Harmonic Analysis

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Harmonic Analysis. 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 ). With the trigonometric identity: - PowerPoint PPT Presentation

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Harmonic AnalysisThe observed flow u may be represented as the sum of M harmonics:

u = u0 + jM=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 1b1 = A1 sin 1

so u is the harmonic representationThe squared errors between the observed current u and the harmonic representation may be expressed as 2 :2 = N [u - u ]2 = u 2 - 2uu + u 2 Then:

2 = N {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., 2/ u0 , 2/ a1 , 2/ b1 :2/ u0 = N { -2u +2u0 + 2a1 sin (1t ) + 2b1 cos (1t ) } = 02/ a1 = N { -2u sin (1t ) +2u0 sin (1t ) + 2b1 sin (1t ) cos (1t ) + 2a1 sin2(1t ) } = 02/ b1 = N {-2u cos (1t ) +2u0 cos (1t ) + 2a1 sin (1t ) cos (1t ) + 2b1 cos2(1t ) } = 0N { -2u +2u0 + 2a1 sin (1t ) + 2b1 cos (1t ) } = 0N {-2u sin (1t ) +2u0 sin (1t ) + 2b1 sin (1t ) cos (1t ) + 2a1 sin2(1t ) } = 0N { -2u cos (1t ) +2u0 cos (1t ) + 2a1 sin (1t ) cos (1t ) + 2b1 cos2(1t ) } = 0Rearranging:

N { u = u0 + a1 sin (1t ) + b1 cos (1t ) } N { u sin (1t ) = u0 sin (1t ) + b1 sin (1t ) cos (1t ) + a1 sin2(1t ) }N { u cos (1t ) = u0 cos (1t ) + a1 sin (1t ) cos (1t ) + b1 cos2(1t ) }And in matrix form:N u cos (1t ) N cos (1t ) N sin (1t ) cos (1t ) N cos2(1t ) b1 N u N N sin (1t ) N cos (1t ) u0N u sin (1t ) = N sin (1t ) N sin2(1t ) N 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 + a12 )Pay attention to the arc tangent function used. For example, in IDL you should use atan (b1,a1) and in MATLAB, you should use atan2For M = 2 harmonics (e.g. diurnal and semidiurnal constituents): u = u0 + A1 sin (1t + 1) + A2 sin (2t + 2)

N cos (1t ) N sin (1t ) cos (1t ) N cos2(1t ) N cos (1t ) sin (2t ) N cos (1t ) cos (2t ) N N sin (1t ) N cos (1t ) N sin (2t ) N cos (2t ) N sin (1t ) N sin2(1t ) N sin (1t ) cos (1t ) N sin (1t ) sin (2t ) N sin (1t ) cos (2t ) Matrix A is then:N sin (2t ) N sin (1t ) sin (2t ) N cos (1t ) sin (2t ) N sin2(2t ) N sin (2t ) cos (2t ) N cos (2t ) N sin (1t ) cos (2t ) N cos (1t ) cos (2t ) N sin (2t ) cos (2t ) N cos2 (2t ) Remember that: X = A-1 B

and B =N u cos (1t ) N u sin (2t )N u cos (2t ) N u N u sin (1t )u0

a1

b1

a2

b2X =Goodness of Fit:

[< uobs > - upred] 2-------------------------------------

[ - uobs] 2Root mean square error:

[1/N (uobs - upred) 2] Fit with M2 only

Fit with M2, K1

Fit with M2, S2, K1

Rayleigh Criterion: record frequency 1 2Fit with M2, S2, K1,M4, M6

M2S2K1Tidal Ellipse ParametersMajor axis: Mminor axis: mellipticity = m / MPhase Orientation

Tidal Ellipse Parameters

ua, va, up, vp are the amplitudes and phases of the east-west and north-south components of velocityamplitude of the clockwise rotary component

amplitude of the counter-clockwise rotary component

phase of the clockwise rotary component

phase of the counter-clockwise rotary componentThe characteristics of the tidal ellipses are:Major axis = M = Qcc + Qcminor axis = m = Qcc - Qcellipticity = m / MPhase = -0.5 (thetacc - thetac)Orientation = 0.5 (thetacc + thetac)Ellipse Coordinates:

M2S2K1

Two Years of Tide Data at Trident Pier, Florida (Cape Caaveral)Use U-tide routine

SA = Solar annualSSA = Solar SemiannualMSM = Lunar synodic monthly (29.53 d)MM = Lunar Monthly (27.55 d)MSF = Lunisolar synodic fortnightly (14.76 d)MF = Lunisolar fortnightly (13.66 d)

SA = Solar annualSSA = Solar SemiannualMSM = Lunar synodic monthly (29.53 d)MM = Lunar Monthly (27.55 d)MSF = Lunisolar synodic fortnightly (14.76 d)MF = Lunisolar fortnightly (13.66 d)Complex DemodulationTime series X(t) taken as nearly periodic plus non-periodic Z(t), still varying in time. Amplitude A and phase of the nearly periodic signal are allowed to be time-dependent but vary slowly compared to the frequency .X(t) = A(t) cos(t +(t))+ Z(t)

Demodulate by multiplying times

Varies slowly, independent of Varies at frequency 2Varies at frequency Low-pass filter to remove frequencies at or above

Varies slowly, independent of (low-pass filter smooths this term denoted by )Separate A and

Sea level at Cape Caaveral, Floridam2 years of dataX(t) = A(t) cos(t +(t))+ Z(t)

Ensenada de la Paz

Ensenada de La Paz, Mexico

Amplitude of complex demodulated series at semidiurnal and diurnal frequencies

YAVAROS BAY, MEXICODworak, J. A., and J. Gomez-Valdes (2005), J. Geophys. Res., 110, C01007, doi:10.1029/2003JC001865.

Dworak, J. A., and J. Gomez-Valdes (2005), J. Geophys. Res., 110, C01007, doi:10.1029/2003JC001865.

Station M

Puerto Morelos Coral Reef Lagoon27Sabrina ParraPacific OceanAtlantic OceanGulf of MexicoCaribbean SeaNorth AmericaMexicoYucatan Peninsula

Pargos SpringNorthern InletCentral InletSouthern InletPuerto Morelos LagoonCoronado et al. 2007Here we have North America. Puerto Morelos coral reef lagoon is located on the Yucatan Peninsula, just south of Cancun. As we zoom in, here we have the lagoon. Coronado showed that the circulation is mostly forced by waves entering the lagoon over the shallow coral reefs and strong outflow through the 3 main inlets: Northern, Central and Southern. The buoyant jet discharge we studied is called Pargos and is located here, inside the lagoon. To understand the effects of waves and tides in the lagoon we deployed these instruments.27