Passive, active, and digital filters (3ed., CRC, 2009) - tiera.ru

Two-Dimensional IIR Filters 23-33Condition 1 is usually satisfied in practice. Therefore, a local minimum point x* can be found by forcingl i to approach l* i (i ¼ 1,2,...,m) and making E(x*) j sufficiently small. These two constraints can besimultaneously satisfied by applying the following algorithm.ALGORITHM 3:Charalambous Minimax AlgorithmStep 1. Set j ¼ 0 and l i ¼ 1 for i ¼ 1, 2, . . . , m. Initialize x.Step 2. Minimize function J(x,j, l) to obtain x.Step 3. SetS ¼ X i2I 1l i j i (x, j) þ X i2I 2J i (x, j)and update l i and j as8l i J i (x, j)=S if J i (x, j) 0, l i 0>:0 if J i (x, j) < 0j ¼ Xml i E i (x)i¼1Step 4. Stop ifE M (x) j eE M (x)otherwise go to step 2.The parameter e is a prescribed termination tolerance. When the algorithm converges, conditions 2 and 3are satisfied and x ¼ x*. The unconstrained optimization in step 2 can be accomplished by applying aquasi-Newton algorithm.Example 23.6 [23]Design a 2-D circularly symmetric filter with the same specifications as in Example 23.5, using algorithm 3.Solution1. Construct the ideal discrete amplitude response of the filter. Since the passband and stopbandcontours are circles, the sample points can be placed on arcs of a set of circles centered at theorigin. Five circles with radiir 1 ¼ 0:3v p , r 2 ¼ 0:6v p , r 3 ¼ 0:8v p , r 4 ¼ 0:9v p , and r 5 ¼ v pare placed in the passband and five circles with radiir 6 ¼ v a , r 7 ¼ v a þ0:1(p v a ), r 8 ¼ v a þ0:2(p v a ), r 9 ¼ v a þ0:55(p v a ), and r 10 ¼ pare placed in the stopband.