Electronics and Communication Engineering (ECE) Exam > Electronics and Communication Engineering (ECE) Questions > A white noise signal having two sided spectr...
Start Learning for Free
A white noise signal having two sided spectral density of 5*10-6 V2/Hz is applied to a simple RC low pass filter whose 3 dB cutoff frequency is 5 kHz. The mean squared value of noise obtained at the output of filter would be
- a)0.3927 V2
- b)39.27*10-6 V2
- c)39.27 V2
- d)39.27*10-3 V2
Correct answer is option 'D'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Most Upvoted Answer
A white noise signal having two sided spectral density of 5*10-6 V2/H...
White noise with 2 sided PSD 5*10-6 V2/Hz.
RC low pass filter has 3 dB cutoff 5 kHz.
Mean Squared value of the output power can be found using following way,
E(y2)=Ry(0)
This will give the Mean squared output power to be 39.27*10-3 V2
Free Test
FREE
| Start Free Test |
Community Answer
A white noise signal having two sided spectral density of 5*10-6 V2/H...
To find the mean squared value of noise at the output of the RC low pass filter, we need to consider the transfer function of the filter and the power spectral density of the input white noise signal.
1. Transfer Function of the RC Low Pass Filter:
The transfer function of an RC low pass filter is given by:
H(f) = 1 / (1 + j2πf/fcut)
where fcut is the 3 dB cutoff frequency of the filter.
2. Power Spectral Density of the Input Signal:
The power spectral density (PSD) of a white noise signal is constant across all frequencies. In this case, the two-sided PSD is given as 5*10^-6 V^2/Hz.
3. Mean Squared Value of Noise at the Output:
The mean squared value of the noise at the output of the filter can be found by integrating the PSD of the input signal multiplied by the squared magnitude of the transfer function of the filter over all frequencies.
Let's calculate this step by step:
- First, we square the magnitude of the transfer function:
|H(f)|^2 = 1 / (1 + (2πf/fcut)^2)
- Then, we multiply it by the PSD of the input signal:
PSD_output(f) = PSD_input(f) * |H(f)|^2
= (5*10^-6 V^2/Hz) * (1 / (1 + (2πf/fcut)^2))
- Now, we integrate the PSD_output(f) over all frequencies:
Mean Squared Value = ∫ PSD_output(f) df (from -∞ to +∞)
Since the PSD of the white noise signal is constant across all frequencies, we can take it out of the integral:
Mean Squared Value = PSD_input * ∫ |H(f)|^2 df (from -∞ to +∞)
= (5*10^-6 V^2/Hz) * ∫ 1 / (1 + (2πf/fcut)^2) df (from -∞ to +∞)
- Solving the integral, we get:
Mean Squared Value = (5*10^-6 V^2/Hz) * (π/fcut)
Substituting the value of fcut = 5 kHz = 5000 Hz:
Mean Squared Value = (5*10^-6 V^2/Hz) * (π/5000 Hz)
= (5*10^-6 V^2/Hz) * (3.14/5000)
= 3.14*10^-9 V^2
= 3.14*10^-3 * 10^-6 V^2
= 3.14*10^-3 * 10^-6 * 10^3 V^2
= 3.14*10^-9 * 10^3 V^2
= 3.14*10^-6 V^2
= 39.27*10^-6 V^2
Therefore, the mean squared value of the noise obtained at the output of the RC low pass filter is 39.27*10^-6 V^2, which is option D.
1. Transfer Function of the RC Low Pass Filter:
The transfer function of an RC low pass filter is given by:
H(f) = 1 / (1 + j2πf/fcut)
where fcut is the 3 dB cutoff frequency of the filter.
2. Power Spectral Density of the Input Signal:
The power spectral density (PSD) of a white noise signal is constant across all frequencies. In this case, the two-sided PSD is given as 5*10^-6 V^2/Hz.
3. Mean Squared Value of Noise at the Output:
The mean squared value of the noise at the output of the filter can be found by integrating the PSD of the input signal multiplied by the squared magnitude of the transfer function of the filter over all frequencies.
Let's calculate this step by step:
- First, we square the magnitude of the transfer function:
|H(f)|^2 = 1 / (1 + (2πf/fcut)^2)
- Then, we multiply it by the PSD of the input signal:
PSD_output(f) = PSD_input(f) * |H(f)|^2
= (5*10^-6 V^2/Hz) * (1 / (1 + (2πf/fcut)^2))
- Now, we integrate the PSD_output(f) over all frequencies:
Mean Squared Value = ∫ PSD_output(f) df (from -∞ to +∞)
Since the PSD of the white noise signal is constant across all frequencies, we can take it out of the integral:
Mean Squared Value = PSD_input * ∫ |H(f)|^2 df (from -∞ to +∞)
= (5*10^-6 V^2/Hz) * ∫ 1 / (1 + (2πf/fcut)^2) df (from -∞ to +∞)
- Solving the integral, we get:
Mean Squared Value = (5*10^-6 V^2/Hz) * (π/fcut)
Substituting the value of fcut = 5 kHz = 5000 Hz:
Mean Squared Value = (5*10^-6 V^2/Hz) * (π/5000 Hz)
= (5*10^-6 V^2/Hz) * (3.14/5000)
= 3.14*10^-9 V^2
= 3.14*10^-3 * 10^-6 V^2
= 3.14*10^-3 * 10^-6 * 10^3 V^2
= 3.14*10^-9 * 10^3 V^2
= 3.14*10^-6 V^2
= 39.27*10^-6 V^2
Therefore, the mean squared value of the noise obtained at the output of the RC low pass filter is 39.27*10^-6 V^2, which is option D.
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 white noise signal having two sided spectral density of 5*10-6 V2/Hz is applied to a simple RC low pass filter whose 3 dB cutoff frequency is 5 kHz. The mean squared value of noise obtained at the output of filter would bea)0.3927 V2b)39.27*10-6 V2c)39.27 V2d)39.27*10-3 V2Correct answer is option 'D'. Can you explain this answer?
Question Description
A white noise signal having two sided spectral density of 5*10-6 V2/Hz is applied to a simple RC low pass filter whose 3 dB cutoff frequency is 5 kHz. The mean squared value of noise obtained at the output of filter would bea)0.3927 V2b)39.27*10-6 V2c)39.27 V2d)39.27*10-3 V2Correct answer is option 'D'. 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 white noise signal having two sided spectral density of 5*10-6 V2/Hz is applied to a simple RC low pass filter whose 3 dB cutoff frequency is 5 kHz. The mean squared value of noise obtained at the output of filter would bea)0.3927 V2b)39.27*10-6 V2c)39.27 V2d)39.27*10-3 V2Correct answer is option 'D'. 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 white noise signal having two sided spectral density of 5*10-6 V2/Hz is applied to a simple RC low pass filter whose 3 dB cutoff frequency is 5 kHz. The mean squared value of noise obtained at the output of filter would bea)0.3927 V2b)39.27*10-6 V2c)39.27 V2d)39.27*10-3 V2Correct answer is option 'D'. Can you explain this answer?.
A white noise signal having two sided spectral density of 5*10-6 V2/Hz is applied to a simple RC low pass filter whose 3 dB cutoff frequency is 5 kHz. The mean squared value of noise obtained at the output of filter would bea)0.3927 V2b)39.27*10-6 V2c)39.27 V2d)39.27*10-3 V2Correct answer is option 'D'. 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 white noise signal having two sided spectral density of 5*10-6 V2/Hz is applied to a simple RC low pass filter whose 3 dB cutoff frequency is 5 kHz. The mean squared value of noise obtained at the output of filter would bea)0.3927 V2b)39.27*10-6 V2c)39.27 V2d)39.27*10-3 V2Correct answer is option 'D'. 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 white noise signal having two sided spectral density of 5*10-6 V2/Hz is applied to a simple RC low pass filter whose 3 dB cutoff frequency is 5 kHz. The mean squared value of noise obtained at the output of filter would bea)0.3927 V2b)39.27*10-6 V2c)39.27 V2d)39.27*10-3 V2Correct answer is option 'D'. Can you explain this answer?.
Solutions for A white noise signal having two sided spectral density of 5*10-6 V2/Hz is applied to a simple RC low pass filter whose 3 dB cutoff frequency is 5 kHz. The mean squared value of noise obtained at the output of filter would bea)0.3927 V2b)39.27*10-6 V2c)39.27 V2d)39.27*10-3 V2Correct answer is option 'D'. 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 white noise signal having two sided spectral density of 5*10-6 V2/Hz is applied to a simple RC low pass filter whose 3 dB cutoff frequency is 5 kHz. The mean squared value of noise obtained at the output of filter would bea)0.3927 V2b)39.27*10-6 V2c)39.27 V2d)39.27*10-3 V2Correct answer is option 'D'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
A white noise signal having two sided spectral density of 5*10-6 V2/Hz is applied to a simple RC low pass filter whose 3 dB cutoff frequency is 5 kHz. The mean squared value of noise obtained at the output of filter would bea)0.3927 V2b)39.27*10-6 V2c)39.27 V2d)39.27*10-3 V2Correct answer is option 'D'. Can you explain this answer?, a detailed solution for A white noise signal having two sided spectral density of 5*10-6 V2/Hz is applied to a simple RC low pass filter whose 3 dB cutoff frequency is 5 kHz. The mean squared value of noise obtained at the output of filter would bea)0.3927 V2b)39.27*10-6 V2c)39.27 V2d)39.27*10-3 V2Correct answer is option 'D'. Can you explain this answer? has been provided alongside types of A white noise signal having two sided spectral density of 5*10-6 V2/Hz is applied to a simple RC low pass filter whose 3 dB cutoff frequency is 5 kHz. The mean squared value of noise obtained at the output of filter would bea)0.3927 V2b)39.27*10-6 V2c)39.27 V2d)39.27*10-3 V2Correct answer is option 'D'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A white noise signal having two sided spectral density of 5*10-6 V2/Hz is applied to a simple RC low pass filter whose 3 dB cutoff frequency is 5 kHz. The mean squared value of noise obtained at the output of filter would bea)0.3927 V2b)39.27*10-6 V2c)39.27 V2d)39.27*10-3 V2Correct answer is option 'D'. 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