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This circuit looks like a combination of resistors, inductors, and capacitors, and you're tasked with finding the transfer function \( \frac{V_o(s)}{V_i(s)} \) in the Laplace domain.

Here's a step-by-step outline of how we can solve this:

### 1. **Impedances in the Laplace Domain**:
   - For the resistor, the impedance is \( R = 6 \, \Omega \).
   - For the inductor, the impedance is \( Z_L = sL \), where \( L \) is the inductance. So, for the two inductors:
     - The impedance of the first inductor is \( s \times 1 = s \).
     - The impedance of the second inductor is \( 2s \).
   - For the capacitor, the impedance is \( Z_C = \frac{1}{sC} \). The capacitor's impedance is given as \( \frac{1}{s} \), meaning \( C = 1 \, \text{F} \).

### 2. **Equivalent Impedance of the Circuit**:
   We need to analyze how these components are connected.
   
   - The voltage source is on the left with \( V_i(s) \).
   - There is a series combination of \( 3 \, \Omega \) resistance, \( s \, \Omega \) inductance, and \( 2s \, \Omega \) inductance.
   - The output voltage \( V_o(s) \) is across the parallel combination of:
     - A 6 \( \Omega \) resistor.
     - A \( \frac{1}{s} \, \Omega \) capacitor.
     
   Let's first find the equivalent impedance of the parallel section:
   \[
   Z_{parallel} = \left( \frac{1}{6} + \frac{s}{1} \right)^{-1} = \frac{6}{1 + 6s}
   \]
   
### 3. **Overall Impedance**:
   Now, the total impedance \( Z_{total} \) in series with the voltage source will be the sum of the series components and the parallel impedance:
   \[
   Z_{total} = 3 + s + 2s + \frac{6}{1 + 6s} = 3 + 3s + \frac{6}{1 + 6s}
   \]

### 4. **Transfer Function**:
   The transfer function \( \frac{V_o(s)}{V_i(s)} \) is given by the voltage divider formula:
   \[
   \frac{V_o(s)}{V_i(s)} = \frac{Z_{parallel}}{Z_{total}}
   \]
   Substituting the values:
   \[
   \frac{V_o(s)}{V_i(s)} = \frac{\frac{6}{1 + 6s}}{3 + 3s + \frac{6}{1 + 6s}}
   \]

Now, we can simplify this expression to get the transfer function in a more usable form. I'll proceed with this simplification.The transfer function \( \frac{V_o(s)}{V_i(s)} \) simplifies to:

\[
\frac{V_o(s)}{V_i(s)} = \frac{2}{(s + 1)(6s + 1) + 2}
\]

This is the expression for the ratio of the output voltage to the input voltage in the Laplace domain.

Would you like further details on this, or any specific questions about the steps? Here are five questions you might explore:

1. How would you analyze the behavior of this transfer function as \( s \to 0 \) (DC analysis)?
2. What is the steady-state gain of this system?
3. How would this circuit behave for high-frequency signals (as \( s \to \infty \))?
4. How can you convert this transfer function back to the time domain?
5. What is the physical interpretation of the poles and zeros of this transfer function?

**Tip**: Understanding the poles and zeros of a transfer function helps in predicting the stability and frequency response of a circuit.

Find the transfer function Vo(s)/Vi(s) for the given circuit involving resistors, inductors, and capacitors in the Laplace domain.

This detailed solution explores the transfer function Vo(s)/Vi(s) of a complex RLC circuit. It covers key circuit analysis techniques, including the calculation of impedances and the use of the voltage divider rule in the Laplace domain. Suitable for undergraduate electrical engineering students.

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