[sdiy] Better waveforms of our nature

[Donald Tillman]

> On Oct 18, 2016, at 6:29 AM, Matthias Puech <matthias.puech at gmail.com> wrote:
> 
> Thank you Donald for the interesting read! 
> I've been wondering: is there a known relationship between an arbitrary waveform's spectrum and its integral/derivative's? Are all integral waveforms "mellow" versions of their derivatives? 
> 
> I've asked recently on another DSP mailing list but did not get any answer. It might be very simple maths...

(Yet another reason we're better than the DSP mailing list!)

Other's have answered well, buy here is my take on it...

In math, the derivative of a function is the slope as you go along the curve, and the integral of a function is the area underneath as you go along the curve.  These are reciprocal operations; the derivative-of-an-integral-of-a-derivative is the function itself.

In my chart of waveforms, the mellow waveforms are the integrals of the bright waveforms, and the bright waveforms are the derivatives of the mellow waveforms, *except* that the mellow waveforms have been moved over a quarter cycle so that the fundamental sine component is in phase with the bright waveforms.

And you can check it out; the constant positive and negative slopes on the triangle wave correspond to the flat values on the square wave.  And the area under the constant slope of the sawtooth wave corresponds to the x^2 shape of the parabolic wave.

This isn't necessarily the case; you can set up the phases of the harmonics so the waveforms look different and the derivative/integral relationship doesn't hold.

As far as spectrum goes, a derivative means a high frequency tilt up growing at +6dB/oct.  And an integral means a high frequency tilt down at -6dB/oct.  And you can see that in the spectrum displays.

  -- Don
--
Don Tillman
Palo Alto, California
don at till.com
http://www.till.com