Engineering Mathematics II for Automotive Mechatronics professionals in Kenya requires advanced mathematical tools to analyze and solve differential equations arising in system dynamics, control systems, and signal processing. The Laplace transform is a powerful integral transform method enabling the conversion of time-domain functions into a complex frequency domain, simplifying the solution of differential equations common in automotive electronics and mechatronic control systems. Mastering the derivation and application of Laplace transforms from first principles equips automotive mechatronics technicians to model and analyze transient responses in circuits and mechanical systems efficiently.

7.1 Meaning of Laplace transforms deriving Laplace transforms from first principles

The Laplace transform converts a time-dependent function \( f(t) \), defined for \( t \geq 0 \), into a complex frequency domain function \( F(s) \). This transformation simplifies the handling of differential equations by converting differentiation and integration operations into algebraic manipulations. The Laplace transform is defined by an improper integral involving an exponential decay factor \( e^{-st} \), where \( s \) is a complex parameter typically with a real positive part to ensure convergence.

7.1.1 Definition of the Laplace Transform

The Laplace transform of a function \( f(t) \) is given by the integral

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

where

• \( f(t) \) is the original time-domain function, valid for \( t \geq 0 \),
• \( s \) is a complex frequency parameter \( s = \sigma + j\omega \) with \( \sigma \) real and positive for convergence,
• \( F(s) \) is the image of \( f(t) \) in the complex frequency domain.

This integral transforms a function of time into a function of \( s \), providing a tool to solve linear differential equations by algebraic methods.

Worked Examples

Example 1: Find the Laplace transform of the function \( f(t) = 1 \) for \( t \geq 0 \).

Given: \( f(t) = 1 \), \( t \geq 0 \)

$$ F(s) = \int_0^\infty e^{-st} \cdot 1 \, dt $$

Calculate the integral:

$$ F(s) = [ \frac{-1}{s} e^{-st} ]_0^\infty $$

At upper limit \( t \to \infty \), since \( \text{Re}(s) > 0 \), \( e^{-st} \to 0 \). At lower limit \( t=0 \), \( e^{-s \times 0} = 1 \).

So:

$$ F(s) = 0 - ( \frac{-1}{s} \times 1 ) = \frac{1}{s} $$

Answer:

$$ F(s) = \frac{1}{s}, \quad \text{for } \text{Re}(s) > 0 $$

Example 2: Find the Laplace transform of \( f(t) = t \), \( t \geq 0 \).

Given: \( f(t) = t \)

$$ F(s) = \int_0^\infty e^{-st} t \, dt $$

Use integration by parts:

Let \( u = t \Rightarrow du = dt \), and \( dv = e^{-st} dt \Rightarrow v = \frac{-1}{s} e^{-st} \).

Then:

$$ F(s) = uv \big|_0^\infty - \int_0^\infty v \, du = [ t \cdot \frac{-1}{s} e^{-st} ]_0^\infty + \frac{1}{s} \int_0^\infty e^{-st} dt $$

At \( t \to \infty \), \( t e^{-st} \to 0 \) (since exponential dominates polynomial), and at \( t=0 \), term is zero.

Thus:

$$ F(s) = 0 + \frac{1}{s} \cdot \frac{1}{s} = \frac{1}{s^2} $$

Answer:

$$ F(s) = \frac{1}{s^2}, \quad \text{for } \text{Re}(s) > 0 $$

Example 3: Find the Laplace transform of \( f(t) = e^{at} \), where \( a \) is a constant.

Given: \( f(t) = e^{at} \)

$$ F(s) = \int_0^\infty e^{-st} e^{at} dt = \int_0^\infty e^{-(s, a) t} dt $$

Calculate the integral:

$$ F(s) = [ \frac{-1}{s, a} e^{-(s, a) t} ]_0^\infty = 0 - ( \frac{-1}{s, a} \times 1 ) = \frac{1}{s, a} $$

Answer:

$$ F(s) = \frac{1}{s, a}, \quad \text{for } \text{Re}(s) > a $$

Example 4: Find the Laplace transform of \( f(t) = \sin(\omega t) \), \( \omega > 0 \).

Given: \( f(t) = \sin(\omega t) \)

$$ F(s) = \int_0^\infty e^{-st} \sin(\omega t) dt $$

Use the standard integral result or integration by parts twice:

$$ F(s) = \frac{\omega}{s^2 + \omega^2} $$

Answer:

$$ F(s) = \frac{\omega}{s^2 + \omega^2}, \quad \text{for } \text{Re}(s) > 0 $$

Example 5: Find the Laplace transform of \( f(t) = t^n \), where \( n \) is a positive integer.

Given: \( f(t) = t^n \)

The formula for the Laplace transform is:

$$ F(s) = \int_0^\infty e^{-st} t^n dt = \frac{n!}{s^{n+1}} $$

Answer:

$$ F(s) = \frac{n!}{s^{n+1}}, \quad \text{for } \text{Re}(s) > 0 $$

7.1.2 Derivation of Laplace Transform from First Principles

The Laplace transform is derived by applying the integral transform operator to \( f(t) \), multiplying by an exponential kernel \( e^{-st} \), and integrating over \( t \geq 0 \). This kernel introduces exponential decay, ensuring convergence for a wide class of functions.

The derivation steps are:

1. Start with a function \( f(t) \) defined for \( t \geq 0 \).
2. Multiply \( f(t) \) by \( e^{-st} \), where \( s \) is a complex parameter.
3. Integrate the product over \( t \) from 0 to infinity.
4. Analyze convergence conditions on \( s \).
5. Interpret the resulting function \( F(s) \) as the Laplace transform.

The integral form is:

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

This integral transforms the time domain function into the complex frequency domain function.

Worked Examples

Example 1: Derive the Laplace transform of \( f(t) = 1 \) from first principles.

Given: \( f(t) = 1 \)

$$ F(s) = \int_0^\infty e^{-st} \cdot 1 \, dt $$

Calculate the integral:

$$ F(s) = [ \frac{-1}{s} e^{-st} ]_0^\infty = 0 - ( \frac{-1}{s} \times 1 ) = \frac{1}{s} $$

Example 2: Derive the Laplace transform of \( f(t) = t \) using integration by parts.

Given: \( f(t) = t \)

$$ F(s) = \int_0^\infty e^{-st} t dt $$

Let \( u = t \), \( dv = e^{-st} dt \).

Then \( du = dt \), \( v = \frac{-1}{s} e^{-st} \).

Apply integration by parts:

$$ F(s) = uv \big|_0^\infty - \int_0^\infty v du = [ t \cdot \frac{-1}{s} e^{-st} ]_0^\infty + \frac{1}{s} \int_0^\infty e^{-st} dt $$

Evaluate the boundary terms and integral:

$$ F(s) = 0 + \frac{1}{s} \times \frac{1}{s} = \frac{1}{s^2} $$

Example 3: Derive the Laplace transform of \( f(t) = e^{at} \).

Given: \( f(t) = e^{at} \)

$$ F(s) = \int_0^\infty e^{-st} e^{at} dt = \int_0^\infty e^{-(s, a) t} dt $$

Integrate:

$$ F(s) = [ \frac{-1}{s, a} e^{-(s, a) t} ]_0^\infty = 0 - ( \frac{-1}{s, a} \times 1 ) = \frac{1}{s, a} $$

Example 4: Derive the Laplace transform of \( f(t) = \sin(\omega t) \) using first principles.

Given: \( f(t) = \sin(\omega t) \)

$$ F(s) = \int_0^\infty e^{-st} \sin(\omega t) dt $$

Use integration by parts or known integral of sine with exponential decay:

$$ F(s) = \frac{\omega}{s^2 + \omega^2} $$

Example 5: Derive the Laplace transform of \( f(t) = t^n \) from first principles.

Given: \( f(t) = t^n \)

Use repeated integration by parts or the Gamma function identity:

$$ F(s) = \int_0^\infty e^{-st} t^n dt = \frac{n!}{s^{n+1}} $$

7.1.3 Linearity Property of the Laplace Transform

The Laplace transform is a linear operator, meaning the transform of a sum of functions equals the sum of their transforms, and scalar multiplication factors out of the transform. This property enables the decomposition of complex functions into simpler parts for transformation.

Mathematically:

$$ \mathcal{L}\{a f(t) + b g(t)\} = a \mathcal{L}\{f(t)\} + b \mathcal{L}\{g(t)\} $$

where \( a \) and \( b \) are constants, and \( f(t), g(t) \) are functions of time.

Worked Examples

Example 1: Find the Laplace transform of \( f(t) = 3 + 4t \).

Given: \( f(t) = 3 + 4t \)

Using linearity:

$$ F(s) = 3 \mathcal{L}\{1\} + 4 \mathcal{L}\{t\} $$

From previous transforms:

$$ \mathcal{L}\{1\} = \frac{1}{s}, \quad \mathcal{L}\{t\} = \frac{1}{s^2} $$

Therefore:

$$ F(s) = 3 \times \frac{1}{s} + 4 \times \frac{1}{s^2} = \frac{3}{s} + \frac{4}{s^2} $$

Example 2: Find the Laplace transform of \( f(t) = 5 e^{2t} + 7 \sin(3t) \).

Given: \( f(t) = 5 e^{2t} + 7 \sin(3t) \)

Using linearity:

$$ F(s) = 5 \mathcal{L}\{ e^{2t} \} + 7 \mathcal{L}\{ \sin(3t) \} $$

Recall:

$$ \mathcal{L}\{ e^{at} \} = \frac{1}{s, a}, \quad \mathcal{L}\{ \sin(\omega t) \} = \frac{\omega}{s^2 + \omega^2} $$

Then:

$$ F(s) = 5 \times \frac{1}{s - 2} + 7 \times \frac{3}{s^2 + 9} = \frac{5}{s - 2} + \frac{21}{s^2 + 9} $$

Example 3: Find the Laplace transform of \( f(t) = 2 t^2 - 6 e^{-t} \).

Given: \( f(t) = 2 t^2 - 6 e^{-t} \)

Using linearity:

$$ F(s) = 2 \mathcal{L}\{ t^2 \} - 6 \mathcal{L}\{ e^{-t} \} $$

Recall:

$$ \mathcal{L}\{ t^n \} = \frac{n!}{s^{n+1}}, \quad \mathcal{L}\{ e^{at} \} = \frac{1}{s, a} $$

Since \( n=2 \), \( 2! = 2 \):

$$ F(s) = 2 \times \frac{2}{s^{3}} - 6 \times \frac{1}{s + 1} = \frac{4}{s^3} - \frac{6}{s + 1} $$

Example 4: Find the Laplace transform of \( f(t) = 10 + 3t + 4 \cos(2t) \).

Given: \( f(t) = 10 + 3t + 4 \cos(2t) \)

Using linearity:

$$ F(s) = 10 \mathcal{L}\{1\} + 3 \mathcal{L}\{t\} + 4 \mathcal{L}\{\cos(2t)\} $$

Recall:

$$ \mathcal{L}\{1\} = \frac{1}{s}, \quad \mathcal{L}\{t\} = \frac{1}{s^2}, \quad \mathcal{L}\{\cos(\omega t)\} = \frac{s}{s^2 + \omega^2} $$

Then:

$$ F(s) = 10 \times \frac{1}{s} + 3 \times \frac{1}{s^2} + 4 \times \frac{s}{s^2 + 4} = \frac{10}{s} + \frac{3}{s^2} + \frac{4s}{s^2 + 4} $$

Example 5: Find the Laplace transform of \( f(t) = 2 e^{3t} - 5 t + 7 \sin(t) \).

Given: \( f(t) = 2 e^{3t} - 5 t + 7 \sin(t) \)

Using linearity:

$$ F(s) = 2 \mathcal{L}\{ e^{3t} \} - 5 \mathcal{L}\{ t \} + 7 \mathcal{L}\{ \sin(t) \} $$

Recall:

$$ \mathcal{L}\{ e^{at} \} = \frac{1}{s, a}, \quad \mathcal{L}\{ t \} = \frac{1}{s^2}, \quad \mathcal{L}\{ \sin(\omega t) \} = \frac{\omega}{s^2 + \omega^2} $$

Substitute:

$$ F(s) = 2 \times \frac{1}{s - 3} - 5 \times \frac{1}{s^2} + 7 \times \frac{1}{s^2 + 1} = \frac{2}{s - 3} - \frac{5}{s^2} + \frac{7}{s^2 + 1} $$

Chapter Summary

This chapter introduced the concept of Laplace transforms, explaining their meaning and demonstrating how to derive them from first principles. It then outlined the key properties of Laplace transforms that facilitate their application in engineering problems. The chapter covered the determination of inverse Laplace transforms, especially for simple functions and those expressed using partial fractions. Techniques for solving ordinary differential equations using Laplace transforms were presented, highlighting the method’s efficiency in handling initial conditions. Furthermore, the chapter extended these methods to solve simultaneous differential equations when initial conditions are provided. Through these topics, students gained a comprehensive understanding of applying Laplace transforms to analyze and solve complex engineering mathematics problems.