expm_2009_10_26_01_2up

# expm_2009_10_26_01_2up - 11 1 Matrix Exponential S Lall...

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11 - 1 Matrix Exponential S. Lall, Stanford 2009.10.26.01 11. Matrix Exponential Diagonalization Modal form Matrix exponential Sampling a continuous-time system Stability 11 - 2 Matrix Exponential S. Lall, Stanford 2009.10.26.01 Diagonalization suppose v 1 , . . . , v n is a linearly independent set of eigenvectors of A R n × n : Av i = λ i v i , i = 1 , . . . , n express as A ± v 1 · · · v n ² = ± v 1 · · · v n ² λ 1 . . . λ n deFne T = ± v 1 · · · v n ² and Λ = diag( λ 1 , . . . , λ n ) , so AT = T Λ hence T 1 AT = Λ T invertible since v 1 , . . . , v n linearly independent similarity transformation by T diagonalizes A

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11 - 3 Matrix Exponential S. Lall, Stanford 2009.10.26.01 Modal form Suppose A is diagonalizable by T . Defne new coordinates by x = T ˜ x , so T ˙ ˜ x = AT ˜ x ˙ ˜ x = T 1 AT ˜ x in new coordinate system, system is diagonal ˙ ˜ x = Λ˜ x Called modal form ; trajectories consist oF n independent modes, i.e. , ˜ x i ( t ) = e λ i t ˜ x i (0) iF initial state x (0) is an eigenvector v , resulting motion is simple — always on the line spanned by v For λ R , λ < 0 , mode contracts or shrinks as t For λ R , λ > 0 , mode expands or grows as t 11 - 4 Matrix Exponential S. Lall, Stanford 2009.10.26.01 Matrix exponential ±or a square matrix M , the matrix exponential is defned to be e M = I + M + M 2 2! + · · · Then the solution oF oF ˙ x = Ax , with A R n × n is x ( t ) = e tA x (0) the series For e M converges For all M one can check that e tA x (0)
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## This note was uploaded on 08/23/2010 for the course EE 263 taught by Professor Boyd,s during the Fall '08 term at Stanford.

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expm_2009_10_26_01_2up - 11 1 Matrix Exponential S Lall...

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