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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’ = u 0 + Σ j M =1 A j sin (  j t +  j ). For M = 1 harmonic (e.g. a diurnal or semidiurnal constituent): u’ = u 0 + A 1 sin (  1 t +  1 ).

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Least Squares Fit to Main Harmonics

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  1. Least Squares Fit to Main Harmonics The observed flow u’ may be represented as the sum of M harmonics: u’ = u0 + ΣjM=1Aj 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’

  2. Using u’ = u0 + a1 sin (1t ) + b1 cos (1t ) 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 ) + a12sin2 (1t ) + b12cos2 (1t ) } 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, to find the minimum distance between observed and theoretical values we need to minimize 2 with respect to u0 a1and b1, i.e., δ 2/ δu0 , δ 2/ δa1 , δ 2/ δb1 : δ2/ δu0 = ΣN {-2u +2u0 + 2a1 sin (1t ) + 2b1 cos (1t ) } = 0 δ2/ δa1 = ΣN { -2u sin (1t ) +2u0 sin (1t ) + 2b1 sin (1t ) cos (1t ) + 2a1 sin2(1t ) } = 0 δ2/ δb1 = ΣN {-2u cos (1t ) +2u0 cos (1t ) + 2a1 sin (1t ) cos (1t ) + 2b1 cos2(1t ) } = 0

  3. 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 ) } ΣN u N ΣN sin (1t ) ΣN cos (1t ) u0 ΣN {-2u +2u0 + 2a1 sin (1t ) + 2b1 cos (1t ) } = 0 ΣN u sin (1t ) = ΣN sin (1t ) ΣN sin2(1t ) ΣN sin (1t ) cos (1t ) a1 ΣN u cos (1t ) ΣN cos (1t ) ΣN sin (1t ) cos (1t ) ΣN cos2(1t ) b1 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 X = A-1 B Rearranging: And in matrix form: B = A X

  4. 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+ a12 )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

  5. Matrix A is then: N ΣN sin (1t ) ΣN cos (1t ) ΣN sin (2t ) ΣN cos (2t ) ΣN sin (1t ) ΣN sin2(1t ) ΣN sin (1t ) cos (1t ) ΣN sin (1t ) sin (2t ) ΣN sin (1t ) cos (2t ) ΣN cos (1t ) ΣN sin (1t ) cos (1t ) ΣN cos2(1t ) ΣN cos (1t ) sin (2t ) ΣN cos (1t ) cos (2t ) 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 = u0 a1 b1 a2 b2 ΣN u ΣN u sin (1t ) X = ΣN u cos (1t ) ΣN u sin (2t ) ΣN u cos (2t ) For M = 2 harmonics (e.g. diurnal and semidiurnal constituents): u’ = u0+ A1 sin (1t + 1) + A2 sin (2t + 2)

  6. Goodness of Fit: Σ [< uobs > - upred] 2 ------------------------------------- Σ [<uobs > - uobs] 2 Root mean square error: [1/N Σ (uobs - upred) 2] ½

  7. Fit with M2 only

  8. Fit with M2, K1

  9. Fit with M2, S2, K1 Rayleigh Criterion: record frequency ≤ ω1 – ω2

  10. Fit with M2, S2, K1, M4, M6

  11. Tidal Ellipse Parameters amplitude of the counter-clockwise rotary component phase of the clockwise rotary component phase of the counter-clockwise rotary component Ellipse Coordinates: ua, va, up, vpare the amplitudes and phases of the east-west and north-south components of velocity amplitude of the clockwise rotary component The characteristics of the tidal ellipses are: Major axis = M = Qcc + Qc minor axis = m = Qcc - Qc ellipticity = m / M Phase = -0.5 (thetacc - thetac) Orientation = 0.5 (thetacc + thetac)

  12. M2 S2 K1

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