Electronics and Communication Engineering - Electromagnetic Field Theory - Discussion

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

[A]. 2.5 GHz
[B]. 5 GHz
[C]. 10 GHz
[D]. 12.5 GHz

Answer: Option B

Explanation:

for dominant mode.


Kamlesh said: (Aug 18, 2014)  
Option A is correct.

Phase velocity is c/2 hence f = c/4a.

Vinay said: (Mar 10, 2015)  
Yes option A is correct.

C' = C/2.

Allaa said: (Sep 5, 2015)  
c = c0/sqrt(εr).
f = c/2a.

Ravi said: (May 21, 2016)  
Cut off frequency = c/(2a * sqrt (Er * Ur).

Anjli said: (Jul 5, 2016)  
As f = C'/2a.

And C' = C/2 (because Er = 4).

So f = c/4a.

Pk Rai said: (Aug 4, 2016)  
A is correct.

Sunil Chavan said: (Jun 28, 2017)  
A is the answer.

Ronston said: (Aug 10, 2017)  
What does 'a' stand for? dimension?

Azar said: (Sep 20, 2017)  
fc = C/2a.

fc = Cut-off Frequency.
C = Speed of Light = 3 * 10^8,
a = Width = 3 cm.

fc = 3 * 10^8 / 2 *3 = 0.5 * 10^8 ==> 5 GHz.

Vijay said: (Oct 20, 2017)  
Fc should be divided by 2.

Parvez said: (Nov 1, 2017)  
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.

Ravi said: (Nov 10, 2017)  
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.

John Howard said: (Dec 30, 2017)  
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.

Priti said: (Apr 21, 2019)  
Where c is velocity of light.

a =3,
fc= 3 * 10^8/2 * 3 * 10^-3,
= 5 * 10^9.
= 5 GHz.

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