# Laplace Transform

Surface · z = f(x, y, t)

`z = ∫₀^∞ sin(y · u + t) · exp(−x · u) du`

[Open in the app](https://www.wavelace.com/app#p=97) · [This page](https://www.wavelace.com/presets/laplace-transform)

### 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.

### Read more

- [Laplace transform](https://en.wikipedia.org/wiki/Laplace_transform)
- [Operational calculus](https://en.wikipedia.org/wiki/Operational_calculus)
- [Simpson's rule](https://en.wikipedia.org/wiki/Simpson%27s_rule)
- [Pierre-Simon Laplace](https://en.wikipedia.org/wiki/Pierre-Simon_Laplace)