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

FIR Filters 18-5TABLE 18.1Frequency Response of Linear Phase FIR Filtersh(nT) N H(e ivT )Symmetrical Odd e jv(N 1)T=2 (NP1)=2a(k) cos (vkT)Even ek¼0PN=2jv(N 1)T=2k¼1 1b(k) cos v k 2TAntisymmetrical Odd je jv(N 1)T=2 (NP1)=2c(k) sin (vkT)Even jejv(N 1)T=2PN=2k¼1k¼1 1d(k) sin v k 2T (N 1)TWhere a(o) ¼ h2 , a(k) ¼ c(k) ¼ 2hN 12k T , b(k) ¼ d(k) ¼ 2hN2k T .( )(NX1)=2H(e jvT ) ¼ je jv(N 1)T=2 c(k) sin (vkT)k¼1(18:25)A notable feature of this frequency response is that at frequencies v ¼ 0 and v ¼ p, the frequencyresponse is always zero, independent of c(k).(4) Antisymmetric impulse response and N ¼ even. For this case, the frequency response is the same asthat in (2) except the cosine summations become sine summations multiplied by j as follows:( N=2 X 1 H(e jvT ) ¼ je jv(N 1)T=2 2h(kT) sin v N )1k T2 2k¼0(18:26)Letting d(k) ¼ 2h[N=2k)T], k ¼ 1, 2, . . . , N=2, Equation 18.26 becomes( X N=2 )H(e jvT ) ¼ je jv(N 1)T=2 1d(k) sin v k T2k¼1(18:27)In this case, the frequency response is zero at v ¼ 0, independent of d(k).In summary, the frequency responses of the four possible types of FIR filters with linear phase aregiven in Table 18.1.18.1.3 Locations of Zeros of Linear Phase FIR FiltersThe symmetric and antisymmetric conditions of the impulse response given by Equations 18.13 and18.15 impose certain constraints on the zeros of the transfer function H(z) [2]. For the case where N is anodd value, H(z) can be written as(NX1)=2H(z) ¼ z (N 1)=2k¼0a(k)2 (zk z k ) (18:28)where the sign corresponds to symmetry and antisymmetry in the impulse response respectively, anda(o) and a(k) are defined in Table 18.1.