By Carol Kaye

Kay is a smart participant and an excellent instructor. it is a helpful ebook for electrical Bassists.

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**Extra resources for Electric Bass Lines No. 2 **

**Sample text**

Bn−1 . θ2n −L bn θ2n+1 1 b1 b2 b3 .. 82) give bi in terms of L: θ (L)n−2 (L)n−1 L2 n+2 1 L · · · b1 2! (n−2)! (n−1)! θn+3 b2 (L)n−3 (L)n−2 0 1 L · · · (n−3)! (n−2)! θn+4 .. . = .. .. . . ... · . . . 82) yields the following n-degree polynomial equation in L: n i=0 θn+i+1 i L = 0. i! 84) has n roots for L. In selecting a suitable solution, a rule of thumb is to choose one that leads to the minimal output error between the step response of the estimated model and the given one.

M! 72) n times gives y(t) = −a1 (1) [0,t] y − a2 (2) [0,t] y · · · − an−1 (n−1) [0,t] y − an (n) [0,t] y 1 1 +hb1 (t − L) + hb2 (t − L)2 + · · · + hbn−1 (t − L)n−1 2 (n − 1)! 1 + hbn (t − L)n , n! n (n) (1) bj (−L)j = −a1 y + ht0 y − · · · − an j! [0,t] [0,t] j=1 +ht1 n j=1 + Define n−1 ht2 bj (−L)j−1 + (j − 1)! 2! n n j=2 bj (−L)j−2 + ··· (j − 2)! j−(n−1) ht bj (−L) (n − 1)! j=n−1 1! + htn bn . n! 4 Model Reduction γ(t) = y(t), φT (t) = − (1) [0,t] (n) y, [0,t] bj (−L)j n , j=1 j!

S2 − 4 2s2 − 8 One proceeds, ∆0 (s) = 1, ∆1 (s) = GCD{1, −1, s2 − s − 4, 2s2 − s − 8, s2 − 4, 2s2 − 8} = 1, 1 −1 1 −1 ∆2 (s) = GCD{ 2 , , 2 s + s − 4 2s2 − s − 8 s − 4 2s2 − 8 s2 + s − 4 2s2 − s − 8 } s2 − 4 2s2 − 8 = (s + 2)(s − 2). 2 Polynomial Matrices 27 It follows that λ1 (s) = ∆2 ∆1 = 1, λ2 (s) = = (s + 2)(s − 2). ∆0 ∆1 Thus, one obtains the Smith form: 1 0 Λ(s) = 0 (s + 2)(s − 2) . 3. Let P (s), L(s), and R(s) be polynomial matrices such that P (s) = L(s)R(s). Then R(s) (resp. L(s)) is called a right (resp.