how to extract frequency associated with fft values in python

Question

I used fft function in numpy which resulted in a complex array. How to get the exact frequency values?

Answer 1

np.fft.fftfreq tells you the frequencies associated with the coefficients:

import numpy as np

x = np.array([1,2,1,0,1,2,1,0])
w = np.fft.fft(x)
freqs = np.fft.fftfreq(len(x))

for coef,freq in zip(w,freqs):
    if coef:
        print('{c:>6} * exp(2 pi i t * {f})'.format(c=coef,f=freq))

# (8+0j) * exp(2 pi i t * 0.0)
#    -4j * exp(2 pi i t * 0.25)
#     4j * exp(2 pi i t * -0.25)

---

The OP asks how to find the frequency in Hertz. I believe the formula is frequency (Hz) = abs(fft_freq * frame_rate).

Here is some code that demonstrates that.

First, we make a wave file at 440 Hz:

import math
import wave
import struct

if __name__ == '__main__':
    # http://stackoverflow.com/questions/3637350/how-to-write-stereo-wav-files-in-python
    # http://www.sonicspot.com/guide/wavefiles.html
    freq = 440.0
    data_size = 40000
    fname = "test.wav"
    frate = 11025.0
    amp = 64000.0
    nchannels = 1
    sampwidth = 2
    framerate = int(frate)
    nframes = data_size
    comptype = "NONE"
    compname = "not compressed"
    data = [math.sin(2 * math.pi * freq * (x / frate))
            for x in range(data_size)]
    wav_file = wave.open(fname, 'w')
    wav_file.setparams(
        (nchannels, sampwidth, framerate, nframes, comptype, compname))
    for v in data:
        wav_file.writeframes(struct.pack('h', int(v * amp / 2)))
    wav_file.close()

This creates the file test.wav. Now we read in the data, FFT it, find the coefficient with maximum power, and find the corresponding fft frequency, and then convert to Hertz:

import wave
import struct
import numpy as np

if __name__ == '__main__':
    data_size = 40000
    fname = "test.wav"
    frate = 11025.0
    wav_file = wave.open(fname, 'r')
    data = wav_file.readframes(data_size)
    wav_file.close()
    data = struct.unpack('{n}h'.format(n=data_size), data)
    data = np.array(data)

    w = np.fft.fft(data)
    freqs = np.fft.fftfreq(len(w))
    print(freqs.min(), freqs.max())
    # (-0.5, 0.499975)

    # Find the peak in the coefficients
    idx = np.argmax(np.abs(w))
    freq = freqs[idx]
    freq_in_hertz = abs(freq * frate)
    print(freq_in_hertz)
    # 439.8975

Answer 2

Frequencies associated with DFT values (in python)

By fft, Fast Fourier Transform, we understand a member of a large family of algorithms that enable the fast computation of the DFT, Discrete Fourier Transform, of an equisampled signal.

A DFT converts a list of N complex numbers to a list of N complex numbers, with the understanding that both lists are periodic with period N.

Here we deal with the numpy implementation of the fft.

In many cases, you think of

• a signal x defined in the time domain of length N, sampled at a constant interval dt,
• its DFT X (here specifically X = np.fft.fft(x)), whose elements are sampled on the frequency axis with a sample rate dw.

Some definition

• the period (aka duration) of the signal x, sampled at dt with N samples is is

  T = dt*N
  

• the fundamental frequencies (in Hz and in rad/s) of X, your DFT are

  df = 1/T
  dw = 2*pi/T # =df*2*pi
  

• the top frequency is the Nyquist frequency

  ny = dw*N/2
  

  (and it's not dw*N)

The frequencies associated with a particular element in the DFT

The frequencies corresponding to the elements in X = np.fft.fft(x) for a given index 0<=n<N can be computed as follows:

def rad_on_s(n, N, dw):
    return dw*n if n<N/2 else dw*(n-N)

or in a single sweep

w = np.array([dw*n if n<N/2 else dw*(n-N) for n in range(N)])

if you prefer to consider frequencies in Hz, s/w/f/

f = np.array([df*n if n<N/2 else df*(n-N) for n in range(N)])

Using those frequencies

If you want to modify the original signal x -> y applying an operator in the frequency domain in the form of a function of frequency only, the way to go is computing the w's and

Y = X*f(w)
y = ifft(Y)

Introducing np.fft.fftfreq

Of course numpy has a convenience function np.fft.fftfreq that returns dimensionless frequencies rather than dimensional ones but it's as easy as

f = np.fft.fftfreq(N)*N*df
w = np.fft.fftfreq(N)*N*dw

Because df = 1/T and T = N/sps (sps being the number of samples per second) one can also write

f = np.fft.fftfreq(N)*sps

Answer 3

The frequency is just the index of the array. At index n, the frequency is 2πn / the array's length (radians per unit). Consider:

>>> numpy.fft.fft([1,2,1,0,1,2,1,0])
array([ 8.+0.j,  0.+0.j,  0.-4.j,  0.+0.j,  0.+0.j,  0.+0.j,  0.+4.j,
        0.+0.j])

the result has nonzero values at indices 0, 2 and 6. There are 8 elements. This means

       2πit/8 × 0       2πit/8 × 2       2πit/8 × 6
    8 e           - 4i e           + 4i e
y ~ ———————————————————————————————————————————————
                          8