1.

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))/2I have been asked this question in an interview.This intriguing question comes from Transients in division Waveguides of Electromagnetic Theory

Answer»

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.


Discussion

No Comment Found

Related InterviewSolutions