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