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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 ol H [z] = L h [k]zk 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 .320gei2wT + 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 ighthand s ide is a g eometric s eries, a nd u sing t he r esult
i n B .74, w e h ave _ 1 L [1 _ z  N O e j2"r] N ol 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 ol 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 Noorder comb filter (see Example 12.9). T he second filter
w ith transfer function H2[Z] is a parallel combination o f n + 1 firstorder 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 ighthand 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.
 Spring '13
 Bayliss
 Signal Processing, The Land

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