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

The mode $(m, n)$ of the rectangular guide has $τ^{2} = k^{2} - h_{mn}^{2}$ . Requiring $τ = p π / d$ turns this into a condition on $k$ , that is on the frequency:

k^{2} = ω^{2} μ ε = (\frac{mπ}{a})^{2} + (\frac{nπ}{b})^{2} + (\frac{p π}{d})^{2} . (7.8.7)

The three sides now enter on an equal footing. Only the frequencies that satisfy Eq. (7.8.7) are allowed.

Resonant frequencies of the rectangular cavity

An $a \times b \times d$ rectangular cavity filled with a medium of wave speed $v$ resonates at

f_{mn p} = \frac{v}{2} (\frac{m}{a})^{2} + (\frac{n}{b})^{2} + (\frac{p}{d})^{2}, (7.8.8)

or, equivalently,

f_{mn p}^{2} = f_{co, mn}^{2} + (\frac{p v}{2 d})^{2} . (7.8.9)

The integers $m$ , $n$ , $p$ count the half-period variations of the field along $x$ , $y$ and $z$ .

In plain words

The resonant frequency is the cutoff of the guide mode, "plus" the condition $d = p λ_{g} /2$ , added in quadrature. A box has a whole ladder of such frequencies, one for every allowed triple $(m, n, p)$ .

Figure 7.8.3 shows how the end walls select points on the dispersion curve of the guide. For the air-filled cavity of Problem 7.8.1 ( $a = 3 cm$ , $b = 1.5 cm$ , $d = 4 cm$ ) the TE $_{10}$ mode of the guide has its cutoff at $f_{co, 10} = c /2 a = 5 GHz$ and allows any $τ$ above it. The end walls pick $τ = π / d$ , $2 π / d$ , $3 π / d$ , and the dots are the resonances TE $_{101}$ , TE $_{102}$ and TE $_{103}$ at $6.25$ , $9.01$ and $12.31 GHz$ . Each is a point where the hyperbola $f^{2} = f_{co, 10}^{2} + (v τ /2 π)^{2}$ meets a vertical line.

Discrete resonances on the dispersion curve of the TE_{10} mode, for the cavity of Problem 7.8.1 (a=3\,\mathrm{cm}, d=4\,\mathrm{cm}, air). The guide allows any \tau above its cutoff of 5\,\mathrm{GHz}. The end walls select \tau=p\pi/d, and the dots are the resonances TE_{101}, TE_{102} and TE_{103}. — **Figure 7.8.3.** Discrete resonances on the dispersion curve of the TE $_{10}$ mode, for the cavity of Problem 7.8.1 ( $a = 3 cm$ , $d = 4 cm$ , air). The guide allows any $τ$ above its cutoff of $5 GHz$ . The end walls select $τ = p π / d$ , and the dots are the resonances TE $_{101}$ , TE $_{102}$ and TE $_{103}$ .

Quick check

An air-filled cavity has $a = 3 cm$ and $d = 4 cm$ . Without using Eq. (7.8.8) directly, find $f_{101}$ from the cutoff of the TE $_{10}$ guide mode and the condition $d = λ_{g} /2$ .

$f_{co, 10} = c /2 a = 5 GHz$ and $p v /2 d = c /2 d = 3.75 GHz$ , so by Eq. (7.8.9) $f_{101} = 5^{2} + 3.7 5^{2} = 6.25 GHz$ . The same number follows from Eq. (7.8.8): $15 1/9 + 1/16 = 15 \times 5/12 = 6.25 GHz$ , with $c /2 = 15 GHz cm$ .

7.8.3 Resonant frequencies of the rectangular cavity