Laplace transform

The Laplace transform is an integral operator that converts a function of time $f(t)$ into a function of a complex variable $s$. Used to solve linear ODEs by turning differential equations into algebraic equations.

$\mathcal{L}\{f\}(s)=F(s)={\int }_0^{\infty }{e}^{-st}f(t)dt$

The kernel $e^{-st}$ is a decaying exponential that “weights” the signal over time. As long as the integral converges (more on convergence below), $F(s)$ is well-defined.

When you apply the transform, you move from the time domain (functions of $t$) to the frequency domain (functions of $s$, also called the s-domain). The variable $s$ is generally complex, encoding both decay rate and oscillation frequency.

Why it solves ODEs

The core property: differentiation in time becomes multiplication by $s$ in the s-domain.

$\mathcal{L}\{f^{\prime }\}(s)=sF(s)-f(0)$

$\mathcal{L}\{f^{\prime \prime }\}(s)=s^{2}F(s)-sf(0)-f^{\prime }(0)$

A linear ODE with constant coefficients, like $a{y}^{\prime \prime }+b{y}^{\prime }+cy=g(t)$, transforms into:

$a[s^{2}Y-sy(0)-y^{\prime }(0)]+b[sY-y(0)]+cY=G(s)$

This is just an algebraic equation in $Y(s)$. Solve for $Y(s)$, then invert the transform to get $y(t)$:

$y(t)={\mathcal{L}}^{-1}\{Y(s)\}$

The Laplace transform reduces the ODE to algebra plus a table lookup for the inverse transform. See Method of Laplace transform for the full procedure.

Linearity

The Laplace transform is a linear operator: for any constants $c_1,c_2$ and functions $f,g$:

$\mathcal{L}\{c_{1}f+c_{2}g\}=c_1\mathcal{L}\{f\}+c_2\mathcal{L}\{g\}$

This is what lets you transform linear ODEs term by term.

Convergence

The integral ${\int }_0^{\infty }{e}^{-st}f(t)dt$ doesn’t always converge. The strict definition is

${\int }_0^{\infty }{e}^{-st}f(t)dt:={\lim }_{N\to \infty }{\int }_0^N{e}^{-st}f(t)dt$

For this limit to exist, we need conditions on $f$. The standard sufficient condition: $f$ is piecewise continuous and of exponential order — see Properties of Laplace transform.

If $f$ grows too fast (faster than any exponential), no $s$ makes the integral converge — the Laplace transform doesn’t exist for such $f$.

Common transforms

The most-used pairs:

$f(t)$	$F(s)=\mathcal{L}\{f\}$	Region of convergence
$1$	$1/s$	$\mathrm{Re}(s)>0$
$t^n$ ($n\in {\mathbb{Z}}_{\ge 0}$)	$n!/s^{n+1}$	$\mathrm{Re}(s)>0$
$e^{at}$	$1/(s-a)$	$\mathrm{Re}(s)>a$
$\sin (\omega t)$	$\omega /(s^2+{\omega }^2)$	$\mathrm{Re}(s)>0$
$\cos (\omega t)$	$s/(s^2+{\omega }^2)$	$\mathrm{Re}(s)>0$
$u_c(t)$ (Heaviside step function)	$e^{-cs}/s$	$\mathrm{Re}(s)>0$
$\delta (t-c)$ (Dirac delta function)	$e^{-cs}$	all $s$

The variable $s$ is complex; the convergence condition is on its real part. Writing "$s>0$" works when $s$ is real but is sloppy for the general case — convergence of ${\int }_0^{\infty }{e}^{-st}f(t)dt$ depends on the decay rate $\mathrm{Re}(s)$ of the kernel, not on $s$ itself. The imaginary part of $s$ contributes oscillation only and never affects whether the integral converges.

A larger table is in any DE textbook.

Example application

Compute $\mathcal{L}\{1\}$:

$\mathcal{L}\{1\}(s)={\int }_0^{\infty }{e}^{-st}dt={\left[-\frac{e^{-st}}{s}\right]}_0^{\infty }=\frac{1}{s}$

This works for $s>0$ (otherwise the integral diverges). For $s\le 0$, the Laplace transform of $1$ doesn’t exist.

Compute $\mathcal{L}\{e^{at}\}$:

$\mathcal{L}\{e^{at}\}(s)={\int }_0^{\infty }{e}^{-st}{e}^{at}dt={\int }_0^{\infty }{e}^{(a-s)t}dt={\left[\frac{e^{(a-s)t}}{a-s}\right]}_0^{\infty }=\frac{1}{s-a}$

For $s>a$. If $s<a$, the exponential blows up and the integral diverges.

When to use Laplace

The Laplace transform shines for:

1. IVPs — initial conditions are baked into the algebra naturally.
2. Discontinuous forcing — handles step functions, impulses cleanly via Heaviside step function and Dirac delta function.
3. Convolutions — products in the s-domain correspond to convolutions in time, and vice versa. See Convolution integral.
4. Systems — multiple coupled ODEs become a system of algebraic equations.
5. Transfer functions — input-output behavior of linear systems is encoded in $H(s)$. See Transfer function.

For the inverse direction, see Inverse Laplace transform. For the full solution procedure, see Method of Laplace transform.

Bilateral and complex frequency

The form above is the unilateral Laplace transform — appropriate for causal signals and IVPs. For signals defined on the whole real line, the bilateral form integrates from $-\infty$ to $\infty$ and requires tracking the region of convergence.

The variable $s=\sigma +j\omega$ lives in the complex s-plane; the imaginary part $\omega$ is the frequency from the Fourier transform, and the real part $\sigma$ is the damping factor that lets the transform converge for signals the Fourier integral can’t handle. Stable signals have $X(s)$ defined on a strip containing the imaginary axis, where $X(j\omega )=X(s){|}_{s=j\omega }$ recovers the Fourier transform.

For inversion, partial fractions are the standard technique — see Partial fraction decomposition. For the transfer-function view of LTI systems, see Transfer function and Pole and zero.