Question

2. Two type II filters Hi(ejw) and Hz(elw) can be expressed as Ri(w).exp(01(w)) and R2(w).exp(j02(w)) respectively. Hz(ew) is formed as Hi(ejw). H2(elw). (a) For Hi(elw), give expressions for R1(w) and 01(w). (b) Express the impulse response h3(n) in terms of hi(n) and h2(n). (c) Associated with H3(el) and h3(n) are the quantities M3, R3(w), and 03(w). Express 03(w) and M3 in terms of M. Is M3 even or odd ? (d) Express R3(w) in terms of Ri(w) and R2(w). Is R3(w) even or odd ? (e) What type of filter is h3(n)?

Answer 1

Solution:

AnSwea t (e) be E1pesed a R() A ype filtea o un.ctfon at h.l) ard Ni. e (ja (u)) which can be E1poM iute Hz (e and a type J R Co exp (je2 Co)) ohich, e a funcfon CLI h2 Co and M2. at a type I ilten, the foequency N=4 FIR type I fiteo a) Fo nesponse ength Ceven -dength ymmeto) can be waitten thoe thie -ysco3/2ju hiejhie +hoe e-2/2jus ho e anjahoe ) 2j He e e. 32ju 2ho Cos 2hi Cos e R, (a)e Jo Co) wheae, (u) 2h, cos R C)2 ho cos ( + e,(o atype-FIR filteoi ot ungth RIC)e In genenal tt (e3) H (et)R (aw)e lo, ()

N/21 R (w) (n) cos (M-o)) -1 even ungth Antf type filteJ (b) Foa we can woute as Sym metac oll ous -2 sa ht e -3ju ho e ju hoe e h e - hoe e -e e - 3hjCjho stn ))+ ( 2j ht sto (4 )). c+ah, stn w) e 32 e J플|aho Sin ()w e Pz(wle J((a)) Cu he e 2. + 2h Sn () P2 ()2 he So (

FTR dteo o length N So, type iu (o+) IMM- 2 C)1yp FIR'S h.(A)- h ((M-n) O<n<m whe oie and type N add Terieges an od tofegea ,-CM-k) MH (21) ha(onm-nJos ns mis HC) h (M-n) M ΞΣz H(z.) that s and, H z Then Hi)w have factoay (1-z.z-) CI-22 -z1) (1)MH (1) HI() (M-) MH2 (z) H3 (z)- Hi (z) * z (2) H2 (2)-hM-z n 0 Z S0, E1ptefon -2m -2M - Z HI (H2 (z I2T

d)For m1 and m2 the value of the term will be odd.Same for R3(w).

Comment-- frequency response of the system is in series.So the convolution of two system is the multiplication.