从 ifft 信号中提取数据值

Question

如何从Fourier transformation中提取filtered signal的values？由于ifft returns 一个复杂的数字，很难进一步计算。

我的python代码：

import numpy as np

# Create a simple signal with two frequencies
dt = 0.001
t = np.arange(0,1,dt)
f = np.sin(2*np.pi*50*t) + np.sin(2*np.pi*120*t) # Sum of 2 frequencies
f_clean = f
noise = 2.5*np.random.randn(len(t))
f = f + noise              # Add some noise

## Compute the Fast Fourier Transform (FFT)

n = len(t)
fhat = np.fft.fft(f,n)                     # Compute the FFT
PSD = fhat * np.conj(fhat) / n             # Power spectrum (power per freq)
freq = (1/(dt*n)) * np.arange(n)           # Create x-axis of frequencies in Hz
L = np.arange(1,np.floor(n/2),dtype='int') # Only plot the first half of freqs

## Use the PSD to filter out noise
indices = PSD > 100       # Find all freqs with large power
PSDclean = PSD * indices  # Zero out all others
fhat = indices * fhat     # Zero out small Fourier coeffs. in Y
ffilt = np.fft.ifft(fhat) # Inverse FFT for filtered time signal

这里ffilt是returns一个complex number的滤波信号。我想用这个信号进行数学计算，但不确定提取值的过程。

Answer 1

下面您可以看到如何从复数中获取实数值，特别是它的实部和虚部，以及它的大小。

a = ffilt[10]
# (1.09200370931126+4.0997278010904346e-17j)

# Get real part    
a.real
# 1.09200370931126

# Get imaginary part
a.imag
# 4.0997278010904346e-17
    
# Get magnitude 
abs(a)
# 1.09200370931126 
# (the imaginary part is close to zero, so the magnitude 
# is almost equal to the real part)

您可以对整个数组执行上述操作。

ffilt.real
ffilt.imag
abs(ffilt)
    
# Bonus: phase
np.angle(ffilt)

我不知道你接下来的计算是什么，但你可能想要使用星等 (abs)。

从 ifft 信号中提取数据值

Extracting data values from the ifft signal

python

fft

complex-numbers