Question 17.PE.MCQ.20: What is the upper cut-off frequency where the voltage gain i......

(c) The high frequency response of an amplifier is given by

A_{\mathrm{v(high)}}={\frac{A_{\mathrm{v}}}{\left(1+j{\frac{f}{f_{\mathrm{H}}}}\right)}}

The frequency (f) at which the gain decreases by 1 dB the gain is given by

20\log{\frac{A_{\mathrm{v(high)}}}{A_{\mathrm{v}}}}=-1

Therefore,         {\frac{A_{\mathrm{v(high)}}}{A_{\mathrm{v}}}}=0.89

Therefore,

0.89={\frac{1}{\sqrt{1+\left({\frac{f}{f_{\mathrm{H}}}}\right)^{2}}}}

Hence,  {\sqrt{1+\left({\frac{f}{f_{\mathrm{H}}}}\right)^{2}}}=1.22

or               \left({\frac{f}{f_{\mathrm{H}}}}\right)^{2}=0.259

Therefore,  f=51{\mathrm{~kHz}}