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

Therefore t he s ystem r esponse t o s uch s inusoids

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Unformatted text preview: e r esonance (peaking) becomes p ronounced as bl a pproaches 1. Fig. 12.5c shows a c anonical r ealization of t his filter [see Eq. (12.20)J. • =1 25.5 o C omputer E xample C 12.2 Using M ATLAB, find t he f requency response o f t he b andpass filter in E xample 12.2 for hi = 0.96. g amma=0.96; n um=[1 0 - 1]; d en=[1 - sqrt(2)*gamma g amma 2j; W =-pi:pi/l00:pi; H =freqz(num,den,W); m ag=abs(H); p hase= 1 80 / pi*unwrap( a ngle(H»; s ubplot(2,1,1) p lot(W,mag) s ubplot(2,1,2) p lot(W,phase) A 6.41 o 2501t 0 0_ , ,/4 (a) ( b) 0 • ( c) F ig. 1 2.5 D esigning a b andpass filter i n E xample 12.2. place a pole i n i ts v icinity (near ei "/4). B ecause this is a complex pole we also n eed i ts c onjugate n ear e - i ,,/4, a s i ndicated i n F ig. 12.5a. Let u s c hoose t hese poles Al a nd A2 as Al = bl~"/4 a nd E xample 1 2.3: N otch ( Bandstop) F ilter D esign a second-order notcl1 filter t o h ave zero transmission a t 250 Hz a nd a s harp recovery o f g ain t o u nity o n b oth s ides o f 250 Hz. T he h ighest significant frequency t o b e p rocessed is F h = 500 Hz. I n t his c ase, T ::; 1 /2Fh = 1 0- 3 . L et u s choose T = 1 0-3 . For t he f requency 250 Hz, w T = 21r(250)T = 11"/2. T hus, t he frequency 250 Hz is represented by a p oint eiwT = ei "/2 = j o n t he u nit circle, a s d epicted in Fig. 12.6a. Since we need zero transmission a t t his frequency, we m ust p lace a zero a t z = ei "/2 = j a nd i ts c onjugate a t z = e - i ,,/2 = - j. We also require a s harp recovery of g ain o n b oth sides of frequency 250 Hz. T o a ccomplish t his goal, we place two poles close t o t he t wo zeros i n o rder t o c ancel o ut t he effect o f t he z eros as we move away from t he p oint j ( corresponding t o f requency 250 Hz). For t his r eason, let u s u se poles a t ± ja w ith a < 1 for stability. T he closer t he p oles a re t o t he z eros ( the closer t he a t o 1), t he f aster is t he g ain recovery o n e ither s ide o f 250 Hz. T he r esulting t ransfer f unction is A2 = 1'Yle-i "/4 w here 1,1 < 1 for stability. T he closer t he value o f bl is t o t he u nit circle, t he m ore s harply p eaked is t he r esponse a round w = 25011". We also have a zeros a t ± 1. Hence H[z = K (z - j)(z + j ) = K z2 + 1 J ( z-ja)(z+ja) Z 2+ a2 T he d c gain ( gain a t w = 0, o r z = 1 ) o f t his filter is H[IJ=K~ l +a (12.20) B ecause we require a dc g ain o f unity, we m ust s elect K therefore F or convenience we s hall choose K = 1. T he a mplitude r esponse is given by jwT IH[e _ lei2wT - 11 leiwT _ h lei"/41IeiwT -1'Yle i,,/41 = 2 1 ";a . T he t ransfer f unction is (12.22) JI- a nd a ccording t o E q. (12.9) Now, u sing E q.(12.9), we o btain H eiwT (12.21) 1 [ 11 2 = (1 + a2)2 4 i2wT (e + 1 )(e-i2wT + 1) (e]2wT + a 2)(e-j2wT + a 2) (1 + a 2)2(1 + cos 2wT) 2(1 + a 4 + 2a 2 cos 2 wT) , 12 Frequency Response a nd D igital Filters 728 729 12.3 Digital Filters 1 2.3 Digital Filters D igital filters a nd some of their advantages were discussed in Sec. 8.5. Digital filters can be classified as either r ecursive o r n onrecursive. ( a) Recursive Filters T he t erms recursive a nd nonre...
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