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Unformatted text preview: Chap.11 Spectral Representation 11.1 Factorization and Innovations This section discusses a signal modeling method to represent a real WSS RP ( ) t X as the output of a minimumphase system ( ) L s with an input of white noise process ( ) t i . If ( ) t X can be treated in this way, it is called a regular process. Definition: A LTI system ( ) L s is minimumphase if both ( ) L s and its inverse system 1 ( ) ( ) s L s = are causal and stable. Here, stable means that their impulse responses ( ) t l and ( ) t are of finite energy. ( ) L s is minimumphase iff both ( ) L s and ( ) s are analytic in the right half plane of s , i.e., all their poles are in the left half plane of s . A RP is regular if it is linearly equivalent to a whitenoise process ( ) t i in the sense that ( ) ( ) ( ) t t d = i X and ( ) ( ) R = ii ( ) ( ) ( ) t t d = X i l { } 2 2 ( ) ( ) E t t dt L = < X l ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) R R = = = XX ii l l l l ( ) ( ) d = + l l and ( ) ( ) R R = XX XX ( ) ( ) ( ) S s L s L s = and 2 ( )  ( )  S L = ( ) L s is called the innovations filter of ( ) t X , 1 ( ) ( ) s L s = is called the whitening filter of ( ) t X , and ( ) t i is called the innovations of ( ) t X . The problem of determining ( ) L s can be stated as follows: Given a positive even function ( ) S , find a minimumphase system ( ) L s such that 2  ( ) ( ) L j S = . The PaleyWiener condition 2 log ( )  1 S d  < + says that an ( ) S which satisfies the above condition has a solution. Note that ( ) S cannot contain line spectrum and cannot be bandlimited (which implies predictable instead of regular). This problem is, in general, difficult to be solved. However, for the special case that ( ) L s is a rational function of s , ( ) S is a rational function of 2 . It is hence easily to find ( ) S s and decompose it to obtain ( ) L s . This is proved in the following: Since 2 * ( )  ( )  ( ) ( ) ( ) ( ) S L j L j L j L j L j = = = ( ) ( ) S S = This implies that 2 2 ( ) ( ) ( ) A S B = . Changing variable by / s j = , we obtain 2 2 ( ) ( ) ( ) A s S s B s = . So the poles and zeros of ( ) S s appear in the format of symmetric pairs j s s = and j s (and * j s L * j s for complex j s ). Fig. 112(a) shows an example. In this case, we can extract all poles and zeros in the left half plane and use them to form ( ) L s . To be more specific, we first factorize ( ) S s ( ) ( ) ( ) ( ) ( ) N s N s S s D s D s = ( ) ( ) L s L s = , where ( ) ( ) N s D s contains all poles and zeros in the left half plane (see Fig. 112(b))....
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 Fall '08
 SinHorngChen

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