Introduction / Context:
In a rectangular waveguide operating in the dominant TE10 mode, signals propagate with phase velocity greater than the speed of light in vacuum. Computing vp from the cutoff and operating frequencies is a standard microwave engineering task.
Given Data / Assumptions:
- Broad wall dimension a = 2 cm = 0.02 m (dominant TE10 mode).
- Operating frequency f = 9 GHz.
- Waveguide is air-filled (c ≈ 3 × 10^8 m/s).
Concept / Approach:
For TE10: fc = c / (2a). The phase velocity is vp = c / √(1 − (fc/f)^2). This exceeds c for all f > fc.
Step-by-Step Solution:
Step 1: Compute cutoff: fc = 3×10^8 / (2×0.02) = 3×10^8 / 0.04 = 7.5 GHz.Step 2: Ratio r = fc/f = 7.5/9 ≈ 0.8333; r^2 ≈ 0.6944.Step 3: √(1 − r^2) = √(1 − 0.6944) = √0.3056 ≈ 0.5528.Step 4: vp = c / 0.5528 ≈ 3×10^8 / 0.5528 ≈ 5.43×10^8 m/s.Step 5: Nearest option: 5 × 10^8 m/s.
Verification / Alternative check:
Group velocity vg = c √(1 − (fc/f)^2) ≈ 1.66×10^8 m/s; vp × vg = c^2—relationship holds.
Why Other Options Are Wrong:
3×10^8: Equals c; vp in a waveguide above cutoff must exceed c.1.5×10^8 and 2×10^8: Too low; these are closer to group velocities.9×10^8: Too high for the given ratio.
Common Pitfalls:
Using λg formulas without computing fc; confusing vp and vg magnitudes.
Final Answer:
5 × 10^8 m/s
Discussion & Comments