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

8242 s in w t hrejwtj l hrkje iktw 0682 s in 2 wt

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Unformatted text preview: unction H [z] o f t he f ilter is f irst o btained b y t aking t he z -transform o f h[k] i n E q. (12.98): N o-l H [z] = L h [k]z-k k=O _ 0.1146 Z5 z6 + 0.0792z 5 + 0.3209z 4 + 0.4285z 3 + 0 .3209z 2 + 0 .0792z - 0.1146 z6 a nd H[e iwT ] = - 0.1146 + 0 .0792e- iwT + 0 .320ge-i2wT + 0 .4285e- j3wT = e- j3wT + 0 .320ge- j4wT + 0 .0792e- j5wT _ 0 .1146e- j6wT [0.4285 + 0.6418 cos w T + 0.1584 cos 2 wT - 0.2292 cos 3wT] T he s econd s um o n t he r ight-hand s ide is a g eometric s eries, a nd u sing t he r esult i n B .7-4, w e h ave _ 1 L [1 _ z - N O e j2"r] N o-l H [z] - -N T he m agnitude o f t his r esponse (Fig. 12.21) shows t hat t he realized filter values m atch e xactly t he d esired response a t t he No s ample points. T he t ime delay a dds a l inear phase - 3Tw t o t he filter characteristic. • o Hr r =O 1- e j"h!.!:. NO z -l o C omputer E xample C I2.11 Using MATLAB, find h[kJ a nd t he c orresponding H[e iwT ] for t he frequency sampling filter in E xample 12.12. N O=7; H =[1 1 0 0 0 0 1 ]; f or i =I:NO H ence, Z No - 1 H [z]=--NNoz 0 ~ H,[z] z Hr N o-l L r =O Z- e j"h!.!:. NO ' ----v----" H2[Z] (12.100) 4 7 76 12 Frequency Response a nd Digital Filters Observe t hat we do n ot need t o perform I DFT (or I FFT) c omputations t o o btain t he desired filter transfer function. All we need is t he values o f t he frequency samples H r , which are given. Equation (12.100) shows t hat t he desired filter is realized as a cascade of two filters with transfer functions Hl[Z] a nd H2[Z], Also, Hl[Z] is t he t ransfer function of a n No-order comb filter (see Example 12.9). T he second filter w ith transfer function H2[Z] is a parallel combination o f n + 1 first-order filters, . 2 .,..r w hose poles lie on t he u nit circle a t eJ/ilQ (r = 0, 1, 2 "", No - 1). For t he lowpass or bandpass filters m any coefficients Hr a ppearing in H2[Z] a re zero. Recall t hat in Example 12.12 (low pass filter) four o ut of seven coefficients are zero. Thus, in practice the final filter is usually much simpler t han i t appears in Eq. (12.100). As a r esult, this method m ay require a fewer number of computations (multiplications a nd additions) compared t o those in t he filter obtained by t he d irect method (using I DFT). T he poles o f t he frequency sampling filter are complex in general because t hey lie o n t he unit circle. Therefore, we m ust combine t he c onjugate poles t o form q uadratic t ransfer functions t o realize them. All these points will be clarified by designing the filter in Example 12.12 by this method. • E xample 1 2.13 R edo Example 12.12 using t he m ethod o f frequency s ampling filter. + 1 = 7. Also In t his case n = 6, No = n a nd S ubstituting t hese values i n t he t ransfer function H[z] o f t he d esired filter in Eq. (12.100), we o btain Z7 _ 1 H [z] = 7T [z z_1 + z e- j6 ,,/7 z _ e j2 ,,/7 ' -v--' , H tlz] + z e j6"/7] z _ e - j2,,/7 , H ;lz) We c ombine t he l ast t wo t erms o n t he r ight-hand side corresponding t o c...
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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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