part35

part35 - Univ. Of Maryland at College Park, ECE Department...

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Univ. Of Maryland at College Park, ECE Department ENEE324 Fall 2002 Instructor : M. Olfat Solution to Assignment #: 13 Posted on: 12/03/02 [ ] [ ] [ ] [ ] 0 EXYZ E X E Y EZ ++ = ++= a) From Eq. 5.3 we have: ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 va r va r va r va r 2co v , v , v, XY ZXY Z X Y X Z YZ =++ +++ ( 29 11 111 2 2 0 23 44  + + +-=   b) ( 29 ( 29 ( 29 ( 29 r va r va r va r3 ZXYZ =++= [ ] [ ] nn nii ii E SE X E Xn m ==  ===   ∑∑ ( 29 ( 29 ( 29 1 sum of diag. sum of off-diag. elements of element of K covariance matrix K r va r co n n k jk k S X XX = =+ 14243 14 44244 43 2 2 2 2 12 2 2 2 22 1 n n X n C s r s r s rs r ss r s r - - - = K K MO K ( 29 ( 29 r 21 n S s rs =+- Problem 1:[GAR]5.1 Problem 2:[GAR]5.2
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[ ] [ ] 11 nn nii ii E SE X E Xn m ==  ===   ∑∑ The same argument as problem 2 2 2 2 2 12 2 2 2 22 1 n n X n C s r s r s rs r ss r s r - - - = K K MO K ( 29 1 2 2 1 01 1 va r 1 j j k n j kj S r s r s rs r - -- = - = + =+ - 1 2 111 n n n rr s rs r rr -  = +-  - --  a) ( 29 Z Ms ab = ++ b) ( 29 Z is partial fraction expansion, where b ab - = , a ba ab - = ( 29 Z a bab a b a a bb =- ( 29 ( 29 1 0 tt ZZ f t M se et b a a b ab - = - = - a) ˆ 0 Y = ii) ˆ 0 Y = iii) ˆ YX b) i) ( 29 1 1 2 py -= for 1,1 y ˆˆ 1 or 1 YY = ( 29 = 0 y = ˆ 0 Y ⇒= ( 29 1 1 2 = 1,1 y 1 = Problem 3: [GAR]5.3 Problem 4: [GAR]5.7 Problem 5: [GAR]4.89
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ii) ( 29 1 1 3 py -= for 1,0,1 y =- choice is arbitrary ( 29 1 0 3 = 1,0,1 y ˆ 0 Y = iii) ( 29 11 1 y = ˆ 1 Y ⇒= ( 29 01 = 0 y = ˆ 0 Y ( 29 = for 1 y ˆ 1 Y c) X -1 0 1 0 0 0 0 0 0 1 0 -1 Entries in table are EYX   .
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This note was uploaded on 11/29/2009 for the course EE 131A taught by Professor Lorenzelli during the Fall '08 term at UCLA.

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part35 - Univ. Of Maryland at College Park, ECE Department...

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