Question

Question Two. Consider a two-channel filter bank consisting of filters ho[n] and hi[n], such that ho[n] = 0 for n < 0 and n > L-1 and also hi[n] = 0 for n < 0 and n > L-1. Assume that ho[n] is a symmetric filter and that hi[n] is an an- tisymmetric filter. Let Ha(2), H6(2), Hc(2), Ha(z) be filters of a four-channel iterated filter bank obtained from ho[n] and hi[n] as: H.(z) = H (2)H,(22), Hy(z) = Ho(2)H1(22), H.(z) = H1(2)H.(zº), Ha(z) = H1(2)H1(22). Note that we use H.(z) to denote the z-transform of ho[n], and analogously for other filters. a) Is it possible to have (ho[n], ho[n – 2k]) = 8[k], k e Z, if all samples of ho[n] are non-zero for 0 <n<L-1 and L is an odd number? (4 marks) b) Express the length Ly of filters ha[n], hv[n], he[n], and ha[n] in terms of (5 marks) c) Is ha[n] a symmetric or antisymmetric filter or neither. (4 marks) d) Is hv[n] a symmetric or antisymmetric filter or neither. (4 marks) e) Is he[n] a symmetric or antisymmetric filter or neither. (4 marks) f) Is ha[n] a symmetric or antisymmetric filter or neither. (4 marks)

Answer 1

Holz) = Sholm) = * n=0 ho[n] = 0, n<0 and n > L-1, L is odd ho[n] = ho[L-1-n] {Symmetric} M = (L-1)/2 k=M-1 H ejr) = e-jMS 2ho[k]cos((M k=0 - k) 2) + ho [M]) h[n] = 0, n<0 and n > L-1, L is odd ho[n] = -ho[L-1-n] {Anti-Symmetric} M = (L-1)/2 k=M-1 Hz(54) = je-M101 219[k]sin((M – k)N) + ha{M) k=0 h[M] = 0 <ho[n], ho[n-2k]> = Ho(z).Ho(z)z<2k = 1 => Ho(z) = +zk => hol-k] = +1 Conditions: 1) O s-ks L-1, L is odd number, Lmin = 1 k = 0 (Only possibility)

b)

H. (2) = H (2) H (22

$H_a(z) = (\sum_{n=0}^{L-1}h_0[n]z^{-n})(\sum_{m=0}^{L-1}h_0[m]z^{-2m})$

L-1 1-1 H, (α) =ΣΣΑo[n]to[n]=-1--2η Π=0 n=0

L-11-1 Ha(z) = 2 ho[n]ho[m]2=(n+2m n=0 m =0

0 < n < L-1

0 < m < L-1

0 < n + 2m < 3L - 3

Length, L1 = 3L-2

L-1 L-1 Η(Ε) – Σωμής "Σίνη) - ) n=0 Τη=0

$H_b(z) = \sum_{n=0}^{L-1}\sum_{m=0}^{L-1}h_0[n]h_1[m]z^{-n}z^{-2m}$

$H_b(z) = \sum_{n=0}^{L-1}\sum_{m=0}^{L-1}h_0[n]h_1[m]z^{-(n+2m)}$

0 < n < L-1

0 < m < L-1

0 < n + 2m < 3L - 3

Length, L1 = 3L-2

$H_c(z) = (\sum_{n=0}^{L-1}h_1[n]z^{-n})(\sum_{m=0}^{L-1}h_0[m]z^{-2m})$

$H_c(z) = \sum_{n=0}^{L-1}\sum_{m=0}^{L-1}h_1[n]h_0[m]z^{-n}z^{-2m}$

$H_c(z) = \sum_{n=0}^{L-1}\sum_{m=0}^{L-1}h_1[n]h_0[m]z^{-(n+2m)}$

0 < n < L-1

0 < m < L-1

0 < n + 2m < 3L - 3

Length, L1 = 3L-2

$H_d(z) = (\sum_{n=0}^{L-1}h_1[n]z^{-n})(\sum_{m=0}^{L-1}h_1[m]z^{-2m})$

$H_d(z) = \sum_{n=0}^{L-1}\sum_{m=0}^{L-1}h_1[n]h_1[m]z^{-n}z^{-2m}$

$H_d(z) = \sum_{n=0}^{L-1}\sum_{m=0}^{L-1}h_1[n]h_1[m]z^{-(n+2m)}$

0 < n < L-1

0 < m < L-1

0 < n + 2m < 3L - 3

Length, L1 = 3L-2

Assuming real-coefficients, of ho[n] and h1 [n] H. (2) = Ho(-12)* => H.(2) is conjugate symmetric H.(292) = Ho(-22)* => H.(22) is conjugate symmetric H (2) = H,(-12)* => H (2) is conjugate symmetric H (292) = H,(-22)* => H4(292) is conjugate symmetric Ha(2) = H. (2).H.(22) => Conjugate symmetric H (2) =H.(12).H (292) => Conjugate symmetric H (2) =H7(2).H.(292) => Conjugate symmetric H (2) = H, (2).H (292) => Conjugate symmetric