Basic idea is to convert the data to frequency domain using discrete fourier transformation, and using that we can find the major frequency in the data.Before going to the details lets look the equation of a basic Sin wave

where T is the time period of wave

t is the instantaneous time.

Pi Ofcourse 22/7 ,

Lets plot this wave. I am using mathematica to plot the wave.

Here is the wave form

T = 20

Plot[ Sin[2*Pi/T*t], {t, 0, 50}] . (We can verify T =20 from the graph.)

Pi Ofcourse 22/7 ,

Lets plot this wave. I am using mathematica to plot the wave.

Here is the wave form

T = 20

Plot[ Sin[2*Pi/T*t], {t, 0, 50}] . (We can verify T =20 from the graph.)

So the frequency of this wave is 1/T . that is 1/20.

Ok. this is no big deal unless you are very weak in mathematics. Now lets add some random noise to it

n = 1500;

T = 20;

SampledData= Table[Sin[2 Pi* x/T] + RandomReal[.8], {x, n}] ;

ListLinePlot[SampledData]

I hope now you cannot identify the frequency from this ;) .Lets find the T period from this , it must approximate to 20.Before that lest look the equation of one dimensional DFT

Where N is the total number of elements and k/N will give the frequency (because Sin (2*Pi*f * n ) .Ok , now i am going to do DFT with data generated with equation Sin[2 Pi* x/T] and I am going to find the power spectrum(magnitude of complex numbers)

DftData = Abs[Fourier[SampledData]];

lets plot the power spectrum graph

In graph you can see two spikes. The rightmost spike represent the nyquest frequency( i will explain it later). lets find the first position where magnitude is highest.

pos = Position[f, Max[f]][[1, 1]]

it will be at 76.

Now we can find the frequency by substituting this value for k in equation k/N .

that is 76/1500 = .0506. which approximates to 1/20.

If you want to find the timer period just find the reciprocal.

T = 1500/76 = 19.73!! approximates to 20.

Hey you just learnt a great thing!!.

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