I was playing with filters in SciPy and noticed something that I'm not that familiar with:

I'm using the code:

import numpy as np
from scipy.signal import butter, lfilter, freqz
import matplotlib.pyplot as plt

def butter_lowpass(cutoff, fs, order=5):
    nyq = 0.5 * fs
    normal_cutoff = cutoff / nyq
    b, a = butter(order, normal_cutoff, btype='low', analog=False)
    return b, a

def butter_lowpass_filter(data, cutoff, fs, order=5):
    b, a = butter_lowpass(cutoff, fs, order=order)
    y = lfilter(b, a, data)
    return y

# Filter requirements.
order = 6
fs = 30.0       # sample rate, Hz
cutoff = 3.667  # desired cutoff frequency of the filter, Hz

# Get the filter coefficients so we can check its frequency response.
b, a = butter_lowpass(cutoff, fs, order)

# Plot the frequency response.
w, h = freqz(b, a, worN=8000)

taken from: https://stackoverflow.com/a/25192640/4959635

And doing some plots of the given freqz results:


enter image description here


enter image description here

So why must I take abs to get "the plot that I want"? And what's the first one with negative amplitudes?


Frequency response has two parts: amplitude response and the phase response. Both of these are represented as a complex signal when you get the response from freqz. In order to plot the amplitude response you need to use abs. Otherwise I doubt it only shows you the real part which I think is what you see in the first figure. Note that when dealing with the amplitude response, in most cases you will do a further step and convert it to $\textrm{dB}$ scale.

  • $\begingroup$ Ah of course. So frequency response is always defined in the complex plane with magnitude and phase and to get a cartesian plot one must take the norm which for complex numbers $z=a+bi$ is $|z|=\sqrt{a^2+b^2}$. $\endgroup$
    – mavavilj
    Aug 30 '16 at 10:54
  • $\begingroup$ Yes that is right. $\endgroup$
    – msm
    Aug 30 '16 at 11:34

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