Control ENG HW_Part_50

# Control ENG HW_Part_50 - Soln. Balance at each node (1) =...

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Soln. Balances at each node (1) = R – C ------------------(a) (2) = 2 (1) = 2(R – C) ------------------(b) (3) = (2) – (4) = 2(R – C) – (4) -------------------(c) (4) = (3)/s = (2(R – C) – (4))/s -------------------(d) (5) = (4) – C -------------------(e) C = 2(5) -------------------(f) Solving for (4) using (d) s (4) = 2(R – C) – (4) (4) = 2(R – C) / (s +1) Using (e) (6) = 2(R – C) / (s +1) – C ( ) - - + = C C R s C 1 2 2 ( ) 7 3 4 ) 1 ( 2 4 ) 1 ( 4 + = + + + + = s R C s s C R Q – 12.4. Derive the transfer function Y/X for the control system shown

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Unformatted text preview: Soln. Balance at each node (1) = (5) + X -----------------(a) (2) = (1) (4) -----------------(b) (3) = (2)/s ------------------(c) Y = (3)/s ------------------(d) (5) = 2 (3) ------------------(e) (4) = 25Y ------------------(f) From (b) (4) = (1) (2) = (1) s (3) from (c) = (1) s 2 Y from (d) = (5) + X - s 2 Y from (a) = 2 (3) + X - s 2 Y from (e) = 2 s Y + X - s 2 Y From (f) Y = (2 s Y + X - s 2 Y)/25 X = Y( 25 2s + s 2 ) 25 2 1 2 +-= s s X Y...
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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_50 - Soln. Balance at each node (1) =...

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