Electronics and Communication Engineering - Exam Questions Papers - Discussion
Discussion Forum : Exam Questions Papers - Exam Paper 1 (Q.No. 4)
4.
A rectangular waveguide, in dominant TE mode, has dimensions 10 cm x 15 cm. The cut off frequency is
Answer: Option
Explanation:
For TE10 mode
Discussion:
8 comments Page 1 of 1.
Thenmozhi said:
5 years ago
@Gopal.
What is "c" here?
What is "c" here?
(1)
NAVYA REKHA said:
6 years ago
@Gauri.
Here a is wider band& b is narrow band means a=15,b=10.
So,f=3*10^8 m/2*15 cm,
3*10^8/0.3=1ghz.
Here a is wider band& b is narrow band means a=15,b=10.
So,f=3*10^8 m/2*15 cm,
3*10^8/0.3=1ghz.
Rajesh said:
6 years ago
Dominant mode is TE10 only when dimensions are a>b but here in the question he gave a<b so dominant mode will be TE01 so Cut-off frequency is C/(2b).
Gauri said:
6 years ago
Yes, answer should be 1.5 GHz.
(1)
Gopal said:
7 years ago
@Alok.
I agree with your answer.
For Dominant mode TE10.
Cutoff frequency f = c/2a.
Put all the value and we get our answer.
I agree with your answer.
For Dominant mode TE10.
Cutoff frequency f = c/2a.
Put all the value and we get our answer.
Santhosh said:
8 years ago
Please explain clearly. I can't understand.
Alok said:
9 years ago
Answer will be 1.5 GHz.
For TE mode we will consider a = 10 cm and f= (3*10^10) / (2*a (cm)).
For TE mode we will consider a = 10 cm and f= (3*10^10) / (2*a (cm)).
Vasundhara said:
9 years ago
In general dimensions of rectangle are represented as l:b. So a = 10cm, but here they have considered a = 15cm. Can any one help me to understand this?
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