Question 6 of 8: Two Lowest Resonances of a Rectangular Cavity
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. 07-Elec-A7 Electromagnetics, National Examinations, December 2016 — 3 hours, closed book (one of two approved Casio or Sharp calculators). Eight questions, all of equal value, and the paper states that any five questions constitute a complete paper. All eight are solved here, because this set is a study resource rather than a timed sitting. Page-1 aids as printed: ε0 = 8.85 × 10−12 F/m and μ0 = 4π × 10−7 H/m.
Reference texts for 22-Elec-A7 Electromagnetics. D. M. Pozar, Microwave Engineering, 4th ed. (transmission lines, stubs, waveguides and cavities); M. N. O. Sadiku, Elements of Electromagnetics, 7th ed. (Maxwell’s equations, plane waves, magnetostatic forces); W. H. Hayt & J. A. Buck, Engineering Electromagnetics, 9th ed. (transmission-line transients and polarisation); F. T. Ulaby & U. Ravaioli, Fundamentals of Applied Electromagnetics, 7th ed.; C. A. Balanis, Antenna Theory: Analysis and Design, 4th ed. (short elements and radiated power density).
Constants used throughout. The paper pins its own numbers: Questions 1 and 2 both state a propagation velocity of 3 × 108 m/s and Question 1 uses a 377 Ω line, so this solution works with c = 3.00 × 108 m/s and η0 = 377 Ω and reserves the printed page-1 aids for the per-unit-length quantities of Question 3. Using 2.998 × 108 m/s instead shifts every answer by less than 0.1 % and changes no conclusion.
Question numbering. The eighth question (the short horizontal current element) carries no printed number on page 3 of the original — it follows Question 7 as an unnumbered block. It is numbered Question 8 here, which is consistent with the paper’s own statement that the exam holds more than five questions of equal value.
Question 6: Two Lowest Resonances of a Rectangular Cavity (20 marks)
Given. A rectangular box with perfectly conducting walls and an air-filled interior, whose three inside dimensions are all of the same order.
Given data
Quantity
Symbol
Value
Width (x-direction)
a
2.0 cm
Height (y-direction)
b
1.0 cm
Length (z-direction)
d
2.3 cm
Interior filling
εr, μr
1.0 (air)
Wave speed inside
c
3.00 × 108 m/s
Find. The two lowest frequencies at which the box will resonate, together with the modes responsible.
Figure 6.1 — Cavity dimensions and the four lowest resonances; the two lowest, TE(101) and TE(102), are highlighted.
Approach. Impose a standing-wave condition along all three axes at once, enumerate the index triples that correspond to modes that actually exist, and rank the resulting frequencies numerically rather than trusting a remembered ordering.
State the resonance condition. A rectangular cavity supports a standing wave whenever an integer number of half wavelengths fits along each axis, which gives
$$f_{mnp}=\frac{c}{2}\sqrt{\left(\frac{m}{a}\right)^{2}+\left(\frac{n}{b}\right)^{2}+\left(\frac{p}{d}\right)^{2}}$$
with \(m,n,p\) counting half-wave variations along \(x,y,z\) respectively.
Apply the existence rules before computing anything. A TEmnp mode requires \(p\ge1\) with \(m\) and \(n\) not both zero; a TMmnp mode requires \(m\ge1\) and \(n\ge1\). In every case at least two of the three indices must be non-zero, so triples such as (1,0,0) and (0,0,1) are not modes at all. Skipping this step is what produces spurious “lowest” frequencies far below the true one.
Reason about which indices are cheap. The reciprocal dimensions are \(1/a=50\ \text{m}^{-1}\), \(1/b=100\ \text{m}^{-1}\) and \(1/d=43.5\ \text{m}^{-1}\). Any index placed on the 1 cm height costs twice as much as one placed on the 2 cm width, so the least expensive admissible pair is one half-wave along \(x\) and one along \(z\).
Evaluate the lowest mode. That pair is TE101:
$$f_{101}=\frac{3.00\times10^{8}}{2}\sqrt{\left(\frac{1}{0.020}\right)^{2}+\left(\frac{1}{0.023}\right)^{2}}=\left(1.50\times10^{8}\right)\sqrt{2500+1890}=\boxed{9.939\ \text{GHz}}$$
Test the candidates for second place. Three triples compete, and each must be evaluated rather than guessed:
$$f_{102}=\left(1.50\times10^{8}\right)\sqrt{2500+7561}=15.046\ \text{GHz}$$
$$f_{011}=f_{201}=\left(1.50\times10^{8}\right)\sqrt{10\,000+1890}=16.356\ \text{GHz},\qquad f_{110}=\left(1.50\times10^{8}\right)\sqrt{2500+10\,000}=16.771\ \text{GHz}$$
Adding a second half-wave along the 2.3 cm length is cheaper than introducing the first half-wave along the 1 cm height, so
$$\boxed{f_{\text{2nd}}=f_{102}=15.046\ \text{GHz}\ \ (\text{TE}_{102})}$$
Note the degeneracy that falls out of the geometry. Because \(a=2b\) exactly, \(2/a=1/b\) and the TE201 and TE011 modes land on precisely the same frequency, 16.356 GHz. This is a genuine degeneracy of the box, not a rounding coincidence, and in a real cavity it makes that resonance sensitive to any manufacturing asymmetry.
Sanity-check the answer against the dominant-mode guide. The same cross-section used as a waveguide would have \(f_{c,\text{TE10}}=c/2a=7.50\) GHz, so a cavity built from it must resonate above that — 9.939 GHz duly clears it, and the corresponding guide wavelength \(\lambda_{g}=2d=4.6\) cm gives back the same frequency, confirming that TE101 is one half guide-wavelength long.
