雙通道完全重構(gòu)濾波器設(shè)計(jì)濾波器

1、Design of Two-Channel PR Filter BanksAbstract： In this paper, we will address the concepts of two-channel PR filter banks and perfect-reconstruction (PR) property. The basic principle and design method of two-channel PR filter banks will be discussed in detail. Then the filter coefficients, time dom

2、ain error, the amplitude response and amplitude distortion will be showed in figures. Finally, the design results will be analyzed and summarized.1. Two-Channel Filter Banks1.1 Basic principle of the Two-Channel Filter BankFig. 1-1 shows a two-channel filter bank, popularly called the quadrature mir

3、ror filter (QMF) bank. The input signal x(n) is first filtered by two filters H0(z) and H1(z), typically lowpass and highpass filters. Each subband xk(n) signal is therefore bandlimited to a total bandwidth of . The subband signals are decimated by a factor of 2 to produce vk(n). Each decimated sign

4、al vk(n) is then coded in such a way that the special properties of the subband are exploited .At the receiver end, the received signals are decoded and then produce the signals yk(n) by passing through the upsampler. These signals are filtered by the filters F0(z) and F1(z), producing the output si

5、gnal x (n).( H0(z) and F0(z) respectively are the analysis and synthesis filters of the k-th channel, where k=0,1 .)Fig. 1-1 The two-channel QMF bank.From Fig. 1-1 we haveXk(z)=H(z)X(z)，k=0,1. (1-1)The z-transforms of the decimated signals vk(n) areVk(z)=1éXkz+Xk-zù，k=0,1. (1-2) ú

6、5;û2ê()()The second term above represents aliasing. The z-transforms of Fk(z) is Vk(z2), so that11Yk(z)=Vk(z2)Xz+X-z=()()kk2Hk(z)X(z)+Hk(z)X(-z),k=0,1.(1-3) 2The reconstructed signal is(z)=F(Z)Y(Z)+F(Z)Y(Z)=1H(z)F(z)+H(z)F(z)X(z)+X0011001121 (1-4) H0(-z)F0(z)+H1(-z)F1(z)X(-z)2The term X(-z

7、) takes into account aliasing due to the downsamplers and imaging due to the upsamplers. We refer to this just as the aliasing term or alias component.We will see that the reconstructed signal x (n) differs from x(n) due to three reasons: aliasing, amplitude distortion, phase distortion. It can be s

8、hown that the filters can be designed in such a way that all of these distortions are reduced or eliminated.1.2 DistortionsFrom (1-4), it is clear that we can cancel aliasing by choosing the filters such that the quantityH0(-z)F0(z)+H1(-z)F1(z)is zero.Now (1-4) can be written as(z)=T(z)X(z)+A(z)X(-z

9、). (1-5) XWe say that T(z) is the distortion transfer function or “overall” transfer function of thealias-free system and A(z) is the gain of the aliasing term. Here1H0(z)F0(z)+H1(z)F1(z). (1-6) 21H0(-z)F0(z)+H1(-z)F1(z)While A(z)=. (1-7) 2T(z)=j()Letting Tej=Teje，we have ()()X(ej)=T(ej)ej()X(ej). (

10、1-8)(ej)suffers from “amplitude distortion”. Similarly Unless is T(z) allpass, we say that X(ej)suffers unless T(z) has linear phase (that is, ()=a+b for constant a and b), Xfrom phase distortion. 2. The concept of PR filter banksA filter bank is called perfect-reconstruction (PR) system if the reco

11、nstructed signal x n is a scaled and delayed version of the input , namely,x n =cx(nn0). (1-9)Where c is a nonzero constant and n0 is a positive integer.If a QMF bank is free from aliasing, amplitude distortion, and phase distortion, it is said to have the PR property. For a PR QMF bank we havez =cz

12、n0X(z). (1-10) Xfor all possible inputs x(n). This, of course, ignores the coding/decoding error and filter round off noise.3. Design of Two-Channel PR Filter BanksFor a two-channel critically decimated filter bank, the PR condition can be given byH0(z)H1(-z)-H0(-z)H1(z)=cz-n0 (1-11)The analysis and

13、 synthesis filters are related asH0(-z)F0(z)+H1(-z)F1(z)=0. For the linear-phase filter bank, H0(z) and H1(z) are chosen respectively to be symmetric and anti-symmetric, which is an even number. The system delay should be equal toN0+N1-1, where N0 and N1 are respectively the lengths of the lowpass a

14、nd highpass filters. 2The coefficients of filters H0(z) and H1(z) are obtained by solving the following constrained optimizationmin=a(1-|H0(e)|)d+(1-a)|H0(ej)|2d+2h0pjs(1-a)|H1(e)|d+a(1-|H1(e)|)d0pj2j2 sSubject to (1-11), (1-12)Where p and s are the passband and stopband cut off frequencies of H0(z)

15、 and H1(z) , respectively, is a weighting constant from 0 to 1, and h is the vector containing the free variables in the impulse response.In the time domain ,the constraint can be ash0(n)(-1)nh1(n)-h1(n)(-1)nh0(n)=C(n-n0).(1-13)Letting H0(z)=H1(-z) (that is, h0(n)=(-1)nh1(n)in the time domain), then the constrained optimization will turn as the following formpmin=a(1-|H0(ej)|)2d+(1-a)|H0(ej)|2d h0ss.t. h0(n)h0(n)-(-1)nh0(n)(-1)nh0(n)=c(n-n0). (1-14) Solving this constrained op