1. **Problem Statement:** Find the transfer function $G(s) = \frac{V_o(s)}{V_i(s)}$ for the circuits labeled (a) and (b). 2. **General Approach:** For circuits with resistors and capacitors, use the impedance in the Laplace domain: - Resistor: $R$ - Capacitor: $\frac{1}{sC}$ 3. **Circuit (a) Analysis:** - Input series resistor: $R_1 = 500\times10^3\ \Omega$ - Input capacitor: $C_1 = 2\times10^{-6}\ F$ - Output series resistor: $R_2 = 100\times10^3\ \Omega$ - Output capacitor: $C_2 = 2\times10^{-6}\ F$ 4. **Impedances:** - $Z_{C1} = \frac{1}{sC_1} = \frac{1}{s\times 2\times10^{-6}} = \frac{1}{2\times10^{-6}s}$ - $Z_{C2} = \frac{1}{sC_2} = \frac{1}{2\times10^{-6}s}$ 5. **Node $v_1$ voltage:** - Voltage divider at input: $$V_1(s) = V_i(s) \times \frac{Z_{C1}}{R_1 + Z_{C1}} = V_i(s) \times \frac{\frac{1}{2\times10^{-6}s}}{500\times10^3 + \frac{1}{2\times10^{-6}s}}$$ 6. **Output voltage $V_o$:** - The output stage is a voltage divider from $v_1$ to $V_o$: $$V_o(s) = V_1(s) \times \frac{Z_{C2}}{R_2 + Z_{C2}} = V_1(s) \times \frac{\frac{1}{2\times10^{-6}s}}{100\times10^3 + \frac{1}{2\times10^{-6}s}}$$ 7. **Transfer function $G(s)$:** $$G(s) = \frac{V_o(s)}{V_i(s)} = \left( \frac{\frac{1}{2\times10^{-6}s}}{500\times10^3 + \frac{1}{2\times10^{-6}s}} \right) \times \left( \frac{\frac{1}{2\times10^{-6}s}}{100\times10^3 + \frac{1}{2\times10^{-6}s}} \right)$$ 8. **Simplify denominators by multiplying numerator and denominator by $2\times10^{-6}s$:** $$G(s) = \frac{1}{1 + 500\times10^3 \times 2\times10^{-6} s} \times \frac{1}{1 + 100\times10^3 \times 2\times10^{-6} s} = \frac{1}{(1 + 1s)(1 + 0.2s)}$$ 9. **Final transfer function for (a):** $$G_a(s) = \frac{1}{(1 + s)(1 + 0.2 s)}$$ --- 10. **Circuit (b) Analysis:** - Input resistor: $R_1 = 400\times10^3\ \Omega$ - Parallel branch to ground: resistor $R_p = 600\times10^3\ \Omega$ and capacitor $C_p = 4\times10^{-6}\ F$ - Output resistor: $R_2 = 600\times10^3\ \Omega$ - Output capacitor: $C_2 = 4\times10^{-6}\ F$ - Feedback resistor from output to node before output capacitor: $R_f = 110\times10^3\ \Omega$ 11. **Impedances:** - $Z_{C_p} = \frac{1}{sC_p} = \frac{1}{4\times10^{-6}s}$ - $Z_{C_2} = \frac{1}{4\times10^{-6}s}$ 12. **Parallel impedance at node $v_1$ to ground:** $$Z_p = \left( \frac{1}{R_p} + \frac{1}{Z_{C_p}} \right)^{-1} = \left( \frac{1}{600\times10^3} + s \times 4\times10^{-6} \right)^{-1}$$ 13. **Voltage at $v_1$:** $$V_1(s) = V_i(s) \times \frac{Z_p}{R_1 + Z_p}$$ 14. **Output stage:** - Node before output capacitor is $v_2$. - Voltage divider from $v_1$ to $v_2$: $$V_2(s) = V_1(s) \times \frac{Z_{C_2}}{R_2 + Z_{C_2}} = V_1(s) \times \frac{\frac{1}{4\times10^{-6}s}}{600\times10^3 + \frac{1}{4\times10^{-6}s}}$$ 15. **Feedback branch:** - Feedback resistor $R_f$ connects $v_2$ to output $V_o$. - The output voltage $V_o$ is related to $V_2$ by the feedback: $$V_o(s) = V_2(s) + I_f(s) R_f$$ - Current through $R_f$ is: $$I_f(s) = \frac{V_o(s) - V_2(s)}{R_f}$$ 16. **Using ideal op-amp rule (negative input equals positive input at ground):** - The node $v_1$ is virtual ground, so $V_1(s) \approx 0$. - This implies the feedback forces: $$V_o(s) = - \frac{R_f}{R_2} V_1(s)$$ 17. **Combining all:** - The transfer function is: $$G_b(s) = \frac{V_o(s)}{V_i(s)} = - \frac{R_f}{R_2} \times \frac{Z_p}{R_1 + Z_p}$$ 18. **Substitute values:** - $\frac{R_f}{R_2} = \frac{110\times10^3}{600\times10^3} = \frac{11}{60}$ - $Z_p = \left( \frac{1}{600\times10^3} + 4\times10^{-6} s \right)^{-1}$ 19. **Final transfer function for (b):** $$G_b(s) = - \frac{11}{60} \times \frac{\left( \frac{1}{600\times10^3} + 4\times10^{-6} s \right)^{-1}}{400\times10^3 + \left( \frac{1}{600\times10^3} + 4\times10^{-6} s \right)^{-1}}$$