Transcription

1 Low-Pass Filter _012 Triple Butterworth Filter to avoid numerical instability for high order filter. In [5]: import numpy as np import matplotlib.pyplot as plt %matplotlib notebook fsz = (7,5) # figure size import scipy.signal as ss Suppose you need to design a Butterworth low-pass filter with -3 db cutoff at 50 Hz and stopband attenuation of -40 db at 75 Hz on a signal with sampling rate F s = 8000 Hz. This requires a fairly high filter order because the transition bandwidth of 25 Hz is only a small fraction (25/4000 = ) of the maximum frequency that can be represented at the given F s. In [2]: Fs = 8000 tt = np.arange(fs)/float(fs)-1/2.0 # time axis sec uimp = np.zeros(tt.size) ix0 = np.argmin(np.abs(tt)) uimp[ix0] = Fs # unit impulse In [3]: # Trying a filter order of 15 fl = 50 f_ord = 15 b,a = ss.butter(f_ord,2*fl/float(fs)) ht = ss.lfilter(b,a,uimp) # DT Butterworth filter In [6]: fsz2 = (fsz[0],fsz[1]/2.0) plt.figure(1, figsize=fsz2) plt.plot(tt,ht,'-b') plt.xlabel('t [sec]') plt.ylabel('h(t)') strt1 = 'Butterworth h(t), $f_l=${} Hz, ord={}'.format(fl,f_ord) strt1 = strt1 + ', $F_s=${} Hz'.format(Fs) plt.title(strt1) 1 of 11 10/2/18, 6:11 PM

2 This is clearly not the right impulse response of a low-pass filter. It is in fact the impulse response of an unstable filter. Because of the scaling of the y-axis (1e220), we can assume that the filtering was done using double precision floating point numbers (float64). To find out what's wrong, let's look at the poles and zeros of the filter. In [7]: f_poles = np.roots(a) f_zeros = np.roots(b) u_circ = np.exp(1j*2*np.pi*np.arange(361)/360.0) # unit circle plt.figure(3,fsz) plt.plot(np.real(u_circ),np.imag(u_circ),':k') plt.plot(np.real(f_poles),np.imag(f_poles),'xr',label='poles') plt.plot(np.real(f_zeros),np.imag(f_zeros),'ob',label='zeros') plt.gca().set_aspect('equal', adjustable='box') plt.xlabel('real Part') plt.ylabel('imaginary Part') strt3 = 'Butterworth Poles/Zeros, $f_l=${} Hz, ord={}'.format(fl,f_ord) strt3 = strt3 + ', $F_s=${} Hz'.format(Fs) plt.title(strt3) plt.legend(loc=1) It is easy to see that there are poles outside the unit circle which cause the instability of the filter. Also since f L = 50 = 1/ a b, the poles seem to be too far spread out. The most likely reason for this is that the computation of the and coefficients of the filter would have required a higher precision that float64. Through trial and error we found that for the given parameters ( f L and F s ) the highest filter order that results in a stable filter is 9. So let's try that. 2 of 11 10/2/18, 6:11 PM

3 In [8]: # Trying a filter order of 9 fl = 50 f_ord = 9 b,a = ss.butter(f_ord,2*fl/float(fs)) ht = ss.lfilter(b,a,uimp) # DT Butterworth filter In [9]: plt.figure(5, figsize=fsz2) plt.plot(tt,ht,'-b') plt.xlabel('t [sec]') plt.ylabel('h(t)') strt5 = 'Butterworth h(t), $f_l=${} Hz, ord={}'.format(fl,f_ord) strt5 = strt5 + ', $F_s=${} Hz'.format(Fs) plt.title(strt5) Now we can take a look at the frequency response of the filter. 3 of 11 10/2/18, 6:11 PM

4 In [10]: def FTapprox(tt,xt,dlim): """ DFT/FFT approximation to Fourier transform of x(t) with time axis tt. >>>>> ff, Xf, absxf, argxf, Df = FTapprox(tt,xt,dlim) <<<<< where ff: Frequency axis absxf: X(f) argxf: angle(x(f)) in degrees Df: Frequency resolution tt: Time axis xt: Sampled CT waveform dlim = [f1,f2,llim]: display limits f1,f2: Lower/upper frequency limits llim = 0: display Xfabs linear and absolute llim > 0: same as llim=0 but phase is masked for Xfabs<llim llim < 0: display 20*log_{10}( X(f)/max( X(f))) in db with lower display limit llim db. phase is masked for display values < llim db """ # ***** prepare x(t), swap pos/neg parts along time axis ***** N = tt.size Fs = int(round((n-1)/(tt[-1]-tt[0]))) ixtp = np.where(tt>=0)[0] ixtn = np.where(tt<0)[0] xtp = np.hstack((xt[ixtp],xt[ixtn])) # ***** compute X(f), make frequency axis ***** Xf = 1/float(Fs)*np.fft.fft(xtp) ff = Fs/float(N)*np.arange(N) if dlim[0] < 0: ixfp = np.where(ff<fs/2.0)[0] ixfn = np.where(ff>=fs/2.0)[0] ff = np.hstack((ff[ixfn]-fs, ff[ixfp])) Xf = np.hstack((xf[ixfn],xf[ixfp])) # ***** compute X(f), arg[x(f)], trim to [f1,f2) ***** absxf = np.abs(xf) maxxf = np.amax(absxf) # max of X(f) ixf = np.where(np.logical_and(ff>=dlim[0],ff<dlim[1]))[0] ff = ff[ixf] Xf = Xf[ixf] absxf = absxf[ixf] argxf = np.angle(xf) # ***** Mask phase ***** if dlim[2] > 0: ix = np.where(absxf<dlim[2])[0] argxf[ix] = 0 elif dlim[2] < 0: absxf = absxf/float(maxxf) # normalized magnitude ix = np.where(absxf==0)[0] absxf[ix] = 1e-300 # avoid log of zero absxf = 20*np.log10(absXf) ix = np.where(absxf<dlim[2])[0] absxf[ix] = dlim[2] argxf[ix] = 0 # mask phase return ff, Xf, absxf, argxf, Fs/float(N) In [11]: f_disp, llim = 200, 1e-6 dlim = [-f_disp, f_disp, llim] ff,hf,abshf,arghf,df = FTapprox(tt,ht,dlim) N = tt.size 4 of 11 10/2/18, 6:11 PM

5 In [12]: plt.figure(7,figsize=fsz) plt.subplot(211) plt.plot(ff,abshf,'-b') plt.ylabel('$ H(f) $') strt7 = 'Butterworth $H(f)$, $f_l=${} Hz, ord={}'.format(fl, f_ord) strt7 = strt7 + ', $F_s=${} Hz'.format(Fs) plt.title(strt7) plt.subplot(212) plt.plot(ff,180/np.pi*arghf,'-r') plt.ylabel('arg[$h(f)$] [deg]') plt.xlabel('f [Hz]') One of the defining characteristics of Butterworth filters is that their magnitude is maximally flat in the passband. This is clearly not the case in the above magnitude frequency response. Let's try what happens if we reduce the filter order to 8. In [13]: # Trying a filter order of 8 fl = 50 f_ord = 8 b,a = ss.butter(f_ord,2*fl/float(fs)) ht = ss.lfilter(b,a,uimp) # DT Butterworth filter In [14]: ff,hf,abshf,arghf,df = FTapprox(tt,ht,dlim) 5 of 11 10/2/18, 6:11 PM

6 In [15]: plt.figure(9,figsize=fsz) plt.subplot(211) plt.plot(ff,abshf,'-b') plt.ylabel('$ H(f) $') strt9 = 'Butterworth $H(f)$, $f_l=${} Hz, ord={}'.format(fl, f_ord) strt9 = strt9 + ', $F_s=${} Hz'.format(Fs) plt.title(strt9) plt.subplot(212) plt.plot(ff,180/np.pi*arghf,'-r') plt.ylabel('arg[$h(f)$] [deg]') plt.xlabel('f [Hz]') By design, the -3 db frequency is 50 Hz. But what about the -40 db attenuation? At which frequency does it occur? We need to take a look at the magnitude of the frequency response in db. In [16]: f_dispdb, llimdb = 200, -80 dlimdb = [-f_dispdb, f_dispdb, llimdb] ffdb,hf,abshfdb,arghfdb,df = FTapprox(tt,ht,dlimdB) 6 of 11 10/2/18, 6:11 PM

7 In [17]: plt.figure(11,figsize=fsz) plt.subplot(211) plt.plot(ffdb,abshfdb,'-b') plt.ylabel('$ H(f) $ [db]') strt11 = 'Butterworth $H(f)$, $f_l=${} Hz, ord={}'.format(fl, f_ord) strt11 = strt11 + ', $F_s=${} Hz'.format(Fs) plt.title(strt11) plt.subplot(212) plt.plot(ffdb,180/np.pi*arghfdb,'-r') plt.ylabel('arg[$h(f)$] [deg]') plt.xlabel('f [Hz]') Zooming in, we find that the -40 db frequency is approximately 89 Hz. To make the transition from passband to stopband more steep we need to add more filtering. To avoid numerical instability, we need to use more than one filter to increase the filter order. Ideally, we could compute the poles and zeros from the and polynomials and then distribute them over the filters. But since the problem here seems to be the and polynomials themselves, we just have to specify -3 db frequencies and orders 8 for two or more filters in cascade. a b a b In [18]: # parameters for cascade of two Butterworth filters fl1, f_ord1 = 57, 8 fl2, f_ord2 = 51, 7 In [19]: b1,a1 = ss.butter(f_ord1,2*fl1/float(fs)) # DT Butterworth filter 1 h1t = ss.lfilter(b1,a1,uimp) b2,a2 = ss.butter(f_ord2,2*fl2/float(fs)) # DT Butterworth filter 2 h2t = ss.lfilter(b2,a2,h1t) 7 of 11 10/2/18, 6:11 PM

8 In [20]: plt.figure(13, figsize=fsz2) plt.plot(tt,h2t,'-b') plt.xlabel('t [sec]') plt.ylabel('h(t)') strt13 = 'Two Butterworth h(t), $f_{{l1}}=${} Hz, ord$_1$={}'.format(fl1,f_ ord1) strt13 = strt13 + ', $f_{{l2}}=${} Hz, ord$_2$={}'.format(fl2,f_ord2) strt13 = strt13 + ', $F_s=${} Hz'.format(Fs) plt.title(strt13) In [21]: ffdb,h2f,absh2fdb,argh2fdb,df = FTapprox(tt,h2t,dlimdB) 8 of 11 10/2/18, 6:11 PM

9 In [22]: plt.figure(15,figsize=fsz) plt.subplot(211) plt.plot(ffdb,absh2fdb,'-b') plt.ylabel('$ H(f) $ [db]') strt15 = 'Two Butterworth $H(f)$, $f_{{l1}}=${} Hz, ord$_1$={}'.format(fl1, f_ord1) strt15 = strt15 + ', $f_{{l2}}=${} Hz, ord$_2$={}'.format(fl2,f_ord2) strt15 = strt15 + ', $F_s=${} Hz'.format(Fs) plt.title(strt15) plt.subplot(212) plt.plot(ffdb,180/np.pi*argh2fdb,'-r') plt.ylabel('arg[$h(f)$] [deg]') plt.xlabel('f [Hz]') f L1 = 57 1 = 8 f L2 = 51 = 7 By trial and error, we found a cascade of two Butterworth filters, with Hz, ord, and Hz, ord2 with -3 db frequency of Hz and -40 db frequency Hz. The total filter order is thus To compensate for the filter delay, we measure the delay of the peak of the unit impulse response and then pad the filter input with zeros corresponding to that delay. After filtering the same amount of samples is removed from the beginning of the filter response. Small adjustments to the delay can then be made based on the phase of the frequency response. In [23]: t_dly = 28.5e-3 # Filter delay in seconds n_dly = int(np.round(t_dly*fs)) h1t_dly = ss.lfilter(b1,a1,np.hstack((uimp,np.zeros(n_dly)))) h2t_dly = ss.lfilter(b2,a2,h1t_dly) h2t_dly = h2t_dly[n_dly:] # delay compensated output 9 of 11 10/2/18, 6:11 PM

10 In [24]: plt.figure(17, figsize=fsz2) plt.plot(tt,h2t_dly,'-b') plt.xlabel('t [sec]') plt.ylabel('h(t)') strt17 = 'Two Butterworth h(t), $f_{{l1}}=${} Hz, ord$_1$={}'.format(fl1,f_ ord1) strt17 = strt17 + ', $f_{{l2}}=${} Hz, ord$_2$={}'.format(fl2,f_ord2) strt17 = strt17 + ', $F_s=${} Hz'.format(Fs) plt.title(strt17) In [25]: ffdb,h2f_dly,absh2fdb_dly,argh2fdb_dly,df = FTapprox(tt,h2t_dly,dlimdB) 10 of 11 10/2/18, 6:11 PM

11 In [26]: plt.figure(19,figsize=fsz) plt.subplot(211) plt.plot(ffdb,absh2fdb_dly,'-b') plt.ylabel('$ H(f) $ [db]') strt19 = 'Two Butterworth $H(f)$, $f_{{l1}}=${} Hz, ord$_1$={}'.format(fl1, f_ord1) strt19 = strt19 + ', $f_{{l2}}=${} Hz, ord$_2$={}'.format(fl2,f_ord2) strt19 = strt19 + ', $F_s=${} Hz'.format(Fs) plt.title(strt19) plt.subplot(212) plt.plot(ffdb,180/np.pi*argh2fdb_dly,'-r') plt.ylabel('arg[$h(f)$] [deg]') plt.xlabel('f [Hz]') After some fine tuning, the phase reponse near dc is essentially flat, indicating a proper delay compensation of the filter. In [ ]: 11 of 11 10/2/18, 6:11 PM