现有（合成）信号与滤波信号之间的大FFT幅度差

Question

我认为，滤波信号的FFT图上的噪声区域中的幅度应该更低，然后现在。可能是scipy.fftpack.lfilter()中出现的小数字偏差/错误。

我试图为现有信号添加噪音，但没有结果。

为什么噪声区域中滤波信号（绿色）的FFT幅度如此之高？

更新
300 dB的FFT幅度是非物理的 - 很明显，这是由于Python环境中的64位浮点数。

滤波后的信号（绿色）的FFT具有如此低的dB（~67 dB），因为所有信号都有~4000个样本（图片上不是“每秒”，没有错误，没有关键），采样率= 200个样本/秒。 1频率箱= 200/4000/2 = 0.025Hz，显示2000个箱。

如果我们采用更长的信号，我们会获得更高的每个频率分辨率（即40 000样本，1个频率分箱= 200/40 000/2 = 0.0025 Hz）。而且我们得到滤波信号的FFT约为87 dB。

（数字67 dB和87 dB似乎是非物理的，因为初始信号SNR为300dB，但我尝试在现有信号上添加一些噪声并获得相同的数字）

如果要获取信号中采样数的非依赖FFT图像，则应使用窗口化FFT和滑动窗口来计算整个信号FFT。

'''
Created on 13.02.2013, python 2.7.3

@author: 
'''
from numpy.random import normal
from numpy import sin, pi, absolute, arange, round
#from numpy.fft import fft
from scipy.fftpack import fft, ifft
from scipy.signal import kaiserord, firwin, lfilter, freqz
from pylab import figure, clf, plot, xlabel, ylabel, xlim, ylim, title, grid, axis, show, log10,\
                  subplots_adjust, subplot

def filter_apply(filename):
    pass

def sin_generator(freq_hz = 1000, sample_rate = 8000, amplitude = 1.0, time_s = 1):
    nsamples = round(sample_rate * time_s)
    t = arange(nsamples) / float(sample_rate.__float__())
    signal = amplitude * sin(2*pi*freq_hz.__float__()*t)
    return signal, nsamples

def do_fir(signal, sample_rate):
    return signal

#-----------------make a signal---------------
freq_hz = 10.0
sample_rate = 400
amplitude = 1.0
time_s = 10
a1, nsamples = sin_generator(freq_hz, sample_rate, amplitude, time_s)
a2, nsamples = sin_generator(50.0, sample_rate, 0.5*amplitude, time_s)
a3, nsamples = sin_generator(150.0, sample_rate, 0.5*amplitude, time_s)
mu, sigma = 0, 0.1 # mean and standard deviation
noise = normal(mu, sigma, nsamples)
signal = a1 + a2 + a3 # + noise

#----------------create low-pass FIR----
# The Nyquist rate of the signal.
nyq_rate = sample_rate / 2.0
# The desired width of the transition from pass to stop,
# relative to the Nyquist rate.  We'll design the filter
# with a 5 Hz transition width.
width = 5.0/nyq_rate

# The desired attenuation in the stop band, in dB.
ripple_db = 60.0

# Compute the order and Kaiser parameter for the FIR filter.
N, beta = kaiserord(ripple_db, width)
print 'N = ',N, 'beta = kaiser param = ', beta

# The cutoff frequency of the filter.
cutoff_hz = 30.0

# Use firwin with a Kaiser window to create a lowpass FIR filter.
# Length of the filter (number of coefficients, i.e. the filter order + 1)
taps = firwin(N, cutoff_hz/nyq_rate, window=('kaiser', beta))

# Use lfilter to filter x with the FIR filter.
filtered_signal = lfilter(taps, 1.0, signal)

#----------------plot signal----------------------
hh,ww=2,2  
figure(figsize=(12,9))  
subplots_adjust(hspace=.5)
#figure(1)
subplot(hh,ww,1)

# existing signal
x = arange(nsamples) / float(sample_rate)

# The phase delay of the filtered signal.
delay = 0.5 * (N-1) / sample_rate

# original signal
plot(x, signal, '-bo' , linewidth=2) 

# filtered signal shifted to compensate for 
# the phase delay.
plot(x-delay, filtered_signal, 'r-' , linewidth=1)

# Plot just the "good" part of the filtered signal. 
# The first N-1 samples are "corrupted" by the
# initial conditions.
plot(x[N-1:]-delay, filtered_signal[N-1:], 'g', linewidth=2)

xlabel('time (s)')
ylabel('amplitude')
axis([0, 1.0/freq_hz*2, -(amplitude*1.5),amplitude*1.5]) # two periods of freq_hz
title('Signal (%d samples)' % nsamples)
grid(True)


#-------------- FFT of the signal
subplot(hh,ww,2)

signal_fft=fft(signal)
filtered_fft =fft(filtered_signal[N-1:])

# existing signal 
y = 20*log10( ( abs( signal_fft/nsamples )*2.0)/max( abs( signal_fft/nsamples )*2.0) )# dB Amplitude
x = arange(nsamples)/float(nsamples)*float(sample_rate)

# filtered signal
y_filtered = 20*log10( (abs(filtered_fft/ (nsamples - N + 1) )*2.0)/max(abs(signal_fft/ (nsamples - N + 1) )*2.0) )# dB Amplitude
x_filtered = arange(nsamples - N + 1)/float(nsamples - N + 1)*float(sample_rate)
yy = fft(ifft(filtered_fft))

plot(x,y, linewidth=1) 
plot(x_filtered, y_filtered, 'g', linewidth=2)
xlim(0, sample_rate/2) # compensation of mirror (FFT imaginary part)
xlabel('freq (Hz)')
ylabel('amplitude, (dB)')
title('Signal (%d samples)' % nsamples)
grid(True)


#--------------FIR ampitude response
subplot(hh,ww,3)

w, h = freqz(taps, worN=8000)
#plot((w/pi)*nyq_rate, absolute(h), linewidth=2)
plot((w/pi)*nyq_rate, 20*log10(absolute(h)/1.0),'r', linewidth=1)
xlabel('Frequency (Hz)')
#ylabel('Gain -blue')
ylabel('Gain (dB)')
title('Frequency Response')
#ylim(-0.05, 1.05)
grid(True)


#--------------FIR coeffs
subplot(hh,ww,4)
plot(taps, 'bo-', linewidth=2)
title('Filter Coefficients (%d taps)' % N)
grid(True)

show()

Answer 1

我认为-300dB的噪音非常低。请记住，这是一个对数刻度，所以我们讨论的是64位分辨率下的几个数字。

使用双精度浮点数（64位），你不会得到任何更低。