By Farid Golnaraghi, Benjamin C. Kuo

Computerized keep an eye on platforms presents engineers with a clean new controls booklet that areas certain emphasis on mechatronics. It follows a innovative method by means of truly together with a actual lab. furthermore, readers will locate authoritative assurance of contemporary layout instruments and examples. present mechatronics functions construct motivation to benefit the fabric. vast use of digital lab software program is additionally built-in through the chapters. Engineers will achieve a robust comprehend of regulate platforms with the aid of glossy examples and routines.

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**Example text**

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. 82. These are not functions of K. 4536, pole-zero cancellation occurs. 02s 2 ) = 100 s + 12 s + 70 s + 100 3 2 State diagram by direct decomposition: State equations: x&1 x& = 2 x&3 0 0 −100 x1 0 0 1 x2 + 0 u −70 −12 x3 1 1 0 (b) Characteristic equation of closed-loop system: Roots of characteristic equation: 62 s 3 + 12 s + 70 s + 200 = 0 2 − 5.

563 s + 10 +1 + 5 . 235 s + 2s + 2 2 x& ( t ) = Ax ( t ) + B u ( t ) State equations: 0 0 0 −5 A= 0 0 0 0 0 1 −2 0 1 1 B= 0 1 0 0 0 −2 5-25 (a) G (s) = Y (s) U ( s) = 10 + 1. 3 30 e U ( s) 90U( s) −1 T ( s ) + T ( s ) + + D D ∆( s) s + 2 ( s + 2)( s + 20) ( s + 2)( s + 5)( s + 20) (s + 5)( s + 20) 1 Y ( s) = ∆( s ) = 1 + Y ( s) −0 . 2 s + 2 )( s + 20 (s − ( s + 19 . 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 .

5623 (2) Transfer function relation: −1 −1 s 1 −1 X( s) = ( sI − A ) B U ( s) = 0 ∆ (s ) 1 s 2 + 3s + 2 0 U ( s) = 1 −1 ∆ (s ) −s 1 0 0 −1 s + 3 s 2 s +3 1 0 1 1 s ( s + 3 ) s 0 U ( s) = s U (s ) ∆ (s ) 2 2 −2 s − 1 s 1 s ∆( s ) = s + 3 s + 2 s + 1 3 2 (3) Output transfer function: 1 1 s = = C ( s ) ( sI − A) B = [1 0 0 ] 3 2 U (s ) ∆( s ) 2 s + 3 s + 2 s + 1 s Y ( s) (b) (1) Eigenvalues of A: 1 −1 − 1, − 1.