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B) Transfer function: Y (s) AB ] = s = U ( s) s 2 +3 +2s−3 1 = s −1 Since there is pole-zero cancellation in the input-output transfer function, the system is either uncontrollable or unobservable or both. In this case, the state variables are already defined, and the system is uncontrollable as found out in part (a). 5-37 (a) α = 1, (b) 2 , or 4 . These values of α will cause pole-zero cancellation in the transfer function. The transfer function is expanded by partial fraction expansion, Y (s) α −1 α R( s) = 3( s + 1) − 2( s By parallel decomposition, the state equations are: x& ( t )  −1 0 0  A =  0 −2 0   0 −4   0 The system is uncontrollable for (c) α = 1, or −2 + 2) + α −4 6( s +4) = Ax ( t ) + B r ( t ) , α −1  B = α − 2    α − 4  output equation: y ( t ) D= = C x ( t ).

For K φ K (1 + R2C s ) = R1C s (K b K i + Ra J L s ) + Kφ KKe N (1 + R2C s ) = 36 / 2 π pulse s / rad ωm = pulse s / sec ω m = 1800 = = 5 . 73 pul 200 RPM = 200( ses / rad. 2 π / 60 ) rad / sec N ( 36 / 2 π ) 200( 2 π / 60 ) RPM, 120 = = 120 N pulse s / sec N ( 36 / 2 π ) 1800( 2 π / 60 ) = 1080 N. Thus, N = 9. 4-20 (a) Differential equations:  dθ m − dθ L   dt dt dt   dt 2  dθ dθ L  =  J d θ L + B dθ L  + T K (θ m − θ L ) + B  m −   L 2 L  L  dt dt   dt dt  d θm 2 Ki ia = Jm (b) 2 + Bm dθ m + K ( θm − θ L ) + B  Take the Laplace transform of the differential equations with zero initial conditions, we get ( ) Ki I a ( s ) = J m s + Bm s + Bs + K Θm ( s ) + ( Bs + K ) Θ L ( s ) ( Bs + K ) Θ Solving for Θm ( s ) and 2 m ΘL ( s ) Θ m (s ) = Θ L (s ) = ( ) ( s ) − ( Bs + K ) Θ L ( s ) = J L s + BL sΘ L (s ) + TL ( s) 2 from the last two equations, we have Ki J m s + ( Bm + B ) s + K 2 Bs + K J L s + ( BL + B ) s + K 2 I a (s ) + Θ m (s ) − Bs + K J m s + ( Bm + B ) s + K 2 Θ L (s ) TL ( s ) J L s + ( BL + B ) s + K 2 Signal flow graph: (c) Transfer matrix: 2 1  Ki  J L s + ( BL + B ) s + K  Θ m ( s )  = Θ (s )  ∆ (s )  Ki ( Bs + K )  L   o 31   Ia ( s )   2  Jm s + (B m + B ) s + K  −TL ( s )  Bs + K ∆ o ( s ) = J L Jm s + [ J L 3 (B + B) + J m ( BL + B ) ] s + [ BL Bm + ( BL + BM ) B + (J 3 m +J m ) K]s L + K ( BL + B ) s 2 4-21 (a) Nonlinear differential equations: dx ( t ) dt With R a = 0, φ (t ) = = dv ( t ) v(t ) = −k ( v ) − g ( x ) + dt e(t ) =K K v(t ) f = i (t ) f = K i (t ) f f K = − Bv f (t ) i (t ) (t ) f (t ) i (t ) Then, f a + a = b = K φ ( t ) ia ( t ) = i (b) State equations: K e (t ) i K 2 b K .

2 c θ 3 x , and 3 2 = θ) dt f L sin dt dθ 2 and f y from the equations above, and sin x 2 d + Mc 2 d 2 = 2 dt M + L sin θ dt J u(t ) − M bg = b rotation about CG: (b) State equations: K K b 2 f (t ) e(t 0 = dx . 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 .

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