We can continue this process on the N 1 N 1 submatrix on the lower right We

We can continue this process on the n 1 n 1 submatrix

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We can continue this process on the ( N - 1) × ( N - 1) submatrix on the lower right. We take x 2 = 0 0 ˜ a 32 ˜ a 42 . . . ˜ a N 2 , y 2 = 0 0 r 2 0 . . . 0 , with r 2 = k x 2 k 2 , and H 2 = I - 2 u 2 u T 2 , u 2 = x 2 - y 2 k x 2 - y 2 k 2 . Then A 3 = H 2 A 2 H 2 = a 11 r 1 0 0 · · · 0 r 1 ˜ a 22 r 2 0 · · · 0 0 r 2 b a 33 b a 34 · · · b a 3 N 0 0 b a 43 b a 44 · · · b a 4 N . . . . . . . . . . . . . . . . . . 0 0 b a N 3 b a N 4 · · · b a NN Repeating this procedure N - 1 times gives us a tri-diagonal matrix T = A N = H N H N - 1 · · · H 1 AH 1 H 2 · · · H N = Q T AQ . where Q = H 1 H 2 · · · H N . Note that since all of the H n are orthonormal, so is Q . 23 Georgia Tech ECE 6250 Fall 2019; Notes by J. Romberg and M. Davenport. Last updated 13:22, November 13, 2019
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Technical Details: Solving Toeplitz systems and the Levinson-Durbin algorithm We start by looking at how to solve Hv = y for a very particular right-hand side. Consider h 0 h 1 · · · h N - 1 h 1 h 0 h 1 · · · h N - 2 h 2 h 1 h 0 · · · h N - 3 . . . h N - 1 · · · · · · h 0 v [1] v [2] . . . . . . v [ N ] = h 1 h 2 . . . h N - 1 h N . (7) Here the first N - 1 entries of the “observation” vector y match the last N - 1 entries in the first column of H . This system is special- ized, but not at all contrived — ( 7 ) are called the Yule-Walker equations , and appear in many different places in statistical signal processing. Moreover, solving systems with general right-hand sides solve systems of the form ( 7 ) as an intermediate step. The crux of the Levinson-Durbin algorithm relies upon a seemingly innocuous fact. Let J be the N × N exchange matrix (also called the counter identity ): J = 0 0 · · · 0 0 1 0 0 · · · 0 1 0 0 0 · · · 1 0 0 . . . . . . 1 0 · · · 0 . Applying J to a vector x reverses the entries in x . For example, 0 0 1 0 1 0 1 0 0 5 - 1 7 = 7 - 1 5 . 24 Georgia Tech ECE 6250 Fall 2019; Notes by J. Romberg and M. Davenport. Last updated 13:22, November 13, 2019
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It should be clear that J T = J and J 2 = JJ = I . Applying J to the left of a matrix reverses all of its columns, while applying J to the right reverses the rows. In particular, if H is a symmetric Toeplitz matrix, then J H = 0 0 · · · 0 0 1 0 0 · · · 0 1 0 0 0 · · · 1 0 0 . . . . . . 1 0 · · · 0 h 0 h 1 · · · h N - 1 h 1 h 0 h 1 · · · h N - 2 h 2 h 1 h 0 · · · h N - 3 . . . h N - 1 · · · · · · h 0 = h N - 1 h N - 2 · · · h 0 h N - 2 h N - 3 h N - 4 · · · h 1 h N - 3 h N - 4 h N - 5 · · · h 2 .
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