Electronics and Communication Engineering - Electromagnetic Field Theory - Discussion
Discussion Forum : Electromagnetic Field Theory - Section 1 (Q.No. 10)
10.
A rectangular metal waveguide filled with a dielectric of relative permittivity εr = 4, has the inside dimensions 3 x 1.2 cm, the cut off frequency for the dominant mode is
Answer: Option
Explanation:
for dominant mode.
Discussion:
15 comments Page 2 of 2.
Parvez said:
8 years ago
Vp = c /sqrt(μrεr), here μr =1 ; εr = 4 (given).
So, fc = Vp/2a.
i.e. fc =c/(2a*sqrt(μrεr)).
fc = 2.5 GHz.
So, fc = Vp/2a.
i.e. fc =c/(2a*sqrt(μrεr)).
fc = 2.5 GHz.
Ravi said:
8 years ago
5 GHz is correct answer.
fc = 1.5*10^8 sqrt{(m/a)^2+(n/b)^2} .
Here m=1, n=0 (dominant mode in a rectangular waveguide is TE1,0).
So,1.5*10^8 * m/a = 1.5*10^8 * 1/0.03 = 5*10^9 Hz.
Here a is width.
fc = 1.5*10^8 sqrt{(m/a)^2+(n/b)^2} .
Here m=1, n=0 (dominant mode in a rectangular waveguide is TE1,0).
So,1.5*10^8 * m/a = 1.5*10^8 * 1/0.03 = 5*10^9 Hz.
Here a is width.
John Howard said:
8 years ago
Option A is correct.
λc=2a; fc=v/λc or fc=(c/√(εr))/λc = 2.5 GHZ. v=1.5 10^8 m/s and λc=0.06 m.
λc=2a; fc=v/λc or fc=(c/√(εr))/λc = 2.5 GHZ. v=1.5 10^8 m/s and λc=0.06 m.
(1)
Priti said:
6 years ago
Where c is velocity of light.
a =3,
fc= 3 * 10^8/2 * 3 * 10^-3,
= 5 * 10^9.
= 5 GHz.
a =3,
fc= 3 * 10^8/2 * 3 * 10^-3,
= 5 * 10^9.
= 5 GHz.
Midhun Joy said:
3 years ago
Option A is the correct answer.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers