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

Recap

TEM wave: two conductors allow $E_{z} = H_{z} = 0$ with $τ = k$ and no cutoff. $E_{T} = - \nabla_{T} Φ$ with $\nabla_{T}^{2} Φ = 0$ , the electrostatic problem of the cross-section. $E = \frac{V _{0}}{ρ l n ( b / a )} e^{- j k z} a_{ρ}$ , $H = \frac{1}{η} a_{z} \times E$ , Eq. (7.5.5).
Line constants, Eq. (7.5.8): $Z_{0} = \frac{η}{2 π} ln \frac{b}{a} = \frac{60}{ε _{r}} ln \frac{b}{a}$ , $L^{'} = \frac{μ}{2 π} ln \frac{b}{a}$ , $C^{'} = \frac{2 π ε}{l n ( b / a )}$ , $L^{'} / C^{'} = Z_{0}$ , $1/ L^{'} C^{'} = v$ : the line model is exact.
Power $P = ∣ V_{0} ∣^{2} /2 Z_{0}$ , largest field at the inner conductor, $P_{m a x} = \frac{π a ^{2} E _{br}^{2}}{η} ln \frac{b}{a}$ .
Loss, Eq. (7.5.13): $α_{c} = \frac{R _{s}}{2 η} \frac{1/ a + 1/ b}{l n ( b / a )} \propto f$ , $α_{d} = k tan δ /2 \propto f$ , mostly in the inner conductor.
Optima at fixed $b$ : minimum loss $b / a = 3.59$ ( $76.7 Ω$ in air), maximum power $1.65$ ( $30 Ω$ ), highest voltage $e$ ( $60 Ω$ ). $50 Ω$ is the compromise.
Higher modes: TE and TM with $J_{m}$ and $Y_{m}$ . TE $_{11}$ first, $h_{c} \approx 2/ (a + b)$ , $f_{c} \approx v / [π (a + b)]$ , Eq. (7.5.17): the upper frequency limit of every cable and connector.

7.5.7 Recap