--- title: "Rectangular waveguide modes and critical parameters" subject: "RF and Microwave Engineering" chapter: 4 key: "semester_7/rf_and_microwave_engineering:rectangular-waveguide-modes-parameters" source: https://www.bctengineer.com/q/semester_7/rf_and_microwave_engineering/rectangular-waveguide-modes-parameters site: BCT Engineering --- # Rectangular waveguide modes and critical parameters RF and Microwave Engineering · Chapter 4: RF and Microwave Components and Devices - Asked in: 2069 Bhadra (old course), 8 marks - Video: [Rectangular Waveguide (Basics, Modes, Cut-Off Frequency, Wave Impedance, Phase & Group Velocity)](https://www.youtube.com/watch?v=Y3szgmhuXeg) — Engineering Funda - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/rf_and_microwave_engineering/rectangular-waveguide-modes-parameters - Whole chapter: https://www.bctengineer.com/md/semester_7/rf_and_microwave_engineering/chapter-4 ## Question Describe rectangular waveguide based on modes of propagation and other critical parameters. ## Short answer Rectangular waveguide is a hollow metal pipe supporting TE and TM modes above cut-off, acting as a high-pass filter. The dominant TE10 mode has cut-off frequency fc=c/(2a), guide wavelength λg=λ0/√(1−(fc/f)²), phase velocity vp > c, and group velocity vg < c. Parameters depend on mode indices m and n and waveguide dimensions a (broad) and b (narrow). ## Answer A **rectangular waveguide** is a hollow metal pipe of inner width *a* and height *b* (a > b, often a ≈ 2b). Being a single conductor, it cannot carry TEM waves; it carries **TE** and **TM** modes above their cut-off frequencies, acting as a high-pass filter. ```text y b +-----------------+ | E (TE10) ↑ | propagation along z +-----------------+ 0 a -> x ``` #### Modes of propagation - **TEmn (H modes):** Ez = 0, Hz = H0 cos(mπx/a) cos(nπy/b). m, n = 0, 1, 2 … but not both zero. Lowest: TE10. - **TMmn (E modes):** Hz = 0, Ez = E0 sin(mπx/a) sin(nπy/b). m, n ≥ 1. Lowest: TM11. - *m* and *n* are the numbers of half-wave field variations along the broad (a) and narrow (b) walls. - **Dominant mode:** TE10 (lowest fc). **Degenerate modes:** TEmn and TMmn with the same m, n share the same fc. #### Critical parameters (air-filled) | Parameter | Expression | |---|---| | Cut-off frequency | fc = (c/2)√((m/a)² + (n/b)²) | | Cut-off wavelength | λc = 2/√((m/a)² + (n/b)²); TE10: 2a | | Phase constant | β = (2π/λ0)√(1 − (fc/f)²) | | Guide wavelength | λg = λ0/√(1 − (fc/f)²) | | Phase velocity | vp = c/√(1 − (fc/f)²) > c | | Group velocity | vg = c√(1 − (fc/f)²) < c | | TE wave impedance | Z_TE = η/√(1 − (fc/f)²) | | TM wave impedance | Z_TM = η√(1 − (fc/f)²) | Also 1/λ0² = 1/λg² + 1/λc² and vp·vg = c². **Example (WR-90, a = 2.286 cm, b = 1.016 cm, f = 10 GHz, TE10):** fc = 6.562 GHz, λg = 3.976 cm, vp = 3.976 × 10⁸ m/s, vg = 2.264 × 10⁸ m/s, Z_TE ≈ 500 Ω. TE20 cuts off at 13.12 GHz, so only TE10 propagates. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why can a hollow rectangular waveguide not carry TEM waves? A. It is too wide B. It is a single conductor C. It is made of copper D. It has no cut-off Answer: B. TEM needs two conductors; a single hollow conductor supports only TE and TM modes above cut-off. (Key point: Why no TEM) 2. Which index ranges are correct? A. TE: m, n ≥ 1; TM: not both zero B. TE: m, n = 0,1,2.. not both zero; TM: m, n ≥ 1 C. Both: m = n = 0 allowed D. TE: m, n ≥ 1; TM: m or n zero allowed Answer: B. TMmn needs m, n ≥ 1 or Ez vanishes; TEmn may have one index zero. (Key point: Mode indices) 3. Above cut-off, vp and vg compare with c as: A. vp < c and vg > c B. vp > c and vg < c, with vp·vg = c² C. Both equal c D. Both greater than c Answer: B. vp = c/√(1 − (fc/f)²) > c and vg = c√(1 − (fc/f)²) < c. (Key point: Phase and group velocity) 4. The guide wavelength λg is related to λ0 by: A. λg = λ0√(1 − (fc/f)²) B. λg = λ0/√(1 − (fc/f)²), which is larger than λ0 C. λg = λ0 D. λg = λ0·fc/f Answer: B. λg = λ0/√(1 − (fc/f)²) > λ0; also 1/λ0² = 1/λg² + 1/λc². 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