Revisions to why does the additive synthesis method for a triangle wave require amplitude scaling by 8/pi^2?

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edited Jul 10, 2023 at 12:51

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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 See here.

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

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.

See here.

Converted equations to MathJax

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edited Jul 10, 2023 at 12:43

lennon310

3.6k
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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

Converted equations to MathJax

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edit approved Jul 10, 2023 at 12:43

ZaellixA

1.5k
2
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givenGiven 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...


1/(1*1)
1/(5*5)
1/(9*9)
1/(13*13)
1/(17*17)
1/(21*21)
1/(25*25)

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

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

butBut as it seems the inverted/phaseshifted-for-pi$\pi$ 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 ~1.22 $\sim 1.22$ - which is about the inverse of 8pi squared$8 \pi^{2}$, or ~0.82$ \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

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...


1/(1*1)
1/(5*5)
1/(9*9)
1/(13*13)
1/(17*17)
1/(21*21)
1/(25*25)

... you could think that the amplitude of the sum should only reach around ~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 ~1.22 - which is about the inverse of 8pi squared, or ~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

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