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**Additional info for Automatic Control Systems**

**Example text**

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.

51 2 K = 5476 j 1614. 6 (b) Transfer function: TAO ( s ) 8 ± j 1275 (b) Transfer function: Ω( s ) α ( s) = K1 K 4 e −τ D s Js + ( JK L + B ) s + K 2 B + K3 K 4 e 2 − τ Ds (c) Characteristic equation: Js + ( JK L + B ) s + K 2 B + K 3 K 4e 2 (d) Transfer function: Ω( s ) α ( s) − τD s =0 K1 K 4 ( 2 − τ D s ) ≅ ∆ ( s) Characteristic equation: ∆ ( s ) ≅ J τ D s + ( 2 J + JK 2τ D + Bτ D ) s + ( 2 JK 2 + 2B − τ D K 2 B − τ D K 3 K4 ) s + 2 ( K 2 B + K 3 K 4 ) = 0 3 2 4-19 (a) Transfer function: Ec ( s) G ( s) = = E (s ) 1 + R2C s 1 + ( R1 + R 2 ) Cs (b) Block diagram: (c) Forward-path transfer function: Ωm ( s) = E (s ) [1 + ( R 1 K (1 + R2C s ) + R 2 ) Cs ] ( K b Ki + Ra JL s ) (d) Closed-loop transfer function: Ωm ( s) Fr ( s ) (e) = Gc ( s) = [1 + ( R 1 E c ( s) = Kφ K (1 + R2C s ) + R2 ) Cs] ( K b K i + Ra J L s ) + Kφ KK e N (1 + R2C s ) (1 + R C s ) E (s ) Forward-path transfer function: Ωm ( s) E (s ) 2 RCs 1 = K (1 + R2C s ) RCs ( K b K i + Ra J L s ) 1 Closed-loop transfer function: 30 Ωm ( s) Fr ( s ) (f) Ke = 36 pulse s / rev f = 120 pulse s / sec = e f r ω NK ω m = 120 Thus, N = 1.

T) = I + At + A t + L = = −t t 3 5 2 4 2! 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.