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

Five further exercises on the circular guide, in the spirit of Problems 7.4.1–7.4.2, the worked examples and the home exercises, graded from easy to harder. They use only the results listed at the head of this subsection. (Five more, on the coaxial line, close Section 7.5.)

Exercise 7.4.1

Reading a mode from its field

In an air-filled circular waveguide the longitudinal electric field is

E_{z} = 50 J_{1} (240 ρ) sin ϕ e^{- j 180 z} V/m (ρ, z in m),

and the wall $ρ = a$ is the first zero of $J_{1} (240 ρ)$ away from the axis. (a) Is the mode TE or TM? Give $m$ and $n$ . (b) Find the radius $a$ . (c) Find the frequency, the cutoff frequency, $λ$ and $λ_{g}$ . (d) Find $Z_{T M}$ (or $Z_{T E}$ ) and write $E_{ρ}$ and $H_{ϕ}$ . (e) Which other modes propagate at this frequency, and how many field patterns does the guide carry in all?

Try it yourself first.

Strategy

A nonzero $E_{z}$ means TM. The order of the Bessel function and of $sin m ϕ$ gives $m$ , the number of zeros along a radius gives $n$ . The two numbers in the formula are $h = 240$ rad/m and $τ = 180$ rad/m, and $k^{2} = h^{2} + τ^{2}$ fixes the frequency.

$E_{z} \neq = 0$ : a TM mode. $J_{1}$ and $sin ϕ$ : $m = 1$ . The wall is the first zero of $J_{1} (240 ρ)$ away from the axis: $n = 1$ . The mode is TM $_{11}$ , in its $sin ϕ$ version.

$ha = x_{11} = 3.832$ , so $a = 3.832/240 = 15.97$ mm.

$k = 24 0^{2} + 18 0^{2} = 300$ rad/m and

f = \frac{k c}{2 π} = \frac{300 \times 3 \times 1 0 ^{8}}{2 π} = 14.32 GHz, f_{c} = \frac{h c}{2 π} = 11.46 GHz, λ = \frac{2 π}{k} = 20.9 mm, λ_{g} = \frac{2 π}{τ} = 34.9 mm .

$s = τ / k = 0.6$ and $Z_{T M} = 377 \times 0.6 = 226.2 Ω$ . From Eq. (7.4.13) with $H_{z} = 0$ , $E_{ρ} = - (j τ / h^{2}) \partial E_{z} / \partial ρ$ , so

E_{ρ} H_{ϕ} = - \frac{j τ}{h} 50 J_{1}^{'} (240 ρ) sin ϕ e^{- j 180 z} = - j 37.5 J_{1}^{'} (240 ρ) sin ϕ e^{- j 180 z} V/m, = \frac{E _{ρ}}{Z _{T M}} = - j 0.1658 J_{1}^{'} (240 ρ) sin ϕ e^{- j 180 z} A/m .

A mode propagates if its zero is below $k a = 300 \times 0.01597 = 4.79$ : TE $_{11}$ ( $1.841$ ), TM $_{01}$ ( $2.405$ ), TE $_{21}$ ( $3.054$ ), TE $_{01}$ and TM $_{11}$ ( $3.832$ ), TE $_{31}$ ( $4.201$ ). TM $_{21}$ ( $5.136$ ) is cut off. Six modes. TM $_{01}$ and TE $_{01}$ have one pattern each, the other four two: $10$ field patterns.

Answer

TM_{11}, a = 15.97 mm f = 14.32 GHz, f_{c} = 11.46 GHz λ = 20.9 mm, λ_{g} = 34.9 mm Z_{T M} = 226.2 Ω 6 modes, 10 field patterns

Check

$H_{ϕ}$ from its own formula in Eq. (7.4.17): $- \frac{j ω ε _{0}}{h} 50 = - j \frac{k}{η _{0} h} 50 = - j \frac{300}{377 \times 240} \times 50 = - j 0.1658$ . And $f / f_{c} = 300/240 = 1.25$ gives $s = 1 - 0.64 = 0.6$ .

(a) TM_{11} (\cos\phi version). — (a) TM $_{11}$ ( $cos ϕ$ version).

(b) J_1(3.832\rho/a)/0.582. — (a) TM $_{11}$ ( $cos ϕ$ version).

Remark

TM $_{11}$ shares its cutoff with TE $_{01}$ because $J_{0}^{'} = - J_{1}$ . A guide that is meant to carry the low-loss TE $_{01}$ therefore always offers TM $_{11}$ as a partner with exactly the same phase velocity, into which bends convert power without any phase mismatch to stop it (Figure 7.4.12).

cf. Problem 7.4.1

Exercise 7.4.1: Reading a mode from its field