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ps7sol

# ps7sol - MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department...

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Unformatted text preview: MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Aeronautics and Astronautics 16.060: Principles of Automatic Control Fall 2003 PROBLEM SET 7 Solutions Problem 1 (1): K > 0 (1): K < 0 −6 −4 −2 0 2 −5 0 5 −2 0 2 4 6 −5 0 5 (2): K > 0 (2): K < 0 −6 −4 −2 0 2 −5 0 5 −4 −2 0 2 4 6 −5 0 5 (3): K > 0 (3): K < 0 −0.5 0 0.5 −1 0 1 −12 −10 −8 −6 −4 −2 0 −4 −2 0 2 (4): K > 0 (4): K < 0 0.4 1 0.2 0 0 −1 −0.2 −0.4 −3 −2.5 −2 −1.5 −1 −0.5 0 −4 −2 0 2 1 Problem 2 Problem 2.1 • 3 asymptotes with direction ± 60 , 180 3 = − 1 − 1 − 2 4 = • 3 − 3 2 • Let ± j be the points of intersection of the root locus 1 with the imaginary axis. To satisfy the angle condition, must satisfy the equation: 0 − 2 tan − 1 ( ) − tan − 1 ( ) = − 180 2 By trial and error, we get 2 . 2. Now use the mag- −1 −2 nitude condition: −3 K crit = 1 2 + 2 . 2 2 1 2 + 2 . 2 2 2 2 + 2 . 2 2 = 17 . 4 −5 −4 −3 −2 −1 0 · · • We can also solve for K crit algebraically. The characteristic equation is: ( s + 1) 2 ( s + 2) + K = s 3 + 4 s 2 + 5 s + 2 + K = Now substitute s = j and K = K crit : ( j ) 3 + 4( j ) 2 + 5( j ) + 2 + K crit = ( − 4 2 + 2 + K crit...
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ps7sol - MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department...

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