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core30 min studyLesson 285 in Physics

Transmission-line telegrapher equations

Derive signed voltage-current evolution from a distributed line, obtain propagation speed, and separate attenuation from phase delay.

Before you begin
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The question for today

What you’ll understand

  • Derive both telegrapher equations from a consistently oriented segment.
  • Calculate lossless wave speed from per-length parameters.
  • Interpret complex propagation constant and the low-loss attenuation approximation.

A wire pair can be locally described by inductance and capacitance even when its total length is too large for one common voltage. Dividing it into short segments preserves local circuit balances while allowing voltage and current to depend on position. Taking the segment length toward zero produces transmission-line equations. This bridge between circuits and fields explains why a perfect conductor can still have a position dependent voltage and why a line can delay signals without dissipating them.

Orient the segment before taking a limit

Let z increase from source toward load and define positive current along that direction on the signal conductor, with a return conductor completing the circuit. A short segment has series resistance and inductance proportional to its length, plus shunt leakage conductance and capacitance proportional to that same length. The voltage at its input exceeds the voltage at its output by the series drop. Thus the spatial voltage derivative carries a minus sign under the passive convention.

Current decreases along the segment when some flows through its shunt conductance or charges its distributed capacitance. The current derivative therefore has the same negative orientation. These signs are not arbitrary decorations: a forward wave must carry positive voltage and current together for positive forward power. Reversing the definition of current consistently changes the formulas. Mixing a load-facing voltage convention with a source-facing current convention can falsely turn attenuation into growth or reverse reflected-wave power.

∂v∂z=−R′i−L′∂i∂t,∂i∂z=−G′v−C′∂v∂t\frac{\partial v}{\partial z}=-R^{\prime}i-L^{\prime}\frac{\partial i}{\partial t},\qquad \frac{\partial i}{\partial z}=-G^{\prime}v-C^{\prime}\frac{\partial v}{\partial t}
v is in volts and i in amperes. Per-length parameters R′, L′, G′, C′ have units Ω/m, H/m, S/m, F/m. The uniform distributed model describes a single TEM or suitable quasi-TEM mode with linear constitutive behavior.↔ Scroll wide equations horizontally.

Lossless storage produces traveling waves

Set resistance and leakage to zero and differentiate the first equation in position. Substituting the time derivative of the second yields a wave equation with propagation speed set by the product of per-length inductance and capacitance. A similar equation holds for current. A single forward wave has a fixed voltage-current ratio determined by the ratio of those same parameters. Their product controls speed while their ratio controls impedance, so equal propagation speeds do not imply equal characteristic impedances.

For an ideal uniform two-conductor line in a homogeneous medium, the TEM description can be exact for that mode. In inhomogeneous structures such as microstrip, a quasi-TEM model is an approximation whose parameters can depend on frequency. Higher modes and discontinuity fields are outside a single voltage-current description. The local segment limit does not automatically validate that reduction: transverse geometry, material response, and the frequency band must still support the assumed mode.

∂2v∂z2=L′C′∂2v∂t2,vp=1L′C′,Z0=L′C′\frac{\partial^2v}{\partial z^2}=L^{\prime}C^{\prime}\frac{\partial^2v}{\partial t^2},\qquad v_p=\frac1{\sqrt{L^{\prime}C^{\prime}}},\qquad Z_0=\sqrt{\frac{L^{\prime}}{C^{\prime}}}
These lossless formulas require R′ = G′ = 0. Phase speed vp is in m/s and characteristic impedance Z₀ in ohms. The wave equation admits forward and backward traveling solutions, with boundary conditions setting their amplitudes.↔ Scroll wide equations horizontally.

Attenuation and phase have different roles

Using the positive-time phasor convention gives a forward voltage proportional to an exponential with minus gamma z. Gamma's real part is the attenuation constant and its imaginary part is the phase constant. For a passive line choose the branch with nonnegative attenuation and the forward phase convention. Voltage amplitude falls with attenuation; power in a uniform traveling wave falls with twice that exponent. A phase change alone is not power loss.

When series resistance is much smaller than inductive reactance per length and shunt conductance is much smaller than capacitive susceptance per length, expand the square root to obtain the low-loss attenuation formula. Both conductor loss and dielectric leakage contribute positively. If these inequalities fail, use the full complex expression and generally complex characteristic impedance. Constant per-length parameters are themselves a band-limited model; skin effect and dielectric dispersion can invalidate a broadband extrapolation even when a single-frequency calculation is sound.

γ=α+jβ=(R′+jωL′)(G′+jωC′),α≃R′2Z00+G′Z002,Z00=L′/C′\gamma=\alpha+j\beta=\sqrt{(R^{\prime}+j\omega L^{\prime})(G^{\prime}+j\omega C^{\prime})},\qquad \alpha\simeq\frac{R^{\prime}}{2Z_{00}}+\frac{G^{\prime}Z_{00}}2,\quad Z_{00}=\sqrt{L^{\prime}/C^{\prime}}
α is in Np/m and β in rad/m. The approximation requires R′ ≪ ωL′ and G′ ≪ ωC′; Z₀₀ is the lossless impedance. For a forward wave V(z) = V(0)e^{−γz}, positive α gives decay.↔ Scroll wide equations horizontally.

Work it through

Extract speed and impedance separately

A lossless line has L′ = 250 nH/m and C′ = 100 pF/m. Find its phase speed, characteristic impedance, and delay through 3 m.

  1. The product is (250×10−9)(100×10−12)=2.5×10−17 s2/m2(250\times10^{-9})(100\times10^{-12})=2.5\times10^{-17}\,\mathrm{s^2/m^2}. Its square root is 5.0×10−9 s/m5.0\times10^{-9}\,\mathrm{s/m}.

  2. Therefore vp=2.0×108 m/sv_p=2.0\times10^8\,\mathrm{m/s}. The parameter ratio is L′/C′=2500 Ω2L^{\prime}/C^{\prime}=2500\,\Omega^2, giving Z0=50 ΩZ_0=50\,\Omega.

  3. Travel time is length divided by speed: 3/(2.0×108)=15 ns3/(2.0\times10^8)=15\,\mathrm{ns}. This delay exists even though the line has no resistive or leakage loss.

The result

The phase speed is 2.0 × 10⁸ m/s, characteristic impedance is 50 Ω, and one-way delay through 3 m is 15 ns.

The living notebook

Explore the model

Interactive lab

Interactive lab · waves

Phase delay without amplitude loss

Reset and set BOTH Phase difference to 0 degrees and Amplitude ratio to 1; note identical component peaks. Then Reset and set BOTH Phase difference to −90 degrees and Amplitude ratio to 1; predict shifted peaks with unchanged component amplitudes. The two waves have fixed equal frequency and travel in the same direction. This illustrates phase offset, not a lossy-line solution, adjustable speed, or pulse propagation.

Phase delay without amplitude lossTwo waves share the same frequency and wave number. Playback advances t in sin(2x − t) + a sin(2x − t + phase). Current readings: Sum at x = 0: 0.866; Resultant amplitude: 1.732; Phase difference: 60°. All parameters are adjustable with the labeled controls below.−3.14−1.5701.573.14−3.4−1.701.73.4x · radiansAmplitude
SumFirst componentSecond component
Sum at x = 0
0.866
Resultant amplitude
1.732
Phase difference
60°
t = 0 s

Two waves share the same frequency and wave number. Playback advances t in sin(2x − t) + a sin(2x − t + phase). Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA shifted phase and a reduced amplitude are distinct effects in a propagation model.

Pause & reason

Check your understanding

For a passive forward wave V(z) = V(0)e^{−(α+jβ)z}, what is the role of α?

Your turn

Try it yourself

A weakly lossy line has lossless impedance 50 Ω, R′ = 2 Ω/m, and G′ = 0.0004 S/m at a frequency where the low-loss inequalities hold. Estimate α and the voltage-amplitude ratio after 10 m.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Add conductor and leakage contributions, then exponentiate minus attenuation times length.

Compare with the solution

The conductor term is 2/(2×50)=0.020 m−12/(2\times50)=0.020\,\mathrm{m^{-1}}. Leakage contributes 0.0004(50)/2=0.010 m−10.0004(50)/2=0.010\,\mathrm{m^{-1}}. Hence α≃0.030 Np/m\alpha\simeq0.030\,\mathrm{Np/m}. The amplitude ratio over 10 m is e−0.30≃0.741e^{-0.30}\simeq0.741. In this low-loss traveling-wave approximation, the corresponding power ratio is approximately e−0.60≃0.549e^{-0.60}\simeq0.549. The lossless speed formula alone would not reveal these reductions.

Keep the idea

Take it with you

  • Distributed storage creates delay even with perfect conductors and dielectrics.
  • Per-length parameter products set speed, while their ratios set impedance.
  • A passive propagation constant separates positive attenuation from phase advance.

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