带通滤波器 ValueError：数字滤波器临界频率必须为 0 < Wn < 1

Question

I am trying to use bandpass filter on an ecg signals, this is the code:

from scipy.signal import butter

def bandpass_filter(self, data, lowcut, highcut, signal_freq, filter_order):
        nyquist_freq = 0.5 * signal_freq
        low = lowcut / nyquist_freq
        high = highcut / nyquist_freq
        b, a = butter(filter_order, [low, high], btype='band', analog=False)
        y = lfilter(b,a, data)
        return y

def detect_peaks(self):
    self.filtered_ecg_measurements = self.bandpass_filter(ecg_measurements,
                                         lowcut=self.filter_lowcut,
                                         highcut=self.filter_highcut,
                                         signal_freq=self.signal_frequency,
                                         filter_order=self.filter_order)
    self.signal_frequency = 250  
    self.filter_lowcut = 0.0
    self.filter_highcut = 15.0
    self.filter_order = 1

This error appears every time I try to run this function:

Traceback (most recent call last):
  File "D:/Project/code/untitled/test.py", line 297, in <module>
    log_data=True, plot_data=True, show_plot=False)
  File "D:/Project/code/untitled/test.py", line 98, in __init__
    self.detect_peaks()
  File "D:/Project/code/untitled/test.py", line 135, in detect_peaks
    filter_order=self.filter_order)
  File "D:/Project/code/untitled/test.py", line 256, in bandpass_filter
    b, a = butter(filter_order, [low, high], btype='band', analog=False)
  File "C:\Users\AppData\Roaming\Python\Python36\site-packages\scipy\signal\filter_design.py", line 2394, in butter
    output=output, ftype='butter')
  File "C:\Users\AppData\Roaming\Python\Python36\site-packages\scipy\signal\filter_design.py", line 1959, in iirfilter
    raise ValueError("Digital filter critical frequencies "
ValueError: Digital filter critical frequencies must be 0 < Wn < 1

the error text is:

Digital filter critical frequencies must be 0 < Wn < 1

Answer 1

It may be because of the input of parameter fs . It has to be larger than any of 2 * Wn . As in scipy/signal/filter_design.py source code:

if fs is not None:
    if analog:
        raise ValueError("fs cannot be specified for an analog filter")
    Wn = 2*Wn/fs

And later:

if not analog:
    if numpy.any(Wn <= 0) or numpy.any(Wn >= 1):
        raise ValueError("Digital filter critical frequencies must be 0 < Wn < 1")

Answer 2

As Darkoob12 pointed out in the comments to the question, lowcut cannot equal zero. If lowcut is set to zero, it will produce the error message seen by OP.

Answer 3

scipy.signal.butter(N, Wn, btype='low', analog=False, output='ba')

Wn : array_like

A scalar or length-2 sequence giving the critical frequencies. For a Butterworth filter, this is the point at which the gain drops to 1/sqrt(2) that of the passband (the “-3 dB point”). For digital filters, Wn is normalized from 0 to 1, where 1 is the Nyquist frequency, pi radians/sample. (Wn is thus in half-cycles / sample.) For analog filters, Wn is an angular frequency (eg rad/s).

Your exception is here where low/high is not [0,1]

b, a = butter(filter_order, [low, high], btype='band', analog=False)

for filter_type == 1 , Wn is float other than nparray.

https://github.com/scipy/scipy/blob/v0.19.1/scipy/signal/filter_design.py#L2226-L2297

>>> N, Wn = signal.buttord([20, 50], [14, 60], 3, 40, True)
>>> b, a = signal.butter(N, Wn, 'band', True)