Let's solve **Question 3.1** step-by-step. We are to **derive the step response for the capacitor voltage** ( v_C(t) ) for the given circuit. --- ### **1️⃣ Circuit description** The circuit consists of: * ( R_1 = 10\,k\Omega ) * ( R_2 = 10\,k\Omega ) * ( L = 200\,H ) * ( C = 10\,\mu F ) * Input: ( v(t) = u(t) ) (unit step) The circuit topology: * ( R_1 ) is in series with the parallel combination of ( R_2 ), ( L ), and ( C ). --- ### **2️⃣ Equivalent impedance of the parallel branch** The parallel impedance ( Z_p(s) ) of ( R_2 ), ( L ), and ( C ) is: [ \frac{1}{Z_p(s)} = \frac{1}{R_2} + \frac{1}{sL} + sC ] So: [ Z_p(s) = \frac{1}{\dfrac{1}{R_2} + \dfrac{1}{sL} + sC} ] --- ### **3️⃣ Voltage across capacitor in Laplace domain** Let ( V_C(s) ) be the voltage across the capacitor. By voltage divider rule: [ V_C(s) = V_{in}(s) \cdot \frac{Z_p(s)}{R_1 + Z_p(s)} ] Given ( V_{in}(s) = \dfrac{1}{s} ) (since ( v(t) = u(t) )). --- ### **4️⃣ Substitute ( Z_p(s) ) into the expression** [ V_C(s) = \frac{1}{s} \cdot \frac{Z_p(s)}{R_1 + Z_p(s)} = \frac{1}{s} \cdot \frac{1}{1 + \dfrac{R_1}{Z_p(s)}} ] Compute ( \dfrac{R_1}{Z_p(s)} ): [ \frac{R_1}{Z_p(s)} = R_1 \left( \frac{1}{R_2} + \frac{1}{sL} + sC \right) ] \[ = \frac{R_1}{R_2} + \frac{R_1}{sL} + R_1 sC \] Therefore, [ V_C(s) = \frac{1}{s\left[1 + \frac{R_1}{R_2} + \frac{R_1}{sL} + R_1 sC \right]} \] --- ### **5️⃣ Simplify the expression** Multiply numerator and denominator by ( sL ): [ V_C(s) = \frac{L}{s \left[ L(1 + \frac{R_1}{R_2})s + R_1 L C s^2 + R_1 \right]} ] Simplify terms: [ V_C(s) = \frac{1}{R_1 C s^3 + \left(1 + \frac{R_1}{R_2}\right) s^2 + \frac{R_1}{L} s} ] --- ### **6️⃣ Substitute numerical values** Given: * ( R_1 = R_2 = 10\,k\Omega = 10^4\,\Omega ) * ( L = 200\,H ) * ( C = 10\,\mu F = 10 \times 10^{-6}\,F ) Substitute: \[ V_C(s) = \frac{1}{10^4 \times 10 \times 10^{-6} s^3 + (1 + 1) s^2 + \frac{10^4}{200} s} \] Calculate each term: \[ 10^4 \times 10 \times 10^{-6} = .1 \] \[ \frac{10^4}{200} = 50 \] Thus: \[ V_C(s) = \frac{1}{.1 s^3 + 2 s^2 + 50 s} \] --- ### **7️⃣ Simplify denominator** Factor out ( s ): \[ V_C(s) = \frac{1}{s(.1 s^2 + 2 s + 50)} \] Rewrite: \[ V_C(s) = \frac{1}{s \left( .1 s^2 + 2 s + 50 \right)} \] Divide numerator and denominator by .1: \[ V_C(s) = \frac{10}{s (s^2 + 20 s + 500)} \] --- ### **8️⃣ Determine poles and response** The quadratic: \[ s^2 + 20 s + 500 = \] Discriminant: \[ \Delta = 20^2 - 4 \times 1 \times 500 = 400 - 200 = -160 \] Roots: \[ s = -10 \pm j \sqrt{160} = -10 \pm j 40 \] (Actually, earlier, the approximate value was \(20.615\), but with exact calculations, the imaginary part is: \[ \sqrt{500 - 100} = \sqrt{400} = 20 \] This suggests a correction: the roots are: \[ s = -10 \pm j 20 \] since: \[ s = -\frac{b}{2a} \pm j \frac{\sqrt{4ac - b^2}}{2a} \] with \( a=1 \), \(b=20\), \(c=500\): \[ s = -10 \pm j \sqrt{500 - 100} = -10 \pm j 20 \] which matches earlier. --- ### **9️⃣ Partial fraction expansion** Express: \[ V_C(s) = \frac{10}{s (s + 10)^2 + 20^2} \] In standard form: \[ V_C(s) = \frac{A}{s} + \frac{B s + C}{(s + 10)^2 + 20^2} \] --- ### **10️⃣ Inverse Laplace transform** Using known transforms: \[ \mathcal{L}^{-1} \left\{\frac{1}{s}\right\} = 1 \] \[ \mathcal{L}^{-1} \left\{\frac{\omega_d}{(s + a)^2 + \omega_d^2}\right\} = e^{-a t} \sin (\omega_d t) \] \[ \mathcal{L}^{-1} \left\{\frac{s + a}{(s + a)^2 + \omega_d^2}\right\} = e^{-a t} \cos (\omega_d t) \] Constants \( a = 10 \), \( \omega_d = 20 \). Final expression: \[ v_C(t) = 1 - e^{-10 t} \left[ \cos(20 t) + \frac{10}{20} \sin(20 t) \right] \] which simplifies to: \[ v_C(t) = 1 - e^{-10 t} \left[ \cos(20 t) + .5 \sin(20 t) \right] \] for \( t \geq \). --- ### **11️⃣ Final answer** \[ \boxed{ v_C(t) = \left[ 1 - e^{-10 t} \left( \cos(20 t) + .5 \sin(20 t) \right) \right] u(t) } \] --- ### **12️⃣ Interpretation** - Steady-state voltage: \( v_C(\infty) = 1\,V \). - Oscillation frequency: \( \omega_d = 20\,rad/s \). - Damping factor: \( e^{-10 t} \) envelope. - Response is underdamped due to complex conjugate poles. This completes the derivation.

# Step Response for Capacitor Voltage \( v_C(t) \) ## 1️⃣ Circuit Description The circuit consists of: - \( R_1 = 10\,k\Omega \) - \( R_2 = 10\,k\Omega \) - \( L = 200\,H \) - \( C = 10\,\mu F \) - Input: \( v(t) = u(t) \) (unit step) **Topology**: \( R_1 \) is in series with the parallel combination of \( R_2 \), \( L \), and \( C \). --- ## 2️⃣ Equivalent Impedance of the Parallel Branch The parallel impedance \( Z_p(s) \) of \( R_2 \), \( L \), and \( C \) is: \[ \frac{1}{Z_p(s)} = \frac{1}{R_2} + \frac{1}{sL} + sC \] Thus, \[ Z_p(s) = \frac{1}{\dfrac{1}{R_2} + \dfrac{1}{sL} + sC} \] --- ## 3️⃣ Voltage Across Capacitor in Laplace Domain Let \( V_C(s) \) be the voltage across the capacitor. Using the voltage divider rule: \[ V_C(s) = V_{in}(s) \cdot \frac{Z_p(s)}{R_1 + Z_p(s)} \] Given \( V_{in}(s) = \frac{1}{s} \) (since \( v(t) = u(t) \)). --- ## 4️⃣ Substitute \( Z_p(s) \) into the Expression \[ V_C(s) = \frac{1}{s} \cdot \frac{Z_p(s)}{R_1 + Z_p(s)} = \frac{1}{s} \cdot \frac{1}{1 + \frac{R_1}{Z_p(s)}} \] Calculate \( \frac{R_1}{Z_p(s)} \): \[ \frac{R_1}{Z_p(s)} = R_1 \left( \frac{1}{R_2} + \frac{1}{sL} + sC \right) \] \[ = \frac{R_1}{R_2} + \frac{R_1}{sL} + R_1 sC \] Thus, we have: \[ V_C(s) = \frac{1}{s\left[1 + \frac{R_1}{R_2} + \frac{R_1}{sL} + R_1 sC \right]} \] --- ## 5️⃣ Simplify the Expression Multiply numerator and denominator by \( sL \): \[ V_C(s) = \frac{L}{s \left[ L(1 + \frac{R_1}{R_2})s + R_1 L C s^2 + R_1 \right]} \] This simplifies to: \[ V_C(s) = \frac{1}{R_1 C s^3 + \left(1 + \frac{R_1}{R_2}\right) s^2 + \frac{R_1}{L} s} \] --- ## 6️⃣ Substitute Numerical Values Given: - \( R_1 = R_2 = 10\,k\Omega = 10^4\,\Omega \) - \( L = 200\,H \) - \( C = 10\,\mu F = 10 \times 10^{-6}\,F \) Substituting these values: \[ V_C(s) = \frac{1}{10^4 \times 10 \times 10^{-6} s^3 + (1 + 1) s^2 + \frac{10^4}{200} s} \] Calculating each term: \[ 10^4 \times 10 \times 10^{-6} = 0.1 \] \[ \frac{10^4}{200} = 50 \] Thus: \[ V_C(s) = \frac{1}{0.1 s^3 + 2 s^2 + 50 s} \] --- ## 7️⃣ Simplify Denominator Factor out \( s \): \[ V_C(s) = \frac{1}{s(0.1 s^2 + 2 s + 50)} \] Rewriting gives: \[ V_C(s) = \frac{10}{s(s^2 + 20 s + 500)} \] --- ## 8️⃣ Determine Poles and Response The denominator quadratic is: \[ s^2 + 20 s + 500 = 0 \] The discriminant is: \[ \Delta = 20^2 - 4 \times 1 \times 500 = 400 - 200 = -160 \] Thus, the roots are: \[ s = -10 \pm j\sqrt{160} = -10 \pm j20 \] This indicates the response is **underdamped**. --- ## 9️⃣ Partial Fraction Expansion Express: \[ V_C(s) = \frac{10}{s[(s + 10)^2 + 20^2]} \] This can be rewritten as: \[ V_C(s) = \frac{A}{s} + \frac{Bs + C}{(s + 10)^2 + 20^2} \] --- ## 10️⃣ Inverse Laplace Transform Using known transforms: \[ \mathcal{L}^{-1} \left\{\frac{1}{s}\right\} = 1 \] \[ \mathcal{L}^{-1} \left\{\frac{s + a}{(s + a)^2 + \omega_d^2}\right\} = e^{-a t} \cos (\omega_d t) \] \[ \mathcal{L}^{-1} \left\{\frac{\omega_d}{(s + a)^2 + \omega_d^2}\right\} = e^{-a t} \sin (\omega_d t) \] From solving for constants, we find: \[ v_C(t) = 1 - e^{-10 t} \left[\cos(20 t) + 0.5 \sin(20 t)\right] , u(t) \] --- ## ✅ Final Answer \[ \boxed{ v_C(t) = \left[1 - e^{-10 t}\left(\cos(20 t) + 0.5 \sin(20 t)\right)\right] u(t) } \] --- ## 12️⃣ Interpretation - **Steady-state voltage**: \( v_C(\infty) = 1\,V \) - **Time constant**: \( \frac{1}{10} = 0.1\,s \) - **Oscillation frequency**: \( 20\,rad/s \) - **Response**: Underdamped due to complex conjugate poles.