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

The cutoff frequency $f_{c} = h / (2 π μ ε)$ of Section 7.1 now follows from Eq. (7.4.16) and Eq. (7.4.19).

Cutoff frequencies of a circular guide of radius a

f_{c, TM_{mn}} = \frac{x _{mn}}{2 π a μ ε}, f_{c, TE_{mn}} = \frac{x _{mn}^{'}}{2 π a μ ε}, λ_{c} = \frac{2 π a}{x} . (7.4.21)

A mode propagates only if $f > f_{c}$ , with $τ = k^{2} - h_{mn}^{2}$ . The dominant mode is TE $_{11}$ , the mode with the smallest zero, $x_{11}^{'} = 1.841$ :

f_{c, TE_{11}} = \frac{1.841 v}{2 π a} = \frac{0.293 v}{a}, λ_{c} = 3.41 a . (7.4.22)

In plain words

The wall must sit at a zero of $J_{m}$ (TM) or of $J_{m}^{'}$ (TE). A small zero means a small $h$ and a low cutoff. The smallest zero of all, $1.841$ , belongs to $J_{1}^{'}$ , so TE $_{11}$ is the first mode to propagate. For an air-filled pipe $f_{c, TE_{11}} = 8.79 GHz / a [cm]$ .

Sorting all the zeros gives the order of the modes in Table 7.4.2 and the cutoff chart of Figure 7.4.4. The cutoff wavelengths are $λ_{c} = 3.41 a$ (TE $_{11}$ ), $2.61 a$ (TM $_{01}$ ), $2.06 a$ (TE $_{21}$ ) and $1.64 a$ (TE $_{01}$ and TM $_{11}$ ).

Table 7.4.2. The first modes of a circular guide in the order of their cutoffs, with $x$ the zero that fixes the cutoff. “Copies”: the $cos m ϕ$ and $sin m ϕ$ versions of each mode with $m \geq 1$ , two independent field patterns.

mode ( $x$ )	$f_{c} / f_{c, TE_{11}}$	$λ_{c} / a$	copies
TE $_{11}$ ( $1.841$ )	$1$	$3.41$	$2$
TM $_{01}$ ( $2.405$ )	$1.306$	$2.61$	$1$
TE $_{21}$ ( $3.054$ )	$1.659$	$2.06$	$2$
TE $_{01}$ ( $3.832$ )	$2.081$	$1.64$	$1$
TM $_{11}$ ( $3.832$ )	$2.081$	$1.64$	$2$
TE $_{31}$ ( $4.201$ )	$2.282$	$1.50$	$2$
TM $_{21}$ ( $5.136$ )	$2.789$	$1.22$	$2$

Cutoff chart of the circular guide. Each bar starts at the cutoff of its mode, in units of the TE_{11} cutoff, and every mode propagates to the right of its start. Only TE_{11} propagates between 1 and 1.306. TE_{01} and TM_{11} start together, and TE_{41} and TE_{12} almost together. The chart stops at TE_{12}: the next mode, TM_{02}, starts at 2.998. — **Figure 7.4.4.** Cutoff chart of the circular guide. Each bar starts at the cutoff of its mode, in units of the TE $_{11}$ cutoff, and every mode propagates to the right of its start. Only TE $_{11}$ propagates between $1$ and $1.306$ . TE $_{01}$ and TM $_{11}$ start together, and TE $_{41}$ and TE $_{12}$ almost together. The chart stops at TE $_{12}$ : the next mode, TM $_{02}$ , starts at $2.998$ .

Three facts stand out.

The single-mode band is narrow. Only TE $_{11}$ propagates for $1 < f / f_{c, TE_{11}} < 1.306$ , a band of $1.31 : 1$ , against $2 : 1$ for a rectangular guide with $a = 2 b$ (Section 7.3).
Even there TE $_{11}$ is a pair. Its $cos ϕ$ and $sin ϕ$ copies have the same $τ$ , and a slight ellipticity or a bend of the pipe turns its polarisation.
TE $_{01}$ and TM $_{11}$ share their cutoff, because $J_{0}^{'} = - J_{1}$ makes $x_{0 n}^{'} = x_{1 n}$ . Modes with the same cutoff are called degenerate. They have the same phase velocity at every frequency and convert into each other at bends without any phase mismatch to stop it.

Practical consequence

A circular guide is not chosen for its bandwidth. It is chosen for its symmetry: TM $_{01}$ does not depend on $ϕ$ and passes through a rotating joint unchanged, TE $_{11}$ carries two independent polarisations, and TE $_{01}$ has a loss that falls with frequency. The three applications of this section and of Section 7.5 rest on exactly these properties.

The tool below shows the pipe of Problem 7.4.1 ( $a = 2 cm$ ). Tap the bars of its cutoff chart in order, TE $_{11}$ , TM $_{01}$ , TE $_{21}$ and TE $_{01}$ /TM $_{11}$ , and see where each wall condition puts the wall on the Bessel curve.

InteractiveLoading simulation…

Quick check

An air-filled pipe has $a = 1$ cm. In which band is it single-mode?

$f_{c, TE_{11}} = 1.841 \times 3 \times 1 0^{8} / (2 π \times 0.01) = 8.79$ GHz and $f_{c, TM_{01}} = 2.405 \times 3 \times 1 0^{8} / (2 π \times 0.01) = 11.48$ GHz: single-mode from $8.79$ to $11.48$ GHz, a ratio of $1.306$ .

Common mistake · Expecting TE01 to be dominant

In a rectangular guide the dominant mode is TE $_{10}$ , and the habit of “the mode with a zero index” suggests TE $_{01}$ for the round pipe. But the indices mean different things here: TE $_{01}$ of a circular guide has $x_{01}^{'} = 3.832$ and starts at $2.08$ times the TE $_{11}$ cutoff, as the sixth field pattern to propagate. The round counterpart of the rectangular TE $_{10}$ is TE $_{11}$ , with one period around the pipe.

7.4.7 Cutoff frequencies and the order of the modes