cutoff freq of butterworth filter

157 visualizzazioni (ultimi 30 giorni)
joo
joo il 26 Nov 2012
Commentato: George James il 24 Nov 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

Risposta accettata

Wayne King
Wayne King il 26 Nov 2012
Modificato: Wayne King il 27 Nov 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.
  7 Commenti
Kenny
Kenny il 7 Nov 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 il 24 Nov 2018
yes, I wish somebody could answer this. Ive never found an answer before. I second this question

Accedi per commentare.

Più risposte (1)

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

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by