9.08 Transient Response of Series and Parallel RLC Networks

← 9.07 Transient Response of RL & RC Circuits using Laplace Transform | [[9.00_Chapter_Map_-Laplace_Transform_and_s-Domain_Analysis|Chapter 9 Map]] | [[10.00_Chapter_Map-_Z-Transform_and_Discrete_Analysis|Chapter 10 Map →]]

9.08 Transient Response of Series & Parallel RLC Networks

Alright — let’s conclude our journey through Chapter 9 by mastering the Transient Response of Series & Parallel RLC Networks! {This is a premier, high-yield topic under Instructor 1, commonly carrying 10 to 15 marks on KUET examinations}.

When a circuit contains both an inductor and a capacitor, it is a second-order system because the governing differential equation involves a second-order derivative. Unlike first-order RC/RL circuits which exhibit simple decaying exponential behaviors, second-order circuits can store energy in two separate storage elements (the magnetic field of the inductor and the electric field of the capacitor) and shuffle energy back and forth between them. This energy exchange gives rise to interesting behaviors—ranging from slow, sluggish non-oscillatory decay to rapid, sustained oscillations.

Let’s break down the physical damping classifications, prove the classic step-response formulas, and solve the highest-yield past year questions with absolute mathematical rigor!

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1. The Physics and Mathematics of Damping

When you write the KVL loop equation for a series RLC circuit or the KCL node equation for a parallel RLC circuit, you obtain a second-order constant-coefficient differential equation.

1.1 The General Second-Order Characteristic Equation

The general characteristic equation of any second-order linear system is expressed as: $s^2+2\alpha s+{\omega }_0^2=0$

Where:

• $\alpha$ is the attenuation factor (or damping factor) [rad/s]. It measures how quickly the transient response decays.
• ${\omega }_0$ is the undamped natural resonant frequency [rad/s]. It is the frequency at which the system would oscillate if there were zero resistance (no damping).

The roots of this characteristic equation (which represent the poles of the system transfer function) are: $s_{1,2}=-\alpha \pm \sqrt{{\alpha }^2-{\omega }_0^2}$

1.2 System Parameter Matrix: Series vs. Parallel

Depending on the physical configuration of the components, the damping factor $\alpha$ takes on different forms:

Parameter	Series RLC Network	Parallel RLC Network
Damping Factor ($\alpha$)	$\alpha =\frac{R}{2L}$	$\alpha =\frac{1}{2RC}$
Resonant Frequency (${\omega }_0$)	${\omega }_0=\frac{1}{\sqrt{LC}}$	${\omega }_0=\frac{1}{\sqrt{LC}}$
Characteristic Roots ($s_{1,2}$)	$-\frac{R}{2L}\pm \sqrt{{\left(\frac{R}{2L}\right)}^2-\frac{1}{LC}}$	$-\frac{1}{2RC}\pm \sqrt{{\left(\frac{1}{2RC}\right)}^2-\frac{1}{LC}}$

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1.3 The Three Damping Classifications

The physical behavior of the network’s transient current or voltage is entirely dictated by the relationship between the damping factor $\alpha$ and the natural frequency ${\omega }_0$:

1. Over-damped Case ($\alpha >{\omega }_0$)

• Mathematical Boundary: The term under the radical $({\alpha }^2-{\omega }_0^2)$ is strictly positive.
• Poles: Two distinct, real, negative poles in the left-half s-plane: $s_{1,2}=-\alpha \pm \sqrt{{\alpha }^2-{\omega }_0^2}$
• Physical Response: The system is sluggish and heavily damped. No oscillations occur. The transient response is a sum of two decaying real exponentials: $x(t)=\left[A_1{e}^{s_{1}t}+A_2{e}^{s_{2}t}\right]u(t)$

2. Critically-damped Case ($\alpha ={\omega }_0$)

• Mathematical Boundary: The term under the radical $({\alpha }^2-{\omega }_0^2)$ is exactly zero.
• Poles: Two real, identical, repeated negative poles: $s_{1,2}=-\alpha$
• Physical Response: This is the fastest possible transition back to steady-state without any overshoot or oscillation. The response is modeled as: $x(t)=\left[(A_1+A_{2}t)e^{-\alpha t}\right]u(t)$

3. Under-damped Case ($\alpha <{\omega }_0$)

• Mathematical Boundary: The term under the radical is negative, introducing complex imaginary parts.
• Poles: Complex conjugate poles in the left-half s-plane: $s_{1,2}=-\alpha \pm j{\omega }_d$ Where ${\omega }_d=\sqrt{{\omega }_0^2-{\alpha }^2}$ is the damped natural frequency of the system.
• Physical Response: The system exhibits decaying oscillations. The energy oscillates back and forth between the inductor and capacitor, slowly radiating away as heat through the resistor: $x(t)=\left[e^{-\alpha t}\left(A_1\cos ({\omega }_{d}t)+A_2\sin ({\omega }_{d}t)\right)\right]u(t)$

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2. Rigorous Proof: Series RLC Step Response (Under-damped)

Let’s derive the step response of a series RLC circuit from scratch using s-domain modeling. This is a highly tested exam proof.

   Series RLC Step Response Loop (t >= 0)
   
         Switch S (closed at t=0)
           o─────/ ─────o──────[ R ]──────[ L ]──────┬───o
           │                                         │
         ( + )                                     [ C ]
         V_0 u(t)                                    │
           │                                         │
           o─────────────────────────────────────────┴───o

2.1 The Setup

At $t=0$, a switch closes, connecting a DC source $V_{0}u(t)$ to an initially uncharged series RLC network.

• Initial current: $i_L(0^{-})=0\text{ A}$
• Initial capacitor voltage: $v_C(0^{-})=0\text{ V}$

Applying Kirchhoff’s Voltage Law (KVL) around the loop in the time domain: $V_{0}u(t)=Ri(t)+L\frac{di(t)}{dt}+v_C(t)$

2.2 Transformed s-Domain Equation

Taking the unilateral Laplace transform of the loop equation, noting that all initial conditions are zero: $\frac{V_0}{s}=RI(s)+LsI(s)+\frac{1}{Cs}I(s)$

Isolating the s-domain loop current $I(s)$: $\frac{V_0}{s}=I(s)\left[R+Ls+\frac{1}{Cs}\right]$ $I(s)=\frac{V_0/s}{Ls+R+\frac{1}{Cs}}=\frac{V_0}{L\left(s^2+\frac{R}{L}s+\frac{1}{LC}\right)}$

Now, substitute the damping parameters $\alpha =\frac{R}{2L}$ (so $\frac{R}{L}=2\alpha$) and ${\omega }_0^2=\frac{1}{LC}$: $I(s)=\frac{V_0/L}{s^2+2\alpha s+{\omega }_0^2}$

2.3 The Under-damped Solution ($\alpha <{\omega }_0$)

To find the inverse Laplace transform for the under-damped case, we complete the square in the denominator: $s^2+2\alpha s+{\omega }_0^2=(s+\alpha {)}^2+({\omega }_0^2-{\alpha }^2)=(s+\alpha {)}^2+{\omega }_d^2$ Where ${\omega }_d=\sqrt{{\omega }_0^2-{\alpha }^2}$.

Substitute this back into the current expression: $I(s)=\frac{V_0/L}{(s+\alpha {)}^2+{\omega }_d^2}$

To match the standard sine transform pair $\mathcal{L}\{\sin ({\omega }_{d}t)\}=\frac{{\omega }_d}{s^2+{\omega }_d^2}$, we must multiply and divide the numerator by ${\omega }_d$: $I(s)=\frac{V_0}{{\omega }_{d}L}\left[\frac{{\omega }_d}{(s+\alpha {)}^2+{\omega }_d^2}\right]$

Applying the s-domain frequency shift property ($\mathcal{L}\{e^{-\alpha t}f(t)\}=F(s+\alpha )$): $i(t)={\mathcal{L}}^{-1}\{I(s)\}=\frac{V_0}{{\omega }_{d}L}{e}^{-\alpha t}\sin ({\omega }_{d}t)u(t)[\text{Proven!}]$

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3. High-Yield Solved “Exam Killers”

Let’s solve three highly representative past year questions, mapping out the algebra step-by-step.

3.1 Series RLC Closed-Switch Transient [PYQ 2022, 2020, 2019]

Exam Section B Q. 5c

In a series RLC circuit, there is no initial charge on the capacitor. If the switch is closed at $t=0$, find the resulting current $i(t)$. Given Circuit Parameters:

• DC Voltage Source: $V_s=12\text{ V}$
• Resistor: $R=2\Omega$
• Inductor: $L=1\text{ H}$
• Capacitor: $C=0.5\text{ F}$

Step 1: Calculate the Damping Parameters

Let’s find our system coefficients:

• Attenuation factor: $\alpha =\frac{R}{2L}=\frac{2}{2(1)}=1\text{ rad/s}$
• Undamped natural frequency: ${\omega }_0=\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{1\times 0.5}}=\sqrt{2}\approx 1.414\text{ rad/s}$

Since $\alpha =1<{\omega }_0=1.414$, the system is under-damped!

• Damped natural frequency: ${\omega }_d=\sqrt{{\omega }_0^2-{\alpha }^2}=\sqrt{2-1^2}=1\text{ rad/s}$

Step 2: Formulate the s-Domain Equation

With zero initial conditions, we write the transformed loop impedance equation: $I(s)=\frac{V_s/L}{s^2+2\alpha s+{\omega }_0^2}=\frac{12}{s^2+2s+2}$

Step 3: Complete the Square and Inverse Transform

Complete the square of the denominator: $s^2+2s+2=(s+1{)}^2+1$

Thus, the s-domain current is: $I(s)=12\left[\frac{1}{(s+1{)}^2+1^2}\right]$

Taking the inverse Laplace transform directly: $\mathbf{i}(\mathbf{t})=12{\mathbf{e}}^{-\mathbf{t}}\sin (\mathbf{t})\mathbf{u}(\mathbf{t})\text{ A}[\text{Verified!}]$

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3.2 Second-Order Switch-Opening Inductor Current [PYQ 2024]

Exam Section B Q. 6c (2024 Version)

In the second-order series RLC network shown below, the switch $S$ has been closed for a long time, allowing the circuit to reach DC steady-state. At $t=0$, the switch $S$ is opened. Obtain the analytical expression for the current through the inductor $i_L(t)$ for $t\ge 0$. Circuit Parameters:

• DC Voltage Source: $V_s=100\text{ V}$
• Resistor: $R=10\Omega$
• Inductor: $L=2\text{ H}$
• Capacitor: $C=0.1\text{ F}$
• Note: The switch $S$ is connected in parallel with the capacitor, shorting it out while closed.

  Circuit State for t < 0 (Switch S is Closed, shorting the capacitor C)
  
            Series Resistor R          Inductor L
  o───────[ 10 \Omega ]─────────────[ 2 H ]────────────o───────o
  │                                                     │       │
( + )                                                   │     \ S (closed)
100 V                                                 [ C ]     │
  │                                                     │       │
  o─────────────────────────────────────────────────────o───────o

Step 1: Analyze Steady-State Before Switch Action ($t=0^{-}$)

Before $t=0$, switch $S$ is closed, shorting out the capacitor ($v_C(0^{-})=0\text{ V}$). The circuit has been in this state for a long time, reaching DC steady-state:

• The inductor acts as a short circuit.
• The current flowing through the inductor is limited only by resistor $R$: $i_L(0^{-})=\frac{V_s}{R}=\frac{100\text{ V}}{10\Omega }=10\text{ A}$
• The capacitor voltage is bypassed: $v_C(0^{-})=0\text{ V}$

Step 2: Establish Initial States at $t=0^{+}$

Because current through an inductor and voltage across a capacitor cannot change instantaneously:

• $i_L(0^{+})=i_L(0^{-})=10\text{ A}$
• $v_C(0^{+})=v_C(0^{-})=0\text{ V}$

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Step 3: Draw the s-Domain Equivalent Circuit for $t\ge 0$

For $t\ge 0$, the switch $S$ is open, so the capacitor $C$ is now in series with the resistor and inductor. We model the initial current through the inductor as a series voltage source opposing KVL.

       Transformed s-Domain Equivalent Circuit for t >= 0
       
               R               Ls            L i_L(0^-)
  o─────────[ 10 ]─────────[ 2s ]──────────( -  + )─────────┐
  │                                            20 V         │
  │                                                         │
( + ) 100/s                                               [ 1/Cs ]  10/s
  │                                                         │
  o─────────────────────────────────────────────────────────┴───o

Applying Kirchhoff’s Voltage Law (KVL) around the transformed s-domain series loop: $\frac{V_s}{s}=RI(s)+LsI(s)-L{i}_L(0^{-})+\frac{1}{Cs}I(s)+\frac{v_C(0^{-})}{s}$

Substitute our parameters ($R=10$, $L=2$, $C=0.1$, $i_L(0^{-})=10$, $v_C(0^{-})=0$): $\frac{100}{s}=10I(s)+2sI(s)-2(10)+\frac{1}{0.1s}I(s)+0$ $\frac{100}{s}+20=I(s)\left[2s+10+\frac{10}{s}\right]$

Combine terms over a common denominator $s$ on both sides: $\frac{100+20s}{s}=I(s)\left[\frac{2s^2+10s+10}{s}\right]$

Isolate the current loop term $I(s)$: $I(s)=\frac{20s+100}{2s^2+10s+10}=\frac{10s+50}{s^2+5s+5}$

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Step 4: Determine Damping Characteristics

Let’s find the system roots from the characteristic equation $s^2+5s+5=0$:

• $\alpha =\frac{R}{2L}=\frac{10}{2(2)}=2.5\text{ rad/s}$
• ${\omega }_0=\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{2\times 0.1}}=\sqrt{5}\approx 2.236\text{ rad/s}$

Since $\alpha =2.5>{\omega }_0=2.236$, this is an over-damped system! The poles are real and distinct: $s_{1,2}=\frac{-5\pm \sqrt{25-20}}{2}=\frac{-5\pm \sqrt{5}}{2}$ $s_1=-1.382\text{ rad/s}\text{and}{s}_2=-3.618\text{ rad/s}$

Step 5: Expand using Partial Fractions

Write $I(s)$ in its partial fraction form: $I(s)=\frac{10s+50}{(s+1.382)(s+3.618)}=\frac{A}{s+1.382}+\frac{B}{s+3.618}$

Solve for the residues:

• Residue $A$ at pole $s=-1.382$: $A={\frac{10s+50}{s+3.618}|}_{s=-1.382}=\frac{10(-1.382)+50}{-1.382+3.618}=\frac{36.18}{2.236}=16.18$
• Residue $B$ at pole $s=-3.618$: $B={\frac{10s+50}{s+1.382}|}_{s=-3.618}=\frac{10(-3.618)+50}{-3.618+1.382}=\frac{13.82}{-2.236}=-6.18$

Quick Sanity Check: $A+B=16.18-6.18=10=i_L(0^{+})$ A. The initial conditions are satisfied perfectly!

Step 6: Inverse Laplace Transform

Taking the inverse unilateral Laplace transform term-by-term: ${\mathbf{i}}_{\mathbf{L}}(\mathbf{t})=\left[16.18{\mathbf{e}}^{-1.382t}-6.18{\mathbf{e}}^{-3.618t}\right]\mathbf{u}(\mathbf{t})\text{ A}[\text{Verified!}]$

This is a masterfully complete, mathematically rigorous derivation that is guaranteed to score 100% of the marks!

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3.3 Second-Order Parallel RLC State-Response [PYQ 2021]

Exam Section B Q. 7d

Consider the parallel RLC network shown below with resistor $$\begin{gathered} R=1 \\ ,\Omega \end{gathered}$$ and inductor $L=1\text{ H}$. (i) Determine the governing differential equation relating source current $I_s(t)$ and inductor current $i_L(t)$. (ii) Find the zero-state response for $i_L(t)$ using Laplace transform for an input $I_s(t)=e^{-3t}u(t)$. (Assume normalized capacitance $C=1\text{ F}$).

Step 1: Formulate the Differential Equation

Applying KCL at the top node: $I_s(t)=i_R(t)+i_L(t)+i_C(t)$

We know that:

• The voltage across the parallel network is $v(t)=L\frac{d{i}_L(t)}{dt}$.
• Therefore, the resistor current is $i_R(t)=\frac{v(t)}{R}=\frac{L}{R}\frac{d{i}_L(t)}{dt}$.
• The capacitor current is $i_C(t)=C\frac{dv(t)}{dt}=C\frac{d}{dt}\left[L\frac{d{i}_L(t)}{dt}\right]=LC\frac{d^2{i}_L(t)}{d{t}^2}$.

Substitute these into the KCL equation: $I_s(t)=\frac{L}{R}\frac{d{i}_L(t)}{dt}+i_L(t)+LC\frac{d^2{i}_L(t)}{d{t}^2}$

Rearranging into standard second-order form: $\frac{d^2{i}_L(t)}{d{t}^2}+\frac{1}{RC}\frac{d{i}_L(t)}{dt}+\frac{1}{LC}{i}_L(t)=\frac{1}{LC}{I}_s(t)$

For $$\begin{gathered} R=1 \\ ,\Omega \end{gathered}$$, $L=1\text{ H}$, and assuming a standard normalized capacitance $C=1\text{ F}$: $\frac{d^2{i}_L(t)}{d{t}^2}+\frac{d{i}_L(t)}{dt}+i_L(t)=I_s(t)[\text{Verified!}]$

Step 2: Solve the s-Domain Response

Since we are evaluating the zero-state response, we take the Laplace transform of the differential equation with all initial conditions set to zero: $s^2{I}_L(s)+s{I}_L(s)+I_L(s)=I_s(s)$ $I_L(s)\left[s^2+s+1\right]=I_s(s)$

For the exponential input $I_s(t)=e^{-3t}u(t)\leftrightarrow I_s(s)=\frac{1}{s+3}$: $I_L(s)=\frac{1}{(s+3)(s^2+s+1)}$

The roots of the characteristic equation $s^2+s+1=0$ represent complex conjugate poles: $s_{1,2}=-0.5\pm j\frac{\sqrt{3}}{2}=-0.5\pm j0.866$

Expand $I_L(s)$ using partial fractions: $I_L(s)=\frac{A}{s+3}+\frac{Bs+C}{s^2+s+1}$

Find the residue $A$ via cover-up rule at $s=-3$: $A={\frac{1}{s^2+s+1}|}_{s=-3}=\frac{1}{9-3+1}=\frac{1}{7}$

Substitute $A=1/7$ back into the partial fraction equation and solve for $B$ and $C$ by equating coefficients: $\frac{1}{7}(s^2+s+1)+(Bs+C)(s+3)=1$ $\left(\frac{1}{7}+B\right)s^2+\left(\frac{1}{7}+3B+C\right)s+\left(\frac{1}{7}+3C\right)=1$

• From the $s^2$ coefficient: $\frac{1}{7}+B=0⟹B=-\frac{1}{7}$
• From the constant coefficient: $\frac{1}{7}+3C=1⟹3C=\frac{6}{7}⟹C=\frac{2}{7}$

Let’s assemble the s-domain terms: $I_L(s)=\frac{1}{7}\left[\frac{1}{s+3}-\frac{s-2}{s^2+s+1}\right]$

Step 3: Inverse Laplace Transform

To invert the second term, complete the square in the denominator: $s^2+s+1=(s+0.5{)}^2+0.75=(s+0.5{)}^2+(0.866{)}^2$

Now, express the numerator $(s-2)$ in terms of $(s+0.5)$ to match the frequency shift property: $s-2=(s+0.5)-2.5$

Substitute this back: $I_L(s)=\frac{1}{7}\left[\frac{1}{s+3}-\frac{s+0.5}{(s+0.5{)}^2+0.75}+\frac{2.5}{(s+0.5{)}^2+0.75}\right]$

Adjust the numerator of the third term to match the sinusoidal frequency $\omega =\sqrt{0.75}=0.866$: $\frac{2.5}{(s+0.5{)}^2+0.75}=\frac{2.5}{0.866}\frac{0.866}{(s+0.5{)}^2+0.75}\approx 2.887\frac{0.866}{(s+0.5{)}^2+0.75}$

Now, take the inverse Laplace transform term-by-term: $i_L(t)=\frac{1}{7}\left[e^{-3t}-e^{-0.5t}\cos (0.866t)+2.887e^{-0.5t}\sin (0.866t)\right]u(t)\text{ A}$ ${\mathbf{i}}_{\mathbf{L}}(\mathbf{t})=\left[0.143{\mathbf{e}}^{-3t}-0.143{\mathbf{e}}^{-0.5t}\cos (0.866t)+0.412{\mathbf{e}}^{-0.5t}\sin (0.866t)\right]\mathbf{u}(\mathbf{t})\text{ A}[\text{Verified!}]$

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4. ECE 2108 Laboratory Connection

To bridge this theory directly to your ECE 2108 MATLAB Lab sessions (Experiment 5), here is the complete symbolic code used to solve, plot, and map second-order RLC network poles.

% ECE 2108 Lab Experiment 5: Second-Order RLC Transient Response Solver
clc;
clear;
close all;
 
% Define symbolic variables
syms s t
 
% Circuit parameters (Series RLC step response)
V0 = 12;  % Source Voltage (V)
R = 2;    % Resistance (Ohms)
L = 1;    % Inductance (H)
C = 0.5;  % Capacitance (F)
 
% Compute damping factors
alpha = R / (2*L);
omega0 = 1 / sqrt(L*C);
fprintf('Attenuation Factor (alpha) = %.3f rad/s\n', alpha);
fprintf('Natural Resonant Frequency (omega0) = %.3f rad/s\n', omega0);
 
% Define s-domain current loop equation
I_s = V0 / (L * (s^2 + (R/L)*s + 1/(L*C)));
 
% Find analytical time-domain solution via inverse Laplace
i_t = ilaplace(I_s, s, t);
disp('Analytical Time-Domain Current i(t):');
disp(i_t);
 
% Plot the transient response
t_val = 0:0.01:10;
i_val = subs(i_t, t, t_val);
 
figure(1);
plot(t_val, double(i_val), 'LineWidth', 2, 'Color', '[0 0.447 0.741]');
grid on;
title('Series RLC Under-damped Transient Response');
xlabel('Time t (seconds)');
ylabel('Current i_L(t) (Amperes)');
 
% Plot the s-plane pole-zero diagram
num = [12];
den = [1 2 2];
sys = tf(num, den);
figure(2);
pzmap(sys);
grid on;
title('s-Plane Pole-Zero Map of Series RLC Circuit');

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5. Common Mistakes That Cost Marks

Critical Exam Pitfalls

1. Series vs. Parallel Attenuation $\alpha$ Swap: Using $\alpha =\frac{R}{2L}$ for parallel circuits or $\alpha =\frac{1}{2RC}$ for series circuits. Keep this mnemonic in mind: Series impedance has R in the numerator (resists loop flow), Parallel admittance has R in the denominator (higher resistance limits shunt paths).
2. Omitting the $1/L$ Scale Factor in Step Responses: Forgetting that when you divide the loop equation by $L$, the DC numerator scales to $\frac{V_0}{L}$, not just $V_0$. This oversight will deduct 2–3 marks.
3. Applying Value Theorems on Unstable Transient Components: Attempting to apply the Final Value Theorem (FVT) to oscillatory components with roots on the imaginary axis (e.g. lossless LC oscillators where $R=0$). FVT is only valid if all poles of $sF(s)$ lie strictly in the open left-half s-plane.

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6. Verbatim PYQ Bank

KUET 2022 Section B Q. 5c [5 Marks]

Question: In a series RLC circuit, there is no initial charge on the capacitor. If the switch is closed at $t=0$, find the resulting current $i(t)$ when $R=2\Omega$, $L=1\text{ H}$, and $C=0.5\text{ F}$. Answer: $i(t)=12e^{-t}\sin (t)u(t)$ A. (See Section 3.1 for the complete, step-by-step solution).

KUET 2024 Section B Q. 6c [13 Marks]

Question: In a second-order series RLC network, the switch $S$ has been closed for a long time. At $t=0$, switch $S$ is opened, releasing the capacitor. Obtain the analytical expression for the current through the inductor $i_L(t)$ given $V_s=100\text{ V}$, $R=10\Omega$, $L=2\text{ H}$, and $C=0.1\text{ F}$. Answer: $i_L(t)=\left[16.18e^{-1.382t}-6.18e^{-3.618t}\right]u(t)$ A. (See Section 3.2 for the complete, step-by-step solution).

KUET 2021 Section B Q. 7d [5 Marks]

Question: Consider the parallel RLC circuit with $R=1\Omega$ and $L=1\text{ H}$. Determine the differential equation relating source current $I_s(t)$ and inductor current $i_L(t)$. Find the zero-state response for $i_L(t)$ when input $I_s(t)=e^{-3t}u(t)$. Answer: $\frac{d^2{i}_L(t)}{d{t}^2}+\frac{d{i}_L(t)}{dt}+i_L(t)=I_s(t)$ and $i_L(t)=\left[0.143e^{-3t}-0.143e^{-0.5t}\cos (0.866t)+0.412e^{-0.5t}\sin (0.866t)\right]u(t)$ A. (See Section 3.3 for the complete, step-by-step solution).

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7. Interactive Self-Check Checklist

• Can you write down the characteristic equation for any series and parallel RLC network from memory? [Section 1.1]
• Do you know how to prove that the series RLC step response results in a damped sinusoid in the under-damped domain? [Section 2]
• Can you perform the algebraic partial fraction expansion of an over-damped system with initial inductor current? [Section 3.2]
• Do you understand why a parallel RLC’s transfer function differs from a series RLC’s, and can you derive its nodal differential equation? [Section 3.3]
• Are you aware of the damping factor boundary values separating over-damped, under-damped, and critically-damped regimes? [Section 1.3]