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 = 25kHzI have been asked this question by my college director while I was bunking the class.Question is from Effects of Feedback in division General and Feedback Amplifiers of Analog Circuits

Consider an open loop circuit with lower cutoff frequency 5kHz and upp

1.	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 = 25kHzI have been asked this question by my college director while I was bunking the class.Question is from Effects of Feedback in division General and Feedback Amplifiers of Analog Circuits
Answer» RIGHT answer is (c) Lower cutoff = 2kHz, Upper cutoff = 25kHz For EXPLANATION I would SAY: 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.

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