Laplace Transform
z = ∫₀^∞ sin(y · u + t) · exp(−x · u) du
Open in the app The dials and keys named below are the app's.
What it draws
The height is an integral, computed at every mesh point: z = ∫₀^∞ sin(y·u + t) · e^(−x·u) du. Read x as s and y as ω, the frequency of the signal f(u) = sin(ω·u + φ); the clock t is the phase φ. The sheet is the Laplace transform F(s) = ∫₀^∞ f(u) · e^(−s·u) du of a sinusoid, over every s and ω.
For s > 0 the integral has a closed form, (ω · cos φ + s · sin φ) / (s² + ω²). At φ = 0 that is ω / (s² + ω²), the transform of sin(ω·u). At φ = π/2 it is s / (s² + ω²), the transform of cos(ω·u). The surface rocks between the two every 2π ≈ 6.28 seconds at Speed 1.
Where it exists
The factor e^(−s·u) makes the integral finite. For s > 0 it kills the signal as u grows; for s ≤ 0 it does not, and the integral has no value. The app integrates numerically out to a fixed reach and leaves a point blank when the integrand has not died away by then. The left half of the sheet is a hole, with a cliff along s = 0, the abscissa of convergence, and the region of convergence is the shape of the mesh.
At s = ω = 0 the closed form has a pole, and the sheet spikes at the corner of the hole. The crest of ω / (s² + ω²) runs along the diagonal ω = s, with a trough along ω = −s, since the sine transform is odd in ω. The cosine transform is even, one ridge along ω = 0 falling off as 1/s; the rocking is one sheet giving way to the other.
History
Pierre-Simon Laplace used integrals of this form in his work on probability in the 1780s. Oliver Heaviside solved the equations of telegraph circuits in the 1890s by treating d/dt as a symbol, without proof. The transform, which turns a derivative into a product with s, made his operational calculus rigorous.
Try
- Top: the cliff is the line s = 0, and the crest and trough along the diagonals fade and return as the ridge along ω = 0 takes over.
- Damp the signal, integral(sin(y·u + t) · exp(−(x + 1)·u), u, 0, inf): the transform of e^(−u) · sin(ω·u + φ) converges for s > −1, so the cliff moves one unit to the left.
- Stop at the clock, integral(sin(y·u) · exp(−x·u), u, 0, t), then Rewind and Play: the right half settles onto the transform as the window grows, the left half runs off the plot.
- Swap the axes, integral(sin(x·u + t) · exp(−y·u), u, 0, inf): the hole is now the near half.