回答此问题可获得 20 贡献值,回答如果被采纳可获得 50 分。
<p>所以,在我最后两个问题之后,我来谈谈我的实际问题。也许有人在我的理论程序中发现了错误,或者我在编程上做了些错事。在</p>
<p>我使用<code>scipy.signal</code>(使用firwin函数)在Python中实现带通滤波器。我的原始信号包括两个频率(w_1=600Hz,w_2=800Hz)。可能会有更多的频率所以我需要一个带通滤波器。在</p>
<p>在这个例子中,我想过滤掉大约600hz的频带,所以我取了600+/-20Hz作为截止频率。当我实现滤波器并使用<code>lfilter</code>在时域中再现信号时,正确的频率被过滤了。在</p>
<p>为了消除相移,我用<code>scipy.signal.freqz</code>绘制了频率响应图,返回值为firwin的h作为分子,1作为预定义的denumerator。
如freqz文档中所述,我还绘制了相位(=doc中的角度),并且能够查看频率响应图,以获得滤波信号频率600hz的相移。在</p>
<p>所以相位延迟t</p>
<p>tΒp=—(Tetha(w))/(w)</p>
<p>不幸的是,当我把这个相位延迟加到滤波信号的时间数据中时,它并没有得到与原始600hz信号相同的相位。在</p>
<p>我加了密码。奇怪的是,在消除部分代码以保持最小值之前,过滤后的信号以正确的振幅开始-现在情况更糟了。在</p>
<pre><code>################################################################################
#
# Filtering test
#
################################################################################
#
from math import *
import numpy as np
from scipy import signal
from scipy.signal import firwin, lfilter, lti
from scipy.signal import freqz
import matplotlib.pyplot as plt
import matplotlib.colors as colors
################################################################################
# Nb of frequencies in the original signal
nfrq = 2
F = [60,80]
################################################################################
# Sampling:
nitper = 16
nper = 50.
fmin = np.min(F)
fmax = np.max(F)
T0 = 1./fmin
dt = 1./fmax/nitper
#sampling frequency
fs = 1./dt
nyq_rate= fs/2
nitpermin = nitper*fmax/fmin
Nit = int(nper*nitpermin+1)
tps = np.linspace(0.,nper*T0,Nit)
dtf = fs/Nit
################################################################################
# Build analytic signal
# s = completeSignal(F,Nit,tps)
scomplete = np.zeros((Nit))
omg1 = 2.*pi*F[0]
omg2 = 2.*pi*F[1]
scomplete=scomplete+np.sin(omg1*tps)+np.sin(omg2*tps)
#ssingle = singleSignals(nfrq,F,Nit,tps)
ssingle=np.zeros((nfrq,Nit))
ssingle[0,:]=ssingle[0,:]+np.sin(omg1*tps)
ssingle[1,:]=ssingle[0,:]+np.sin(omg2*tps)
################################################################################
## Construction of the desired bandpass filter
lowcut = (60-2) # desired cutoff frequencies
highcut = (60+2)
ntaps = 451 # the higher and closer the signal frequencies, the more taps for the filter are required
taps_hamming = firwin(ntaps,[lowcut/nyq_rate, highcut/nyq_rate], pass_zero=False)
# Use lfilter to get the filtered signal
filtered_signal = lfilter(taps_hamming, 1, scomplete)
# The phase delay of the filtered signal
delay = ((ntaps-1)/2)/fs
plt.figure(1, figsize=(12, 9))
# Plot the signals
plt.plot(tps, scomplete,label="Original signal with %s freq" % nfrq)
plt.plot(tps-delay, filtered_signal,label="Filtered signal %s freq " % F[0])
plt.plot(tps, ssingle[0,:],label="original signal %s Hz" % F[0])
plt.grid(True)
plt.legend()
plt.xlim(0,1)
plt.xlabel('Time (s)')
plt.ylabel('Amplitude')
# Plot the frequency responses of the filter.
plt.figure(2, figsize=(12, 9))
plt.clf()
# First plot the desired ideal response as a green(ish) rectangle.
rect = plt.Rectangle((lowcut, 0), highcut - lowcut, 5.0,facecolor="#60ff60", alpha=0.2,label="ideal filter")
plt.gca().add_patch(rect)
# actual filter
w, h = freqz(taps_hamming, 1, worN=1000)
plt.plot((fs * 0.5 / np.pi) * w, abs(h), label="designed rectangular window filter")
plt.xlim(0,2*F[1])
plt.ylim(0, 1)
plt.grid(True)
plt.legend()
plt.xlabel('Frequency (Hz)')
plt.ylabel('Gain')
plt.title('Frequency response of FIR filter, %d taps' % ntaps)
plt.show()'
</code></pre>