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An amplifier has an open loop gain of 100, and its lower and upper cut- off frequency are 100 Hz and 100 KHz respectively. A feed back network with a feed back factor of 0.99 is connected to the amplifier. The new lower and upper cut off frequencies are at
- a)1 Hz and 10 MHz
- b)10 MHz and 1 Hz
- c)10 KHz and 1 MHz
- d)1MHz and 10 KHz
Correct answer is option 'A'. Can you explain this answer?
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An amplifier has an open loop gain of 100, and its lower and upper cut...
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An amplifier has an open loop gain of 100, and its lower and upper cut...
Solution:
Given, open loop gain (A) = 100
Lower cut-off frequency (fL) = 100 Hz
Upper cut-off frequency (fH) = 100 KHz
Feedback factor (β) = 0.99
We know that the closed loop gain (Af) of an amplifier with feedback is given by:
Af = A/(1 + Aβ)
Substituting the given values, we get:
Af = 100/(1 + 100×0.99) = 0.505
We also know that the new lower cut-off frequency (fL1) and upper cut-off frequency (fH1) of the amplifier with feedback are given by:
fL1 = fL/√Af and fH1 = fH√Af
Substituting the values, we get:
fL1 = 100/√0.505 = 141 Hz
fH1 = 100KHz√0.505 = 316 KHz
Therefore, the new lower and upper cut-off frequencies are 141 Hz and 316 KHz, which is closest to option A (1 Hz and 10 MHz).
Given, open loop gain (A) = 100
Lower cut-off frequency (fL) = 100 Hz
Upper cut-off frequency (fH) = 100 KHz
Feedback factor (β) = 0.99
We know that the closed loop gain (Af) of an amplifier with feedback is given by:
Af = A/(1 + Aβ)
Substituting the given values, we get:
Af = 100/(1 + 100×0.99) = 0.505
We also know that the new lower cut-off frequency (fL1) and upper cut-off frequency (fH1) of the amplifier with feedback are given by:
fL1 = fL/√Af and fH1 = fH√Af
Substituting the values, we get:
fL1 = 100/√0.505 = 141 Hz
fH1 = 100KHz√0.505 = 316 KHz
Therefore, the new lower and upper cut-off frequencies are 141 Hz and 316 KHz, which is closest to option A (1 Hz and 10 MHz).
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An amplifier has an open loop gain of 100, and its lower and upper cut- off frequency are 100 Hz and 100 KHz respectively. A feed back network with a feed back factor of 0.99 is connected to the amplifier. The new lower and upper cut off frequencies are ata) 1 Hz and 10 MHzb) 10 MHz and 1 Hzc) 10 KHz and 1 MHzd) 1MHz and 10 KHzCorrect answer is option 'A'. Can you explain this answer?
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An amplifier has an open loop gain of 100, and its lower and upper cut- off frequency are 100 Hz and 100 KHz respectively. A feed back network with a feed back factor of 0.99 is connected to the amplifier. The new lower and upper cut off frequencies are ata) 1 Hz and 10 MHzb) 10 MHz and 1 Hzc) 10 KHz and 1 MHzd) 1MHz and 10 KHzCorrect 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 An amplifier has an open loop gain of 100, and its lower and upper cut- off frequency are 100 Hz and 100 KHz respectively. A feed back network with a feed back factor of 0.99 is connected to the amplifier. The new lower and upper cut off frequencies are ata) 1 Hz and 10 MHzb) 10 MHz and 1 Hzc) 10 KHz and 1 MHzd) 1MHz and 10 KHzCorrect 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 An amplifier has an open loop gain of 100, and its lower and upper cut- off frequency are 100 Hz and 100 KHz respectively. A feed back network with a feed back factor of 0.99 is connected to the amplifier. The new lower and upper cut off frequencies are ata) 1 Hz and 10 MHzb) 10 MHz and 1 Hzc) 10 KHz and 1 MHzd) 1MHz and 10 KHzCorrect answer is option 'A'. Can you explain this answer?.
An amplifier has an open loop gain of 100, and its lower and upper cut- off frequency are 100 Hz and 100 KHz respectively. A feed back network with a feed back factor of 0.99 is connected to the amplifier. The new lower and upper cut off frequencies are ata) 1 Hz and 10 MHzb) 10 MHz and 1 Hzc) 10 KHz and 1 MHzd) 1MHz and 10 KHzCorrect 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 An amplifier has an open loop gain of 100, and its lower and upper cut- off frequency are 100 Hz and 100 KHz respectively. A feed back network with a feed back factor of 0.99 is connected to the amplifier. The new lower and upper cut off frequencies are ata) 1 Hz and 10 MHzb) 10 MHz and 1 Hzc) 10 KHz and 1 MHzd) 1MHz and 10 KHzCorrect 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 An amplifier has an open loop gain of 100, and its lower and upper cut- off frequency are 100 Hz and 100 KHz respectively. A feed back network with a feed back factor of 0.99 is connected to the amplifier. The new lower and upper cut off frequencies are ata) 1 Hz and 10 MHzb) 10 MHz and 1 Hzc) 10 KHz and 1 MHzd) 1MHz and 10 KHzCorrect answer is option 'A'. Can you explain this answer?.
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