Guided Wavelength & Electrical Length

Fig 1 — The same frequency has a shorter wavelength on a line than in free space. Electrical length measures a line in wavelengths (or degrees) of the guided wave

Introduction to Wavelength in Transmission Lines

An RF designer rarely asks "how long is this line in millimetres?" The useful question is "how long is it in wavelengths?" A 30 mm trace is a negligible stub at 100 MHz, a quarter-wave impedance transformer at about 1.4 GHz on FR-4, and a resonant half-wave filter element at twice that frequency. What decides its behaviour is its electrical length — its length measured in guided wavelengths, or equivalently the phase shift a wave suffers in crossing it.

This article connects physical length and phase. It builds on the phase constant β from Part 5 and prepares for Part 7, where the electrical length decides the input impedance of a terminated line, and for the quarter-wave and half-wave circuits of Part 12.

Free-Space Wavelength

In free space an electromagnetic wave travels at c = 299 792 458 m/s, and its wavelength is the distance it travels in one period:

Free-space wavelength

λ_{0} = \frac{c}{f} k_{0} = \frac{2 π}{λ_{0}} = \frac{ω}{c}

A convenient form: λ₀ [mm] = 299.79 / f [GHz], or λ₀ [m] = 299.79 / f [MHz].

λ₀ is 2.07 m at 145 MHz, 125 mm at 2.4 GHz, 30 mm at 10 GHz and 3.9 mm at 77 GHz. It sets the scale of antennas (Introduction to Antennas) and of free-space propagation, but on a transmission line the wave travels more slowly, and its wavelength is shorter.

Guided Wavelength

The guided wavelength λ_g is the distance along the line over which the phase of a travelling wave changes by 2π — the spatial period of the wave on the line. It follows from the phase velocity:

Guided wavelength

λ_{g} = \frac{v_{p}}{f} = \frac{2 π}{β} = \frac{λ_{0}}{\sqrt{ε_{eff}}}

For a TEM line in a uniform dielectric εeff = εr; for microstrip and CPW, (εr + 1)/2 < εeff < εr; for a hollow waveguide λg is longer than λ₀ (Part 2).

Since β = ω√(L′C′) for a low-loss line, λ_g depends on the line's L′ and C′ — in other words on its dielectric, and for quasi-TEM lines also on its geometry. That is why a 25 Ω and a 100 Ω microstrip on the same board have different guided wavelengths (Example 3 of Part 9 and Fig 9 there).

Velocity Factor

Cable manufacturers express the same information as a velocity factor — the ratio of the phase velocity to c, which is also the ratio λ_g/λ₀:

Velocity factor

VF = \frac{v_{p}}{c} = \frac{λ_{g}}{λ_{0}} = \frac{1}{\sqrt{ε_{eff}}}

Line / dielectric	εr or εeff	Velocity factor	Delay per metre
Air line, open-wire feeder	≈ 1.0	0.95–1.0	3.3–3.5 ns
Foam PE coax (LMR-400, RG-6)	1.3–1.5	0.82–0.88	3.8–4.1 ns
Solid PTFE coax (RG-142, semi-rigid)	2.1	0.69–0.70	4.8 ns
Solid PE coax (RG-58, RG-213)	2.25	0.66	5.05 ns
Microstrip on RO4003C (50 Ω)	≈ 2.75	≈ 0.60	≈ 5.5 ns
Microstrip on FR-4 (50 Ω)	≈ 3.25	≈ 0.55	≈ 6.0 ns
Stripline in FR-4	≈ 4.2	≈ 0.49	≈ 6.8 ns

The velocity factor is the number to use when cutting a cable to an electrical length: a quarter-wave coax stub is VF × λ₀/4 long, not λ₀/4. Forgetting it is a classic mistake that leaves stubs 30–50% too long.

Effective Dielectric Constant and Guided Wavelength

On a line whose field fills a single dielectric, εeff = εr exactly. On microstrip and coplanar lines the field is partly in air, and εeff is the permittivity of an imaginary uniform medium that would give the same phase velocity. Part 9 derived εeff = C/C_a from the line capacitances with and without the dielectric, and wrote it as εeff = 1 + q(εr − 1) with a filling factor q of about 0.6–0.8.

Three consequences follow for guided wavelength:

λ_g depends on the line width. Wider (low-Z₀) lines have more of their field in the substrate and a shorter λ_g.
λ_g depends on frequency. Dispersion raises εeff with frequency, so λ_g shrinks a little faster than 1/f (Example 8).
λ_g depends on the material tolerance — and on other things near the line, such as solder mask, conformal coating or a nearby enclosure lid, all of which change εeff.

Electrical Length

The electrical length of a line of physical length ℓ is the phase shift βℓ a wave suffers in travelling along it, usually expressed in degrees or in wavelengths:

Electrical length

θ = β ℓ = \frac{2 π ℓ}{λ_{g}} rad = 360^{\circ} \frac{ℓ}{λ_{g}} = \frac{360^{\circ} ℓ f \sqrt{ε_{eff}}}{c}

τ = \frac{θ}{ω} = \frac{ℓ}{v_{p}}

θ is proportional to both length and frequency. The delay τ = θ/ω = ℓ/v_p is the same quantity in time.

Electrical length is the quantity that appears in every transmission-line formula. The input impedance, the reflection coefficient seen at the input, the phase of S₂₁ and the resonant frequency of a stub depend on ℓ only through θ. Two lines of different physical length behave identically if they have the same Z₀ and the same θ — the basis of scaling a design from one substrate or frequency to another.

Electrically short and electrically long lines

Electrical length also decides how a piece of line should be modelled. Part 1 showed that the error of a lumped model grows as θ²/3, which gives a practical scale:

Electrical length	Length	Model	Typical use
θ < 10°	< λ_g/36	Lumped element, error < 1%	Short connections, bond wires, vias
10°–30°	λ_g/36 – λ_g/12	Lumped L or C with care (1–10% error)	Short high-Z lines as series inductors, open stubs as capacitors
30°–90°	λ_g/12 – λ_g/4	Transmission line	Matching sections, phase shifters
≥ 90°	≥ λ_g/4	Transmission line; behaviour repeats every 180°	Transformers, resonators, feeders

The "short line as a lumped element" idea is itself a design technique. A section of high-impedance line much shorter than λ_g/4 behaves like a series inductor L ≈ Z₀θ/ω, and a short section of low-impedance line like a shunt capacitor C ≈ θ/(ωZ₀). Stepped-impedance low-pass filters are built entirely from such sections (Part 13).

Phase Shift Along a Transmission Line

A forward wave on a lossless line is V⁺(z) = V₀⁺e^−jβz: at a distance ℓ from the input its phase lags by θ = βℓ. A matched line section therefore has S₂₁ = e^−jθ — unit magnitude, phase −θ — and is a pure phase shifter or delay.

With a reflected wave present, the reflection coefficient itself changes along the line. This is the key to every impedance transformation:

Derivation — how Γ rotates along a line

Put the load at z = 0 and the input at z = −ℓ. At any point the local reflection coefficient is the ratio of backward to forward wave voltages:

Γ (z) = \frac{V_{0}^{-} e^{+ j β z}}{V_{0}^{+} e^{- j β z}} = Γ_{L} e^{j 2 β z}

\Rightarrow

Γ_{i n} = Γ (- ℓ) = Γ_{L} e^{- j 2 θ}

Moving a distance ℓ towards the source rotates Γ clockwise by 2θ without changing its magnitude: the wave goes to the load and comes back, so it accumulates twice the one-way phase. The input impedance follows from Γ_in:

Input impedance through Γ

Z_{i n} = Z_{0} \frac{1 + Γ_{L} e^{- j 2 θ}}{1 - Γ_{L} e^{- j 2 θ}}

This is the rotation on a circle of constant |Γ| that the Smith chart makes visual (Impedance Transformation Along Transmission Lines).

Since Γ rotates by 2θ, the input impedance repeats every time θ increases by 180° — every half wavelength of line. Fig 2 shows this for a 100 Ω load on a 50 Ω line: the input impedance alternates between 100 Ω (θ = 0°, 180°, 360°) and 25 Ω (θ = 90°, 270°), passing through complex values in between.

Fig 2 — Input impedance of a 50 Ω line terminated in 100 Ω, against electrical length. The impedance inverts at every odd multiple of 90° and repeats every 180°

R_in (Ω)X_in (Ω)

Quarter-Wavelength Transmission Lines

When θ = 90°, e^−j2θ = −1 and Γ_in = −Γ_L. Substituting into Z_in:

Quarter-wave impedance inversion

Z_{i n} = Z_{0} \frac{1 - Γ_{L}}{1 + Γ_{L}} = \frac{Z_{0}^{2}}{Z_{L}} (θ = 90^{\circ})

The input impedance is the inverse of the load, scaled by Z₀². A short becomes an open, an open becomes a short, a capacitor becomes an inductor.

This single property makes the quarter-wave line one of the most useful elements in microwave engineering:

Quarter-wave transformer: a line of impedance Z_T = √(Z₀Z_L) matches a resistive load Z_L to Z₀ (Example 10 of Part 9).
Quarter-wave stubs: a shorted λ/4 stub is an open circuit at its input — a DC return or bias feed that is invisible at the design frequency; an open λ/4 stub is a short — a notch or an RF ground.
Impedance inverters in filters, Doherty amplifiers and branch-line couplers.

Part 12 treats quarter-wave transformers and stubs in full, including their bandwidth and multi-section designs.

Half-Wavelength Transmission Lines

When θ = 180°, e^−j2θ = 1, Γ_in = Γ_L and Z_in = Z_L whatever the line's Z₀. A half-wave line is "transparent": it repeats the load impedance at its input. This is used to move a load to a convenient place without changing its impedance (for example, a feeder cut to a multiple of λ/2 so that the antenna's impedance appears at the transmitter), in half-wave resonators and filters, and in the half-wave coaxial balun.

The repetition holds only at the frequencies where θ is a multiple of 180°. The calculator's S-parameter plot of a 50 Ω FR-4 line in Part 9 shows it directly: the small mismatch reflection vanishes at about 4.9 GHz and 9.6 GHz, where the 16.95 mm line is one and two half-wavelengths long.

Scaling a design to another frequency or substrate

Because transmission-line circuits depend only on Z₀ and θ, a working design can be moved to a new frequency or a new laminate by keeping every Z₀ and every θ the same. The widths are re-synthesised for the new substrate, and every length scales as

Length scaling at constant electrical length

ℓ_{2} = ℓ_{1} \frac{f_{1}}{f_{2}} \sqrt{\frac{ε_{eff, 1}}{ε_{eff, 2}}}

Subscripts 1 and 2 refer to the original and new designs. Junction and bend parasitics do not scale exactly, so the final layout is checked with a simulator.

Moving a 2.45 GHz FR-4 design (εeff = 3.24) to 5.8 GHz on RO4003C (εeff = 2.77) multiplies every length by (2.45/5.8)√(3.24/2.77) = 0.457. If the new substrate is also thinner, the widths shrink in proportion to h, since Z₀ depends only on w/h.

Frequency Dependence of Electrical Length

Since θ ∝ f, any circuit that relies on a particular electrical length works exactly only at one frequency. A quarter-wave transformer designed for f₀ is 90° long there but 108° long at 1.2 f₀, and the impedance it presents drifts away from the target. Fig 3 shows the resulting reflection for transformers matching 100 Ω and 200 Ω loads to 50 Ω: the larger the impedance ratio, the narrower the bandwidth.

Fig 3 — |Γ| at the input of a single quarter-wave transformer against normalised frequency, for 100 Ω and 200 Ω loads matched to 50 Ω. The −20 dB bandwidth is about 37% for the 2:1 ratio and 17% for 4:1

100 Ω load, Z_T = 70.7 Ω200 Ω load, Z_T = 100 Ω

Electrical length also varies with three other things that are easy to overlook:

Dispersion — on microstrip εeff rises with frequency, so θ grows slightly faster than f (Example 8).
Material tolerance and temperature — a ±5% spread in εr shifts the frequency of a resonant line by about ±2% on FR-4 microstrip (Example 9). PTFE cables show a sudden phase change of several hundred ppm near 19 °C, where PTFE changes crystal phase, which matters in phase-matched assemblies.
Length of the line in wavelengths — a long line multiplies every small change: a 0.1% change in v_p changes the phase of a 100-wavelength cable by 36°.

Measuring and Trimming Electrical Length

Physical length is easy to measure; electrical length must be measured electrically. Three methods are in everyday use:

Phase slope (group delay). A VNA measures the phase of S₂₁ against frequency; the slope gives the delay τ = −dφ/dω and θ = 360°fτ at any frequency (Example 10). Its "electrical delay" or "port extension" functions subtract a known delay to move the reference plane.
Resonance. A shorted or open line resonates where it is a quarter or half wavelength long; measuring the resonant frequency gives v_p and the velocity factor with high accuracy.
Time-domain reflectometry. The round-trip time to a discontinuity gives the electrical distance to it, which is how cable faults are located.

In phased arrays, beam-forming networks and interferometers, cables are bought as phase-matched sets: each is cut slightly long, its phase measured on a VNA, and trimmed until all agree to within a few degrees at the operating frequency. A useful rule for trimming: removing a length Δℓ shortens the phase by 360°·Δℓ/λ_g, so at 6 GHz on PTFE coax (λ_g = 34.5 mm) each millimetre is worth 10.4°.

For waveguides, the same ideas apply with λ_g = λ₀/√(1 − (f_c/f)²), which is longer than λ₀. Electrical length in a waveguide changes much faster with frequency than in a TEM line, because λ_g itself varies strongly near cutoff, and the group delay τ = ℓ/v_g differs from the phase delay ℓ/v_p (Part 5).

Practical Wavelength and Electrical Length Calculations

The examples use c = 299.79 mm/ns, so λ₀ [mm] = 299.79/f [GHz].

Wavelength on a 2 m Band Feeder

Problem: Find λ₀ at 145 MHz, and λ_g and the length of a quarter-wave stub on RG-213 coax (VF = 0.66).

Free-space and guided wavelength

λ_{0} = \frac{299.79}{145} = 2.068 m

λ_{g} = 0.66 \times 2.068 = 1.365 m

\frac{λ_{g}}{4} = 341 mm

Answer: λ₀ = 2.07 m, λ_g = 1.37 m, λ/4 stub = 341 mm

A stub cut to λ₀/4 = 517 mm would actually be 0.38λ_g long and would not act as a quarter-wave stub at all. In practice the stub is cut slightly long and trimmed while watching its response on an analyser, because connectors add a little electrical length.

Electrical Length of a Dual-Band Trace

Problem: A 250 mm microstrip on FR-4 (εeff = 3.3) carries both 900 MHz and 2.4 GHz signals. Find its electrical length at each frequency.

θ = 360° ℓ f √εeff / c (√3.3 = 1.817)

θ_{0.9} = \frac{360 \times 250 \times 0.9 \times 1.817}{299.79} = 491^{\circ}

θ_{2.4} = \frac{360 \times 250 \times 2.4 \times 1.817}{299.79} = 1309^{\circ}

Answer: 491° (1.36λ_g) at 900 MHz and 1309° (3.64λ_g) at 2.4 GHz

The trace is several wavelengths long at both frequencies, so it must be a matched 50 Ω line: any mismatch at its ends would produce input impedances that differ completely between the two bands.

Quarter-Wave Line for GPS L1

Problem: Find the length of a 50 Ω quarter-wave line on 0.813 mm RO4003C (εeff ≈ 2.76 at 1.575 GHz).

Guided wavelength and λ/4

λ_{g} = \frac{299.79}{1.575 \times \sqrt{2.76}} = \frac{299.79}{2.6166} = 114.6 mm

\frac{λ_{g}}{4} = 28.6 mm

Answer: λ_g = 114.6 mm; the λ/4 line is 28.6 mm long

At 1.575 GHz a quarter-wave line is long enough to need meandering on a compact board. Bends change the electrical length slightly, so a meandered line is checked with the calculator or an EM simulator (Part 13).

Cutting a Coaxial Stub

Problem: A shorted quarter-wave stub of PTFE semi-rigid coax (VF = 0.695) is to provide a DC ground that is invisible at 433.92 MHz. How long is it?

Quarter-wave length

ℓ = VF \times \frac{λ_{0}}{4} = 0.695 \times \frac{299.79}{4 \times 0.43392} = 0.695 \times 172.7 = 120.0 mm

Answer: 120.0 mm

At 433.92 MHz the shorted stub presents an open circuit (Z_in = Z₀²/0 → ∞), so it can be connected across the RF line to ground the DC without affecting the signal — a common way to protect a receiver input against static discharge.

Input Impedance at λ/8, λ/4 and λ/2

Problem: A 50 Ω line is terminated in 100 Ω. Find Z_in when the line is λ/8, λ/4 and λ/2 long.

Using Z_in = Z₀(Z_L + jZ₀ tan θ)/(Z₀ + jZ_L tan θ)

θ = 45^{\circ} : Z_{i n} = 50 \frac{100 + j 50}{50 + j 100} = 50 (0.8 - j 0.6) = 40 - j 30 Ω

θ = 90^{\circ} : Z_{i n} = \frac{50^{2}}{100} = 25 Ω θ = 180^{\circ} : Z_{i n} = 100 Ω

Answer: 40 − j30 Ω, 25 Ω and 100 Ω

These are three points on the curves of Fig 2. Note |Γ| = 1/3 in every case — the line changes the impedance but not the mismatch, which is why the VSWR on a lossless line is the same everywhere.

A Quarter-Wave Transformer Away from f₀

Problem: A 70.71 Ω quarter-wave transformer matches 100 Ω to 50 Ω at f₀. Find Z_in, |Γ| and return loss at 1.1 f₀ and 1.2 f₀.

At 1.1 f₀ (θ = 99°) and 1.2 f₀ (θ = 108°)

Z_{i n} (1.1 f_{0}) = 50.6 + j 5.5 Ω, | Γ | = 0.055

Z_{i n} (1.2 f_{0}) = 52.5 + j 10.9 Ω, | Γ | = 0.109

Answer: return loss 25.2 dB at 1.1 f₀ and 19.3 dB at 1.2 f₀

The match degrades smoothly away from f₀; the 20 dB return-loss band is about ±18% here (Fig 3). Wider bandwidth needs several cascaded quarter-wave sections with smaller impedance steps — the binomial and Chebyshev transformers of Part 12.

Designing a 1 ns Delay Line

Problem: A 1.000 ns delay line is to be made in RO4350B stripline (εr = 3.66). What length is needed, and what is its electrical length at 5 GHz?

Length and phase

ℓ = \frac{c τ}{\sqrt{ε_{r}}} = \frac{299.79 \times 1.000}{1.913} = 156.7 mm

θ = 360^{\circ} \times f τ = 360 \times 5 \times 1 = 1800^{\circ}

Answer: 156.7 mm, which is 1800° (5 wavelengths) at 5 GHz

Stripline is chosen because it is non-dispersive (pure TEM): the delay is 1.000 ns at every frequency, so the line can delay a wideband pulse without distorting it. The same delay in microstrip would vary by a few picoseconds across a wide band.

Dispersion Shifts a Quarter-Wave Resonator

Problem: A 50 Ω quarter-wave resonator for 10 GHz on 0.813 mm RO4003C is designed with the quasi-static εeff = 2.787. At 10 GHz the actual εeff is 2.875 (Kirschning–Jansen). How long is the line, and where does it actually resonate?

Designed length

ℓ = \frac{299.79}{4 \times 10 \times \sqrt{2.787}} = 4.490 mm

Actual resonance — same length, higher εeff

f_{r} \approx 10 \times \sqrt{\frac{2.787}{2.875}} = 9.85 GHz

Answer: ℓ = 4.49 mm, but it resonates near 9.85 GHz — 1.5% low

A 1.5% shift exceeds the bandwidth of many narrowband filters. Designers either use the dispersive εeff at the design frequency (as the calculator does) or simulate the layout with an EM solver.

Effect of εr Tolerance on a Resonant Line

Problem: A resonant 50 Ω microstrip on 1.6 mm FR-4 is designed for εr = 4.3 (εeff = 3.240). The laminate's εr can be anywhere from 4.1 to 4.5. How far can the resonance move?

εeff at the limits (Hammerstad–Jensen, same 3.07 mm strip)

ε_{r} = 4.1 : ε_{eff} = 3.106

ε_{r} = 4.5 : ε_{eff} = 3.375

Resonance scales as 1/√εeff

\frac{f_{r}}{f_{0}} = \sqrt{\frac{3.240}{3.106}} = 1.021 \sqrt{\frac{3.240}{3.375}} = 0.980

Answer: about ±2% — ±49 MHz on a 2.45 GHz resonator

A filter with a 2% bandwidth could miss its band entirely. This is the main reason resonant microwave circuits use laminates with a tight εr tolerance (±0.05) rather than FR-4.

Measuring Electrical Length from the Phase Slope

Problem: The phase of S₂₁ through a cable decreases by 36° for every 10 MHz increase in frequency. Find the delay and, if VF = 0.66, the physical length.

Group delay from the phase slope, τ = −dφ/dω

τ = \frac{36^{\circ} / 360^{\circ}}{10 \times 10^{6} Hz} = 10.0 ns

ℓ = τ VF c = 10.0 \times 0.66 \times 0.2998 = 1.98 m

Answer: τ = 10.0 ns, ℓ ≈ 1.98 m

The phase slope avoids the 360° ambiguity of a single phase reading and is how a VNA's "electrical delay" function works. Removing this delay from a measurement moves the reference plane to the end of the cable.

Summary

λ₀ = c/f; on a line λ_g = λ₀/√εeff = VF·λ₀, with VF = 1/√εeff.
Electrical length θ = βℓ = 360°·ℓ/λ_g = 360°·ℓf√εeff/c; delay τ = ℓ/v_p.
Moving towards the source, Γ rotates by −2θ: Γ_in = Γ_Le^−j2θ, so Z_in repeats every λ/2.
Quarter-wave line: Z_in = Z₀²/Z_L (impedance inversion). Half-wave line: Z_in = Z_L.
θ ∝ f, so length-based circuits are narrowband; dispersion and εr tolerance shift them further.

With Z₀, γ and electrical length in hand, we can connect a load and solve the line completely. Part 7: Terminated Lossless Transmission Lines applies the load boundary condition to derive the reflection coefficient, the general input-impedance formula and the standing-wave pattern.

📘 Go deeper: Electrical length on the Smith chart

Moving along a line is a rotation around the Smith chart by twice the electrical length. Open Smith Chart & Impedance Matching series →

Put It Into Practice

Use the NEXT ANTENNAS calculators to apply the ideas from this article to real lines and circuits.

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Guided Wavelength and Electrical Length