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

Integrate the time-average Poynting vector $S = \frac{1}{2} Re {E \times H^{*}}$ over the gap $0 < y < b$ for a unit width in $x$ . The result $P^{'}$ is a power per unit width $[W/m]$ .

TEM: $S_{z} = - \frac{1}{2} Re {E_{y} H_{x}^{*}} = \frac{1}{2} η ∣ A_{0} ∣^{2}$ , uniform. With the field amplitude $E_{0} = η ∣ A_{0} ∣$ , $P^{'} = b E_{0}^{2} / (2 η)$ .
TM $_{n}$ : $(E_{y} a_{y}) \times (H_{x} a_{x}) = - E_{y} H_{x} a_{z}$ , so $S_{z} = - \frac{1}{2} Re {E_{y} H_{x}^{*}} = \frac{ω ε τ _{n} ∣ C ∣ ^{2}}{2 h _{n}^{2}} cos^{2} \frac{nπ y}{b}$ , and $P^{'} = \frac{ω ε τ _{n} b ∣ C _{2 n} ∣ ^{2}}{4 h _{n}^{2}}$ .
TE $_{n}$ : $S_{z} = \frac{1}{2} Re {E_{x} H_{y}^{*}} = \frac{ω μ τ _{n} ∣ C ∣ ^{2}}{2 h _{n}^{2}} sin^{2} \frac{nπ y}{b}$ , and $P^{'} = \frac{ω μ τ _{n} b ∣ C _{1 n} ∣ ^{2}}{4 h _{n}^{2}}$ .

In terms of the peak transverse field $E_{0}$ (for TM $_{n}$ , $E_{0} = τ_{n} ∣ C ∣/ h_{n}$ , for TE $_{n}$ , $E_{0} = ω μ ∣ C ∣/ h_{n}$ ) all three take a common form,

Power per unit width

P_{T E M}^{'} P_{T E_{n}}^{'} = \frac{b E _{0}^{2}}{2 η}, P_{T M_{n}}^{'} = \frac{b E _{0}^{2}}{4 Z _{T M}}, = \frac{b E _{0}^{2}}{4 Z _{T E}} [W/m] . (7.2.34)

In plain words

The power is the peak field squared over twice the wave impedance, times the gap. The extra $\frac{1}{2}$ of the higher modes is the average of $cos^{2}$ or $sin^{2}$ over the gap. Below cutoff $Z$ is imaginary and $P^{'} = 0$ .

Figure 7.2.11. Profile of the power density

S_{z}

across the gap. The TEM wave carries its power uniformly, TM

_{1}

mainly near the plates, where

E_{y}

and

H_{x}

are largest, and TE

_{1}

in the middle, where its

E_{x}

is largest.

7.2.9 Power flow in a parallel-plate guide