Homework_3_solutions

# A in the linearized deviation variable form prime

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(a) In the linearized deviation variable form (prime indicates deviation variable, and subscript s denotes the steady state value), (10 pts) dX 0 dt ( μ m S s K s + S s - D s ) X 0 + μ m X s K s ( K s + S s ) 2 S 0 - X s D 0 dS 0 dt ( - μ m S s Y ( K s + S s ) ) X 0 - ( μ m X s K s Y ( K s + S s ) 2 + D s ) S 0 + ( S fs - S s ) D 0 + D s S 0 f Plug in steady state values, μ m S s K s + S s - D s = 0; μ m X s K s ( K s + S s ) 2 = 0 . 1125; - μ m S s Y ( K s + S s ) = - 0 . 2; - ( μ m X s K s Y ( K s + S s ) 2 + D s ) = - 0 . 325, S fs - S s = 4 . 5. (b) Transformation to Laplace domain: (10 pts) sX 0 ( s ) = 0 . 1125 S 0 ( s ) - 2 . 25 D 0 ( s ) sS 0 ( s ) = - 0 . 2 X 0 ( s ) - 0 . 325 S 0 ( s ) + 4 . 5 D 0 ( s ) + 0 . 1 S 0 f ( s ) (c) Rewrite the second expression in the Laplace domain, S 0 ( s ) = - 0 . 2 X 0 ( s )+4 . 5 D 0 ( s )+0 . 1 S 0 f ( s ) s +0 . 325 Substitute this in the first and rearrange to obtain, X 0 ( s ) = - 2 . 25 s - 0 . 225 s 2 +0 . 325 s +0 . 0225 D 0 ( s ) + 0 . 01125 s +0 . 325 S 0 f ( s ) The transfer function relating the cell concentration to the dilution rate, therefore, is (10 pts) g ( s ) = - 2 . 25 s - 0 . 225 s 2 +0 . 325 s +0 . 0225 2

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