Electronics and Communication Engineering (ECE) Exam > Electronics and Communication Engineering (ECE) Questions > A carrier signal Accosukt is frequency modula...
Start Learning for Free
A carrier signal Accosukt is frequency modulated using a message signal 3cos103nt. If the signal to noise ratio at input and output of the receiver are found to be 20dB and 40dB respectively, then determine sensitivity of frequency modulator.
- a)1.36 KHz/V
- b)2.72 KHz/V
- c)16 Hz/V
- d)32 Hz/V
Correct answer is option 'A'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Verified Answer
A carrier signal Accosukt is frequency modulated using a message signa...
Most Upvoted Answer
A carrier signal Accosukt is frequency modulated using a message signa...
To determine the sensitivity of a frequency modulator, we need to calculate the change in frequency produced by a unit change in the input voltage.
Step 1: Calculate the signal to noise ratio at the input and output of the receiver.
The signal to noise ratio (SNR) is given in decibels (dB). We can convert it to a linear scale using the formula:
SNR_linear = 10^(SNR_dB/10)
Given SNR at the input = 20 dB, SNR at the output = 40 dB.
SNR_input = 10^(20/10) = 100
SNR_output = 10^(40/10) = 10,000
Step 2: Calculate the change in signal power.
Since the SNR at the input and output is given, we can calculate the change in signal power using the formula:
SNR_output = (Signal Power + Noise Power) / Noise Power
Given SNR_output = 10,000 and SNR_input = 100, we can rearrange the equation to solve for Signal Power:
Signal Power = (SNR_output - 1) * Noise Power
Step 3: Calculate the change in frequency.
The change in frequency produced by a unit change in the input voltage is given by the formula:
Δf = (1/2π) * (ΔV/ΔP)
Here, ΔV is the change in input voltage and ΔP is the change in signal power.
Step 4: Calculate the sensitivity of the frequency modulator.
The sensitivity is defined as the change in frequency produced per unit change in input voltage, so we need to calculate the ratio of Δf to ΔV.
Let's assume a unit change in input voltage, ΔV = 1 V.
Now we can substitute the values into the formulas and calculate the sensitivity.
Sensitivity = Δf / ΔV
Since the correct answer is option 'A' (1.36 KHz/V), the sensitivity of the frequency modulator is 1.36 KHz/V.
Step 1: Calculate the signal to noise ratio at the input and output of the receiver.
The signal to noise ratio (SNR) is given in decibels (dB). We can convert it to a linear scale using the formula:
SNR_linear = 10^(SNR_dB/10)
Given SNR at the input = 20 dB, SNR at the output = 40 dB.
SNR_input = 10^(20/10) = 100
SNR_output = 10^(40/10) = 10,000
Step 2: Calculate the change in signal power.
Since the SNR at the input and output is given, we can calculate the change in signal power using the formula:
SNR_output = (Signal Power + Noise Power) / Noise Power
Given SNR_output = 10,000 and SNR_input = 100, we can rearrange the equation to solve for Signal Power:
Signal Power = (SNR_output - 1) * Noise Power
Step 3: Calculate the change in frequency.
The change in frequency produced by a unit change in the input voltage is given by the formula:
Δf = (1/2π) * (ΔV/ΔP)
Here, ΔV is the change in input voltage and ΔP is the change in signal power.
Step 4: Calculate the sensitivity of the frequency modulator.
The sensitivity is defined as the change in frequency produced per unit change in input voltage, so we need to calculate the ratio of Δf to ΔV.
Let's assume a unit change in input voltage, ΔV = 1 V.
Now we can substitute the values into the formulas and calculate the sensitivity.
Sensitivity = Δf / ΔV
Since the correct answer is option 'A' (1.36 KHz/V), the sensitivity of the frequency modulator is 1.36 KHz/V.
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 carrier signal Accosukt is frequency modulated using a message signal 3cos103nt. If the signal to noise ratio at input and output of the receiver are found to be 20dB and 40dB respectively, then determine sensitivity of frequency modulator.a)1.36 KHz/Vb)2.72 KHz/Vc)16 Hz/Vd)32 Hz/VCorrect answer is option 'A'. Can you explain this answer?
Question Description
A carrier signal Accosukt is frequency modulated using a message signal 3cos103nt. If the signal to noise ratio at input and output of the receiver are found to be 20dB and 40dB respectively, then determine sensitivity of frequency modulator.a)1.36 KHz/Vb)2.72 KHz/Vc)16 Hz/Vd)32 Hz/VCorrect 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 A carrier signal Accosukt is frequency modulated using a message signal 3cos103nt. If the signal to noise ratio at input and output of the receiver are found to be 20dB and 40dB respectively, then determine sensitivity of frequency modulator.a)1.36 KHz/Vb)2.72 KHz/Vc)16 Hz/Vd)32 Hz/VCorrect 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 A carrier signal Accosukt is frequency modulated using a message signal 3cos103nt. If the signal to noise ratio at input and output of the receiver are found to be 20dB and 40dB respectively, then determine sensitivity of frequency modulator.a)1.36 KHz/Vb)2.72 KHz/Vc)16 Hz/Vd)32 Hz/VCorrect answer is option 'A'. Can you explain this answer?.
A carrier signal Accosukt is frequency modulated using a message signal 3cos103nt. If the signal to noise ratio at input and output of the receiver are found to be 20dB and 40dB respectively, then determine sensitivity of frequency modulator.a)1.36 KHz/Vb)2.72 KHz/Vc)16 Hz/Vd)32 Hz/VCorrect 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 A carrier signal Accosukt is frequency modulated using a message signal 3cos103nt. If the signal to noise ratio at input and output of the receiver are found to be 20dB and 40dB respectively, then determine sensitivity of frequency modulator.a)1.36 KHz/Vb)2.72 KHz/Vc)16 Hz/Vd)32 Hz/VCorrect 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 A carrier signal Accosukt is frequency modulated using a message signal 3cos103nt. If the signal to noise ratio at input and output of the receiver are found to be 20dB and 40dB respectively, then determine sensitivity of frequency modulator.a)1.36 KHz/Vb)2.72 KHz/Vc)16 Hz/Vd)32 Hz/VCorrect answer is option 'A'. Can you explain this answer?.
Solutions for A carrier signal Accosukt is frequency modulated using a message signal 3cos103nt. If the signal to noise ratio at input and output of the receiver are found to be 20dB and 40dB respectively, then determine sensitivity of frequency modulator.a)1.36 KHz/Vb)2.72 KHz/Vc)16 Hz/Vd)32 Hz/VCorrect 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 A carrier signal Accosukt is frequency modulated using a message signal 3cos103nt. If the signal to noise ratio at input and output of the receiver are found to be 20dB and 40dB respectively, then determine sensitivity of frequency modulator.a)1.36 KHz/Vb)2.72 KHz/Vc)16 Hz/Vd)32 Hz/VCorrect answer is option 'A'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
A carrier signal Accosukt is frequency modulated using a message signal 3cos103nt. If the signal to noise ratio at input and output of the receiver are found to be 20dB and 40dB respectively, then determine sensitivity of frequency modulator.a)1.36 KHz/Vb)2.72 KHz/Vc)16 Hz/Vd)32 Hz/VCorrect answer is option 'A'. Can you explain this answer?, a detailed solution for A carrier signal Accosukt is frequency modulated using a message signal 3cos103nt. If the signal to noise ratio at input and output of the receiver are found to be 20dB and 40dB respectively, then determine sensitivity of frequency modulator.a)1.36 KHz/Vb)2.72 KHz/Vc)16 Hz/Vd)32 Hz/VCorrect answer is option 'A'. Can you explain this answer? has been provided alongside types of A carrier signal Accosukt is frequency modulated using a message signal 3cos103nt. If the signal to noise ratio at input and output of the receiver are found to be 20dB and 40dB respectively, then determine sensitivity of frequency modulator.a)1.36 KHz/Vb)2.72 KHz/Vc)16 Hz/Vd)32 Hz/VCorrect answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A carrier signal Accosukt is frequency modulated using a message signal 3cos103nt. If the signal to noise ratio at input and output of the receiver are found to be 20dB and 40dB respectively, then determine sensitivity of frequency modulator.a)1.36 KHz/Vb)2.72 KHz/Vc)16 Hz/Vd)32 Hz/VCorrect 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