03 · Advanced

Characteristic impedance

Promote the forward-wave voltage/current ratio to the exact lossy characteristic impedance Z₀=√(Z′/Y′), tie its branch to passive γ, recover the lossless limit, connect Z₀ to geometry through a TEM coax example, derive the low-loss approximation and use the backward-current sign to audit transported power.

190–245 minAdvanced
In this lesson
  • Derive Z₀=Z′/γ=γ/Y′=√(Z′/Y′) from the first-order telegrapher equations.
  • Explain why the physical square-root branch of Z₀ is tied to the selected passive branch of γ.
  • Check the units and distinguish Z₀ from the per-unit-length series impedance Z′.
  • Recover Z₀=√(L′/C′) and invert {Z₀,v_p} to recover L′ and C′ on a lossless line.
  • Explain how line geometry and material properties jointly determine characteristic impedance.
  • Derive the exact TEM coax expression and separate intrinsic field impedance η from line Z₀.
  • Derive the first-order low-loss approximation for complex Z₀.
  • Use peak-phasor line power to show why the backward-wave current sign produces negative +z power.
  • Distinguish characteristic, load, input and intrinsic wave impedance by physical definition.
  • Audit a 50 Ω line numerically in both lossless and weakly lossy regimes.

Chapter 02 already produced the quantity √(Z′/Y′) while relating forward voltage and current. We now promote that ratio to a physical line parameter and make its meaning precise. Characteristic impedance Z₀ is not a resistor inserted into the line and not the same as the bulk field impedance η: it is the complex voltage/current ratio of ONE traveling mode on a uniform transmission line, with its branch tied to the passive propagation constant.

0103.1 — Define Z₀ from the first-order line equations, not by memorization

For the +z contribution, chapter 02 gave i⁺=(γ/Z′)v⁺. Therefore the forward traveling-wave voltage/current ratio is Z′/γ. Because γ²=Z′Y′, the same quantity is γ/Y′ and also √(Z′/Y′).

AoN defines this modal ratio as the characteristic impedance Z₀. Its unit is ohms because Z′/Y′ has units Ω². This definition belongs to a UNIFORM line mode; it does not require a load to be attached.

For the reverse wave, the total line current is still referenced toward +z, so i⁻=−v⁻/Z₀. Consequently V⁻/I⁻=−Z₀ under the fixed current reference. Dropping this sign corrupts later power and reflection formulas.

Figure 3.1 — Z₀ as the modal ratio of a traveling wave

One question: with the current reference fixed toward +z, why is the voltage/current ratio +Z₀ for the forward wave but −Z₀ for the backward wave?

+zV⁺ / I⁺ = Z₀−zV⁻ / I⁻ = −Z₀

Both conductor lines represent the same line mode. The ratio sign comes from keeping the current reference fixed toward +z: reversing propagation changes the modal current sign, not the reference itself.

Z0=Z′/γ=γ/Y′=√(Z′/Y′)
[Z0]=Ω
i+=v+/Z0
i-=−v-/Z0
V+/I+=Z0, V-/I-=−Z0

0203.2 — The square-root branch is tied to the passive propagation branch

The algebraic square root has two signs. Do not choose the sign of Z₀ independently from γ. AoN first chooses the forward passive propagation branch γ with Re(γ)=α≥0, then defines Z₀=Z′/γ.

For an ordinary passive line with L′>0, C′>0 and nonnegative R′,G′, this choice gives the physically useful branch with Re(Z₀)>0. Flipping both γ and Z₀ merely swaps which exponential is called forward; flipping only one breaks the first-order telegrapher equations.

This linked branch choice is more reliable than memorizing 'take the positive square root', especially when Z₀ is complex.

γ=+√(Z′Y′), choose Re(γ)≥0 for the +z passive branch
Z0=Z′/γ
passive ordinary line: Re(Z0)>0

0303.3 — Lossless limit: Z₀ becomes real and frequency-independent

Set R′=G′=0. Then Z′=jωL′ and Y′=jωC′, so their ratio is L′/C′. The characteristic impedance is therefore real and positive: Z₀=√(L′/C′).

Combine this with v_p=1/√(L′C′). The pair {Z₀,v_p} can be inverted to recover the storage parameters: L′=Z₀/v_p and C′=1/(Z₀v_p).

In this ideal nondispersive line model neither Z₀ nor v_p depends on frequency. Real interconnects can depart from that through conductor loss, dielectric dispersion, frequency-dependent field confinement and higher modes.

R′=G′=0 ⇒ Z0=√(L′/C′)
vp=1/√(L′C′)
L′=Z0/vp
C′=1/(Z0 vp)

0403.4 — Geometry and material both enter Z₀: exact TEM coax example

Characteristic impedance is not a material constant. The per-unit-length L′ and C′ depend on the transverse field geometry as well as ε and μ. A clean exact transverse electromagnetic (TEM) example is a homogeneous coaxial line with inner radius a and outer-conductor inner radius b.

For that geometry, L′=(μ/2π)ln(b/a) and C′=2πε/ln(b/a). Their ratio gives Z₀=(1/2π)√(μ/ε)ln(b/a)=η ln(b/a)/(2π), where η=√(μ/ε) is the intrinsic wave impedance of the filling medium.

The formula displays the distinction clearly: η is a FIELD ratio of the medium, while Z₀ multiplies it by a geometry factor. Different line geometries in the same dielectric therefore have different characteristic impedances.

coax: L′=(μ/2π) ln(b/a)
coax: C′=2πε/ln(b/a)
Z0=(1/2π)√(μ/ε) ln(b/a)=η ln(b/a)/(2π)
nonmagnetic: Z0≈(60/√εr) ln(b/a) [Ω]

0503.5 — Exact lossy Z₀ and the controlled low-loss approximation

With loss, Z₀ is generally complex: Z₀=√[(R′+jωL′)/(G′+jωC′)]. A complex voltage/current ratio means the local traveling-wave voltage and current phasors are not exactly in phase.

If R′≪ωL′ and G′≪ωC′, expand the square root to first order. Under AoN's exp(+jωt) convention, the correction is −(j/2)[R′/(ωL′)−G′/(ωC′)] multiplying the lossless value.

This approximation is useful only when both dimensionless loss ratios are small. Near DC, or on a strongly resistive/leaky line, use the exact expression.

Z0=√[(R′+jωL′)/(G′+jωC′)]
low loss: Z0≈√(L′/C′){1−(j/2)[R′/(ωL′)−G′/(ωC′)]}
require: R′/(ωL′)≪1, G′/(ωC′)≪1

0603.6 — Z₀ controls the sign and magnitude of traveling-wave power

For peak phasors, the time-averaged line power crossing a transverse section in the +z bookkeeping direction is P=(1/2)Re{v i*}. Apply this to a single forward wave i⁺=v⁺/Z₀.

The forward power is (1/2)|V⁺|²Re(1/Z₀), positive for a passive line branch. For a reverse wave the fixed +z current reference contributes the minus sign, so the +z-directed power is negative: P⁻=−(1/2)|V⁻|²Re(1/Z₀).

On a lossless line Z₀ is real, reducing these to ±|V|²/(2Z₀). The reverse-current sign from chapter 02 is therefore exactly what makes a reverse wave transport energy toward −z.

P(z)=(1/2) Re{v(z) i*(z)}
P+=(|V+|²/2) Re(1/Z0)>0
P-=−(|V-|²/2) Re(1/Z0)<0
lossless: P+=|V+|²/(2Z0), P-=−|V-|²/(2Z0)

0703.7 — Do not confuse Z₀ with load, input or intrinsic wave impedance

Z₀ is a property of a uniform line mode. A load impedance Z_L is a boundary condition connected at a termination. The input impedance Z_in is the voltage/current ratio seen at a chosen position after forward and backward waves have combined; it therefore depends on Z_L, line length and γ.

Intrinsic wave impedance η=E/H belongs to an electromagnetic mode in a medium. In a TEM line, Z₀ is derived by integrating fields into a line voltage and current, so geometry enters. The coax example made this distinction explicit.

A line can have Z₀=50 Ω while being terminated in 75 Ω, shorted, open-circuited or connected to a complex device. Z₀ does not become the load; the mismatch determines reflected-wave amplitude, which belongs to RF-01.

Z0: traveling-wave ratio of the uniform line mode
ZL: termination boundary condition
Zin(z)=v(z)/i(z) for the TOTAL field
η=E/H [field wave impedance]

0803.8 — Numerical audit: 50 Ω lossless line, then a weakly lossy case

For L′=250 nH/m and C′=100 pF/m, Z₀=√(L′/C′)=50.0 Ω and v_p=2.00×10^8 m/s. These are the same storage values used in chapter 02.

Now add R′=10 Ω/m and G′=2.0 mS/m at 100 MHz. The exact expressions give Z₀≈50.0316−j0.7945 Ω and γ≈0.14998+j3.14199 m⁻¹. The line remains close to 50 Ω but voltage and current acquire a small relative phase.

If the forward peak voltage amplitude is 2.00 V, the average forward power is about 39.96 mW using (1/2)|V⁺|²Re(1/Z₀). Using 50 Ω as if the lossy impedance were exactly real would be a small approximation here, not an identity.

L′=250 nH/m, C′=100 pF/m ⇒ Z0=50.0 Ω
f=100 MHz, R′=10 Ω/m, G′=2.0 mS/m
Z0≈50.0316−j0.7945 Ω
γ≈0.14998+j3.14199 m-1
V+peak=2.00 V ⇒ P+≈39.96 mW

Practice — derive, branch-select and interpret characteristic impedance

  1. Starting from d v̲/dz=−Z′i̲ and a pure forward exponential, derive Z₀=Z′/γ.
  2. Use γ²=Z′Y′ to prove Z′/γ=γ/Y′=√(Z′/Y′), including a unit check.
  3. Explain why the sign of Z₀ cannot be chosen independently from the branch of γ.
  4. For a lossless line with L′=360 nH/m and C′=90 pF/m, compute Z₀ and v_p.
  5. Invert Z₀=√(L′/C′) and v_p=1/√(L′C′) to derive L′ and C′ in terms of Z₀ and v_p.
  6. A homogeneous nonmagnetic coax has ε_r=2.25 and b/a=3. Calculate its ideal TEM Z₀ using 60 ln(b/a)/√ε_r.
  7. Derive the first-order low-loss approximation for Z₀ by expanding the exact ratio.
  8. For a pure backward wave, use i⁻=−v⁻/Z₀ to show that the +z-directed average power is negative on a passive line.
  9. Explain the difference among Z₀, Z_L, Z_in and η without using the word 'impedance' alone as the definition.
  10. Using R′=10 Ω/m, L′=250 nH/m, G′=2 mS/m, C′=100 pF/m at 100 MHz, compare the loss ratios R′/(ωL′) and G′/(ωC′) and judge whether the low-loss approximation is justified.
Worked solutions

1. For v̲⁺=V⁺e^{−γz}, d v̲⁺/dz=−γv̲⁺. The first-order equation gives −γv̲⁺=−Z′i̲⁺, hence v̲⁺/i̲⁺=Z′/γ≡Z₀.

2. Because γ²=Z′Y′, divide by γY′ to get γ/Y′=Z′/γ. Squaring either ratio gives Z′/Y′. Since (Ω/m)/(S/m)=Ω², its square root has units Ω.

3. The forward exponential is selected by a γ branch. The first-order equation then fixes i̲⁺=(γ/Z′)v̲⁺, so Z₀=Z′/γ. Changing only Z₀ would violate that equation; changing both signs swaps the forward/backward labeling.

4. Z₀=√(360 nH/m ÷ 90 pF/m)=√4000≈63.25 Ω. v_p=1/√[(360×10⁻⁹)(90×10⁻¹²)]≈1.757×10^8 m/s.

5. Multiply/divide the two equations: Z₀/v_p=√(L′/C′)√(L′C′)=L′, while 1/(Z₀v_p)=C′.

6. Z₀≈60 ln3/1.5≈43.94 Ω.

7. Write Z′=jωL′[1−jR′/(ωL′)] and Y′=jωC′[1−jG′/(ωC′)]. To first order, their ratio contributes 1−jR′/(ωL′)+jG′/(ωC′); sqrt(1+ε)≈1+ε/2 gives the stated formula.

8. P⁻=(1/2)Re{v̲⁻(i̲⁻)*}=−(1/2)|V⁻|²Re(1/Z₀). For a passive branch Re(1/Z₀)>0, so P⁻<0 in the +z bookkeeping direction: energy moves toward −z.

9. Z₀ is the voltage/current ratio of one traveling line mode; Z_L is the termination relation imposed by the connected load; Z_in is the total local voltage/current ratio after waves combine; η is the E/H field ratio of an electromagnetic mode in a medium.

10. At 100 MHz, ωL′≈157.08 Ω/m and ωC′≈0.06283 S/m. Thus R′/(ωL′)≈0.0637 and G′/(ωC′)≈0.0318. Both are below one but only a few percent; first-order low-loss formulas are useful estimates, while the exact expression is preferable for precision.

Chapter 04 will return to γ and study propagation loss and distortion in detail: low-loss α and β, conductor versus dielectric contributions, the Heaviside distortionless condition, group/phase delay and the limits of a frequency-independent RLGC model.