Assignment EE818 4 Solutions 1. Waveguide modes a) A rectangular metallic waveguide has dimensions 2.104 cm by 1.052 cm. What are the cut-off frequencies for the six lowest modes? b) A circular metallic waveguide has a radius of 0.839 cm (thus it has about the same crosssectional area as the waveguide in part a. What are the cut-off frequencies of the six lowest modes? How do the two waveguides compare? Answer: a) The cut-off frequency for the (m, n) mode of a rectangular waveguide is 1 c = m a
2
+
n b
2
1/2
For the dimensions given the seven lowest modes are TE10 4.48 1010 rad/s = 7.13 GHz TE01 and TE20 8.95 1010 rad/s = 14.2 GHz TE11 and TM11 1.00 1011 rad/s = 15.9 GHz TE21 and TM21 1.79 1011 rad/s = 20.2 GHz b) The cut-off frequency for the (n, l) mode of a circular waveguide is 1 tnl c = a 1 snl c = a for the TM modes
for the TE modes
The lowest modes are simply the ones corresponding to the lowest values of tnl and snl . TE11 6.58 1010 rad/s = 10.5 GHz TM01 8.59 1010 rad/s = 13.7 GHz TE21 1.09 1011 rad/s = 17.4 GHz TE01 and TM11 1.37 1011 rad/s = 21.8 GHz
TE31 1.50 1011 rad/s = 23.9 GHz The cut-off frequencies are in general larger for a circular waveguide compared to a rectangular waveguide of the same cross-section. 2. Practical waveguide dimensions A rectangular waveguide has small dimension b and large dimension a = xb; x > 1. Write down the cutoff frequencies for the TE10 , TE01 , and TE20 modes. For what values of x will the waveguide have as large a frequency range as possible in which only a single mode propagates? Explain why most commercial waveguides have x near 2. Answer: Let a = xb with x 1. The lowest mode is TE10 with cut-off frequency 10 = ( )1/2 a
The next higher mode will either be TE01 (if x 2) or TE20 (if x 2). 01 = ( )1/2 20 = ( )1/2 b 2 a
If x < 2 then 01 < 20 , and the range will extend 10 from to 01 . For x 2, the range is fixed by the value of a and is from 10 to 20 = 210 which is larger than the range for x < 2. For a given a, as x increases, b decreases which lowers the cross-sectional area of the waveguide lowering the power handling capability and increasing the attenuation. Thus, to get the widest single-mode frequency range with the least attenuation and highest power rating, the best choice is x = 2, i.e. a = 2b. 3. Waveguide design Design a rectangular waveguide that will propagate a single mode at 12 GHz. The cutoff frequency needs to be at least 20% below the operating frequency and the next higher mode needs to be at least 10% higher than the operating frequency. Answer: The previous problem showed that the widest single-mode frequency range occurs for a = 2b. The cut-off frequencies are 1 = ( )1/2 2 = ( )1/2 2b b for TE10
for TE10 and TE20
We want 1 1 1 < 2(12 GHz) < 2 0.8 1.1 0.781 cm < b < 1.14 cm So any b in this range with a = 2b meets the design requirements. 4. Maximum power in a waveguide Air breaks down when an electric field exceeds about 3 106 V/m...