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Find $R_1$ and $R_2$ in lossy integrator so that the peak gain is 20 dB and gain is 3dB down from its peak

when $ω$=10000 rad/s . Use capacitance of 0.01 MF Given: Max gain = 20 dB, 3dB frequency. $ω_s=10000 \frac{rad}{s} ,Cr=0.01Mf$

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Component values:

$∴\text{DC gain} = 20 dB= 20 \log_{10⁡}|Av| \\ ∴|Av|=10 \\ \text{But} |Av|=\frac{R_2}{R_1} \\ ∴R_2=10 R_1 \\ \text{Also,} w_a=10000 \frac{rad}{\sec} \\ ∴f_a=\frac{10000}{2π}=1591.55 Hz \\ \text{But} f_a=\frac{1}{2πR_2 C_f} \\ ∴C_f=0.01 \text{MF given} \\ ∴R_2=\frac{1}{2πf_a cf}=\frac{1}{10000 \times 0.01 \times 10^{-6}} \\ ∴R_2=10 kΩ \\ ∴ \text{substituting,} 10 kΩ=10 R_1 \\ ∴R_1=1 kΩ$

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