Chapter 1: What Is the Laplace Transform?

Learn the basics of the mathematical operation that converts a function in the time domain into a function in the complex domain.

1.1 Motivation for the Laplace Transform

Solving differential equations is extremely important in physics and engineering. However, solving differential equations directly is often quite difficult.

A Typical Problem

Consider the following differential equation:

$$\frac{d^2y}{dt^2} + 3\frac{dy}{dt} + 2y = e^{-t}$$

Solving this under the initial conditions $y(0) = 1$, $y'(0) = 0$ requires cumbersome computation by traditional methods.

The power of the Laplace transform lies in the fact that it converts a differential equation into an algebraic equation. This replaces complicated differentiation with simple multiplication.

Differential equation (hard to solve) Algebraic equation (easy to solve) Time-domain solution y(t) s-domain solution Y(s) Laplace transform Solve algebraically Inverse Laplace transform Solve directly (difficult)
Figure 1: Solution route via the Laplace transform. Instead of solving in the time domain directly, we move to the s-domain, solve algebraically, and return via the inverse transform.

As this diagram shows, using the Laplace transform takes a "detour" that actually makes the problem simpler. This is analogous to using logarithms to convert multiplication into addition.

1.2 Definition of the Laplace Transform

Definition: Laplace Transform

For a function $f(t)$ defined on $t \geq 0$, the Laplace transform $\mathcal{L}\{f(t)\}$ is defined by the following integral:

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

Here $s$ is a complex-valued parameter, and $F(s)$ is defined on the region where this integral converges.

Notation summary:

  • $f(t)$: the original function (time domain)
  • $F(s)$: the transformed function ($s$-domain, or complex domain)
  • $\mathcal{L}$: the Laplace transform operator
  • $e^{-st}$: the kernel of the transform

Meaning of the Integral

The Laplace transform is a "weighted integral". We multiply the function $f(t)$ by the damping factor $e^{-st}$ and integrate from $t = 0$ to $t = \infty$. Thanks to this damping factor, the integral converges for many functions.

Example 1.1: Laplace Transform of a Constant Function

We compute the Laplace transform of $f(t) = 1$.

$$\begin{align} \mathcal{L}\{1\} &= \int_0^{\infty} e^{-st} \, dt \\ &= \left[ -\frac{1}{s} e^{-st} \right]_0^{\infty} \\ &= 0 + \frac{1}{s} = \frac{1}{s} \end{align}$$

Here we used the fact that $\lim_{t \to \infty} e^{-st} = 0$ when $\text{Re}(s) > 0$.

Theorem: Linearity of the Laplace Transform

For any constants $a$ and $b$, the following holds:

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

Proof

Computing directly from the definition:

$$\mathcal{L}\{af(t) + bg(t)\} = a \int_0^{\infty} f(t) e^{-st} \, dt + b \int_0^{\infty} g(t) e^{-st} \, dt = a\mathcal{L}\{f(t)\} + b\mathcal{L}\{g(t)\}$$

This follows directly from the linearity of the integral. $\square$

1.3 Intuitive Meaning

The Laplace transform is a tool for analyzing a function from the perspective of "frequency and decay".

Time domain f(t) t f s-domain F(s) Re(s) Im(s) Pole Intuitive interpretation of the Laplace transform 1. Extracts decaying components 2. Separates frequency components 3. Pole locations determine system characteristics 4. Differentiation becomes multiplication by s
Figure 2: Comparison between the time-domain $f(t)$ and the $s$-domain $F(s)$. The ✕ marks on the right indicate poles (zeros are conventionally drawn as ○).

Relation to the Fourier Transform

The Laplace transform and the Fourier transform are closely related. Setting $s = j\omega$ (where $j$ is the imaginary unit) reduces the Laplace transform to the Fourier transform. The Laplace transform can be viewed as an extension of the Fourier transform.

Key Points

  • The Laplace transform is a one-sided transform (we consider only $t \geq 0$)
  • It is well suited for solving initial value problems of differential equations
  • $s$ is in general complex, but in many cases treating it as real causes no issues

1.4 Historical Background

The Laplace transform is named after the French mathematician Pierre-Simon Laplace (1749-1827).

Historical Timeline

  • 1780s: Laplace used similar transforms in his work on probability theory
  • 1880s: Oliver Heaviside introduced the operational calculus for analyzing electrical circuits
  • 1920s: Bromwich and Carson established a rigorous mathematical foundation
  • Present day: Widely used in control engineering, signal processing, and physics

Originally a tool of pure mathematics, the Laplace transform came to play a revolutionary role in engineering in the 20th century. It has become an indispensable tool, especially for analyzing transient phenomena in electrical circuits and for designing automatic control systems.

1.5 Exercises

Problem 1

Find the Laplace transform of $f(t) = 5$.

Solution

By linearity: $\mathcal{L}\{5\} = 5 \cdot \mathcal{L}\{1\} = \dfrac{5}{s}$ (for $\text{Re}(s) > 0$).

Problem 2

Find the condition on $s$ for $\displaystyle\int_0^{\infty} e^{2t} \cdot e^{-st} \, dt$ to converge.

Solution

Since $e^{2t} \cdot e^{-st} = e^{(2-s)t}$, the integral converges as $t \to \infty$ iff the real part of the exponent is negative, i.e., $\text{Re}(2 - s) < 0$, giving $\text{Re}(s) > 2$.

Problem 3

Given $\mathcal{L}\{f(t)\} = F(s)$ and $\mathcal{L}\{g(t)\} = G(s)$, find $\mathcal{L}\{3f(t) - 2g(t)\}$.

Solution

By linearity: $\mathcal{L}\{3f(t) - 2g(t)\} = 3F(s) - 2G(s)$.