# Series RC Circuit with Step Input – Laplace Analysis and Time Response ## Circuit Interpretation - **Source:** V step applied at \( = \) - \( = 100\,\text{k}\Omega \) - \( C_1 = 6\,\mu\text{F} \) - \( C_2 = 3\,\mu\text{F} \) - **Capacitors in series** - Measure voltages: - \( v_1(t) \): across \( C_1 \) - \( v_2(t) \): across \( C_2 \) - **Initial condition:** Both capacitors uncharged --- ## 1. Laplace Analysis (Step-by-Step) ### Step 1: Convert to the \( s \)-Domain **Step input:** \[ V_{in}(s) = \frac{12}{s} \] **Capacitor impedances:** \[ Z_{C1} = \frac{1}{sC_1}, \quad Z_{C2} = \frac{1}{sC_2} \] **Substitute values:** \[ Z_{C1} = \frac{1}{6\times10^{-6}s}= \frac{1.6667\times10^{5}}{s} \] \[ Z_{C2} = \frac{1}{3\times10^{-6}s}= \frac{3.3333\times10^{5}}{s} \] **Total capacitor impedance:** \[ Z_C = Z_{C1} + Z_{C2} \] \[ Z_C = \frac{1.6667\times10^{5}}{s} + \frac{3.3333\times10^{5}}{s} \] \[ Z_C = \frac{5\times10^{5}}{s} \] --- ### Step 2: Total Circuit Impedance \[ Z_{total} = R + Z_C \] \[ Z_{total} = 10^5 + \frac{5\times10^5}{s} \] --- ### Step 3: Circuit Current \( I(s) \) \[ I(s) = \frac{V_{in}(s)}{Z_{total}} \] \[ I(s) = \frac{\frac{12}{s}}{10^5 + \frac{5\times10^5}{s}} \] Multiply numerator and denominator by \( s \): \[ I(s) = \frac{12}{10^5 s + 5\times10^5} \] Factor: \[ I(s) = \frac{12}{10^5(s + 5)} \] \[ I(s) = \frac{1.2\times10^{-4}}{s + 5} \] --- ## 2. Find \( v_1(t) \) and \( v_2(t) \) --- ### Voltage across \( C_1 \) \[ V_1(s) = I(s) Z_{C1} \] \[ V_1(s) = \frac{1.2\times10^{-4}}{s + 5} \times \frac{1.6667\times10^5}{s} \] \[ V_1(s) = \frac{20}{s(s + 5)} \] #### Partial Fraction \[ \frac{20}{s(s + 5)} = \frac{A}{s} + \frac{B}{s + 5} \] \[ 20 = A(s + 5) + B s \] For \( s = \): \[ A = 4 \] For \( s = -5 \): \[ B = -4 \] So, \[ V_1(s) = \frac{4}{s} - \frac{4}{s + 5} \] #### Inverse Laplace \[ v_1(t) = 4(1 - e^{-5t})u(t) \] --- ### Voltage across \( C_2 \) \[ V_2(s) = I(s) Z_{C2} \] \[ V_2(s) = \frac{1.2\times10^{-4}}{s + 5} \times \frac{3.3333\times10^5}{s} \] \[ V_2(s) = \frac{40}{s(s + 5)} \] Partial fraction: \[ \frac{40}{s(s + 5)} = \frac{8}{s} - \frac{8}{s + 5} \] Inverse Laplace: \[ v_2(t) = 8(1 - e^{-5t})u(t) \] --- ## 3. Final Time Responses \[ \boxed{v_1(t) = 4(1 - e^{-5t})\,V} \] \[ \boxed{v_2(t) = 8(1 - e^{-5t})\,V} \] **Check steady state:** \[ v_1(\infty) = 4\,V, \quad v_2(\infty) = 8\,V \] \[ 4 + 8 = 12\,V \] ✔ Correct. **Time constant:** \[ \tau = \frac{1}{5} = .2\,s \] --- ## 4. Transfer Functions **Input:** \( V_{in}(s) = \frac{12}{s} \) --- ### For \( v_1 \) \[ H_1(s) = \frac{V_1(s)}{V_{in}(s)} \] \[ H_1(s) = \frac{20}{s(s + 5)} \div \frac{12}{s} \] \[ H_1(s) = \frac{20}{12(s + 5)} \] \[ \boxed{H_1(s) = \frac{5}{3}\frac{1}{s + 5}} \] --- ### For \( v_2 \) \[ H_2(s) = \frac{V_2(s)}{V_{in}(s)} \] \[ H_2(s) = \frac{40}{s(s + 5)} \div \frac{12}{s} \] \[ \boxed{H_2(s) = \frac{10}{3}\frac{1}{s + 5}} \] --- ## 5. Stability Analysis Pole location from denominator: \[ s + 5 = \] \[ \boxed{s = -5} \] Because the pole is **in the left half plane**, the system is: ✅ **Stable** --- ## 6. Frequency Response Substitute \( s = j\omega \): --- ### For \( v_1 \) \[ H_1(j\omega) = \frac{5/3}{j\omega + 5} \] **Magnitude:** \[ |H_1(j\omega)| = \frac{5/3}{\sqrt{\omega^2 + 25}} \] **Phase:** \[ \angle H_1(j\omega) = -\tan^{-1}\left(\frac{\omega}{5}\right) \] --- ### For \( v_2 \) \[ H_2(j\omega) = \frac{10/3}{j\omega + 5} \] **Magnitude:** \[ |H_2(j\omega)| = \frac{10/3}{\sqrt{\omega^2 + 25}} \] --- ## 7. Bode Plot (Asymptotic Method) **Break frequency:** \[ \omega_p = 5\,\text{rad/s} \] --- ### Low Frequency Gain For \( v_1 \): \[ |H_1()| = \frac{5/3}{5} = \frac{1}{3} \] \[ 20\log_{10}(1/3) = -9.54\,\text{dB} \] For \( v_2 \): \[ |H_2()| = \frac{10/3}{5} = \frac{2}{3} \] \[ 20\log_{10}(2/3) = -3.52\,\text{dB} \] --- ### Slopes - Before \( 5\,\text{rad/s} \): ** dB/dec** - After \( 5\,\text{rad/s} \): **–20 dB/dec** **Phase shifts from:** \[ ^\circ \rightarrow -90^\circ \] --- ## Final Answers ### Time Domain \[ v_1(t) = 4(1 - e^{-5t})\,V \] \[ v_2(t) = 8(1 - e^{-5t})\,V \] ### Transfer Functions \[ H_1(s) = \frac{5}{3(s + 5)} \] \[ H_2(s) = \frac{10}{3(s + 5)} \] ### Pole \[ s = -5 \] **System is stable.** --- *If you want, I can also show:* - **Full hand-drawn style Bode diagram** - **Professor-style exam solution format** - **MATLAB code to plot the Bode diagram**