Methods of approximating sine waves (given other sine waves)












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I'm trying to find a way to approximate sine waves using sine waves that I already have. I have sine waves that are define by:
$$f(n) = 2^{frac{n-49}{12}}times440$$
$$sin(2pi f(n)x), sin(2pi f(n+4)x)$$
What I have already tried is to just use linear interpolation on all the points like so:
$$sin(2pi f(n+c)x) approx frac{csin(2pi f(n)x)+(4-c)sin(2pi f(n+4)x)}{4}$$
I only need to approximate it for one period, but this is still only a mediocre approximation at best. I can only use basic arithmetic, and there is no way for me to calculate the actual values for $f(n)$ and $sin(n)$ on the fly, because I am working with a machine that does not implement floating point arithmetic. (To represent the output values of the sine wave, I'm just mapping it from -1 to 1 to whole numbers in the range 0 to 255). I want to precompute the values of the sine waves for certain n-values to conserve space, but I don't want to do it for all possible n-values (there would be 88 in total). Is there a better approximation, given these limitations?










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  • Why not just scale $x$ by $f(n+c)/f(n)$ and be done with it? You never said what operations are allowed.
    – Andy Walls
    Nov 26 at 20:29










  • I tried explaining better the limitations, I hope its more clear now. Sorry.
    – Kristbp2
    Nov 26 at 20:41
















0














I'm trying to find a way to approximate sine waves using sine waves that I already have. I have sine waves that are define by:
$$f(n) = 2^{frac{n-49}{12}}times440$$
$$sin(2pi f(n)x), sin(2pi f(n+4)x)$$
What I have already tried is to just use linear interpolation on all the points like so:
$$sin(2pi f(n+c)x) approx frac{csin(2pi f(n)x)+(4-c)sin(2pi f(n+4)x)}{4}$$
I only need to approximate it for one period, but this is still only a mediocre approximation at best. I can only use basic arithmetic, and there is no way for me to calculate the actual values for $f(n)$ and $sin(n)$ on the fly, because I am working with a machine that does not implement floating point arithmetic. (To represent the output values of the sine wave, I'm just mapping it from -1 to 1 to whole numbers in the range 0 to 255). I want to precompute the values of the sine waves for certain n-values to conserve space, but I don't want to do it for all possible n-values (there would be 88 in total). Is there a better approximation, given these limitations?










share|cite|improve this question
























  • Why not just scale $x$ by $f(n+c)/f(n)$ and be done with it? You never said what operations are allowed.
    – Andy Walls
    Nov 26 at 20:29










  • I tried explaining better the limitations, I hope its more clear now. Sorry.
    – Kristbp2
    Nov 26 at 20:41














0












0








0







I'm trying to find a way to approximate sine waves using sine waves that I already have. I have sine waves that are define by:
$$f(n) = 2^{frac{n-49}{12}}times440$$
$$sin(2pi f(n)x), sin(2pi f(n+4)x)$$
What I have already tried is to just use linear interpolation on all the points like so:
$$sin(2pi f(n+c)x) approx frac{csin(2pi f(n)x)+(4-c)sin(2pi f(n+4)x)}{4}$$
I only need to approximate it for one period, but this is still only a mediocre approximation at best. I can only use basic arithmetic, and there is no way for me to calculate the actual values for $f(n)$ and $sin(n)$ on the fly, because I am working with a machine that does not implement floating point arithmetic. (To represent the output values of the sine wave, I'm just mapping it from -1 to 1 to whole numbers in the range 0 to 255). I want to precompute the values of the sine waves for certain n-values to conserve space, but I don't want to do it for all possible n-values (there would be 88 in total). Is there a better approximation, given these limitations?










share|cite|improve this question















I'm trying to find a way to approximate sine waves using sine waves that I already have. I have sine waves that are define by:
$$f(n) = 2^{frac{n-49}{12}}times440$$
$$sin(2pi f(n)x), sin(2pi f(n+4)x)$$
What I have already tried is to just use linear interpolation on all the points like so:
$$sin(2pi f(n+c)x) approx frac{csin(2pi f(n)x)+(4-c)sin(2pi f(n+4)x)}{4}$$
I only need to approximate it for one period, but this is still only a mediocre approximation at best. I can only use basic arithmetic, and there is no way for me to calculate the actual values for $f(n)$ and $sin(n)$ on the fly, because I am working with a machine that does not implement floating point arithmetic. (To represent the output values of the sine wave, I'm just mapping it from -1 to 1 to whole numbers in the range 0 to 255). I want to precompute the values of the sine waves for certain n-values to conserve space, but I don't want to do it for all possible n-values (there would be 88 in total). Is there a better approximation, given these limitations?







trigonometry approximation-theory






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edited Nov 26 at 20:41

























asked Nov 26 at 19:17









Kristbp2

11




11












  • Why not just scale $x$ by $f(n+c)/f(n)$ and be done with it? You never said what operations are allowed.
    – Andy Walls
    Nov 26 at 20:29










  • I tried explaining better the limitations, I hope its more clear now. Sorry.
    – Kristbp2
    Nov 26 at 20:41


















  • Why not just scale $x$ by $f(n+c)/f(n)$ and be done with it? You never said what operations are allowed.
    – Andy Walls
    Nov 26 at 20:29










  • I tried explaining better the limitations, I hope its more clear now. Sorry.
    – Kristbp2
    Nov 26 at 20:41
















Why not just scale $x$ by $f(n+c)/f(n)$ and be done with it? You never said what operations are allowed.
– Andy Walls
Nov 26 at 20:29




Why not just scale $x$ by $f(n+c)/f(n)$ and be done with it? You never said what operations are allowed.
– Andy Walls
Nov 26 at 20:29












I tried explaining better the limitations, I hope its more clear now. Sorry.
– Kristbp2
Nov 26 at 20:41




I tried explaining better the limitations, I hope its more clear now. Sorry.
– Kristbp2
Nov 26 at 20:41















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