数字信号 ■编码位数的选择 口将x[n]转换为十进制 x[n=.,0.000,0.101,0.111,0.101,0.000,1.101,1.111,1.101.} x[n=.,0.0,0.625,0.875,0.625,0.0,-0.625,-0.875,-0.625,} ◆存在误差 x(n)={.,0.0,0.6364,0.900,0.6364,0.0000,-0.6364,-0.9000-0.6364,} 口量化误差：幅度值上有误差，有损变换 与二进制编码位数有关 15数字信号  编码位数的选择  将x[n]转换为十进制 { } x n[ ] { ,0.000, ,0.111, ,0.000, ,1.111    0.101 0.101 1.101 1.101 , , } 存在误差 x[ ] n  {  ,0.0, ,0.875, ,0.0, , 0.875 0.625 0.625 0.625  ,0.625, } x n( ) { ,0.0, ,0.900, ,0    0.6364 0.6364 0. .0000, , 0.9000,   6364 0.6364  , }  量化误差：幅度值上有误差，有损变换 与二进制编码位数有关 15