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# 582hw2 - EEE 582 Problem 2.1 HW 2 SOLUTIONS y(n(t...

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EEE 582 HW # 2 SOLUTIONS Problem 2.1 y ( n ) ( t ) + a n ¡ 1 ( t ) y ( n ¡ 1) ( t ) + ¢ ¢ ¢ + a 0 ( t ) y ( t ) = b 0 ( t ) u ( t ) + b 1 ( t ) u (1) ( t ) Let x n ( t ) = y ( n ¡ 1) ( t ) ¡ b 1 ( t ) u ( t ) and x 1 ( t ) = y ( t ) then _ x 1 ( t ) = _ y ( t ) = x 2 ( t ) _ x 2 ( t ) = Ä y ( t ) = x 3 ( t ) . . . _ x n ¡ 2 ( t ) = x n ¡ 1 ( t ) _ x n ¡ 1 ( t ) = x n + b 1 ( t ) u ( t ) _ x n ( t ) = y ( n ) ( t ) ¡ b (1) 1 u ( t ) ¡ b 1 ( t ) u (1) ( t ) = ¡ n ¡ 1 X i =0 a i ( t ) y ( i ) ( t ) + b 0 ( t ) u ( t ) ¡ b (1) 1 ( t ) u ( t ) = ¡ n ¡ 2 X i =0 a i ( t ) x i ( t ) + a n ¡ 1 ( t ) ( x n ( t ) + b 1 ( t ) u ( t )) + ³ b 0 ( t ) ¡ b (1) 1 ( t ) ´ u ( t ) then we can write A ( t ) = 2 6 6 6 6 6 4 0 1 ¢ ¢ ¢ 0 0 0 ¢ ¢ ¢ 0 . . . . . . . . . 0 0 ¢ ¢ ¢ 1 ¡ a 0 ( t ) ¡ a 1 ( t ) ¢ ¢ ¢ ¡ a n ¡ 1 ( t ) 3 7 7 7 7 7 5 ; B ( t ) = 2 6 6 6 6 6 4 0 0 . . . b 1 ( t ) b 0 ( t ) ¡ b (1) 1 ( t ) + a n ¡ 1 ( t ) b 1 ( t ) 3 7 7 7 7 7 5 C ( t ) = £ 1 0 ¢ ¢ ¢ 0 ¤ ; D ( t ) = 0 Problem 2.8 Identity dc-gain means that for a given ~ u , 9 ~ x , such that A ~ x + B ~ u = 0, C ~ x = ~ u , this implies that the matrix · A B C 0 ¸ is invertible. 1. If K 2 IR m £ n is such that ( A + BK ) is invertible, then C ( A + BK ) ¡ 1 B is invertible. Since · A B C 0 ¸ is invertible , for any K , · A + BK B C 0 ¸ is invertible, this from · A + BK B C 0 ¸ = · A B C 0 ¸ · I 0 K I ¸ Then · A + BK B C 0 ¸ · R 1 R 2 R 3 R 4 ¸ = · I 0 0 I ¸ so ( A + BK ) R 1 + BR 3 = I ( A + BK ) R 2 + BR 4 = 0 ) R 2 = ¡ ( A + BK ) ¡ 1 BR 4 CR 1 = 0 CR 2 = I ) ¡ C ( A + BK ) ¡ 1 BR 4 = I hence C ( A + BK ) ¡ 1 B is invertible. 1

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2. We need to show that there exits N such that 0 = ( A + BK )~ x + BN ~ u ~ u = C ~ x The ¯rst equation gives ~ x = ¡ ( A + BK ) ¡ 1 BN ~ u . Thus we need to choose N
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582hw2 - EEE 582 Problem 2.1 HW 2 SOLUTIONS y(n(t...

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