如何获得链式IIR滤波器的b，a（分子/分母）？

Question

假设我们先后应用了3个过滤器：

b, a = iirfilter(...)  # or bilinear(...) or anything else producing b, a
y = lfilter(b, a, x)
b, a = iirfilter(...) 
y = lfilter(b, a, y)
b, a = iirfilter(...) 
y = lfilter(b, a, y)

如何获取与3个过滤器等效的系数b2，a2，这样我们只需一次通过lfilter就可以找到结果： < / p>

y = lfilter(b2, a2, x)

？

---

编辑：卷积似乎无效：

fs = 44100
b2, a2 = iirfilter(2, 2.0/fs * np.asarray([40, 60]), btype='bandstop')  # 50 hz reject
b3, a3 = iirfilter(2, 2.0/fs * np.asarray([85, 115]), btype='bandstop')  # 100 hz reject
b = np.convolve(b2, b3)
a = np.convolve(a2, a3)
w, h = signal.freqz(b, a, worN=10000)

给予：

我尝试使用np.convolve的same，full，valid参数，但没有一个解决问题。

Answer 1

请参见https://dsp.stackexchange.com/questions/38675/how-to-plot-magnitude-and-phase-response-of-2-cascaded-filters-in-matlab

您可以分别对分子和分母进行卷积

import scipy as sp
import scipy.signal as sig

# Individual filters
b1, a1 = sig.iirfilter(...)
b2, a2 = sig.iirfilter(...)

# Cascaded filter
a = sp.convolve(a1, a2)
b = sp.convolve(b1, b2)
y = sig.lfilter(b, a, x)

例如，给定的采样率太高，并且复合滤波器的阶数不够长，以致于无法对彼此靠近的零点进行很大的拒绝。降低采样率，然后内插44.1 kHz。

这里是采样率降低到4410 Hz的结果。

fs = 4410.0
b2, a2 = sig.iirfilter(2, 2.0/fs * sp.asarray([40, 60]), btype='bandstop')  # 50 hz reject
w2, h2 = sig.freqz(b2, a2, worN=4096)

b3, a3 = sig.iirfilter(2, 2.0/fs * sp.asarray([85, 115]), btype='bandstop')  # 100 hz reject
w3, h3 = sig.freqz(b3, a3, worN=4096)

b = sp.convolve(b2, b3)
a = sp.convolve(a2, a3)
w, h = sig.freqz(b, a, worN=4096)

f = w/2.0*fs

然后将IIR滤波器的输出通过10倍插值滤波器，以回到44.1 kHz的采样率。

或，降低过滤器顺序：

fs = 44100.0
b2, a2 = sig.iirfilter(1, 2.0/fs * sp.asarray([40, 60]), btype='bandstop')  # 50 hz reject
w2, h2 = sig.freqz(b2, a2, worN=4096)

b3, a3 = sig.iirfilter(1, 2.0/fs * sp.asarray([85, 115]), btype='bandstop')  # 100 hz reject
w3, h3 = sig.freqz(b3, a3, worN=4096)

b = sp.convolve(b2, b3, 'full')
a = sp.convolve(a2, a3, 'full')
w, h = sig.freqz(b, a, worN=4096)

以44.1 kHz的原始采样率产生