butterworth.c

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/* Copyright (c) Colorado School of Mines, 2006.*/
/* All rights reserved.                       */

/*********************** self documentation **********************/
/*****************************************************************************
BUTTERWORTH - Functions to design and apply Butterworth filters:

bfdesign	design a Butterworth filter
bfhighpass	apply a high-pass Butterworth filter 
bflowpass	apply a low-pass Butterworth filter 

******************************************************************************
Function Prototypes:
void bfhighpass (int npoles, float f3db, int n, float p[], float q[]);
void bflowpass (int npoles, float f3db, int n, float p[], float q[]);
void bfdesign (float fpass, float apass, float fstop, float astop,
	int *npoles, float *f3db);

******************************************************************************
bfdesign:
Input:
fpass		frequency in pass band at which amplitude is >= apass
apass		amplitude in pass band corresponding to frequency fpass
fstop 		frequency in stop band at which amplitude is <= astop
astop		amplitude in stop band corresponding to frequency fstop

Output:
npoles		number of poles
f3db		frequency at which amplitude is sqrt(0.5) (-3 db)

bfhighpass and bflowpass:
Input:
npoles		number of poles (and zeros); npoles>=0 is required
f3db		3 db frequency; nyquist = 0.5; 0.0<=f3db<=0.5 is required
n		length of p and q
p		array[n] to be filtered

Output:
q		filtered array[n] (may be equivalent to p)

******************************************************************************
Notes:
(1) Nyquist frequency equals 0.5

(2) The following conditions must be true:
	(0.0<fpass && fpass<0.5) &&
	(0.0<fstop && fstop<0.5) &&
	(fpass!=fstop) &&
	(0.0<astop && astop<apass && apass<1.0)

(3) if (fpass<fstop)

bfdesign:
Butterworth filter:  compute number of poles and -3 db frequency
for a low-pass or high-pass filter, given a frequency response
constrained at two frequencies.

******************************************************************************
Author:  Dave Hale, Colorado School of Mines, 06/02/89
*****************************************************************************/
/**************** end self doc ********************************/

#include "cwp.h"

void
bfdesign (float fpass, float apass, float fstop, float astop,
	int *npoles, float *f3db)
/*****************************************************************************
Butterworth filter:  compute number of poles and -3 db frequency
for a low-pass or high-pass filter, given a frequency response
constrained at two frequencies.
******************************************************************************
Input:
fpass		frequency in pass band at which amplitude is >= apass
apass		amplitude in pass band corresponding to frequency fpass
fstop 		frequency in stop band at which amplitude is <= astop
astop		amplitude in stop band corresponding to frequency fstop

Output:
npoles		number of poles
f3db		frequency at which amplitude is sqrt(0.5) (-3 db)
******************************************************************************
Notes:
(1) Nyquist frequency equals 0.5

(2) The following conditions must be true:
	(0.0<fpass && fpass<0.5) &&
	(0.0<fstop && fstop<0.5) &&
	(fpass!=fstop) &&
	(0.0<astop && astop<apass && apass<1.0)

(3) if (fpass<fstop)
		a low-pass filter is assumed
	else
		a high-pass filter is assumed
******************************************************************************
Author:  Dave Hale, Colorado School of Mines, 06/02/89
*****************************************************************************/
{
	float wpass,wstop,fnpoles,w3db;

	/* warp frequencies according to bilinear transform */
	wpass = 2.0*tan(PI*fpass);
	wstop = 2.0*tan(PI*fstop);

	/* if lowpass filter, then */
	if (fstop>fpass) {
		fnpoles = log((1.0/(apass*apass)-1.0)/(1.0/(astop*astop)-1.0))
			/ log(pow(wpass/wstop,2.0));
		w3db = wpass/pow((1.0/(apass*apass)-1.0),0.5/fnpoles);

	/* else, if highpass filter, then */
	} else {
		fnpoles = log((1.0/(apass*apass)-1.0)/(1.0/(astop*astop)-1.0))
			/ log(pow(wstop/wpass,2.0));
		w3db = wpass*pow((1.0/(apass*apass)-1.0),0.5/fnpoles);
	}

	/* determine integer number of poles */
	*npoles = 1+(int)fnpoles;

	/* determine (unwarped) -3 db frequency */
	*f3db = atan(0.5*w3db)/PI;
}

void
bfhighpass (int npoles, float f3db, int n, float p[], float q[])
/*****************************************************************************
Butterworth filter:  high-pass
******************************************************************************
Input:
npoles		number of poles (and zeros); npoles>=0 is required
f3db		3 db frequency; nyquist = 0.5; 0.0<=f3db<=0.5 is required
n		length of p and q
p		array[n] to be filtered

Output:
q		filtered array[n] (may be equivalent to p)
******************************************************************************
Author:  Dave Hale, Colorado School of Mines, 06/02/89
*****************************************************************************/
{
	int jpair,j;
	float r,scale,theta,a,b1,b2,pj,pjm1,pjm2,qjm1,qjm2;

	r = 2.0*tan(PI*fabs(f3db));
	if (npoles%2!=0) {
		scale = r+2.0;
		a = 2.0/scale;
		b1 = (r-2.0)/scale;
		pj = 0.0;
		qjm1 = 0.0;
		for (j=0; j<n; j++) {
			pjm1 = pj;
			pj = p[j];
			q[j] = a*(pj-pjm1)-b1*qjm1;
			qjm1 = q[j];
		}
	} else {
		for (j=0; j<n; j++)
			q[j] = p[j];
	}
	for (jpair=0; jpair<npoles/2; jpair++) {
		theta = PI*(2*jpair+1)/(2*npoles);
		scale = 4.0+4.0*r*sin(theta)+r*r;
		a = 4.0/scale;
		b1 = (2.0*r*r-8.0)/scale;
		b2 = (4.0-4.0*r*sin(theta)+r*r)/scale;
		pjm1 = 0.0;
		pj = 0.0;
		qjm2 = 0.0;
		qjm1 = 0.0;
		for (j=0; j<n; j++) {
			pjm2 = pjm1;
			pjm1 = pj;
			pj = q[j];
			q[j] = a*(pj-2.0*pjm1+pjm2)-b1*qjm1-b2*qjm2;
			qjm2 = qjm1;
			qjm1 = q[j];
		}
	}
}

void
bflowpass (int npoles, float f3db, int n, float p[], float q[])
/*****************************************************************************
Butterworth filter:  low-pass
******************************************************************************
Input:
npoles		number of poles (and zeros); npoles>=0 is required
f3db		3 db frequency; nyquist = 0.5; 0.0<=f3db<=0.5 is required
n		length of p and q
p		array[n] to be filtered

Output:
q		filtered array[n] (may be equivalent to p)
******************************************************************************
Author:  Dave Hale, Colorado School of Mines, 06/02/89
*****************************************************************************/
{
	int jpair,j;
	float r,scale,theta,a,b1,b2,pj,pjm1,pjm2,qjm1,qjm2;

	r = 2.0*tan(PI*fabs(f3db));
	if (npoles%2!=0) {
		scale = r+2.0;
		a = r/scale;
		b1 = (r-2.0)/scale;
		pj = 0.0;
		qjm1 = 0.0;
		for (j=0; j<n; j++) {
			pjm1 = pj;
			pj = p[j];
			q[j] = a*(pj+pjm1)-b1*qjm1;
			qjm1 = q[j];
		}
	} else {
		for (j=0; j<n; j++)
			q[j] = p[j];
	}
	for (jpair=0; jpair<npoles/2; jpair++) {
		theta = PI*(2*jpair+1)/(2*npoles);
		scale = 4.0+4.0*r*sin(theta)+r*r;
		a = r*r/scale;
		b1 = (2.0*r*r-8.0)/scale;
		b2 = (4.0-4.0*r*sin(theta)+r*r)/scale;
		pjm1 = 0.0;
		pj = 0.0;
		qjm2 = 0.0;
		qjm1 = 0.0;
		for (j=0; j<n; j++) {
			pjm2 = pjm1;
			pjm1 = pj;
			pj = q[j];
			q[j] = a*(pj+2.0*pjm1+pjm2)-b1*qjm1-b2*qjm2;
			qjm2 = qjm1;
			qjm1 = q[j];
		}
	}
}