Hallo, Geoff hat gesagt: // Geoff wrote:
Just a few things I need to clarify
The frequency of the phasor is determined by the slope of the line,
by the phase increment k.If y=mx +k and wrapping this produces an oscillator, I understand
that adding k can be the phase increment i.e. what point the waveform
will start from,
The phase increment is not where the waveform starts but the amount it advances each time step. So the phase increment would be m in the equation y = mx + k. You can ignore k for phasor~-like signals, it always is zero.
"y = mx + k" is a linear equation - see http://en.wikipedia.org/wiki/Line_(geometry)
Does this mean that in this example the [*~ 3] object represents m
and therefore the frequency of the oscillator and you would add a
[+~ ] to control the phase increment?
The [*~ 3] is a frequency multiplier. If you multiply a phasor signal by x and then wrap it, you get a phasor signal with x times the base frequency. You can see this in the example in that the vline~ produces a burst of three spikes, although it itself is only doing a single ramp.
If you add anything to that signal with [+~] you don't change the frequency, only the offset from the x-axis i.e. the "DC offset" or the start of a sample in a phasor-driven [tabread4~]-sampler.
In practice, in code, we combine the integrator and wrap into a
single unit so that the line does not increase without bound. If it does then
the oscillator will work for a while, but then the line will exceed the
representation range of the CPU.Am I understanding you correctly in thinking that the integrator and
wrap and combined so that rather than leave the integrator to keep
counting higher and higher (unitl the processor can't cope) the wrap
function is somehow combined with it so that the integrator keeps
reseting itself so that the number it counts up to never gets that big?
In Andy's patches the wrap~ is not resetting the integrators, but in the source code for [phasor~] in Pd, the integrator indeed is reset.
Btw: If you're like me and are always having difficulties reading [biquad~] coefficients: In Andy's patch you could also use [rpole~ 1] instead of the biquad.
rpole calculates this:
y[n] = y[n-1] + a[n] * x[n]
so if a == 1 it calculates: y[n] = y[n-1] + x[n] which means, it constantly adds the new input to the old output to produce the next output which is integration.
Ciao
Frank Barknecht Do You RjDj.me? _ ______footils.org__