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

For a TE mode $E_{z} = 0$ and

H_{z} = C J_{m} (h ρ) cos m ϕ e^{- j τ z} . (7.4.18)

Now $E_{z} = 0$ holds everywhere, and the wall condition left is $E_{ϕ} (a, ϕ) = 0$ . With $E_{z} = 0$ , Eq. (7.4.13) gives $E_{ϕ} = (j ω μ / h^{2}) \partial H_{z} / \partial ρ = (j ω μ / h) C J_{m}^{'} (h ρ) cos m ϕ$ , so the condition is $J_{m}^{'} (ha) = 0$ : the product $ha$ is one of the zeros $x_{mn}^{'}$ of the derivative of $J_{m}$ (Figure 7.4.3b).

TE modes of the circular guide

h_{mn} = \frac{x _{mn}^{'}}{a} (zeros of J_{m}^{'}), m = 0, 1, 2, \dots, n = 1, 2, \dots (7.4.19)

The transverse components follow in the same way (factor $e^{- j τ z}$ omitted):

E_{ρ} E_{ϕ} H_{ρ} H_{ϕ} = \frac{j ω μ m}{h ^{2} ρ} C J_{m} (h ρ) sin m ϕ, = \frac{j ω μ}{h} C J_{m}^{'} (h ρ) cos m ϕ, = - \frac{j τ}{h} C J_{m}^{'} (h ρ) cos m ϕ, = \frac{j τ m}{h ^{2} ρ} C J_{m} (h ρ) sin m ϕ . (7.4.20)

$E_{ρ}$ is normal to the wall and may stay nonzero there (it ends on surface charge), while $H_{ρ} \propto J_{m}^{'} (h ρ)$ vanishes at the wall together with $E_{ϕ}$ . For $m = 0$ the root $x = 0$ of $J_{0}^{'}$ is not counted: $h = 0$ would give a constant $H_{z}$ with no transverse field at all, which is not a wave. So again $n \geq 1$ .

sin mφ instead of cos mφ

Replacing $cos m ϕ$ by $sin m ϕ$ in Eq. (7.4.15) or Eq. (7.4.18) gives the same mode turned by $9 0^{\circ} / m$ about the axis, with the same $h$ and therefore the same $τ$ . For $m \geq 1$ every mode comes as a degenerate pair, two independent field patterns (two polarisations) that travel at exactly the same speed. Only the circularly symmetric modes with $m = 0$ are single. This is the price and the gift of the round cross-section: a square guide has the same pairing for its TE $_{10}$ and TE $_{01}$ , a rectangular one with $a \neq = b$ does not.

Quick check

Show that the TE fields Eq. (7.4.20) satisfy every wall condition, not only the one used to find $h$ .

At $ρ = a$ , $J_{m}^{'} (x_{mn}^{'}) = 0$ , so $E_{ϕ} = 0$ and $H_{ρ} = 0$ : no tangential $E$ along $ϕ$ , no normal $H$ . $E_{z} = 0$ everywhere for a TE mode. $E_{ρ}$ (normal) and $H_{ϕ}$ , $H_{z}$ (tangential) may be nonzero on a perfect conductor: they are balanced by surface charge and surface current.

7.4.6 TE modes