Discussion
An amplifier has an open loop gain of 100 and its lower and upper hand frequency of 100 Hz and 100 KHz, respectively. A feedback network with a feedback factor of 0.99 is connected to the amplifier, the new lower and upper cut off frequencies are at
Answer: A
With negative feedback the bandwidth is extended by the factor \(1 + A\beta\), where \(A = 100\) and \(\beta = 0.99\):
\[ 1 + A\beta = 1 + 100 \times 0.99 = 100 \]The upper cut-off frequency rises and the lower one falls by that factor:
\[ f_{H}' = 100\ \mathrm{kHz} \times 100 = 10\ \mathrm{MHz} \]\[ f_{L}' = \frac{100\ \mathrm{Hz}}{100} = 1\ \mathrm{Hz} \]This is option (A). Options B, C and D use factors other than 100, so their cut-off frequencies do not follow from the feedback relations.
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