Download e-book for iPad: Automatic Control Systems, 8th ed. (Solutions Manual) by Benjamin C. Kuo, Farid Golnaraghi

By Benjamin C. Kuo, Farid Golnaraghi

ISBN-10: 0471134767

ISBN-13: 9780471134763

Real-world applications--Integrates real-world research and layout purposes in the course of the textual content. Examples comprise: the sun-seeker approach, the liquid-level keep an eye on, dc-motor regulate, and space-vehicle payload regulate. * Examples and problems--Includes an abundance of illustrative examples and difficulties. * Marginal notes through the textual content spotlight details.

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

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1434 e  −1 (b) (c) Characteristic equati on: 2 . 782 Eigenvalues: (d) ∆ ( s ) = s + 90 . 91 s + 818 Same remark as in part (d) of Problem 5-14. 5-16 (a) Forward-path transfer function: Y ( s) 5 ( K1 + K 2s ) G ( s) = = E ( s ) s [s ( s + 4)( s + 5) + 10 ] M ( s) = Y (s ) R (s ) = G (s) 1 + G(s ) = Closed-loop transfer function: 5 ( K1 + K 2s ) s + 9 s + 20 s + (10 + 5 K 2 ) s + 5 K1 4 3 2 (b) State diagram by direct decomposition: 49 1 s  State equations:  x&1   x&   2 =  x&3   x&   4  0  0   0  −5 K  (c) Final value: Output equation:   x1   0  1 0   x2   0    +   r 0 1   x3   0      −20 −9   x4   1  1 0 0 0 − (10 + 5 K 2 ) r(t ) = u s ( t ), R( s) 0 1 = 5-17 s →0 = 5 K1 5K2 .

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 . 782 Eigenvalues: (d) ∆ ( s ) = s + 90 . 91 s + 818 Same remark as in part (d) of Problem 5-14. 5-16 (a) Forward-path transfer function: Y ( s) 5 ( K1 + K 2s ) G ( s) = = E ( s ) s [s ( s + 4)( s + 5) + 10 ] M ( s) = Y (s ) R (s ) = G (s) 1 + G(s ) = Closed-loop transfer function: 5 ( K1 + K 2s ) s + 9 s + 20 s + (10 + 5 K 2 ) s + 5 K1 4 3 2 (b) State diagram by direct decomposition: 49 1 s  State equations:  x&1   x&   2 =  x&3   x&   4  0  0   0  −5 K  (c) Final value: Output equation:   x1   0  1 0   x2   0    +   r 0 1   x3   0      −20 −9   x4   1  1 0 0 0 − (10 + 5 K 2 ) r(t ) = u s ( t ), R( s) 0 1 = 5-17 s →0 = 5 K1 5K2 .

5-35 (a) S = B 2 S is singular. The system is uncontrollable.

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Automatic Control Systems, 8th ed. (Solutions Manual) by Benjamin C. Kuo, Farid Golnaraghi


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