buttord | R Documentation |
Compute butterworth filter order and cutoff for the desired response characteristics.
buttord(Wp, Ws, Rp, Rs)
Wp , Ws |
pass-band and stop-band edges. For a low-pass or
high-pass filter, |
Rp |
allowable decibels of ripple in the pass band. |
Rs |
minimum attenuation in the stop band in dB. |
Deriving the order and cutoff is based on:
|H(W)|^2 = \frac{1}{1+(W/Wc)^{2n}} = 10^{-R/10}
With some algebra, you can solve simultaneously for Wc
and n
given
Ws
, Rs
and Wp
, Rp
. For high-pass filters, subtracting the band edges
from Fs/2, performing the test, and swapping the resulting Wc
back
works beautifully. For bandpass- and bandstop-filters, this process
significantly overdesigns. Artificially dividing n
by 2 in this case
helps a lot, but it still overdesigns.
An object of class FilterOfOrder
with the following list elements:
n |
filter order |
Wc |
cutoff frequency |
type |
filter type, one of “low”, “high”, “stop”, or “pass” |
This object can be passed directly to butter
to compute filter coefficients.
Original Octave version by Paul Kienzle, pkienzle@user.sf.net. Conversion to R by Tom Short.
Octave Forge https://octave.sourceforge.io/
butter
, FilterOfOrder
, cheb1ord
Fs <- 10000
btord <- buttord(1000/(Fs/2), 1200/(Fs/2), 0.5, 29)
plot(c(0, 1000, 1000, 0, 0), c(0, 0, -0.5, -0.5, 0),
type = "l", xlab = "Frequency (Hz)", ylab = "Attenuation (dB)")
bt <- butter(btord)
plot(c(0, 1000, 1000, 0, 0), c(0, 0, -0.5, -0.5, 0),
type = "l", xlab = "Frequency (Hz)", ylab = "Attenuation (dB)",
col = "red", ylim = c(-10,0), xlim = c(0,2000))
hf <- freqz(bt, Fs = Fs)
lines(hf$f, 20*log10(abs(hf$h)))
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