- 2080 Baisakh · 6+4 marks
Air-filled 3 cm waveguide at 6 GHz; coupler parameters
Short answer
For a=3 cm, TE10 gives λc=6 cm and fc=5 GHz; at 6 GHz, λg=9.045 cm, vp=5.427e8 m/s, vg=1.658e8 m/s, and Z_TE≈682 Ω. Directional coupler performance is described by coupling, directivity, isolation, and insertion loss.
For an air-filled rectangular waveguide with a width of 3 cm and a desired frequency of operation of 6 GHz (for dominant mode), determine cut-off frequency, cut-off wavelength, group velocity, phase velocity, propagation wavelength in the waveguide and the characteristic impedance. Explain the four basic parameters used to describe the performance of a directional coupler.
Part 1: Air-filled waveguide, a = 3 cm, f = 6 GHz, TE10 mode
Constants: c = 3 × 10⁸ m/s, η0 = 120π ≈ 377 Ω.
Cut-off wavelength:
λc = 2a = 2 × 3 = 6 cm
Cut-off frequency:
fc = c/λc = (3 × 10⁸)/(0.06) = 5 × 10⁹ Hz = 5 GHz
Since f = 6 GHz > 5 GHz, TE10 propagates.
Common factor:
λ0 = c/f = 3 × 10⁸ / 6 × 10⁹ = 5 cm
√(1 - (fc/f)²) = √(1 - (5/6)²) = √0.3056 = 0.5528
Guide (propagation) wavelength:
λg = λ0/0.5528 = 5/0.5528 = 9.045 cm
Phase velocity:
vp = c/0.5528 = 5.427 × 10⁸ m/s
Group velocity:
vg = c × 0.5528 = 1.658 × 10⁸ m/s
(check: vp·vg = 9 × 10¹⁶ = c²)
Characteristic (wave) impedance:
Z_TE = η0/0.5528 = 377/0.5528 = 682 Ω
| Quantity | Value |
|---|---|
| fc | 5 GHz |
| λc | 6 cm |
| λg | 9.045 cm |
| vp | 5.427 × 10⁸ m/s |
| vg | 1.658 × 10⁸ m/s |
| Z_TE | 682 Ω |
Answer: fc = 5 GHz, λc = 6 cm, vg = 1.658 × 10⁸ m/s, vp = 5.427 × 10⁸ m/s, λg = 9.045 cm, Z_TE ≈ 682 Ω.
Part 2: Four basic parameters of a directional coupler
Ports: 1 = input, 2 = through, 3 = coupled, 4 = isolated.
(1) Input =================== Through (2)
coupling region
(4) Isolated ---------------- Coupled (3)
- Coupling factor: C = 10 log(P1/P3) dB = −20 log|S31|. The fraction of input power sent to the coupled port (e.g. 10 dB coupler sends 10%).
- Directivity: D = 10 log(P3/P4) dB = 20 log(|S31|/|S41|). How well the coupler separates forward and reverse waves; ideally infinite, typically 30–40 dB.
- Isolation: I = 10 log(P1/P4) dB = −20 log|S41| = C + D. Power leaking to the isolated port.
- Insertion loss: IL = 10 log(P1/P2) dB = −20 log|S21|. Power lost in the main line, including the coupled power.
✓ Answer checked by Claude Opus 5.5 (high effort)
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