Let's solve the problem step by step: --- ## **Step 1: Write the Transfer Function \((j\omega) \)** circuit is a **series RLC circuit** with the output across the capacitor C \): \[ \textTotal impedance: \quad Z_{total = R + j\omega L \frac{1}{jomega C} \] The voltage across the is: \[ V_{out} = V_C = I \cdot Z_C \] where \( Z_C = \frac{1}{j\omega C} \) and \( I = \frac{V_{in}}{Z_{total}} \). So, \[ V_{out} = V_{in} \cdot \frac{Z_C}{Z_{total}} \] Thus, \[ H(j\omega) = \frac{V_{out}}{V_{in}} = \frac{Z_C}{Z_{total}} \] --- ## **Step 2: Substitute the Impedances** \[ H(j\omega) = \frac{\frac{1}{j\omega C}}{R + j\omega L + \frac{1}{j\omega C}} \] Multiply numerator and denominator by \( j\omega C \): \[ H(j\omega) = \frac{1}{j\omega C R + (j\omega C)(j\omega L) + 1} \] Note: - \( (j\omega C)(j\omega L) = (j)^2 (\omega^2) (LC) = -\omega^2 LC \) So: \[ H(j\omega) = \frac{1}{1 + j\omega RC - \omega^2 LC} \] --- ## **Step 3: Magnitude Response \( |H(j\omega)| \)** The magnitude is: \[ |H(j\omega)| = \frac{1}{\sqrt{[1 - \omega^2 LC]^2 + [\omega RC]^2}} \] --- ## **Step 4: Phase Response \( \theta(\omega) \)** The phase is the angle of the denominator (negative because it's in the denominator): \[ \theta(\omega) = -\tan^{-1} \left( \frac{\omega RC}{1 - \omega^2 LC} \right) \] --- ## **Step 5: Final Answers** ### **a) Magnitude Response** \[ \boxed{ |H(j\omega)| = \frac{1}{\sqrt{[1 - \omega^2 LC]^2 + [\omega RC]^2}} } \] --- ### **b) Phase Response** \[ \boxed{ \theta(\omega) = -\tan^{-1} \left( \frac{\omega RC}{1 - \omega^2 LC} \right) } \] --- ### **Summary Table** | Response Type | Expression | |----------------------|------------| | Magnitude \(|H(j\omega)|\) | \(\frac{1}{\sqrt{[1 - \omega^2 LC]^2 + [\omega RC]^2}}\) | | Phase \(\theta(\omega)\) | \(-\tan^{-1} \left( \frac{\omega RC}{1 - \omega^2 LC} \right)\) | Let me know if you need a plot or further explanation!