Mumbai University > Computer Engineering > Sem 3 > Electronic Circuits and Communication Fundamentals

Marks: 5 Marks

Year: May 2014

The peak value in dB is given by

$A(dB)=20 \log_{10} \dfrac{R_f}{R_1\sqrt{1+0}}=20 \\ or,\ \ \ \log_{10} \dfrac{R_f}{R_1} =1$

Thus we have, $R_f/R_1 =10 \\ Or, \ \ \ \ R_1= \dfrac{R_f}{10}$

At $ω=10^4$ rad/s, gain in dB is down by 3 dB from its peak of 20 dB, and thus is 17 dB. Therefore, converting gain to dB in Eq. and substituting for ω, and $R_f/R_1$, we have

$$20 \log_{10} \dfrac{10}{\sqrt{1+[(10^4)10^(-8) R_f]^2}} =17 dB \\ 20 \log_{10} 10-20\log_{10} \sqrt{1+[10^{-4} R_f]^2 }=17 dB$$

This simplifies to

$20 \log_{10} (1+[10^{-4} R_f]^2) = 3 dB \\ 1+[〖10^{-4} R_f]^2=10^{3/10}=2$

Thus we have

$[10^{-4} R_f]^2=1 \\ R_f=10^4Ω=1kΩ \\ R_1=10 kΩ/10=1 kΩ$