How does one go about creating simple approximations of curves of known values (with tolerances)?
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I'm currently working on a synth, and many of the knobs are not linear in nature. Often, knobs will have a simple algorithm by which they transform linear changes into curved ones (such as volume or frequency). In other cases, the curve is a bit more arbitrary, and this is one of those cases.
I know what values I want to hit (roughly), and have a pretty large degree of relative tolerance for some of the values. Here are the values and their tolerances:
(0, 1) //no tolerance
(0.25, 1.75194 +- .00001)
(0.5, 3.5 +- .5)
(0.75, 11 +- 1)
(1, 500 +- 100)
They end up creating a pretty hard curve, and I've been attempting to match it by plugging values into a somewhat similar curve equation:
I know there must be a better way, but I have no idea how to solve a multiple variable equation for multiple known values which each have tolerances. Further, it has to (in my case) be computationally simple.
How can one create an simple equation to approximate a curve of known values which each have tolerances?
So far my steps are:
- Find a simple equation that's already fairly similar.
- Graph the points and equation
- Start messing with the increment rate of x, and any other variables and hope you get close.
But I'm certain there are much better approaches.
Edit: for anyone interested, here's where the values of y and their tolerances were derived from. It is an equation which can go from a triangle wave, to roughly a sine wave, to roughly a square wave. I think it's pretty neat!
approximation curves
asked Dec 14 '18 at 14:47
Seph ReedSeph Reed
1187
$endgroup$
add a comment |
$begingroup$
I'm currently working on a synth, and many of the knobs are not linear in nature. Often, knobs will have a simple algorithm by which they transform linear changes into curved ones (such as volume or frequency). In other cases, the curve is a bit more arbitrary, and this is one of those cases.
I know what values I want to hit (roughly), and have a pretty large degree of relative tolerance for some of the values. Here are the values and their tolerances:
(0, 1) //no tolerance
(0.25, 1.75194 +- .00001)
(0.5, 3.5 +- .5)
(0.75, 11 +- 1)
(1, 500 +- 100)
They end up creating a pretty hard curve, and I've been attempting to match it by plugging values into a somewhat similar curve equation:
I know there must be a better way, but I have no idea how to solve a multiple variable equation for multiple known values which each have tolerances. Further, it has to (in my case) be computationally simple.
How can one create an simple equation to approximate a curve of known values which each have tolerances?
So far my steps are:
- Find a simple equation that's already fairly similar.
- Graph the points and equation
- Start messing with the increment rate of x, and any other variables and hope you get close.
But I'm certain there are much better approaches.
Edit: for anyone interested, here's where the values of y and their tolerances were derived from. It is an equation which can go from a triangle wave, to roughly a sine wave, to roughly a square wave. I think it's pretty neat!
approximation curves
asked Dec 14 '18 at 14:47
Seph ReedSeph Reed
1187
$endgroup$
1
$begingroup$
Looks kinda like an exponential. Why not try fitting an exponential in Excel and see what $R^2$ value you get? If you're feeling particularly ambitious, plot the residuals to make sure you're not missing anything.
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– Adrian Keister
Dec 14 '18 at 14:49
$begingroup$
In terms of the amount computation needed to calculate the value for any particular $x$, a cubic spline would probably be faster than exponential, logarithmic, trig, or even taking of roots. Though it would look more complicated to you since the computer is already programmed to do those "behind the scene". On the other hand, a cubic splne may give you some unwanted behavior, so it may not be the best choice.
$endgroup$
– Paul Sinclair
Dec 14 '18 at 23:37
$begingroup$
I actually managed to find a very suitable equation:(log((1-x)/2)/log(1-((1-x)/2))^.762207. I'll have to take some time to explain how it came about later though.
$endgroup$
– Seph Reed
Dec 15 '18 at 0:50
add a comment |
$begingroup$
I'm currently working on a synth, and many of the knobs are not linear in nature. Often, knobs will have a simple algorithm by which they transform linear changes into curved ones (such as volume or frequency). In other cases, the curve is a bit more arbitrary, and this is one of those cases.
I know what values I want to hit (roughly), and have a pretty large degree of relative tolerance for some of the values. Here are the values and their tolerances:
(0, 1) //no tolerance
(0.25, 1.75194 +- .00001)
(0.5, 3.5 +- .5)
(0.75, 11 +- 1)
(1, 500 +- 100)
They end up creating a pretty hard curve, and I've been attempting to match it by plugging values into a somewhat similar curve equation:
I know there must be a better way, but I have no idea how to solve a multiple variable equation for multiple known values which each have tolerances. Further, it has to (in my case) be computationally simple.
How can one create an simple equation to approximate a curve of known values which each have tolerances?
So far my steps are:
- Find a simple equation that's already fairly similar.
- Graph the points and equation
- Start messing with the increment rate of x, and any other variables and hope you get close.
But I'm certain there are much better approaches.
Edit: for anyone interested, here's where the values of y and their tolerances were derived from. It is an equation which can go from a triangle wave, to roughly a sine wave, to roughly a square wave. I think it's pretty neat!
approximation curves
asked Dec 14 '18 at 14:47
Seph ReedSeph Reed
1187
$endgroup$
I'm currently working on a synth, and many of the knobs are not linear in nature. Often, knobs will have a simple algorithm by which they transform linear changes into curved ones (such as volume or frequency). In other cases, the curve is a bit more arbitrary, and this is one of those cases.
I know what values I want to hit (roughly), and have a pretty large degree of relative tolerance for some of the values. Here are the values and their tolerances:
(0, 1) //no tolerance
(0.25, 1.75194 +- .00001)
(0.5, 3.5 +- .5)
(0.75, 11 +- 1)
(1, 500 +- 100)
They end up creating a pretty hard curve, and I've been attempting to match it by plugging values into a somewhat similar curve equation:
I know there must be a better way, but I have no idea how to solve a multiple variable equation for multiple known values which each have tolerances. Further, it has to (in my case) be computationally simple.
How can one create an simple equation to approximate a curve of known values which each have tolerances?
So far my steps are:
- Find a simple equation that's already fairly similar.
- Graph the points and equation
- Start messing with the increment rate of x, and any other variables and hope you get close.
But I'm certain there are much better approaches.
Edit: for anyone interested, here's where the values of y and their tolerances were derived from. It is an equation which can go from a triangle wave, to roughly a sine wave, to roughly a square wave. I think it's pretty neat!
approximation curves
approximation curves
asked Dec 14 '18 at 14:47
Seph ReedSeph Reed
1187
asked Dec 14 '18 at 14:47
Seph ReedSeph Reed
1187
edited Dec 14 '18 at 14:52
Seph Reed
asked Dec 14 '18 at 14:47
Seph ReedSeph Reed
1187
asked Dec 14 '18 at 14:47
Seph ReedSeph Reed
1187
asked Dec 14 '18 at 14:47
Seph ReedSeph Reed
1187
1187
1
$begingroup$
Looks kinda like an exponential. Why not try fitting an exponential in Excel and see what $R^2$ value you get? If you're feeling particularly ambitious, plot the residuals to make sure you're not missing anything.
$endgroup$
– Adrian Keister
Dec 14 '18 at 14:49
$begingroup$
In terms of the amount computation needed to calculate the value for any particular $x$, a cubic spline would probably be faster than exponential, logarithmic, trig, or even taking of roots. Though it would look more complicated to you since the computer is already programmed to do those "behind the scene". On the other hand, a cubic splne may give you some unwanted behavior, so it may not be the best choice.
$endgroup$
– Paul Sinclair
Dec 14 '18 at 23:37
$begingroup$
I actually managed to find a very suitable equation:(log((1-x)/2)/log(1-((1-x)/2))^.762207. I'll have to take some time to explain how it came about later though.
$endgroup$
– Seph Reed
Dec 15 '18 at 0:50
add a comment |
1
$begingroup$
Looks kinda like an exponential. Why not try fitting an exponential in Excel and see what $R^2$ value you get? If you're feeling particularly ambitious, plot the residuals to make sure you're not missing anything.
$endgroup$
– Adrian Keister
Dec 14 '18 at 14:49
$begingroup$
In terms of the amount computation needed to calculate the value for any particular $x$, a cubic spline would probably be faster than exponential, logarithmic, trig, or even taking of roots. Though it would look more complicated to you since the computer is already programmed to do those "behind the scene". On the other hand, a cubic splne may give you some unwanted behavior, so it may not be the best choice.
$endgroup$
– Paul Sinclair
Dec 14 '18 at 23:37
$begingroup$
I actually managed to find a very suitable equation:(log((1-x)/2)/log(1-((1-x)/2))^.762207. I'll have to take some time to explain how it came about later though.
$endgroup$
– Seph Reed
Dec 15 '18 at 0:50
1
1
$begingroup$
Looks kinda like an exponential. Why not try fitting an exponential in Excel and see what $R^2$ value you get? If you're feeling particularly ambitious, plot the residuals to make sure you're not missing anything.
$endgroup$
– Adrian Keister
Dec 14 '18 at 14:49
$begingroup$
Looks kinda like an exponential. Why not try fitting an exponential in Excel and see what $R^2$ value you get? If you're feeling particularly ambitious, plot the residuals to make sure you're not missing anything.
$endgroup$
– Adrian Keister
Dec 14 '18 at 14:49
$begingroup$
In terms of the amount computation needed to calculate the value for any particular $x$, a cubic spline would probably be faster than exponential, logarithmic, trig, or even taking of roots. Though it would look more complicated to you since the computer is already programmed to do those "behind the scene". On the other hand, a cubic splne may give you some unwanted behavior, so it may not be the best choice.
$endgroup$
– Paul Sinclair
Dec 14 '18 at 23:37
$begingroup$
In terms of the amount computation needed to calculate the value for any particular $x$, a cubic spline would probably be faster than exponential, logarithmic, trig, or even taking of roots. Though it would look more complicated to you since the computer is already programmed to do those "behind the scene". On the other hand, a cubic splne may give you some unwanted behavior, so it may not be the best choice.
$endgroup$
– Paul Sinclair
Dec 14 '18 at 23:37
$begingroup$
I actually managed to find a very suitable equation:
(log((1-x)/2)/log(1-((1-x)/2))^.762207. I'll have to take some time to explain how it came about later though.$endgroup$
– Seph Reed
Dec 15 '18 at 0:50
$begingroup$
I actually managed to find a very suitable equation:
(log((1-x)/2)/log(1-((1-x)/2))^.762207. I'll have to take some time to explain how it came about later though.$endgroup$
– Seph Reed
Dec 15 '18 at 0:50
add a comment |
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$begingroup$
Looks kinda like an exponential. Why not try fitting an exponential in Excel and see what $R^2$ value you get? If you're feeling particularly ambitious, plot the residuals to make sure you're not missing anything.
$endgroup$
– Adrian Keister
Dec 14 '18 at 14:49
$begingroup$
In terms of the amount computation needed to calculate the value for any particular $x$, a cubic spline would probably be faster than exponential, logarithmic, trig, or even taking of roots. Though it would look more complicated to you since the computer is already programmed to do those "behind the scene". On the other hand, a cubic splne may give you some unwanted behavior, so it may not be the best choice.
$endgroup$
– Paul Sinclair
Dec 14 '18 at 23:37
$begingroup$
I actually managed to find a very suitable equation:
(log((1-x)/2)/log(1-((1-x)/2))^.762207. I'll have to take some time to explain how it came about later though.$endgroup$
– Seph Reed
Dec 15 '18 at 0:50