1. What are the dominant frequencies? Fourier transforms decompose a data sequence into a set of discrete spectral estimates – separate the variance of a time series as a function of frequency. A common use of Fourier transforms is to find the frequency components of a signal buried in a time domain signal.

2. FAST FOURIER TRANSFORM (FFT) In practice, if the time series f(t) is not a power of 2, it should be padded with zeros

3. What is the statistical significance of the peaks? Each spectral estimate has a confidence limit defined by a chi-squared distribution

4. Spectral Analysis Approach 1. Remove mean and trend of time series 2. Pad series with zeroes to a power of 2 3. To reduce end effect (Gibbs’ phenomenon) use a window (Hanning, Hamming, Kaiser) to taper the series 4. Compute the Fourier transform of the series, multiplied times the window 5. Rescale Fourier transform by multiplying times 8/3 for the Hanning Window 6. Compute band-averages or block-segmented averages 7. Incorporate confidence intervals to spectral estimates

5. Sea level at Mayport, FL July 1, 2007 (day “0” in the abscissa) to September 1, 2007 m m Raw data and Low-pass filtered data High-pass filtered data 1. Remove mean and trend of time series (N = 1512) 2. Pad series with zeroes to a power of 2 (N = 2048)

6. Cycles per day m 2 /cpd Spectrum of raw data Spectrum of high-pass filtered data

7. Day from July 1, 2007 Value of the Window Hanning Window Hamming Window 3. To reduce end effect (Gibbs’ phenomenon) use a window (Hanning, Hamming, Kaiser) to taper the series

8. Day from July 1, 2007 Value of the Window Hanning Window Hamming Window Kaiser-Bessel, α = 2 Kaiser-Bessel, α = 3 3. To reduce end effect (Gibbs’ phenomenon) use a window (Hanning, Hamming, Kaiser) to taper the series

9. m m Raw series x Hanning Window (one to one) Raw series x Hamming Window (one to one) Day from July 1, 2007 To reduce side-lobe effects 4. Compute the Fourier transform of the series, multiplied times the window

10. m m High-pass series x Hanning Window (one to one) High pass series x Hamming Window (one to one) Day from July 1, 2007 To reduce side-lobe effects 4. Compute the Fourier transform of the series, multiplied times the window

11. High pass series x Kaiser-Bessel Window α=3 (one to one) m Day from July 1, 2007 4. Compute the Fourier transform of the series, multiplied times the window

12. Cycles per day m 2 /cpd Original from Raw Data with Hanning window with Hamming window Windows reduce noise produced by side-lobe effects Noise reduction is effected at different frequencies

13. Cycles per day m 2 /cpd with Hanning window with Hamming and Kaiser- Bessel (α=3) windows

14. 5. Rescale Fourier transform by multiplying: times 8/3 for the Hanning Window times 2.5164 for the Hamming Window times ~8/3 for the Kaiser-Bessel (Depending on alpha)

15. 6. Compute band-averages or block-segmented averages 7. Incorporate confidence intervals to spectral estimates Upper limit: Lower limit: 1-alpha is the confidence (or probability) nu are the degrees of freedom gamma is the ordinate reference value

17. 0.995 0.990 0.975 0.950 0.900 0.750 0.500 0.250 0.100 0.050 0.025 0.010 0.005 7.88 6.63 5.02 3.84 2.71 1.32 0.45 0.10 0.02 0.00 0.00 0.00 0.00 10.60 9.21 7.38 5.99 4.61 2.77 1.39 0.58 0.21 0.10 0.05 0.02 0.01 12.84 11.34 9.35 7.81 6.25 4.11 2.37 1.21 0.58 0.35 0.22 0.11 0.07 14.86 13.28 11.14 9.49 7.78 5.39 3.36 1.92 1.06 0.71 0.48 0.30 0.21 16.75 15.09 12.83 11.07 9.24 6.63 4.35 2.67 1.61 1.15 0.83 0.55 0.41 18.55 16.81 14.45 12.59 10.64 7.84 5.35 3.45 2.20 1.64 1.24 0.87 0.68 20.28 18.48 16.01 14.07 12.02 9.04 6.35 4.25 2.83 2.17 1.69 1.24 0.99 21.95 20.09 17.53 15.51 13.36 10.22 7.34 5.07 3.49 2.73 2.18 1.65 1.34 23.59 21.67 19.02 16.92 14.68 11.39 8.34 5.90 4.17 3.33 2.70 2.09 1.73 25.19 23.21 20.48 18.31 15.99 12.55 9.34 6.74 4.87 3.94 3.25 2.56 2.16 26.76 24.72 21.92 19.68 17.28 13.70 10.34 7.58 5.58 4.57 3.82 3.05 2.60 28.30 26.22 23.34 21.03 18.55 14.85 11.34 8.44 6.30 5.23 4.40 3.57 3.07 29.82 27.69 24.74 22.36 19.81 15.98 12.34 9.30 7.04 5.89 5.01 4.11 3.57 31.32 29.14 26.12 23.68 21.06 17.12 13.34 10.17 7.79 6.57 5.63 4.66 4.07 32.80 30.58 27.49 25.00 22.31 18.25 14.34 11.04 8.55 7.26 6.26 5.23 4.60 34.27 32.00 28.85 26.30 23.54 19.37 15.34 11.91 9.31 7.96 6.91 5.81 5.14 35.72 33.41 30.19 27.59 24.77 20.49 16.34 12.79 10.09 8.67 7.56 6.41 5.70 37.16 34.81 31.53 28.87 25.99 21.60 17.34 13.68 10.86 9.39 8.23 7.01 6.26 38.58 36.19 32.85 30.14 27.20 22.72 18.34 14.56 11.65 10.12 8.91 7.63 6.84 40.00 37.57 34.17 31.41 28.41 23.83 19.34 15.45 12.44 10.85 9.59 8.26 7.43 41.40 38.93 35.48 32.67 29.62 24.93 20.34 16.34 13.24 11.59 10.28 8.90 8.03 42.80 40.29 36.78 33.92 30.81 26.04 21.34 17.24 14.04 12.34 10.98 9.54 8.64 44.18 41.64 38.08 35.17 32.01 27.14 22.34 18.14 14.85 13.09 11.69 10.20 9.26 45.56 42.98 39.36 36.42 33.20 28.24 23.34 19.04 15.66 13.85 12.40 10.86 9.89 46.93 44.31 40.65 37.65 34.38 29.34 24.34 19.94 16.47 14.61 13.12 11.52 10.52 48.29 45.64 41.92 38.89 35.56 30.43 25.34 20.84 17.29 15.38 13.84 12.20 11.16 49.64 46.96 43.19 40.11 36.74 31.53 26.34 21.75 18.11 16.15 14.57 12.88 11.81 50.99 48.28 44.46 41.34 37.92 32.62 27.34 22.66 18.94 16.93 15.31 13.56 12.46 52.34 49.59 45.72 42.56 39.09 33.71 28.34 23.57 19.77 17.71 16.05 14.26 13.12 53.67 50.89 46.98 43.77 40.26 34.80 29.34 24.48 20.60 18.49 16.79 14.95 13.79 55.00 52.19 48.23 44.99 41.42 35.89 30.34 25.39 21.43 19.28 17.54 15.66 14.46 56.33 53.49 49.48 46.19 42.58 36.97 31.34 26.30 22.27 20.07 18.29 16.36 15.13 57.65 54.78 50.73 47.40 43.75 38.06 32.34 27.22 23.11 20.87 19.05 17.07 15.82 58.96 56.06 51.97 48.60 44.90 39.14 33.34 28.14 23.95 21.66 19.81 17.79 16.50 60.27 57.34 53.20 49.80 46.06 40.22 34.34 29.05 24.80 22.47 20.57 18.51 17.19 Probability 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 Degrees of freedom

19. Includes low frequency N=1512

20. Excludes low frequency N=1512

23. 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 ) With the trigonometric identity: sin (A + B) = cosBsinA + cosAsinB u’ = u 0 + a 1 sin (  1 t ) + b 1 cos (  1 t ) taking: a 1 = A 1 cos  1 b 1 = A 1 sin  1

24. The squared errors between the observed current u and the harmonic representation may be expressed as  2 :  2 = Σ N [u - u’ ] 2 = u 2 - 2uu’ + u’ 2 Then:  2 = Σ N {u 2 - 2uu 0 - 2ua 1 sin (  1 t ) - 2ub 1 cos (  1 t ) + u 0 2 + 2u 0 a 1 sin (  1 t ) + 2u 0 b 1 cos (  1 t ) + 2a 1 b 1 sin (  1 t ) cos (  1 t ) + a 1 2 sin 2 (  1 t ) + b 1 2 cos 2 (  1 t ) } Using u’ = u 0 + a 1 sin (  1 t ) + b 1 cos (  1 t ) Then, to find the minimum distance between observed and theoretical values we need to minimize  2 with respect to u 0 a 1 and b 1, i.e., δ  2 / δu 0, δ  2 / δa 1, δ  2 / δb 1 : δ  2 / δ u 0 = Σ N { -2u +2u 0 + 2a 1 sin (  1 t ) + 2b 1 cos (  1 t ) } = 0 δ  2 / δ a 1 = Σ N { -2u sin (  1 t ) +2u 0 sin (  1 t ) + 2b 1 sin (  1 t ) cos (  1 t ) + 2a 1 sin 2 (  1 t ) } = 0 δ  2 / δ b 1 = Σ N {-2u cos (  1 t ) +2u 0 cos (  1 t ) + 2a 1 sin (  1 t ) cos (  1 t ) + 2b 1 cos 2 (  1 t ) } = 0

25. Σ N { -2u +2u 0 + 2a 1 sin (  1 t ) + 2b 1 cos (  1 t ) } = 0 Σ N {-2u sin (  1 t ) +2u 0 sin (  1 t ) + 2b 1 sin (  1 t ) cos (  1 t ) + 2a 1 sin 2 (  1 t ) } = 0 Σ N { -2u cos (  1 t ) +2u 0 cos (  1 t ) + 2a 1 sin (  1 t ) cos (  1 t ) + 2b 1 cos 2 (  1 t ) } = 0 Rearranging: Σ N { u = u 0 + a 1 sin (  1 t ) + b 1 cos (  1 t ) } Σ N { u sin (  1 t ) = u 0 sin (  1 t ) + b 1 sin (  1 t ) cos (  1 t ) + a 1 sin 2 (  1 t ) } Σ N { u cos (  1 t ) = u 0 cos (  1 t ) + a 1 sin (  1 t ) cos (  1 t ) + b 1 cos 2 (  1 t ) } And in matrix form: Σ N u cos (  1 t ) Σ N cos (  1 t ) Σ N sin (  1 t ) cos (  1 t ) Σ N cos 2 (  1 t ) b 1 Σ N u N Σ N sin (  1 t ) Σ N cos (  1 t ) u 0 Σ N u sin (  1 t ) = Σ N sin (  1 t ) Σ N sin 2 (  1 t ) Σ N sin (  1 t ) cos (  1 t ) a 1 B = A X X = A -1 B

26. Finally... The residual or mean is u 0 The phase of constituent 1 is:  1 = atan ( b 1 / a 1 ) The amplitude of constituent 1 is: A 1 = ( b 1 2 + a 1 2 ) ½ Pay attention to the arc tangent function used. For example, in IDL you should use atan (b 1,a 1 ) and in MATLAB, you should use atan2

27. For M = 2 harmonics (e.g. diurnal and semidiurnal constituents): u’ = u 0 + A 1 sin (  1 t +  1 ) + A 2 sin (  2 t +  2 ) Σ N cos (  1 t ) Σ N sin (  1 t ) cos (  1 t ) Σ N cos 2 (  1 t ) Σ N cos (  1 t ) sin (  2 t ) Σ N cos (  1 t ) cos (  2 t ) N Σ N sin (  1 t ) Σ N cos (  1 t ) Σ N sin (  2 t ) Σ N cos (  2 t ) Σ N sin (  1 t ) Σ N sin 2 (  1 t ) Σ N sin (  1 t ) cos (  1 t ) Σ N sin (  1 t ) sin (  2 t ) Σ N sin (  1 t ) cos (  2 t ) Matrix A is then: Σ N sin (  2 t ) Σ N sin (  1 t ) sin (  2 t ) Σ N cos (  1 t ) sin (  2 t ) Σ N sin 2 (  2 t ) Σ N sin (  2 t ) cos (  2 t ) Σ N cos (  2 t ) Σ N sin (  1 t ) cos (  2 t ) Σ N cos (  1 t ) cos (  2 t ) Σ N sin (  2 t ) cos (  2 t ) Σ N cos 2 (  2 t ) Remember that: X = A -1 B and B = Σ N u cos (  1 t ) Σ N u sin (  2 t ) Σ N u cos (  2 t ) Σ N u Σ N u sin (  1 t ) u0a1b1a2b2u0a1b1a2b2 X =

28. Goodness of Fit: Σ [ - u pred ] 2 ------------------------------------- Σ [ - u obs ] 2 Root mean square error: [1/N Σ (u obs - u pred ) 2 ] ½

29. Fit with M 2 only

30. Fit with M 2, K 1

31. Fit with M 2, S 2, K 1

32. Fit with M 2, S 2, K 1, M 4, M 6

33. Tidal Ellipse Parameters u a, v a, u p, v p are the amplitudes and phases of the east-west and north-south components of velocity amplitude of the clockwise rotary component amplitude of the counter-clockwise rotary component phase of the clockwise rotary componentphase of the counter-clockwise rotary component The characteristics of the tidal ellipses are: Major axis = M = Q cc + Q c minor axis = m = Q ac - Q c ellipticity = m / M Phase = -0.5 (theta cc - theta c ) Orientation = 0.5 (theta cc + theta c ) Ellipse Coordinates:

34. M2S2K1M2S2K1

35. Fit with M2 only (dotted) and M2 + M4 (continuous)

36. Fit with M2 only (dotted) and M2 + M6 (continuous)