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Chapter 11_English

# Chapter 11_English - Chapter 11 Spectral Representation Te...

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Chapter 11 Spectral Representation Teacher: Sin-Horng Chen Office: Engineering Bld. #4, Room 805 Tel: ext. 31822 Email: [email protected]

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2 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 Γ = are causal and stable. Here, stable means that their impulse responses ( ) t and ( ) t γ are of finite energy.
3 ( ) L s is minimum-phase iff both ( ) L s and ( ) s Γ are analytic in the right half plane of s , i.e., all their poles are in the left half plane of s .

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4 A RP is regular if it is linearly equivalent to a white-noise process ( ) t i in the sense that 0 ( ) ( ) ( ) t t d γ α α α = - i X and ( ) ( ) R τ δ τ = ii 0 ( ) ( ) ( ) t t d α α α = - X i { } 2 2 0 ( ) ( ) E t t dt = < ∞ X ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) R R τ τ ρ τ δ τ τ τ τ τ = = - = - XX ii 0 ( ) ( ) d τ α α α = + 0 τ and ( ) ( ) R R τ τ = - XX XX
5 ( ) ( ) ( ) S s L s L s = - and 2 ( ) | ( ) | S L ϖ ϖ = ( ) L s is called the innovations filter of ( ) t X , 1 ( ) ( ) s L s Γ = is called the whitening filter of ( ) t X , and ( ) t i is called the innovations of ( ) t X .

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6 The problem of determining ( ) L s can be stated as follows: Given a positive even function ( ) S ϖ , find a minimum-phase system ( ) L s such that 2 | ( ) | ( ) L j S ϖ ϖ = . The Paley-Wiener condition 2 |log ( ) | 1 S d ϖ ϖ ϖ -∞ < ∞ + says that an ( ) S ϖ which satisfies the above condition has a solution. Note that ( ) S ϖ cannot contain line spectrum and cannot be bandlimited (which implies predictable instead of regular).
7 This problem is, in general, difficult to be solved. However, for the special case that ( ) L s is a rational function of s , ( ) S ϖ is a rational function of 2 ϖ . 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 ϖ ϖ ϖ ϖ ϖ ϖ = = = - ( ) ( ) S S ϖ ϖ = - This implies that 2 2 ( ) ( ) ( ) A S B ϖ ϖ ϖ = . Changing variable by / s j ϖ = , 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.

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8
9 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)). Lastly, form ( ) ( ) / ( ) L s N s D s = .

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10 Ex. 11-1 Let 2 2 ( ) N S ϖ α ϖ = + for 0 α . Find ( ) L s . Solution h 2 2 ( ) ( )( ) N N S s s s s α α α = = - - + ( ) N L s s α = + Ex. 11-2 Let 2 4 2 49 25 ( ) 10 9 S ϖ ϖ ϖ ϖ + = + + . Find ( ) L s . 2 4 2 2 2 49 25 (7 5 )(7 5 ) (7 5 )(7 5 ) ( ) 10 9 (1 )(9 ) (1 )(1 )(3 )(3 ) s s s s s S s s s s s s s s s - + - + - = = = - + - - - + - + 7 5 ( ) (1 )(3 ) s L s s s + = + +
11 Ex. 11-3 Let 4 25 ( ) ( 1) S ϖ ϖ = + . Find ( ) L s .

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Chapter 11_English - Chapter 11 Spectral Representation Te...

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