A circuit with an ideal OPAMP is shown. The Bode plot for the magnitude (in dB) of the gain transfer function $A_V(j\omega)=V_{\text{out}}(j\omega)/V_{\text{in}}(j\omega)$ of the circuit is also provided (here, $\omega$ is the angular frequency in rad/s). The values of $R$ and $C$ are ____________.
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For active filters, the low-frequency gain determines the feedback resistor ratio, while the -3 dB frequency gives the $RC$ time constant.
From the Bode magnitude plot, the DC gain (low-frequency gain) is approximately:
\[
20 \log_{10}(|A_V(0)|) \approx 12\ \text{dB}.
\]
Thus,
\[
|A_V(0)| = 10^{12/20} = 10^{0.6} \approx 4.
\]
For a non-inverting amplifier, the low-frequency gain is:
\[
A_V(0) = 1 + \frac{R}{1\ \text{k}\Omega}.
\]
So,
\[
1 + \frac{R}{1000} = 4 \quad \Rightarrow \quad R = 3000\ \Omega = 3\ \text{k}\Omega.
\]
Step 1: Determine the cutoff frequency.
From the Bode plot, the gain drops by 3 dB near:
\[
\log_{10}(\omega_c) \approx 2 \quad \Rightarrow \quad \omega_c \approx 10^2 = 100\ \text{rad/s}.
\]
But the slope clearly continues downward to about:
\[
\log_{10}(\omega_c) \approx 2.5 \Rightarrow \omega_c \approx 300\ \text{rad/s}.
\]
Most accurate reading gives:
\[
\omega_c \approx 300\ \text{rad/s}.
\]
Step 2: Use the standard formula.
For the capacitor in the feedback loop:
\[
\omega_c = \frac{1}{RC}.
\]
Substitute $R = 3000$:
\[
C = \frac{1}{3000 \times 300} \approx 1.1 \times 10^{-6}\ \text{F} \approx 1\ \mu\text{F}.
\]
Thus the correct combination is:
\[
R = 3\ \text{k}\Omega,\quad C = 1\ \mu\text{F}.
\]
Final Answer: $R=3\ \text{k}\Omega,\ C=1\ \mu\text{F}$
