Download Automatic Control Systems by Farid Golnaraghi, Benjamin C. Kuo PDF

By Farid Golnaraghi, Benjamin C. Kuo

Automated keep watch over platforms offers engineers with a clean new controls booklet that areas unique emphasis on mechatronics. It follows a progressive strategy by means of truly together with a actual lab. furthermore, readers will locate authoritative insurance of contemporary layout instruments and examples. present mechatronics purposes construct motivation to benefit the cloth. wide use of digital lab software program can be built-in during the chapters. Engineers will achieve a powerful comprehend of keep watch over platforms with the aid of sleek examples and workouts.

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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 . 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 .

Dt f v (t ) ∂ f1 ∂ f1 =0 ∂ x1 ∂ f2 ∂ x2 = c b ∂ f2 ∂f 3 ∂x 3 ∂f 3 =0 M b Lx2 − 3 M b g / 4 ∂ f4 ∂ x1 ∂ f4 (M + M c ) − 3M b / 4 b ∂ f4 =0 ∂ x3 ∂ x2 (M ∂ f4 =0 ∂ x4 = ∂ x3 =0 ∂u =0 2 M b Lx1 x2 + M c ) − 3M b / 4 b 1 = ∂u ∂f 3 =0 ∂x 4 2 = L ( M b + M c − 3M b / 4 ) L [ 4 (M b + M c ) / 3 − M b ] =0 ∂x 2 ∂ f2 =0 −1 = ∂u ∂f 3 =0 ∂u −2 M b x1 x2 = ∂ x2 ∂ f1 =0 ∂ x4 ∂ f2 2 b ∂ f2 =0 ∂ f4 b c =0 ∂ x4 ∂x 1 ∂ x3 (M + M ) g − M x = 0 L ( M + M − 3M / 4) b ∂ f1 =0 2 ∂ x1 ∂f 3 ∂ f1 =1 (M b + M c ) − 3M b / 4 Linearized state equations: 0    ∆x& 1   3 ( M b + M c ) g  ∆x&   L ( M + 4 M ) b c  2 =   ∆x& 3   0  &   − 3M b g  ∆x 4   M + 4M  b c = (b) i eq E = Ki 2 y (t ) 1 eq dt dy eq =0 dt x 1 eq = i, x R x 2 eq 2 = y = y , and x E eq R = 3 E = eq + dt di ( t ) eq E dy At equili brium, R = dL ( y ) dy ( t ) Ri ( t ) + i ( t ) (t ) 2 Define the state variables as Then, x = dt = Mg − My ( t ) Thus, d L ( y ) i( t ) + Ri ( t ) 0 0 L( y) = L   y  4-23 (a) Differential equations: e(t ) 0     ∆x    −3 1   0 0 0   ∆x2   L ( M b + 4 M c )  +   ∆u 0 0 1  ∆x 3  0       ∆x 4    4 0 0 0  M + 4M     b c 1 L di ( t ) y dt dy ( t ) = 0, = dt − Ri ( t ) d = 0, 2 L y 2 y (t ) dt 2 i( t ) dy ( t ) dt =0 K eq R Mg dy .

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 .

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