DFT和FFT（python）的不同结果？

Question

我正在使用 FFT 在我的博士论文中执行一些转换。由于我需要傅里叶变换在某些频率下，我想到了编写自己的 DFT（我不能使用 FFT，因为它的频率由样本数和速率固定）。

但是我遇到了两种算法的输出之间的差异。

这是一个最小的例子，它再现了 DFT 和 FFT 的计算，并输出了一些对比图：

from scipy.fft import fft, fftshift, fftfreq
import numpy as np
import matplotlib.pyplot as plt

#Setting gaussian signal to perform example
N = 200 #sample number
x = np.linspace (-100,100,N) #axis to perform fft into
delta = x[1]-x[0] #space between samples
sigma = 10 #numerical value
freqs = fftshift(fftfreq(N, d=delta)) #FFT output frequencies. Will also be used in DFT to contrast results

#Calculating signal
signal = np.exp(-x**2/sigma**2)

#Plot input signal, for reference
plt.figure(figsize = (8,5))
plt.plot(x, signal)
plt.title("Input signal")
plt.show()

#Perform fft
output_fft = fftshift(fft(signal))

#Perform DFT in same frequencies as FFT
output_DFT1 = []
for fx in freqs:
    spec = np.sum(signal * np.exp(-1j*2*np.pi * fx * x))
    output_DFT1.append(spec)

#Plotting comparative results

#Real part 
plt.figure(figsize=(8,5))
plt.plot(freqs, np.real(output_fft), label='FFT')
plt.plot(freqs, np.real(output_DFT1), label='DFT1')
plt.legend(loc='upper right')
plt.title("Comparative on FFT and DFT real parts")
plt.show()

#Imaginary part
plt.figure(figsize=(8,5))
plt.plot(freqs, np.imag(output_fft), label='FFT')
plt.plot(freqs, np.imag(output_DFT1), label='DFT1')
plt.legend(loc='upper right')
plt.title("Comparative on FFT and DFT imag parts")
plt.show()

这是我的输出图像： Real part comparison Imaginary part comparison

为什么结果不同？

Answer 1

我在this Wikipedia 中遵循了关于 DFT 的 Eq.1，结果与 scipy 中的两种方法相似； fft 和 dft。代码如下：

from scipy.fft import fft, fftshift, fftfreq
import numpy as np
import matplotlib.pyplot as plt

#Setting gaussian signal to perform example
N = 200 #sample number
x = np.linspace (-100,100,N) #axis to perform fft into
delta = x[1]-x[0] #space between samples
sigma = 10 #numerical value
freqs = fftshift(fftfreq(N, d=delta)) #FFT output frequencies. Will also be used in DFT to contrast results
freqs_no_shift = fftfreq(N, d=delta)

#Calculating signal
signal = np.exp(-x**2/sigma**2)

#Plot input signal, for reference
plt.figure(figsize = (8,5))
plt.plot(x, signal)
plt.title("Input signal")
plt.show()

#Perform fft
output_fft = fftshift(fft(signal))

#Perform DFT in same frequencies as FFT
output_DFT1 = []
for idx, fx in enumerate(freqs_no_shift):
    spec = 0
    for i, ampi in enumerate(signal):
        spec += ampi * np.exp(-1j*2*np.pi*idx*i/len(signal))
    output_DFT1.append(spec)
output_DFT1 = fftshift(np.array(output_DFT1))

from scipy.linalg import dft
m = dft(len(signal))
dft2 = m @ signal
dft2 = fftshift(dft2)

#Plotting comparative results

#Real part 
plt.figure(figsize=(8,5))
plt.plot(freqs, np.real(output_fft), label='FFT')
plt.plot(freqs, np.real(output_DFT1), label='DFT1')
plt.plot(freqs, np.real(dft2), label='DFT2')
plt.legend(loc='upper right')
plt.title("Comparative on FFT and DFT real parts")
plt.show()

#Imaginary part
plt.figure(figsize=(8,5))
plt.plot(freqs, np.imag(output_fft), label='FFT')
plt.plot(freqs, np.imag(output_DFT1), label='DFT1')
plt.plot(freqs, np.imag(dft2), label='DFT2')
plt.legend(loc='upper right')
plt.title("Comparative on FFT and DFT imag parts")
plt.show()

希望对您的问题有所帮助。