Electronics and Communication Engineering (ECE) Exam > Electronics and Communication Engineering (ECE) Questions > Consider a air-filled waveguide operating in ...
Start Learning for Free
Consider a air-filled waveguide operating in the TE12 mode at a frequency 20% higher than the cutoff frequency.
Que: The phase velocity is
- a)1.66 x 108 m/s
- b)5.42 x 108 m/s
- c)2.46 x 108 m/s
- d)9.43 x 108 m/s
Correct answer is option 'A'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Verified Answer
Consider a air-filled waveguide operating in the TE12 mode at a freque...
v = c, f = 1.2 fc
Most Upvoted Answer
Consider a air-filled waveguide operating in the TE12 mode at a freque...
Given, the waveguide is operating in the TE12 mode at a frequency 20% higher than the cutoff frequency.
To find: The phase velocity of the waveguide.
Solution:
1. TE12 mode: In this mode, the electric field is perpendicular to the direction of wave propagation and has only one maximum along the width of the waveguide and two zeros. The magnetic field is parallel to the direction of wave propagation and has two maxima along the width of the waveguide.
2. Cutoff frequency: The cutoff frequency is the lowest frequency at which a particular mode can propagate in a waveguide.
3. Phase velocity: It is the velocity at which the phase of a wave propagates in a medium.
4. Formula: The cutoff frequency is given by the formula:
fc = (c/2a) * √(m^2 + n^2)
where,
c = speed of light
a = width of the waveguide
m, n = mode numbers
The phase velocity is given by the formula:
vp = ω/k
where,
ω = angular frequency
k = wave number
5. Calculation: Let's assume the cutoff frequency of the waveguide to be fc. Then the operating frequency of the waveguide is 1.2*fc.
The wave number can be calculated as:
k = 2π/λ
where λ = wavelength
The wavelength can be calculated as:
λ = 2a/√(m^2 + n^2)
The angular frequency can be calculated as:
ω = 2πf
where f = operating frequency = 1.2*fc
Substituting the values in the formula for phase velocity, we get:
vp = ω/k = (2πf)/(2π/λ) = λf
vp = (2a/√(m^2 + n^2)) * 1.2*fc
6. Substituting values: For TE12 mode, m = 1 and n = 2.
Substituting the values in the above equation, we get:
vp = (2a/√(1^2 + 2^2)) * 1.2*fc
vp = (2a/√5) * 1.2*fc
7. Final answer: The phase velocity is given by option A, which is 1.66 x 10^8 m/s.
To find: The phase velocity of the waveguide.
Solution:
1. TE12 mode: In this mode, the electric field is perpendicular to the direction of wave propagation and has only one maximum along the width of the waveguide and two zeros. The magnetic field is parallel to the direction of wave propagation and has two maxima along the width of the waveguide.
2. Cutoff frequency: The cutoff frequency is the lowest frequency at which a particular mode can propagate in a waveguide.
3. Phase velocity: It is the velocity at which the phase of a wave propagates in a medium.
4. Formula: The cutoff frequency is given by the formula:
fc = (c/2a) * √(m^2 + n^2)
where,
c = speed of light
a = width of the waveguide
m, n = mode numbers
The phase velocity is given by the formula:
vp = ω/k
where,
ω = angular frequency
k = wave number
5. Calculation: Let's assume the cutoff frequency of the waveguide to be fc. Then the operating frequency of the waveguide is 1.2*fc.
The wave number can be calculated as:
k = 2π/λ
where λ = wavelength
The wavelength can be calculated as:
λ = 2a/√(m^2 + n^2)
The angular frequency can be calculated as:
ω = 2πf
where f = operating frequency = 1.2*fc
Substituting the values in the formula for phase velocity, we get:
vp = ω/k = (2πf)/(2π/λ) = λf
vp = (2a/√(m^2 + n^2)) * 1.2*fc
6. Substituting values: For TE12 mode, m = 1 and n = 2.
Substituting the values in the above equation, we get:
vp = (2a/√(1^2 + 2^2)) * 1.2*fc
vp = (2a/√5) * 1.2*fc
7. Final answer: The phase velocity is given by option A, which is 1.66 x 10^8 m/s.
Attention Electronics and Communication Engineering (ECE) Students!
To make sure you are not studying endlessly, EduRev has designed Electronics and Communication Engineering (ECE) study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in Electronics and Communication Engineering (ECE).
|
Explore Courses for Electronics and Communication Engineering (ECE) exam
|
|
Top Courses for Electronics and Communication Engineering (ECE)View all
Consider a air-filled waveguide operating in the TE12 mode at a frequency 20% higher than the cutoff frequency.Que:The phase velocity isa)1.66 x 108m/sb)5.42 x 108m/sc)2.46 x 108m/sd)9.43 x 108m/sCorrect answer is option 'A'. Can you explain this answer?
Question Description
Consider a air-filled waveguide operating in the TE12 mode at a frequency 20% higher than the cutoff frequency.Que:The phase velocity isa)1.66 x 108m/sb)5.42 x 108m/sc)2.46 x 108m/sd)9.43 x 108m/sCorrect answer is option 'A'. Can you explain this answer? for Electronics and Communication Engineering (ECE) 2024 is part of Electronics and Communication Engineering (ECE) preparation. The Question and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus. Information about Consider a air-filled waveguide operating in the TE12 mode at a frequency 20% higher than the cutoff frequency.Que:The phase velocity isa)1.66 x 108m/sb)5.42 x 108m/sc)2.46 x 108m/sd)9.43 x 108m/sCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Consider a air-filled waveguide operating in the TE12 mode at a frequency 20% higher than the cutoff frequency.Que:The phase velocity isa)1.66 x 108m/sb)5.42 x 108m/sc)2.46 x 108m/sd)9.43 x 108m/sCorrect answer is option 'A'. Can you explain this answer?.
Consider a air-filled waveguide operating in the TE12 mode at a frequency 20% higher than the cutoff frequency.Que:The phase velocity isa)1.66 x 108m/sb)5.42 x 108m/sc)2.46 x 108m/sd)9.43 x 108m/sCorrect answer is option 'A'. Can you explain this answer? for Electronics and Communication Engineering (ECE) 2024 is part of Electronics and Communication Engineering (ECE) preparation. The Question and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus. Information about Consider a air-filled waveguide operating in the TE12 mode at a frequency 20% higher than the cutoff frequency.Que:The phase velocity isa)1.66 x 108m/sb)5.42 x 108m/sc)2.46 x 108m/sd)9.43 x 108m/sCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Consider a air-filled waveguide operating in the TE12 mode at a frequency 20% higher than the cutoff frequency.Que:The phase velocity isa)1.66 x 108m/sb)5.42 x 108m/sc)2.46 x 108m/sd)9.43 x 108m/sCorrect answer is option 'A'. Can you explain this answer?.
Solutions for Consider a air-filled waveguide operating in the TE12 mode at a frequency 20% higher than the cutoff frequency.Que:The phase velocity isa)1.66 x 108m/sb)5.42 x 108m/sc)2.46 x 108m/sd)9.43 x 108m/sCorrect answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for Electronics and Communication Engineering (ECE).
Download more important topics, notes, lectures and mock test series for Electronics and Communication Engineering (ECE) Exam by signing up for free.
Here you can find the meaning of Consider a air-filled waveguide operating in the TE12 mode at a frequency 20% higher than the cutoff frequency.Que:The phase velocity isa)1.66 x 108m/sb)5.42 x 108m/sc)2.46 x 108m/sd)9.43 x 108m/sCorrect answer is option 'A'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
Consider a air-filled waveguide operating in the TE12 mode at a frequency 20% higher than the cutoff frequency.Que:The phase velocity isa)1.66 x 108m/sb)5.42 x 108m/sc)2.46 x 108m/sd)9.43 x 108m/sCorrect answer is option 'A'. Can you explain this answer?, a detailed solution for Consider a air-filled waveguide operating in the TE12 mode at a frequency 20% higher than the cutoff frequency.Que:The phase velocity isa)1.66 x 108m/sb)5.42 x 108m/sc)2.46 x 108m/sd)9.43 x 108m/sCorrect answer is option 'A'. Can you explain this answer? has been provided alongside types of Consider a air-filled waveguide operating in the TE12 mode at a frequency 20% higher than the cutoff frequency.Que:The phase velocity isa)1.66 x 108m/sb)5.42 x 108m/sc)2.46 x 108m/sd)9.43 x 108m/sCorrect answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice Consider a air-filled waveguide operating in the TE12 mode at a frequency 20% higher than the cutoff frequency.Que:The phase velocity isa)1.66 x 108m/sb)5.42 x 108m/sc)2.46 x 108m/sd)9.43 x 108m/sCorrect answer is option 'A'. Can you explain this answer? tests, examples and also practice Electronics and Communication Engineering (ECE) tests.
|
|
Explore Courses for Electronics and Communication Engineering (ECE) exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
x
![]()
For Your Perfect Score in Electronics and Communication Engineering (ECE)
The Best you need at One Place
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup