Question

我试图在R中复制下图:(改编自http://link.springer.com/article/10.1007/PL00011669）

DFT figure

该图的基本概念是显示DFT的前几个分量，在时域中绘制，然后仅使用相对于原始数据的这些分量（X'）显示时域中的重建波（ X）。我想略微修改上图，使得所有显示的行都叠加在一个图上。

我一直在尝试使用60 Hz采样的一些实际数据来调整数字。例如：

## 3 second sample where: time is in seconds and var is the variable of interest
temp = data.frame(time=seq(from=0,to=3,by=1/60),
            var = c(0.054,0.054,0.054,0.072,0.072,0.072,0.072,0.09,0.09,0.108,0.126,0.126,
                  0.126,0.126,0.126,0.144,0.144,0.144,0.144,0.144,0.162,0.162,0.144,0.126,
                  0.126,0.108,0.144,0.162,0.18,0.162,0.126,0.126,0.108,0.108,0.126,0.144,
                  0.162,0.144,0.144,0.144,0.144,0.162,0.162,0.126,0.108,0.09,0.09,0.072,
                  0.054,0.054,0.054,0.036,0.036,0.018,0.018,0.018,0.018,0,0.018,0,
                  0,0,-0.018,0,0,0,-0.018,0,-0.018,-0.018,0,-0.018,
                  -0.018,-0.018,-0.018,-0.036,-0.036,-0.054,-0.054,-0.072,-0.072,-0.072,-0.072,-0.072,
                  -0.09,-0.09,-0.108,-0.126,-0.126,-0.126,-0.144,-0.144,-0.144,-0.162,-0.162,-0.18,
                  -0.162,-0.162,-0.162,-0.162,-0.144,-0.144,-0.144,-0.126,-0.126,-0.108,-0.108,-0.09,
                  -0.072,-0.054,-0.036,-0.018,0,0,0,0,0.018,0.018,0.036,0.054,
                  0.054,0.054,0.054,0.054,0.054,0.054,0.054,0.054,0.054,0.072,0.054,0.072,
                  0.072,0.072,0.072,0.072,0.072,0.054,0.054,0.054,0.036,0.036,0.036,0.036,
                  0.036,0.054,0.054,0.072,0.09,0.072,0.036,0.036,0.018,0.018,0.018,0.018,
                  0.036,0.036,0.036,0.036,0.018,0,-0.018,-0.018,-0.018,-0.018,-0.018,0,
                  -0.018,-0.036,-0.036,-0.018,-0.018,-0.018,-0.036,0,0,-0.018,-0.018,-0.018,-0.018))

##plot the original data
ggplot(temp, aes(x=time, y=var))+geom_line()

我相信我可以使用fft()来最终实现这一目标，但是从fft()的输出到我的目标的跳跃有点不清楚。

我意识到这个问题有点类似于：How do I calculate amplitude and phase angle of fft() output from real-valued input?但我对上面特定数据的实际代码更感兴趣。

请注意我对时间序列分析相对较新，所以你可以提供任何清晰度w.r.t.将fft（）的输出放在上下文中，或者您可以推荐的任何可以有效完成此任务的包将不胜感激。

谢谢

Answer 1

Matlab是您最好的工具，具体功能只是fft（）。要使用它，首先要确定时域数据的几个基本参数：

1，持续时间（T），等于3s。

2，采样间隔T_s，等于1/60秒。

3，频域旋转f_s，其等于两个相邻傅立叶基础之间的频率差。您可以根据需要定义f_s。然而，最小可能的f_s等于1 / T = 0.333Hz。因此，如果您想要更好的频域旋转（更小的f_s），则需要更长的时域数据。

4，最大频率f_M，根据香农抽样理论，等于1 /（2T_s）= 30。

5，DFT长度N，等于2 * f_M / f_s。

然后找出您想要用来近似数据的四个傅里叶基的特定频率。例如，3,6,9和12 Hz。所以f_s = 3 Hz。然后N = 2 * f_M / f_s = 20。

您的Matlab代码如下所示：

var=[0.054,0.054,0.054 ...]; % input all your data points here
f_full=fft(var,20); % Do 20-point fft
f_useful=f_full(2:5); % You are interested with the lowest four frequencies except DC

这里f_useful包含四个傅立叶基的四个复系数。要重建var，请执行以下操作：

% Generate basis functions
dt=0:1/60:3;
df=[3:3:12];
basis1=exp(1j*2*pi*df(1)*dt);
basis2=exp(1j*2*pi*df(2)*dt);
basis3=exp(1j*2*pi*df(3)*dt);
basis4=exp(1j*2*pi*df(4)*dt);

% Reconstruct var
var_recon=basis1*f_useful(1)+...
basis2*f_useful(2)+...
basis3*f_useful(3)+...
basis4*f_useful(4);
var_recon=real(var_recon);

% Plot both curves
figure;
plot(var);
hold on;
plot(var_recon);

将此代码改编为您的论文：）

Answer 2

从Signal Processing调整我自己的帖子。我认为它仍然适用于Python中的那些。

我不是这个主题的专家，但有一些有用的例子可供分享。

您保留的傅里叶组件越多，您就越接近模仿原始信号。

此示例显示了将10,20，......保存到 n 组件时会发生什么。假设x和y是您的数据向量。

import numpy
from matplotlib import pyplot as plt

n = len(y)
COMPONENTS = [10, 20, n]

for c in COMPONENTS:
    colors = numpy.linspace(start=100, stop=255, num=c)
    for i in range(c):
        Y = numpy.fft.fft(y)
        numpy.put(Y, range(i+1, n), 0.0)
        ifft = numpy.fft.ifft(Y)
        plt.plot(x, ifft, color=plt.cm.Reds(int(colors[i])), alpha=.70)

    plt.title("First {c} fourier components".format(c=c))
    plt.plot(x,y, label="Original dataset", linewidth=2.0)
    plt.grid(linestyle='dashed')
    plt.legend()
    plt.show()

对于本书的数据集，最多可保留4,10和 n 组件：

对于您的数据集，最多可保留4,10和 n 组件：

从R中的离散傅里叶变换重建信号

2 个答案: