Electronics and Communication Engineering (ECE) Exam > Electronics and Communication Engineering (ECE) Questions > A standard air filled rectangular waveguide w...
Start Learning for Free
A standard air filled rectangular waveguide with dimensions a = 8cm , b = 4cm operates at 3.4 GHz. For the dominant mode of wave propagation, the phase velocity of the signal is vp . The value (rounded off to two decimal places) of vp /c , where c denotes the velocity of light is _______.
Correct answer is '1.198'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Most Upvoted Answer
A standard air filled rectangular waveguide with dimensions a = 8cm , ...
Given : a = 8cm, b = 4cm,
For a standard rectangular waveguide a > b
So, cut-off frequency for dominant mode (TE10) is
Given operating frequency is, f = 3.4 GHz
So, phase velocity in rectangular waveguide,
Hence, the value of vp /c is 1.198.
Question_Type: 4
Free Test
FREE
| Start Free Test |
Community Answer
A standard air filled rectangular waveguide with dimensions a = 8cm , ...
To find the phase velocity of the signal in a rectangular waveguide, we first need to calculate the cut-off frequency. The cut-off frequency is defined as the frequency below which the waveguide does not support any propagating modes. In this case, we are interested in the dominant mode of wave propagation.
1. Calculate the cut-off frequency:
The cut-off frequency for the dominant mode in a rectangular waveguide is given by the formula:
fc = c / (2 * sqrt(a^2 + b^2))
where c is the speed of light and a and b are the dimensions of the waveguide.
Substituting the given values:
a = 8 cm = 0.08 m
b = 4 cm = 0.04 m
c = speed of light = 3 x 10^8 m/s
fc = (3 x 10^8 m/s) / (2 * sqrt((0.08 m)^2 + (0.04 m)^2))
= (3 x 10^8 m/s) / (2 * sqrt(0.0064 + 0.0016))
= (3 x 10^8 m/s) / (2 * sqrt(0.008))
= (3 x 10^8 m/s) / (2 * 0.0894)
= (3 x 10^8 m/s) / 0.1789
≈ 1.676 x 10^9 Hz
2. Calculate the phase velocity:
The phase velocity is given by the formula:
vp = c / sqrt(1 - (fc / f)^2)
where f is the operating frequency.
Substituting the given value:
f = 3.4 GHz = 3.4 x 10^9 Hz
vp = (3 x 10^8 m/s) / sqrt(1 - ((1.676 x 10^9 Hz) / (3.4 x 10^9 Hz))^2)
= (3 x 10^8 m/s) / sqrt(1 - (0.4935)^2)
= (3 x 10^8 m/s) / sqrt(1 - 0.2436)
= (3 x 10^8 m/s) / sqrt(0.7564)
= (3 x 10^8 m/s) / 0.8698
≈ 3.448 x 10^8 m/s
3. Calculate vp / c:
vp / c = (3.448 x 10^8 m/s) / (3 x 10^8 m/s)
≈ 1.149
Rounding off to two decimal places, vp / c ≈ 1.15
Therefore, the correct answer is 1.15 (not 1.198 as mentioned).
1. Calculate the cut-off frequency:
The cut-off frequency for the dominant mode in a rectangular waveguide is given by the formula:
fc = c / (2 * sqrt(a^2 + b^2))
where c is the speed of light and a and b are the dimensions of the waveguide.
Substituting the given values:
a = 8 cm = 0.08 m
b = 4 cm = 0.04 m
c = speed of light = 3 x 10^8 m/s
fc = (3 x 10^8 m/s) / (2 * sqrt((0.08 m)^2 + (0.04 m)^2))
= (3 x 10^8 m/s) / (2 * sqrt(0.0064 + 0.0016))
= (3 x 10^8 m/s) / (2 * sqrt(0.008))
= (3 x 10^8 m/s) / (2 * 0.0894)
= (3 x 10^8 m/s) / 0.1789
≈ 1.676 x 10^9 Hz
2. Calculate the phase velocity:
The phase velocity is given by the formula:
vp = c / sqrt(1 - (fc / f)^2)
where f is the operating frequency.
Substituting the given value:
f = 3.4 GHz = 3.4 x 10^9 Hz
vp = (3 x 10^8 m/s) / sqrt(1 - ((1.676 x 10^9 Hz) / (3.4 x 10^9 Hz))^2)
= (3 x 10^8 m/s) / sqrt(1 - (0.4935)^2)
= (3 x 10^8 m/s) / sqrt(1 - 0.2436)
= (3 x 10^8 m/s) / sqrt(0.7564)
= (3 x 10^8 m/s) / 0.8698
≈ 3.448 x 10^8 m/s
3. Calculate vp / c:
vp / c = (3.448 x 10^8 m/s) / (3 x 10^8 m/s)
≈ 1.149
Rounding off to two decimal places, vp / c ≈ 1.15
Therefore, the correct answer is 1.15 (not 1.198 as mentioned).
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
|
|
Similar Electronics and Communication Engineering (ECE) Doubts
Top Courses for Electronics and Communication Engineering (ECE)View all
A standard air filled rectangular waveguide with dimensions a = 8cm , b = 4cm operates at 3.4 GHz. For the dominant mode of wave propagation, the phase velocity of the signal is vp . The value (rounded off to two decimal places) of vp /c , where c denotes the velocity of light is _______.Correct answer is '1.198'. Can you explain this answer?
Question Description
A standard air filled rectangular waveguide with dimensions a = 8cm , b = 4cm operates at 3.4 GHz. For the dominant mode of wave propagation, the phase velocity of the signal is vp . The value (rounded off to two decimal places) of vp /c , where c denotes the velocity of light is _______.Correct answer is '1.198'. 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 A standard air filled rectangular waveguide with dimensions a = 8cm , b = 4cm operates at 3.4 GHz. For the dominant mode of wave propagation, the phase velocity of the signal is vp . The value (rounded off to two decimal places) of vp /c , where c denotes the velocity of light is _______.Correct answer is '1.198'. 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 A standard air filled rectangular waveguide with dimensions a = 8cm , b = 4cm operates at 3.4 GHz. For the dominant mode of wave propagation, the phase velocity of the signal is vp . The value (rounded off to two decimal places) of vp /c , where c denotes the velocity of light is _______.Correct answer is '1.198'. Can you explain this answer?.
A standard air filled rectangular waveguide with dimensions a = 8cm , b = 4cm operates at 3.4 GHz. For the dominant mode of wave propagation, the phase velocity of the signal is vp . The value (rounded off to two decimal places) of vp /c , where c denotes the velocity of light is _______.Correct answer is '1.198'. 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 A standard air filled rectangular waveguide with dimensions a = 8cm , b = 4cm operates at 3.4 GHz. For the dominant mode of wave propagation, the phase velocity of the signal is vp . The value (rounded off to two decimal places) of vp /c , where c denotes the velocity of light is _______.Correct answer is '1.198'. 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 A standard air filled rectangular waveguide with dimensions a = 8cm , b = 4cm operates at 3.4 GHz. For the dominant mode of wave propagation, the phase velocity of the signal is vp . The value (rounded off to two decimal places) of vp /c , where c denotes the velocity of light is _______.Correct answer is '1.198'. Can you explain this answer?.
Solutions for A standard air filled rectangular waveguide with dimensions a = 8cm , b = 4cm operates at 3.4 GHz. For the dominant mode of wave propagation, the phase velocity of the signal is vp . The value (rounded off to two decimal places) of vp /c , where c denotes the velocity of light is _______.Correct answer is '1.198'. 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 A standard air filled rectangular waveguide with dimensions a = 8cm , b = 4cm operates at 3.4 GHz. For the dominant mode of wave propagation, the phase velocity of the signal is vp . The value (rounded off to two decimal places) of vp /c , where c denotes the velocity of light is _______.Correct answer is '1.198'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
A standard air filled rectangular waveguide with dimensions a = 8cm , b = 4cm operates at 3.4 GHz. For the dominant mode of wave propagation, the phase velocity of the signal is vp . The value (rounded off to two decimal places) of vp /c , where c denotes the velocity of light is _______.Correct answer is '1.198'. Can you explain this answer?, a detailed solution for A standard air filled rectangular waveguide with dimensions a = 8cm , b = 4cm operates at 3.4 GHz. For the dominant mode of wave propagation, the phase velocity of the signal is vp . The value (rounded off to two decimal places) of vp /c , where c denotes the velocity of light is _______.Correct answer is '1.198'. Can you explain this answer? has been provided alongside types of A standard air filled rectangular waveguide with dimensions a = 8cm , b = 4cm operates at 3.4 GHz. For the dominant mode of wave propagation, the phase velocity of the signal is vp . The value (rounded off to two decimal places) of vp /c , where c denotes the velocity of light is _______.Correct answer is '1.198'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A standard air filled rectangular waveguide with dimensions a = 8cm , b = 4cm operates at 3.4 GHz. For the dominant mode of wave propagation, the phase velocity of the signal is vp . The value (rounded off to two decimal places) of vp /c , where c denotes the velocity of light is _______.Correct answer is '1.198'. 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