# New PDF release: kuo and golnaragh automatic control systems 8ed - solutions

By Benjamin Kuo

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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. 1434 e −1 (b) (c) Characteristic equati on: 2 .

D) Eigenvalues of A: −1, −1, −1 The matrix A is already in Jordan canonical form. Thus, the DF transformation matrix T is the identity matrix I. 9597 5-11 (a) 1 −2 0 0 S = [B AB] = S = B 1 −1 1 2 AB A B = 2 −2 2 3 −3 3 S = [B AB] = S is singular. (b) S is singular. (c) 2 2 2 +2 2 2+ 2 S is singular. (d) 45 −1 0 1 1 −3 1 −2 4 AB A B = 0 0 0 1 −4 14 S = B 5-12 (a) 2 S is singular. Rewrite the differential equations as: d θm 2 dt 2 B d θm 2 =− J dt 2 − K J θm + State variables: x = θ , x m 1 Ki 2 J = dia ia dθ =− Kb dθ m dt m x , dt 3 La dt − Ra La ia + K a Ks (θ La r −θm ) = ia State equations: Output equation: dx1 dt 0 dx2 = − K dt J dx K K 3 − a s dt La 1 − B J − Kb La 0 x1 Ki x2 + 0 θ r J x K K Ra 3 a s − La La 0 y = 1 0 0 x (b) Forward-path transfer function: s Θm ( s ) K G ( s) = = [1 0 0 ] J E (s ) 0 −1 s+ B J Kb La 0 Ki − J Ra s+ L a −1 0 KiK a 0 = K ∆ o ( s) a La ∆ o ( s ) = J La s + ( BLa + Ra J ) s + ( KLa + Ki Kb + Ra B) s + KRa = 0 3 2 Closed-loop transfer function: s Θm ( s ) K M ( s) = = [1 0 0 ] J Θr ( s ) KaKs La = −1 s+ B J Kb La K i Ka K s 0 Ki − J R s+ a La −1 0 K s G( s ) 0 = K K 1 + K s (s ) a s La JLa s + ( BLa + Ra J ) s + ( KLa + Ki Kb + Ra B) s + K i K a K s + KRa 3 2 5-13 (a) 46 = x1 A= 0 1 −1 0 A = 2 − 1 0 0 −1 A = 3 0 − 1 1 0 A = 4 1 0 0 1 (1) Infinite series expansion: 3 5 t t t2 t4 1 − + − L t − + − L 1 2 2 2!

Taking the inverse Laplace transform −1 37 on both sides of the equation gives the desired relationship for 5-3 (a) Characteristic equation: Eigenvalues: s ∆( s ) = = −0 . 5 − j 1. 323 , φ( t ) . j 1. 333e −4 t −t ∆ ( s ) = ( s + 3) = 0 2 −4 t −4 t Eigenvalues: = −3, − 3 s State transition matrix: e −3 t φ ( t) = 0 (d) Characteristic equation: ∆( s ) = −3 t − 9 = 0 Eigenvalues: 2 s e 0 s = −3 , 3 s = − State transition matrix: e3 t φ ( t) = 0 (e) Characteristic equation: e 0 −3 t ∆ ( s ) = s + 4 = 0 Eigenvalues: 2 j2, j2 State transition matrix: cos2 t − sin2 t φ ( t) = (f) Characteristic equation: ∆( s ) = s 3 s i n 2t cos2t + 5 s + 8 s + 4 = 0 Eigenvalues: 2 s = − 1, − 2 , −2 State transition matrix: e− t φ ( t) = 0 0 (g) Characteristic equation: ∆( s ) = s 3 0 e 0 + 15 e φ ( t) = 0 0 −5 t 5-4 State transition equation: x (t ) = φ (t )x( t ) + −2 t + 75 s + 125 = 0 2 s te −2t e 0 −2 t te e −5 t −5 t 0 Eigenvalues: s = − 5, − 5, −5 te −5 t e 0 −5 t ∫ φ (t − τ )Br (τ )d τ t 0 (a) 38 φ (t ) for each part is given in Problem 5-3.

### kuo and golnaragh automatic control systems 8ed - solutions manual by Benjamin Kuo

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