4. For the given circuit and the given input, use convolution integral to a) find vo(t) for t≥ 0. b) find vo at t = 0.25 s. v(1) (+ 6 kn ww + v(1): 4 ΚΩ ww 100 μF ww 8 kΩ ; + v (1) - v(1) (V) 9 0.3 t (s)

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### Problem 4: Convolution Integral in Electrical Circuits

**Objective:**
For the given circuit and the given input, use the convolution integral to:
a) Find \( v_o(t) \) for \( t \geq 0 \).
b) Find \( v_o \) at \( t = 0.25 \text{ s} \).

---

**Circuit Diagram:**
The circuit consists of the following components connected in sequence:
- A voltage source \( v_i(t) \)
- A resistor of \( 6 \text{ k}\Omega \) is in series with an inductor of \( 1 \text{ H} \)
- A parallel branch consisting of:
  - A \( 4 \text{ k}\Omega \) resistor in series with a capacitor of \( 100 \mu \text{F} \)
  - An \( 8 \text{ k}\Omega \) resistor
- The voltage \( v_o(t) \) is measured across the parallel branch containing the \( 4 \text{ k}\Omega \) resistor and the capacitor

---

**Input Diagram:**
The input \( v_i(t) \) is a step function:

- \( v_i(t) \) is 0 for \( 0 \leq t < 0.3 \text{ s} \)
- \( v_i(t) \) is 9 V for \( 0.3 \leq t \text{ s} \)
  
This can be represented as:

- A horizontal line at 0 V from time \( 0 \) to \( 0.3 \text{ s} \)
- A horizontal line at 9 V from time \( 0.3 \text{ s} \) onwards

The graph has \( v_i(t) \) on the vertical axis (in volts) and time \( t \) on the horizontal axis (in seconds).

---

**Steps to Solve Using the Convolution Integral:**

1. **Identify the system impulse response \( h(t) \):**
    - Determine \( h(t) \) based on the circuit's configuration.
    
2. **Express the input \( v_i(t) \) as a piecewise function:**
    - \( v_i(t) = 0 \) for \( 0 \leq t < 0.3 \text{ s} \)
    - \( v_i(t)
Transcribed Image Text:### Problem 4: Convolution Integral in Electrical Circuits **Objective:** For the given circuit and the given input, use the convolution integral to: a) Find \( v_o(t) \) for \( t \geq 0 \). b) Find \( v_o \) at \( t = 0.25 \text{ s} \). --- **Circuit Diagram:** The circuit consists of the following components connected in sequence: - A voltage source \( v_i(t) \) - A resistor of \( 6 \text{ k}\Omega \) is in series with an inductor of \( 1 \text{ H} \) - A parallel branch consisting of: - A \( 4 \text{ k}\Omega \) resistor in series with a capacitor of \( 100 \mu \text{F} \) - An \( 8 \text{ k}\Omega \) resistor - The voltage \( v_o(t) \) is measured across the parallel branch containing the \( 4 \text{ k}\Omega \) resistor and the capacitor --- **Input Diagram:** The input \( v_i(t) \) is a step function: - \( v_i(t) \) is 0 for \( 0 \leq t < 0.3 \text{ s} \) - \( v_i(t) \) is 9 V for \( 0.3 \leq t \text{ s} \) This can be represented as: - A horizontal line at 0 V from time \( 0 \) to \( 0.3 \text{ s} \) - A horizontal line at 9 V from time \( 0.3 \text{ s} \) onwards The graph has \( v_i(t) \) on the vertical axis (in volts) and time \( t \) on the horizontal axis (in seconds). --- **Steps to Solve Using the Convolution Integral:** 1. **Identify the system impulse response \( h(t) \):** - Determine \( h(t) \) based on the circuit's configuration. 2. **Express the input \( v_i(t) \) as a piecewise function:** - \( v_i(t) = 0 \) for \( 0 \leq t < 0.3 \text{ s} \) - \( v_i(t)
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