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

The transverse-field formulas Eq. (7.4.2) can be written in a form that does not depend on the coordinates. With the transverse gradient $\nabla_{T} = a_{x} \partial / \partial x + a_{y} \partial / \partial y$ , the four equations are the components of

E_{T} H_{T} = - \frac{1}{h ^{2}} (j τ \nabla_{T} E_{z} - j ω μ a_{z} \times \nabla_{T} H_{z}), = - \frac{1}{h ^{2}} (j τ \nabla_{T} H_{z} + j ω ε a_{z} \times \nabla_{T} E_{z}) . (7.4.12)

(Check the $x$ component: $a_{z} \times \nabla_{T} H_{z} = a_{y} \partial_{x} H_{z} - a_{x} \partial_{y} H_{z}$ , so $- j ω μ a_{z} \times \nabla_{T} H_{z}$ contributes $+ j ω μ \partial_{y} H_{z}$ to the bracket of $E_{x}$ , as in Eq. (7.4.2).) In cylindrical coordinates

\nabla_{T} = a_{ρ} \frac{\partial}{\partial ρ} + a_{ϕ} \frac{1}{ρ} \frac{\partial}{\partial ϕ}, a_{z} \times a_{ρ} = a_{ϕ}, a_{z} \times a_{ϕ} = - a_{ρ},

so that $a_{z} \times \nabla_{T} H_{z} = a_{ϕ} \partial H_{z} / \partial ρ - a_{ρ} (1/ ρ) \partial H_{z} / \partial ϕ$ . Collecting the $a_{ρ}$ and $a_{ϕ}$ parts of Eq. (7.4.12):

Transverse fields in cylindrical coordinates

E_{ρ} = - \frac{1}{h ^{2}} (j τ \frac{\partial E _{z}}{\partial ρ} + \frac{j ω μ}{ρ} \frac{\partial H _{z}}{\partial ϕ}), E_{ϕ} = - \frac{1}{h ^{2}} (\frac{j τ}{ρ} \frac{\partial E _{z}}{\partial ϕ} - j ω μ \frac{\partial H _{z}}{\partial ρ}), H_{ρ} = - \frac{1}{h ^{2}} (j τ \frac{\partial H _{z}}{\partial ρ} - \frac{j ω ε}{ρ} \frac{\partial E _{z}}{\partial ϕ}), H_{ϕ} = - \frac{1}{h ^{2}} (\frac{j τ}{ρ} \frac{\partial H _{z}}{\partial ϕ} + j ω ε \frac{\partial E _{z}}{\partial ρ}) . (7.4.13)

In plain words

Nothing new happens: the same four recipes as in a rectangular guide, with $x$ replaced by $ρ$ and $y$ by $ρϕ$ , the arc length around the axis. At $ϕ = 0$ , where $a_{ρ} = a_{x}$ and $a_{ϕ} = a_{y}$ , the four lines are literally the $E_{x}$ , $E_{y}$ , $H_{x}$ , $H_{y}$ formulas of Eq. (7.4.2).

For a TM mode ( $H_{z} = 0$ ) the formulas give $E_{ρ} / H_{ϕ} = - E_{ϕ} / H_{ρ} = τ / (ω ε)$ , and for a TE mode ( $E_{z} = 0$ ) they give $E_{ρ} / H_{ϕ} = - E_{ϕ} / H_{ρ} = ω μ / τ$ :

Z_{T M} Z_{T E} = \frac{E _{ρ}}{H _{ϕ}} = - \frac{E _{ϕ}}{H _{ρ}} = \frac{τ}{ω ε}, = \frac{E _{ρ}}{H _{ϕ}} = - \frac{E _{ϕ}}{H _{ρ}} = \frac{ω μ}{τ} . (7.4.14)

These are the wave impedances Eq. (7.4.3) of every guide. The phase constant $τ$ , the guide wavelength, the phase and group velocities follow from $f_{c}$ exactly as in Section 7.1. Only $f_{c}$ depends on the shape of the guide, and the rest of this section is about finding it.

7.4.4 Transverse fields in cylindrical coordinates