blockdiagram10 - s s s s Y s R s Y s s s + + =-+ + (6)...

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Block Diagrams Q UESTION 10 For the block diagram shown in Figure 10.1, determine the poles of the closed–loop transfer function. 1 s 2 s 5 2 R(s) Y(s) + - - - + + Figure 10.1 Denoting the output of the left summing junction by E 1 ( s ) ( 29 ( 29 ( 29 ( 29 1 2 5 E s R s E s Y s = - - (1) Denoting the output of the right summing junction by E 2 ( s ) ( 29 ( 29 ( 29 ( 29 2 1 1 1 2 E s E s E s Y s s = + - (2) Writing an equation for the output ( 29 ( 29 2 2 Y s sE s = (3) Combining equations (1) and (2) ( 29 ( 29 ( 29 ( 29 ( 29 2 2 2 1 5 s E s R s E s Y s Y s s + = - - - (4) Solving equation (4) explicitly for E 2 ( s ) ( 29 ( 29 ( 29 2 2 1 3 1 11 5 11 5 s s E s R s Y s s s + + = - + + (5) Multiplying equation (5) by 2 s and combining the resulting equation with equation (3)
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Block Diagrams ( 29 ( 29 ( 29 2 2 4 2 6 2 11 5 11 5
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Unformatted text preview: s s s s Y s R s Y s s s + + =-+ + (6) Rearranging equation (6), the closed–loop transfer function is ( 29 ( 29 2 2 4 2 6 13 5 Y s s s R s s s + = + + (7) The transfer function in equation (8) appears to be second order; however, the numerator and denominator have a common factor of 2 s + 1. Therefore, the true closed–loop transfer function is ( 29 ( 29 2 3 5 Y s s R s s = + (8) The closed–loop poles are found by setting the closed–loop characteristic equation to zero 5 3 5 3 s s + = → = -(9) 2...
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This note was uploaded on 02/01/2012 for the course MECH ENG 279 taught by Professor Landers during the Spring '11 term at Missouri S&T.

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blockdiagram10 - s s s s Y s R s Y s s s + + =-+ + (6)...

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