And we are able to see that the stronger signals are the 1Hz, 2Hz and 3Hz which makes some sense. The wages have a monthly or two-week cycle that reflects on the first harmonics (1 Hz here corresponds to 1 week cycle, 2 Hz means every 2 weeks, etc.).

Let’s detrended it know, ie, find the linear trend and work with the residuals:

Adding these concepts we get the general form of the Fourier Series:

Example: fs = 8000 samples per second, N = 16000 samples. Therefore, x lasts two seconds long.

This concept deals with translations over the y-axis. In this case corresponds to an additive constant signal.

Ignoring the outliers on the extreme left it appears that the highest spike is near x = 18

In the Cycles textbox the values after the colons mean the starting point of that cycle (in degrees), ie, the cycle’s delay. So :180 means that that cycle starts at the initial rotation of 180 degrees, or \(\pi\) radians.

Let’s check that last eg. Notice that this plot is equal to the blue line in the animation:

A Fourier Transform converts a wave from the time domain into the frequency domain. There is a set of sine waves that, when sumed together, are equal to any given wave. These sine waves each have a frequency and amplitude. A plot of frequency versus strength (amplitude) on an x-y graph of these sine wave components is a frequency spectrum (we’ll see one briefly). Ie, the trajectory can be translated to a set of frequency spikes.

An inverse Fourier Transform (IFT) converts the frequency domain components back into the original time wave. This is given by the next equation:

Let’s assume that we don’t know the functional form of trajectory, we only have its contents, the period and the sampling time points:

Joseph Fourier showed that any periodic wave can be represented by a sum of simple sine waves. This sum is called the Fourier Series. The Fourier Series only holds while the system is linear. If there is, eg, some overflow effect (a threshold where the output remains the same no matter how much input is given), a non-linear effect enters the picture, breaking the sinusoidal wave and the superposition principle.

which allow us to replace phase shifts with cosines.

Take the FFT (however, make sure the FFT size is twice as big as the window, with the second half being padded with zeroes. If the FFT size is only the size of the window, you will effectively be taking the circular autocorrelation, which is not what you want. see https://secure.wikimedia.org/wikipedia/en/wiki/Discrete_Fourier_transform#Circular_convolution_theorem_and_cross-correlation_theorem )

Find the largest peak in the iFFT. This is the strongest periodicity of the waveform. You can actually be a little more clever in which peak you pick, but for most purposes this should be enough. To find the frequency, you just take f=1/T.

To learn about using complex numbers in R check www.johnmyleswhite.com/notebook/2009/12/18/using-complex-numbers-in-r/

So this takes care of strength and delay. What about speed? That is given by the cycles’ sequence. The first cycle is stationary (0 Hz, ie, the DC component), then for every next cycle, the frequency increases (1Hz, 2Hz, 3Hz,…).

How does one convert this into a frequency (number of oscillations in 1 second)?

Note that unless the wavelength of the data is an exact integer submultiple of the FFT length, the actual peak will be between bins, thus distributing energy among multiple nearby FFT result bins. So you may have to interpolate to better estimate the frequency peak. Common interpolation methods to find a more precise frequency estimate are 3-point parabolic and Sinc convolution (which is nearly the same as using a zero-padded longer FFT).

In general, it can be implementation (of the DFT) dependent. You should create a test pure sine wave at a known frequency and then make sure the calculation gives the same number.

A new artificial magnetic material has been developed by researchers, …

We can check if there’s some harmonic hidden in it (there is one, 3Hz harmonics):

A MATLAB example is similar, but they are choosing the length of 1000 (an even number) for the example:

There are factors that facilitate the formation of putatively toxic …

Now the frequency is located at the largest peak.

This result should make intuitive sense; the DC component is simply the average value of the signal. A similar analysis provides the "a" coefficients:

Given a trajectory the fourier transform (FT) breaks it into a set of related cycles that describes it. Each cycle has a strength, a delay and a speed. These cycles are easier to handle, ie, compare, modify, simplify, and if needed, they can be used to reconstruct the original trajectory.

Each wave component \(a_k \times sin(kwt + \rho_k)\) is also called a harmonic.

The FFT is just a faster implementation of the DFT. The FFT algorithm reduces an n-point Fourier transform to about

You do indeed have a peak at about 18Hz. However, the data is very noisy and nothing like a sine wave. I would recommend trying to get better data. Trying to estimate an exact sine wave from Fourier is difficult even if the data is good. Look here for an extensive discussion on this topic.

\[sin(a+b) = sin(a)cos(b) + cos(a)sin(b)\]

### finding Frequency of Sinewave data using Fourier

The trajectory is processed thru a set of filters:

Replace all coefficients of the FFT with their square value (real^2+imag^2). This is effectively taking the autocorrelation.

Sometimes it’s easier to deal with one description, sometimes with the other.

I want to find the frequency $f$ using a fourier transform

Let’s try with one eg. We’ll make a trajectory given the following complex wave

And we find a peak at the 3 Hz harmonics, as expected!

In code: (I've written it in python, but it should be equally simple in c#):

Another feature of the fourier series is phase shift. Phase shifts are translations in the x-axis for a given wave component. These shifts are measured in angles (radians).

Also, if we are trying to identify signals of \(n \times F\) Hz frequencies, the time series length should be divisible by \(F\), ie, we must do something like detrended.trajectory[-(1:(length(detrended.trajectory) %% F))].