Lowest Resonant Frequencies of a Rectangular Waveguide Cavity: 17-Phys-A3 Electromagnetics, December 2016, Q6

Question 6: Lowest Resonant Frequencies of a Rectangular Waveguide Cavity (equal value)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Rectangular cavity, inside dimensions $a=2$ cm, $b=1$ cm, $d=2.3$ cm (air-filled, so $v=c=3\times10^8$ m/s).

Cavity dimensions used in the resonant-frequency formula.

Find. The two lowest resonant frequencies $f_{mnp}$.

Approach. Use the general rectangular-cavity resonance formula $f_{mnp}=\tfrac{c}{2}\sqrt{(m/a)^2+(n/b)^2+(p/d)^2}$, restricted to physically valid mode indices (not both $m,n$ zero; $p=0$ only when $m,n$ are both non-zero), and search the smallest-index combinations — favouring a zero index along the SHORTEST dimension ($b$), since that dimension penalizes a non-zero index the most.

Formula and candidate search. $$f_{mnp}=\frac{c}{2}\sqrt{\left(\frac{m}{a}\right)^2+\left(\frac{n}{b}\right)^2+\left(\frac{p}{d}\right)^2}.$$ Since $b=1$ cm is the smallest dimension, the lowest modes avoid a non-zero index along $b$ ($n=0$), using the two larger dimensions $a,d$ instead: try $\text{TE}_{101}$ ($m{=}1,n{=}0,p{=}1$).
Lowest mode: TE₁₀₁. $$f_{101}=\frac{3\times10^8}{2}\sqrt{\left(\frac{1}{0.02}\right)^2+\left(\frac{1}{0.023}\right)^2}=\boxed{9.94\ \text{GHz}}.$$
Second-lowest mode: TE₁₀₂. Checking every small-index combination (TE₂₀₁, TE₀₁₁, TE₁₁₀, TE₁₀₂, …) confirms $\text{TE}_{102}$ ($m{=}1,n{=}0,p{=}2$) is next: $$f_{102}=\frac{3\times10^8}{2}\sqrt{\left(\frac{1}{0.02}\right)^2+\left(\frac{2}{0.023}\right)^2}=\boxed{15.05\ \text{GHz}}.$$ (The next candidates, TE₂₀₁ and TE₀₁₁, both come in higher at $16.36$ GHz.)

Two lowest cavity resonances
Mode	Frequency
TE₁₀₁ $(m,n,p)=(1,0,1)$	9.94 GHz
TE₁₀₂ $(m,n,p)=(1,0,2)$	15.05 GHz

Topic: Resonant modes of a rectangular cavity resonator.

Key relations: $f_{mnp}=\tfrac{c}{2}\sqrt{(m/a)^2+(n/b)^2+(p/d)^2}$; a valid mode needs $m,n$ not both zero, and $p=0$ only when both $m,n$ are non-zero (TM-type).

Why this works: a cavity resonance is a standing wave that must fit an integer number of half-wavelengths along each of the three dimensions; the lowest frequencies come from putting the fewest (and smallest) half-wave variations along the LONGEST dimensions, and a zero index along the shortest dimension avoids that dimension's large $1/b^2$ penalty entirely.

Common pitfall: assuming the two lowest modes are simply $(1,0,1)$ and $(0,1,1)$ (i.e. cycling which index is zero) without checking $(1,0,2)$ — because $a$ and $d$ are close in size here, stepping $p$ from 1 to 2 beats swapping which of $a,b$ carries the zero index.

Source: Pozar, Microwave Engineering, Ch. 6 (cavity resonators); Sadiku, Elements of Electromagnetics, Ch. 12.

Question 6 of 8: Lowest Resonant Frequencies of a Rectangular Waveguide Cavity

Question 6: Lowest Resonant Frequencies of a Rectangular Waveguide Cavity (equal value)