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Consider an open loop circuit with lower cutoff frequency 5kHz and upper cutoff frequency 20kHz. If negative feedback is applied to the same, choose correct option stating the new cutoff frequencies.
- a)Lower cutoff = 5kHz, Upper cutoff = 20kHz
- b)Lower cutoff = 2kHz, Upper cutoff = 18kHz
- c)Lower cutoff = 2kHz, Upper cutoff = 25kHz
- d)Lower cutoff = 10kHz, Upper cutoff = 25kHz
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
Consider an open loop circuit with lower cutoff frequency 5kHz and upp...
Introduction:
In an open-loop circuit, the frequency response is determined by the components used in the circuit. The lower cutoff frequency and upper cutoff frequency define the range of frequencies over which the circuit operates effectively. When negative feedback is applied to the circuit, it can alter the frequency response characteristics.
Effect of Negative Feedback:
Negative feedback in an amplifier circuit can help improve stability, linearity, and reduce distortion. It also affects the frequency response of the circuit.
Explanation:
When negative feedback is applied, the circuit becomes a closed-loop system. The feedback network introduces a new pole and zero to the transfer function of the circuit, which can change the cutoff frequencies.
Let's consider the given open-loop circuit with a lower cutoff frequency of 5kHz and an upper cutoff frequency of 20kHz. When negative feedback is applied, the closed-loop gain is given by:
Acl = Aol / (1 + βAol)
Where Acl is the closed-loop gain, Aol is the open-loop gain, and β is the feedback factor.
Lower Cutoff Frequency:
The closed-loop lower cutoff frequency is determined by the pole introduced by the feedback network. In this case, we can assume that the feedback network is designed to introduce a pole well below 5kHz.
Therefore, the lower cutoff frequency of the closed-loop circuit will be lower than 5kHz. This rules out option 'a'.
Upper Cutoff Frequency:
The closed-loop upper cutoff frequency is determined by the zero introduced by the feedback network. In this case, we can assume that the feedback network is designed to introduce a zero above 20kHz.
Therefore, the upper cutoff frequency of the closed-loop circuit will be higher than 20kHz. This rules out option 'a'.
New Cutoff Frequencies:
Considering the above analysis, the correct option is 'c' where the lower cutoff frequency is 2kHz and the upper cutoff frequency is 25kHz. This is because the negative feedback network introduces a pole below 5kHz and a zero above 20kHz.
Therefore, the correct answer is option 'c' - Lower cutoff = 2kHz, Upper cutoff = 25kHz.
In an open-loop circuit, the frequency response is determined by the components used in the circuit. The lower cutoff frequency and upper cutoff frequency define the range of frequencies over which the circuit operates effectively. When negative feedback is applied to the circuit, it can alter the frequency response characteristics.
Effect of Negative Feedback:
Negative feedback in an amplifier circuit can help improve stability, linearity, and reduce distortion. It also affects the frequency response of the circuit.
Explanation:
When negative feedback is applied, the circuit becomes a closed-loop system. The feedback network introduces a new pole and zero to the transfer function of the circuit, which can change the cutoff frequencies.
Let's consider the given open-loop circuit with a lower cutoff frequency of 5kHz and an upper cutoff frequency of 20kHz. When negative feedback is applied, the closed-loop gain is given by:
Acl = Aol / (1 + βAol)
Where Acl is the closed-loop gain, Aol is the open-loop gain, and β is the feedback factor.
Lower Cutoff Frequency:
The closed-loop lower cutoff frequency is determined by the pole introduced by the feedback network. In this case, we can assume that the feedback network is designed to introduce a pole well below 5kHz.
Therefore, the lower cutoff frequency of the closed-loop circuit will be lower than 5kHz. This rules out option 'a'.
Upper Cutoff Frequency:
The closed-loop upper cutoff frequency is determined by the zero introduced by the feedback network. In this case, we can assume that the feedback network is designed to introduce a zero above 20kHz.
Therefore, the upper cutoff frequency of the closed-loop circuit will be higher than 20kHz. This rules out option 'a'.
New Cutoff Frequencies:
Considering the above analysis, the correct option is 'c' where the lower cutoff frequency is 2kHz and the upper cutoff frequency is 25kHz. This is because the negative feedback network introduces a pole below 5kHz and a zero above 20kHz.
Therefore, the correct answer is option 'c' - Lower cutoff = 2kHz, Upper cutoff = 25kHz.
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Community Answer
Consider an open loop circuit with lower cutoff frequency 5kHz and upp...
Negative feedback decreases lower cutoff frequencies and increases the higher cutoff frequency.
fHF = fH(1+Aβ)
fLF = fL/(1+Aβ)
Total bandwidth is thus increased.
fHF = fH(1+Aβ)
fLF = fL/(1+Aβ)
Total bandwidth is thus increased.
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Consider an open loop circuit with lower cutoff frequency 5kHz and upper cutoff frequency 20kHz. If negative feedback is applied to the same, choose correct option stating the new cutoff frequencies.a)Lower cutoff = 5kHz, Upper cutoff = 20kHzb)Lower cutoff = 2kHz, Upper cutoff = 18kHzc)Lower cutoff = 2kHz, Upper cutoff = 25kHzd)Lower cutoff = 10kHz, Upper cutoff = 25kHzCorrect answer is option 'C'. Can you explain this answer?
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Consider an open loop circuit with lower cutoff frequency 5kHz and upper cutoff frequency 20kHz. If negative feedback is applied to the same, choose correct option stating the new cutoff frequencies.a)Lower cutoff = 5kHz, Upper cutoff = 20kHzb)Lower cutoff = 2kHz, Upper cutoff = 18kHzc)Lower cutoff = 2kHz, Upper cutoff = 25kHzd)Lower cutoff = 10kHz, Upper cutoff = 25kHzCorrect answer is option 'C'. 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 an open loop circuit with lower cutoff frequency 5kHz and upper cutoff frequency 20kHz. If negative feedback is applied to the same, choose correct option stating the new cutoff frequencies.a)Lower cutoff = 5kHz, Upper cutoff = 20kHzb)Lower cutoff = 2kHz, Upper cutoff = 18kHzc)Lower cutoff = 2kHz, Upper cutoff = 25kHzd)Lower cutoff = 10kHz, Upper cutoff = 25kHzCorrect answer is option 'C'. 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 an open loop circuit with lower cutoff frequency 5kHz and upper cutoff frequency 20kHz. If negative feedback is applied to the same, choose correct option stating the new cutoff frequencies.a)Lower cutoff = 5kHz, Upper cutoff = 20kHzb)Lower cutoff = 2kHz, Upper cutoff = 18kHzc)Lower cutoff = 2kHz, Upper cutoff = 25kHzd)Lower cutoff = 10kHz, Upper cutoff = 25kHzCorrect answer is option 'C'. Can you explain this answer?.
Consider an open loop circuit with lower cutoff frequency 5kHz and upper cutoff frequency 20kHz. If negative feedback is applied to the same, choose correct option stating the new cutoff frequencies.a)Lower cutoff = 5kHz, Upper cutoff = 20kHzb)Lower cutoff = 2kHz, Upper cutoff = 18kHzc)Lower cutoff = 2kHz, Upper cutoff = 25kHzd)Lower cutoff = 10kHz, Upper cutoff = 25kHzCorrect answer is option 'C'. 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 an open loop circuit with lower cutoff frequency 5kHz and upper cutoff frequency 20kHz. If negative feedback is applied to the same, choose correct option stating the new cutoff frequencies.a)Lower cutoff = 5kHz, Upper cutoff = 20kHzb)Lower cutoff = 2kHz, Upper cutoff = 18kHzc)Lower cutoff = 2kHz, Upper cutoff = 25kHzd)Lower cutoff = 10kHz, Upper cutoff = 25kHzCorrect answer is option 'C'. 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 an open loop circuit with lower cutoff frequency 5kHz and upper cutoff frequency 20kHz. If negative feedback is applied to the same, choose correct option stating the new cutoff frequencies.a)Lower cutoff = 5kHz, Upper cutoff = 20kHzb)Lower cutoff = 2kHz, Upper cutoff = 18kHzc)Lower cutoff = 2kHz, Upper cutoff = 25kHzd)Lower cutoff = 10kHz, Upper cutoff = 25kHzCorrect answer is option 'C'. Can you explain this answer?.
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ample number of questions to practice Consider an open loop circuit with lower cutoff frequency 5kHz and upper cutoff frequency 20kHz. If negative feedback is applied to the same, choose correct option stating the new cutoff frequencies.a)Lower cutoff = 5kHz, Upper cutoff = 20kHzb)Lower cutoff = 2kHz, Upper cutoff = 18kHzc)Lower cutoff = 2kHz, Upper cutoff = 25kHzd)Lower cutoff = 10kHz, Upper cutoff = 25kHzCorrect answer is option 'C'. Can you explain this answer? tests, examples and also practice Electronics and Communication Engineering (ECE) tests.
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