 Ω

Answer: W ≈ 0.54 mm

The same 50 Ω microstrip on a 0.5 mm RO4350B layer (the same dielectric thickness to the nearest ground) would be about 1.1 mm wide. Stripline is narrower because it has a ground on both sides, which doubles the capacitance per unit width. The guided wavelength is λ/√3.66 at every frequency.

CPW Impedance by Conformal Mapping

Problem: A CPW on a thick alumina substrate (εr = 9.8) has a 100 µm centre strip and 60 µm gaps. Find εeff and Z₀. What gap gives 50 Ω?

Step 1 — Modulus and elliptic integrals

k = \frac{100}{100 + 120} = 0.4545 k^{'} = \sqrt{1 - {0.4545}^{2}} = 0.8907

K (k) = 1.6630 K (k^{'}) = 2.2424 \frac{K (k^{'})}{K (k)} = 1.3485

Step 2 — εeff and Z₀

ε_{eff} = \frac{9.8 + 1}{2} = 5.4 Z_{0} = \frac{30 π}{\sqrt{5.4}} \times 1.3485 = 40.56 \times 1.3485 = 54.7 Ω

Answer: εeff = 5.4, Z₀ ≈ 54.7 Ω; a gap of about 44 µm gives 50 Ω

Narrowing the gaps raises the capacitance and lowers Z₀; repeating the calculation for s = 44 µm gives 50 Ω. The same impedance could also be reached with a 50 µm strip and narrower gaps, scaling w and s together — the extra freedom of CPW. Elliptic integrals are available in every maths library; good approximations for K(k′)/K(k) are given in Part 10.

WR-90 Waveguide: Modes, Guide Wavelength and Impedance

Problem: WR-90 has inner dimensions 22.86 × 10.16 mm (air). List the first four mode cutoffs, and find the guide wavelength, group velocity and TE₁₀ wave impedance at 10 GHz.

Cutoffs, f_c = (c/2)√((m/a)² + (n/b)²)

{TE}_{10} : 6.557 {TE}_{20} : 13.11 {TE}_{01} : 14.75 {TE}_{11}, {TM}_{11} : 16.15 GHz

At 10 GHz, √(1 − (6.557/10)²) = 0.755

λ_{g} = \frac{29.98}{0.755} = 39.7 mm v_{g} = 0.755 c Z_{TE} = \frac{376.7}{0.755} = 499 Ω

Answer: TE₁₀ 6.56, TE₂₀ 13.11, TE₀₁ 14.75, TE₁₁/TM₁₁ 16.15 GHz; λg = 39.7 mm, vg = 0.755c, Z_TE = 499 Ω

A slot or coupling hole spaced by λg/2 in this guide is 19.9 mm apart, not the 15 mm that a free-space half wavelength would suggest. The standard WR-90 band (8.2–12.4 GHz) avoids both the steep dispersion just above 6.56 GHz and the TE₂₀ mode at 13.1 GHz.

Single-Mode Band of a Circular Waveguide

Problem: A 20 mm inner-diameter circular waveguide (air) is used as an antenna feed horn throat. Over what frequency range does it carry only the TE₁₁ mode?

Cutoffs with a = 10 mm, c/(2πa) = 4.771 GHz

f_{{TE}_{11}} = 1.841 \times 4.771 = 8.78 GHz

f_{{TM}_{01}} = 2.405 \times 4.771 = 11.47 GHz

Answer: 8.78–11.47 GHz (a ratio of only 1.31:1)

A dual-polarised X-band feed operating at 10–11 GHz fits this window. If the feed must work above 11.5 GHz the TM₀₁ mode, which has a null on axis, must be kept from being excited by keeping the structure symmetric — the reason circular feeds are machined with great care.

Designing an SIW for a 24 GHz Radar

Problem: Design an SIW on 0.508 mm RO4003C (εr = 3.55) for a 24 GHz radar, with the TE₁₀ cutoff at 18 GHz. Use 0.4 mm vias at 0.8 mm pitch. Find the via-row spacing, the single-mode band and the guide wavelength at 24 GHz.

Step 1 — Equivalent width for f_c = 18 GHz

a_{eff} = \frac{299.79}{2 \times 18 \times \sqrt{3.55}} = \frac{299.79}{67.83} = 4.420 mm

Step 2 — Via-row spacing (check s = 2d ✓)

a = a_{eff} + \frac{d^{2}}{0.95 s} = 4.420 + \frac{0.16}{0.76} = 4.63 mm

Step 3 — Band and guide wavelength

{TE}_{20} : 2 \times 18 = 36 GHz

λ_{g} = \frac{299.79 / (24 \sqrt{3.55})}{\sqrt{1 - (18 / 24)^{2}}} = \frac{6.630}{0.6614} = 10.0 mm

Answer: via rows 4.63 mm apart, single-mode 18–36 GHz, λg = 10.0 mm at 24 GHz

Operating at 24 GHz puts the guide at 1.33 f_c, in the recommended range. The via pitch of 0.8 mm is λg/12.5, comfortably fine enough to keep leakage between the vias negligible.

Summary

Lines with two or more conductors carry TEM (or quasi-TEM) waves from DC upward; single-conductor waveguides carry only TE and TM modes above a cutoff.
Coax: Z₀ = (60/√εr) ln(b/a). Air coax has minimum loss at 77 Ω, maximum power at 30 Ω; 50 Ω is the compromise. Its upper limit is the TE₁₁ cutoff, f_c ≈ c/[π(a + b)√εr].
Parallel-wire and twisted pair are balanced lines: Z₀ = (η₀/π√εr) cosh⁻¹(D/2a). Twisting cancels interference and crosstalk.
Microstrip (quasi-TEM, open), stripline (TEM, shielded), CPW (all conductors on one side) and GCPW (CPW with a back ground and via fences) are the planar lines of RF boards and MMICs.
Rectangular waveguide: f_c = c/(2a√εr) for TE₁₀, λg = λ/√(1 − (f_c/f)²), very low loss and very high power. Circular guide supports dual polarisation and rotary joints.
SIW builds a waveguide into a PCB with via walls, with a_eff = a − d²/(0.95s).
Choose a line by asking: between boxes or on a board, highest frequency, loss, power, isolation, balance and cost.

All the TEM and quasi-TEM lines in this article can be described by the same four per-unit-length parameters. Part 3: The RLGC Model and Telegrapher's Equations derives the RLGC model from a short section of line and obtains the telegrapher's equations that govern every line in this article.

📘 Go deeper: Fields inside the lines

Every line in this article is a solution of Maxwell's equations with different boundary conditions. Open Maxwell's Equations series →

Put It Into Practice

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Transmission Lines Microstrip & Stripline Impedance Calculator Field-solver backed design of microstrip, embedded microstrip, stripline, GCPW and differential pairs: width, εeff, Z₀, loss, dispersion and tolerance. Open calculator → Antenna & Signal Wavelength / Frequency Calculator Convert between frequency and wavelength in free space and in dielectrics. Open calculator → Impedance Matching Smith Chart Visualiser Plot impedances, move along lines and stubs, load Touchstone files and design matching networks. Open calculator → All Tools All RF Calculators Browse every NEXT ANTENNAS calculator for RF, microwave and antenna design. Open calculator →

Types of Transmission Lines