Funs

a rational function oscillator

The short story

Funs’ outputs two related waves. The wave shapes is determined by three parameters: a, b and c.

On the display on the panel the wave shapes are shown (wave 1 in yellow, wave 2 in red). On each wave 7 points are marked. The left- and rightmost points are fixed: these are a zero-point of the wave.

Parameter c determines the other zero-point.

Parameter a takes values between 0 and c and determines the position of the maximum of the wave.

Parameter b takes values between a and c determines the position where the wave equals ½√2 (in proportion to the maximum).

If the knobs for these three parameters are all at noon, the resulting waves are a good approximation of a sine.

Other features

The FM jack provides linear through-0 FM.

The pitch knob can be quantized in octaves or semitones via the context menu. This only affects the knob, not the V/octave input.

Funs can work with polyphony. The number of channels is determined be the number of channels coming in at the V/oct jack. (The two waves are flipped for even numbered channels.)

To mittigate the worst aliasing, the tree parameters are restricted in their range. The range is bigger for lower frequencies and higher sample rates. There is an option in the context menu to bypass these restricions. This can result in harsh sounding wave shapes because of aliasing.

The pitch knob can be quantized in octaves or semitones. This only affects the knob, not the V/octave input. The FM jack provides linear through-0 FM.

The long story

We start with a function g : [0, ½] → [0, 1]. This is going to be half a wave. We impose the following five conditions on it:

g(0) = 0

g(½) = 0

g(a) = 1

g’(a) = 0

g(b) = ½√2

where a and b are parameters which are going to determine the shape of out wave. In particular, a determines the maximum of the wave. Note that g(x) = sin(2πx) is a solution for the case a = ¼ and b = ⅜.

We take g to be a rational function though, namely one of numerator and denominator degree 2, i.e. it’s of the form

g(x) =  

p₀ + p₁x + p₂x²

q₀ + q₁x + q₂x²

The solution is the function

g(x) =  

x(2x – 1)(a – b)²

a²[2(√2 – 1)b² – (√2 – 1)b + x(2x – 1)] + bx[–2a(2(√2 – 1)b + 2x – √2) + b(2√2x – 1) – (√2 – 1)x]

To make this into a full wave function g̅, we combine two half waves:

g̅(x) =  

g(x)	for 0 ≤ x < ½
– g(1 – x)	for ½ ≤ x < 1

Then we introduce a second function, h : [0, 1] → [0, 1]. This one is going to be our phase distortion, i.e.: the final wave is going to be something like f(x) = g̅(h(x)). We impose the following four conditions on this one:

h(0) = 0

h(1) = 1

h’(0) = g’(1)

h(c) = ½

The third condition makes sure there are no cusps in h. The parameter c determines the shape the distortion. Since g̅ crosses 0 at x = ½, c determines the zero-crossing of our final wave f. For c = ½ we are going to have no distortion, i.e. h(x) = x.

We take h to be a rational function as well, namely of the form

h(x) =  

p₀ + p₁x

q₀ + q₁x + q₂x²

Then we get two solutions:

h_±(x) = 

2(c³ – 2c² + c)x

x(2c³ – c(x – 3)) + c²(2x² – 7x + 1) ± √[c(4c³ – 12c² + 13c – 4)] (x – 1)(c – x)

This is why we end up with two wave outputs on our module. The final wave forms are then

f_±(x) = g̅(h_±(x))