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# 2 3 4 s s s s ka ks 6 11 6 2 3 4 ka ks s s s s using

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2 3 4 = + + + + s s s s 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 c o l um n . 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 k60 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 5432 () 2 2 4 11 10 qs s s s s s = ++++ + . Determine the system is stable or unstable by using Routh- Hurwitz criterion: (8 Marks) 2 2 4 s s s s s =+ + + + + 5 4 3 2 1 1 1 0 12 1 1 241 0 6 10 10 s s s sc sd s ε c 1 = 41 2 d 1 = 1 1 61 0 c c = 12 = 12 0 12 ⎛⎞ ×− ⎜⎟ ⎝⎠ = 6 There are two signs change in 1 st column, two roots of q(s) lie in right hand plane. This system is unstable.
7. An elevator control system has the structure shown in figure. Determine the gain at which the system will become stable and the system is marginally stable system. (16 Marks) 2 (1 ) () (3 3 ) K Gs ss s + = ++ , H (s) =1 2 ) 3 ) K GH s s + = Characteristics equation: 1( ) 0 GH s += 2 ) 10 3 ) K s + 32 33 1 0 sss K + + = 3 2 1 0 13 31 8 3 1 s sK K s + + For Stable System, sign may not be change in the first column. So, 8 01 0 3 81 18 K and K K and K K >+ > <> −< < For Marginally Stable System, the gain is 8 0 3 8 K K = = Y(s) Vertical position + R (s) - ) K + Controller 2 1 3 ) s + + Elevator dynamics Desired vertical position

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Chapter (7) 8. Consider a feedback system with a loop transfer function GH(s) = ) 3 )( 2 ( ) 1 ( + + + s s s s k (a)Find the asymptotes and draw then in s-plane. (b)Find the root locus breakaway point and the gain K for this point. (16 Marks) Solution GH(s) = ) 3 )( 2 ( ) 1 ( + + + s s s s k 1+kp(s) = 0 (a) 1+ ) 3 )( 2 ( ) 1 ( + + + s s s s k = 0 Asymptote center, z p zero pole A n n = σ = 1 3 ) 1 ( ) 3 ( ) 2 ( 0 + + = 2 4 2 1 5 = =-2 Asymptote angle ,.... 2 , 1 , 0 , 180 * 1 2 = + = q n n q z p A φ = 180 * 1 3 1 0 * 2 + = 180 * 2 1 = 0 9 & (b) Breakaway point, 0 ) 3 )( 2 ( ) 1 ( 1 = + + + + s s s s k ω j α = k 0 = k 2 1 3 0 = k = k 2 = A Breakaway=-2.5 pt
1 6 5 ) 1 ( 2 3 = + + + s s s s k k = ) 1 ( ) 6 5 ( 2 3 + + + s s s s 0 ) 1 ( 1 * ) 6 5 ( ) 6 10 3 )( 1 ( 2 2 3 2 = + + + + + = s s s s s s s ds dk 0 6 5 6 10 3 6 10 3 2 3 2 2 3 = + + + s s s s s s s s 0 6 10 8 2 2 3 = + + + s s s s = -2.5, -0.76 0 ± breakaway point s = -2.5 breakaway point maz k Routh-Hurwitz 0 6 5 ) 1 ( 1 2 3 = + + + + s s s s k s s s 6 5 2 3 + + +ks+1=0 0 1 2 3 s s s s k k 5 5 30 5 1 + 6 1 0 0 k + 0 5 30 0 5 5 30 = + = + k k 5k = -30 k = -6 U(s) = 5s 2 +k = 5s 6 2 = 1 . 1 ± 9. A unity feedback system has a plant G(s) = 2 (2 5 ) k ss s + + (16 Marks) (a)Find the asymptotes and draw them in the s-plane (b)Find the angle of departure from the complex poles s -2.1 -2.2 k 0.12 0.29 0.37 04 -2.3 0.48 0.39 -2.4 -2.5 -2.6

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(c)Determine the gain when two roots lie on the imaginary axis (d)Sketch the root locus. 9. Solution 0 ) 2 1 )( 2 1 ( 1 0 ) 5 2 ( 1 0 ) ( 1 ) 5 2 ( ) ( 1 ) ( ) 5 2 ( ) ( 2 2 2 = + + + + = + + + = + + + = = + + = j s j s s k s s s k s kp s s s k s GH s H s s s k s G (a) Asymptote center 0 3 ) 2 1 ( ) 2 1 ( 0 + + + = = j j n n z p zero pole A σ = 67 . 0
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2 3 4 s s s s ka ks 6 11 6 2 3 4 ka ks s s s s Using Routh...

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