如何提取.wav文件特定频率范围？ [英] How to extract a specific frequency range from a .wav file?

问题描述

我的真的新的无害化处理，所以也许我的问题将是微不足道的。
我想要做的是提取特定频率范围（比方说150-400赫兹）从一个wav文件，使用R.换句话说，我要创建另一个波形文件（WAVE2）只包含的频率成分，我指定（150〜400赫兹，或者还有什么）。

I'm really new on sound processing, so maybe my question will be trivial. What I want to do is to extract a specific frequency range (let's say 150-400 Hz) from a wav file, using R. In other words, I want to create another wave file (wave2) that contains only the frequency component that I specify (150 to 400 Hz, or what else).

我看了网上的东西，我发现了，这可以用FFT分析来完成，而这里的来的问题。

I read something on the net, and I discovered out that this can be done with a FFT analysis, and here's come the problems.

假设我在这篇code：

Suppose I've this code:

library(sound)
s1 <- Sine(440, 1)
s2 <- Sine(880, 1)
s3 <- s1 + s2

s3.s <- as.vector(s3$sound)
  # s3.s is now a vector, with length 44100; 
  # bitrate is 44100 (by default)
  # so total time of s3 is 1sec.

  # now I calculate frequencies
N <- length(s3.s)   # 44100
k <- c(0:(N-1))
Fs <- 44100         # sampling rate
T <- N / Fs
freq <- k / T
x <- fft(s3.s) / N

plot(freq[1:22050], x[1:22050], type="l") # we need just the first half of FFT computation

我们得到的情节是：

The plot we obtain is:

那么，有两个峰值。如果我们想知道它们对应什么频率，只要找到：

Well, there are two peaks. If we want to know to what frequency they correspond, just find:

order(Mod(x)[1:22050], decreasing=T)[1:10]
[1] 441 881 882 880 883 442 440 879 884 878

前两个值是真正接近的频率，我用来创建我的声音：

First two values are really near to the frequency I've used to create my sound:

        real     computed
 Freq1: 440   |  441 
 Freq2: 880   |  881 

所以，现在问题来了：如何进行，如果我想从我的声音范围中删除的频率，比如（1，500）？以及如何选择（并保存）只有范围（1，500）？
我顾不上什么，是我的新的声音（与删除的频率）将接近到简单的事正弦（频率= 880，持续时间= 1）（我知道，它不能是酷似如此！）。
那可能吗？

So, now comes the problem: how to proceed, if I want to delete from my sound the frequencies in the range, say, (1, 500) ? And how to select (and save) only the range (1, 500) ? What I attend, is that my new sound (with deleted frequencies) will be something near to simple Sine(freq=880, duration=1) (I know, it cannot be exactly like so!). Is that possible?

我是pretty确保 FFT（DATA，逆= TRUE）正是我需要的。但我不知道，但和我不知道如何着手。

I'm pretty sure that fft(DATA, inverse = TRUE) is what I need. But I'm not sure, and however I don't know how to proceed.

推荐答案

也许我忽略了一点，但不要你已经有答案？从您的文章：

Maybe I missed the point, but don't you already have your answer? From your post:

order(Mod(x)[1:22050], decreasing=T)[1:10]
[1] 441 881 882 880 883 442 440 879 884 878 

只要收集超过500的所有值：

Simply collect all values above 500:

junk <- order(Mod(x)[1:22050], decreasing=T)[1:10]
(junk1 <- junk[junk > 500])
[1] 881 882 880 883 879 884 878

要生成新的信号简单地重复你做了什么来构建原始信号：

To generate the new signal simply repeat what you did to build the original signal:

junk2 <- Sine(0, 1)    
for (i in 1:length(junk1)) {     
    junk2 <- junk2 + Sine(junk1[i], 1)    
}    
junk2.s <- as.vector(junk2$sound)    

要保持低于500的值：

To keep the values below 500:

(junk3 <- junk[junk <= 500])
[1] 441 442 440

这篇关于如何提取.wav文件特定频率范围？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！