Laplace Transform — Introduction

Definition and Basic Transforms (university freshman level)

Introduction Overview

Time domain f(t) Laplace transform s-domain (complex) F(s) Re Im Linear ODEs become algebraic equations in s
\(\displaystyle\int_{0}^{\infty} f(t)\, e^{-st}\, dt\)
Figure 1: The concept of the Laplace transform. A time-domain function $f(t)$ is converted into an $s$-domain function $F(s)$. The ✕ marks in the $s$-domain are example poles: when a Laplace transform or a transfer function is a rational function, such "poles" appear. They are discussed in later chapters.

In this introduction, we learn the basic idea of the Laplace transform. Our goal is to build an intuitive understanding of what it means to "transform a function of time into a function of a complex variable."

Learning Goals

  • Understand the definition of the Laplace transform
  • Understand the role of the exponential function $e^{-st}$
  • Compute the Laplace transform of basic functions
  • Learn to use a transform table
  • Get to know where the Laplace transform is applied

Contents

  1. Chapter 1 What Is the Laplace Transform?

    Motivation, definition, and intuitive meaning of the transform

  2. Chapter 2 Laplace Transform of Exponential Functions

    Meaning of $e^{-st}$ and convergence conditions

  3. Chapter 3 Basic Transforms

    Transforms of $1$, $t$, $t^n$, and $e^{at}$

  4. Chapter 4 Laplace Transform of Trigonometric Functions

    Transforms of $\sin(at)$ and $\cos(at)$

  5. Chapter 5 Transform Table and How to Use It

    A list of key transforms and how to look up inverse transforms

  6. Chapter 6 Overview of Applications

    Applications to differential equations, circuit analysis, and control theory

Prerequisites

  • Basics of calculus (computing integrals)
  • Properties of exponential and logarithmic functions
  • Basics of complex numbers (helpful but not required)

Frequently Asked Questions

What is the Laplace transform?

The Laplace transform converts a function $f(t)$ ($t\geq 0$) into a complex function $F(s)=\int_0^\infty f(t)e^{-st}dt$. Because derivatives turn into algebraic expressions involving multiplication by $s$ together with the initial values ($\mathcal{L}\{f'(t)\}=sF(s)-f(0)$), ODEs can be handled as algebraic equations. It is closely related to the Fourier transform, but with this one-sided definition substituting $s=j\omega$ gives $\int_0^\infty f(t)e^{-j\omega t}dt$, which is the Fourier transform of the causal signal obtained by setting $f(t)=0$ for $t<0$ (when the imaginary axis lies in the region of convergence). For the two-sided Laplace transform, whose integral runs from $-\infty$, the Fourier transform is obtained directly as the values on the imaginary axis $s=j\omega$.

What are the main applications of the Laplace transform?

Control engineering (transfer functions $H(s)=Y(s)/U(s)$, Nyquist stability, frequency response), signal processing (filter analysis), electric circuits (RLC transient response, impedance), and systematic solution of initial value problems for ODEs. It provides a unified framework for linear time-invariant (LTI) system analysis.

How does the Laplace transform differ from the Fourier transform?

The Fourier transform $\hat{f}(\omega)=\int_{-\infty}^\infty f(t)e^{-j\omega t}dt$ is, in its elementary integral form, introduced for functions that are integrable over the whole real line (it extends much further — to $L^2$ in the sense of Plancherel, and to distributions — but that is beyond this introduction). The Laplace transform $F(s)=\int_0^\infty f(t)e^{-st}dt$ includes an exponential damping factor $e^{-\sigma t}$ (where $s=\sigma+j\omega$), making it applicable to functions that grow exponentially and to causal signals starting at $t=0$.

How is the Laplace transform related to the z-transform?

The z-transform is the counterpart of the Laplace transform for discrete-time signals. If a continuous-time system is sampled with period $T$, the correspondence $z=e^{sT}$ appears, and the left half of the $s$-plane maps to the interior of the unit circle in the $z$-plane. This is why the continuous-time stability condition (poles in the left half-plane) becomes $|z|<1$ in discrete time.