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22-Elec-A7 Electromagnetics · May 2016

Question 4 of 8: Shortest and longest guide wavelength at 20 GHz

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Exams, May 2016 — 07-Elec-A7 Electromagnetics. Three hours, closed book; one of two approved calculators permitted. Eight questions, all of equal value (20 marks each). The rubric states that any five questions constitute a complete paper and that only the first five presented are marked — all eight are worked here, because this set is a study resource rather than a sat examination.

Constants. The paper prints $\varepsilon_0 = 8.85\times10^{-12}\ \text{F/m}$ and $\mu_0 = 4\pi\times10^{-7}\ \text{H/m}$ as aids. Its own numbers imply the classroom values $c = 3.00\times10^{8}\ \text{m/s}$ and $\eta_0 = 120\pi = 377\ \Omega$: the offset in Question 3 is exactly $\lambda/8$ only if $\lambda = 3.00\ \text{cm}$ at 10 GHz, and the 2.25 cm guide of Question 4 places a cutoff exactly on 20 GHz only for the same value. Those are the constants used throughout; the more precise values shift every field and guide wavelength here by less than 0.3 %, and the one place where the difference matters is flagged in Question 4.

Reference texts. D. M. Pozar, Microwave Engineering, 4th ed., §2.1–2.7 (transmission-line theory, standing waves, impedance transformation). M. N. O. Sadiku, Elements of Electromagnetics, 7th ed., ch. 11 (transmission lines) and ch. 12 (waveguides). W. H. Hayt and J. A. Buck, Engineering Electromagnetics, 9th ed., ch. 10–11 (transmission lines, uniform plane waves). F. T. Ulaby and U. Ravaioli, Fundamentals of Applied Electromagnetics, 8th ed., ch. 2 and ch. 7 (line transients, wave polarisation). C. A. Balanis, Antenna Theory: Analysis and Design, 4th ed., ch. 4 (the infinitesimal/short current element).

Question 4: Shortest and longest guide wavelength at 20 GHz (20 marks)

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. An air-filled rectangular guide of broad dimension $a = 2.25\ \text{cm}$ and narrow dimension $b = 1\ \text{cm}$ is excited at $20\ \text{GHz}$, for which the free-space wavelength is $\lambda_0 = c/f = 1.500\ \text{cm}$. (The paper writes the frequency as “$2\times10^{8}$ Hz”; 20 GHz is $2\times10^{10}$ Hz, and every other number in the question is consistent with 20 GHz.)

Find. The largest and smallest guide wavelengths among all modes that actually propagate at this frequency.

Approach. Rank every mode by cutoff frequency, keep only those whose cutoff lies below 20 GHz, then apply the guide-wavelength formula — the mode furthest below cutoff gives the shortest guide wavelength and the mode nearest cutoff gives the longest.

  1. Write the cutoff law. For an air-filled rectangular guide every TE$_{mn}$ mode, and every TM$_{mn}$ mode with $m,n \ge 1$, has $$f_{c,mn} = \frac{c}{2}\sqrt{\left(\frac{m}{a}\right)^{2} + \left(\frac{n}{b}\right)^{2}}$$
  2. Rank the modes. Evaluating this with $a = 0.0225\ \text{m}$ and $b = 0.0100\ \text{m}$ gives the table below. Note that TM modes require both indices non-zero, so TE$_{11}$ and TM$_{11}$ are degenerate and share one cutoff.
  3. Apply the guide-wavelength formula. Above cutoff $$\lambda_g = \frac{\lambda_0}{\sqrt{1 - (f_c/f)^{2}}}$$ which always exceeds $\lambda_0$, and diverges as $f_c \to f$.
  4. Take the extremes. The dominant mode is furthest below cutoff and so is the most tightly wound: $$\lambda_{g,\min} = \frac{1.500}{\sqrt{1 - (6.667/20)^{2}}} = \boxed{1.591\ \text{cm}\ (\text{TE}_{10})}$$
  5. And the mode closest to cutoff. Of the modes that genuinely propagate, the degenerate TE$_{11}$/TM$_{11}$ pair sits highest, at 16.415 GHz, giving $$\lambda_{g,\max} = \frac{1.500}{\sqrt{1 - (16.415/20)^{2}}} = \boxed{2.626\ \text{cm}\ (\text{TE}_{11}/\text{TM}_{11})}$$
Mode census for a 2.25 cm × 1 cm guide at 20 GHz
ModeCutoff expression$f_c$ (GHz)At 20 GHz$\lambda_g$ (cm)
TE$_{10}$$c/2a$$6.667$propagates$1.591$
TE$_{20}$$c/a$$13.333$propagates$2.012$
TE$_{01}$$c/2b$$15.000$propagates$2.268$
TE$_{11}$, TM$_{11}$$\tfrac{c}{2}\sqrt{a^{-2}+b^{-2}}$$16.415$propagates$2.626$
TE$_{30}$$3c/2a$$20.000$exactly at cutoff$\infty$
TE$_{21}$, TM$_{21}$$\tfrac{c}{2}\sqrt{4a^{-2}+b^{-2}}$$20.069$evanescent—
a = 2.25 cmb = 1 cmguide cross-sectionf (GHz)TE21 / TM21 fc = 20.069 GHz (evanescent)TE30 fc = 20.000 GHz (AT CUTOFF)TE11 / TM11 fc = 16.415 GHz (propagates)TE01 fc = 15.000 GHz (propagates)TE20 fc = 13.333 GHz (propagates)TE10 fc = 6.667 GHz (propagates)signal 20 GHzTE30 lands exactly on the signal frequency: beta = 0, so it stores energy rather than carrying it.
Figure 4.1 — cutoff ladder for the guide. Four modes clear 20 GHz; TE$_{30}$ lands exactly on it and TE$_{21}$ misses by 69 MHz.

The interesting feature of this guide is the entry that is neither propagating nor comfortably evanescent. With $a = 2.25\ \text{cm}$ the third-order mode has $f_c = 3c/2a = 20.000\ \text{GHz}$ exactly, which is plainly deliberate on the examiner's part. At its own cutoff a mode has $\beta = \sqrt{k^{2} - k_c^{2}} = 0$: the guide wavelength is infinite, the group velocity is zero, and no power is carried. TE$_{30}$ therefore does not belong in the propagating set, and the longest genuine guide wavelength is the 2.626 cm of the TE$_{11}$/TM$_{11}$ pair.

Check: this exclusion is exact only for the paper’s own $c = 3.00\times10^{8}\ \text{m/s}$. Using $c = 2.998\times10^{8}\ \text{m/s}$ moves the TE$_{30}$ cutoff to 19.986 GHz — 14 MHz below the signal — which would technically admit it with $\lambda_g \approx 40.3\ \text{cm}$. Such a mode carries almost no power and is unusable in practice, and a change in $a$ of one part in a thousand flips the answer either way; the four robust modes shift by less than 0.3 % between the two constants. Both readings are given so that the boundary is explicit.

It is also worth noting which mode closes the single-mode band. Here $a/b = 2.25 \gt 2$, so the second mode to appear is TE$_{20}$ at 13.333 GHz rather than TE$_{01}$ at 15 GHz: the useful single-mode range of this guide is 6.667 to 13.333 GHz, and at 20 GHz the guide is being run heavily overmoded.

Final results
QuantityValue
Free-space wavelength$\lambda_0 = 1.500\ \text{cm}$
Propagating modes at 20 GHzTE$_{10}$, TE$_{20}$, TE$_{01}$, TE$_{11}$/TM$_{11}$
Shortest guide wavelength$\mathbf{\lambda_g = 1.591\ \text{cm}}$ (TE$_{10}$)
Longest guide wavelength$\mathbf{\lambda_g = 2.626\ \text{cm}}$ (TE$_{11}$/TM$_{11}$)
Boundary caseTE$_{30}$ at $f_c = 20.000\ \text{GHz}$ ($\beta = 0$, excluded)
Single-mode band of this guide$6.667$ to $13.333\ \text{GHz}$