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

我使用scipy.signal(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

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lfilter在时域中再现信号时，正确的频率被过滤了。在

为了消除相移，我用scipy.signal.freqz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

所以相位延迟t

tΒp=—(Tetha(w))/(w)

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

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################################################################################

#

# 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()'