我知道可以通过将rnorm()
的输出视为时间序列来实现白噪声。有关如何模拟粉红噪音的任何建议吗?
答案 0 :(得分:10)
包tuneR
具有noise
功能,可生成白色或粉红色噪音的波形对象:
require(tuneR)
w <- noise(kind = c("white"))
p <- noise(kind = c("pink"))
par(mfrow=c(2,1))
plot(w,main="white noise")
plot(p,main="pink noise")
编辑:我意识到上面的方法不生成向量(doh)。将其转换为向量的残酷方法是添加以下代码:
writeWave(p,"p.wav")#writes pink noise on your hard drive
require(audio)#loads `audio` package to use `load.wave` function
p.vec <- load.wave("path/to/p.wav")#this will load pink noise as a vector
答案 1 :(得分:0)
正如@mbq所说,您可以使用p @ left来获取向量,而不是保存并读取wav文件。另一方面,您可以直接使用在tuneR中生成时间序列的函数:
min: function() {
return minqty;
}
那很完美:
TK95 <- function(N, alpha = 1){
f <- seq(from=0, to=pi, length.out=(N/2+1))[-c(1,(N/2+1))] # Fourier frequencies
f_ <- 1 / f^alpha # Power law
RW <- sqrt(0.5*f_) * rnorm(N/2-1) # for the real part
IW <- sqrt(0.5*f_) * rnorm(N/2-1) # for the imaginary part
fR <- complex(real = c(rnorm(1), RW, rnorm(1), RW[(N/2-1):1]),
imaginary = c(0, IW, 0, -IW[(N/2-1):1]), length.out=N)
# Those complex numbers that are to be back transformed for Fourier Frequencies 0, 2pi/N, 2*2pi/N, ..., pi, ..., 2pi-1/N
# Choose in a way that frequencies are complex-conjugated and symmetric around pi
# 0 and pi do not need an imaginary part
reihe <- fft(fR, inverse=TRUE) # go back into time domain
return(Re(reihe)) # imaginary part is 0
}