Laplace Transform

The definition

The Laplace transform of a function f(t), defined for t ≥ 0, is the function F(s) given by:

F(s) = L{f(t)} = ∫₀^∞ f(t) · e^(−st) dt

Here s is a complex number, s = σ + iω. The factor e^(−st) damps the integrand, so the integral converges for any f that doesn't grow faster than some exponential. The set of s where the integral converges is called the region of convergence.

Read the integrand as a weighted sum: f(t) is multiplied by every complex exponential e^(−st), and the integral collects the result. The transform measures how much each exponential mode contributes to f. Pierre-Simon Laplace introduced it in the 1780s while studying probability; Oliver Heaviside re-engineered it a century later as an operational calculus for telegraph cables.

Why it converts derivatives to multiplications

The defining trick is integration by parts. For the derivative f'(t):

L{f'(t)} = ∫₀^∞ f'(t) e^(−st) dt
        = [f(t) e^(−st)]₀^∞ + s · ∫₀^∞ f(t) e^(−st) dt
        = −f(0) + s · F(s)
        = sF(s) − f(0)

Differentiation in t becomes multiplication by s, plus a correction term carrying the initial condition. Apply twice:

L{f''(t)} = s²F(s) − s · f(0) − f'(0)

An ODE that involves f, f', f'' becomes a polynomial in s — algebra. The initial conditions are absorbed automatically rather than tacked on at the end.

Transform-pair table

f(t), t ≥ 0	F(s) = L{f(t)}	Region of convergence
1 (unit step u(t))	1/s	Re(s) > 0
t	1/s²	Re(s) > 0
tⁿ (n ≥ 0 integer)	n! / s^(n+1)	Re(s) > 0
e^(at)	1/(s − a)	Re(s) > a
sin(ωt)	ω/(s² + ω²)	Re(s) > 0
cos(ωt)	s/(s² + ω²)	Re(s) > 0
e^(at) sin(ωt)	ω/((s−a)² + ω²)	Re(s) > a
δ(t) (Dirac delta)	1	all s
u(t − a) (delayed step)	e^(−as) / s	Re(s) > 0

Two derivations worth doing once by hand:

L{e^(at)}. Compute directly:

F(s) = ∫₀^∞ e^(at) e^(−st) dt = ∫₀^∞ e^(−(s−a)t) dt = 1/(s − a)

The integral converges only if Re(s − a) > 0, i.e. Re(s) > a. The pole at s = a is the algebraic fingerprint of exponential growth at rate a.

L{sin(ωt)}. Use sin(ωt) = (e^(iωt) − e^(−iωt)) / (2i) and the previous result with a = ±iω:

L{sin(ωt)} = (1/(2i)) · [1/(s − iω) − 1/(s + iω)]
           = (1/(2i)) · [(s + iω) − (s − iω)] / (s² + ω²)
           = ω / (s² + ω²)

Solving an ODE — y'' + y = 0

Take a textbook problem: y'' + y = 0, y(0) = 1, y'(0) = 0. Take the Laplace transform of both sides:

L{y''} + L{y} = 0
[s² Y(s) − s · y(0) − y'(0)] + Y(s) = 0
[s² Y(s) − s · 1 − 0] + Y(s) = 0
(s² + 1) Y(s) = s
Y(s) = s / (s² + 1)

From the table, s/(s² + 1) is the transform of cos(t). So y(t) = cos(t), satisfying both initial conditions. No characteristic equation, no auxiliary work to fit constants — initial conditions enter the algebra directly.

The pattern generalises. Any linear ODE with constant coefficients becomes a rational function in s; partial-fraction expand it, then look each piece up in a table.

Laplace vs Fourier vs Z transforms

	Laplace	Fourier	Z-Transform
Domain	Continuous time, t ≥ 0	Continuous time, t ∈ ℝ	Discrete time, n ≥ 0 (or ℤ)
Variable	s = σ + iω, complex	ω, real frequency	z, complex
Definition	∫₀^∞ f(t) e^(−st) dt	∫_{−∞}^∞ f(t) e^(−iωt) dt	∑ x[n] z^(−n)
Handles initial conditions	Yes (built in)	No (assumes steady state)	Yes
Handles exponential growth	Yes (σ damps it)	No (purely oscillatory)	Yes (|z| > 1 in ROC)
Region of convergence	Right half-plane Re(s) > σ₀	The line σ = 0	Annulus r₁ < |z| < r₂
Typical use	Solving ODEs, transients	Steady-state spectra	Digital filters, sampled systems
Reduction	—	Laplace at s = iω	Laplace of sampled signal, with z = e^(sT)

The Fourier transform is the Laplace transform restricted to s = iω — it sees only the imaginary axis. Laplace adds the σ direction, so it can damp exponentially growing inputs and resolve transient responses that Fourier silently smears across all time. The Z-transform is the discrete cousin: replace integrals with sums, e^(−st) with z^(−n).

Operational properties

Property	Time domain	s-domain
Linearity	a f(t) + b g(t)	a F(s) + b G(s)
First derivative	f'(t)	s F(s) − f(0)
n-th derivative	f^(n)(t)	sⁿ F(s) − sⁿ⁻¹ f(0) − ... − f^(n−1)(0)
Integration	∫₀ᵗ f(τ) dτ	F(s) / s
Time shift	f(t − a) u(t − a)	e^(−as) F(s)
Frequency shift	e^(at) f(t)	F(s − a)
Scaling	f(at), a > 0	(1/a) F(s/a)
Convolution	(f ∗ g)(t) = ∫₀ᵗ f(τ) g(t−τ) dτ	F(s) · G(s)
Final value	lim_{t→∞} f(t)	lim_{s→0} s F(s) (if poles are stable)
Initial value	lim_{t→0⁺} f(t)	lim_{s→∞} s F(s)

The convolution-to-multiplication rule is the workhorse of control theory. A linear system's output is the convolution of its impulse response with the input; in the s-domain that becomes multiplication by the transfer function H(s) = Y(s)/U(s).

Inverting the transform

The formal inverse is a Bromwich contour integral:

f(t) = (1/(2πi)) ∫_{σ−i∞}^{σ+i∞} F(s) e^(st) ds

where σ is chosen to the right of every pole of F. In practice, you almost never compute this contour integral directly. Instead, use partial fractions to break F(s) into recognisable pieces. For example:

F(s) = (2s + 3) / (s² + 3s + 2)
     = (2s + 3) / [(s + 1)(s + 2)]
     = 1/(s + 1) + 1/(s + 2)
⇒ f(t) = e^(−t) + e^(−2t)

The poles s = −1 and s = −2 dictate the time-domain behaviour: each pole at s = a contributes a term e^(at). Real poles give pure exponentials, complex-conjugate pairs give damped sinusoids, repeated poles give terms like t · e^(at).

Where the Laplace transform shows up

• Electrical circuits. A resistor obeys V = RI; an inductor V = L · dI/dt, which becomes L · sI(s) − L · I(0) in the s-domain; a capacitor I = C · dV/dt, which becomes C · sV(s) − C · V(0). Kirchhoff's laws turn into algebraic equations in s. The classical RLC circuit's second-order ODE collapses to a quadratic in s, whose roots determine over-, under-, or critical damping.
• Control systems. Engineers design controllers by shaping pole-zero diagrams in the s-plane. A pole in the right half-plane (Re(s) > 0) means the system response grows exponentially — instability. The whole apparatus of root locus, Bode plots, and Nyquist criteria lives in the Laplace domain.
• Mechanical vibrations. Mass-spring-damper systems give equations m y'' + c y' + k y = F(t). The Laplace transform yields Y(s) = F(s) / (m s² + c s + k); the denominator is the system's characteristic polynomial, the same one a mechanical engineer reads to find natural frequency and damping ratio.
• Probability and queues. Probability-density transforms (moment-generating functions, characteristic functions) are essentially Laplace and Fourier transforms; they linearise convolution, and convolutions are how independent random variables sum.

Common mistakes

• Forgetting the initial-condition terms in derivatives. L{f'} = sF(s) − f(0), not sF(s). Drop the f(0) and you'll get a homogeneous solution that ignores the initial state.
• Treating the Laplace transform as if it knew about t < 0. The one-sided transform only sees t ≥ 0. f(t) for negative t simply doesn't exist in this framework — multiplied implicitly by the unit step u(t).
• Inverting without checking the region of convergence. The same algebraic F(s) can come from different f(t) depending on the ROC. For one-sided signals the rightmost pole bounds the ROC, but for bilateral signals you must specify it.
• Using L{f · g} = L{f} · L{g}. No. Convolution maps to multiplication; pointwise multiplication maps to a complicated convolution in the s-domain.
• Misreading the time-shift property. L{f(t − a)} = e^(−as) F(s) is only true if you also multiply by u(t − a) to enforce f = 0 for t < a. Otherwise you're transforming a different function.