SOX采样率误差

Question

我使用下面的转换代码使用SOX将所有原始文件转换为16 KHZ、16位、mono格式的wav文件。我使用了以下代码

我使用了以下代码

#!/bin/bash
OutDir=converted 
for input in "$@"   
   do sox -S $input -r 16000 -c 1 -b 16 -w -s -t raw "$OutDir/$(basename $input)" 
done

虽然我提到了采样率-r 16000，但还是观察到了以下误差

bsnayak@ubuntu:~/MLLR/input$ ls
an251-fash-b.raw  an254-fash-b.raw  cen1-fash-b.raw  cen4-fash-b.raw  cen7-fash-b.raw  convert.sh
an253-fash-b.raw  an255-fash-b.raw  cen2-fash-b.raw  cen5-fash-b.raw  converted
bsnayak@ubuntu:~/MLLR/input$ ./convert.sh /home/bsnayak/MLLR/input/*.raw 
sox FAIL formats: bad input format for file `/home/bsnayak/MLLR/input/an251-fash-b.raw': sampling rate was not specified
sox FAIL formats: bad input format for file `/home/bsnayak/MLLR/input/an253-fash-b.raw': sampling rate was not specified
sox FAIL formats: bad input format for file `/home/bsnayak/MLLR/input/an254-fash-b.raw': sampling rate was not specified
sox FAIL formats: bad input format for file `/home/bsnayak/MLLR/input/an255-fash-b.raw': sampling rate was not specified
sox FAIL formats: bad input format for file `/home/bsnayak/MLLR/input/cen1-fash-b.raw': sampling rate was not specified
sox FAIL formats: bad input format for file `/home/bsnayak/MLLR/input/cen2-fash-b.raw': sampling rate was not specified
sox FAIL formats: bad input format for file `/home/bsnayak/MLLR/input/cen4-fash-b.raw': sampling rate was not specified
sox FAIL formats: bad input format for file `/home/bsnayak/MLLR/input/cen5-fash-b.raw': sampling rate was not specified
sox FAIL formats: bad input format for file `/home/bsnayak/MLLR/input/cen7-fash-b.raw': sampling rate was not specified

请随时提出建议。

使用下面的命令解决：

#!/bin/bash

SAVEIF=$IFS
IFS=$(echo -en "\n\b")

for file in $(ls *raw)
do
  name=${file%%.raw}

  sox -S -V -r 16k -e signed -c 1 -b 16  $name.raw $name.wav
done
IFS=$SAVEIFS

Answer 1

sox的命令行要求文件选项位于应用它们的文件的名称之前。您没有在$input之前设置任何参数，所以sox不知道输入文件的格式。

假设输入文件是16k，16位签名和1通道，那么重新安排文件参数在输入文件之前将使您通过您的错误。还需要通过附加一个.wav扩展来指示输出格式为wav：

sox -S  -r 16000 -c 1 -b 16 -w -s -t raw $input "$OutDir/$(basename $input).wav"

默认情况下，输入格式将以输出格式复制，除非您通过在输出文件名之前添加参数显式地覆盖它。