Problem 7.2.1: Propagating TE modes of a polyethylene-filled guide

Interactive: Parallel-Plate Waveguide Modes

Problem 7.2.1

Propagating TE modes of a polyethylene-filled guide

Specify the propagating TE modes for a parallel-plate waveguide with a thickness of $10 cm$ and a filling dielectric material with $ε_{r} = 2.25$ (polyethylene) for $2.5 GHz$ .

Find the cutoff frequencies $f_{co, 1}$ , $f_{co, 2}$ , $f_{co, 3}$ and the propagating TE modes.
Find $τ_{n}$ , $λ_{g}$ and $Z_{T E}$ of each propagating TE mode.
How fast does the first non-propagating TE mode decay (Np/m and dB/m)?
Which other modes propagate?

Try it yourself first.

Strategy

"Thickness" is the plate separation $b$ . The filling fixes $v = c / ε_{r}$ and $η = η_{0} / ε_{r}$ . A mode propagates when $f > f_{co, n} = n v /2 b$ , and then $τ_{n} = k^{2} - (nπ / b)^{2}$ . Below cutoff the same square root, taken the other way round, is the attenuation constant.

1. Bookkeeping

$b = 0.1 m$ , $μ_{r} = 1$ : $v = 3 \times 1 0^{8} /1.5 = 2 \times 1 0^{8} m/s$ , $η = 120 π /1.5 = 80 π = 251.3 Ω$ , $λ = v / f = 8 cm$ and $k = 2 π f / v = 25 π = 78.54 rad/m$ .

2. Cutoff frequencies (a)

f_{co, 1} f_{co, 2} = \frac{3 \times 1 0 ^{8}}{2.25} \frac{1}{2 ( 0.1 )} = 1 GHz, = 2 GHz, f_{co, 3} = 3 GHz .

At $2.5 GHz$ only TE $_{1}$ and TE $_{2}$ are above cutoff.

3. Propagating TE modes (b)

$h_{n} = nπ / b = 10 π n rad/m$ .

TE_{1} : τ_{1} λ_{g 1} Z_{T E 1} TE_{2} : τ_{2} λ_{g 2} Z_{T E 2} = π 2 5^{2} - 1 0^{2} = 72.0 rad/m, = \frac{2 π}{72.0} = 8.73 cm, = η \frac{k}{τ _{1}} = 80 π \times \frac{78.54}{72.0} = 274.2 Ω, = π 2 5^{2} - 2 0^{2} = 15 π = 47.1 rad/m, = 13.3 cm, = 80 π \times \frac{25 π}{15 π} = 418.9 Ω.

4. The first evanescent mode (c)

TE $_{3}$ is below cutoff:

α_{3} = (30 π)^{2} - (25 π)^{2} = π 275 = 52.1 Np/m = 52.1 \times 8.686 = 452 dB/m .

After $1 cm$ its field has fallen to $e^{- 0.521} = 0.59$ of its initial value, after $10 cm$ to $e^{- 5.21} = 0.55 %$ .

5. The other modes (d)

The TEM wave propagates at every frequency, and TM $_{1}$ , TM $_{2}$ have the same cutoffs as TE $_{1}$ , TE $_{2}$ .

Answer

TE_{1}, TE_{2} propagate (plus TEM, TM_{1}, TM_{2}), λ_{g 1} = 8.73 cm, λ_{g 2} = 13.3 cm, Z_{T E 1} = 274.2 Ω, Z_{T E 2} = 418.9 Ω, α_{3} = 52.1 Np/m = 452 dB/m

Tap $n = 1$ , $2$ and $3$ in the mode table below to read $τ_{n}$ , $λ_{g}$ and $Z_{T E}$ of part (b) and the decay of TE $_{3}$ in part (c).

InteractiveLoading simulation…

Check

$s_{1} = τ_{1} / k = 0.9165$ and $s_{2} = 0.6$ , so $Z_{T E 1} = η / s_{1} = 274.2 Ω$ and $Z_{T E 2} = η /0.6 = 418.9 Ω$ . Both guide wavelengths exceed $λ = 8 cm$ , and both impedances exceed $η = 251.3 Ω$ , as they must for TE modes above cutoff.

(a) Mode chart, f_{co,1}=1\,\mathrm{GHz}. — (a) Mode chart, $f_{co, 1} = 1 GHz$ .

(b) The \omega–\tau curves f=\frac{v}{2\pi}\sqrt{\tau^2+(n\pi/b)^2}. — (a) Mode chart, $f_{co, 1} = 1 GHz$ .

Read the graphs

The closer a mode is to its cutoff, the smaller its $τ$ and the longer its $λ_{g}$ . TE $_{2}$ works at only $1.25 f_{co, 2}$ , so its $τ$ is just $60 %$ of $k$ and its $λ_{g}$ is $1.67 λ$ .