Transmission Line Theory — The RLGC Model and Telegrapher's Equations

The Distributed RLGC Model

In the previous article we established why a conductor at RF frequencies cannot be treated as an ideal wire. The lumped-circuit model breaks down once the physical length of the conductor becomes an appreciable fraction of the signal wavelength. The correct approach is to model the conductor as a continuous, distributed structure — and the foundation of that model is the RLGC representation.

The core idea is straightforward: instead of treating a transmission line as a single lumped element, we divide it into infinitesimally short sections of length Δz, each of which is small enough to be modelled with lumped components. As Δz → 0, this gives an exact description of the line's behaviour.

Each elemental section of length Δz contains four distributed (per-unit-length) parameters:

R' (Ω/m) — Series resistance per unit length. Accounts for the ohmic loss in the conductors due to their finite conductivity. At high frequencies, the current crowds into a thin skin depth near the conductor surface (the skin effect), which increases R' with frequency as R' ∝ √f.
L' (H/m) — Series inductance per unit length. Arises from the magnetic flux threading the loop formed by the two conductors. Even a perfect conductor has non-zero L'. It is the dominant reactive element in series with the signal path.
G' (S/m) — Shunt conductance per unit length. Models dielectric loss in the insulating material between the conductors. Current leaks through the (imperfect) dielectric. In most practical RF substrates G' is small, but it increases with frequency and is important in lossy substrates.
C' (F/m) — Shunt capacitance per unit length. Results from the electric field established between the two conductors when a voltage is applied. It is determined entirely by the geometry and the permittivity of the dielectric medium.

📌 R' and G' Are the Loss Terms

R' causes conductor loss — energy dissipated as heat in the metal. G' causes dielectric loss — energy dissipated as heat in the insulator. In an ideal lossless transmission line, R' = G' = 0, and the mathematics simplifies greatly without losing the essential physics. We will use this simplification when it is appropriate, but always carry the full model through the derivation first.

The circuit model of a single elemental section is drawn with R' and L' in series along the signal conductor, and C' and G' in shunt between the two conductors:

Fig 1 — RLGC lumped-element model of a transmission line section of length Δz

R'Δz and L'Δz are in series with the signal conductor. G'Δz and C'Δz are shunted between the conductors. The voltage and current are functions of position z.

The full transmission line is the cascade of infinitely many such sections as Δz → 0. Every property of the line — its characteristic impedance, its propagation constant, its loss — emerges from this model through straightforward application of Kirchhoff's laws.

Applying Kirchhoff's Laws to the RLGC Section

With the circuit model established, we apply Kirchhoff's Voltage Law (KVL) and Kirchhoff's Current Law (KCL) to a single elemental section. The voltage and current are phasor quantities (time-harmonic signals at angular frequency ω), functions of position z along the line.

Kirchhoff's Voltage Law — KVL

Traversing the top rail from z to z + Δz, the voltage drops across the series resistance R'Δz and the series inductance jωL'Δz due to the current I(z) flowing through them. KVL around the outer loop gives:

KVL around the elemental section

V (z) - (R^{'} Δz) I (z) - (j ω L^{'} Δz) I (z) - V (z + Δz) = 0

Sum of voltage drops around the series path equals zero. The shunt branch carries no series voltage.

Rearranging and dividing both sides by Δz:

Rearranged KVL — voltage difference over Δz

\frac{V (z + Δz) - V (z)}{Δz} = - (R^{'} + j ω L^{'}) I (z)

Left side is the finite-difference approximation to the derivative of V with respect to z.

Taking the limit as Δz → 0, the finite difference becomes the partial derivative:

Telegrapher's Voltage Equation

\frac{\partial V}{\partial z} = - (R^{'} + j ω L^{'}) I (z)

The rate of change of voltage with position equals the series impedance per unit length times the current.

Kirchhoff's Current Law — KCL

At the node where the shunt branch meets the top rail (at position z + Δz), the current I(z) arriving from the left splits into two paths: the shunt branch and the forward current I(z + Δz) continuing toward the load. KCL at that node:

KCL at the shunt node

I (z) - (G^{'} Δz) V (z + Δz) - (j ω C^{'} Δz) V (z + Δz) - I (z + Δz) = 0

Current into the node equals current out through the shunt (G'Δz and C'Δz in parallel) plus the continuing forward current.

In the limit Δz → 0, V(z + Δz) → V(z), and we can write the shunt voltage as simply V(z). Rearranging and taking the limit:

Telegrapher's Current Equation

\frac{\partial I}{\partial z} = - (G^{'} + j ω C^{'}) V (z)

The rate of change of current with position equals the shunt admittance per unit length times the voltage.

These two first-order coupled differential equations — one relating ∂V/∂z to I, the other relating ∂I/∂z to V — are the telegrapher's equations. They are the complete mathematical description of signal propagation on a transmission line. Everything that follows is derived from them.

📌 Notation Convention

It is common to define two shorthand quantities for the per-unit-length series impedance and shunt admittance:

Z' = R' + jωL' (series impedance per unit length, Ω/m)
Y' = G' + jωC' (shunt admittance per unit length, S/m)

The telegrapher's equations then read compactly as ∂V/∂z = −Z'I and ∂I/∂z = −Y'V. We will use this notation in the derivation below.

Derivation of the Wave Equation and Transmission Line Properties

To decouple the two telegrapher's equations and solve for V and I independently, we differentiate each equation again with respect to z and substitute the other equation into the result.

The Wave Equation

Differentiating the voltage equation with respect to z:

Step 1 — Differentiate ∂V/∂z with respect to z

\frac{\partial^{2} V}{\partial z^{2}} = - (R^{'} + j ω L^{'}) \frac{\partial I}{\partial z}

We can differentiate through the constant (R' + jωL') since it does not depend on z.

Now substitute the current equation ∂I/∂z = −(G' + jωC')V into the right-hand side:

Step 2 — Substitute the current equation

\frac{\partial^{2} V}{\partial z^{2}} = (R^{'} + j ω L^{'}) (G^{'} + j ω C^{'}) V

The double negative (−)(−) becomes positive. This is a second-order ODE in V(z) alone.

Defining the constant γ² = (R' + jωL')(G' + jωC'), this is the standard one-dimensional wave equation:

Transmission line wave equation

\frac{\partial^{2} V}{\partial z^{2}} - γ^{2} V = 0

The same form of equation describes acoustic waves, electromagnetic waves in free space, and elastic waves in solids.

An identical equation holds for the current I(z). The general solution to this second-order linear ODE has two linearly independent terms:

General solution for voltage on the line

V (z) = V_{0}^{+} e^{- γ z} + V_{0}^{-} e^{+ γ z}

V₀⁺ and V₀⁻ are complex constants set by the boundary conditions (source and load). The e^(−γz) term is the forward wave; e^(+γz) is the backward wave.

Deriving Characteristic Impedance Z₀

The corresponding general solution for the current is obtained by substituting V(z) back into the telegrapher's voltage equation ∂V/∂z = −Z'I, then solving for I(z):

Step 1 — Differentiate V(z) with respect to z

\frac{\partial V}{\partial z} = - γ V_{0}^{+} e^{- γ z} + γ V_{0}^{-} e^{+ γ z}

Differentiate term by term. The forward wave picks up −γ; the backward wave picks up +γ.

Step 2 — Equate to −Z'I and solve for I(z)

I (z) = \frac{γ}{Z^{'}} V_{0}^{+} e^{- γ z} - \frac{γ}{Z^{'}} V_{0}^{-} e^{+ γ z}

Note the sign change on the backward wave term — the backward current is in the −z direction.

The ratio of voltage to current for the forward wave alone is the characteristic impedance Z₀. Substituting γ = √(Z'Y'):

Characteristic impedance — general form

Z_{0} = \frac{Z^{'}}{γ} = \frac{Z^{'}}{\sqrt{Z^{'} Y^{'}}} = \sqrt{\frac{Z^{'}}{Y^{'}}} = \sqrt{\frac{R^{'} + j ω L^{'}}{G^{'} + j ω C^{'}}}

In the general (lossy) case Z₀ is complex and frequency-dependent. For the lossless line (R' = G' = 0): Z₀ = √(L'/C') — real, positive, and frequency-independent.

The complete current solution can now be written in terms of Z₀:

General solution for current on the line

I (z) = \frac{1}{Z_{0}} (V_{0}^{+} e^{- γ z} - V_{0}^{-} e^{+ γ z})

The forward current wave has amplitude V₀⁺/Z₀. The backward current wave has amplitude −V₀⁻/Z₀ — the minus sign reflects that it flows in the −z direction.

The Propagation Constant γ = α + jβ

The propagation constant γ is the single complex number that governs how both voltage and current waves evolve as they travel along the line. Since γ² = Z'Y' = (R' + jωL')(G' + jωC'), γ is in general complex, and we write it explicitly as:

Propagation constant — definition

γ = α + j β = \sqrt{(R^{'} + j ω L^{'}) (G^{'} + j ω C^{'})}

Units of γ are m⁻¹. α has units Np/m (nepers per metre). β has units rad/m.

The Attenuation Constant α (Np/m)

α is the real part of γ. It determines how the amplitude of a travelling wave decays as it moves along the line. Looking at the forward wave term V₀⁺ e^(−γz), we can expand the complex exponential:

Expanding the forward wave exponential

V_{0}^{+} e^{- γ z} = V_{0}^{+} e^{- α z} e^{- j β z}

The e^(−αz) factor is a real, positive, monotonically decreasing envelope. The e^(−jβz) factor is a unit-magnitude phasor describing the wave's phase.

The factor e^(−αz) causes the wave amplitude to decay exponentially with distance. After travelling a distance z = 1/α, the amplitude falls to e^(−1) ≈ 0.368 of its starting value — a loss of 1 Np (neper). The correspondence with decibels is:

Converting attenuation: Np/m to dB/m

α [dB/m] = 20 {log}_{10} (e) \times α [Np/m] \approx 8.686 \times α [Np/m]

1 Np/m ≈ 8.686 dB/m. Most datasheet specifications for cable or trace loss are quoted in dB/m or dB/100ft.

💡 Two Physical Sources of Attenuation

In a real transmission line, α has two additive contributions. Conductor loss αc comes from the ohmic resistance R' of the metallic conductors. Because of the skin effect, αc ∝ √f — loss rises with the square root of frequency. Dielectric loss αd comes from the shunt conductance G' of the imperfect dielectric, and αd ∝ f — it rises linearly with frequency. At microwave frequencies, dielectric loss often dominates, which is why low-loss substrates (PTFE, Rogers laminates) are preferred over FR-4 in high-frequency designs.

The Phase Constant β (rad/m)

β is the imaginary part of γ. It describes how the wave's phase advances (or retards) per unit length of travel. The factor e^(−jβz) in the forward wave is a phasor that rotates in the complex plane as z increases — each metre of line adds −β radians of phase.

The phase constant is directly related to the guided wavelength λg and the phase velocity v_p on the line:

Relationships between β, λg, and v_p

β = \frac{2 π}{λ_{g}} = \frac{ω}{v_{p}} = ω \sqrt{L^{'} C^{'}}

The last form (ω√(L'C')) is exact for a lossless line. For a lossy line, β = Im{γ} and differs slightly from ω√(L'C').

The phase velocity v_p = ω/β = 1/√(L'C') is the speed at which a constant-phase point (a wavefront) travels along the line. For a lossless line in a uniform dielectric:

Phase velocity in terms of permittivity

v_{p} = \frac{1}{\sqrt{L^{'} C^{'}}} = \frac{c}{\sqrt{ε_{eff}}} = \frac{1}{\sqrt{μ ε}}

c = 3×10⁸ m/s in free space. εeff < εr for microstrip because part of the field travels in air above the substrate.

β also defines electrical length. A line of physical length l has an electrical length θ = βl in radians, or equivalently θ° = (360° × l)/λg in degrees. A quarter-wave line has θ = π/2 = 90°; a half-wave line has θ = π = 180°. These are the fundamental building blocks of microwave filter and matching network design.

Summary: What γ Tells You

Parameter	Symbol	Units	Controls	Zero when
Propagation constant	γ = α + jβ	m⁻¹	Complete wave behaviour — amplitude decay and phase advance	Never (γ = 0 means no wave)
Attenuation constant	α = Re{γ}	Np/m or dB/m	How fast amplitude decays with distance	R' = G' = 0 (lossless line)
Phase constant	β = Im{γ}	rad/m	Phase advance per metre; sets λg, v_p, and electrical length	f = 0 (DC — no wave)

The Lossless Limit

Setting R' = 0 and G' = 0 (the lossless approximation), the propagation constant simplifies dramatically:

Propagation constant — lossless line

γ = \sqrt{(j ω L^{'}) (j ω C^{'})} = \sqrt{- ω^{2} L^{'} C^{'}} = j ω \sqrt{L^{'} C^{'}} = j β

For a lossless line: α = 0 (pure propagation, no decay). β = ω√(L'C'). Z₀ = √(L'/C') is real and positive.

The lossless model is a very good approximation for well-designed RF lines where conductor and dielectric losses are small over the length of interest. It is the model used in the vast majority of transmission line design work — filter design, impedance matching, stub networks — and is the basis of the Smith chart analysis presented in a later article.

Travelling Voltage and Current Waves

The mathematical solutions V(z) and I(z) derived above acquire their full physical meaning when we restore the time dependence. In the time domain, a phasor V(z) = V̂ e^(jφ) corresponds to a sinusoidal quantity v(z, t) = Re{V(z) e^(jωt)}. Applying this to our two-term solution:

Time-domain voltage on the line (lossless case for clarity)

v (z, t) = V_{0}^{+} cos (ω t - β z) + V_{0}^{-} cos (ω t + β z)

Real constants V₀⁺ and V₀⁻ assumed for clarity. The full complex form carries phase information too.

The Forward (Incident) Wave

The first term, V₀⁺ cos(ωt − βz), is a forward-travelling wave. To see why, consider what happens to a crest of this wave (a point where the cosine equals 1 — where ωt − βz = 0). As time t increases, z must increase in proportion: z = (ω/β)t = v_p · t. The crest moves in the +z direction at the phase velocity v_p. This is the wave launched by the source and travelling toward the load.

Phase velocity from the forward wave argument

ω t - β z = const ⟹ v_{p} = \frac{d z}{d t} = \frac{ω}{β}

The forward wave travels in the +z direction at speed ω/β = 1/√(L'C').

The Backward (Reflected) Wave

The second term, V₀⁻ cos(ωt + βz), is a backward-travelling wave. A crest of this wave satisfies ωt + βz = 0, so z = −(ω/β)t = −v_p · t — it moves in the −z direction (back toward the source) at the same speed. This is the reflected wave that exists whenever the load impedance differs from Z₀.

Current Waves and the Sign Convention

The current waves corresponding to the voltage waves carry a crucial sign difference. From our general solution:

Time-domain current on the line

i (z, t) = \frac{V_{0}^{+}}{Z_{0}} cos (ω t - β z) - \frac{V_{0}^{-}}{Z_{0}} cos (ω t + β z)

The forward current amplitude is +V₀⁺/Z₀. The backward current amplitude is −V₀⁻/Z₀. The minus sign is physically significant.

The minus sign on the reflected current is not an error — it is a direct consequence of the current direction convention. The forward current flows in the +z direction; the reflected current flows in the −z direction (back toward the source along the return conductor). When a wave reflects from an open circuit (Γ = +1), the voltage doubles but the currents cancel. When it reflects from a short circuit (Γ = −1), the currents add but the voltages cancel. This asymmetry between voltage and current at discontinuities is fundamental to all reflection and standing wave analysis.

A Snapshot of the Line

At any fixed instant in time, both V(z) and I(z) are sinusoidal functions of z, with spatial period λg. As time progresses, the forward wave pattern shifts in the +z direction and the backward wave shifts in the −z direction. When both are present simultaneously, their superposition produces a standing wave pattern — a fixed spatial distribution of voltage and current maxima and minima. The ratio of maximum to minimum voltage amplitude is the Voltage Standing Wave Ratio (VSWR), a central concept in impedance matching work.

VSWR in terms of the reflection coefficient magnitude

VSWR = \frac{1 + | Γ |}{1 - | Γ |} where Γ = \frac{V_{0}^{-}}{V_{0}^{+}} = \frac{Z_{L} - Z_{0}}{Z_{L} + Z_{0}}

|Γ| = 0 → VSWR = 1 (perfect match, no reflected wave). |Γ| = 1 → VSWR = ∞ (total reflection: open or short circuit).

📌 What Comes Next

We now have the complete mathematical framework: the RLGC model, the telegrapher's equations, the propagation constant γ = α + jβ, the characteristic impedance Z₀, and the general form of the voltage and current waves. The next article in this series applies these results in practice — examining how the load impedance sets the boundary conditions, deriving the input impedance of a terminated line as a function of electrical length, and introducing the reflection coefficient as a complex quantity that maps naturally onto the Smith chart.

Questions & Worked Examples

Apply the theory from this article to realistic RF engineering problems. Work through each question fully before reading the solution.

01

Calculate γ, α, and β for a Lossy Coaxial Cable

Problem: A coaxial cable at 1 GHz has the following per-unit-length parameters: R' = 0.8 Ω/m, L' = 250 nH/m, G' = 0.1 mS/m, C' = 100 pF/m. Calculate the propagation constant γ, the attenuation constant α in both Np/m and dB/m, and the phase constant β. What is the phase velocity?

Step 1 — Series impedance per unit length Z'

Z^{'} = R^{'} + j ω L^{'} = 0.8 + j (2 π \times 10^{9} \times 250 \times 10^{- 9}) = 0.8 + j 1570.8 Ω/m

At 1 GHz, ωL' ≫ R' — the reactive (inductive) term dominates the series impedance.

Step 2 — Shunt admittance per unit length Y'

Y^{'} = G^{'} + j ω C^{'} = 0.0001 + j (2 π \times 10^{9} \times 100 \times 10^{- 12}) = 0.0001 + j 0.6283 S/m

Similarly, ωC' ≫ G' — the capacitive term dominates the shunt admittance.

Step 3 — Compute γ² = Z'Y'

γ^{2} = (0.8 + j 1570.8) (0.0001 + j 0.6283)

Expanding the product

γ^{2} = (0.8 \times 0.0001) + j (0.8 \times 0.6283) + j (1570.8 \times 0.0001) + j^{2} (1570.8 \times 0.6283)

Collecting real and imaginary parts (j² = −1)

γ^{2} = (0.00008 - 986.9) + j (0.5026 + 0.1571) \approx - 986.9 + j 0.660

The real part of γ² is large and negative — this is characteristic of a propagating wave. The imaginary part (the loss term) is small in comparison.

Step 4 — Take the square root: γ = √(γ²)

γ \approx 0.000334 + j 31.41 m⁻¹

√(−986.9 + j0.660) ≈ j31.41 to leading order, with a small real part from the loss terms. Use the polar form: |γ²| = √(986.9² + 0.66²) ≈ 986.9, arg = π − arctan(0.660/986.9) ≈ π. Then γ = √986.9 × e^(jπ/2) ≈ j31.41. The small real part (α) requires careful handling of the complex square root.

Step 5 — Extract α and β, convert α to dB/m

α \approx 0.000334 Np/m \to 8.686 \times 0.000334 \approx 2.9 mdB/m β \approx 31.41 rad/m

Phase velocity

v_{p} = \frac{ω}{β} = \frac{2 π \times 10^{9}}{31.41} \approx 2.0 \times 10^{8} m/s = 0.667 c

The signal travels at about 2/3 of the speed of light — consistent with a PTFE-filled coaxial cable (εr ≈ 2.25, v_p = c/√2.25 = 0.667c).

Answer: α ≈ 0.00334 Np/m (2.9 mdB/m) | β ≈ 31.41 rad/m | v_p ≈ 0.667c

The attenuation is extremely low — less than 3 millidecibels per metre — indicating a high-quality low-loss cable. Over 100 m the total loss is only about 0.3 dB, which is typical of a good PTFE-dielectric coax. The phase velocity of 0.667c gives a velocity factor (VF) of 0.667, which is commonly printed on commercial cable datasheets.

02

Compute Z₀ and Verify the Lossless Approximation

Problem: Using the same coaxial cable as Q1, calculate the characteristic impedance Z₀ exactly (using the full lossy formula) and approximately (using the lossless formula Z₀ = √(L'/C')). Compare the two results and assess the validity of the lossless approximation at 1 GHz.

Exact Z₀ — general lossy formula

Z_{0} = \sqrt{\frac{Z^{'}}{Y^{'}}} = \sqrt{\frac{0.8 + j 1570.8}{0.0001 + j 0.6283}}

Compute Z'/Y' — divide numerator and denominator

\frac{0.8 + j 1570.8}{0.0001 + j 0.6283} \approx 2499.7 + j 1.26 \approx 2499.7 Ω²

The imaginary part (j1.26) is negligible compared to the real part (2499.7). This is the hallmark of a low-loss line at high frequency.

Exact Z₀

Z_{0} = \sqrt{2499.7} \approx 49.997 Ω ∠ 0.029°

The exact Z₀ is very nearly real and equal to 50 Ω, with a phase angle of only 0.029°.

Approximate Z₀ — lossless formula

Z_{0} \approx \sqrt{\frac{L^{'}}{C^{'}}} = \sqrt{\frac{250 \times 10^{- 9}}{100 \times 10^{- 12}}} = \sqrt{2500} = 50 Ω

The lossless approximation gives exactly 50 Ω — indistinguishable from the exact result for practical purposes.

Answer: Z₀(exact) ≈ 49.997 Ω ∠0.029° | Z₀(lossless) = 50.000 Ω | Error: < 0.006%

At 1 GHz, where ωL' ≫ R' and ωC' ≫ G', the lossless approximation is excellent. The error is under 0.006%. For almost all practical RF design work above 100 MHz, Z₀ = √(L'/C') is sufficient. The full complex formula for Z₀ matters mainly at low frequencies (audio to HF) or in highly lossy media.

03

Wave Amplitudes at the Source and at a Mismatched Load

Problem: A lossless 50 Ω transmission line (length l = λ/4 at 2.4 GHz) connects a source to a load Z_L = 100 Ω. The source delivers a forward voltage wave of amplitude V₀⁺ = 10 V. Find: (a) the reflection coefficient Γ at the load, (b) the amplitude of the reflected voltage wave V₀⁻, (c) the amplitude of the reflected current wave, (d) the VSWR on the line.

(a) Reflection coefficient at the load

Γ = \frac{Z_{L} - Z_{0}}{Z_{L} + Z_{0}} = \frac{100 - 50}{100 + 50} = \frac{50}{150} = \frac{1}{3} \approx 0.333

Γ is real and positive (0 < Γ < 1) because Z_L > Z₀. The reflected wave is in phase with the incident wave at the load.

(b) Reflected voltage wave amplitude

V_{0}^{-} = Γ \times V_{0}^{+} = \frac{1}{3} \times 10 = 3.33 V

One third of the incident voltage amplitude is reflected back toward the source.

I_{0}^{-} = - \frac{V_{0}^{-}}{Z_{0}} = - \frac{3.33}{50} = - 66.7 mA

The minus sign indicates the reflected current flows in the −z direction (back toward source). The magnitude is 66.7 mA.

(d) VSWR

VSWR = \frac{1 + | Γ |}{1 - | Γ |} = \frac{1 + 0.333}{1 - 0.333} = \frac{1.333}{0.667} = 2.0

VSWR = 2.0 : 1 — exactly at the conventional 2:1 threshold. Return loss = −20 log(1/3) ≈ 9.5 dB.

Answer: Γ = 1/3 | V₀⁻ = 3.33 V | I₀⁻ = −66.7 mA (backward) | VSWR = 2.0 : 1

The power reflected is |Γ|² = (1/3)² = 11.1% of the incident power. The power delivered to the load is 1 − 11.1% = 88.9%. Note that the quarter-wave length does not affect Γ or VSWR — these are determined entirely by the load and line impedances. The λ/4 length does, however, transform the 100 Ω load to a different input impedance: Z_in = Z₀²/Z_L = 50²/100 = 25 Ω — the quarter-wave transformer equation, a topic for the next article.

04

Guided Wavelength and Electrical Length on Microstrip

Problem: A microstrip trace on Rogers RO4350B (εeff = 3.03 at 5 GHz) is designed as a quarter-wave resonator at 5 GHz. (a) What physical length l is required? (b) What is β on this line? (c) If the same physical length is used at 2.4 GHz (εeff ≈ 3.10), what is its electrical length in degrees?

(a) Guided wavelength and quarter-wave length at 5 GHz

λ_{g} = \frac{c}{f \sqrt{ε_{eff}}} = \frac{3 \times 10^{8}}{5 \times 10^{9} \times \sqrt{3.03}} = \frac{3 \times 10^{8}}{5 \times 10^{9} \times 1.741} \approx 34.46 mm

Quarter-wave physical length

l = \frac{λ_{g}}{4} = \frac{34.46}{4} \approx 8.62 mm

This is a very compact resonator — less than 9 mm for a 5 GHz quarter-wave on RO4350B.

(b) Phase constant β at 5 GHz

β = \frac{2 π}{λ_{g}} = \frac{2 π}{0.03446} \approx 182.3 rad/m

β can also be computed as β = (2πf/c)√εeff = (2π×5×10⁹/3×10⁸)×1.741 ≈ 182.3 rad/m — same result.

θ = \frac{360 ° \times l \times f \times \sqrt{ε_{eff}}}{c} = \frac{360 \times 0.00862 \times 2.4 \times 10^{9} \times \sqrt{3.10}}{3 \times 10^{8}}

Evaluate

θ = \frac{360 \times 0.00862 \times 2.4 \times 10^{9} \times 1.761}{3 \times 10^{8}} \approx 43.7 °

The same trace that is λ/4 (90°) at 5 GHz is only 43.7° at 2.4 GHz — less than a quarter-wave. A dual-band design cannot simply use the same length for both frequencies.

Answer: l = 8.62 mm | β = 182.3 rad/m at 5 GHz | θ = 43.7° at 2.4 GHz

The frequency scaling is nearly linear here — 43.7° ÷ 90° ≈ 48.6%, while 2.4/5 = 48%. The small deviation arises because εeff changes slightly between 5 GHz and 2.4 GHz (dispersion). This is a critical insight for dual-band and wideband antenna and filter design: electrical length scales approximately with frequency, but dispersion in the substrate means it is not perfectly proportional.