# FFT of tidal data - problems identifying Mf component in MATLAB

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David Young
2015-2-9

编辑：David Young
2015-2-11

The apparent fortnightly frequency is in fact a beat pattern - there is little energy at this frequency and the low-frequency noise masks any peak there in the FFT. I think this is the point that Youssef KHMOU is making. [ Edit: I initially thought there was no energy there. In fact, there is a component in the gravitational potential with a fortnightly frequency, but its amplitude is relatively small.]

In fact, your data do show good agreement with the expected peaks. The code below demonstrates it - if you run it you will see that the red tick marks at the twice-daily frequencies line up very nicely with the peaks in your data. Only the largest diurnal component is visible above the noise, but you also have strong thrice-daily and four-times-daily peaks.

Note that it's useful to plot only the low-frequency end of the spectrum, and taking logs makes the lower peaks much easier to see. Also removing the mean helps visibility and accuracy.

Incidentally, the FFT isn't a good way to estimate the amplitude and phase of tidal components, if that's what you aim to do. It's much better to do a direct least-squares fit of waves at the known frequencies.

%%Get data

d = load('Tide_data');

tide = d.NH_tide_data;

t = d.NH_tide_data_time;

%%Take FFT and plot with correct frequencies on x-axis

deltat = mean(diff(t)); % have checked that t has uniform increments

N = length(tide);

deltaf = 1 / (N * deltat); % frequency step in day^(-1)

f = (0:N-1)*deltaf; % frequency vector, correct origin and scale

ftide = fft(tide - mean(tide)); % fft

nplot = floor(N/20); % choose how much to plot

plot(f(2:nplot), log(abs(ftide(2:nplot))));

%%Compare with astronomical tidal periods

% periods in hours of M2 K1 S2 O1 M1 N2 Q1 L2 2N2 J1 OO1

% (main diurnal and twice-daily components)

tideperiodsHours = [12.421 23.934 12.0 25.819 24.833 12.658 26.868 ...

12.192 12.905 23.098 22.306];

tidefreqs = 24./tideperiodsHours; % main frequencies in day^(-1)

hold on

for k = 1:length(tidefreqs)

freq = tidefreqs(k);

ind = round(freq/deltaf) + 1;

tideval = log(abs(ftide(ind))); % position tick mark vertically

plot([freq freq], tideval + [0.1 0.5], 'r-');

end

hold off

[ Edit: additional information about tidal component estimation added.]

The FFT is fundamentally a bad method for estimating tidal component amplitudes, because it fits sinusoids whose periods divide exactly into the time span of the series. These periods won't coincide with the tidal periods, so errors are introduced.

The simplest alternative is to fit a model made of sinusoids at the known tidal frequencies, and this works very well. The least-squares approach is known as "classical harmonic analysis", for example in this PowerPoint presentation by Russ Herman. It has the additional advantage that missing data are not a problem. A basic function that carries out harmonic analysis is attached.

You can check out the results by estimating the amplitudes, then seeing how a resynthesised time series, plotted below in red, compares with the original, plotted in blue:

%%Compare fitted model with original data

% estimate tidal amplitudes

tidenomean = tide - mean(tide);

[a, b] = harmonicFit(t, tidenomean, tidefreqs);

disp(hypot(a, b)); % display amplitudes

% synthesise time series using only these frequencies

synth = harmonicSynth(a, b, tidefreqs, t);

% plot initial section only for clarity

nplot = 500;

plot(t(1:nplot), tidenomean(1:nplot), 'b-', ...

t(1:nplot), synth(1:nplot), 'r-');

There are visible mismatches. I suspect this is because my array of frequencies does not include the 3-cycles-per-day and higher frequencies which are prominent in the Newhaven data. You can also compare in the frequency domain, by looking at the power spectrum of the residuals, like this:

% Look at spectrum of residuals

fresid = fft(tidenomean.' - synth);

nplot = floor(N/20);

plot(f(2:nplot), log(abs(fresid(2:nplot))));

Adding more frequencies would clearly improve the fit.

However, harmonic analysis is still relatively unsophisticated, and there's a big literature in the area, going back to George Darwin. One way to improve on the simple function attached is to get error estimates, and a good paper dealing with this is here. Pawlowicz has MATLAB code relating to this here. It may also be well worth checking out some of the other toolboxes around, such as this one on the FEX.

Finally, I think it's worth having a look at the nineteenth-century version of my attached file harmonicSynth.m - it's much nicer to look at and is here.

### 更多回答（3 个）

Youssef Khmou
2015-2-9

编辑：Youssef Khmou
2015-2-9

From the result in the second figure; you can remark that there are two frequency components. which means that there are two wave forms with different magnitudes and infinitesimal difference of frequencies, without using the data, the following simulation produces the same pattern ;

Fs=20;t=0:1/Fs:40-1/Fs;

y=sin(2*pi*4*t)+0.2*sin(2*pi*4.2*t);

plot(t,y)

figure; plot((abs(fft(y))))

basic solution consists of making the region of interest smaller, in the given example it is :

axis([100 200 0 450]);

##### 2 个评论

Youssef Khmou
2015-2-9

编辑：Youssef Khmou
2015-2-9

how do you recognize the fortnightly frequency in the first graph? try :

spectrogram(y);

which is short term transform,

Erik S.
2015-2-9

Hi,

Have you tried to plot the power spectral density instead? If you provide the sampling frequency you can get better insight to which frequencies you have.

You get the following graph by pwelch( NH_tide_data-mean(NH_tide_data) );

##### 1 个评论

David Young
2015-2-9

qj fan
2017-6-26

Dear David Young,

i have a question. There are some daily recorded series for temperature, and show evidently seasonal trend. how can i get the main frequency and get rid of the periodic trend by fft. one way is the following algorithm:

Smoothing Filter :

1. Let the power-law noise be represented by x and the sinusoidal trend by t. Therefore the noise with the trend is given by y = x + t; with Fourier transform F (f ).

2. Let the frequency in the Fourier transform corresponding to the periodic trend occur at f = fk.

3. Replace the power at the frequency fk by a smoothing filter

|F∗(fk)| = 0.5(|F(fk−1)| + |F(fk+1)|) |F∗(f)| = |F(f)| ,f ̸= fk

Assign random phases so as to satisfy the conjugacy constraints.

4. Determine the inverse Fourier transform to obtain the filtered data yf .

because fft need sample frequency, i don't know how to solve it.

would you like to give me some suggestion.

##### 0 个评论

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