ECE 485 Lesson 2

ECE 485 Lesson 2 - ECE 485 Advanced Engineering Mathematics...

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Unformatted text preview: ECE 485 Advanced Engineering Mathematics Lesson 2 System of Linear Algebraic Equations Gauss and Gauss-Jordan Elimination Methods Digital Filter Unit impulse response Unit sample response Digital Filter h(n) (n) h(n) 1 for n = 0 ( n) = 0 for n 0 Unit impulse sequence or unit sample 2 Matrix Form of Input-Output Relationship Toeplitz Matrix 0 y ( 0 ) h( 0 ) y (1) h(1) h( 0 ) x( 0 ) = y ( 2 ) h( 2 ) h(1) x(1) h( 2 ) y ( 3) 0 3 Numerical Example from Lesson 1 h ( 0 ) = 3, h ( 1) = 2, h ( 2 ) = 1 x ( 0 ) = 1, x ( 1) = 1 ( 0) y 3 y 2 ( 1) = ( 2) y 1 y 0 ( 3) 0 3 3 1 5 = 1 2 3 1 1 4 Determination of Input from the Output system of linear algebraic equations 3 2 1 0 0 3 3 x ( 0 ) 5 = 2 x ( 1) 3 1 1 3x ( 0 ) = 3 2 x ( 0 ) + 3 x ( 1) = 5 x ( 0 ) + 2 x ( 1) = 3 x ( 1) = 1 x ( 0) = 1 x ( 1) = 1 5 Gaussian Elimination Augmented matrix 3 2 1 0 0 3 2 1 1 0 0 0 3 2 1 5 3 2 3 0 3 1 1 0 2 3 1 1 1 0 1 1 0 1 1 0 3 2 3 1 5 -1 -1 0 3 -6 -6 0 1 1 1 0 2 3 1 1 Row-echelon form 0 0 0 0 6 Back Substitution 1 0 0 0 x(0) x(1) x ( 0 ) + 2 x ( 1) = 3 x ( 0) = 2 - 2 1 = 1 7 3 1 0 0 x ( 1) = 1 2 1 0 0 x ( 0 ) = 3 - 2 x ( 1) Gauss-Jordan Elimination solution x(0) x(1) x(0) x(1) 1 0 Row-echelon form 0 0 2 1 0 0 3 1 0 0 1 0 0 0 0 1 0 0 1 1 Reduced row-echelon form 0 0 Gauss-Jordan elimination method does not require back substitution 8 ...
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