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cutoff freq of butterworth filter

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joo
joo el 26 de Nov. de 2012
Comentada: George James el 24 de Nov. de 2018
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

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Wayne King
Wayne King el 26 de Nov. de 2012
Editada: Wayne King el 27 de Nov. de 2012
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.
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Kenny
Kenny el 7 de Nov. de 2018
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.
George James
George James el 24 de Nov. de 2018
yes, I wish somebody could answer this. Ive never found an answer before. I second this question

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BHAVATHA BALIGA
BHAVATHA BALIGA el 10 de Nov. de 2017
If I hav a pre generated wave how to get it's cut off and sampling frequency??

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