1. **Problem Statement:** Find the transfer function $H(s) = \frac{V_o(s)}{V_i(s)}$ of the given operational amplifier circuit. 2. **Circuit Description:** The input voltage $V_i(t)$ passes through a 100k\Omega resistor ($R_1 = 100000\ \Omega$) and a 1\mu F capacitor ($C_1 = 1 \times 10^{-6} F$) to the inverting input of the op-amp. The non-inverting input is grounded. The feedback loop from output $V_o(t)$ to the inverting input has a 100k\Omega resistor ($R_f = 100000\ \Omega$) and a 1\mu F capacitor ($C_f = 1 \times 10^{-6} F$) in series. 3. **Assumptions and Formula:** For an ideal op-amp in a negative feedback configuration, the voltage at the inverting input is virtually zero (virtual ground). The transfer function is given by the ratio of impedances in the feedback and input paths: $$H(s) = -\frac{Z_f(s)}{Z_1(s)}$$ where $Z_1(s)$ is the impedance of the input branch and $Z_f(s)$ is the impedance of the feedback branch. 4. **Calculate Impedances:** - Impedance of resistor: $R$ - Impedance of capacitor: $\frac{1}{sC}$ Input branch impedance $Z_1(s)$ is resistor $R_1$ in series with capacitor $C_1$: $$Z_1(s) = R_1 + \frac{1}{sC_1} = 100000 + \frac{1}{s \times 10^{-6}} = 100000 + \frac{1}{10^{-6}s}$$ Feedback branch impedance $Z_f(s)$ is resistor $R_f$ in series with capacitor $C_f$: $$Z_f(s) = R_f + \frac{1}{sC_f} = 100000 + \frac{1}{s \times 10^{-6}} = 100000 + \frac{1}{10^{-6}s}$$ 5. **Simplify the transfer function:** $$H(s) = -\frac{Z_f(s)}{Z_1(s)} = -\frac{100000 + \frac{1}{10^{-6}s}}{100000 + \frac{1}{10^{-6}s}}$$ Since numerator and denominator are identical, the transfer function simplifies to: $$H(s) = -1$$ 6. **Interpretation:** The circuit acts as an inverting amplifier with a gain of -1 over all frequencies, due to identical impedances in input and feedback paths. **Final answer:** $$\boxed{H(s) = -1}$$