lecture_EE1C2_lec6_laplace_transformation_2026-01-05

📌 Overview

🚀 The Goal: Escaping the Time Domain

The Problem: In the time domain, circuits with multiple energy storage elements (L, C) create Differential Equations. $y^{\prime \prime }(t)+3y^{\prime }(t)+2y(t)=f(t)$ These are hard because derivatives are mixed together.

The Dream Insight: There is one function that is incredibly easy to differentiate: the Exponential $e^{st}$. $\frac{d}{dt}(e^{st})=s\cdot e^{st}$ Derivatives turn into simple multiplication.

The Strategy: If we can break ANY complicated signal $f(t)$ into a massive pile of simple exponential building blocks ($e^{st}$), we can turn difficult Calculus problems into simple Algebra problems.

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🎯 Learning Objectives

• Grasp the Intuition: Understand Laplace as a “frequency probe” that detects exponential components.
• Master the Dictionary: Memorize the transform pairs ($u(t)$, $e^{-at}$, $\sin$, $\cos$).
• Construct Signals: Build complex time-domain shapes using steps and ramps.
• Solve Circuits: Convert high-order circuits to impedances and solve algebraically.

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➗ Formulas & Definitions

1. The Transformation

$F(s)=\mathcal{L}\{f(t)\}={\int }_{0^{-}}^{\infty }f(t)e^{-st}dt$ Note: We use $0^{-}$ to capture impulses at $t=0$.

2. The Golden Table (Essential)

Signal Name	Time Domain f(t)	Laplace Domain F(s)	ROC
Impulse	$\delta (t)$	$1$	All $s$
Step	$u(t)$	$\frac{1}{s}$	$\text{Re}(s)>0$
Exponential	$e^{-at}u(t)$	$\frac{1}{s+a}$	$\text{Re}(s)>-a$
Ramp	$t\cdot u(t)$	$\frac{1}{s^2}$	$\text{Re}(s)>0$
Power	$t^{n}u(t)$	$\frac{n!}{s^{n+1}}$	$\text{Re}(s)>0$
T-Exponential	$t{e}^{-at}u(t)$	$\frac{1}{(s+a{)}^2}$	$\text{Re}(s)>-a$
Cosine	$\cos (\omega t)u(t)$	$\frac{s}{s^2+{\omega }^2}$	$\text{Re}(s)>0$
Sine	$\sin (\omega t)u(t)$	$\frac{\omega }{s^2+{\omega }^2}$	$\text{Re}(s)>0$
Phase Sine	$\sin (\omega t+\theta )u(t)$	$\frac{s\sin \theta +\omega \cos \theta }{s^2+{\omega }^2}$	$\text{Re}(s)>0$
Phase Cosine	$\cos (\omega t+\theta )u(t)$	$\frac{s\cos \theta -\omega \sin \theta }{s^2+{\omega }^2}$	$\text{Re}(s)>0$
Damped Sine	$e^{-at}\sin (\omega t)u(t)$	$\frac{\omega }{(s+a{)}^2+{\omega }^2}$	$\text{Re}(s)>-a$
Damped Cosine	$e^{-at}\cos (\omega t)u(t)$	$\frac{s+a}{(s+a{)}^2+{\omega }^2}$	$\text{Re}(s)>-a$

3. Operational Properties

• Linearity: $a{f}_1(t)+b{f}_2(t)\leftrightarrow a{F}_1(s)+b{F}_2(s)$
• Scaling: $f(at)\leftrightarrow \frac{1}{a}F(\frac{s}{a})$
• Time Shift: $f(t-a)u(t-a)\leftrightarrow e^{-as}F(s)$
• Frequency Shift: $e^{-at}f(t)\leftrightarrow F(s+a)$
• Time Derivative:
  • $f^{\prime }(t)\leftrightarrow sF(s)-f(0^{-})$
  • $f^{\prime \prime }(t)\leftrightarrow s^{2}F(s)-sf(0^{-})-f^{\prime }(0^{-})$
• Time Integration: ${\int }_0^{t}f(\tau )d\tau \leftrightarrow \frac{F(s)}{s}$
• Freq. Derivative: $tf(t)\leftrightarrow -\frac{d}{ds}F(s)$
• Initial Value: $f(0^{+})={\lim }_{s\to \infty }sF(s)$
• Final Value: $f(\infty )={\lim }_{s\to 0}sF(s)$ (If steady state exists)
• Convolution: $f_1(t)\ast f_2(t)\leftrightarrow F_1(s)\cdot F_2(s)$

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💡 Intuitive stepwise derivation

1. The Assumption: Everything is Exponentials

We assume every signal $f(t)$ is just a “soup” of complex exponentials. $f(t)\approx \sum C_k{e}^{s_{k}t}$

• $s=\sigma +j\omega$ is the “complex frequency” (rate of spin and growth).

Visualizing the Components: A Sine wave (oscillating) is actually built of two counter-rotating exponentials (Euler’s Identity): $\cos (\omega t)=\frac{1}{2}{e}^{j\omega t}+\frac{1}{2}{e}^{-j\omega t}$

Complex Plane View (The Spinners): How does $e^{j\omega t}$ look if we look “down the barrel” of time? It’s a rotating vector. $\cos (\omega t)$ is the sum of two counter-rotating vectors that cancel out vertically (imaginary part) and add up horizontally (real part).

2. The Concept of “Probing”

How do we find out which frequencies $s_k$ are inside $f(t)$? We use a Probe: The exact opposite spinner $e^{-st}$ and integrate over time.

• Case A (Mismatch): The integral remains finite (averages out).
• Case B (Match): $e^{s_{k}t}\cdot e^{-st}=1$. The integral “explodes” to infinity (Pole).

Visualizing the “Explosion”: The Laplace Transform $F(s)$ blows up exactly where the probe frequency $s$ matches a component in the signal.

3. Reconstruction (The Inverse)

To find $f(t)$ from $F(s)$, we analyze the “strength” of these poles.

Step 1: Partial Fraction Expansion

Break the system into simple “Golden Table” components. $F(s)=\frac{N(s)}{(s-p_1)(s-p_2)}=\frac{K_1}{s-p_1}+\frac{K_2}{s-p_2}$

Step 2: The Cover-Up Method

Imagine we have a function with two poles. To find the strength of Pole 1, we effectively “cover up” the term $(s-p_1)$ in the denominator and evaluate the rest at $s=p_1$.

$F(s)=\frac{10}{(s+1)(s+3)}$

To find the coefficient for $e^{-t}$ (at $s=-1$):

1. Cover $(s+1)$.
2. Evaluate $\frac{10}{s+3}$ at $s=-1$: $\frac{10}{-1+3}=5$.
3. Result: $5e^{-t}$.

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✍️ Notes / Example exercises

Exercise 1: Piecewise Signal Construction 🧩

Task: Find the Laplace transform of a ramp that starts at $t=1$, reaches 5 at $t=2$, and then stays at 5.

Step-by-Step Plan:

1. Start Ramp: $5(t-1)u(t-1)$ (Slope 5, starts at 1).
2. Stop Ramp (Flatten): To make the slope 0 at $t=2$, add a ramp with slope -5 starting at $t=2$: $-5(t-2)u(t-2)$.
3. Apply Time-Shift Property: $f(t-a)u(t-a)\leftrightarrow e^{-as}F(s)$.
4. Final Result: $F(s)=\frac{5}{s^2}{e}^{-s}-\frac{5}{s^2}{e}^{-2s}=\frac{5}{s^2}(e^{-s}-e^{-2s})$.

Exercise 2: Solving a 3rd Order Circuit ⚡

Task: Input $u(t)$, $H(s)=\frac{6}{s^3+6s^2+11s+6}$. Find $v_{out}(t)$.

1. Equation: $V_{out}(s)=\frac{1}{s}\cdot \frac{6}{(s+1)(s+2)(s+3)}$.
2. Expansion: $\frac{K_0}{s}+\frac{K_1}{s+1}+\frac{K_2}{s+2}+\frac{K_3}{s+3}$.
3. Residues (Cover-up):
   • $K_0=\frac{6}{1\cdot 2\cdot 3}=1$
   • $K_1=\frac{6}{-1\cdot 1\cdot 2}=-3$
   • $K_2=\frac{6}{-2\cdot -1\cdot 1}=3$
   • $K_3=\frac{6}{-3\cdot -2\cdot -1}=-1$
4. Time Domain: $v_{out}(t)=(1-3e^{-t}+3e^{-2t}-e^{-3t})u(t)$.

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🔗 Resources

• Presentation:
• Inverse Laplace Formalism (The Bromwich Integral):

3. Special Case: Completing the Square ⬛

When the denominator has complex roots (quadratic doesn’t factor nicely), we force it into the $\sin /\cos$ templates from the Golden Table.

Example Task: Inverse Laplace of $V(s)=\frac{s+2}{s^2+4s+20}$

1. Identify the Target: Looks like a damped cosine $\frac{s+a}{(s+a{)}^2+{\omega }^2}$.
2. Complete the Square:
   • Take half of middle term ($4/2=2$).
   • Write as $(s+2{)}^2$.
   • Expand $(s+2{)}^2=s^2+4s+4$.
   • Adjust the constant: $20=4+16$.
3. Rewrite: $V(s)=\frac{s+2}{(s+2{)}^2+4^2}$.
4. Lookup Table: $a=2,\omega =4$.
5. Result: $v(t)=e^{-2t}\cos (4t)u(t)$.

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❓ Post lecture

⚠️ Method Constraints & Rare Cases

1. Region of Convergence (ROC): The integral only exists if $\text{Re}(s)$ is large enough.
   • Rule: The integration path $\gamma$ must be to the right of all poles.
2. Complex Poles:
   • Do not use PFE with $A$ and $B$ if roots are complex.
   • Fix: Complete the square in the denominator. Use Sine/Cosine forms.
3. Initial Conditions:
   • For $f^{\prime }(t)$, you must subtract $f(0^{-})$. This handles stored energy ($V_c,I_l$).
4. Final Value Theorem Limit:
   • Only valid if the limit $sF(s)$ exists and the system is stable (poles in LHP).
   • If poles are on $j\omega$ axis (like a pure sine), the “final value” doesn’t exist (it keeps oscillating!).

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📖 Homework

• SGH5: Practice table lookups.
• Challenge: Solve for $v(t)$ if $V(s)=\frac{s+2}{s^2+4s+20}$. (Done in example above ☝️)
• Next Step: Apply to RLC circuits with non-zero initial conditions.