Skip to main content
3 of 4
Converted equations to MathJax
lennon310
  • 3.6k
  • 19
  • 25
  • 27

Given that the max amplitude of the normalized cosine is known (-1 to 1) and given that the additively constructed triwave contains cosines with weighted max amplitudes appearing at the same time such as...

$$ \frac{1}{1 \cdot 1}, \frac{1}{5 \cdot 5}, \frac{1}{9 \cdot 9}, \frac{1}{13 \cdot 13}, \frac{1}{17 \cdot 17}, \frac{1}{21 \cdot 21}, \frac{1}{25 \cdot 25}, \ldots $$

You could think that the amplitude of the sum should only reach around $\sim 1.068$ and $0.81$ for correction is like $3$ times more.

But as it seems the inverted/phaseshifted-for-$\pi$ partials did also count in because if you include the third and the fifth partials in this game you end up with a max amplitude of $\sim 1.22$ - which is about the inverse of $8 \pi^{2}$, or $ \sim 0.82$.

so in other words, one can explain that using simple arithmetics already.

https://www.wolframalpha.com/input?i=1%2F%28+%281%2F%281*1%29%29%2B%281%2F%283*3%29%29%2B%281%2F%285*5%29%29%2B%281%2F%287*7%29%29%2B%281%2F%289*9%29%29%2B%281%2F%2811*11%29%29%2B%281%2F%2813*13%29%29%2B%281%2F%2815*15%29%29%2B%281%2F%2817*17%29%29%2B%281%2F%2819*19%29%29%2B%281%2F%2821*21%29%29%2B%281%2F%2823*23%29%29%2B%281%2F%2825*25%29%29%2B%281%2F%2827*27%29%29%2B%281%2F%2829*29%29%29+%29