Consider a rectangular waveguide, with transverse dimensions a = 10.668 mm and b = 4.318 mm. The waveguide is uniformly
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Consider a rectangular waveguide, with transverse dimensions a = 10.668 mm and b = 4.318 mm. The waveguide is uniformly
Consider a rectangular waveguide, with transverse dimensions a = 10.668 mm and b = 4.318 mm. The waveguide is uniformly filled with a non-magnetic dielectric with €= 2. We wish to use this waveguide to propagate a wave at frequency f = 13 GHz. Which propagation modes can be sustained by this waveguide at 13 GHz? Explain in detail how you arrive at the answer. [10 marks] The cutoff frequencies for the propagation modes of a rectangular waveguide are given by the well-known formula: femn V = 2 + -C) [2 marks for identifying the correct formula to use] Since there is a dielectric in the waveguide, the velocity to be used in the formula is that for the propagation of a wave in a uniform medium having the same dielectric constant: 3х108 V = = = 2.12 x 108 m/s Ver Va [2 marks] With this, and the formula above, we can start calculating a few cutoff frequencies: TE modes: Mode Cutoff frequency (GHz) TE 10 9.942 TE20 19.88 TE01 24.56 TE11 26.5 TM modes: Mode Cutoff frequency (GHz) TM11 26.5 TE21 31.6 There's no point going any higher, since it is clear that, at 13 GHz, the only mode that propagates is the TE10. [6 marks for correctly calculating the cutoff frequency of the TE10 mode, and identifying that all the other modes have a cutoff higher than 13 GHz]
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