Solve the telegrapher equations
Move the time-domain line equations into AoN phasor form, define the series impedance and shunt admittance per unit length, decouple the system, derive γ²=Z′Y′, interpret γ=α+jβ, and obtain forward/backward voltage and current waves with the correct reverse-current sign.
- Apply AoN's exp(+jωt) phasor convention while retaining spatial dependence.
- Define Z′=R′+jωL′ and Y′=G′+jωC′ with correct units.
- Derive the second-order voltage and current wave equations from the coupled first-order line equations.
- Define γ from γ²=Z′Y′ and choose the passive branch with α=Reγ≥0.
- Interpret α and β from exp(−γz) and verify +z phase propagation.
- Write the general forward/backward voltage solution and explain how boundary conditions set its amplitudes.
- Derive the current solution from the first-order equation and preserve the backward-wave minus sign.
- Recover β=ω√(L′C′) and v_p=1/√(L′C′) for a lossless line.
- Use underlined lowercase phasors consistently with ADR-0018.
- Solve numerical phase/delay examples and distinguish phase change from attenuation.
Chapter 01 produced two coupled partial differential equations in z and t. We now solve them for sinusoidal steady state without changing AoN's electrical notation: instantaneous line quantities remain lowercase v(z,t), i(z,t), while their harmonic phasors are underlined lowercase quantities. The result is the bridge from a distributed circuit model to genuine forward and backward waves.
0102.1 — Move to phasors without changing the spatial problem
Assume a single angular frequency ω and write each real instantaneous quantity as the real part of a complex phasor amplitude multiplied by exp(+jωt). The phasors still depend on z: harmonic steady state removes the explicit time derivative, not the spatial dependence.
Therefore ∂/∂t becomes multiplication by jω, while d/dz remains a spatial derivative. This is exactly the same phasor logic used in AC-01, now applied to a continuum rather than a finite lumped network.
0202.2 — The harmonic telegrapher equations become two coupled ODEs
Insert the phasor forms into the time-domain equations from chapter 01. Define the series impedance per unit length Z′=R′+jωL′ and the shunt admittance per unit length Y′=G′+jωC′.
The line equations are now ordinary differential equations in z. Their units remain Z′ in Ω/m and Y′ in S/m, so Z′i has V/m and Y′v has A/m.
Do not confuse Z′ with the characteristic impedance of the line. Z′ is a SERIES impedance per unit length; the modal voltage/current ratio is derived later.
0302.3 — Decouple the equations and obtain two wave equations
Differentiate the voltage equation once more with respect to z. For a uniform line Z′ and Y′ are independent of z. Substitute d i/dz from the second equation; the current disappears and voltage satisfies a second-order equation.
Performing the same operation on current gives the identical propagation factor. The common coefficient is the product Z′Y′, with units 1/m².
Define γ by γ²=Z′Y′. The square root has two algebraic signs; the physical branch for a passive forward-decaying line is chosen so Re(γ)=α≥0.
0402.4 — Read attenuation and phase directly from γ
For the +z solution exp(−γz), write γ=α+jβ. Then exp(−γz)=exp(−αz)exp(−jβz). The real nonnegative α attenuates amplitude in nepers per metre; β is the phase constant in radians per metre.
With the physical factor exp(+jωt), constant phase satisfies ωt−βz=constant, so the exp(−jβz) factor moves toward +z. This sign check matches EM-04's plane-wave convention.
The −z solution exp(+γz) is not 'unphysical' merely because it grows with increasing z on an infinite coordinate axis. It represents a wave referenced in the opposite direction and is used on finite lines. Boundary conditions decide which combination is present.
Figure 2.1 — Two propagation solutions on the same z axis
With the exp(+jωt) time convention, exp(−γz) is the +z solution when γ=α+jβ with α≥0; exp(+γz) is the opposite solution. Arrows indicate phase-propagation direction, not an imposed current direction.
The total voltage is the superposition of both contributions. Boundary conditions set their relative amplitudes; the line itself does not force only one wave to exist.
0502.5 — The general voltage solution is a superposition of two waves
The second-order voltage equation has two independent exponential solutions. Denote their complex amplitudes by V⁺ and V⁻. These constants are set by source/load boundary conditions, not by the differential equation alone.
For a passive line with α>0, the +z contribution decays as z increases. The −z term is naturally interpreted by referencing distance from the load or by remembering that it propagates toward decreasing z.
A pure traveling wave is a special boundary-condition case. Reflections in RF-01 will make both terms present and create position-dependent interference.
0602.6 — Current follows from the FIRST-order equation and acquires the opposite sign for the reverse wave
Do not guess the current solution independently. Insert the voltage solution into d v/dz=−Z′ i. Differentiating exp(−γz) gives −γ, while differentiating exp(+γz) gives +γ.
Therefore the forward current coefficient has the same sign as V⁺, whereas the backward current coefficient has the opposite sign relative to V⁻. This sign is essential for power direction and later reflection formulas.
The ratio Z′/γ=γ/Y′=√(Z′/Y′) emerges naturally. Chapter 03 will name, branch-select and interpret this modal voltage/current ratio as characteristic impedance Z₀.
0702.7 — Lossless line: finite propagation appears from L′ and C′ alone
Set R′=G′=0. Then γ²=(jωL′)(jωC′)=−ω²L′C′. Choose the passive/positive-phase branch γ=jβ with β=ω√(L′C′). There is no attenuation: α=0.
Phase velocity is v_p=ω/β=1/√(L′C′). Thus the same per-unit-length inductive and capacitive storage introduced in chapter 01 directly set the signal propagation speed.
This result is one of the clearest reasons transmission-line theory is not merely 'many tiny capacitors and inductors': their continuum limit represents electromagnetic energy storage and propagation.
0802.8 — Numerical audit with L′=250 nH/m and C′=100 pF/m
Take a pedagogical lossless line with L′=250 nH/m and C′=100 pF/m. Their product is 2.5×10⁻¹⁷ s²/m², so v_p=1/√(L′C′)=2.00×10⁸ m/s.
At 1.00 GHz, β=ω√(L′C′)=31.416 rad/m and λ_g=2π/β=0.200 m. A 10 cm section is therefore exactly half a guided wavelength, or 180° electrically, in this ideal example.
The corresponding forward phase factor across 10 cm is exp(−jπ)=−1: the phasor voltage changes sign even though a lossless line has not dissipated energy. Phase change is not attenuation.
Practice — solve and interpret the harmonic line equations
- Starting from the time-domain voltage telegrapher equation, derive its phasor form using exp(+jωt).
- Show dimensionally that Z′Y′ has units 1/m² and γ has units 1/m.
- Derive d²v/dz²=γ²v without assuming a voltage solution in advance.
- For γ=0.08+j12 rad/m, state attenuation constant, phase constant and the amplitude ratio after 2 m for the +z wave.
- Use constant phase to prove that exp(+jωt)exp(−jβz) travels toward +z.
- Differentiate the general voltage solution and derive the current solution, including the minus sign on the backward-wave current.
- For R′=G′=0, derive β and v_p from γ².
- A lossless line has L′=400 nH/m and C′=100 pF/m. Find v_p and λ_g at 500 MHz.
- For the chapter numerical line, what is the phasor phase change over 5 cm at 1 GHz?
- Why must boundary conditions be supplied after solving the differential equation? Give one physical example of a boundary condition.
Worked solutions
1. Write v=Re{v̲e^{jωt}}, i=Re{i̲e^{jωt}}. Then ∂i/∂t maps to jωi̲, giving d v̲/dz=−R′i̲−jωL′i̲=−(R′+jωL′)i̲.
2. Z′ is Ω/m and Y′ is S/m=1/(Ω·m), so Z′Y′ has 1/m². Its square root γ has 1/m.
3. Differentiate d v̲/dz=−Z′i̲. For a uniform line Z′ is constant: d²v̲/dz²=−Z′ d i̲/dz=Z′Y′v̲=γ²v̲.
4. α=0.08 Np/m and β=12 rad/m. The forward-wave amplitude factor over 2 m is exp(−αz)=exp(−0.16)≈0.852.
5. Constant phase gives ωt−βz=C, hence z=(ω/β)t−C/β. Increasing t increases z when ω,β>0.
6. Differentiate v̲=V⁺e^{−γz}+V⁻e^{+γz}; substitute into i̲=−(1/Z′)dv̲/dz. The result is (γ/Z′)[V⁺e^{−γz}−V⁻e^{+γz}].
7. γ²=−ω²L′C′, so choose γ=jω√(L′C′). Hence β=ω√(L′C′), α=0 and v_p=ω/β=1/√(L′C′).
8. √(L′C′)=√(4×10⁻¹⁷)=6.3249 ns/m, so v_p≈1.581×10⁸ m/s. At 500 MHz, λ_g=v_p/f≈0.3162 m.
9. The 10 cm section is 180°, so 5 cm is 90°. The forward phasor factor is exp(−jπ/2)=−j.
10. The second-order equation fixes the allowed spatial form but not V⁺ and V⁻. A load impedance at z=ℓ, a specified source voltage/current at z=0, or a matched termination supplies boundary information.
Chapter 03 will formalize the voltage/current ratio that already emerged here: characteristic impedance Z₀. We will derive its exact lossy form, lossless limit, physical units, passive square-root branch and the relation between forward/backward current signs and transported power.