solution04 - 1/. The engine block is a first order system...

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1/. The engine block is a first order system with τ = 17,935 sec and k = 0.07 ”C/W. Let t = 0 be the time at which the car is brought home. At this time T θ is 110 & (-15) = 125 ”C . The system’s response to this initial condition is 17935 / / 125 ) 0 ( ) ( t t e e T t T = = τ θ The system input is ) 25200 ( ( 5 . 302 )) hours 7 ( ( 5 . 302 = = t u t u q IN The system’s response to this delayed step input is () () () 17935 / ) 25200 ( 17935 / ) 25200 ( / ) 25200 ( 1 175 . 21 1 ) 07 . 0 )( 5 . 302 ( 1 ) ( : 25200 for 0 ) ( : 25200 for = = = >= = < t t t e e e Ck t T t t T t The system’s overall response is the sum of its response to the initial conditions and its response to the input. () 17935 / ) 25200 ( 17935 / 17935 / 1 175 . 21 125 ) ( : 25200 for 125 ) ( : 25200 for + = >= = < t t t e e t T t e t T t At 8:00 in the morning ( t = 9 hours = 32400 seconds) () C 53 . 12 53 . 27 ) 15 ( 53 . 27 1 175 . 21 125 17935 / ) 25200 32400 ( 17935 / 32400 ° = + = + = = + = T T T e e T E 2/. = = = t t X t t X t d e e Ck d e k e C d t h x t y 0 / ) ( / 0 / ) ( / 0 ) 1 ( ) 1 ( ) ( ) ( ) ( λ
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() () = = = + + X t t t t X t t t t X t e e X X e Ck e X X e Ck d e e Ck t y / / / ) / ) ( / ( / ) ( 0 0 ) / ) ( / ( / ) ( 1 ) ( τ λ The output is always less than the output for a step input of magnitude C (note that the rightmost term will always be positive, regardless of whether X is less than or greater than τ ). This is reasonable as the input is always less a step input of magnitude C . As X
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This note was uploaded on 07/16/2009 for the course SYSC 3600 taught by Professor John bryant during the Winter '08 term at Carleton CA.

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solution04 - 1/. The engine block is a first order system...

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