{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

Info iconThis preview shows pages 49–55. Sign up to view the full content.

View Full Document Right Arrow Icon
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)
Background image of page 49

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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.
Background image of page 50
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
Background image of page 51

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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
Background image of page 52
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
Background image of page 53

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
(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
Background image of page 54
Image of page 55
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page49 / 168

2 3 4 s s s s ka ks 6 11 6 2 3 4 ka ks s s s s Using Routh...

This preview shows document pages 49 - 55. Sign up to view the full document.

View Full Document Right Arrow Icon bookmark
Ask a homework question - tutors are online