By Farid Golnaraghi, Benjamin C. Kuo

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**Example text**

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 .