# solutions11 - Matlab Solutions Time Series (2DD23) Week 11...

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Unformatted text preview: Matlab Solutions Time Series (2DD23) Week 11 Exercise 9.1a y t = 2 x t , so T { x t } = 2 x t . T { λ 1 x 1 , t + λ 2 x 2 , t } = 2 (λ 1 x 1 , t + λ 2 x 2 , t ) = 2 λ 1 x 1 , t + 2 λ 2 x 2 , t = λ 1 ( 2 x 1 , t ) + λ 2 ( 2 x 2 , t ) = λ 1 ( T { x 1 , t } ) + λ 2 ( T { x 2 , t } ) So the system is linear. Now define: x 1 , t := x t- τ . Then T { x 1 , t } = 2 x t- τ = y t- τ , so the system is time invariant. Exercise 9.1b y t = . 7 x t- 3 , so T { x t } = . 7 x t- 3 . T { λ 1 x 1 , t + λ 2 x 2 , t } = . 7 (λ 1 x 1 , t + λ 2 x 2 , t )- 3 = λ 1 ( . 7 x 1 , t ) + λ 2 ( . 7 x 2 , t )- 3 This does not equal λ 1 ( . 7 x 1 , t- 3 ) + λ 2 ( . 7 x 2 , t- 3 ) . Counter example: take λ 1 = λ 2 = 1 and x 1 , t = x 2 , t = 1 . Then . 7 + . 7- 3 = - 1 . 6 , and . 7- 3 + . 7- 3 = - 4 . 6 . This is not the same, so the system is not linear. Now define: x 1 , t := x t- τ . Then T { x 1 , t } = . 7 x 1 , t- 3 = . 7 x t- τ- 3 = y t- τ , so the system is time invariant. Exercise 9.1c y t = . 5 x t + . 3 x t- 1 , so T { x t } = . 5 x t + . 3 x t- 1 . T { λ 1 x 1 , t + λ 2 x 2 , t } = . 5 (λ 1 x 1 , t + λ 2 x 2 , t ) + . 3 (λ 1 x 1 , t- 1 + λ 2 x 2 , t- 1 ) = λ 1 ( . 5 x 1 , t + . 3 x 1 , t- 1 ) + λ 2 ( . 5 x 2 , t . 3 x 2 , t- 1 ) = λ 1 ( T { x 1 , t } ) + λ 2 ( T { x 2 , t } ) So the system is linear....
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## This note was uploaded on 09/23/2009 for the course MATH 200 taught by Professor Gur during the Spring '09 term at Accreditation Commission for Acupuncture and Oriental Medicine.

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solutions11 - Matlab Solutions Time Series (2DD23) Week 11...

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