Automatic Control Systems, 8th ed. (Solutions Manual) by Benjamin C. Kuo, Farid Golnaraghi

By Benjamin C. Kuo, Farid Golnaraghi

Real-world applications--Integrates real-world research and layout purposes in the course of the textual content. Examples contain: the sun-seeker approach, the liquid-level regulate, dc-motor regulate, and space-vehicle payload regulate. * Examples and problems--Includes an abundance of illustrative examples and difficulties. * Marginal notes through the textual content spotlight details.

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T + t + t L 1 + t + t + L −e + e   3! 5 ! 2 ! 4! 58 θr, then the state equations are in the form of CCF. 22  ( sI − A ) = −1 For a unit-step function input, u ( t ) s =1 / s. 479 e  (c) Characteristic equation: (d) ∆( s ) = s 2 + 80 . 65 s + 322 From the state equations we see that whenever there is to increase the effective value of Ra by (1 + K ) Rs . =0 . 58 Ra there is ( 1 + K ) Rs . Thus, the purpose of R is s This improves the time constant of the system. 18 θr. The equations are in the form of CCF with v as the input.

1e (s −( s + 20 Ω( s ) = −0 . 1e (s 5-29 (a) 0 . 1e −0 . 1 e (s (s ) T D 30 e (s)+ + 5) (s T (s) D + 2 )( s + 20 −0 . 2 s (s (s + 5) (s ) + 90( s + 2 )U ( s ) + 2 )( s + 20 ) + 0 . 1e 30 e + − 0. 2 s −0 . 2 s −0 . 1e There should not be any incoming branches to a state variable node other than the s −1 −0 . 2 s branch. Thus, we should create a new node as shown in the following state diagram. Notice that there is a loop with gain −1 after all the s (b) State equations: dx 1 dt = 17 x 2 1 + 1 x 2 dx 2 2 dt = 15 2 x − 1 1 2 x 2 + 1 r R (s ) = 2.

0 x 5-18 (a) State variables: x 1 = y, x 2 = dy , dt x 3 = d 2 y 2 dt x& ( t ) = Ax ( t ) + B r ( t ) State equ ations: 0 1 0 A= 0 0 1    −1 −3 −3   0 B =  0   1  (b) State transition matrix: s2 + 3 s + 3 s + 3 1   s −1 0  1  −1  2 Φ ( s ) = ( sI − A ) =  0 s −1  = −1 s + 3s s     ∆ (s ) 2  − s  1 3 s + 3  −3 s − 1 s   1 1 1 1 2 1  + + +   2 3 ( s + 1) 2 ( s + 1)3 ( s + 1 )3   s + 1 ( s + 1) ( s + 1)   −1 1 1 2 s = + −  2 s + 1 ( s + 1) ( s + 1 )3 ( s + 1) 3 ( s + 1) 3     2 −s −3 2 s   +  ( s + 1 )3 ( s + 1) 2 ( s + 1)3 ( s + 1)3  −1 ∆ (s ) = s3 + 3 s2 + 3s + 1 = ( s + 1) (1 + t + t 2 / 2 ) e− t  2 −t φ ( t) =  −t e / 2  ( −t + t 2 / 2 ) e −t (t + t ) e (1 + t − t ) e −t 2 2 2 t e 3  ( t − t / 2 ) e  (1 − 2t + t 2 / 2 ) e −t  2 −t t e /2 −t −t 2 −t (d) Characteristic equation: ∆ ( s ) = s + 3 s + 3 s + 1 = 0 3 − 1, −1, −1 Eigenvalues: 5-19 (a) State variables: 2 x 1 = y, x 2 = dy dt State equations:  dx1 (t )   dt   0 1   x1 ( t)   0   =   +   r ( t)  dx2 ( t)   − 1 − 2   x2 ( t)   1   dt  State transition matrix:  s+ 2  ( s + 1) 2  s −1  −1  Φ ( s ) = ( sI − A ) =  1 s + 2  =  − 1  ( s + 1) 2  −1  ( s + 1)  s  ( s + 1)2  1 2 51  (1 + t ) e−t φ (t ) =   − te −t   (1 − t ) e  te −t −t Characteristic equation: = y, −y= x (b) State variables: x 1 ∆ (s ) = ( s +1) = 0 2 x 2 = y dy + dt State equations: dx dt 1 = dy dt = x 2 dx − x1 2 2 dt d = 2 y dt dy + 2 dt = −y − dy +r = −x2 +r dt  dx1   dt   −1 2   x1  0   =   + r  dx2   0 −1  x2  1   dt  State transition matrix:  1 s + 1 −2   s + 1 Φ (s ) =  =  0 s + 1   0  ( s + 1)  1   s +1  −2 −1 2 (c) Characteristic equation: ∆ (s ) = ( s +1) = 0 φ (t ) = e −t 0  − te e −t −t    2 5-20 (a) State transition matrix: ω  s − σ sI − A =  −ω s − σ  ( sI − A ) −1 which is the same as in part (a).

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Automatic Control Systems, 8th ed. (Solutions Manual) by Benjamin C. Kuo, Farid Golnaraghi
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