Certainly! Here’s a **complete rewrite** of the provided solution, maintaining the logical flow and clarity, and concluding with the final answer as requested. --- ## ? Circuit Analysis in the Laplace Domain ### **Given Circuit Structure:** - **Series Element:** $R_1$ - **Parallel Branch (across $R_1$):** - **Branch 1:** Inductor $L$ - **Branch 2:** Series combination of $R_2$ and Capacitor $C$ --- ### **Step 1: Express the Input Current** \[ I(s) = \frac{V_{in}(s)}{Z_{eq}(s)} \] --- ### **Step 2: Find the Equivalent Impedance** The total impedance is: \[ Z_{eq}(s) = R_1 + \left[ sL \parallel \left( R_2 + \frac{1}{Cs} \right) \right] \] Using the parallel impedance formula: \[ Z_{eq}(s) = R_1 + \frac{sL \left( R_2 + \frac{1}{Cs} \right)}{sL + R_2 + \frac{1}{Cs}} \] Multiply numerator and denominator by $Cs$ to clear the fraction: \[ Z_{eq}(s) = R_1 + \frac{sL \left( R_2 Cs + 1 \right)}{s^2 L C + R_2 Cs + 1} \] Combine into a single denominator: \[ Z_{eq}(s) = \frac{s^2 R_1 L C + R_1 R_2 C s + R_1 + sL (R_2 C s + 1)}{s^2 L C + R_2 C s + 1} \] \[ \Rightarrow Z_{eq}(s) = \frac{s^2 (R_1 L C + R_2 L C) + s(R_1 R_2 C + L) + R_1}{s^2 L C + R_2 C s + 1} \tag{1} \] --- ### **Step 3: Current Through $R_2$ Branch** Let $I_1(s)$ be the current through the branch containing $R_2$ and $C$ (in series with $L$ in parallel): \[ I_1(s) = I(s) \cdot \frac{sL}{sL + R_2 + \frac{1}{Cs}} \] Multiplying numerator and denominator by $Cs$ gives: \[ I_1(s) = I(s) \cdot \frac{s^2 C L}{s^2 C L + R_2 C s + 1} \] --- ### **Step 4: Output Voltage Expression** The output voltage across $R_2$ is: \[ V_{out}(s) = R_2 \cdot I_1(s) = R_2 \cdot I(s) \cdot \frac{s^2 C L}{s^2 C L + R_2 C s + 1} \] Now substitute $I(s)$ from Step 1: \[ V_{out}(s) = V_{in}(s) \cdot \frac{s^2 R_2 C L}{Z_{eq}(s) \left[ s^2 C L + R_2 C s + 1 \right]} \] --- ### **Step 5: Transfer Function Derivation** \[ \frac{V_{out}(s)}{V_{in}(s)} = \frac{s^2 R_2 C L}{Z_{eq}(s) \left( s^2 C L + R_2 C s + 1 \right)} \] Substitute $Z_{eq}(s)$ from equation (1): \[ T(s) = \frac{V_{out}(s)}{V_{in}(s)} = \frac{s^2 R_2 C L}{s^2 (R_1 L C + R_2 L C) + s (R_1 R_2 C + L) + R_1} \] --- ## ? Frequency Response Analysis Substitute $s = j\omega$: \[ T(j\omega) = \frac{ -\omega^2 R_2 C L }{ -\omega^2 (R_1 L C + R_2 L C) + j\omega (R_1 R_2 C + L) + R_1 } \] --- ### **Case 1: $\omega = 0$** \[ T(0) = \frac{0}{R_1} = 0 \] --- ### **Case 2: $\omega \to \infty$** At high frequencies, the highest order terms dominate: \[ T(\infty) = \frac{R_2}{R_1 + R_2} \] --- ### **Conclusion** \[ \boxed{ T(0) = 0, \qquad T(\infty) = \frac{R_2}{R_1 + R_2} } \] \[ \boxed{ \text{Therefore, the circuit functions as a high-pass filter.} \quad \text{Ans.} } \] ---