Electromagnetic Fields and Waves: Notes, Solved Problems and Interactive Lab

Because the transverse electric field has zero curl, the line integral of $E$ between the conductors in one cross-section does not depend on the path, and the voltage is well defined:

V (z) = \int_{a}^{b} E_{ρ} d ρ = \frac{V _{0}}{ln ( b / a )} e^{- j k z} \int_{a}^{b} \frac{d ρ}{ρ} = V_{0} e^{- j k z} . (7.5.6)

The current follows from Ampère's law around the inner conductor (Section 4.1): the circle of radius $ρ$ encloses the current $I$ of the inner conductor, and

I (z) = \oint H_{ϕ} ρ d ϕ = 2 π ρ H_{ϕ} = \frac{2 π V _{0}}{η ln ( b / a )} e^{- j k z}, (7.5.7)

independent of $ρ$ , as it must be. The outer conductor carries $- I$ on its inner surface.

Characteristic impedance and line constants

Z_{0} = \frac{V}{I} = \frac{η}{2 π} ln \frac{b}{a} = \frac{60}{ε _{r}} ln \frac{b}{a} [Ω], L^{'} = \frac{μ}{2 π} ln \frac{b}{a}, C^{'} = \frac{2 π ε}{ln ( b / a )} . (7.5.8)

In plain words

$Z_{0}$ is the ratio of voltage to current of a wave travelling one way. It is set by the filling ( $η$ ) and by the geometry through the logarithm of $b / a$ only: scaling the whole cable up or down does not change $Z_{0}$ .

$C^{'}$ is the capacitance per unit length of the cylindrical capacitor, Eq. (3.4.29) in Section 3.4, and $L^{'}$ is the external inductance per unit length of the coaxial cable (Home exercise 4.2.1 in Section 4.2). Two checks connect the field picture with circuit theory:

\frac{L ^{'}}{C ^{'}} = \frac{μ}{ε} \frac{ln ( b / a )}{2 π} = Z_{0}, \frac{1}{L ^{'} C ^{'}} = \frac{1}{μ ε} = v .

The transmission-line model with $L^{'}$ and $C^{'}$ is exact for the TEM wave: the line equations of circuit theory and Maxwell's equations give the same voltage, current and speed.

Figure 7.5.3 shows $Z_{0}$ against $b / a$ . A $50 Ω$ line needs $b / a = 2.30$ in air, $3.49$ with polyethylene and $3.35$ with PTFE. $75 Ω$ in polyethylene needs $b / a = 6.52$ . Because $Z_{0}$ depends only logarithmically on $b / a$ , a wide range of impedances is impractical: $10 Ω$ in air needs $b / a = 1.18$ , a gap of a few per cent of the radius, and $150 Ω$ needs $b / a = 12.2$ , a very thin inner conductor.

Characteristic impedance of a coaxial line against b/a, for air (60\ln(b/a)), PTFE (\varepsilon_r=2.1) and polyethylene (\varepsilon_r=2.25, 40\ln(b/a)). The dots mark the 50\ \Omega designs (b/a=2.30, 3.35, 3.49) and the 75\ \Omega polyethylene cable (6.52). — **Figure 7.5.3.** Characteristic impedance of a coaxial line against $b / a$ , for air ( $60 ln (b / a)$ ), PTFE ( $ε_{r} = 2.1$ ) and polyethylene ( $ε_{r} = 2.25$ , $40 ln (b / a)$ ). The dots mark the $50 Ω$ designs ( $b / a = 2.30$ , $3.35$ , $3.49$ ) and the $75 Ω$ polyethylene cable ( $6.52$ ).

Quick check

An RG-58 type cable has $2 a = 0.81$ mm and a polyethylene dielectric of diameter $2 b = 2.95$ mm. Find $Z_{0}$ and the velocity factor $v / c$ .

$ln (b / a) = ln (2.95/0.81) = 1.293$ and $Z_{0} = 40 \times 1.293 = 51.7 Ω$ , nominally a $50 Ω$ cable. $v / c = 1/ 2.25 = 0.667$ : a signal needs $5.0$ ns per metre.

7.5.2 Voltage, current and characteristic impedance