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ExamSolution copy - EE 562a Midterm Solution October, 2009...

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Unformatted text preview: EE 562a Midterm Solution October, 2009 1 1. Lets approach this problem algebraically first, and then review it geometrically. There are many references to the vector b in different situations, so lets generally define b = cos sin . to compress notation, note that we are giving the arguments of the sine and cosine function in degrees rather than the standard radians. (Be careful if you are evaluating these numerically using a computer.) The problem statement tells you that E {| b t 30 x ( u ) | 2 } = 10 = b t 30 K x b 30 and E {| b t 75 x ( u ) | 2 } = 6 = b t 75 K x b 75 Since there are three unknowns in the K x matrix (using symmetry), the above two equations by themselves are not enough to determine K x . You must also use the fact that the maximum mean-squared projection is achieved when using b 30 , and hence b 30 = e 1 is a normalized eigenvector of K x with eigenvalue 1 = 10. Furthermore, the remaining orthonormal eigen- vector must be e 2 = b 120 or its negative. Hence there is really only one unknown in K x , namely the second eigenvalue 2 . K x = h b 30 . . . b 120 i 10 2 h b 30 . . . b 120 i t Now using information about the second measurement gives 6 = b t 75 h b 30 . . . b 120 i 10 2 h b 30 . . . b 120 i t b 75 since all the inner products of b vectors involve 45 angles, each of the inner products is 1 / 2. The above equation reduces to 6 = (10 + 2 ) / 2 = 2 = 2 . So the results are as follows: (a) E {| x ( u ) | 2 } = Tr { K x } = 1 + 2 = 12 . (b) Summarizing, e 1 = b 30 = 1 2 3 1 , 1 = 10, e 2 = b 120 = 1 2- 1 3 , 2 = 2. (c) Because m x = , it follows that the covariance matrix and correlation matrix of x ( u ) are identical. Using the finite-dimensional equivalent of Mercers expansion gives R x = 1 b 30 b t 30 + 2 b 120 b t 120 = 10 1 4 3 3 3 1 + 2 1 4 1- 3- 3 3 = 8 2 3 2 3 4 (d) The eigenvector with minimum eigenvalue points the way. The minimizing choices for b are b 120 and the corresponding angles are 120 and- 60 = 300 . 2 EE 562a Midterm Solution October 15, 2009 (e) Here are two ways to write the answer. b : E {| y ( u ) | 2 } = b t K x b = 8cos 2 + 4sin 2 + 4 3cos sin b : E {| y ( u ) | 2 } = b t K x b = b t h b 30 . . . b 120 i 10 0 2 h b 30 . . . b 120 i t b = 10cos 2 ( - 30 ) + 2cos 2 ( - 120 ) and you can get other equivalent expressions by applying trig identities. Now lets take a geometric view. Basic geometry (the pythagorean theorem) tells us that the sums of squares of the projections of a vector x ( u ) onto an orthonormal basis is | x ( u ) | 2 ....
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ExamSolution copy - EE 562a Midterm Solution October, 2009...

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