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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?
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Consider a air-filled waveguide operating in the TE12 mode at a freque...
v = c, f = 1.2 fc
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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.
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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?
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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?.
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