Ka ks 6 11 6 2 3 4 ka ks s s s s using routh hurwitz

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ka ks 0 6 11 6 2 3 4 = + + + + + ka ks s s s s Using Routh Hurwitz, 0 1 2 3 4 s s s s s a k c b 6 1 0 0 6 11 a k k + 0 0 0 0 a k b= 6 60 6 6 66 6 ) 6 ( 66 k k k = = + c = b k k b a 6 ) 6 ( + The system is stable, no sign change in st 1 column. 0 b and 0 c Desired position Y(s) Data head position ) 3 )( 2 ( 1 + + s s s ) 1 ( ) ( + + s a s k - + R (s)
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6 60 k 0 and b ka k b 6 ) 6 ( + 0 60-k 0 and b(k+6)-6ka 0 60 k and b(k+6) kia 6 k 60 and b 6 6 + k ka k 60 and 6 6 6 60 + k ka k k 60 and (60-k)(k+6) ka 36 k 60 and a k k k + 36 ) 6 )( 60 ( if k=40, a 639 . 0 6. A system has a characteristic equation 5 4 3 2 ( ) 2 2 4 11 10 q s s s s s s = + + + + + . Determine the system is stable or unstable by using Routh- Hurwitz criterion: (8 Marks) 5 4 3 2 ( ) 2 2 4 11 10 q s s s s s s = + + + + + 5 4 3 2 1 1 1 0 1 2 11 2 4 10 6 10 10 s s s s c s d s ε c 1 = 4 12 ε ε d 1 = 1 1 6 10 c c ε = 12 ε = 12 6 10 12 ε ε ε × − = 6 There are two signs change in 1 st column, two roots of q(s) lie in right hand plane. This system is unstable.
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