# Gaussian Blur

Wave · y = f(x, t)

`y = ∫₋₂² sign(sin(3(x − u) − t)) · exp(−4u²) du`

[Open in the app](https://www.wavelace.com/app#p=108) · [This page](https://www.wavelace.com/presets/gaussian-blur)

### What it draws

Inside the integral, `sign(sin(3·(x − u) − t))` is a square wave, +1 or −1 and nothing between, with wavenumber 3 and so a wavelength of `2π/3 ≈ 2.09`, shifted along by `u`. The other factor, `e^(−4u²)`, is a bell with standard deviation `1/√8 ≈ 0.354`. Integrating their product over `u` replaces the square wave at each `x` by its average under the bell centred there. That is a convolution, and a convolution with a bell is a Gaussian blur, the same operation an image editor applies to soften a picture. The limits `±2` cut the bell where it is already `e^(−16) ≈ 10⁻⁷` high, so they cost nothing.

The blur takes the corners off and the height down. Without it the square wave weighed by the bell would stand at `∫e^(−4u²) du = √π/2 ≈ 0.886`; the blurred crests reach about 0.65, just under three quarters of that, and the flat tops are gone. The `−t` inside the sine carries the whole pattern to the right, a crest advancing a third of a unit each second at `Speed` 1, so one wavelength passes in `2π ≈ 6.28` seconds.

### Why the corners go

A square wave is a sum of odd harmonics, `(4/π)·(sin v + sin 3v/3 + sin 5v/5 + …)`, with `v = 3·(x − u) − t` here. So the harmonics have wavenumbers 3, 9, 15 and up. Blurring with a Gaussian multiplies each harmonic by a factor of its own, `e^(−w²/16)` for wavenumber `w`, because the transform of a bell is another bell. The fundamental keeps `e^(−9/16) ≈ 0.57` of its height. The third harmonic keeps `e^(−81/16)`, under 1%, and the rest are gone. What survives is very nearly the fundamental alone, `(4/π)·0.886·0.57 ≈ 0.64` high, which is why the blurred square wave looks like a sine. The sharp edges were the high harmonics; the blur is a low-pass filter, and they are what it removes.

### Try

- Blur a single edge instead, `integral(sign(x − u)·exp(−4u²), u, −2, 2)`: the step becomes a smooth ramp, the error function, climbing from `−0.886` to `0.886` and already at 0.75 half a unit from the edge.
- Slow the wave down, `integral(sign(sin((x − u) − t))·exp(−4u²), u, −2, 2)`: with a wavelength of `2π ≈ 6.28` the same bell only rounds the corners, and the flat top at 0.886 shows. The crests now travel a unit a second.
- Widen the bell, `integral(sign(sin(3·(x − u) − t))·exp(−u²), u, −2, 2)`: the standard deviation doubles to 0.707, the fundamental keeps only `e^(−9/4) ≈ 0.11` of its height, and the wave all but vanishes.
- Drop the blur, `0.886·sign(sin(3x − t))`: the square wave the integral started from, at the height the bell's weight gives it, with the vertical edges the blur removed.

### Read more

- [Gaussian blur](https://en.wikipedia.org/wiki/Gaussian_blur)
- [Convolution](https://en.wikipedia.org/wiki/Convolution)
- [Square wave](https://en.wikipedia.org/wiki/Square_wave)
- [Gaussian filter](https://en.wikipedia.org/wiki/Gaussian_filter)
- [Error function](https://en.wikipedia.org/wiki/Error_function)