Automatic Control Systems by Farid Golnaraghi, Benjamin C. Kuo

By Farid Golnaraghi, Benjamin C. Kuo

Computerized keep watch over structures offers engineers with a clean new controls booklet that areas designated emphasis on mechatronics. It follows a progressive strategy by means of truly together with a actual lab. additionally, readers will locate authoritative insurance of contemporary layout instruments and examples. present mechatronics functions construct motivation to benefit the cloth. broad use of digital lab software program is additionally built-in through the chapters. Engineers will achieve a robust comprehend of regulate structures with assistance from glossy examples and routines.

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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.

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