Question 4 of 7: WR-90 rectangular waveguide at 10 GHz
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. 16-Elec-A7 Electromagnetics, December 2019 — 3 hours, closed book, approved calculator only. Seven questions of equal value (20 marks each); the paper states that five completed questions constitute a complete paper. All seven are solved here, because the set is intended as a study resource.
Reference texts (22-Elec-A7 Electromagnetics). F. T. Ulaby & U. Ravaioli, Fundamentals of Applied Electromagnetics, 8th ed.; M. N. O. Sadiku, Elements of Electromagnetics, 7th ed.; W. H. Hayt & J. A. Buck, Engineering Electromagnetics, 9th ed.; D. M. Pozar, Microwave Engineering, 4th ed.; C. A. Balanis, Antenna Theory: Analysis and Design, 4th ed.
Constants used throughout. $c = 3.00\times 10^{8}\ \text{m/s}$, $\varepsilon_0 = 8.854\times 10^{-12}\ \text{F/m}$, $\mu_0 = 4\pi\times 10^{-7}\ \text{H/m}$, $\eta_0 = 376.7\ \Omega$. Where a boundary case could turn on the choice between $c = 3\times 10^{8}$ and $c = 2.998\times 10^{8}$ m/s, both readings are stated.
Figure reading. In Figure 1(b) the maxima sit at z = 0, 1.5 and 3.0 m and the minima at z = 0.75 and 2.25 m, so the wavelength is 3.00 m. Three independent checks confirm that reading — the quarter-wave maximum-to-minimum spacing, the fact that the stated 3.75 m line is then exactly 1.25 wavelengths, and the fact that z = l then lands on a voltage minimum where the quarter-wave transformer result must give $Z_{\text{in}} = Z_0/S$.
Question 4: WR-90 rectangular waveguide at 10 GHz (20 marks)
Given. A standard X-band WR-90 guide, air-filled, operated at 10 GHz.
Given data
Quantity
Symbol
Value
Broad (wide) wall
$a$
$22.86\ \text{mm}$
Narrow wall
$b$
$10.16\ \text{mm}$
Aspect ratio
$a/b$
$2.25$
Filling
$\varepsilon_r,\ \mu_r$
$1,\ 1$ (free space)
Operating frequency
$f$
$10\ \text{GHz}$
Free-space wavelength
$\lambda_0$
$3.00\ \text{cm}$
Find. The dominant mode and its cut-off, then the phase velocity, guide wavelength and wave impedance at 10 GHz, and finally the next three cut-off frequencies with their mode types.
WR-90 cross-section with the TE10 electric field: a half-sine across the wide wall a, uniform across the narrow wall b.
Approach. Everything follows from the cut-off formula and the single dispersion factor $\sqrt{1-(f_c/f)^2}$: the phase velocity and guide wavelength divide by it, the wave impedance divides by it, and the group velocity multiplies by it.
Write the general cut-off formula and identify the dominant mode. For a rectangular guide filled with a medium of wave speed $u$, $$f_{c,mn} = \frac{u}{2}\sqrt{\left(\frac{m}{a}\right)^{2} + \left(\frac{n}{b}\right)^{2}}, \qquad u = c \ \text{(air-filled)}.$$ The smallest non-zero index pair on the larger dimension gives the lowest cut-off, and since $a \gt b$ that is TE$_{10}$: $$f_{c,10} = \frac{c}{2a} = \frac{3.00\times 10^{8}}{2(0.02286)} = \boxed{6.562\ \text{GHz (TE}_{10}\text{)}}.$$ (With the exact $c = 2.998\times 10^{8}$ m/s this is 6.557 GHz — a 0.07 % difference that changes nothing here.)
Form the dispersion factor once and reuse it. At $f = 10$ GHz, $$\frac{f_c}{f} = \frac{6.562}{10} = 0.6562, \qquad \sqrt{1 - (f_c/f)^{2}} = \sqrt{1 - 0.4306} = 0.7546.$$ The guide is comfortably above cut-off, so the mode propagates.
Phase velocity. $$v_p = \frac{c}{\sqrt{1-(f_c/f)^{2}}} = \frac{3.00\times 10^{8}}{0.7546} = \boxed{3.976\times 10^{8}\ \text{m/s}}$$ which exceeds $c$ — correctly so, because a guided mode is a zig-zagging plane wave whose phase fronts sweep along the axis faster than the wave itself moves. Energy travels at the group velocity $v_g = c\sqrt{1-(f_c/f)^2} = 2.264\times 10^{8}$ m/s, and $v_p v_g = c^{2}$ exactly.
Guide wavelength. The free-space wavelength at 10 GHz is $\lambda_0 = c/f = 3.00$ cm, and the guide stretches it by the same factor: $$\lambda_g = \frac{\lambda_0}{\sqrt{1-(f_c/f)^{2}}} = \frac{3.00}{0.7546} = \boxed{3.976\ \text{cm}}$$ which equals $v_p/f$, as it must.
Wave impedance. For a TE mode the transverse E and H are related by $$Z_{\text{TE}} = \frac{\eta_0}{\sqrt{1-(f_c/f)^{2}}} = \frac{376.7}{0.7546} = \boxed{499.2\ \Omega}.$$ It is larger than $\eta_0$, and rises without bound as $f \to f_c$ — the mode becomes all E and no propagating H at cut-off. (A TM mode would instead have $Z_{\text{TM}} = \eta_0\sqrt{1-(f_c/f)^2}$.)
Rank the higher-order modes. Substituting index pairs into the cut-off formula and sorting: TE$_{20}$ at $c/a = 13.123$ GHz, TE$_{01}$ at $c/2b = 14.764$ GHz, and the degenerate TE$_{11}$/TM$_{11}$ pair at $(c/2)\sqrt{(1/a)^2+(1/b)^2} = 16.156$ GHz. So $$\boxed{\text{TE}_{20} = 13.123\ \text{GHz (TE only)},\ \text{TE}_{01} = 14.764\ \text{GHz (TE only)},\ \text{TE}_{11}/\text{TM}_{11} = 16.156\ \text{GHz (both)}}$$ Note two things. First, TM modes require both indices non-zero (a TM$_{m0}$ mode would have no field at all), which is why only the $(1,1)$ entry is doubled. Second, because $a/b = 2.25 \gt 2$ it is TE$_{20}$, not TE$_{01}$, that closes the single-mode band — a ratio below 2 would swap them.
Confirm single-mode operation. Only TE$_{10}$ (6.562 GHz) lies below the 10 GHz operating point; the next mode is 3.1 GHz above it. WR-90 is therefore genuinely single-mode at 10 GHz, which is exactly why it is the standard X-band guide.
Cut-off ladder for WR-90. Only TE10 falls below the 10 GHz operating line, so the guide is single-mode there; TE20 closes the band because a/b > 2.
