jeffrey@penn:~/blog$ glow laplace-transform.md

The Laplace transform

November 1, 2026 #signal-processing #math #tutorial #control

The Fourier transform fails for signals that grow, like $e^{t}u(t)$ . The Laplace transform fixes this by multiplying the signal by a decaying exponential $e^{-\sigma t}$ before transforming it.

Definition

X(s) = \mathcal{L}\{x(t)\} = \int_{-\infty}^{\infty} x(t)\, e^{-st}\, dt, \qquad s = \sigma + j\omega .

The complex variable $s$ carries both a decay rate $\sigma$ and a frequency $\omega$ . The set of $s$ where the integral converges is the region of convergence (ROC).

Example: a decaying exponential

For $x(t) = e^{-at}u(t)$ ,

X(s) = \int_{0}^{\infty} e^{-(s+a)t}\, dt = \frac{1}{s+a}, \qquad \operatorname{Re}(s) > -a .

The transform has a pole at $s = -a$ , and the ROC is everything to the right of it.

Relation to the Fourier transform

Setting $\sigma = 0$ gives back the Fourier transform:

X(\omega) = X(s)\big|_{s = j\omega},

as long as the $j\omega$ axis is inside the ROC. For a causal system this is why stability is decided by the poles: the system is stable exactly when every pole is in the left half-plane ( $\operatorname{Re}(s) < 0$ ).

Properties

Property	Time domain	$s$ domain
Linearity	$a x_1(t) + b x_2(t)$	$a X_1(s) + b X_2(s)$
Time shift	$x(t - t_0)$	$e^{-s t_0} X(s)$
Derivative	$\dot{x}(t)$	$s X(s) - x(0^-)$
Integral	$\int_{0^-}^{t} x(\tau)\, d\tau$	$X(s)/s$
Convolution	$(x * h)(t)$	$X(s) H(s)$

The derivative rule turns differential equations into algebra. For example $\dot{y} + a y = x$ with $y(0^-) = 0$ becomes $(s + a) Y(s) = X(s)$ , so the transfer function is $H(s) = \frac{1}{s + a}$ .

For sampled, discrete-time signals the same role is played by the Z transform.

---
title: "The Laplace transform"
date: 2026-11-01
tags: [signal-processing, math, tutorial, control]
---

The [Fourier transform](/blog/fourier-transform/) fails for signals that grow, like $e^{t}u(t)$. The Laplace transform fixes this by multiplying the signal by a decaying exponential $e^{-\sigma t}$ before transforming it.

## Definition

$$
X(s) = \mathcal{L}\{x(t)\} = \int_{-\infty}^{\infty} x(t)\, e^{-st}\, dt, \qquad s = \sigma + j\omega .
$$

The complex variable $s$ carries both a decay rate $\sigma$ and a frequency $\omega$. The set of $s$ where the integral converges is the **region of convergence** (ROC).

## Example: a decaying exponential

For $x(t) = e^{-at}u(t)$,

$$
X(s) = \int_{0}^{\infty} e^{-(s+a)t}\, dt = \frac{1}{s+a}, \qquad \operatorname{Re}(s) > -a .
$$

The transform has a **pole** at $s = -a$, and the ROC is everything to the right of it.

## Relation to the Fourier transform

Setting $\sigma = 0$ gives back the Fourier transform:

$$
X(\omega) = X(s)\big|_{s = j\omega},
$$

as long as the $j\omega$ axis is inside the ROC. For a causal system this is why stability is decided by the poles: the system is stable exactly when every pole is in the left half-plane ($\operatorname{Re}(s) < 0$).

## Properties

| Property | Time domain | $s$ domain |
|----------|-------------|------------|
| Linearity | $a x_1(t) + b x_2(t)$ | $a X_1(s) + b X_2(s)$ |
| Time shift | $x(t - t_0)$ | $e^{-s t_0} X(s)$ |
| Derivative | $\dot{x}(t)$ | $s X(s) - x(0^-)$ |
| Integral | $\int_{0^-}^{t} x(\tau)\, d\tau$ | $X(s)/s$ |
| Convolution | $(x * h)(t)$ | $X(s) H(s)$ |

The derivative rule turns differential equations into algebra. For example $\dot{y} + a y = x$ with $y(0^-) = 0$ becomes $(s + a) Y(s) = X(s)$, so the transfer function is $H(s) = \frac{1}{s + a}$.

For sampled, discrete-time signals the same role is played by the [Z transform](/blog/z-transform/).