零极点累试法设计滤波器_图文

1、Double Notch Filter DesignXun Wang10112746101.IntroductionA lot of signals in real situation are often associated with noise in signal processing.In order to reduce or eliminate the influence of noise and extract useful signals,we use a filter to decrease the scale of noise signal.Filter can be defi

2、ned as a system which can produce the response which we need by inputing the original signal to the system.The requirement of response can be exist in time domain or frequency domain.The function of filter is mainly decided by the system function.The pole location have a significant effect on the pr

3、operties of amplitude,the position of the peak of wave and extent of acuity.The zero location mainly affect the position of trough of wave and the extent of hollow.We can qualitatively draw the shape of amplitude according to the analysis of pole and zero points.This theory provide us a method to di

4、rectly design the filter:At first we can set the position of poles according to the properties of amplitude,and make sure what the system function is.Comparing it with the one we need,if the system function can not satisfy our requirement,we can revise it by adding or deleting the number of zero poi

5、nts,or by move the positions of zero points.This method is called pole-zero placement filter design method.It can be used not only in designing IIR digital filters but also in designing FIR digital filters.Filter design2.2.FilterIn order to obtain the system function,we must try many times to set th

6、e the poles and zeros of the system and observe if the properties can satisfy the requirement.If it does not,we must try these steps again and again to make the filter satisfy our requirement.In fact it is a process of approaching.We should obey some rules when using this method.(1we should consider

7、 which kind of filter it belongs to.If we need to design a FIR filter,we should not set poles,and just need to set zeros.By doing this we can design a non-recursive filter.It is possible that it becomes to recursive if we set some poles in the designing process.(2If we want to filter a point of freq

8、uency,we can set a zero at this point of digital frequency.(3The poles must be in the unit circle according to the requirement of a stable system.(4Aiming at accomplish the structure of filter,the poles are zeros are both should be set symmetrically(other than they are in the real axis.It means that

9、 the a couple of zeros must be conjugate.The situation is the same as poles.(5If we want to make the curve of frequency deeper or sharper at a certain point,we can set more than one zeros or poles near this point.Proccess:Step1:analysis of the signal in frequency domainBy analysing the frequency of

10、the signal,we can make sure which frequency we should to filter.The value of an ideal notch filter should be 0at the point of frequency which we need to eliminate and be 1at the other frequency.It just be like this:(=010,H (2-1The sampling period is 81921,so the sampling frequency is 8192Hz.The freq

11、uency f1=697Hz and f2=1209Hz are the points of frequency which we should eliminate.Step2:setting the zero point of the filter.Because the property of digital filter in frequency domain is its Z-transfer in unit-circle of unit impulse response,we just need to set zero at the points in unit circle whi

12、ch correspond to the location of the frequency which you need to eliminate.17.020851.0281926971=×=×=(2-2(a295.02819212092=×=(2-2(bSo we set four zeros at295.0*(-exp ,295.0*(exp c ,0.17*(-exp c ,0.17*exp(j 4321j c j j c =The system function is42244321z 1.2z -(1z 1.7208z -1(-(-(-(-(z z

13、c z c z c z c z K z H +=4-3-2-1-4234z z 1092.2z 5206.4z 1092.2-1z 1z 1092.2-z 5206.4z 1092.2-z +=+=(2-3K is a constant.we can suppose it is 1now.We can use matlab to draw the curve of this system function.Input matlab code:clear all;close all;clc;BB=exp(697/8192*2*pi*iexp(-(697/8192*2*pi*iexp(1209/8

14、192*2*pi*iexp(-(1209/8192*2*pi*i;B=poly(BB;A=1;W=-3*pi:0.01:3*pi;H=freqz(B,A,W;subplot(211,plot(W/pi,abs(H,'LineWidth',2;xlabel('omega/pi'ylabel('|H(ejomega|'title('amplitude-frequency response'axis(-22-0.215;grid;subplot(212,plot(W/pi,angle(H,'LineWidth',2;xl

15、abel('omega/pi'ylabel('theta(omega'title('phase-frequency response'axis(-22-44;grid;set(gcf,'color','w'We can get the curve: Figure1the amplitude-frequency response andphase-frequency response curves Figure2The affection of the two zeros in curveThe two figure

16、s present the properties of filter without poles.From them we can find that the two points are really.0.It means that the noise at f1and f2can295.0and17really be filtered.But we can also find some serious flaws.There are more noise at high frequency band and the transition band is long,so it is not

17、an perfect filter.Step3:setting the zero point of the filterAfter we finishing the processing of setting the zeros,we begin to set the poles.The poles must be in the unit cycle.The offset effect will be more obvious if the poles are closer to zeros,then the stop-band will be more narrow and the tran

18、sition band will be steeper.In order to prove this theorem,we set the two poles295.0(-exp ,295.0(exp ,17.0(-exp ,17.0(exp 24231211=j d j d j d j d In this equation,1and 2are variables.So the system function can be expressed as(222121222122221213214234222221122243214321z 2.17208.1-z 065.2z 2.11.7208-

19、z 1z 1092.2-z 5206.4z 2109.2-z z 2.1-z (z 7208.1-z (1z 2.1-z (1z 1.7208-z (d -z (d -z (d -z (d -z (c -z (c -z (c -z (c -z (z (H +=+=(2-4Because we are detecting the relationship between 21,and |H(z|,we must change the value of 1and 2and observe the change of curves.In order to simplify the process o

20、f designing,I set 21=.The matlab code isclear all;close all;clc;c=1;d=2;for a=0.96:0.01:0.99;AA=a*exp(0.17*pi*ia*exp(-0.17*pi*ia*exp(0.295*pi*ia*exp(-0.295*pi*i;A=poly(AA;BB=exp(697/8192*2*pi*iexp(-(697/8192*2*pi*iexp(1209/8192*2*pi*iexp(-(1209/8192*2*pi*i;B=poly(BB;W=-3*pi:0.00001:3*pi;H=freqz(B,A,

21、W;subplot(4,2,c,semilogy(W*8192/(2*pi,abs(H;xlabel('f'ylabel('|H(ejomega|'title('amplitude-frequency response'axis(04500-200150;grid;subplot(4,2,d,plot(W*8192/(2*pi,angle(H,'LineWidth',2;xlabel('f'ylabel('theta(omega'title('phase-frequency response

22、'axis(04500-44;grid;set(gcf,'color','w'c=c+2;d=d+2;EndThe result is F igure3amplitude-frequency and phase-frequencycurves of filter with variable polesWe can find that the first set of data is suitable to us.We can zoom the first figure: Figure4The amplitude-frequency response wh

23、en1=2=0.96It is obvious that the noise at0.17and0.295are really filtered and there is no noise at high frequency band.The transtion band can be steep enough considering the amplitude-frequency curve of noise signal in x. Figure5amplitude-frequency curve of noise signal Experimental results3.3.Experi

24、mentalThe amplitude-frequency and phase-frequency spectrums of original sound x are: Figure6Amplitude-frequency and phase-frequencyspectrums of original sound xBy passing the sound x through the filter I designed,we can get the output signal f1. The program is like this:load x.matBB=exp(697/8192*2*p

25、i*iexp(-(697/8192*2*pi*iexp(1209/8192*2*pi*iexp(-(1209/8192*2*pi*i;B=poly(BB;AA=0.96*exp(697/8192*2*pi*i0.96*exp(-(697/8192*2*pi*i0.96*exp(1209/8192*2*pi*i0.96*exp(-(1209/8192*2*pi*i;A=poly(AA;f1=filter(B,A,x;freqz(f1The spectrums of the output signalare Figure 7Amplitude-frequency and phase-frequen

26、cyspectrums of output signalFrom this figure I think the first set of data is the most suitable because the magnitude at the noise frequency points is decreased.The system function of filter is4-3-2-1-4-3-2-1-z .849302.5843z -z 7464.3z 2.8041-1z z 1092.2-z 6520.4z 9210.2-1z (H +=(2-5So4xn-3xn-1092.2

27、-2xn-5206.41xn-1092.2xn-4yn-8493.03yn-4358.2-2yn-6474.31yn-4180.2yn-X(zz z 1092.2-z 5206.4z 1092.2-1(Y(zz 8493.0z 4358.2-z 4674.3z 4180.2-1(H(zX(zY(z4-3-2-1-4-3-2-1-+=+=+=(2-6Equation (2-6is the difference equation.We can use matlab to calculate the poles and zeros of H(zThe program:BB=exp(697/8192*

28、2*pi*iexp(-(697/8192*2*pi*iexp(1209/8192*2*pi*iexp(-(1209/8192*2*pi*i;B=poly(BB;AA=0.96*exp(697/8192*2*pi*i0.96*exp(-(697/8192*2*pi*i0.96*exp(1209/8192*2*pi*i0.96*exp(-(1209/8192*2*pi*i;A=poly(AA;z1=roots(B;p1=roots(A;z=vpa(z1,8,p=vpa(p1,8,zplane(z1,p1;set(gcf,'color','w'The result i

29、sz=.60000307+.79999770*i.60000307-.79999770*i.86047642+.50949027*i.86047642-.50949027*ip=.57600294+.76799779*i.57600294-.76799779*i.82605736+.48911066*i.82605736-.48911066*iFigure8The locations of poles and zeros of filter's system functionWe can also get the colormaps of spectrum of the origina

30、l and the output signals Figure9spectrum of the original signal Figure10spectrum of the output signalComparing figure9and figure10,we can find that figure9is brighter than figure10.It is because the original signal contains the noise signal in a large scope of frequency band.After passing the filter,most of the noise signal is filtered.The final magnitude-frequency curve is: Figure 11 The final magnitude-frequency curve 4.Conclusion Designing digital filter by pole-zero placement filter design method