Let's go through the problem step-by-step: --- # Step-by-Step Solution ## Given Data: - Input voltage: \( v_g(t) \) with \( v_g(t) = 50 \, \text{V} \times u(t) \) - Circuit components: - \( R = 100\, \Omega \) - \( R_{th} = 250\, \Omega \) - \( C = .05\, \text{F} \) - \( V_{th} = 10\, \text{V} \) - Output voltage: \( v_o(t) \) - The transfer function: \( \frac{V_o(s)}{V_g(s)} \) --- ## 1. **Derive the transfer function \( \frac{V_o(s)}{V_g(s)} \)** ### Circuit Analysis: The circuit appears to be a resistor-capacitor network with a dependent source or a certain configuration. Assuming the circuit described is a typical high-pass or low-pass filter, with the input \( V_g \) and output across the capacitor or resistor. ### Step 1. Write the impedance: - Resistor \( R \): \( R \) - Capacitor \( C \): \( Z_C = \frac{1}{sC} \) --- ### Step 2. Write the voltage divider: Suppose the output \( V_o \) is across the capacitor \( C \): \[ V_o(s) = V_{C}(s) = V_g(s) \times \frac{Z_C}{R + Z_C} \] which gives: \[ \frac{V_o(s)}{V_g(s)} = \frac{\frac{1}{sC}}{R + \frac{1}{sC}} = \frac{1}{sC} \times \frac{1}{R + \frac{1}{sC}} \] Multiply numerator and denominator by \( sC \): \[ \frac{V_o(s)}{V_g(s)} = \frac{1}{1 + s R C} \] --- ### **Transfer Function:** \[ \boxed{ \frac{V_o(s)}{V_g(s)} = \frac{1}{1 + s R C} } \] --- ## 2. **Calculate the magnitude and phase of \( V_o(t) \) when \( v_g(t) = 50 u(t) \)** Since \( v_g(t) \) is a step input of 50 V: \[ V_g(s) = \frac{50}{s} \] The output in Laplace domain: \[ V_o(s) = \frac{1}{1 + s R C} \times \frac{50}{s} \] --- ## 3. **Find \( v_o(t) \) (time domain response)** ### Step 1: Inverse Laplace Transform: Expressed as: \[ V_o(s) = \frac{50}{s (1 + s R C)} \] Use partial fraction: \[ \frac{50}{s (1 + s R C)} = \frac{A}{s} + \frac{B}{1 + s R C} \] Multiply both sides by denominator: \[ 50 = A (1 + s R C) + B s \] Set \( s= \): \[ 50 = A \times 1 \Rightarrow A=50 \] Set \( s = - \frac{1}{R C} \): \[ 50 = B \times \left( - \frac{1}{R C} \right) \Rightarrow B = -50 R C \] ### Time domain: \[ v_o(t) = A \times u(t) + B \times e^{-\frac{t}{R C}} u(t) = 50 \left( 1 - e^{-\frac{t}{R C}} \right) \] --- ## 4. **Final expression for \( v_o(t) \):** \[ \boxed{ v_o(t) = 50 \left( 1 - e^{-\frac{t}{R C}} \right) \quad \text{for } t \geq } \] where: \[ R C = 100 \times .05 = 50\, \text{seconds} \] --- ## **Summary:** | Step | Description | Result | |--------|--------------|---------| | 1 | Transfer function \( V_o(s)/V_g(s) \) | \(\frac{1}{1 + s R C}\) | | 2 | Time domain response \( v_o(t) \) | \(50 \left( 1 - e^{-\frac{t}{50}} \right)\) V | --- # Final Answer: \[ \boxed{ v_o(t) = 50 \left( 1 - e^{-\frac{t}{50}} \right) \text{ V} } \] This describes the transient response of the circuit to a step input of 50 V.