lecture_27

# lecture_27 - MIT OpenCourseWare http/ocw.mit.edu 2.004...

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MIT OpenCourseWare http://ocw.mit.edu 2.004 Dynamics and Control II Spring 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms .

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+ - R ( s ) C ( s ) K G ( s ) G ( s ) c o n t r o l l e r p l a n t c p H ( s ) Massachusetts Institute of Technology Department of Mechanical Engineering 2.004 Dynamics and Control II Spring Term 2008 Lecture 27 1 Reading: Nise: Chapter 8 1 Root Locus Development In Lecture 26 we saw some simple examples of root locus plots. We now look at the general method of generating root loci. Recall that the closed-loop characteristic equation is: 1 + KG ( s ) = 0 where G ( s ) = G c ( s ) G p ( s ) H ( s ) is the open-loop transfer function. We now ask ourselves: how can we tell if an arbitrary point s = σ + lies on the root locus? In other words we seek conditions that determine whether s is a root of the characteristic equation. From above, s is a root if KG ( s ) = 1 + j 0 . In polar form this may be expressed as KG ( s ) = KG ( s ) e j ( KG ( s )) | j (2 n | +1) π = 1 × e for n = 0 , 1 , 2 ,... = cos((2 n + 1) π ) + j sin((2 n + 1) π ) = 1 + j 0 . This tells us that for any point s = σ + on the root locus | KG ( s ) | = 1 and ( G ( s )) = (2 n + 1) π which generates two important conditions: 1 copyright c D.Rowell 2008 27–1
| | The Angle Condition: ( G ( s )) = (2 n + 1) π The Magnitude Condition: KG ( s ) = 1 | | In practice, the angle condition is used to determine whether a point s lies on the root locus, and if it does, the magnitude condition is used to determine the gain K associated with that point, since K = 1 / G ( s ) . | | Example 1 Given the open-loop system 1 G ( s ) = ( s + 2)( s + 4) determine whether the points s = 1, s = 3 . 5, s = 3 + j 5 are on the root locus. For s = 1 : K K KG ( s ) = ( 1 + 2)( 1 + 4) 3 KG ( s ) = 1 for any K > 0, so we conclude s = 1 is not on the root-locus. For s = 3 . 5 : K K KG ( s ) = ( 3 . 5 + 2)( 3 . 5 + 4) 0 . 75 KG ( s ) = 1 for K = 0 . 75, so we conclude s = 3 . 5 lies on the root-locus. For

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lecture_27 - MIT OpenCourseWare http/ocw.mit.edu 2.004...

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