I have a problem for obtain the frequency of oscillations with Fourier,
For example if I have
tf = 60; NN = (tf + 1)/0.001; dt = tf/NN; data = Table[Sin[t], {t, 0., tf, 0.001}]; fft = Fourier[data]; peak1 = First[Position[Abs[fft], Max[Abs[fft]]]]; The frequency is obtained would be $2\pi k/ N dt$, where k is the position of the first peak. If I evaluate this expression I do not obtain the frequency.
OP's apparent intention:
tf = 60; NN = (tf)/0.001; dt = tf/NN; data = Table[Sin[t], {t, 0., tf, 0.001}]; fft = Fourier[data]; peak1 = First@First[Position[Abs[fft], Max[Abs[fft]]]] freq = (peak1 - 1)/tf (* 11 0.166667 *)
It's closest estimate possible, since the adjacent estimates have a larger error:
1/(2 Pi) - (peak1 - {0, 1, 2})/tf // N (* {-0.0241784, -0.00751172, 0.00915494} *)
Here is the method from the docs for Fourier (under "Applications"), which uses a more sophisticated approach:
Min[TakeLargest[Abs@fft, 2]]; peaks = Position[Abs[fft], x_ /; x >= %]; pos = First@First[peaks] (* 11 *)
n = Length@fft; fr = Abs[Fourier[data*Exp[2 Pi I (pos - 2) N[Range[0, n - 1]]/n], FourierParameters -> {0, 2/n}]]; frpos = Position[fr, Max[fr]][[1, 1]]; period = N[n/(pos - 2 + 2 (frpos - 1)/n)] dt (* 6.29278 *)
frequency = 1/period 1./(2 Pi) (* 0.158912 0.159155 *)
2 Piin formula forFourier. Shouldn't the frequency be(peak1 -1)/tf? $\endgroup$Sin[2 pi f]for frequency f. You also need to take a look at the FourierParameters option of Fourier and finally, you should account for your sampling rate too. $\endgroup$Fourierfor an example of how to determine the frequency. The shortcoming of that example is that it uses adt(ordx) of1, and doesn't show how to modify the various steps. However, it's not hard to figure out. $\endgroup$