Chap11_English - Chap.11 Spectral Representation 11.1...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 minimum-phase 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 minimum-phase 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 minimum-phase 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 white-noise 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 minimum-phase system ( ) L s such that 2 | ( )| ( ) L j S = . The Paley-Wiener 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. 11-2(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. 11-2(b))....
View Full Document

Page1 / 25

Chap11_English - Chap.11 Spectral Representation 11.1...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online