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

Problem 7.8.1

Modes of a rectangular cavity

An air-filled rectangular cavity has $a = 3 cm$ , $b = 1.5 cm$ and $d = 4 cm$ .

Find the resonant frequencies and the modes up to $12.5 GHz$ .
Which mode is dominant, and how far above it is the next resonance? Which modes are degenerate?
A coaxial probe is inserted through the centre of the top wall ( $x = a /2$ , $z = d /2$ ). Which of the modes of (a) can it excite?

Adapted from: M. N. O. Sadiku, Elements of Electromagnetics, Problem 12.30.

Try it yourself first.

Strategy

In air $f_{mn p} = 15 (m /3)^{2} + (n /1.5)^{2} + (p /4)^{2} GHz$ with the sides in cm. Since $(1/ b)^{2} = 0.4444 c m^{- 2}$ alone already gives $10 GHz$ , every mode with $n \geq 1$ lies at $10 GHz$ or above, so the search is short. For (c), a vertical probe couples only to $E_{y}$ at its tip, so it drives a mode only if $E_{y} \neq = 0$ there.

Bookkeeping

$(1/ a)^{2} = 0.1111$ , $(1/ b)^{2} = 0.4444$ , $(1/ d)^{2} = 0.0625 c m^{- 2}$ .

Modes with $n = 0$ (TE only, $m, p \geq 1$ ):

TE_{101} TE_{102} TE_{201} TE_{103} TE_{202} : 15 0.1111 + 0.0625 = 15 \times \frac{5}{12} = 6.25 GHz, : 15 0.1111 + 0.25 = 9.01 GHz, : 15 0.4444 + 0.0625 = 10.68 GHz, : 15 0.1111 + 0.5625 = 12.31 GHz, : 2 \times 6.25 = 12.50 GHz .

Modes with $n = 1$ :

TE_{011} TM_{110} TE_{111}, TM_{111} TE_{012} : 15 0.4444 + 0.0625 = 10.68 GHz, : 15 0.5556 = 11.18 GHz, : 15 0.6181 = 11.79 GHz, : 15 0.4444 + 0.25 = 12.50 GHz .

The next ones, TE $_{112}$ and TM $_{112}$ at $13.46 GHz$ and TM $_{210}$ at $14.14 GHz$ , are beyond the range. TM $_{mn p}$ needs $m, n \geq 1$ , so no TM mode has $n = 0$ .

$b$ is the smallest side, so the dominant mode is TE $_{101}$ at $6.25 GHz$ . The next resonance, TE $_{102}$ at $9.01 GHz$ , is $2.76 GHz$ ( $44 %$ ) higher. Degenerate pairs: TE $_{011}$ and TE $_{201}$ (because $a = 2 b$ , so $(2/ a)^{2} = (1/ b)^{2}$ ), TE $_{111}$ and TM $_{111}$ (always, for $m, n, p \geq 1$ ), TE $_{012}$ and TE $_{202}$ (again $a = 2 b$ ).

From Eqs. (7.8.11) and (7.8.10),

E_{y} E_{y} \propto m sin \frac{mπ x}{a} cos \frac{nπ y}{b} sin \frac{p π z}{d} (TE_{mn p}), \propto n p sin \frac{mπ x}{a} cos \frac{nπ y}{b} sin \frac{p π z}{d} (TM_{mn p}) .

At the top wall $cos (nπ) = \pm 1$ , and at $x = a /2$ , $z = d /2$ the product $sin \frac{mπ}{2} sin \frac{p π}{2}$ vanishes unless $m$ and $p$ are both odd. Hence

excited: TE $_{101}$ ( $6.25 GHz$ ), TE $_{111}$ and TM $_{111}$ ( $11.79 GHz$ ), TE $_{103}$ ( $12.31 GHz$ ),
not excited: TE $_{102}$ , TE $_{201}$ , TE $_{202}$ (a node of $E_{y}$ at the probe), TE $_{011}$ , TE $_{012}$ ( $m = 0$ : no $E_{y}$ ), TM $_{110}$ ( $p = 0$ : only $E_{z}$ ).

Answer

TE_{101} 6.25, TE_{102} 9.01 GHz TE_{011} / TE_{201} 10.68, TM_{110} 11.18 GHz TE_{111} / TM_{111} 11.79, TE_{103} 12.31 GHz TE_{012} / TE_{202} 12.50 GHz dominant TE_{101} the centre probe sees nothing else below 11.79 GHz

Check

$f_{101}^{2} = f_{co, 10}^{2} + (c /2 d)^{2} = 5^{2} + 3.7 5^{2} = 39.06 GH z^{2}$ , $f_{101} = 6.25 GHz$ : the guide cutoff of TE $_{10}$ plus one half guide wavelength along $d$ , as in Eq. (7.8.9).

(a) Mode chart, line at the dominant resonance.

(b) TE_{101}, 6.25\,\mathrm{GHz}. — (a) Mode chart, line at the dominant resonance.

Remark

Within the range of (a) the probe also drives TE $_{103}$ at $12.31 GHz$ , since $sin (3 π /2) = - 1$ : three half-waves along $z$ put a maximum of $E_{y}$ back at the centre. Seen from the probe the cavity is clean from $6.25 GHz$ up to $1.89$ times that frequency, a wider spurious-free range than the mode chart alone suggests (Figure 7.8.22).

Problem 7.8.1: Modes of a rectangular cavity