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.

Introduction (high-school level)

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:

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

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 $e^{-st}$ and integrate from $t = 0$ to $t = \infty$. Writing $s = \sigma + j\omega$,

$$e^{-st} = e^{-\sigma t}\,e^{-j\omega t}$$

so $e^{-\sigma t}$ decays when $\sigma > 0$ and grows when $\sigma < 0$. Taking $\sigma$ large enough makes this weight a damping factor, and the integral then converges for many functions.

Example 1.1: Laplace Transform of a Constant Function

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

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

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 ○). Note that poles arise for meromorphic $F(s)$ and for system functions; an arbitrary $F(s)$ need not have isolated poles.

Relation to the Fourier Transform

The Laplace transform and the Fourier transform are closely related, but the one-sided transform $\int_0^{\infty} f(t)e^{-st}\,dt$ defined in this chapter gives, on setting $s = j\omega$ (where $j$ is the imaginary unit),

$$\int_0^{\infty} f(t)\,e^{-j\omega t}\,dt$$

which is not the ordinary (two-sided) Fourier transform $\int_{-\infty}^{\infty} f(t)e^{-j\omega t}\,dt$ itself. The two agree for a causal signal, i.e. when $f(t) = 0$ for $t < 0$, and 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 $s = j\omega$. In that sense 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
  • In elementary transform calculations one can often manipulate $s$ as if it were real, but $s$ is properly a complex variable; this becomes essential once poles, oscillation and stability are discussed

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
  • Late 19th to early 20th century: through Bromwich's inversion integral, Carson's engineering formulation and related work, the Laplace transform method took its present form
  • 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: $\mathcal{L}\{3f(t) - 2g(t)\} = 3F(s) - 2G(s)$.

Frequently Asked Questions

What kind of mathematical operation is the Laplace transform?

The Laplace transform is an integral transform that turns a time-domain function $f(t)$ into a function of the complex variable $s$, $F(s)=\int_0^{\infty}f(t)e^{-st}\,dt$. It turns differentiation into multiplication and a differential equation into an algebraic one, so initial value problems can be solved algebraically.

How does the Laplace transform differ from the Fourier transform?

The Fourier transform lives on the imaginary axis $s=i\omega$ and analyses the frequency content of a signal. The Laplace transform is defined for complex $s=\sigma+i\omega$, incorporates initial conditions naturally, and its region of convergence makes causal and stable systems convenient to handle.

Why do differential equations become easier with the Laplace transform?

The derivative $f'(t)$ becomes the algebraic expression $sF(s)-f(0)$. The differential equation $y'+ay=g(t)$ turns into the first-order equation $(s+a)Y(s)=G(s)+y(0)$; solving for $Y(s)=(G(s)+y(0))/(s+a)$ and inverting gives the solution $y(t)$.