Topic: Microwave Engineering
A waveguide of dimensions a = 15 mm and $b=7.5$ mm is used as a high-pass filter. If the stop band attenuation required at 8 GHz is ~109.2 dB, what is the length of the filter? (assume conductor losses to be zero, approximate $\pi=3.14$ and 1 Np ~8.69 dB) $(\log_{10}e=0.4343)$
- A. 100 mm
- B. 869 mm
- C. 86.9 mm
- D. 54.6 mm
Answer: A. 100 mm
Explanation: The dominant mode cutoff frequency is $f_c = c/(2a) = 3\times10^{10}/(2\times1.5) = 10$ GHz. At $f = 8$ GHz (which is below cutoff, hence in the stopband), the attenuation constant is $\alpha = \frac{2\pi f_c}{c} \sqrt{1 - (f/f_c)^2}$. Substituting the values: $\alpha = \frac{2\pi \times 10^{10}}{3\times10^{10}} \sqrt{1 - 0.64} = \frac{2\pi}{3} \times 0.6 = 0.4\pi$ Np/cm = $40\pi$ Np/m. In dB/m, this is $40\pi \times 8.69 \approx 1092$ dB/m. The required total attenuation is 109.2 dB. Therefore, the length required is $109.2 / 1092 = 0.1$ m = 100 mm.