By Benjamin C. Kuo, Farid Golnaraghi

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**Additional info for Automatic Control Systems, 8th ed. (Solutions Manual)**

**Example text**

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).