Control ENG HW_Part_22

# Control ENG HW_Part_22 - an impulse function of magnitude...

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[ ]( ) 1 2 1 6 4 4 1 ) ( 2 3 + + + = s s s s s H From initial value theorem ) ( ) 0 ( 3 3 s sH Lim H S = = ) 1 6 4 )( 1 2 ( 4 2 + + + s s s Lim S = ) 1 6 4 ( ) 1 2 ( 4 2 3 s s s s Lim S + + + H 3 (0) = 0 From final value theorem ) ( ) ( 3 0 3 s sH Lim H S = = ) 1 6 4 )( 1 2 ( 4 2 0 + + + s s s Lim S H 3 (∞) = 4

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7.5 Three identical tanks are operated in series in a non-interacting fashion as shown in fig P7.5 . For each tank R=1, τ = 1. If the deviation in flow rate to the first tank in
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Unformatted text preview: an impulse function of magnitude 2, determine (a) an expression for H(s) where H is the deviation in level in the third tank. (b) sketch the response H(t) (c) obtain an expression for H(t) solution : writing energy balance equation for all tanks dt dh A q q 1 1 =-dt dh A q q 2 2 1 =-...
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## This note was uploaded on 11/13/2011 for the course COP 4355 taught by Professor Koslov during the Spring '10 term at University of Florida.

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Control ENG HW_Part_22 - an impulse function of magnitude...

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