一、信号函数

假设采集128个点

数学表达

$\chi[n]=\cos\left(2\pi\times\frac{2n}{1024}+\frac{\pi}{3}\right)+0.5\cos\left(2\pi\times\frac{5n}{1024}\right), n = 0,1,2,...,127$

Python表达

import numpy as np

N = 128
n = np.arange(N)
y = np.cos(2 * np.pi * 2 * (n / N) + np.pi / 3) + 0.5 * np.cos(2 * np.pi * 5 * (n / N))
print(n.shape, y.shape)

输出(128,) (128,)

信号茎叶图

from matplotlib import pyplot as plt
plt.stem(n, y) #茎叶图
plt.show()

128点茎叶图

二、官方FFT函数验证

模长茎叶图

r1 = np.fft.fft(y) #计算FFT
mag = abs(r1) #取模
#画图
plt.figure(figsize = (20, 5))
plt.stem(n, mag) #茎叶图

image.png

因为图像是左右对称的所以我们只看前一半。
输出数组，小于0.0001的值处理为0

mag_new = []
for i in range(int(len(mag)/2)):
    num = mag[i]
    if num < 0.0001:
        num = 0
    mag_new.append(num)
    print("mag[%d] = %d" % (i, num))

输出

mag[0] = 0
mag[1] = 0
mag[2] = 64
mag[3] = 0
mag[4] = 0
mag[5] = 32
mag[6] = 0
mag[7] = 0
mag[8] = 0
mag[9] = 0
mag[10] = 0
mag[11] = 0
mag[12] = 0
mag[13] = 0
mag[14] = 0
mag[15] = 0
mag[16] = 0
mag[17] = 0
mag[18] = 0
mag[19] = 0
mag[20] = 0
mag[21] = 0
mag[22] = 0
mag[23] = 0
mag[24] = 0
mag[25] = 0
mag[26] = 0
mag[27] = 0
mag[28] = 0
mag[29] = 0
mag[30] = 0
mag[31] = 0
mag[32] = 0
mag[33] = 0
mag[34] = 0
mag[35] = 0
mag[36] = 0
mag[37] = 0
mag[38] = 0
mag[39] = 0
mag[40] = 0
mag[41] = 0
mag[42] = 0
mag[43] = 0
mag[44] = 0
mag[45] = 0
mag[46] = 0
mag[47] = 0
mag[48] = 0
mag[49] = 0
mag[50] = 0
mag[51] = 0
mag[52] = 0
mag[53] = 0
mag[54] = 0
mag[55] = 0
mag[56] = 0
mag[57] = 0
mag[58] = 0
mag[59] = 0
mag[60] = 0
mag[61] = 0
mag[62] = 0
mag[63] = 0

可以看出只在 mag[2], mag[5]时存在数值

振幅

把傅里叶变换后的模长除以 $\frac{N}{2}$ 就是各个分量的振幅

mag_new = np.asarray(mag_new, dtype=float)
print(mag_new / (N/2))

输出

[0.  0.  1.  0.  0.  0.5 0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.
 0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.
 0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.
 0.  0.  0.  0.  0.  0.  0.  0.  0.  0. ]

频率

假设1秒采样128点, 采样频率128HZ
所以傅里叶变换后的分辨率为1HZ
mag[2] => 2HZ
mag[5] => 5HZ

相位

phase = np.pi/np.angle(r1)
print(phase[2], phase[5])

3.0000000000000013 -1430726521854752.5
可以看出x[2]的相位为 $\frac{\pi}{3}$ ,x[2]的相位为 $\frac{\pi}{-1430726521854752}$ => 0

二、朴素算法DFT计算

设F(x)是把上面128个点 $y(n_i), i=0,1,...,127$ 当作系数的多项式
求 $\omega_n^k$ 带入F(x)的点值表达式

Python实现DFT与FFT记录