Let's analyze and solve the problem step-by-step: --- ## **Given Data and Circuit Description:** - **Input voltage:** \( v_g(t) \) with v_g(t = 50 \, \text{V} \) (assuming steady DC for transfer function derivation) - **Resistances:** - R_1 100\, \Omega \) - \( R_2 = 250\, \Omega \) - **Capacitor:** - \( C = .05\, \text{F} \) - **Operational Amplifier:** - Ideal with infinite gain, so \( V_{+} = V_{-} \) (virtual short) - The circuit shows a transfer function involving \( V_o \) and \( V_{g} \). --- ### **Step 1: Derive the transfer function \( V_o/V_g \)** --- ### **Step 1.1: Identify the circuit configuration** The circuit resembles a classic integrator or filter with an op-amp, resistor, capacitor, and input voltage source. --- ### **Step 1.2: Write node equations** - **Node at the inverting input (\( V_{-} \))**: Since the op-amp is ideal and negative feedback is present, the voltage at the inverting input is virtually equal to the non-inverting input, which is grounded: \[ V_{+} = \, V \quad \Rightarrow \quad V_{-} \approx \, V \] - **Current through \( R_1 \):** \[ i_{R_1} = \frac{V_g - V_{-}}{R_1} = \frac{V_g}{R_1} \] - **Current through capacitor \( C \):** \[ i_C = C \frac{dV_o}{dt} \] - **Current through \( R_2 \):** From the circuit, \( V_o \) is across \( R_2 \), and the current through \( R_2 \): \[ i_{R_2} = \frac{V_o - V_{+}}{R_2} = \frac{V_o}{R_2} \] Since the op-amp input draws no current, the sum of currents at the inverting node: \[ i_{R_1} = i_C + i_{R_2} \] --- ### **Step 1.3: Write the node equation:** \[ \frac{V_g}{R_1} = C \frac{dV_o}{dt} + \frac{V_o}{R_2} \] or, \[ \boxed{ C \frac{dV_o}{dt} + \frac{V_o}{R_2} = \frac{V_g}{R_1} } \] --- ### **Step 1.4: Transfer function \( V_o(s)/V_g(s) \):** Taking Laplace transform: \[ C s V_o(s) + \frac{V_o(s)}{R_2} = \frac{V_g(s)}{R_1} \] \[ V_o(s) ( C s + \frac{1}{R_2} ) = \frac{V_g(s)}{R_1} \] \[ \Rightarrow \frac{V_o(s)}{V_g(s)} = \frac{1/R_1}{C s + 1/R_2} \] --- ### **Final transfer function:** \[ \boxed{ \frac{V_o(s)}{V_g(s)} = \frac{1/R_1}{C s + 1/R_2} } \] --- ## **Step 2: Use \( V_g(t) = 50 \, V \) (DC step input)** Since \( v_g(t) \) is a step of 50V, in Laplace: \[ V_g(s) = \frac{50}{s} \] The transfer function is: \[ V_o(s) = \frac{1/R_1}{C s + 1/R_2} \times \frac{50}{s} \] --- ### **Step 3: Find \( V_o(t) \) (time domain response)** Express: \[ V_o(s) = \frac{50}{s} \times \frac{1/R_1}{C s + 1/R_2} \] Plugging in the known values: \[ R_1 = 100\, \Omega, \quad R_2 = 250\, \Omega, \quad C = .05\, F \] Calculate: \[ \frac{1}{R_1} = \frac{1}{100} = .001 \] \[ \frac{1}{R_2} = \frac{1}{250} = .004 \] So, \[ V_o(s) = 50 \times .001 \times \frac{1}{s (.05 s + .004)} \] Simplify numerator: \[ V_o(s) = .05 \times \frac{1}{s (.05 s + .004)} \] --- ### **Step 4: Inverse Laplace Transform** Express as partial fractions: \[ V_o(s) = \frac{.05}{s (.05 s + .004)} \] Factor denominator: \[ .05 s + .004 = .05 (s + .08) \] Then: \[ V_o(s) = \frac{.05}{s \times .05 (s + .08)} = \frac{1}{s (s + .08)} \] --- ### **Step 5: Partial fraction expansion:** \[ \frac{1}{s(s + .08)} = \frac{A}{s} + \frac{B}{s + .08} \] Multiply both sides by denominator: \[ 1 = A (s + .08) + B s \] Set \( s = \): \[ 1 = A \times .08 \quad \Rightarrow \quad A = \frac{1}{.08} = 12.5 \] Set \( s = -.08 \): \[ 1 = B \times (-.08) \quad \Rightarrow \quad B = - \frac{1}{.08} = -12.5 \] --- ### **Step 6: Inverse Laplace to time domain** \[ V_o(t) = 12.5 \cdot u(t) - 12.5 e^{-.08 t} u(t) \] where \( u(t) \) is the unit step function. ### **Final expression:** \[ \boxed{ V_o(t) = 12.5 \left(1 - e^{-.08 t}\right) \quad \text{V} } \] --- ## **Summary:** | **Step** | **Description** | **Result** | | --- | --- | --- | | 1 | Derive transfer function \( V_o/V_g \) | \(\frac{1/R_1}{C s + 1/R_2}\) | | 2 | Plug in \( V_g(t) = 50\, V \) (DC step) | \( V_g(s) = 50/s \) | | 3 | Find \( V_o(s) \) | \( \frac{.05}{s (.05 s + .004)} \) | | 4 | Partial fractions | \( \frac{1}{s(s + .08)} \) | | 5 | Time domain response | \( V_o(t) = 12.5 (1 - e^{-.08 t}) \quad V \) | --- ## **Final Answer:** \[ \boxed{ \boxed{ V_o(t) = 12.5 \left(1 - e^{-.08 t}\right) \quad \text{V} } } \] This is the time response of the circuit to a step input of 50V.

## **Step-by-Step Explanation** ### **Step 1: Derive the Transfer Function** Identify circuit components and configurations. Write node equations for currents through resistors and capacitors. Establish a relationship between input and output voltages to derive the transfer function. ### **Step 2: Use the Transfer Function** Substitute the step input voltage \( V_g(t) \) into the derived transfer function. This transforms the voltage in the Laplace domain, preparing for time-domain analysis. ### **Step 3: Find \( V_o(t) \)** Utilize the Laplace transformation to express \( V_o(s) \) using the known values of resistances and capacitance. Simplify the expression for easier partial fraction decomposition. ### **Step 4: Inverse Laplace Transform** Apply partial fraction expansion to separate the terms. Use the inverse Laplace transform to convert back to the time domain, yielding the final output voltage expression. ### **Final Result** The final output voltage \( V_o(t) \) is expressed as a function of time, representing how the circuit responds to the step input voltage over time.