Question

我想使用R执行傅立叶变换到时间序列。我想：

获得第5至第18次谐波的总和
绘制每一波
并输出为csv文件。

这是指向数据的链接： Link to Data

这是我的初始代码。

dat   <- read.csv("Baguio.csv",header=FALSE)
y     <- dat$V1
ssp   <-spectrum(y)
t     <- 1:73
per   <- 1/ssp$freq[ssp$spec==max(ssp$spec)]
reslm <- lm(y ~ sin(2*pi/per*t)+cos(2*pi/per*t))
rg    <- diff(range(y))

#blue dashed line
plot(y~t,ylim=c(min(y)-0.1*rg,max(y)+0.1*rg))
lines(fitted(reslm)~t,col=4,lty=2)

#green line 2nd harmonics
reslm2 <- lm(y ~ sin(2*pi/per*t)+cos(2*pi/per*t)+sin(4*pi/per*t)+cos(4*pi/per*t))
lines(fitted(reslm2)~t,col=3)

有没有办法简化这段代码？如果我必须达到18次谐波，则等式变得非常长。另外，我仍然不知道如何在这里添加谐波。

非常感谢，

Answer 1

更容易的解决方案是使用快速傅里叶变换（fft）

dat   <- read.csv("Baguio.csv", header=FALSE)
y     <- dat$V1
t     <- 1:73
rg    <- diff(range(y))

nff = function(x = NULL, n = NULL, up = 10L, plot = TRUE, add = FALSE, main = NULL, ...){
  #The direct transformation
  #The first frequency is DC, the rest are duplicated
  dff = fft(x)
  #The time
  t = seq(from = 1, to = length(x))
  #Upsampled time
  nt = seq(from = 1, to = length(x)+1-1/up, by = 1/up)
  #New spectrum
  ndff = array(data = 0, dim = c(length(nt), 1L))
  ndff[1] = dff[1] #Always, it's the DC component
  if(n != 0){
    ndff[2:(n+1)] = dff[2:(n+1)] #The positive frequencies always come first
    #The negative ones are trickier
    ndff[length(ndff):(length(ndff) - n + 1)] = dff[length(x):(length(x) - n + 1)]
  }
  #The inverses
  indff = fft(ndff/73, inverse = TRUE)
  idff = fft(dff/73, inverse = TRUE)
  if(plot){
    if(!add){
      plot(x = t, y = x, pch = 16L, xlab = "Time", ylab = "Measurement",
        main = ifelse(is.null(main), paste(n, "harmonics"), main))
      lines(y = Mod(idff), x = t, col = adjustcolor(1L, alpha = 0.5))
    }
    lines(y = Mod(indff), x = nt, ...)
  }
  ret = data.frame(time = nt, y = Mod(indff))
  return(ret)
}

然后我们需要调用res，将时间序列传递为x，将谐波数量传递为n，然后将上采样（因此我们在原始点旁边绘制时间点）作为up。

png("res_18.png")
res = nff(x = y, n = 18L, up = 100L, col = 2L)
dev.off()

要获得第5到第18次谐波的总和，它只是系列之间的差异

sum5to18 = nff(x = y, n = 18L, up = 10L, plot = FALSE)
sum5to18$y = sum5to18$y - nff(x = y, n = 4L, up = 10L, plot = FALSE)$y
png("sum5to18.png")
plot(sum5to18, pch = 16L, xlab = "Time", ylab = "Measurement", main = "5th to 18th harmonics sum", type = "l", col = 2)
dev.off()

添加参数add和col，我们也可以使用特定颜色绘制多个波形

colors = rainbow(36L, alpha = 0.3)
nff(x = y, n = 36L, up = 100L, col = colors[1])
png("all_waves.png")
for(i in 1:18){
  ad = ifelse(i == 1, FALSE, TRUE)
  nff(x = y, n = i, up = 100L, col = colors[i], add = ad, main = "All waves up to 18th harmonic")
}
dev.off()

有没有办法提取每个系列的数据然后保存为csv文件。所以在这个例子中，我应该有18个csv文件用于18波。

我编辑了代码以允许0次谐波（基本上是一个平均值），所以现在你提取单独的波为：

sep = array(data = NA_real_, dim = c(7300L, 2 + 18), dimnames = list(NULL, c("t", paste0("H", 0:18))))
sep[,1:2] = as.matrix(nff(x = y, n = 0, up = 100L, plot = FALSE))

for(i in 1:18L){
  sep[,i+2] = nff(x = y, n = i, up = 100L, plot = FALSE)$y - nff(x = y, n = i-1, up = 100L, plot = FALSE)$y
}

然后您可以使用write.table编写csv文件。

对R中的时间序列进行傅立叶分析

1 个答案: