Tidal prediction

Hello, im trying to perform tidal prediction. in order to do that i need to analyse the raw data from the mareograph and to get the amplitude and the phase of the first 20 frequencies (those ones have the most impact in the data).

i've used the following functions: Y=fft(X); %X is the raw data Ph=angle(Y); Amp=abs(Y);

the problem is that the values that im getting for the amplitudes are enormous (the original data has values up to 0.4 meters, the amplitude has valuse of tens sometimes hundreds of meters) wich is wrong.

how can i fix it? what am i doing wrong?

thanx for the help .....

[Alexandria]

I want to perfom a tidal prediction, what data do you need to use?

Answer 1

[Wayne King]

Hi Dany, you are most likely not scaling the estimates by the length of the input vector. For example.

t = 0:0.001:1-0.001;
x = cos(2*pi*100*t-pi/4)+0.5*randn(size(t));
xdft = fft(x);
xdft = xdft(1:length(x)/2+1);
xdft(2:end-1) = 2*xdft(2:end-1);
xdft = xdft./length(x);
fprintf('Amplitude at 100 Hz is %3.2f.\n',abs(xdft(101)))     
fprintf('Phase at 100 Hz is %2.3f radians.\n',angle(xdft(101)))

[Daniel Shub]

But then Parseval's theorem doesn't hold... Further, if you take the first 20 components, and do the inverse FFT, the magnitudes will be really screwed up. Scaling might be closer to what Dany wants though.

[Wayne King]

right, to make DFT as matlab implements it as a unitary operator, you have to multiply the output by 1/sqrt(length(x)). For the L2 norm:

x = randn(8,1);

norm(x,2)

xdft = 1/sqrt(length(x))*fft(x);

norm(xdft,2)

but 2/length(x) here for the positive frequencies gives you the MLE estimates of the sine wave amplitudes

[Dany]

Wayne,

thank you for your help, it seems ok now.

Answer 2

[Daniel Shub]

The concept of the first 20 frequencies doesn't make sense. The frequencies to which the first 20 components correspond depends on your sample rate and the number of samples in your signal/FFT. You might want to calculate the PSD instead of the FFT. Also, if all you want is the power in the low frequencies you might want to just lowpass filter your signal. Finally you might want to consider FREQZ instead of FFT.

There is no reason that the FFT of a waveform with a maximum magnitude of 0.4 cannot have values much much larger than 0.4. Consider

Amp = abs(fft(0.4*rand(1e5, 1)));

[Dany]

i will, thank you