Automatic Control Systems 8ed by Kuo and Golnaragh -Solutions Manual

Automatic Control Systems

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Unformatted text preview: Chapter 2 MATHEMATICAL FOUNDATION 2-1 (a) Poles: s = 0, 0, Zeros: s = -1, -10; (b) Poles: s = -2, -2; -1 cancel each other. -2, , , . Zeros: s = 0. The pole and zero at s = (c) Poles: s = 0, -1 + j, -1 - j; -2. (d) Poles: s = 0, -1, -2, . Zeros: s = 2-2 (a) G (s) = (d) G ( s) (b) 5 (c) G ( s) = (e) ( s + 5) 1 s 2 2 (s = 4s 2 +4 ) + 1 s+ 2 G (s) = 4 s 2 + 4s +8 = +4 G (s) e k =0 kT ( s + 5 ) = 1 1 -e -T ( s+5 ) 2-3 (a) g ( t ) = u s ( t ) - 2u s (t - 1) + 2 u s( t - 2) - 2 u s ( t - 3) + L G (s ) = 1 s (1 - 2e - s + 2e-2 s - 2e -3s + L ) = 1 s s 1+ e ( 1-e -s -s ) gT (t ) = u s (t ) - 2us ( - 1) + us (t - 2) t GT (s ) = 0 t 2 2 (1 - 2e - s + e -2s ) = ( 1 - e - s ) 1 s 1 g(t ) = k =0 g T (t - 2k )us (t - 2k ) G (s) = s k =0 1 (1 -e -s ) e 2 -2 ks = 1- e -s -s s (1 + e ) (b) g ( t) = 2tu s ( t ) - 4(t - 0.5) u s (t - 0.5) + 4(t - 1) us (t - 1) - 4(t - 1.5)us (t - 1.5) + L G ( s) = g T 2 s 2 (1 - 2e -0.5 s + 2e -s - 2e -1.5 s ( - 0.5 s ) + L) = 2 -0.5 s s (1 + e ) 2 1-e 2 (t ) = 2 tu s ( t ) - 4 ( t - 0 . 5) u s ( t - 0 . 5) + 2( t - 1 ) u s ( t - 1 ) 2 s 2 0 t 1 GT ( s ) = (1 - 2e-0.5 s + e- s ) = s 2 (1 - e-0.5 s ) 2 g (t ) = k=0 g T ( t - k )us ( t - k ) G(s ) = k=0 s2 ( 2 1 -e -0.5 s ) 2 e - ks = ( -0.5 s ) 2 -0.5s s (1 + e ) 2 1-e 2-4 g(t ) = ( t + 1 ) u s ( t ) - ( t - 1 ) u s ( t - 1 ) - 2 u s ( t - 1 ) - ( t - 2 ) u s ( t - 2 ) + ( t - 3) u s ( t - 3) + u s ( t - 3) G ( s) = 1 s 2 (1 - e - s - e -2 s + e -3 s ) + s (1 -2e - s + e -3 s ) 1 1 6( s 1 3( s 1 2( s 2-5 (a) Taking the Laplace transform of the differential equation, we get 1 1 2 ( s + 5s + 4) F ( s) = F (s) = s +2 ( s + 1)( s + 2 )( s f (t ) +4) = + 4) + + 1) - + 2) = 1 6 e -4 t + 1 3 e -t - 1 2 e -2 t t 0 sX (s) (b) sX ( s ) 1 - x1( 0 ) = X (s) 2 x (0) 1 =1 2 - x 2 ( 0 ) = -2 X 1 ( s ) - 3 X 2 ( s ) + 1 s x (0) 2 =0 Solving for X1 (s) and X2 (s), we have X 1 ( s) = = s s( s 2 + 3 s +1 -1 + 1 )( s + 2 ) = = 1 2s + 1 s +1 1 s - 1 2( s + 2) X (s) 2 -1 s (s + 1 )( s + 2 ) -t +1 + +2 x (t ) 2 Taking the inverse Laplace transform on both sides of the last equation, we get x (t ) 1 = 0 .5 + e - 0 .5 e -2 t t 0 = -e -t +e -2 t t 0 2 2-6 (a) G (s) = 1 3s - 1 2( s + 2) + 1 3( s + 3) g (t ) = 1 3 - 1 2 e -2 t + 1 3 e -3 t t 0 (b) G (s) = -2 . 5 s +1 50 s + 5 (s + 1) 2 + 2 .5 s +3 g(t ) = -2 . 5 e -t + 5 te -t + 2 .5 e -3 t t 0 (c) G (s ) = ( 1 s - 20 s +1 s - 30s + 20 s +4 1 s 2 ) 2 e -s g (t ) = 50 - 20e [ - (t -1) - 30cos2(t - 1) - 5sin2(t - 1) us (t - 1) (d) G (s) = - -1 -0.5 t s 2 +s+2 = + 1 s g (t ) = 1 + 1.069e [ sin1.323t + sin (1.323t - 69.3o ) ] = 1 + e-0.5 t (1.447sin1.323t - cos1.323t ) t +s +2 - s s 2 +s+2 Taking the inverse Laplace transform, t0 (e) g(t ) = 0 .5 t 2 e -t 0 2-7 -1 2 0 A = 0 -2 3 -1 -3 -1 0 0 B = 1 0 0 1 u (t ) = u1( t) u ( t) 2 2-8 (a) Y (s ) R (s ) (c) Y (s ) R (s ) = s ( s + 2) s + 10 s + 2 s + s + 2 4 3 2 (b) = 3s + 1 s + 2 s +5s + 6 (d) 3 2 Y (s) R (s ) = 5 s + 10 s + s + 5 1+ 2e 2 -s 4 2 Y (s ) R (s ) = 2s + s + 5 3 4 5 6 7 8 9 10 11 12 13 Chapter 4 MATHEMATICAL MODELING OF PHYSICAL SYSTEMS 4-1 (a) Force equations: f ( t) = M 1 d y1 dt 2 2 + B1 dy1 dt + B3 dy1 - dy 2 + K ( y - y ) 1 2 dt dt 2 dy1 - dy2 + K ( y - y ) + M d y2 + B dy2 B3 1 2 2 2 2 dt dt dt dt Rearrange the equations as follows: d y1 dt 2 2 2 =- = (B 1 + B 3 ) dy1 dt - M1 + B3 dy2 M 1 dt - K M1 K (y 1 1 - y2 ) + f M1 d y2 dt (i) State diagram: 2 B3 dy1 M 2 dt (B 2 + B3 ) dy2 2 + M dt M2 (y - y2 ) Since y 1 - y2 appears as one unit, the minimum number of integrators is three. State equations: Define the state variables as x = y - y , 1 1 2 x 2 = dy dt 2 , x 3 = dy dt 1 . dx1 dt = - x2 + x3 dx2 dt = K M2 x1 - x (B 2 + B3 ) M2 x2 + = dy dt B3 M2 2 x3 x dx3 dt = =- x K M1 x1 + dy dt 1 B3 M1 . x2 - (B 1 + B3 ) M1 x3 + 1 M f (ii) State variables: dx dt dx dt 1 = y , 2 x 2 , 3 y , 1 4 = State equations: 1 = x2 = x dx dt dx 2 =- = K M 2 x 1 - B M B 2 + B3 2 x K M 3 2 + x K M 2 x B 3 + B M 3 x 4 3 4 K M 1 4 x dt 1 + x 2 - 1 M 3 - 1 + B3 M 1 2 x 4 + 1 M 1 f 1 State diagram: 14 Transfer functions: Y1 ( s ) F (s ) Y2 ( s ) F ( s) = = s M 1 M 2 s + [( B1 + B 3 ) M 2 + ( B2 + B3 ) M 1 ] s + [K ( M 1 + M 2 ) + B1 B2 + B2 B3 + B1 B3 ] s + ( B 1 + B2 ) K 3 2 { { M 2 s + ( B2 + B3 ) s + K 2 } } s M1 M 2 s + [( B1 + B3 ) M 2 + ( B2 + B3 ) M1 ] s + [ K ( M1 + M 2 ) + B1 B2 + B2 B3 + B1 B3 ] s + ( B1 + B2 ) K 3 2 B3 s + K (b) Force equations: d y1 dt 2 2 =- (B 1 + B2 ) dy1 M dt + B2 dy2 M dt + 1 M f dy2 dt = dy1 dt - K B2 y2 (i) State diagram: Define the outputs of the integrators as state variables, x 1 = K M x y , 2 x 2 = dy dt 1 . State equations: dx dt 1 =- K B 2 x 1 + x2 dx dt x 2 =- y , 2 x 1 - B 1 x M 2 + = B 1 M dy dt x f (ii) State equations: State variables: dx dt 1 1 = 2 = y , 1 x 1 3 . =- K B 2 x 1 + x3 dx dt 2 = x3 dx dt 3 =- K M x 1 - 1 M 3 + 1 M f Transfer functions: 15 Y1 ( s ) B2 s + K = 2 F (s ) s MB2 s + ( B1 B2 + KM ) s + ( B1 + B2 ) K (c) Force equations: dy1 dt = dy2 dt + 1 B1 f d y2 dt 2 2 Y2 ( s) F ( s) = B2 M B2 s + ( B1 B2 + KM ) s + ( B1 + B2 ) K 2 =- (B 1 + B2 ) dy2 M dt + B1 dy2 M dt + B1 dy1 M dt - K M y2 (i) State diagram: State equations: Define the outputs of integrators as state variables. dx dt 1 = x2 dx dt 2 =- x K M x 1 - B 2 x M x 2 + y , 2 1 M x f (ii) State equations: state variables: dx dt 1 1 = y , 1 2 = 3 = - dy dt 2 2 . = x3 + 1 B 1 f dx dt 2 = x3 dx dt 3 =- K M x B 2 x M 3 + 1 M f State diagram: Transfer functions: Y1 ( s ) F (s ) = Ms + ( B1 + B 2 ) s + K 2 B1 s Ms + B2 s + K 2 ( ) Y2 ( s ) F (s ) = 1 Ms + B 2 s + K 2 4-2 (a) Force equations: 16 y 1 = 1 K 2 ( f + Mg ) + y 2 d 2 y 2 2 dt =- B dy M dt 2 - K 1 + K2 M y 2 + K 2 y M 1 State diagram: State equations: Define the state variables as: x = y , 1 2 x 2 = dy dt 1 2 . dx dt 1 = x2 s 2 dx dt 2 =- K 1 x M 1 - B M x 2 + ( f M + Mg ) Transfer functions: Y (s) 1 F ( s) = + Bs + K 1 + K 2 2 Y (s) 2 K ( Ms 2 + Bs + K 1 ) B 1 dy 1 F (s) = 1 Ms 2 + Bs + K 1 (b) Force equations: dy1 dt = 1 B1 [ f ( t) + Mg] + dy2 dt - K1 B1 (y 1 - y2 ) d y2 dt 2 2 = dy K B B dy - + ( y - y ) - ( y - y )- M dt dt M M M dt 2 1 2 2 1 2 1 2 2 State diagram: (With minimum number of integrators) To obtain the transfer functions Y ( s ) / F ( s ) and Y ( s ) / F ( s ), we need to redefine the state variables as: x 1 = y , 2 x 2 = dy 2 1 / dt , and x 3 = 2 y . 1 State diagram: 17 Transfer functions: Y1 ( s ) F (s ) = s Ms + ( B1 + B 2 ) s + K 1 2 2 [MBs + ( B B 1 1 2 + MK1 )] Y2 ( s ) F (s ) = s [M B1 s + ( B1 B2 + MK1 ) ] 2 Bs + K 1 4-3 (a) Torque equation: 2 d B d =- + 2 dt J dt State diagram: 1 J T (t ) State equations: dx dt 1 = x2 dx dt 2 =- B J x 2 + 1 J T Transfer function: ( s ) T (s) = s ( Js 1 + B) d 2 dt (b) Torque equations: d 1 2 dt 2 =- K J ( 1 - 2 ) + 1 J T K ( 1 - 2 ) = B State diagram: (minimum number of integrators) State equations: dx dt 1 =- K B x 1 + x2 dx dt 2 =- x K J x 1 + 1 J T State equations: Let x = , 1 2 2 = 1, = x and x 3 = 3 d dt 1 . dx dt 1 =- K B x 1 + K B x dx 2 2 dx 3 dt dt = K J x 1 - K J x 2 + 1 J T State diagram: 18 Transfer functions: 1 ( s ) T ( s) = s BJs + JKs + BK 2 ( Bs + K ) 2 (s ) T ( s) = s BJs + JKs + BK 2 ( K ) (c) Torque equations: T ( t ) = J1 d 1 2 dt 2 + K ( 1 - 2 ) K ( 1 - 2 ) = J 2 d 2 2 dt 2 State diagram: State equations: state variables: dx dt 1 x 1 =2, K J 2 x 2 = dx d dt 3 2 , x 3 = 1, dx 4 x 4 = x d dt 1 . x = x2 dx dt 2 =- K J 2 x 1 + x 3 dt = x 4 dt = K J 1 1 - K J 1 3 + 1 J 1 T Transfer functions: 1 ( s ) T ( s) = s 2 J1 J2 s + K ( J1 + J 2 ) 2 J 2s + K 2 2 (s ) T ( s) = K s 2 J1 J 2 s + K ( J1 + J 2 ) 2 (d) Torque equations: d m 2 T ( t) = J m dt 2 + K 1 ( m - 1 ) + K 2 ( m - 2 ) K 1 ( m - 1 ) = J 1 d 1 2 dt 2 K2 (m - 2 ) = J2 d 2 2 dt 2 19 State diagram: State equations: x = 1 m - 1, x 2 = d dt 1 , x 3 = d dt m , x 4 = m -2 , x 5 = d dt 2 . dx dt 1 = -x2 + x3 dx dt 2 = K J 1 x dx 1 3 1 dt =- K J 1 x 1 - K J 2 x 4 + 1 J m T dx dt 4 = x 3 - x5 dx dt 5 = K J 2 x 4 m m 2 Transfer functions: 1 ( s ) T ( s) 2 ( s) T (s ) = s 2 s 4 + ( K1 J2 J m + K2 J 1 J m + K1 J 1 J 2 + K 2 J 1 J 2 ) s 2 + K1 K 2 ( J m + J 1 + J 2 ) s 4 + ( K 1 J 2 J m + K 2 J 1 J m + K 1 J 1J 2 + K 2 J 1J 2 ) s 2 + K 1 K 2 ( J m + J 1 + J 2 ) K2 ( J1 s + K1 ) 2 K 1 (J 2 s + K 2 ) 2 = s 2 (e) Torque equations: d 2 m dt 2 =- K1 Jm ( m - 1 ) - K2 Jm ( m - 2 ) + 1 Jm T d 1 2 dt 2 = K1 J1 ( m - 1 ) - B1 d 1 J1 dt d 2 2 dt 2 = K2 J2 ( m - 1 ) - B 2 d 2 J 2 dt State diagram: 20 State variables: State equations: dx 1 dt dx dt K1 J 1 x 1 = m - 1, x 2 = d dt 1 , x 3 = d dt m , x 4 = m -2 , x 5 = d dt 2 . = -x 2 + x 3 2 = x 1 - B1 J 1 x dx 3 2 dt =- K1 J m x 1 - K J 2 x 4 + 1 J m T dx dt 4 = x3 - x5 dx dt 5 = K2 J 2 x 4 - B2 J 2 x 5 m Transfer functions: 1 ( s ) T ( s) = 2 K1 J 2 s + B2 s + K 2 2 ( ) 3 2 2 ( s ) T (s ) ( s ) 4 m 1 = 2 K 2 J1 s + B s + K1 1 2 ( ) 2 2 1 2 1 2 m ( s ) 2 1 m 1 ( s ) = s { J 1 J2 Jm s + J + [( B1K 2 (B + B ) s + [ ( K J + K J ) J + ( K + K ) J J + B B J ] s + B K ) J + B K J + B K J ] s + K K ( J + J + J )} 1 2 1 m 1 2 2 2 1 1 1 2 m 1 2 4-4 System equations: Tm ( t) = Jm d m 2 dt 2 + Bm e o d m dt E 20 L + K ( m - L ) K ( m - L ) = J L d L 2 dt 2 + Bp d L dt Output equation: State diagram: = Transfer function: L (s ) Tm ( s) Eo ( s ) Tm ( s) = = s J m J L s + Bm J L + Bp J m s + Jm K + J L K + Bm B p s + Bm K 3 2 ( ) K ( ) s J m J L s + Bm J L + Bp J m s + Jm K + J L K + Bm B p s + Bm K 3 2 ( ) KE / 2 0 ( ) 4-5 (a) Tm ( t) = Jm N1 N3 N2N4 d 1 2 dt 1 2 + T1 T1 = N3 N4 N1 N2 T2 2 T3 = d 3 dt 2 N3 N4 T4 T4 = J L d 1 2 d 3 2 dt 2 T2 = T3 2 = N1 N2 1 3 = T2 = T4 = N3 N4 JL 2 N1 N3 d 21 Tm = J m 2 + T4 = J m + JL 2 dt N 2N 4 N 2 N 4 dt N1 N 3 21 (b) Tm = Jm 2 = N1 N2 d 1 2 dt 1 2 + T1 3 = T2 = J2 N 1N 3 N2N4 1 d 2 2 dt 2 + T3 2 T4 = ( J3 + J L ) d 2 dt 2 d 3 2 dt 2 T1 = + N3 N4 N1 N2 T2 d 3 2 T3 = N3 N4 T4 T2 = J 2 + N3 N4 T4 = J 2 d 2 2 dt 2 (J + JL ) 3 dt 2 2 d 3 d 2 2 N3 Tm ( t) = Jm + J 2 dt 2 + N ( J3 + J 4 ) dt 2 = Jm 2 dt N2 4 d 1 2 N1 2 2 d 21 N1 N1 N3 + J2 + N N ( J3 + J L ) dt 2 N2 2 4 4-6 (a) Force equations: f ( t) = K h ( y1 - y2 ) + Bh (b) State variables: x dy1 - dy 2 dt dt x 2 K h ( y1 - y2 ) + Bh 2 2 dy1 - dy 2 = M d y 2 + B dy 2 t 2 dt dt dt dt 1 = y 1 - y2, = dy dt State equations: dx dt 1 =- K B h h x 1 + 1 B h f (t ) dx dt 2 =- B t x M 2 + 1 M f (t ) 4-7 (a) T m = Jm d 2 m 2 dt 2 +T1 2 T 2 =JL d 2 L 2 dt +TL T 1 = N N 1 T 2 = nT 2 m N 1 = L N 2 nTm - n TL 2 2 Tm = Jm L n d m dt 2 + nJL d L dt 2 + nTL = J m + nJ + nT L L L n 2 Thus, L = 2 Jm + n JL 2 2 Set = 0. (T m - 2 nTL ) J m + n J L - 2nJL nTm - n J L = 0 ( ) ( ) Or, n + J T m L n J T - J J m L =0 L m Optimal gear ratio: n =- J T m L 2J T + J mTL 2 2 + 4 J m J LT m L m 2 where the + sign has been chosen. L m 2J T (b) When T L = 0 , the optimal gear ratio is n = Jm / J L 4-8 (a) Torque equation about the motor shaft: Relation between linear and rotational displacements: 22 T m = Jm d 2 m 2 dt + Mr 2 d 2 m 2 dt + Bm d dt m y = r m (b) Taking the Laplace transform of the equations in part (a), with zero initial conditions, we have Tm ( s) = Jm + Mr Transfer function: ( 2 )s 2 m ( s) + Bm s m (s ) Y ( s) = r m ( s) Y ( s) Tm ( s) = s J m + Mr ( r r )s + B m 4-9 (a) Tm = Jm 2 d m 2 dt 2 2 + r ( T1 - T2 ) Thus, Tm = J m T1 = K2 r m - r p = K 2 ( r m - y ) d m 2 ( ) T2 = K1 ( y - r m ) d y dt 2 2 T1 - T2 = M d y dt dt 2 + r ( K1 + K2 )( r m - y ) M = ( K1 + K2 )( r m - y ) (c) State equations: dx1 = rx3 - x2 dt (d) Transfer function: Y ( s) Tm ( s) dx2 dt = K1 + K 2 M x1 dx3 dt = - r ( K1 + K 2 ) Jm x1 + 1 Jm Tm = s 2 Jm Ms + ( K1 + K2 ) ( Jm + rM ) 2 2 r ( K1 + K 2 ) (e) Characteristic equation: 2 s J m Ms + ( K1 + K2 ) ( Jm + rM ) = 0 4-10 (a) Torque equations: Tm ( t) = Jm d m 2 dt 2 + Bm d m dt + K ( m - L ) K ( m - L ) = J L d L 2 dt 2 + BL d L dt State diagram: 23 (b) Transfer functions: L ( s) Tm ( s) K ( s) m ( s) Tm ( s) J L s2 + BL s + K (s ) = = ( s) = s J mJ L s3 + ( Bm J L + BL J m ) s 2 + ( KJ m + KJ L + Bm BL ) s + Bm K (c) Characteristic equation: (d) Steady -state performance: ( s ) = 0 T (t ) m = T m = consta nt. T (s) m 2 = T m . s lim m ( t) = lim s m ( s) = lim t s 0 s 0 J m J L s + ( Bm J L + BL J m ) s + ( KJ m + KJ L + Bm BL ) s + Bm K 3 2 J L s + BL s + K = 1 Bm Thus, in the steady state, m = L. m and (e) The steady-state values of L do not depend on J m and J . L 4-11 (a) State equations: d L =L dt d dt L = K J 2 m - K J 2 L 1 d dt t = t K1 Jm t + d dt t = K J 1 m - Tm K J 1 t L L d m dt = m d m dt =- Bm Jm m - (K + K2 ) Jm t t m + K2 Jm L + 1 Jm (b) State diagram: (c) Transfer functions: 24 L ( s ) Tm ( s) = K 2 J t s + K1 2 ( ) t(s ) T m( s) ( s) = K1 J L s + K 2 2 ( ) 4 m (s ) Tm ( s ) ( s) 5 = J t J L s + ( K1 J L + K 2 J t ) s + K1 K 2 4 2 ( s) 3 ( s ) = s [ J mJ Ls + B mJ LJ t s + ( K1 J L J t + K 2J L J t + K 1Jm J L + K 2J m J t ) s + Bm J L ( K1 + K 2 ) s + K1 K2 ( J L + J t + J m ) s + BmK 1K 2 ] = 0 2 (d) Characteristic equation: ( s ) = 0 . 4-12 (a) m (s ) TL ( s) Thus, r = 0 K 1H i (s ) -K 1 H i (s ) 1 + K1 H e ( s ) + R + L s B + Js H e ( s ) + R + L s B + Js =0 a a a a = -1 ( s) ( s) H e (s ) = - H i (s ) Ra + La s H i (s) H e (s) K1 K i ( s) + = - ( Ra + L a s ) (b) m (s ) r ( s) ( s ) = 1 + K1 H e ( s ) + = 1+ m (s ) r ( s) TL =0 TL =0 = (R a + La s ) ( B + Js ) (R a + La s )( B + Js ) + K1 K b K1 H i ( s) R a + La s + (R K1 K i K b H e( s) a + La s ) ( ( B + Js ) (R a a + La s )( B + Js ) K1 K b (R a + La s ) ( B + Js ) 1 Kb H e ( s) K 1 Ki = (R + La s ) ( B + Js ) + Ki Kb + K1 Ki Kb H e ( s) K1 K i 4-13 (a) Torque equation: (About the center of gravity C) 2 d J = T s d 2 sin + F d d 1 2 dt Thus, J d 2 F d a 1 = J 1 = J d 2 K F d 1 sin 2 dt = T s d 2 + K F d 1 2 dt - K F d 1 = T s d 2 (b) (c) Js 2 ( s ) - K F d 1 ( s ) = T s d 2 ( s ) d 2 With C and P interchanged, the torque equation about C is: Ts ( d1 + d 2 ) + F d 2 = J 2 dt 2 Ts ( d 1 + d 2 ) + K F d 2 = J ( s ) (s ) = d 2 Js (s ) - K F d 2 ( s ) = Ts ( d1 + d 2 ) ( s ) Ts ( d1 + d 2 ) Js - K F d 2 2 dt 2 25 4-14 (a) Cause-and-effe ct equations: dia dt 2 e = r -o Tm = K ii a nKL Jm e = K s e e a = Ke =- Ra La ia + 1 La (e + a - eb ) Tm - d m dt 2 2 =- = Bm d m J m dt 1 J ( n m - o ) T2 = Tm n 2 = n m d o dt 2 KL JL ( 2 - o ) x 1 State variables: State equations: dx dt dx dt 1 = o, K J x 2 = o , nK J x 3 =m, x 4 = m, x 5 = ia = x2 nK J dx dt 2 =- 2 L L x + 1 L L x dx 3 3 dt = dx x 4 4 =- L x 1 - n KL J m x 3 - Bm J m x 4 + Ki J m x 5 5 m dt =- KK L a s x 1 - Kb L a x 4 - Ra L a x 5 + KK L a s r (b) State diagram: (c) Forward-path transfer function: o ( s) KK s K i nK L = 4 3 2 2 e ( s ) s J mJ L La s + J L ( Ra J m + Bm J m + Bm La ) s + n K LL a J L + K L J m La + Bm R a J L s + ( ) (n R K J 2 a L L + R a KL J m + B mK L L a s + K i K bK L + R a B mK L KK s K i nK L 4 ) Closed-loop transfer function: o ( s) r ( s ) = J mJ LL a s + J 5 L (R J a m + Bm J m + B m La ) s + n K LL a J L + K L J m L a + B mR a J L s + 2 3 ( ) (n R K J 2 a L L + Ra KL J m + Bm KL L a s + ( K i K b K L + Ra BmK L ) s + nKK s K i K L 2 ) (d) K L = , o = 2 = n m . J L is re flecte d to m otor s ide so J T = J m +n 2 J . L State equations: 26 d dt m =- B J m T m + K J i T i d a m dt = m di dt a =- R L a a i a + KK L a s r - KK L a s n m - K L b a m State diagram: Forward-path transfer function: o ( s) e ( s ) o ( s) r ( s ) From part (c), when K L = KKs K i n s J T L a s + ( Ra J T + Bm L a ) s + Ra Bm + K i K b 2 Closed-loop transfer function: = JT L a s + ( R a J T + Bm La ) s + ( Ra Bm + Ki Kb ) s + KK s K i n 3 2 KK sK in = , all t he ter ms wit hout K L in o (s ) / e ( s ) and o ( s ) / r ( s ) can b e negl ected. The same results as above are obtained. 4-15 (a) System equations: dv dt ea = Ra ia + ( La + Las ) dia dt di s dt 0 = Rsis + ( L s + L as ) di s dt di a dt f = K ii a = M T (b) + B Tv - Las + eb - L as Take the Laplace transform on both sides of the last three equations, with zero initial conditions, we have Ki I a ( s ) = ( MT s + BT ) V ( s ) Ea ( s ) = [ Ra + ( La + Las ) s ] I a ( s) - Las sI s ( s ) + K b V( s) 0 = - Las sI a ( s) + [ Rs + s ( Ls + Las ) ] I s ( s ) Rearranging these equations, we get V (s ) = Ia (s ) = Ki M T s + BT 1 I a (s ) Y (s ) = V (s ) s = s ( M T s + BT ) Ki Ia (s) L as s Ra + ( La + L as ) s [E a ( s ) + Las sI s ( s ) - KbV ( s ) ] Is (s ) = Ra + ( L a + L as ) s I a ( s) Block diagram: 27 (c) Transfer function: Y ( s) E a (s ) K i [R s + ( L s + L as ) s ] = s [R a + ( L a + L as ) s ][ R s + ( L s + L as ) s ]( M T s + BT ) + K i K b [R s + ( L a + L as ) s] - Lass 2 2 (M T s + BT ) 4-16 (a) Cause-and-effect equations: ea - eb Ra eb = Kb m e = r - L Tm = K i ia e = K s e d m dt = 1 Jm K K s = 1 V/rad Tm - Bm Jm - KL Jm ea = Ke d L dt 15 . 5 1000 ia = = KL JL ( m - L ) ( m - L ) b = 15 . 5 V / KRPM = 2 / 60 = 0 .148 V / rad / sec State equations: d L dt = L d L dt = KL JL m - KL JL L d m dt = m d m dt =- Bm Jm m - KL Jm L+ 1 Ki Jm Ra ( KK s e - K b m ) (b) State diagram: (c) Forward-path transfer function: G ( s) = K i K Ks KL s J mJ L Ra s + ( Bm Ra + K i K b ) J L s + R a K L ( J L + J 3 2 m )s + K L ( Bm Ra + K i K b ) J m Ra J L = 0 .03 1 .15 0 . 05 = 0 . 001725 Bm Ra J L = 10 1 .15 0 .05 = 0 . 575 Ki K bJ L = 21 0 .148 0 . 05 = 0 .1554 28 R K J a L L = 1.15 50000 0 . 05 = 2875 R K J a L K L ( Bm Ra + K i K b ) = 50000(10 1.15 + 21 0.148) = 730400 G ( s) = m = 1 .15 50000 0 . 03 = 1725 K KK K i s L = 21 1 50000 K = 105000 0K s s + 423.42 s + 2.6667 10 s + 4.2342 10 3 2 6 ( 608.7 10 K 6 8 ) (d) Closed-loop transfer function: L ( s) r ( s) G( s) 1 + G ( s) M (s) M (s ) = = = K i K Ks K L J mJ LR a s + ( B m R a + K i K b ) J L s + R a K L ( J L + J m ) s2 + K L ( Bm Ra + K iK b ) s + K i K Ks K L 4 3 = 6 . 087 s 4 10 2 8 K 8 + 423 .42 s 3 + 2 .6667 10 K s 6 s + 4.2342 10 s + 6 . 087 10 = 5476 405 8 K Characteristic equation roots: K s s s =1 = 2738 s j 1273 . 5 s K = - 1.45 = - 159 . 88 j 1614. 6 = j 1000 = -211 . 7 = j1223 .4 s = -617 .22 j 1275 = - 131 . 05 4-17 (a) Block diagram: (b) Transfer function: TAO ( s ) Tr ( s ) = ( 1 + s ) (1 + s ) + K c s KM KR = m 3.51 20 s + 12 s + 4.51 2 KR 4-19 (a) Block diagram: 29 (b) Transfer function: ( s ) ( s) = Js + ( JK L + B ) s + K 2 B + K3 K 4 e 2 - D s K1 K 4 e - D s - Ds (c) Characteristic equation: 2 Js + ( JK L + B ) s + K 2 B + K 3 K 4e ( s ) ( s) K1 K 4 ( 2 - D s ) ( s) =0 (d) Transfer function: Characteristic equation: ( s ) J D s + ( 2 J + JK 2 D + B D ) s + ( 2 JK 2 + 2B - D K 2 B - D K 3 K4 ) s + 2 ( K 2 B + K 3 K 4 ) = 0 3 2 4-19 (a) Transfer function: G ( s) = (b) Block diagram: Ec ( s) E (s ) = 1 + ( R1 + R 2 ) Cs 1 + R2C s (c) Forward-path transfer function: m ( s) E (s ) (d) Closed-loop transfer function: m ( s) Fr ( s ) (e) = = [1 + ( R 1 + R 2 ) Cs ] ( K b Ki + Ra JL s ) K K (1 + R2C s ) K (1 + R2C s ) [1 + ( R = E (s ) 1 Gc ( s) = E c ( s) (1 + R C s ) 2 + R2 ) Cs] ( K b K i + Ra J L s ) + K KK e N (1 + R2C s ) RCs 1 Forward-path transfer function: m ( s) E (s ) Closed-loop transfer function: = RCs ( Kb Ki + Ra JL s ) 1 K (1 + R2C s ) 30 m ( s) Fr ( s ) Ke = K K (1 + R2C s ) R1C s (K b K i + Ra J L s ) + K KKe N (1 + R2C s ) / 2 pulse s / rad = 36 = 120 = NK pulse s / rev pulse s / sec e = 36 = 5 . 73 pul = 200( ses / rad. 2 / 60 ) rad / sec (f) f f r m = pulse s / sec 200 RPM m = 120 = N ( 36 / 2 ) 200( 2 / 60 ) 120 = 120 N pulse s / sec Thus, N = 1. For m = 1800 RPM, = N ( 36 / 2 ) 1800( 2 / 60 ) = 1080 N. Thus, N = 9. 4-20 (a) Differential equations: d m - d L dt dt dt dt 2 d d L = J d L + B d L + T K ( m - L ) + B m - L 2 L L dt dt dt dt Ki ia = Jm d m 2 2 + Bm d m + K ( m - L ) + B (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 m ( s ) + ( Bs + K ) L ( s ) 2 ( ) ( Bs + K ) Solving for m ( s ) - ( Bs + K ) L ( s ) = J L s + BL s L (s ) + TL ( s) 2 ( ) m ( s ) and L ( s ) from the last two equations, we have m (s ) = L (s ) = Signal flow graph: J m s + ( Bm + B ) s + K 2 Ki I a (s ) + m (s ) - Bs + K J m s + ( Bm + B ) s + K 2 L (s ) Bs + K J L s + ( BL + B ) s + K 2 J L s + ( BL + B ) s + K 2 TL ( s ) (c) Transfer matrix: 2 1 K i J L s ...
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