Chapter 3 · Laplace Transforms

Laplace Transform: Definition & Properties

From Time Domain to s-Domain

Dr. Mohamed Mabrok · Qatar University

The Big Picture: Chapter 3 Roadmap

Definition &
Properties
(THIS LECTURE)

L{f(t)} = F(s)

Inverse
Transform

Solving ODEs

Step Functions &
Convolution

Key Idea

      The Laplace Transform converts time-domain problems into s-domain algebraic problems—much easier to solve!

Why Laplace? The Motivation

The Central Idea

      Differential equations are hard to solve. But what if we could transform them into algebraic equations?
    
      \text{ODE} \xrightarrow{\mathcal{L}} \text{Algebraic Equation} \xrightarrow{\text{Solve}} \text{Solution} \xrightarrow{\mathcal{L}^{-1}} \text{Time Domain}

⚡

Circuit Analysis

Solving RLC circuits with initial conditions.

🎮

Control Systems

Designing feedback controllers and stability analysis.

📡

Signal Processing

Analyzing and filtering signals in real time.

Time vs s-Domain

      Time domain (t): How things evolve over time | s-domain (s): Frequency and damping information

The Laplace Transform: Definition

Formal Definition

      $$\mathcal{L}\{f(t)\} = F(s) = \int_0^{\infty} e^{-st} f(t)\, dt$$
    
      where $s$ is a complex parameter and the integral converges for $\text{Re}(s) > \sigma_c$ (the abscissa of convergence).

Key Features

      Input: $f(t)$, a function of time $t$ 

      Output: $F(s)$, a function of complex parameter $s$ 

      Notation: Lowercase $f(t) \to$ Uppercase $F(s)$ (standard convention) 

      Lower limit: Integral starts at 0 (initial conditions at $t=0$)

Important Note

      The exponential kernel $e^{-st}$ decays as $t \to \infty$, which allows the integral to converge for sufficiently large $\text{Re}(s)$.

Worked Example

Find L{1} from Definition

STEP 1 Set up the integral

\mathcal{L}\{1\} = \int_0^{\infty} e^{-st} \cdot 1 \, dt

STEP 2 Integrate (standard form)

= \left[-\frac{e^{-st}}{s}\right]_0^{\infty}

STEP 3 Evaluate limits

= \left(0 - \left(-\frac{1}{s}\right)\right) = \frac{1}{s}

\mathcal{L}\{1\} = \frac{1}{s}, \quad s > 0

Convergence Condition

      For the integral to converge, we need $\text{Re}(s) > 0$, i.e., $s > 0$ (in the real case).

Worked Example

Find L{e^{at}} from Definition

STEP 1 Set up the integral

\mathcal{L}\{e^{at}\} = \int_0^{\infty} e^{-st} \cdot e^{at} \, dt = \int_0^{\infty} e^{(a-s)t} \, dt

STEP 2 Integrate with exponent $a-s$

= \left[\frac{e^{(a-s)t}}{a-s}\right]_0^{\infty}

STEP 3 Evaluate limits (need $s > a$)

= \left(0 - \frac{1}{a-s}\right) = \frac{1}{s-a}

\mathcal{L}\{e^{at}\} = \frac{1}{s-a}, \quad s > a

Computing f(t)=t^n and sin(bt)

Power Functions

\mathcal{L}\{t^n\} = \frac{n!}{s^{n+1}}

Example: $\mathcal{L}\{t\} = \frac{1}{s^2}$
Example: $\mathcal{L}\{t^2\} = \frac{2}{s^3}$

Method: Integration by parts (repeatedly for $t^n$)

Trigonometric Functions

\mathcal{L}\{\sin(bt)\} = \frac{b}{s^2+b^2}

And also: $\mathcal{L}\{\cos(bt)\} = \frac{s}{s^2+b^2}$

Method: Integration by parts twice, or use Euler's formula $e^{ibt} = \cos(bt) + i\sin(bt)$

Key Observation

      Computing from the definition is labor-intensive. That's why we use a Table of Laplace Transforms!

Linearity Property

Linearity Theorem

      $$\mathcal{L}\{af(t) + bg(t)\} = a\mathcal{L}\{f(t)\} + b\mathcal{L}\{g(t)\} = aF(s) + bG(s)$$
    
      The Laplace transform distributes over addition and scalar multiplication.

Why This Matters

      We don't need to compute complex transforms from scratch. We can break them into pieces, use the table, and combine results!

Worked Example

      $$\mathcal{L}\{3e^{2t} + 5\sin(4t)\}$$
    
      $= 3\mathcal{L}\{e^{2t}\} + 5\mathcal{L}\{\sin(4t)\}$ 

      $= 3 \cdot \frac{1}{s-2} + 5 \cdot \frac{4}{s^2+16}$ 

      $= \frac{3}{s-2} + \frac{20}{s^2+16}$

Standard Laplace Transform Table

Unit Step

f(t) = 1 \quad \Rightarrow \quad F(s) = \frac{1}{s}

Power: t^n

f(t) = t^n \quad \Rightarrow \quad F(s) = \frac{n!}{s^{n+1}}

Exponential

f(t) = e^{at} \quad \Rightarrow \quad F(s) = \frac{1}{s-a}

Product: t^n e^{at}

f(t) = t^ne^{at} \quad \Rightarrow \quad F(s) = \frac{n!}{(s-a)^{n+1}}

Sine

f(t) = \sin(bt) \quad \Rightarrow \quad F(s) = \frac{b}{s^2+b^2}

Cosine

f(t) = \cos(bt) \quad \Rightarrow \quad F(s) = \frac{s}{s^2+b^2}

Damped Sine

f(t) = e^{at}\sin(bt) \quad \Rightarrow \quad F(s) = \frac{b}{(s-a)^2+b^2}

Damped Cosine

f(t) = e^{at}\cos(bt) \quad \Rightarrow \quad F(s) = \frac{s-a}{(s-a)^2+b^2}

Pro Tip

      Memorize the basic transforms: $1, t, e^{at}, \sin, \cos$. The rest follow from properties!

First Shifting Theorem (s-Shift)

The Theorem

      $$\mathcal{L}\{e^{at}f(t)\} = F(s-a)$$
    
      In words: Multiplying by $e^{at}$ in the time domain shifts the s-variable by $a$ in the s-domain.

Intuition

      If $\mathcal{L}\{f(t)\} = F(s)$, then to find $\mathcal{L}\{e^{at}f(t)\}$, just replace every $s$ with $(s-a)$ in $F(s)$.

Example Preview

      $$\mathcal{L}\{e^{3t}\cos(2t)\} = \frac{s-3}{(s-3)^2+4}$$
    
      We know $\mathcal{L}\{\cos(2t)\} = \frac{s}{s^2+4}$, so replace $s$ with $(s-3)$.

Worked Examples: s-Shifting

Example 1: e^{3t} cos(2t)

      Base transform: $\mathcal{L}\{\cos(2t)\} = \frac{s}{s^2+4}$
    
      $$\mathcal{L}\{e^{3t}\cos(2t)\} = \frac{s-3}{(s-3)^2+4}$$
    
      Replace $s \to (s-3)$ everywhere.

Example 2: e^{-2t} t^3

      Base transform: $\mathcal{L}\{t^3\} = \frac{3!}{s^4} = \frac{6}{s^4}$
    
      $$\mathcal{L}\{e^{-2t}t^3\} = \frac{6}{(s+2)^4}$$
    
      Replace $s \to (s-(-2)) = (s+2)$ everywhere.

Transform of Derivatives (Crucial!)

First Derivative

      $$\mathcal{L}\{f'(t)\} = sF(s) - f(0)$$

Second Derivative

      $$\mathcal{L}\{f''(t)\} = s^2F(s) - sf(0) - f'(0)$$

General Pattern (n-th derivative)

      $$\mathcal{L}\{f^{(n)}(t)\} = s^nF(s) - s^{n-1}f(0) - s^{n-2}f'(0) - \cdots - f^{(n-1)}(0)$$

Why This Matters

      This is why Laplace transforms solve DEs! Derivatives become algebraic operations (multiplication by $s$). Initial conditions are automatically incorporated.

When Does the Laplace Transform Exist?

Sufficient Conditions (Dirichlet Conditions)

      The Laplace transform of $f(t)$ exists if:
    
      1. $f(t)$ is piecewise continuous on $[0, \infty)$ (finitely many jump discontinuities) 

      2. $f(t)$ is of exponential order: $|f(t)| \leq Me^{ct}$ for large $t$ and some constants $M, c$

Functions That Do NOT Have Laplace Transforms

      $e^{t^2}$ (grows too fast), $\frac{1}{t}$ (discontinuity at $t=0$), Dirac delta compositions

Exponential Order Intuition

      The function can grow, but not faster than an exponential. Since $e^{-st}$ decays exponentially, the integrand must eventually decay for convergence.

Worked Example

Find L{t^2 - 3e^{-t} + 4sin(2t)}

STEP 1 Apply linearity

\mathcal{L}\{t^2 - 3e^{-t} + 4\sin(2t)\} = \mathcal{L}\{t^2\} - 3\mathcal{L}\{e^{-t}\} + 4\mathcal{L}\{\sin(2t)\}

STEP 2 Look up in table

= \frac{2}{s^3} - 3 \cdot \frac{1}{s+1} + 4 \cdot \frac{2}{s^2+4}

STEP 3 Simplify

= \frac{2}{s^3} - \frac{3}{s+1} + \frac{8}{s^2+4}

Worked Example

Find L{e^{-t}(3cos(4t) - 2sin(4t))}

STEP 1 Apply linearity first

= 3\mathcal{L}\{e^{-t}\cos(4t)\} - 2\mathcal{L}\{e^{-t}\sin(4t)\}

STEP 2 Base transforms (before shifting)

\mathcal{L}\{\cos(4t)\} = \frac{s}{s^2+16}, \quad \mathcal{L}\{\sin(4t)\} = \frac{4}{s^2+16}

STEP 3 Apply s-shift: replace s with (s+1)

= 3 \cdot \frac{s+1}{(s+1)^2+16} - 2 \cdot \frac{4}{(s+1)^2+16}

STEP 4 Combine over common denominator

= \frac{3(s+1) - 8}{(s+1)^2+16} = \frac{3s-5}{(s+1)^2+16}

Worked Example

If L{f(t)} = (s+1)/(s^2+4), Find L{f'(t)} when f(0)=2

STEP 1 Apply the derivative property

\mathcal{L}\{f'(t)\} = sF(s) - f(0) = s \cdot \frac{s+1}{s^2+4} - 2

STEP 2 Simplify the first term

= \frac{s(s+1)}{s^2+4} - 2 = \frac{s^2+s}{s^2+4} - 2

STEP 3 Convert 2 to same denominator

= \frac{s^2+s}{s^2+4} - \frac{2(s^2+4)}{s^2+4} = \frac{s^2+s - 2s^2 - 8}{s^2+4}

STEP 4 Collect terms

\mathcal{L}\{f'(t)\} = \frac{-s^2+s-8}{s^2+4}

Key Takeaways & Summary

Definition

$\mathcal{L}\{f(t)\} = \int_0^\infty e^{-st}f(t)dt$

Linearity

$\mathcal{L}\{af+bg\} = aF+bG$

s-Shifting

$\mathcal{L}\{e^{at}f\} = F(s-a)$

Derivatives

$\mathcal{L}\{f'\} = sF-f(0)$

Common Mistakes to Avoid

Not using the table: Always look up transforms instead of computing from definition.
Forgetting initial conditions: The derivative rule includes $-f(0)$ terms—don't skip them!
Confusing shifting: s-shift: $e^{at}f(t) \to F(s-a)$. Don't flip the sign!
Checking convergence: Always verify $\text{Re}(s) > \sigma_c$ for your result.

Next Up

      Now that we know how to compute Laplace transforms, we'll learn the Inverse Laplace Transform to get solutions back in the time domain, and then solve differential equations!