22-Elec-B9 Electromagnetic Field, Transmission Lines, Antennas, and Radiation · May 2013
Question 4 of 8: Resonant Frequencies of a 1 × 2 × 3 cm Rectangular Cavity
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams, May 2013 — 07-Elec-B9,
Electromagnetic Field, Transmission Lines, Antennas, and Radiation.
Three hours, closed book; an approved Casio or Sharp calculator is permitted.
Eight questions are printed and five constitute a complete paper,
each of equal value (20 marks); the first five appearing in the answer book are
the ones marked. The cover page supplies only two aids,
$\varepsilon_0 = 8.85\times 10^{-12}\ \text{F/m}$ and
$\mu_0 = 4\pi\times 10^{-7}\ \text{H/m}$, plus a Smith chart on page 4.
Candidates are urged to state any interpretation assumptions with their answers.
All eight questions are solved below, because this set is a study
resource rather than a three-hour sitting.
Reference texts. D. M. Pozar, Microwave Engineering,
4th ed. (transmission lines, the Smith chart, waveguides and cavities);
M. N. O. Sadiku, Elements of Electromagnetics, 7th ed. and
W. H. Hayt & J. A. Buck, Engineering Electromagnetics, 9th ed.
(plane waves, oblique incidence, radiation); C. A. Balanis,
Antenna Theory: Analysis and Design, 4th ed. (short elements, arrays,
apertures); F. T. Ulaby & U. Ravaioli, Fundamentals of Applied
Electromagnetics, 8th ed. (radar equation, ground-plane imaging). SI units
throughout; the paper's own $c = 3\times 10^{8}\ \text{m/s}$ and
$Z_0 = 377\ \Omega$ are used so that the printed aids reproduce exactly.
Check: reading of the printed data. Some symbols are printed loosely. Question 3 prints
“10GH” and “1KWs total energy”; these are read as a
10 GHz carrier and a total pulse energy of
$1\ \text{kW}\cdot\text{s} = 1000\ \text{J}$. Question 6 prints
“15 MH” (15 MHz) and the aid
$R_{\text{rad}} = (2\pi/3)Z_0 (l/\lambda)^2$. Question 7 prints
“10GH” (10 GHz) and the aid
$\theta_0 = 1.2\lambda/d$. Question 8 prints
“an 600 angle” ($60^\circ$).
Question 4: Resonant Frequencies of a 1 × 2 × 3 cm Rectangular
Cavity (20 marks)
Given. A closed rectangular cavity with perfectly conducting
walls and inside dimensions $a = 3\ \text{cm}$ (along $x$),
$b = 2\ \text{cm}$ (along $y$) and $d = 1\ \text{cm}$ (along $z$), air
filled, with $c = 3\times 10^{8}\ \text{m/s}$.
Find. Every resonant frequency below 20 GHz, with the modes
that produce it.
Figure 4.1 — The cavity. The
labelling of the axes is arbitrary because the resonance formula is symmetric in
the three dimensions; only the mode names change.
Approach. Apply the rectangular-cavity resonance formula,
impose the rule that at most one of the three mode indices may be zero, and
enumerate every index triple whose frequency falls below 20 GHz.
State the resonance condition.
A standing wave must fit an integer number of half wavelengths along each edge,
which gives
$$f_{mnp} = \frac{c}{2}\sqrt{\left(\frac{m}{a}\right)^{2}
+ \left(\frac{n}{b}\right)^{2} + \left(\frac{p}{d}\right)^{2}} .$$
Insert the dimensions.
With $a, b, d$ in centimetres and $c/2 = 15\ \text{GHz}\cdot\text{cm}$,
$$f_{mnp} = 15\sqrt{\left(\frac{m}{3}\right)^{2}
+ \left(\frac{n}{2}\right)^{2} + p^{2}}\ \ \text{GHz}.$$
Impose the existence rule and the 20 GHz ceiling.
A non-trivial field pattern needs at least two non-zero indices: $\text{TE}_{mnp}$
requires $p \ge 1$ with $m$ and $n$ not both zero, and $\text{TM}_{mnp}$ requires
$m \ge 1$ and $n \ge 1$ with $p \ge 0$. The frequency limit becomes
$$\left(\frac{m}{3}\right)^{2} + \left(\frac{n}{2}\right)^{2} + p^{2}
\lt \left(\frac{20}{15}\right)^{2} = 1.778 ,$$
which immediately kills every triple with $p \ge 2$.
Enumerate the surviving triples.
Sweeping $m$, $n$ and $p$ over the small remaining range leaves ten index
triples, and because TE and TM coexist whenever all three indices are non-zero
they represent twelve distinct modes. Sorted by frequency they give
$$\boxed{\ 9.014,\ 12.500,\ 15.811,\ 16.771,\ 17.500,\ 18.028,\
19.526\ \text{GHz}\ }$$
— seven distinct resonant frequencies, because three of them are
doubly degenerate.
Identify the dominant mode and the first exclusion.
The lowest resonance is $\text{TM}_{110}$ at 9.014 GHz; the smallest dimension (1 cm)
carries no variation at all in that mode, which is exactly why it is the cheapest
pattern to excite. The first triples excluded are $\text{TM}_{320}$ and $\text{TE}_{021}$ at
21.21 GHz and $\text{TM}_{410}$ at 21.36 GHz, comfortably above the ceiling.
Degeneracy is the interesting feature of this cavity: because the edge lengths
are in the ratio 3 : 2 : 1, several different index triples produce identical
sums and therefore identical frequencies. A physical cavity with slightly
imperfect dimensions splits each degenerate pair into two closely spaced
resonances, which is why tuning screws are provided on practical cavities and why
degenerate-mode filters are designed deliberately in some microwave
components.
Final results — Question 4:
resonances below 20 GHz ($a = 3$ cm, $b = 2$ cm, $d = 1$ cm)
