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

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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...
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## This note was uploaded on 04/14/2013 for the course ENG 350 taught by Professor Bayliss during the Spring '13 term at Northwestern.

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