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

A TEM wave has $E_{z} = H_{z} = 0$ . Write its fields as $E = E_{T} (ρ, ϕ) e^{- j τ z}$ , $H = H_{T} (ρ, ϕ) e^{- j τ z}$ , with $E_{T}$ and $H_{T}$ in the cross-section. Split the nabla operator into $\nabla = \nabla_{T} + a_{z} \partial / \partial z$ and apply Maxwell's equations in the source-free filling.

Faraday's law. $\nabla \times E = - j ω μ H$ has two parts. The part along $a_{z}$ is $(\nabla_{T} \times E_{T}) \cdot a_{z} = - j ω μ H_{z} = 0$ , so

\nabla_{T} \times E_{T} = 0. (7.5.1)

The transverse part comes from the $z$ derivative alone: $a_{z} \times (- j τ E_{T}) = - j ω μ H_{T}$ , that is

H_{T} = \frac{τ}{ω μ} a_{z} \times E_{T} . (7.5.2)

Ampère's law. $\nabla \times H = j ω ε E$ gives in the same way $- j τ a_{z} \times H_{T} = j ω ε E_{T}$ . Inserting Eq. (7.5.2) and using $a_{z} \times (a_{z} \times E_{T}) = - E_{T}$ ,

E_{T} = - \frac{τ}{ω ε} a_{z} \times H_{T} = \frac{τ ^{2}}{ω ^{2} μ ε} E_{T} ⟹ τ^{2} = ω^{2} μ ε = k^{2} .

The TEM wave travels with $τ = k$ , the wavenumber of a plane wave in the filling: $h = 0$ , no cutoff, $v_{p} = v_{g} = v = 1/ μ ε$ , no dispersion. With $τ = k$ , Eq. (7.5.2) becomes $H_{T} = (1/ η) a_{z} \times E_{T}$ , exactly as for a plane wave.

Gauss's law. $\nabla \cdot E = \nabla_{T} \cdot E_{T} + \partial E_{z} / \partial z = \nabla_{T} \cdot E_{T} = 0$ .

A field with zero curl, Eq. (7.5.1), is the gradient of a potential, $E_{T} = - \nabla_{T} Φ$ , and zero divergence then gives

\nabla_{T}^{2} Φ = 0 : (7.5.3)

the transverse field of a TEM wave is the electrostatic field of the two conductors, solved in the cross-section. This is why a TEM wave needs two conductors (Section 7.4): with one conductor the only solution of Eq. (7.5.3) is a constant.

The coaxial field. By symmetry $Φ = Φ (ρ)$ , and Laplace's equation in polar coordinates reads

\frac{1}{ρ} \frac{d}{d ρ} (ρ \frac{d Φ}{d ρ}) = 0 ⟹ ρ \frac{d Φ}{d ρ} = B ⟹ Φ = A + B ln ρ .

With the inner conductor at $Φ (a) = V_{0}$ and the outer one at $Φ (b) = 0$ ,

Φ (ρ) E_{T} = V_{0} \frac{ln ( b / ρ )}{ln ( b / a )}, = - \frac{d Φ}{d ρ} a_{ρ} = \frac{V _{0}}{ρ ln ( b / a )} a_{ρ} . (7.5.4)

TEM wave of the coaxial line

E H H = \frac{V _{0}}{ρ ln ( b / a )} e^{- j k z} a_{ρ}, = \frac{V _{0}}{η ρ ln ( b / a )} e^{- j k z} a_{ϕ}, = \frac{1}{η} a_{z} \times E, (7.5.5)

with $k = ω μ ε$ and $η = μ / ε$ of the filling, for $a < ρ < b$ .

In plain words

Take the static field of a cylindrical capacitor and the static field of a straight current, put them together in the same cross-section, and let the whole pattern slide along the line at the speed of light in the dielectric. Both fields fall as $1/ ρ$ : they are strongest at the inner conductor.

Watch the radial $E$ and the circular $H$ of the precision air line of Problem 7.5.1 ( $b / a = 2.30$ ) travel down the line below, then change $b / a$ and see $Z_{0}$ follow $ln (b / a)$ .

InteractiveLoading simulation…

(b) |E| across the gap. — (a) The field lines.

The same pattern at every frequency

The transverse pattern Eq. (7.5.5) does not depend on the frequency. At dc it is the electrostatic field of Section 3.2 and the magnetostatic field of Chapter 4, at $10$ GHz it is the same picture travelling at $v$ . This is what makes the coaxial line a broadband medium: from dc up to the first higher-order cutoff (Section 7.5.6), a pulse of any shape passes undistorted, apart from the losses.

7.5.1 The TEM wave of the coaxial line