cutoff freq of butterworth filter

149 次查看（过去 30 天）

joo 2012-11-26

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/54875-cutoff-freq-of-butterworth-filter

评论： George James 2018-11-24

Cutoff frequency is that frequency where the magnitude response of the filter is sqr(1/2). For butter, the normalized cutoff frequency Wn must be a number between 0 and 1, where 1 corresponds to the Nyquist frequency, π radians per sample.

[b,a]=butter(n,Wn)

my fs=40 fc=9

so my wn= 9/40 or wn=9/(40/2) ?

can you explain? than you very juch

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

请先登录，再回答此问题。

采纳的回答

Wayne King 2012-11-26

2
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/54875-cutoff-freq-of-butterworth-filter#answer_66597

编辑：Wayne King 2012-11-27

在 MATLAB Online 中打开

To convert from frequency in Hz to normalized frequency divide the desired cutoff frequency in Hz by 1/2 the sampling rate.

So if your sampling rate is 40 Hz, then a cutoff frequency of 9 Hz is

9/(40/2)

For normalized frequency multiplication by pi is implied so 0.5 is really 0.5*pi or pi/2 radians/sample. Note that pi/2 radians/sample is 1/4 cycle/sample multiply that by 40 samples/sec and you get 10 cycles/sec or 10 Hz. Therefore a normalized frequency of 0.5(pi) is equal to 10 cycles/sec with a sampling frequency of 40 Hz.

7 个评论
显示 5更早的评论隐藏 5更早的评论

Kenny 2018-11-7

Hello Wayne. Many years late here. But .... could you please explain "For normalized frequency multiplication by pi is implied" ? Why is multiplication by 'pi' implied? What's the origins of that? Thanks Wayne.