lecture_EE1C2_lec7_applied_laplace_transformation_2026-01-12

📌 Overview

This lecture bridges the gap between abstract Laplace math and practical circuit analysis. We learn how to transform entire circuits into the s-domain, treating inductors and capacitors as impedances with attached sources representing their initial energy state. This turns differential equation problems into simpler algebraic problems.

Key Concepts:

• s-domain Models: Replacing $R,L,C$ with impedances $R,sL,1/sC$.
• Initial Conditions: Representing $v_C(0)$ and $i_L(0)$ as voltage or current sources.
• Circuit Analysis: Using KCL, KVL, Nodal, and Mesh analysis directly in the s-domain.
• Transfer Functions: $H(s)=Y(s)/X(s)$ describing system behavior.

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🎯 Learning Objectives

• Model Components: Construct s-domain equivalent circuits for Resistors, Capacitors, and Inductors.
• Handle Initial Conditions: Correctly place and orient the DC sources representing initial stored energy ($v(0),i(0)$).
• Solve Circuits: Calculate voltages and currents in the s-domain and transform back to time-domain.
• Analyze Systems: Determine Transfer Functions and identify system stability via poles and zeros.

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➗ Formulas & Definitions

Element Transformations

Component	Time Domain	s-Domain Impedance ($Z$)	Series Model (Voltage Source)	Parallel Model (Current Source)
Resistor	$v(t)=Ri(t)$	$R$	Just $R$	Just $R$
Inductor	$v(t)=L\frac{di}{dt}$	$sL$	Inductor $sL$ in series with $V_{src}=-Li(0)$	Inductor $sL$ in parallel with $I_{src}=-i(0)/s$
Capacitor	$i(t)=C\frac{dv}{dt}$	$\frac{1}{sC}$	Capacitor $\frac{1}{sC}$ in series with $V_{src}=\frac{v(0)}{s}$	Capacitor $\frac{1}{sC}$ in parallel with $I_{src}=-Cv(0)$

Transfer Function

$H(s)=\frac{Y(s)}{X(s)}=\frac{\text{Output}}{\text{Input}}$

• Impulse Response: $h(t)={\mathcal{L}}^{-1}\{H(s)\}$ (since $X(s)=1$ for $\delta (t)$)
• Step Response: $y(t)={\mathcal{L}}^{-1}\{H(s)\cdot \frac{1}{s}\}$

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💡 Intuitive Stepwise Derivation

Why do we add sources? It comes directly from the differentiation property of Laplace transforms.

1. The Inductor Case

Time domain law: $v(t)=L\frac{di(t)}{dt}$ Apply Laplace transform: $V(s)=L[sI(s)-i(0)]$ $V(s)=\underset{\text{Voltage drop across Z}}{\underbrace{sL\cdot I(s)}}-\underset{\text{Voltage source}}{\underbrace{Li(0)}}$

• Interpretation: The total voltage $V(s)$ is the sum of a drop across an impedance $sL$ and a voltage source $-Li(0)$.
• Visual: A resistor-like element $sL$ in series with a battery $Li(0)$.

2. The Capacitor Case

Time domain law: $i(t)=C\frac{dv(t)}{dt}$ Apply Laplace transform: $I(s)=C[sV(s)-v(0)]$ Rearrange for V (Series Model): $V(s)=\frac{1}{sC}I(s)+\frac{v(0)}{s}$

• Interpretation: The voltage is the drop across impedance $\frac{1}{sC}$ plus a voltage source $\frac{v(0)}{s}$.

3. Visual Models

Using circuitikz to visualize the transformation.

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✍️ Notes / Example Exercises

🧠 Step-by-Step Solving Strategy

1. $\mathbf{t}<0$ Analysis (Steady State)

   • Treat Capacitors as Open Circuits.
   • Treat Inductors as Short Circuits.
   • Calculate $v_C(0^{-})$ and $i_L(0^{-})$. These are your “memory” values.

2. $\mathbf{t}>0$ Circuit Construction

   • Replace sources with their Laplace transforms (e.g., $u(t)\to 1/s$, $\delta (t)\to 1$, $\cos (\omega t)\to s/(s^2+{\omega }^2)$).
   • Replace elements with s-domain models (Impedance + Initial Condition Source).
   • Tip: Use Series models for Mesh analysis (KVL) and Parallel models for Nodal analysis (KCL).

3. Solve Algebraically

   • Use standard techniques (Nodal, Mesh, Superposition, Source Transformation) to find the desired variable $Y(s)$.

4. Inverse Laplace

   • Use Partial Fraction Expansion to convert $Y(s)$ back to $y(t)$.

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📝 Example 1: RC Circuit Step Response

Given:

• Series RC circuit with Voltage source $V_s$.
• Switch closes at $t=0$.
• Initial capacitor voltage $v_c(0)=V_0$.

\usepackage{circuitikz}
\usetikzlibrary{positioning}
\ctikzset{bipoles/length=1.0cm}
\begin{document}
\begin{circuitikz}[american, voltage shift=0.5]
    % Time Domain
    \coordinate (start_td) at (0,0);
    \draw (start_td) to[V, l=$V_s u(t)$] ++(0,2) coordinate (top_td);
    \draw (top_td) to[spst, l=$t=0$] ++(2,0) coordinate (sw_td)
          to[R, l=$R$] ++(2,0) coordinate (res_td)
          to[C, l=$C$, v=$v_c(0){=}V_0$] ++(0,-2) coordinate (bot_td)
          -- (start_td);
    \node [above=0.5cm of top_td] {Time Domain};
 
    % s-Domain
    \coordinate (start_s) at (6,0);
    \draw (start_s) to[V, l=$\frac{V_s}{s}$] ++(0,2) coordinate (top_s);
    \draw (top_s) -- ++(2,0) coordinate (wire_s)
          to[R, l=$R$] ++(2,0) coordinate (res_s)
          to[C, l=$\frac{1}{sC}$] ++(0,-1) coordinate (cap_s)
          to[V, l_=$\frac{V_0}{s}$, invert] ++(0,-1) coordinate (bot_s) % Inverted for drop
          -- (start_s);
    \node [above=0.5cm of top_s] {s-Domain};
\end{circuitikz}
\end{document}

Step 1: s-Domain Circuit

• Source: $V_s/s$ (Step function).
• Resistor: $R$.
• Capacitor: $\frac{1}{sC}$ in series with source $\frac{V_0}{s}$ (opposing current if discharge, aiding if charge? Polarity matters. Here we assume drop + source drop).

Step 2: Analysis (KVL) $\frac{V_s}{s}=R\cdot I(s)+\frac{1}{sC}\cdot I(s)+\frac{V_0}{s}$

Step 3: Solve for I(s) $\frac{V_s-V_0}{s}=I(s)\left(R+\frac{1}{sC}\right)$ $I(s)=\frac{V_s-V_0}{s\left(R+\frac{1}{sC}\right)}=\frac{V_s-V_0}{R\left(s+\frac{1}{RC}\right)}$

Step 4: Inverse Transform $i(t)=\frac{V_s-V_0}{R}{e}^{-\frac{t}{RC}}u(t)$

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📝 Example 2: Series-Parallel RLC System

Question: Find the output voltage $v_o(t)$ across the parallel LC tank for $t>0$.

• Input: $v_s(t)=10u(t)V$
• Components: $R=2\Omega$ in series with a parallel combination of $L=1H$ and $C=0.5F$.
• Initial Conditions: Zero energy ($v_C(0)=0,i_L(0)=0$).

Step 1: Impedances

• Source: $V_s(s)=10/s$
• $Z_R=2$
• $Z_L=sL=s(1)=s$
• $Z_C=\frac{1}{sC}=\frac{1}{s(0.5)}=\frac{2}{s}$
• No initial condition sources needed (Zero state).

Step 2: Equivalent Impedance ($Z_p$) Calculate the impedance of the parallel L-C branch: $Z_p=Z_L||Z_C=\frac{Z_L\cdot Z_C}{Z_L+Z_C}=\frac{s\cdot \frac{2}{s}}{s+\frac{2}{s}}=\frac{2}{\frac{s^2+2}{s}}=\frac{2s}{s^2+2}$

Step 3: Voltage Divider $V_o(s)=V_s(s)\cdot \frac{Z_p}{R+Z_p}$ $V_o(s)=\frac{10}{s}\cdot \frac{\frac{2s}{s^2+2}}{2+\frac{2s}{s^2+2}}$

Step 4: Simplify Algebra $V_o(s)=\frac{10}{s}\cdot \frac{2s}{2(s^2+2)+2s}$ Multiply top and bottom by $(s^2+2)$: $V_o(s)=\frac{10}{s}\cdot \frac{2s}{2s^2+4+2s}$ Cancel $s$ and divide by 2: $V_o(s)=10\cdot \frac{1}{s^2+s+2}=\frac{10}{s^2+s+2}$

Step 5: Inverse Transform Analyze the denominator roots (poles): $s^2+s+2=0⟹s=\frac{-1\pm \sqrt{1-8}}{2}=-0.5\pm j\sqrt{1.75}\approx -0.5\pm j1.32$ This is an underdamped response (complex poles). Standard form for damped sine: $\frac{\omega }{(s+a{)}^2+{\omega }^2}$ Complete the square: $s^2+s+2=(s+0.5{)}^2+1.75$ Let $a=0.5$ and $\omega =\sqrt{1.75}\approx 1.323$. $V_o(s)=\frac{10}{(s+0.5{)}^2+(\sqrt{1.75}{)}^2}$ Adjust numerator to match $\omega$: $V_o(s)=\frac{10}{\sqrt{1.75}}\cdot \frac{\sqrt{1.75}}{(s+0.5{)}^2+1.75}$ $V_o(s)\approx 7.56\cdot \frac{1.323}{(s+0.5{)}^2+1.323^2}$ Inverse transform: $v_o(t)=7.56e^{-0.5t}\sin (1.323t)u(t)V$

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📝 Example 3: RLC Circuit with Initial Conditions

Question: The switch in the circuit below has been closed for a long time and opens at $t=0$. Find the current $i(t)$ through the inductor for $t>0$.

• Components:
  • Source $V_s=10V$.
  • Resistor $R_1=2\Omega$ (Input side).
  • Inductor $L=1H$ in series with Resistor $R_2=2\Omega$.
  • Capacitor $C=0.5F$ in parallel with the L-R2 branch.

`

\usepackage{circuitikz}
\usetikzlibrary{positioning}
\ctikzset{bipoles/length=1.0cm}
\begin{document}
\begin{circuitikz}[american, voltage shift=0.5]
    % t<0 Circuit
    \coordinate (start) at (0,0);
    \draw (start) to[V, l=$10V$] ++(0,3) coordinate (top);
    \draw (top) to[R, l=$R_1(2\Omega)$] ++(3,0) coordinate (sw_node);
    \draw (sw_node) to[spst, l=$t=0$, mirror] ++(1.5,0) coordinate (main_node);
    
    % Parallel Branches
    % Branch 1: L + R2
    \draw (main_node) -- ++(1.5,0) coordinate (br1_top);
    \draw (br1_top) to[L, l=$L(1H)$, i=$i(t)$] ++(0,-1.5) coordinate (L_mid)
          to[R, l=$R_2(2\Omega)$] ++(0,-1.5) coordinate (br1_bot);
          
    % Branch 2: C
    \draw (br1_top) -- ++(2,0) coordinate (br2_top);
    \draw (br2_top) to[C, l=$C(0.5F)$, v=$v_C(t)$] ++(0,-3) coordinate (br2_bot);
    
    % Return path
    \draw (br2_bot) -- (br1_bot) -- ++(-3,0) coordinate (gnd_node) -- (start);
    \node [above=0.2cm of top] {Time Domain ($t<0$)};
 
    % s-Domain Circuit (t>0)
    % Shift right relative to existing nodes
    \coordinate (s_start) at (8,1.5);
    
    % Loop: L, R2, C
    \draw (s_start) to[L, l=$sL$] ++(3,0) coordinate (s_mid1)
          to[V, l=$Li(0)$] ++(1.5,0) coordinate (s_corner1);
          
    \draw (s_corner1) to[R, l=$R_2$] ++(0,-2) coordinate (s_corner2);
    
    \draw (s_corner2) to[V, l=$\frac{v_C(0)}{s}$] ++(-2,0) coordinate (s_mid2)
          to[C, l=$\frac{1}{sC}$] ++(-2.5,0) coordinate (s_corner3);
          
    \draw (s_corner3) -- (s_start);
    \draw (s_start) node[above]{$I(s) \rightarrow$};
    \node [above=1cm of s_mid1] {s-Domain ($t>0$)};
 
\end{circuitikz}
\end{document}

Step 1: Steady State Analysis ($t=0^{-}$)

• Inductor acts as Short Circuit.
• Capacitor acts as Open Circuit.
• Circuit is: 10V source $\to$ $R_1(2\Omega )$ $\to$ $R_2(2\Omega )$ $\to$ Ground.
• Inductor Current $i_L(0^{-})$: $i_L(0)=\frac{10V}{R_1+R_2}=\frac{10}{2+2}=2.5A$
• Capacitor Voltage $v_C(0^{-})$: Since C is in parallel with the L-R2 branch (and L is short), $v_C$ equals the voltage across $R_2$. $v_C(0)=i_L(0)\cdot R_2=2.5A\cdot 2\Omega =5V$

Step 2: s-Domain Circuit ($t>0$) The switch opens, isolating the L-R2-C loop.

• Inductor: Impedance $sL=s$. Series source $V_{L0}=L\cdot i_L(0)=1\cdot 2.5=2.5V$.
  • Polarity: Source opposes current drop, follows passive sign convention with impedance. The voltage drop is $V=sLI-Li(0)$. So source is a “rise” in direction of current.
• Capacitor: Impedance $\frac{1}{sC}=\frac{1}{0.5s}=\frac{2}{s}$. Series source $V_{C0}=\frac{v_C(0)}{s}=\frac{5}{s}$.
  • Polarity: $V=\frac{I}{sC}+\frac{v_C(0)}{s}$. Source adds to voltage drop.
• Resistor: $R_2=2\Omega$.

Step 3: Solve for Current I(s) Apply KVL to the loop (Clockwise): $\sum V_{drops}=0$ $\underset{\text{Inductor Model}}{\underbrace{(s\cdot I(s)-2.5)}}+\underset{\text{Resistor}}{\underbrace{(2\cdot I(s))}}+\underset{\text{Capacitor Model? Check Polarity}}{\underbrace{(\frac{2}{s}I(s)-\frac{5}{s})}}=0$ Wait, let’s check KVL and signs carefully.

• $v_C(t)$ was defined top to bottom (+ at top).
• $i(t)$ flows down through L. So loop current $i(t)$ flows Clockwise.
• Inductor: Drop $sLI(s)-Li(0)$.
• Resistor: Drop $RI(s)$.
• Capacitor: Current enters top (+). $V_C(s)=\frac{1}{sC}I(s)+\frac{v_C(0)}{s}$.
  • Note: If $v_C(0)$ is positive at top, the source $\frac{v_C(0)}{s}$ adds to the potential at the top?
  • Equation $V=ZI+V_{src}$.
  • KVL Sum of drops: $V_L+V_R+V_C=0$? No, it’s a closed loop with no external source. The sum of voltages across components must be zero.
  • $(sI-2.5)+2I+(\frac{2}{s}I+\frac{5}{s})=0$?
  • Wait, Capacitor Voltage $v_C$ matches $v_{R2}$ at $t=0$. Polarity: Top is +, Bottom is -.
  • So KVL: Drop across L + Drop across R + Drop across C = 0?
  • Yes.
  • $(sLI-L{i}_0)+(R_{2}I)+(\frac{1}{sC}I-\frac{v_C(0)}{s})$?
  • Let’s recall Capacitor model: $V=\frac{1}{sC}I+\frac{v_C(0)}{s}$.
  • Wait, does the source oppose or aid?
  • If capacitor discharges, current leaves + terminal.
  • Our defined current $I$ enters + terminal (clockwise).
  • So Capacitor acts as load. $V=ZI+V_{src}$.
  • Is $V_{src}$ opposing?
  • Think: At $t=0$, $I$ might be negative? No.
  • Let’s look at net voltage driving the loop.
  • Stored energy acts as source.
  • Inductor $i_0$ tries to push current.
  • Capacitor $v_0$ tries to push current (discharge).
  • KVL: $\text{Impedance Drops}=\text{Source Rises}$.
  • $I(s)(sL+R+\frac{1}{sC})=Li(0)-\frac{v_C(0)}{s}$?
  • Let’s check directions.
  • $i_L(0)$ is Down. Loop current $I$ is Clockwise (Down). So Inductor source aids current. ($+Li(0)$ on RHS).
  • $v_C(0)$ is + on Top. Loop current enters Top. Capacitor acts as source opposing entry? No, it pushes current Out of Top? No, Out of Bottom?
  • Capacitor acts like battery + on Top.
  • If loop current enters +, it goes against the battery. So battery opposes current.
  • So on RHS (driving force), it should be $-\frac{v_C(0)}{s}$.
  • Result: $I(s)(Z_{total})=Li(0)-\frac{v_C(0)}{s}$.
  • $I(s)(s+2+\frac{2}{s})=2.5-\frac{5}{s}$.

Step 4: Solve Algebra $I(s)\frac{s^2+2s+2}{s}=\frac{2.5s-5}{s}$ $I(s)=\frac{2.5s-5}{s^2+2s+2}$

Step 5: Inverse Laplace

• Roots of $s^2+2s+2$: $(s+1{)}^2+1$. Poles at $-1\pm j1$. Underdamped.
• Rewrite numerator to match shift $(s+1)$: $2.5s-5=2.5(s+1)-2.5-5=2.5(s+1)-7.5$
• Substitute: $I(s)=\frac{2.5(s+1)}{(s+1{)}^2+1}-\frac{7.5}{(s+1{)}^2+1}$
• Transform terms:
  • $\frac{s+1}{(s+1{)}^2+1}\to e^{-t}\cos (t)$
  • $\frac{1}{(s+1{)}^2+1}\to e^{-t}\sin (t)$
• Result: $i(t)=[2.5e^{-t}\cos (t)-7.5e^{-t}\sin (t)]u(t)\text{ A}$

📝 Exercise: RLC Circuit (Series)

Question: Find $v_o(t)$ for $t>0$ in a series RLC circuit where $v_s(t)=10u(t)$, $R=2\Omega$, $L=1H$, $C=0.5F$, and zero initial conditions.

Work & Plan:

1. Transform Components:
   • $V_s(s)=10/s$
   • $Z_R=2$
   • $Z_L=s$
   • $Z_C=2/s$
   • Init conditions $=0$ (no extra sources).
2. Setup Equation (Voltage Divider): $V_o(s)=V_s(s)\cdot \frac{Z_C}{Z_R+Z_L+Z_C}$ $V_o(s)=\frac{10}{s}\cdot \frac{2/s}{2+s+2/s}$
3. Simplify: $V_o(s)=\frac{10}{s}\cdot \frac{2}{s^2+2s+2}=\frac{20}{s(s^2+2s+2)}$
4. Partial Fractions: $\frac{20}{s(s^2+2s+2)}=\frac{A}{s}+\frac{Bs+C}{s^2+2s+2}$
   • Find A: $s\to 0⟹A=20/2=10$.
   • Solve remaining terms…
5. Inverse Laplace:
   • Recognize shifted sine/cosine forms for the quadratic part (underdamped).

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🔗 Resources

• Presentation:

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❓ Post lecture

• Practice deriving the parallel vs series models until it’s second nature.
• Review Partial Fraction Expansion (Case 2: Complex Roots) as it appears frequently in RLC circuits.

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📖 Homework