Least Squares Fit to Main Harmonics
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Transcript of Least Squares Fit to Main Harmonics
Least Squares Fit to Main Harmonics
The observed flow u’ may be represented as the sum of M harmonics:
u’ = u0 + ΣjM
=1 Aj sin (j t + j)
For M = 1 harmonic (e.g. a diurnal or semidiurnal constituent):
u’ = u0 + A1 sin (1t + 1)
With the trigonometric identity: sin (A + B) = cosBsinA + cosAsinB u’ = u0 + a1 sin (1t ) + b1 cos (1t )
taking:a1 = A1 cos 1
b1 = A1 sin 1
so u’ is the ‘harmonic representation’
The squared errors between the observed current u and the harmonic representation may be expressed as 2 :
2 = ΣN [u - u’ ]2 = u 2 - 2uu’ + u’ 2
Then:
2 = ΣN {u 2 - 2uu0 - 2ua1 sin (1t ) - 2ub1 cos (1t ) + u02 + 2u0a1 sin (1t ) +
2u0b1 cos (1t ) + 2a1 b1 sin (1t ) cos (1t ) + a12 sin2 (1t ) +
b12 cos2 (1t ) }
Using u’ = u0 + a1 sin (1t ) + b1 cos (1t )
Then, to find the minimum distance between observed and theoretical values we need to minimize
2 with respect to u0 a1 and b1, i.e., δ 2/ δu0 , δ 2/ δa1 , δ 2/ δb1 :
δ2/ δu0 = ΣN { -2u +2u0 + 2a1 sin (1t ) + 2b1 cos (1t ) } = 0
δ2/ δa1 = ΣN { -2u sin (1t ) +2u0 sin (1t ) + 2b1 sin (1t ) cos (1t ) + 2a1 sin2(1t ) } = 0
δ2/ δb1 = ΣN {-2u cos (1t ) +2u0 cos (1t ) + 2a1 sin (1t ) cos (1t ) + 2b1 cos2(1t ) } = 0
ΣN { -2u +2u0 + 2a1 sin (1t ) + 2b1 cos (1t ) } = 0
ΣN {-2u sin (1t ) +2u0 sin (1t ) + 2b1 sin (1t ) cos (1t ) + 2a1 sin2(1t ) } = 0
ΣN { -2u cos (1t ) +2u0 cos (1t ) + 2a1 sin (1t ) cos (1t ) + 2b1 cos2(1t ) } = 0
Rearranging:
ΣN { u = u0 + a1 sin (1t ) + b1 cos (1t ) }
ΣN { u sin (1t ) = u0 sin (1t ) + b1 sin (1t ) cos (1t ) + a1 sin2(1t ) }
ΣN { u cos (1t ) = u0 cos (1t ) + a1 sin (1t ) cos (1t ) + b1 cos2(1t ) }
And in matrix form:
ΣN u cos (1t ) ΣN cos (1t ) ΣN sin (1t ) cos (1t ) ΣN cos2(1t ) b1
ΣN u N ΣN sin (1t ) Σ N cos (1t ) u0
ΣN u sin (1t ) = ΣN sin (1t ) ΣN sin2(1t ) ΣN sin (1t ) cos (1t ) a1
B = A X X = A-1 B
Finally...
The residual or mean is u0
The phase of constituent 1 is: 1 = atan ( b1 / a1 )
The amplitude of constituent 1 is: A1 = ( b12 + a1
2 )½
Pay attention to the arc tangent function used. For example, in IDL you should use atan (b1,a1) and in MATLAB, you should use atan2
For M = 2 harmonics (e.g. diurnal and semidiurnal constituents):
u’ = u0 + A1 sin (1t + 1) + A2 sin (2t + 2)
ΣN cos (1t ) ΣN sin (1t ) cos (1t ) ΣN cos2(1t ) ΣN cos (1t ) sin (2t ) ΣN cos (1t ) cos (2t )
N ΣN sin (1t ) Σ N cos (1t ) ΣN sin (2t ) Σ N cos (2t )
ΣN sin (1t ) ΣN sin2(1t ) ΣN sin (1t ) cos (1t ) ΣN sin (1t ) sin (2t ) ΣN sin (1t ) cos (2t )
Matrix A is then:
ΣN sin (2t ) ΣN sin (1t ) sin (2t ) ΣN cos (1t ) sin (2t ) ΣN sin2(2t ) ΣN sin (2t ) cos (2t )
ΣN cos (2t ) ΣN sin (1t ) cos (2t ) ΣN cos (1t ) cos (2t ) ΣN sin (2t ) cos (2t ) ΣN cos2 (2t )
Remember that: X = A-1 B
and B =ΣN u cos (1t )
ΣN u sin (2t )
ΣN u cos (2t )
ΣN u
ΣN u sin (1t )
u0
a1
b1
a2
b2
X =
Goodness of Fit:
Σ [< uobs > - upred] 2
-------------------------------------
Σ [<uobs > - uobs] 2
Root mean square error:
[1/N Σ (uobs - upred) 2] ½
Fit with M2 only
Fit with M2, K1
Fit with M2, S2, K1
Rayleigh Criterion: record frequency record frequency ≤ ≤ ωω11 – – ωω22
Fit with M2, S2, K1,M4, M6
Tidal Ellipse Parameters
21
)sin(221 22
ppaaaac uvvuvuQ
ua, va, up, vp are the amplitudes and phases of the east-west and north-south components of velocity
amplitude of the clockwise rotary component
21
)sin(221 22
ppaaaacc uvvuvuQ amplitude of the counter-clockwise rotary component
papa
papac vvuu
vvuu
sincos
cossintan 1 phase of the clockwise rotary component
papa
papacc vvuu
vvuu
sincos
cossintan 1 phase of the counter-clockwise rotary component
The characteristics of the tidal ellipses are: Major axis = M = Qcc + Qc
minor axis = m = Qcc - Qc
ellipticity = m / MPhase = -0.5 (thetacc - thetac)Orientation = 0.5 (thetacc + thetac)
Ellipse Coordinates:
time frequency; harmonic
norientatio
sincoscossin
sinsincoscos
t
tmtMy
tmtMx
M2
S2
K1