从R中的离散傅里叶变换重构信号

我试图在R中复制下图：（改编自）

该图的基本概念是显示DFT的前几个分量，绘制在时域中，然后仅使用这些分量（X'）相对于原始数据（X）在时域中显示重构波。我想稍微修改上图，使所有显示的线都覆盖在一个绘图上

我一直在尝试用60Hz采样的一些真实数据来调整这个数字。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（）

最终实现这个目标，但是从

fft（）

的输出到我的目标的飞跃有点不清楚

我意识到这个问题有点类似于：但是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，因此，如果您能提供w.r.t.将fft（）的输出置于上下文中的任何清晰信息，或者您能推荐的能够有效完成此任务的任何软件包，我们将不胜感激

---

谢谢你

Matlab是你最好的工具，具体功能就是fft（）。要使用它，首先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 s

3，频域旋转fs，等于两个相邻傅里叶基之间的频率差。您可以根据自己的需要定义f_。然而，最小可能的f_s等于1/T=0.333 Hz。因此，如果您想要更好的频域旋转（更小的fs），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，它等于1/（2T_s）=30

5，DFT长度N，等于2*f_M/f_s

然后找出你想要用来近似数据的四个傅里叶基的具体频率。例如，3、6、9和12赫兹。因此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_有用包含四个傅里叶基的四个复系数。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：）

改编我自己的文章。我认为它仍然与Python中的相关

---

我不是这方面的专家，但有一些有用的例子可以分享

保留的傅里叶分量越多，模拟原始信号的距离就越近

这个例子展示了当你保持10，20，…最多n个组件时会发生什么。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，保留最多4、10和n个组件：

---

对于数据集，最多保留4、10和n个组件：

---

谢谢，丹尼尔，这很有帮助。