1 ENEE 624: Advanced Digital Signal Processing Project 1 Spring 2003 Submitted by Zhu Ji Taught by Prof. H. Papadopoulos Problem 1.1: Perfect Reconstruction FIR QMF Bank Design a. Design a two-channel QMF bank with the following properties: 1. All four filters are FIR filters; 2. H0(z) is a lowpass filter with ωs=0.6π and δs=0.01; 3. The overall system is a perfect reconstruction (PR) system. b. Repeat part(a) for δs=0.001. I. SOLUTION In this part, we utilize the power symmetric filters to obtain the required power symmetric filters. As we know, the transition function and aliasing terms of a two-channel QMF bank can be represented in the frequency domain as )(zT = ( )(0zH )(0zF + )(1zH )(1zF )/2 (1) )(zA = ( )(0zH − )(0zF + )(1zH−)(1zF )/2 (2) Assume now that H0(z) is power symmetric, that is, the following formula holds 1)()(~)()(~0000=−−+ zHzHzHzH (3) By comparing this with (2), we see that if the filter H1(z) is chosen as )(~)(01zHzzHN−−=− (4) for some odd N, then (2) reduces to 0.5z-N, that is, we have a perfect reconstruction system. In order for this system to be practical, H0(z) has to be FIR. Further, we see that the synthesis filters are given by )(~)(),(~)(1100zHzzFzHzzFNN −−== (5)2 Then, only the filter H0(z) remains to be designed. The power symmetric property means that the zero-phase filter )()(~)(00zHzHzH = is a half-band filter. Note that )(ωjeH has to be nonnegative. The design steps for the real coefficient case are as follows: 1. Design a nonnegative zero-phase FIR half band filter ∑−=−=NNnnznhzH )()( of order 2N. We design the required filter in a tricky way. First, we design a Type 2 linear phase filter ∑=−=NnnzngzG0)()( by defining the passband to be pθω≤≤0 and transition band to be πωθ≤≤p. This means that N is odd and g(n) is real. Note that ε is the peak passband ripple. There is no stopband. Then, define the transfer function 2/)]([)(2zGzzFN+=−. This is a Type 1 linear phase filter. Suppose we define )()(ˆzFzzHN= . Then, )(ˆzH is a zero-phase half-band filter. Its length is 2N+1, and since N is odd, we have 2N+1=4J+2+1 for some integer J. Clearly )(ˆjweH is real. Define )(ˆ5.0)( zHzH +=ε, i.e., =+≠=0)(ˆ5.00),(ˆ)(nnhnnhnhε (6) Therefore, we obtain a nonnegative zero-phase FIR half band filter. The stopband edges of H(z) and G(z) have the following relationship. )1(2spωθ−= (7) 2. Compute a spectral factor H0(z) of the filter H(z). In principle, this can be done by computing the zeros of H(z) and assigning an appropriate subset to H0(z). In our approach, we design a minimum phase equiripple FIR filter H0(z) by choosing all the zeros inside the unit circle. It is worth mentioning that the peak stopband ripples of H(z) and H0(z) should satisfy 2/2εδ= . Finally, we obtain the H0(z) to make the overall system a PR system. II. SIMULATION RESULTS Using the above algorithm, a two-channel QMF bank can be designed to achieve the required features. Figure 1 shows the amplitude response of H0(z) with ωs=0.6π and δs=0.01.3 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-70-60-50-40-30-20-10010Normalized Frequency (pi rad/sample)Magnitude (dB) Figure 1. the amplitude response of H0(z) with ωs=0.6π and δs=0.01. 0 10 20 30 40 50 60-2.5-2-1.5-1-0.500.511.522.5 Figure 2. the comparison of the reconstructed Xa and original signal Xa for the first 60 samples. The order of H0(z) with ωs=0.6π and δs=0.01 is 33, and the filter coefficients are listed as follows: 1.0241e-001 3.3473e-001 4.7555e-001 2.7654e-001 -8.9125e-002 -1.8930e-001 1.3594e-002 1.2750e-001 1.8930e-003 -8.8371e-002 -1.9518e-003 6.1528e-002 -1.7024e-003 -4.1798e-002 5.3539e-003 2.6724e-002 -7.6631e-003 -1.5265e-002 8.3578e-003 6.9271e-003 -7.7128e-003 -1.2768e-003 6.1185e-003 -2.0198e-003 -4.1447e-003 3.4758e-003 2.2253e-003 -3.6024e-003 -7.3648e-004 3.0372e-003 -1.7165e-004 -2.8298e-003 2.4679e-003 -7.5501e-0044 The PR property can be examined using the provided signal Xa in file testsigs.mat. Figure 2 shows the comparison of the reconstructed Xa and original signal Xa for the first 60 samples. We find that the designed QMF bank achieves PR property with a delay of 33, which is the order of the analysis bank. Further, we compute the maximal difference between the reconstructed Xa and the signal Xa, which is 7.8920e-010. 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-100-80-60-40-200Normalized Frequency (pi rad/sample)Magnitude (dB) Figure 3. The amplitude response of H0(z) with ωs=0.6π and δs=0.001. 0 10 20 30 40 50 60 70 80 90 100-3-2-10123 Figure 4. The comparison of the reconstructed Xa and original signal Xa for the first 100 samples.5 Similarly, Figure 3 and Figure 4 show the corresponding results when δs=0.001. In this case, the order of H0(z) is 53, and the filter coefficients are listed as follows: 3.6371e-002 1.7482e-001 3.7952e-001 4.3430e-001 1.8759e-001 -1.4175e-001 -1.8347e-001 3.9141e-002 1.3985e-001 -6.0085e-003 -1.0409e-001 -1.8877e-003 7.7435e-002 2.5646e-004 -5.7003e-002 4.6461e-003 4.0607e-002 -1.0003e-002 -2.7015e-002 1.4442e-002 1.5640e-002 -1.7317e-002 -6.2723e-003 1.8429e-002 -1.1289e-003 -1.7884e-002 6.5683e-003 1.6000e-002 -1.0101e-002 -1.3213e-002 1.1880e-002 1.0000e-002 -1.2162e-002 -6.8040e-003 1.1288e-002 3.9748e-003 -9.6469e-003 -1.7378e-003 7.6236e-003 1.8098e-004 -5.5533e-003 7.2453e-004 3.6892e-003 -1.0941e-003 -2.1828e-003 1.0817e-003 1.0986e-003 -8.5694e-004 -4.1494e-004 5.6323e-004 6.4503e-005 -3.5007e-004 2.0353e-004 -4.2344e-005 The maximal difference between the reconstructed Xa and the original signal Xa is 3.9453e-006. c. Consider compressing signals by quantizing the outputs of the decimated analysis filter bank. Choose a