EET 309 CHAPTER4 - 4.0ROOTLOCUS OBJECTIVE Determination of...

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    OBJECTIVE Determination  of  root  from  the  characteristic  equation  by  using  graphical  solution. Rules on sketching the root locus. Analysis of closed-loop using root locus 4.0 ROOT LOCUS
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    Root locus concept Root locus concept Consider Open-loop transfer function And Where If And KG ( s ) H ( s ) R ( s ) + E ( s ) Y ( s ) B ( s ) ) ( ) ( s H s KG ) ( ) ( ) ( 1 1 s P s Z s G = = + = + + + = m i i m ) z s ( ) z s ).... ( z s )( z s ( ) s ( Z 1 2 1 1 = + = + + + = n j j n p s p s p s p s s P 1 2 1 1 ) ( ) ).... ( )( ( ) (
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    Closed-loop transfer function ) ( ) ( ) ( ) ( ) ( 1 1 1 s KZ s P s KZ s R s Y + = ~ Number and position of zeros for open-loop and closed-loop are the same ~ Position of poles for the closed-loop depend on the position of poles, zeros and K . Characteristic equation is 0 ) ( ) ( 1 1 = + s KZ s P . If 1 ) ( s H Let ) ( ) ( ) ( 1 1 s P s Z s G = and ) ( ) ( ) ( 2 2 s P s Z s H = Its open-loop transfer function ) ( ) ( ) ( ) ( ) ( ) ( 2 1 2 1 s P s P s Z s Z K s H s KG = ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( 2 1 2 1 2 1 s Z s KZ s P s P s P s KZ s R s Y + = and its closed-loop as ~ Position of poles for the closed-loop depend on the position of poles, zeros and K of the open-loop. transfer function. Root locus concept Root locus concept Where H(s) =1
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    As K varies the closed-loop poles follow similarly and form a locus. For that we define Root locus is a locus of characteristic equation as K varies from 0 to  . Referring to a characteristic equation 0 ) ( ) ( 1 = + s H s KG Let say = = + + = n j j m i i p s z s K s H s KG 1 1 ) ( ) ( ) ( ) ( 1 ) ( ) ( - = s H s KG Re-arrange -1 is a complex number π jr e 1 1 = - where .... , , , r 5 3 1 ± ± ± = Root locus concept Root locus concept
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    Magnitude condition Magnitude condition 1 ) ( ) ( = s H s KG Re-arrange m n m i i n j j z s z s z s p s p s p s z s p s K + + + + + + = + + = = = ...... ...... 2 1 2 1 1 1 where n p s p s p s + + + ...... 2 1 is the magnitude from a test point to open-loop poles n p p p - - - ...... , 2 1 and m z s z s z s + + + ...... 2 1 the magnitude from a test point to open-loop zeros n z z z - - - ...... , 2 1 From the magnitude condition, we can determine K . Let s be the test point, the magnitude are n n n ,..... 1 m m m × × ..... 1 and from open-loop zeros and poles respectively. Hence the magnitude condition is 1 ..... ..... 1 1 = × × × × m n m m n n
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      ϖ j σ s-plane n 1 n n m m m 1 s Magnitude condition Magnitude condition
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    Angle condition Angle condition     Revisiting the complex number of -1 π jr e 1 1 = - Its angle , 5 , 3 , 1 , ) ( ) ( ) ( ) ( 1 1 ± ± ± = = + - + = = = r r p s z s s H s KG n j j m i i Expand [ ] [ ] ..... , 5 , 3 , 1 ) ( ..... ) ( ) ( ( ..... ) ( ) ( 2 1 2 1 ± ± ± = = + + + + + + - + + + + + + r r p s p s p s z s z s z s n m where ( 29 ( 29 ( 29
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This note was uploaded on 03/04/2011 for the course EET 309 taught by Professor Mariahahmad during the Spring '11 term at University of Malaya.

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EET 309 CHAPTER4 - 4.0ROOTLOCUS OBJECTIVE Determination of...

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