The algorithm used to guess the coefficients is as follows:
We know f (t) at some sampling points ti and want to find a, ω and φ such that f (t) = a cos (ω t + φ).
From the analytical expression, we can compute two primitives :
If2 (t) = ∫ f2 = a2 × [t + S (t)] / 2 If'2 (t) = ∫ f'2 = a2 ω2 × [t - S (t)] / 2 where S (t) = sin (2 (ω t + φ)) / (2 ω)
We can remove S between these expressions :
If'2 (t) = a2 ω2 t - ω2 If2 (t)
The preceding expression shows that If'2 (t) is a linear combination of both t and If2 (t): If'2 (t) = A × t + B × If2 (t)
From the primitive, we can deduce the same form for definite integrals between t1 and ti for each ti :
If2 (ti) - If2 (t1) = A × (ti - t1) + B × (If2 (ti) - If2 (t1))
We can find the coefficients A and B that best fit the sample to this linear expression by computing the definite integrals for each sample points.
For a bilinear expression z (xi, yi) = A × xi + B × yi, the coefficients A and B that minimize a least square criterion ∑ (zi - z (xi, yi))2 are given by these expressions:
∑yiyi ∑xizi - ∑xiyi ∑yizi A = ------------------------ ∑xixi ∑yiyi - ∑xiyi ∑xiyi ∑xixi ∑yizi - ∑xiyi ∑xizi B = ------------------------ ∑xixi ∑yiyi - ∑xiyi ∑xiyi
In fact, we can assume both a and ω are positive and compute them directly, knowing that A = a2 ω2 and that B = - ω2. The complete algorithm is therefore:
for each ti from t1 to tn-1, compute: f (ti) f' (ti) = (f (ti+1) - f(ti-1)) / (ti+1 - ti-1) xi = ti - t1 yi = ∫ f2 from t1 to ti zi = ∫ f'2 from t1 to ti update the sums ∑xixi, ∑yiyi, ∑xiyi, ∑xizi and ∑yizi end for |-------------------------- \ | ∑yiyi ∑xizi - ∑xiyi ∑yizi a = \ | ------------------------ \| ∑xiyi ∑xizi - ∑xixi ∑yizi |-------------------------- \ | ∑xiyi ∑xizi - ∑xixi ∑yizi ω = \ | ------------------------ \| ∑xixi ∑yiyi - ∑xiyi ∑xiyi
Once we know ω, we can compute:
fc = ω f (t) cos (ω t) - f' (t) sin (ω t) fs = ω f (t) sin (ω t) + f' (t) cos (ω t)
It appears that fc = a ω cos (φ)
and fs = -a ω sin (φ)
, so we can use these expressions to compute φ. The best estimate over the sample is given by averaging these expressions.
Since integrals and means are involved in the preceding estimations, these operations run in O(n) time, where n is the number of measurements.
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