Question

The cut off frequency of a waveguide with resonant cavity is given by

(a) V√((m/a)^2 + (n/b)^2 + (p/d)^2)/2

(b) V√((m/a)^2 + (n/b)^2 + (p/d)^2)

(c) 2V√((m/a)^2 + (n/b)^2 + (p/d)^2)

(d) V√((m/a) + (n/b) + (p/d))/2

I have been asked this question in an interview.

This intriguing question comes from Transients in division Waveguides of Electromagnetic Theory

Answer 1

Right option is (a) V√((m/a)^2 + (n/b)^2 + (p/d)^2)/2

The best explanation: The cut off frequency of the waveguide of dimensions a x b with the resonant cavity of dimension d is given by fc = √((m/a)^2 + (n/b)^2 + (p/d)^2)/2. Here m and n are the orders of the waveguide, p is the order of the cavity and v is the velocity.