How to convert complex numbers back into "normal" numbers after performing FFT

Question

Say, I have an example signal consists of three cosines where each of which represents 4, 6 and 8 frequency band. Now, I throw this signal into frequency domain with the use of FFT, and in frequency domain I cut off unwantend 6 Hz band. Finally, I want to inverse the signal from the frequency domain back into time domain. But when I simply use numpy.fft.ifft I get array of complex numbers, which is not the best result for further analysis of the signal. How can I inverse FFT after performing bandpassing so I'll get whole information carried by real and imaginary part as one number? I looked into z = sqrt(real^2 + imaginary^2) thing, but it's not the "thing".

Below I provide a working example. I'll be grateful for your help.

import numpy as np
from scipy.fftpack import fftfreq

# Define signal.
Fs = 128  # Sampling rate.
Ts = 1 / Fs  # Sampling interval.
Time = np.arange(0, 10, Ts)  # Time vector.
signal = np.cos(4*np.pi*Time) + np.cos(6*np.pi*Time) + np.cos(8*np.pi*Time)


def spectrum(sig, t):
    """
    Represent given signal in frequency domain.
    :param sig: signal.
    :param t: time scale.
    :return:
    """
    f = fftfreq(sig.size, d=t[1]-t[0])
    y = np.fft.fft(sig)
    return f, np.abs(y)


def bandpass(f, sig, min_freq, max_freq):
    """
    Bandpass signal in a specified by min_freq and max_freq frequency range.
    :param f: frequency.
    :param sig: signal.
    :param min_freq: minimum frequency.
    :param max_freq: maximum frequency.
    :return:
    """
    return np.where(np.logical_or(f < min_freq, f > max_freq), 0, sig)

freq, spec = spectrum(signal, Time)
signal_filtered = np.fft.ifft(bandpass(freq, spec, 5, 7))

print(signal_filtered)

"""
print(signal_filtered) result:

[  2.22833798e-15 +0.00000000e+00j   2.13212081e-15 +6.44480810e-16j
   1.85209996e-15 +1.23225456e-15j ...,   1.41336488e-15 -1.71179288e-15j
   1.85209996e-15 -1.23225456e-15j   2.13212081e-15 -6.44480810e-16j]
"""

unutbu unutbu · Accepted Answer · 2015-04-09T17:34:34

If you want to cut out the frequencies between 5 and 7, then you'll want to keep frequencies where

(f < min_freq) | (f > max_freq)

which is equivalent to

np.logical_or(f < min_freq, f > max_freq)

Therefore, use

return np.where(np.logical_or(f < min_freq, f > max_freq), sig, 0)

instead of

return np.where(np.logical_or(f < min_freq, f > max_freq), 0, sig)

since the second argument to np.where contains the value returned by np.where when the condition is True.

With this one change, your code yields

[ 3.00000000 +0.00000000e+00j  2.96514652 +1.24442385e-15j
  2.86160515 +2.08976636e-15j ...,  2.69239924 +4.71763845e-15j
  2.86160515 +5.88163496e-15j  2.96514652 +6.82134642e-15j]

Note that if your signal is real, you could use rfft to take the discrete Fourier transform of a real sequence, and irfft to take its inverse, and rfftfreq to generate the frequencies.

For example,

from __future__ import division
import numpy as np
import scipy.fftpack as fftpack

# Define signal.
Fs = 128  # Sampling rate.
Ts = 1 / Fs  # Sampling interval.
Time = np.arange(0, 10, Ts)  # Time vector.
signal = np.cos(4*np.pi*Time) + np.cos(6*np.pi*Time) + np.cos(8*np.pi*Time)


def spectrum(sig, t):
    """
    Represent given signal in frequency domain.
    :param sig: signal.
    :param t: time scale.
    :return:
    """
    f = fftpack.rfftfreq(sig.size, d=t[1]-t[0])
    y = fftpack.rfft(sig)
    return f, np.abs(y)


def bandpass(f, sig, min_freq, max_freq):
    """
    Bandpass signal in a specified by min_freq and max_freq frequency range.
    :param f: frequency.
    :param sig: signal.
    :param min_freq: minimum frequency.
    :param max_freq: maximum frequency.
    :return:
    """
    return np.where(np.logical_or(f < min_freq, f > max_freq), sig, 0)

freq, spec = spectrum(signal, Time)
signal_filtered = fftpack.irfft(bandpass(freq, spec, 5, 7))

print(signal_filtered)

yields

[ 3.          2.96514652  2.86160515 ...,  2.69239924  2.86160515
  2.96514652]

Note you must use scipy's fftpack here; do not mix SciPy's implementation with NumPy's.

How to convert complex numbers back into "normal" numbers after performing FFT

2 Answers