Signal Processing and Linear Systems-B.P.Lathi copy

10 sec 76 2 t he 5 s ubsteps are s tep 2 1 f ind ws

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: er described by the t ransfer function H [z] [ Eq. (12.69) or (12.71)] and t he corresponding impulse response h[k] [Eq. (12.73)]. We now show t hat if h[k] is e ither symmetric (Fig. 12.17a) o r a ntisymmetric (Fig. 12.17b) a bout i ts c enter point, t he filter phase response is a linear function of w. We consider a case where n is even. To avoid t oo much a bstractness, we choose some convenient value for n , s ay n = 4, t o d emonstrate our point. It will t hen b e easier t o u nderstand t he generalization t o t he n th-order case. F or n = 4, t he impulse response in Eq. (12.73) reduces t o h[k] = h[O]I5[k] + h[l]l5[k - 1] + h[2]I5[k - 2] + h[3]I5[k - 3] + h[4]I5[k - 4] T he t ransfer function H[z] in Eq. (12.71b) reduces t o = + h [l] + h[2] + h[3] + h[4] z z-2 (h[0]Z2 Z2 z3 (12.75a) Z4 + h [l]z + h[2] + h[3]z-1 + h[4]Z-2) (12.75b) T he q uantity inside t he parenthesis is real; i t may b e positive over some ban~s of frequencies and negative over other bands. This quantity represents t he a mphtude response IH[ejwTlI.t T he p hase response is given by LH[e jwT ] = - 2wT T he phase response is a linear function of w. T he t ime d ela! is ~he negative of t he slope of LH[e jwT ] w ith respect t o w, which is 2T seconds I II t his case [see Eq. (4.59)]. . b I f h[k] is antisymmetric a bout i ts center point, t hen t he a ntlsymmetry a o ut t he center point requires t hat h[k] = 0 a t t he center pointj: [see Fig. 12.17b]. Thus, in this case h[l] = - h[3], h[2] = 0 a nd t he frequency response reduces t o H[ejwT] = e-j2wT ( h [0] (e j2wT _ e - j2wT ) + h[l](e = H [d WT ] = e - j2wT (h[0]e j2WT + h[l]e jwT + h[2] + h[3]e- jwT 11 S ymmetry c onditions f or l inear p hase f requency r esponse i n n onrecursive = 2 je- j2wT ( h[l] sin wT Therefore, the frequency response is I 4 f ilters. h[O] = - h[4], H[z] = h[O] I 3 jwT _ e- jWT )) + h[O] sin 2WT) 2e j (f - 2wT ) ( h[l] sin wT + h[O] sin 2WT) + h[4]e- j2WT ) (12.76) Thus, the phase response in this case is LH[e jwT ] = ::.. - 2wT 2 I f h[k] is s ymmetric a bout its center point (k = 2 in t his case), t hen h[O] = h[4J, h [l] = h[3] = e - j2wT (h[2] + 2h[1] cos wT + 2h[0] cos 2WT) t Strictly s peaking, IH[ejWTll c annot b e negative. Recall, however, t hat th~ o nly restri~tion o n e a mplitude is t hat i t c annot be complex. I t h as t o b e real; i t c an b e ~osltlve o r negatIve. s hould have used some o ther n otation s uch as A (w) t o.denote th~ a mplitude response. ~ut. thIS would c reate t oo m any related functions causing pOSSible confUSIOn. ~oth~r a lte:natlve IS t o i ncorporate t he n egative sign o f t he a mplitude i n t he p hase ~esponse, which. will b e lI~cr~ (~r d ecreased) by 11" over t he b and w here t he a mplitude r esponse IS negative. ThIS a lternative Will s till m aintain t he p hase linearity. . .., j :Antisymmetry p roperty req~ires ~hat h[k] = - h[-k] a t t he c enter p oint also. ThiS conditIOn IS possible only if h[k] = 0 a t thiS p omt. w. a nd t he freq...
View Full Document

Ask a homework question - tutors are online