Re: Finding a sine wave
- To: mathgroup at smc.vnet.net
- Subject: [mg95512] Re: Finding a sine wave
- From: dh <dh at metrohm.com>
- Date: Tue, 20 Jan 2009 05:48:56 -0500 (EST)
- References: <gkppq1$dmn$1@smc.vnet.net>
Hi Hug,
I think a direct method is much simpler than numerical iterations.
Your model is: c0 Sin[c1 t+c2]
You have 3 data points for 3 times. Therefore we can get rid of c0 by
division e.g. the first 2 data points by the last. This leaves 2 data
points and the model: Sin[c1 t+c2]. The data values must be between
-1..1, this shows that there is a constrain on the original values. Now
we take the ArcSin of the data values, this gives a model: c1 t + c2.
Now the fit is trivial. c0 can be obtained from the original data
values. Here is an example:
d0 = {0.45, 0.86};
t = {0.5, 1.2}
d = ArcSin /@ d0;
cs = {c1, c2} /.
Solve[{c1 t[[1]] + c2 == d[[1]], c1 t[[2]] + c2 == d[[2]]}, {c1,
c2}][[1]];
Plot[Sin[cs.{t, 1}], {t, 0, 2 Pi/cs[[1]]},
Prolog -> {Red, Point[Transpose[{t, d0}]]}]
hope this helps, Daniel
Hugh Goyder wrote:
> An experiment will give me three coordinates which lie on a sine wave.
> I have to find the sine wave efficiently. There are three unknowns the
> sine wave amplitude, A, the wavelength of the sine wave, L, and the
> phase, ph. I also know that the wavelength is larger than the interval
> containing my measurement points. I think this condition removes
> possible multiple solutions.
>
> Below I give two methods which need improving. The first method uses
> FindRoot. This methods works about 60% of the time. I give six
> examples where it fails. The failure may be due to poor initial
> guesses. In the second method I use FindFit. This is not quite the
> correct method because I have an equal number of equations and
> unknowns. Some of the failures here, I think, are due to there being
> no error for the algorithm to work with. I give examples of failures.
> I have also tried FindInstance, Reduce and NSolve but I don't think
> these are appropriate.
>
> Here are some questions
>
> 1. I would really like a symbolic solution rather than an iterative
> one. Is such a solution possible?
> 2. Can anyone improve on the methods below to make them more robust?
> 3. I have some control over my x-locations. How can I work out best x
> locations given an estimate of the wavelength L?
>
> Many thanks for all answers.
>
>
> xx = Sort[Join[{0}, RandomReal[{0, 10}, 2]]]
>
> yy = RandomReal[{-25, 25}, 3]
>
> sol = FindRoot[{yy[[1]] - A*Sin[ph],
> yy[[2]] - A*Sin[2*Pi*(xx[[2]]/L) + ph],
> yy[[3]] - A*Sin[2*Pi*(xx[[3]]/L) + ph]}, {{A,
> Max[Abs[yy]]}, {L, Max[xx]}, {ph, 0}}]
>
> Plot[Evaluate[A*Sin[2*Pi*(x/L) + ph] /. sol], {x, 0, 10},
> Epilog -> {PointSize[0.02], (Point[#1] & ) /@
> Transpose[{xx, yy}]}, PlotRange -> {{0, 10}, {-25, 25}}]
>
> FindRootFailures = {{{0, 3.781264462608982,
> 3.797055100562352}, {-22.948348087068737, 1.4581744078038472,
> -14.242676740704574}}, {{0, 4.424069570670131,
> 8.861716396743098}, {-6.444538773775843, -10.787309362608688,
> 15.579334094330942}}, {{0, 4.424069570670131,
> 8.861716396743098}, {-5.944536471563289, 10.627536107497393,
> -21.294232497202316}}, {{0, 1.3680556047878967,
> 8.546115267250002}, {4.128528863849845, 2.9017293848933923,
> -22.51610539815371}}, {{0, 1.3738869718371616,
> 5.689309423462079}, {20.553993773523437, -16.972841620064592,
> 16.61185061676568}}, {{0, 1.0408828831133632,
> 8.484645515699821}, {-3.7267589861478045, 4.1016610850387,
> -24.600635061804443}}};
>
> (({xx, yy} = #1;
> sol = FindRoot[{yy[[1]] - A*Sin[ph],
> yy[[2]] - A*Sin[2*Pi*(xx[[2]]/L) + ph],
>
> yy[[3]] - A*Sin[2*Pi*(xx[[3]]/L) + ph]}, {{A,
> Max[Abs[yy]]}, {L, Max[xx]}, {ph, 0}}];
> Plot[Evaluate[A*Sin[2*Pi*(x/L) + ph] /. sol], {x, 0, 10},
>
> Epilog -> {PointSize[0.02], (Point[#1] & ) /@
> Transpose[{xx, yy}]},
> PlotRange -> {{0, 10}, {-25, 25}}]) & ) /@
> FindRootFailures
>
> xx = Sort[Join[{0}, RandomReal[{0, 10}, 2]]]
>
> yy = RandomReal[{-25, 25}, 3]
>
> sol = FindFit[
> Transpose[{xx, yy}], {A*Sin[2*Pi*(x/L) + ph],
> Max[Abs[xx]] < L}, {A, L, ph}, x]
>
> Plot[Evaluate[A*Sin[2*Pi*(x/L) + ph] /. sol], {x, 0, 10},
> Epilog -> {PointSize[0.02], (Point[#1] & ) /@
> Transpose[{xx, yy}]}, PlotRange -> {{0, 10}, {-25, 25}}]
>
> FindFitFailures = {{{0, 2.463263668380835,
> 4.3190892163093615}, {-1.8407827676541144, -8.736574079785198,
> 12.661520984622932}}, {{0, 3.894521446091823,
> 9.937403619870642}, {12.381369822165155, 15.840399165432128,
> -6.914634137727327}}, {{0, 8.087369725271945,
> 8.343899312282815}, {24.73795103895976, -13.396248713970305,
> 7.150470311065216}}, {{0, 1.5480031866178834,
> 9.575255260205617}, {-8.163720246784278, 13.373468882892958,
> -18.018462091502098}}, {{0, 0.6152784485896601,
> 0.818296772602134}, {-1.858698140836046, 3.695113783904491,
> -1.3186989232026658}}, {{0, 4.743093154057316,
> 6.028314583406327}, {-16.60893277597446, -17.413392198343093,
> -18.54081837965986}}};
>
> (({xx, yy} = #1;
> sol = FindFit[
> Transpose[{xx, yy}], {A*Sin[2*Pi*(x/L) + ph],
> Max[Abs[xx]] < L}, {A, L, ph}, x];
> Plot[Evaluate[A*Sin[2*Pi*(x/L) + ph] /. sol], {x, 0, 10},
>
> Epilog -> {PointSize[0.02], (Point[#1] & ) /@
> Transpose[{xx, yy}]},
> PlotRange -> {{0, 10}, {-25, 25}}]) & ) /@
> FindFitFailures
>
>
>
>