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 + ( BL + B ) s + K m ( s ) = (s ) (s ) Ki ( Bs + K ) L o Ia ( s ) 2 Jm s + (B m + B ) s + K -TL ( s ) Bs + K 31 o ( s ) = J L Jm s + [ J L 3 (B m + B) + J m ( BL + B ) ] s + [ BL Bm + ( BL + BM 3 ) B + (J m +J L ) K]s 2 + K ( BL + B ) s 4-21 (a) Nonlinear differential equations: dx ( t ) dt With R a = v(t ) e(t ) dv ( t ) dt = -k ( v ) - g ( x ) + = K i (t ) f f f (t ) = - Bv (t ) + a f (t ) = 0, (t ) = K v(t ) b 2 =K f i (t ) f = K f a i (t ) Then, i (t ) = e(t 0 K K b 2 f v (t ) f (t ) = K ( t ) ia ( t ) = i a K e (t ) i K 2 b K f v (t ) 2 . Thus, dt dv ( t ) = - Bv (t ) + K K K b 2 i f v (t ) 2 e (t ) (b) State equations: i ( t ) as input. dx ( t ) dt = v (t ) dv ( t ) dt = - Bv (t ) + K K i f a i (t ) 2 (c) State equati ons: ( t ) as input. f (t ) = K iK f a i (t ) 2 ia ( t ) = if (t ) = K K ( t ) K i f f dx ( t ) dt = v (t ) dv ( t ) dt = - Bv (t ) + 2 (t ) 4-22 (a) Force and torque equations: Broom: vertical direction: f d b v 2 - M bg = x (t ) M d b 2 ( L cos dt ) horizontal direction: rotation about CG: f x = M + L sin 2 dt J d b 2 2 dt = 2 f L sin y - f L cos x 2 Car: horizontal direction: u(t ) = f x + Mc x 2 d x (t ) 2 J x (b) State equations: Eliminating f x Define the state variables as x 1 = , = dt d dt b = M L b 3 x 4 , 3 = x , and = dx . dt and f y from the equations above, and sin x 1 x and 1 cos x 2 1 1. dx1 dt dx3 dt (c) Linearization: = x2 = x4 dx2 dt dx4 dt = = (M c L [ 4 (M b + M c ) / 3 - M b ] 2 + M b ) gx1 - M b Lx2 x1 - u ( t ) u ( t) + M b Lx2 x1 - 3 M b gx1 / 4 (M b + M c ) - 3M b / 4 32 f1 x1 f2 x1 f2 x4 f 3 x 1 f4 x1 f4 x3 =0 = f1 x2 b c =1 2 2 f1 x3 b =0 f2 x2 = f1 x 4 =0 -2 M b x1 x2 f1 u =0 =0 f2 x3 =0 (M + M ) g - M x = 0 L ( M + M - 3M / 4) b c b L ( M b + M c - 3M b / 4 ) =0 =0 = f2 u f 3 x 2 2 = -1 L [ 4 (M b + M c ) / 3 - M b ] f 3 x 3 =0 f4 x2 = f 3 x 4 =0 2 M b Lx1 x2 f 3 u =0 =0 (M M b Lx2 - 3 M b g / 4 b + M c ) - 3M b / 4 f4 x4 =0 (M = b + M c ) - 3M b / 4 1 + M c ) - 3M b / 4 =0 f4 u (M b Linearized state equations: 0 3( M + M ) g & x 1 b c x L ( M + 4 M ) &2 b c = & x 3 0 & - 3M b g x 4 M + 4M b c 0 x -3 1 0 0 0 x 2 L ( M b + 4 M c ) + u 0 0 1 x 3 0 x 4 4 0 0 0 M + 4M b c 1 0 0 4-23 (a) Differential equations: e(t ) L( y) = L y = Ri ( t ) + i ( t ) dL ( y ) dy ( t ) dy di ( t ) At equili brium, dt dy eq = Ri ( t ) + d L ( y ) i( t ) dt Ki 2 2 dt + L di ( t ) y dt = Ri ( t ) d 2 - L y 2 i( t ) dy ( t ) dt + L di ( t ) y dt My ( t ) = Mg - i (t ) y (t ) = 0, K Mg dy ( t ) dt = 0, y (t ) 2 dt =0 Thus, eq = E eq R x dt =0 x y eq = = E eq R dy . dt x (b) Define the state variables as 1 = i, x 2 = E y , and x eq 3 Then, x 1 eq = E eq R 2 eq = K Mg R 3 eq =0 The differential equations are written in state equation form: dx dt 1 =- R L x x 1 2 + x x 1 3 x + x 2 e 2 L = f dx 1 2 dt = x 3 = f dx 2 3 dt = g - K x1 M x 2 2 2 = f 3 (c) Linearization: 33 f 1 x 1 =- R L x 2 eq + x x 3 eq 2 eq =- E eq K Mg f 1 x 2 f 2 x 1 f 3 x 2 L =- R L x 1 eq - x x 1 3 x 2 2 + E eq L =0 f 1 x 3 = x x 1 eq 2 eq = Mg K f 1 e f 3 x 1 = x 2 eq L = 1 L K Mg E eq R 2 Rg E eq =0 = f 2 x 2 2 K x 1 eq M x 3 2 eq 2 =0 = 2 Rg E eq f 2 x 3 Mg K =1 f 2 e f 3 e =0 =- 2 K x 1 eq M x 2 2 eq =- =0 The linearized state equations about the equilibrium point are written as: Eeq K - L Mg A = 0 2 Rg - E eq 4-24 (a) Differential equations: M1 d 2 0 0 2 Rg Eeq Mg K Mg K 0 0 & x = A x + B e Eeq K RL Mg B = 0 0 y1 (t ) 2 dt d 2 = M 1g -B dy 1 ( t ) dt dy - (t ) Ki 2 2 (t ) y1 ( t ) + Ki Ki 2 2 (t ) y2 (t ) 1 - y1 ( t ) 2 M y 2 (t ) 2 2 dt = M 2g 2 -B dy dt 2 dt 1 - = (t ) 2 y2 (t ) - y1 (t ) y , 2 Define the state variables as x 1 = y , 1 x 2 = , x 3 x 4 = dy dt 2 . The state equations are: dx1 dt = x2 M1 dx2 dt = M 1 g - Bx2 - dx 1 Ki x1 2 2 + Ki 2 2 dx3 dt 4 (x dx dt 3 3 - x1 ) = 0, 2 2 = x4 M2 =0 dx4 dt = M 2 g - Bx4 - = 0. Ki 2 2 (x 3 - x1 ) At equilibrium, dt = 0, dx dt 2 = 0, 2 dx dt = 0. Thus , x 2 eq and x 2 4 eq M 1g - Solving for I, with X KI X1 2 + KI (X 3 - X 1) =0 M 2g - KI (X 3 - X1) 2 =0 1 = 1 , we have 1/2 M + M2 Y2 = X 3 = 1 + 1 M2 (b) Nonlinear state equations: ( M1 + M 2) g I= K 1/2 34 dx1 dt = x2 dx2 dt = g- B M1 x2 - K M 1x1 2 i + 2 Ki 2 dx3 2 M 1 ( x 3 - x1 ) dt = x4 dx4 dt = g- B M2 x4 - Ki 2 M 2 ( x3 - x1 ) 2 (c) Linearization: f1 =0 x1 f2 x1 = 2 KI 2 3 f1 x2 =0 2 3 f1 x 3 =0 =- B M1 f1 x 4 =0 = f1 i -2 KI 2 =0 f2 3 M 1 x1 + 2 KI f2 x2 f2 x3 M1 ( X 3 - X 1 ) M1 ( X3 - X1 ) f3 x4 x4 =0 f2 i = 2 KI -1 1 2+ 2 M 1 X1 ( X 3 - X 1 ) -2 KI 2 3 f3 x1 =0 f3 x2 =0 f3 x3 = 0 = 1 f3 i = 0 f4 x1 = f4 x2 M2 ( X3 - X1 ) =0 f4 x3 = 2 KI 2 3 f4 x4 M 2 ( X 3 - X1 ) M =- B M2 = 1. =1 f4 i = -2 KI M2 ( X3 - X1) 2 Linearized state equations: 1/2 M 1 = 2, 2 = 1, g = 32 .2, B = 0 .1, K I = 32.2(1 + 2) X = 96.6 X = 9.8285X 1 1 1 1 X1 = 1 9.8285 X 3 = 1 + 1 + 2 X 1 = 2.732 X 1 = Y2 = 2.732 0 2 1 2 KI 1 + 3 3 M1 X1 ( X 3 - X 1 ) A = 0 2 -2 KI 3 M 2 ( X 3 - X1 ) ( ) X 3 - X 1 = 1.732 1 -B M1 0 0 0 -2 KI 2 3 M 1 ( X 3 - X1 ) 0 2 KI 2 M 2 ( X 3 - X 1) 3 1 0 0 115.2 -0.05 = 1 0 0 - 37.18 0 -B M2 0 -18.59 0 0 1 37.18 - 0.1 0 0 0 1 0 2 KI -1 + M 1 X 12 ( X 3 - X 1 ) 2 - 6.552 = B = 0 0 - 6.552 -2 KI 2 M 2 ( X 3 - X 1) 4-25 (a) System equations: 35 Tm = K i ia = ( J m + JL ) T D d m dt + B m m e ea = Ra ia + La b dia dt + K b m a y = n m c y = y ( t - TD ) = d (sec) V = r -b = Ksy E ( s) = KG ( s ) E ( s ) Block diagram: (b) Forward-path transfer function: Y (s ) E (s ) = s {( Ra + La s ) [( Jm + J L ) s + Bm ] + K b Ki } - TD s - TD s K Kin G c ( s ) e - TD s Closed-loop transfer function: Y (s ) R (s ) = s ( Ra + La s ) [( Jm + J L ) s + Bm ] + K bK i s + KG c ( s )K i ne K Kin G c ( s ) e 36 Chapter 5 STATE VARIABLE ANALYSIS OF LINEAR DYNAMIC SYSTEMS 5-1 (a) State variables: State equations: x 1 = y, x 2 = dy dt Output equation: dx1 dt 0 1 x1 0 = + r dx2 -1 -4 x2 5 dt (b) State variables: State equations: x 1 y = [1 0] x1 = x1 x2 = y, x 2 = dy , dt x 3 = d 2 y 2 dt Output equation: dx1 dt 1 0 x1 0 dx2 = 0 0 1 x2 + dt dx -1 -2.5 -1.5 x3 3 dt (c) State variables: State equations: 0 0 r 0.5 dx1 dt dy dt x1 y = [1 0 0 ] x2 = x1 x3 d y dt 2 2 x1 = t 0 y ( ) d , x2 = , x3 = , x4 = & x1 Output equaton: x & 2 = & x3 x &4 0 0 0 1 0 x 0 2+ r 0 0 0 1 x3 0 - 1 - 1 - 3 -5 x 1 4 x 1 1 0 0 x1 0 x1 x 2 y = [1 0 0 0] = x1 x3 x 4 2 (d) State variables: State equations: = y, x 2 = dy , dt x 3 = d y 2 , x dt 4 = d y dt 3 3 & x1 Output equation: x & 2 = & x3 x &4 5-2 0 0 0 1 0 x2 0 + r 0 0 1 x3 0 0 -1 -2.5 0 -1.5 x 1 4 1 0 0 x1 0 x1 x 2 y = [1 0 0 0] = x1 x3 x 4 We shall first show that ( s ) = ( sI - A ) = -1 We multiply both sides of the equation by + 2 + +L 2 s s 2! s ( sI - A ) , and we get I = I. Taking the inverse Laplace transform I A 1 A 2 37 on both sides of the equation gives the desired relationship for ( t ) . 5-3 (a) Characteristic equation: Eigenvalues: s ( s ) = j 1. 323 , sI - A = s2 + s + 2 = 0 - 0 .5 - j 1. 323 = -0 . 5 + State transition matrix: ( t) = cos1.323t + 0.378sin1.323t -1.512sin1.323 t sI - A -0.5t e -1.069sin (1.323t - 69.3 ) 0.756sin1.323t o (b) Characteristic equation: State transition matrix: ( s ) = = s 2 + 5s + 4 = 0 -t Eigenvalues: s = - 4, -1 ( t) = (c) Characteristic equation: State transition matrix: 1.333 e- t - 0.333e-4 t -1.333 e - 1.333 e 2 -t -4 t -0.333e + 1.333e 0.333 e - 0.333e -t -4 t -4 t ( s ) = ( s + 3) = 0 Eigenvalues: s = -3, - 3 e -3 t ( t) = 0 (d) Characteristic equation: State transition matrix: e 0 -3 t ( s ) = s 2 - 9 = 0 Eigenvalues: s = -3 , 3 e3 t ( t) = 0 (e) Characteristic equation: State transition matrix: 2 e 0 -3 t ( s ) = s + 4 = 0 Eigenvalues: s = j2, - j2 ( t) = (f) Characteristic equation: State transition matrix: cos2 t - sin2 t 3 2 cos2t s s i n 2t = - 1, - 2 , -2 ( s ) = s + 5 s + 8 s + 4 = 0 Eigenvalues: e- t ( t) = 0 0 (g) Characteristic equation: ( s ) = s 3 0 e -2 t 0 s 2 te -2t e 0 -2 t + 15 -5 t + 75 s + 125 = 0 -5 t -5 t Eigenvalues: s = - 5, - 5, -5 e ( t) = 0 0 te e 0 te -5 t e 0 -5 t 5-4 State transition equation: x (t ) = (t )x( t ) + (a) (t - )Br ( )d t 0 (t ) for each part is given in Problem 5-3. 38 t 0 ( t - ) Br ( ) d = L 1 1 s + 1 1 0 1 1 ( sI - A) -1 BR( s ) = L-1 ( s ) -2 s 1 0 1 s s+2 2 s(s + s + 2) 1 + 0.378sin1.323 t - cos1.323t -1 = =L t0 -1 + 1.134sin1.323 t + cos1.323t s-2 s s2 + s + 2 ) ( -1 (b) ( t - ) Br ( ) d = L ( sI - A) t -1 0 -1 BR ( s ) = L -1 1 1 s + 5 1 1 1 -4 s 1 1 s ( s) s+6 1.5 1.67 0.167 s( s + 1)( s + 2) -1 s - s + 1 + s + 4 1.5 - 1.67 e- t + 0.167e -4 t -1 =L =L = s-4 -1 1.67 0.667 -1 + 1.67 e -t - 0.667 e - 4t + - s( s + 1)( s + 4) s s +1 s + 4 (c) t0 1 t -1 -1 -1 s + 3 0 ( t - ) Br ( ) d = L ( sI - A) BR( s) = L 0 0 0 -1 = =L 1 t 0.333 (1 - e -3t ) s ( s + 3) (d) 0 1 1 1 s s + 3 0 0 1 t -1 -1 -1 s - 3 ( t - ) Br ( ) d = L ( sI - A) BR( s) = L 0 0 0 = =L 1 s ( s + 3) -1 0 1 1 1 s s + 3 0 0 0.333 (1 - e -3t ) t 0 (e) 39 1 2 t -1 -1 -1 s + 4 ( t - ) Br ( ) d = L ( sI - A) BR( s) = L 0 -2 s 2 + 4 2 s 2 -1 =L = t 0 1 0.5sin2t ( s2 + 4 ) (f) 0 1 s 1 s 2 s + 4 2 1 s + 1 t -1 -1 -1 0 ( t - ) Br ( ) d = L ( sI - A) BR( s) = L 0 0 0 0 1 -1 0.5 (1 - e -2 t ) = =L t0 s( s + 2) 0 0 (g) 0 1 s+ 2 0 0 1 1 1 2 (s + 2 ) s 0 1 s +2 0 1 s + 5 t -1 -1 -1 0 ( t - ) Br ( ) d = L ( sI - A) BR (s ) = L 0 0 1 ( s + 5) 1 s +5 0 2 0 1 1 0 2 ( s + 5) s 1 1 s+5 0 0 0 0 1 -1 0.04 0.04 0.2 -1 -5 t -5 t =L =L 2 s - s + 5 - ( s + 5 ) 2 = 0.04 - 0.04 e - 0.2 te u s ( t) -5t s ( s + 5) 0.2 - 0.2 e 1 0.2 0.2 - s ( s + 5) s s +5 5-5 (a) (b) Not a state transition matrix, since Not a state transition matrix, since ( 0 ) I ( 0 ) I (identity matrix). (identity matrix). (c) ( t ) is a state transition matrix, since ( 0 ) = I and 40 1 [ ( t ) ] = - t 1 - e -1 1 = -t 1 - et e 0 2 t -1 0 e t = ( - t) (d) ( t ) is a state transition matrix, since ( 0 ) = I , and - te e 2t e2 t [ ( t) ]-1 = 0 0 5-6 (a) (1) Eigenvalues of A: 2 .325 , t e / 2 2 2t - te e 2t 0 2t = ( - t) - 0 .3376 - j 0 . 5623 - 0 . 3376 + j 0 . 5623 , (2) Transfer function relation: s 1 -1 X( s) = ( sI - A ) B U ( s) = 0 (s ) 1 -1 s 2 3 -1 s + 3 2 0 0 -1 s 2 + 3s + 2 0 U ( s) = 1 -1 (s ) -s 1 1 0 U ( s) = 1 s U (s ) s ( s + 3) s (s ) 2 2 -2 s - 1 s 1 s s +3 1 0 ( s ) = s + 3 s + 2 s + 1 (3) Output transfer function: 1 1 s = = C ( s ) ( sI - A) B = [1 0 0 ] 3 2 U (s ) ( s ) 2 s + 3 s + 2 s + 1 s Y ( s) -1 1 (b) (1) Eigenvalues of A: - 1, - 1. (2) Transfer function relation: 1 ( s + 1) 2 1 0 1 s + 1 -1 U (s ) X ( s ) = ( sI - A) BU ( s ) = U (s ) = (s ) 0 s + 1 1 1 ( s + 1) (3) Output transfer function: ( s ) = s + 2s + 1 2 1 ( s + 1) 2 Y ( s) 1 1 s+2 -1 = = C ( s ) ( sI - A) B = [1 1] + = 2 2 U (s ) 1 ( s + 1) s + 1 ( s + 1) s +1 (c) (1) Eigenvalues of A: 0, - 1, - 1. (2) Transfer function relation: 1 -1 X( s) = ( sI - A) BU ( s ) = (s ) s + 2s = 1 2 0 0 1 0 U ( s) = 1 s U ( s) s ( s + 2) ) s (s ) -s s 1 s 2 2 s+2 1 0 ( s ) = s s + 2s + 1 2 ( ) 41 (3) Output transfer function: 1 s +1 1 = C ( s ) ( sI - A) B = [1 1 0 ] s = = 2 s ( s + 1 )2 s ( s + 1 ) U (s ) s Y ( s) -1 5-7 dy We write dt = dx dt 1 + dx dt 2 = x 2 + x3 d 2 y dt = dx dt 2 + dx dt 3 = -x1 - 2 x2 - 2 x3 + u dx1 dt d x dy = = dt dt 2 d y dt2 0 1 0 x1 0 0 1 1 x + 0 u 2 -1 -2 -2 x3 1 1 0 0 x = - 1 1 0 x 1 - 1 1 0 0 u 1 (1) x1 1 0 0 x = y = 1 1 0 x & y 0 1 1 Substitute Eq. (2) into Eq. (1), we have (2) dx -1 1 0 = A 1 x + B1u = 0 0 1 x = dt -1 0 -2 5-8 (a) s sI -2 s 0 0 s - A = -1 1 -2 0 = s - 3 s + 2 = s + a 2 s + a1 s + a0 3 2 3 2 a 0 = 2, a 1 = 0, a 2 = -3 -1 a1 M = a2 1 a2 1 0 1 0 0 0 -3 1 = - 3 1 0 1 0 0 S = B AB 0 2 4 A B = 1 2 6 1 1 - 1 2 -2 2 0 P = SM = 0 -1 1 -4 -2 1 42 (b) s sI -2 s 0 0 s - A = -1 1 -1 -1 a2 1 0 = s 3 - 3 s + 2 = s + a 2 s + a1 s + a 0 2 3 2 a 0 = 2, a 1 = 0, a 2 = -3 -1 1 0 0 a1 M = a2 1 0 -3 1 = - 3 1 0 1 0 0 1 2 6 S = B AB A B = 1 3 8 0 0 1 0 -1 1 P = SM = -1 0 1 1 0 0 2 (c) s sI +2 0 1 s -1 +2 2 s 0 0 -A = = s + 7 s + 16 s + 12 = s + a 2 s + a 1 s + a 0 3 2 3 2 a 0 = 12 , a 1 = 16 , a 2 =7 +3 a1 M = a2 1 a2 1 0 1 0 0 16 7 1 = 7 1 0 1 0 0 1 -1 0 S = B AB A B = 1 -2 4 1 -6 23 9 6 1 P = SM = 6 5 1 -3 1 1 2 (d) s sI +1 0 0 s -1 =1 0 s 0 -A = -1 = s + 3 s + 3 s + 1 = 3 2 s 3 + a 2 s + a1 s + a0 2 a 0 = 1, a 1 = 3, a 2 =3 +1 a1 M = a2 1 a2 1 0 1 0 0 3 3 1 = 3 1 0 1 0 0 S = B 2 1 P = SM = 2 3 1 2 (e) sI s 0 1 -1 AB A B = 1 0 - 1 1 -1 1 0 1 1 2 -A = -1 2 s -1 +3 = s 2 + 2 s -1 = s + a1 s + a 0 2 a 0 = - 1, a 1 =2 43 M= a1 1 1 2 1 0 1 0 = P = SM = S = [B AB ] = 0 1 1 -3 1 0 -1 1 5-9 (a) From Problem 5-8(a), 0 -3 1 M = - 3 1 0 1 0 0 Then, C V = CA = 2 CA (b) From Problem 5-8(b), 1 0 1 - 1 2 1 1 2 1 0.5 1 3 Q = (MV ) = 0.5 1.5 4 -0.5 -1 -2 -1 C V = CA = 2 CA (c) From Problem 5-8(c), 16 7 1 M = 7 1 0 1 0 0 1 0 1 -1 - 1 3 1 Q = (MV ) = 2 5 1 0.2308 0.3077 1.0769 0.1538 0.5385 1.3846 -0.2308 -0.3077 -0.0769 C V = CA = 2 CA (d) From Problem 5-8(d), 1 0 0 - 2 1 0 4 -4 0 Since V is singular, the OCF transformation cannot be conducted. 3 3 1 M = 3 1 0 1 0 0 Then, C V = CA = 2 CA (e) From Problem 5-8(e), 1 0 1 - 1 1 -1 1 -2 2 -1 1 0 Q = ( MV ) = 0 1 - 2 1 -1 1 -1 44 M= 2 1 1 0 1, 2.7321, Then, V= C CA 2 = 1 0 1 1 Q = ( MV ) = -1 0 1 1 -3 5-10 (a) Eigenvalues of A: -0.7321 T = [p1 where p , p , and 1 2 p2 0 0.5591 0.8255 p3 ] = 0 0.7637 -0.3022 1 -0.3228 0.4766 3 p are the eigenvectors. (b) Eigenvalues of A: 1, 2.7321, -0.7321 T = [ p1 where p , p , and 1 2 p2 0 0.5861 0.7546 p3 ] = 0 0.8007 -0.2762 1 0.1239 0.5952 p are the eigenvalues. 3 (c) (d) Eigenvalues of A: Eigenvalues of A: -3, -2, -2. A nonsingular DF transformation matrix T cannot be found. -1, -1, -1 The matrix A is already in Jordan canonical form. Thus, the DF transformation matrix T is the identity matrix I. (e) Eigenvalues of A: 0.4142, -2.4142 T = [p2 p2 ] = 0.8629 -0.2811 - 0.5054 0.9597 S is singular. 5-11 (a) S = [B (b) S = B (c) S = [B (d) AB] = AB] = 1 -2 0 0 1 -1 1 2 AB A B = 2 -2 2 3 -3 3 2 2 2 +2 2 2+ 2 S is singular. S is singular. 45 S = B 5-12 (a) 1 -2 4 AB A B = 0 0 0 1 -4 14 2 2 S is singular. Rewrite the differential equations as: d m 2 dt 2 =- B d m J dt 2 - K J m m + x Ki J = ia d dt m dia dt , x 3 =- Kb d m La dt - Ra La ia + K a Ks La ( r -m ) State variables: x = , 1 2 = ia Output equation: State equations: dx1 dt 0 dx2 = - K dt J dx K K 3 - a s dt La 1 - - B J Kb La 0 x1 Ki x2 + 0 r J x K K Ra 3 a s - La La 0 y = 1 0 0 x = x1 (b) Forward-path transfer function: s m ( s ) K G ( s) = = [1 0 0 ] J E (s ) 0 3 -1 s+ B J Kb La 2 0 Ki - J Ra s+ La -1 0 KiK a 0 = K o ( s) a La o ( s ) = J La s + ( BLa + Ra J ) s + ( KLa + Ki Kb + Ra B) s + KRa = 0 Closed-loop transfer function: -1 s m ( s ) K M ( s) = = [1 0 0 ] J r ( s ) KaKs La = 3 2 -1 s+ B J Kb La K i Ka K s 0 Ki - J R s+ a La 0 K s G( s ) 0 = K K 1 + K s (s ) a s La JLa s + ( BLa + Ra J ) s + ( KLa + Ki Kb + Ra B) s + K i K a K s + KRa 5-13 (a) 46 A= 0 1 -1 0 A = 2 - 1 0 0 -1 A = 3 0 - 1 1 0 A = 4 1 0 0 1 (1) Infinite series expansion: 3 5 t t t2 t4 1 - 2 ! + 4 ! - L t - 3! + 5! - L 1 2 2 ( t) = I + A + A t + L = t 3 5 2 4 2! -t + t - t + L 1 - t + t L 3! 5 ! 2! 4 ! cos t = - sin t sin t cos t (2) Inverse Laplace transform: 1 s 1 s -1 ( s ) = ( sI - A ) = 1 s = s 2 + 1 -1 s -1 -1 (t ) = cos t - sin t sin t cos t (b) A= -1 0 0 -2 A = 2 1 0 0 4 A = 3 -1 0 0 - 8 A = 4 -1 0 0 16 (1) Infinite series expansion: 2 3 4 t t t 0 1 - t + 2 ! - 3! + 4 ! + L 1 2 2 ( t) = I + A + A t + L = t 2 3 2! 4t 8t 0 1 - 2t + - +L 2! 3! e -t = 0 e 0 -2 t (2) Inverse Laplace transform: ( s ) = ( sI - A ) = -1 s + 1 0 1 s +1 = s + 2 0 0 -1 1 s + 2 0 (t ) = e- t 0 -2 t e 0 (c) A= 0 1 1 0 A = 2 1 0 0 1 A = 3 0 1 1 0 A = 4 1 0 0 1 (1) Infinite series expansion: 3 5 t t t2 t4 1 + 2 ! + 4 ! + L t + 3! + 5 ! L e- t + et 1 2 2 ( t) = I + A + A t + L = t = -t t 3 5 2 4 2! t + t + t L 1 + t + t + L -e + e 3! 5 ! 2 ! 4! -e + e -t t e +e -t t (2) Inverse Laplace transform: 1 s 1 s -1 -1 ( s ) = ( sI - A ) = -1 s = s 2 - 1 1 s = -1 0.5 - s +1 -0.5 + s +1 s - 1 s + 1 s - 1 0.5 0.5 0.5 + s -1 s +1 s -1 0.5 - 0.5 + 0.5 47 e- t + et ( t) = 0.5 - t t -e + e 5-14 (a) e = K s ( r - y ia = Solve for i -e + e -t t e +e -t t ) ea = e - es eb = K b d y es = Rs ia eu = Kea d y 2 eu - eb Ra + Rs y and d y dt Tm = K i ia = ( J m + J L ) dt 2 a in terms of , we have dt ia = Differential equation: KKs (r - y ) - Kb Rs + Rs + KRs d y dt d y 2 dt 2 = K ii a Jm + J L 1 = x d y - KK s y + KK s y -K b ( J m + J L ) (R a + R s + KRs ) dt Ki 2 State variables: x = , y = d dt y State equations: dx1 0 1 0 dt x1 + = - KK s K i - K b Ki - KK s Ki r x2 dx2 ( J + J ) (R + R + KR ) ( J + J ) (R + R + KR ) ( J + J ) (R + R + KR ) m L a s s m L a s s s dt m L a s 1 x1 0 0 = + r -322.58 -80.65 x2 322.58 We can let v ( t ) = 322 .58 r, then the state equations are in the form of CCF. (b) ( sI - A ) = -1 -1 1 s s + 80.65 1 = 2 322.58 s + 80.65 s + 80.65 s + 322.58 -322.58 s -0.014 0.014 -0.06 - 1.059 s + 76.42 s + 4.22 s + 76.42 + s + 4.22 = 1.0622 0.0587 4.468 - 4.468 s + 76.42 s + 4.22 s + 76.42 - s + 4.22 s -1 For a unit-step function input, u ( t ) =1 / s. 322.2 s ( s + 76.42)( s + 4.22) 1 = ( sI - A)-1 B = s 322.2 s ( s + 76.42)( s + 4.22) 1 + 0.0584 - 1.058 s s + 76.42 s + 4.22 - 4.479 + 4.479 s + 76.42 s + 4.22 48 -76.42t -4.22t -0.014e + 0.01e -0.06e -76.42 t - 1.059e -4.22 t x (t ) = x (0) -76.42 t -4.22t -76.42 t -4.22 t - 4.468e 1.0622 e - 0.0587 e 4.468 e 1 + 0.0584 e -76.42 t - 1.058e -4.22 t = t0 -76.42 t -4.22t + 4.479e -4.479 e (c) Characteristic equation: (d) ( s ) = s 2 + 80 . 65 s + 322 . 58 From the state equations we see that whenever there is to increase the effective value of Ra there is ( 1 + K ) Rs . =0 Thus, the purpose of R is s Ra by (1 + K ) Rs . This improves the time constant of the system. 5-15 (a) State equations: dx1 0 1 0 dt x1 + - KKs Ki - Kb Ki KKs Ki = r x2 dx2 J ( R + R + KR ) J ( R + R + KR ) J ( R + R + KR ) s s s s s s dt 1 x1 0 0 = + r -818.18 -90.91 x2 818.18 .18 Let v = 818 r. The equations are in the form of CCF with v as the input. -1 -1 (b) s -1 1 s + 90.91 ( sI - A ) = 818.18 s + 90.91 = ( s + 10.128 ) ( s + 80.782) -818.18 -10.128t -80.78t -10.128t -80.78t - 0.142 e 0.01415 e - 0.0141e 1.143 e x1 (0) x (t ) = -10.128 t -80.78 t -10.128 t -80.78t + 0.1433e - 0.1433 e + 1.143e - 11.58 e x2 (0) 11.58 e-10.128 t - 11.58 e-80.78t + t0 -10.128 t -80.78t + 0.1433 e 1 - 1.1434 e ( s ) = s + 90 . 91 s + 818 2 1 s (c) Characteristic equati on: Eigenvalues: . 18 =0 - 10.128, -80.782 (d) 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 ) = 4 3 Closed-loop transfer function: s + 9 s + 20 s + (10 + 5 K 2 ) s + 5 K1 2 5 ( K1 + K 2s ) (b) State diagram by direct decomposition: 49 & x1 State equations: Output equation: x & 2 = & x3 x &4 0 0 0 -5 K 1 0 0 - (10 + 5 K 2 ) x1 0 1 0 x2 0 + r 0 1 x3 0 -20 -9 x4 1 0 0 R( s) y = 5 K1 5K2 0 x (c) Final value: r(t ) = u s ( t ), = 1 . s lim y( t) = lim sY ( s ) = lim t s 0 s 0 s + 9 s + 20 s + ( 10 + 5 K 2 ) s + 5 K1 4 3 2 5 ( K1 + K 2s ) =1 5-17 In CCF form, 0 0 A= M 0 - a0 1 0 M 0 - a1 0 1 M 0 - a2 0 0 M 0 - a3 -1 s M 0 a1 L L O L L 0 -1 M 0 a2 n -1 0 M 1 - an -1 0 0 0 M 0 a3 L 0 0 B = M 0 1 0 s 0 sI - A = M 0 a 0 sI L 0 O M 0 -1 L s + an n-2 -A = s n + a n -1 s + a n-2 s + L + a1 s + a 0 Since B has only one nonzero element which is in the last row, only the last column of going to contribute to the last row of adj (s I - A ) is adj (s I - A ) B . The last column of adj (s I - A ) is 1 is obtained from the cofactors of s s 2 ( sI - A ) . Thus, the last column of adj (s I - A ) B L s n -1 ' . 50 5-18 (a) State variables: x 1 = y, x 2 = dy , dt x 3 = d 2 y 2 dt State equ ations: & x ( t ) = Ax ( t ) + B r ( t ) 0 1 0 A= 0 0 1 -1 -3 -3 (b) State transition matrix: -1 0 B = 0 1 s2 + 3 s + 3 s + 3 1 s -1 0 1 -1 2 ( s ) = ( sI - A ) = 0 s -1 = s + 3s s -1 (s ) 2 -s 1 3 s + 3 -3 s - 1 s 1 1 1 1 2 1 + + + 2 3 ( s + 1) 2 ( s + 1)3 ( s + 1 )3 s + 1 ( s + 1) ( s + 1) -1 1 1 2 s = + - 2 s + 1 ( s + 1) ( s + 1 )3 ( s + 1) 3 ( s + 1) 3 2 -s -3 2 s + ( s + 1 )3 ( s + 1) 2 ( s + 1)3 ( s + 1 )3 (1 + t + t 2 / 2 ) e- t 2 -t ( t) = -t e / 2 ( -t + t 2 / 2 ) e -t 3 2 (s ) = s3 + 3 s2 + 3s + 1 = ( s + 1) 3 (t + t ) e (1 + t - t ) e 2 -t 2 -t t e 2 -t (t - t / 2 ) e (1 - 2t + t 2 / 2 ) e -t t e /2 2 2 -t -t (d) Characteristic equation: ( s ) = s + 3 s + 3 s + 1 = 0 Eigenvalues: - 1, -1, -1 x 1 5-19 (a) State variables: = y, x 2 = dy dt State equations: dx1 (t ) dt 0 1 x1 ( t) 0 = + r ( t) dx2 ( t) - 1 - 2 x2 ( t) 1 dt State transition matrix: s+ 2 ( s + 1) 2 s -1 -1 ( s ) = ( sI - A ) = 1 s + 2 = - 1 ( s + 1) 2 -1 ( s + 1) s ( s + 1 )2 1 2 (t ) = (1 + t ) e-t - te -t (1 - t ) e te -t -t 51 Characteristic equation: (s ) = ( s +1) = 0 2 (b) State variables: State equations: dx dt 1 x 1 = y, x 2 = y + dy dt = dy dt = x 2 -y= x 2 - x1 dx dt 2 = d 2 y 2 dt + dy dt = -y - dy dt +r = -x2 +r dx1 dt -1 2 x1 0 = + r dx2 0 -1 x2 1 dt State transition matrix: 1 s + 1 -2 s + 1 (s ) = = 0 s + 1 0 -1 2 ( s + 1) 1 s +1 -2 2 (t ) = e -t 0 - te e -t -t (c) Characteristic equation: (s ) = ( s +1) = 0 which is the same as in part (a). 5-20 (a) State transition matrix: s - sI - A = - s - ( t) = L (b) Eigenvalues of A: -1 ( sI - A ) -1 = s - ( s) 1 s - - ( s ) = s - 2 + + 2 2 ( 2 ) cos t - sin t t ( sI - A )-1 = sin t cos t e + j , - j 5-21 (a) Y (s) 1 U ( s) 1 = = s 1+ s -1 -3 -2 + 2s s +3s + 3s -3 = = 1 s 3 +s +2s+3 2 Y ( s) 2 -3 -2 -3 1 s 3 U (s) 2 1 +s -1 +2s + s + 2s +3 2 = Y ( s) 1 U (s) 1 1 1 (b) State equations [Fig. 5-21(a)]: & x = A x+B u 1 1 1 Output equation: y = C x 0 1 0 A1 = 0 0 1 -3 -2 -1 State equations [Fig. 5-21(b)]: 0 B1 = 0 1 & x = A x+B u 2 2 2 C1 = [1 0 0] = C 2x Output equation: y 2 0 0 -3 A 2 = 1 0 -2 0 1 -1 1 B2 = 0 0 C2 = [0 0 1] 52 Thus, A 2 = A1 ' 5-22 (a) State diagram: (b) State diagram: 5-23 (a) G (s) X (s) = Y (s) U ( s) = 10 s 1 + 8 .5 s -1 -1 -3 -2 X (s) + 20 . 5 s + 15 s -3 Y ( s) X (s) X (s) -3 = 10 X (s) = U ( s ) - 8 .5 s X ( s) - 20 . 5 s -2 X ( s ) - 15 s State diagram: State equation: & x ( t ) = Ax ( t ) + B u ( t ) 53 1 0 0 0 A= 0 1 -15 -20.5 -8.5 (b) G (s) Y ( s) Y (s) U ( s) s -3 0 B = 0 1 s -4 -2 A and B are in CCF = = 10 s -3 + 20 -1 -4 X (s) X (s) X ( s) 1 + 4. 5 s + 3 .5 s X (s) = 10 X ( s) + 20 s = -4. 5 s -1 X ( s ) - 3 .5 s -2 X ( s) +U ( s) State diagram: State equations: & x ( t ) = Ax ( t ) + B u ( t ) 0 0 A= 0 0 (c) G (s) 0 1 0 0 0 1 0 -3.5 -4.5 1 0 0 Y (s) U ( s) 0 0 B= 0 1 = 5s -2 A and B are in CCF = = 5( s s(s + 1) + 5s -3 X (s) -2 + 2 )( s + 10 ) -3 1 + 12 s -1 + 20 s X (s) Y (s) = 5s -2 X (s) + 5s X (s) X (s) = U ( s ) - 12 s -1 X (s) - 20 s -2 X ( s) State equations: & x ( t ) = Ax ( t ) + B u ( t ) 54 0 0 1 0 0 A= 1 0 -20 -12 (d) G ( s) = Y ( s) 0 B = 0 1 1 2 A and B are in CCF Y (s ) U (s ) s -4 = s ( s + 5) s + 2s + 2 X (s) ( ) = s -1 -4 -2 -3 X (s ) X (s ) s -3 -2 1 + 7 s + 12 s + 10 s -1 = X ( s) = U ( s) - 7s X ( s ) - 12 s X (s) - 10 State diagram: State equations: & x ( t ) = Ax ( t ) + B u ( t ) 0 0 0 1 0 0 1 0 A= 0 0 0 1 0 -10 -12 -7 5-24 (a) G (s) 0 0 B= 0 1 A and B are in CCF = Y (s) U ( s) = 10 s 3 + 8 . 5 s + 20 . 5 s + 15 2 = 5 . 71 s + 15 - 6 . 67 s +2 + 0 . 952 s +5 55 State equations: & x ( t ) = Ax ( t ) + B u ( t ) -1.5 0 0 A= 0 -2 0 0 0 -5 (b) G (s) 5.71 B = -6.67 0.952 The matrix B is not unique. It depends on how the input and the output branches are allocated. = Y (s) U ( s) = 10( s s (s 2 + 2) = 1 )( s + 3 . 5) = -4. 5 s + s 0 .49 + 3 .5 + s 4 +1 + 5 . 71 s 2 State diagram: State equation: & x ( t ) = Ax ( t ) + B u ( t ) 0 0 1 0 0 0 0 0 A= 0 0 -1 0 0 0 0 -3.5 0 1 B= 1 1 + 0 . 313 s (b) G (s) = Y ( s) U (s) = 5( s s( s + 1) + 2 )( s + 10 ) = 2 .5 s +2 - 0 . 563 s + 10 State equations: & x ( t ) = Ax ( t ) + B u ( t ) 0 0 0 0 -2 0 A= 0 0 -10 (d) 1 B = 1 1 56 G ( s) = Y (s ) U (s ) = s ( s + 5) s + 2s + 2 2 ( 1 ) = 0.1 s - 0.0118 s+ 5 - 0.0882s + 0.235 s + 2s + 2 2 State equations: & x ( t ) = Ax ( t ) + B u ( t ) 0 0 0 -2 0 0 0 -5 A= 0 0 0 0 5-25 (a) G (s) 0 1 -2 0 1 1 B= 0 1 = Y (s) U ( s) = 10 (s + 1. 5)( s + 2 )( s + 5) State diagram: State equations: & x ( t ) = Ax ( t ) + B u ( t ) -5 1 A = 0 -2 0 0 (b) G ( s) = Y (s ) U (s ) State diagram: 1 -1.5 0 10(s + 2) 0 B= 0 10 = = 10 s + 2 1 2 s ( s + 1)( s + 3.5) s s + 1 s + 3.5 2 57 State equations: & x ( t ) = Ax ( t ) + B u ( t ) 0 0 1 0 0 0 1 1 A= 0 0 -1 1 0 0 0 -3.5 (c) G ( s) = Y (s ) U (s ) State diagram: 0 0 B= 0 10 = = 5 s + 1 1 s( s + 2)( s + 10) s s + 2 s + 10 59 s + 1) State equations: & x ( t ) = Ax ( t ) + B u ( t ) 1 0 0 0 - 10 -1 A= 0 0 -2 (d) G ( s) = State diagram: 0 B = 0 5 Y (s) U (s ) = s ( s + 5) s + 2s + 2 2 ( 1 ) 58 State equations: 0 1 0 0 A= 0 -2 0 0 5-26 (a) G (s) & x ( t ) = Ax ( t ) + B u ( t ) 0 1 -2 0 0 1 -5 0 0 0 B= 0 1 = Y (s) E (s) = 10 s(s + 4 )( s + 5) = 10 s 1+ 9 s -1 -3 X (s) s -2 + 20 X (s) (b) Dynamic equations: & x1 x = & 2 x3 & 1 0 0 0 -10 -20 x1 0 1 x2 + 0 r -9 x3 1 0 y = [10 0 0] x (c) State transition equation: s -1 (1 + 9s -1 + 20 s -2 ) s -2 (1 + 9 s -1 ) s -3 x1 (0) s -3 X 1 (s ) -3 -1 -1 -2 X (s ) = 1 1 -2 1 - 10 s s (1 + 9 s ) s 2 x2 (0) + ( s ) s s (s ) -2 -2 -1 s -1 X 3 (s ) - 10 s - 20 s s x3 (0) 1 2 s s + 9 s + 20 s+9 1 x1 (0) 1 1 = s( s + 9) s x2 (0) + 1 - 10 c (s ) c (s ) 2 s - 10 s - 10 ( 2 s + 1 ) s x3 (0) ( s ) = 1 + 9 s -1 + 20 s -2 + 10 s -3 c( s) = s 3 + 9 s + 20 s + 10 2 59 1.612 0.946 0.114 -0.706 -1.117 -0.169 e -0.708t + 1.692 x(t ) = -1.14 -0.669 -0.081 2.678 4.056 e -2.397t 0.807 0.474 0.057 -4.056 -6.420 -0.972 0.0935 0.171 0.055 + -0.551 -1.009 - 0.325 e -5.895t x(0) 3.249 5.947 1.915 0.1 - 0.161e-0.708t + 0.0706 e-2.397t - 0.00935e-5.895 t -0.708 t - 2.397 t -5.895 t + - 0.169e + 0.055e 0.114 e -0.087 e-0.708t + 0.406 e- 2.397t - 0.325e- 5.895t (d) Output: t 0 y ( t ) = 10 x1 (t ) = 10 1.612e ( -0.708t - 0.706e -2.397t + 0.0935e -5.897 t + 10 1.141 e ( -0.708 t - 0.169 e -2.397 t + 0.0550 e -5.895 t ) x (0) + 10 ( 0.946e - 1.117e + 0.1711e ) x (0) ) x (0) + 1 -1.61e -0.708t + 0.706e -2.397 t - 0.0935e -5.895t -0.708 t -2.397t -5.895t 1 2 3 t0 5-27 (a) Closed-loop transfer function: Y (s) R( s) = s 10 3 + 9 s + 20 s + 10 2 (b) State diagram: (c) State equations: & x1 x = & 2 x3 & 1 0 0 0 -10 -20 (d) State transition equations: [Same answers as Problem 5-26(d)] x1 0 1 x2 + 0 r -9 x3 1 0 (e) Output: [Same answer as Problem 5-26(e)] 5-28 (a) State diagram: (b) State equations: 60 & x1 x & 2 = & x3 x &4 (c) Transfer function relations: From the system block diagram, -2 20 -1 0 x1 0 -1 0 -10 1 0 x2 0 0 u + -0.1 0 -20 1 x3 0 0 T D 0 x 30 0 0 0 - 5 4 Y ( s) = -0.2s 0.3 30 e U ( s) 90U( s) -1 TD ( s) + TD ( s) + + s+2 ( s) ( s + 2)( s + 20) ( s + 2)( s + 5)( s + 20) (s + 5)( s + 20) 1 ( s ) = 1 + Y ( s) 0 . 1e (s -0 . 2 s + 2 )( s + 20 -0 . 2 s = ) T D (s + 2 )( s + 20 ) + 0 .1 e (s - 0. 2 s + 2 )( s + 20 30 e -0 . 2 s ) s = (s - ( s + 19 . 7 ) + 2 )( s + 20 ) + 0 .1e -( s + 20 (s ) -0 . 2 s (s)+ (s + 90( -0 . 2 s + 2 )U ( s ) -0 . 2 s + 5) (s + 2 )( s + 20 ) + 0 . 1e 30 e U ( s) ( s ) = + 2 )( s + 20 ) + 0 .1e T (s) D + (s + 5) (s + 2 )( s + 20 ) + 0 .1e -0 . 2 s 5-29 (a) There should not be any incoming branches to a state variable node other than the s should create a new node as shown in the following state diagram. -1 branch. Thus, we (b) State equations: dx dt 1 Notice that there is a loop with gain -1 after all the s 1 2 x dx 2 2 -1 branches are deleted, so = 2. y = 17 2 x 1 + dt = 15 2 x 1 - 1 2 x 2 + 1 2 r Output equation: = 6 .5 x 1 + 0 .5 x 2 5-30 (a) Transfer function: Y (s ) R (s ) = Ks + 5 s + 1 2 (b) Characteristic equation: 2 ( s + 1) s + 11s + 2 ( ) s s 3 2 ( s + 1) ( s 2 + 11s + 2 = 0 ) Roots of characteristic equation: -1, -0.185, --10.82. These are not functions of K. (c) When K = 1: Y (s) R( s) = + 5s +1 s 2 + 12 + 13 s + 2 State diagram: 61 (d) When K = 4: Y (s ) R (s ) State diagram: = ( s + 1) ( s 4s + 5s + 1 2 2 + 11s + 2 ) ( s + 1) (s = (s + 1)(4 s + 1) 2 + 11s + 2 ) = 4s + 1 s + 11s + 2 2 (e) Y (s) R( s) = Ks (s 2 + 5s =1 s + 1 )( s + 0 .185)( + 10 .82 When K = 4, 2.1914, 0.4536, pole-zero cancellation occurs. ) 5-31 (a) Gp ( s) = Y ( s) U (s ) = ( 1 + 0.5s ) (1 + 0.2s + 0.02s 1 2 ) = 100 s + 12 s + 70 s + 100 3 2 State diagram by direct decomposition: State equations: & x1 x = & 2 x3 & 0 0 -100 x1 0 0 1 x2 + 0 u -70 -12 x3 1 1 0 Roots of characteristic equation: (b) Characteristic equation of closed-loop system: 62 s 3 + 12 s 2 + 70 s + 200 = 0 - 5. 88 , - 3 . 06 + j 4. 965 , - 3 . 06 - j 4. 965 5-32 (a) Gp ( s) = Y ( s) U (s ) (1 + 0.5s ) (1 + 0.133s + 0.0067s 1 - 0.066 s 2 ) = -20( s - 15) s + 22 s + 190 s + 300 3 2 State diagram by direct decomposition: State equations: & x1 x = & 2 x3 & 0 0 -300 x1 0 0 1 x2 + 0 -190 -22 x3 1 1 0 Roots of characteristic equation: Characteristic equation of closed-loop system: s 3 + 22 s 2 + 170 x s + 600 = 0 = m and x -12, -5 + j5, -5 -j5 = D K KK JR d dt K J K J 5-33 (a) State variables: d dt 1 2 State equations: m =- K K b i + K bRa a JR + D J D + i a e D = D R m - D R D (b) State diagram: 63 (c) Open-loop transfer function: m ( s) E (s ) KK i ( J R s + K D ) = JJ RR a s + ( K b J R Ki + K DR a J R + K D JRa ) s + K DK b Ki 2 Closed-loop transfer function: m ( s) r ( s) = JJ R Ra s + ( Kb J R Ki + KD R a J R + KD JRa + Ks K Ki J R ) s + KD K b Ki + Ks K Ki KD 2 Ks KKi ( J R s + KD ) (d) Characteristic equation of closed-loop system: ( s ) = JJ R R a s + ( K D J R K i + K D R a J R + K D JRa + K s KK i J R ) s + K D K b K i + K s K Ki KD = 0 2 ( s ) = Characteristic equation roots: s 2 + 1037 s + 20131 .2 =0 -19.8, -1017.2 & 5-34 (a) State equations: x ( t ) = Ax ( t ) + B r ( t ) A= -b d -2 1 c - a = 2 - 1 B= 0 1 S = [B AB ] = 0 1 1 -1 Since S is nonsingular, the system is controllable. (b) S = [B 5-35 (a) S = B (b) AB AB ] = 0 d 1 - a 1 -1 1 A B = 1 -1 1 1 -1 1 2 The system is controllable for d 0. S is singular. The system is uncontrollable. 64 S = B 1 - 1 1 AB A B = 1 - 2 4 1 -3 9 2 S is nonsingular. The system is controllable. & 5-36 (a) State equations: x ( t ) = Ax ( t ) + B u ( t ) A= -2 3 1 0 B= 1 1 1 0 x S = [B AB ] = 1 1 1 1 0 S is singular. The system is uncontrollable. Output equation: y = = Cx C = 1 V = C ' AC ' ' 1 -2 = 0 3 Y (s) U ( s) V is nonsingular. The system is observable. (b) Transfer function: = s s 2 +3 +2s-3 = s 1 -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) = 1, (b) 2 , or 4 . These values of will cause pole-zero cancellation in the transfer function. -2 + 2) -4 6( s The transfer function is expanded by partial fraction expansion, Y (s) -1 R( s) = 3( s + 1) - 2( s + +4) output equation: y ( t ) & By parallel decomposition, the state equations are: x ( t ) = Ax ( t ) + B r ( t ) , = C x ( t ). 1 6 -1 0 0 A = 0 -2 0 0 0 -4 The system is uncontrollable for -1 B = - 2 - 4 D= 1 - 1 3 2 = 1, or = 2, or = 4. (c) Define the state variables so that -1 0 0 A = 0 -2 0 0 0 - 4 1 3 -1 B= 2 -1 6 D = [ - 1 -2 - 4] The system is unobservable for = 1, or = 2, or = 4. 65 5-38 S = [B AB] = b 1 b ab - 1 S = ab - 1 - b 0 2 The boundary of the region of controllability is described by ab - 1 - b = 0. 2 Regions of controllability: 5-39 S = [B AB] = b1 b1 + b 2 b b2 2 2 S = 0 when b1 b2 - b1 b2 - b2 = 0,or b2 = 0 2 The system is completely controllable when b 0. V = 0 when d 1 0 . V = C ' AC ' ' d2 d1 = d d +d 2 1 2 2 The system is completely observable when d 0. 5-40 (a) State equations: dh 1 K nN K = ( qi - qo ) = I m - o h dt A A A State variable: State equations: x 1 d m dt 3 = d m m =- Ki Kb JRa dt m + Ki K a JRa ei J = Jm +n 2 J L = h, x 2 =m, x = d dt m = m & x = Ax + B e i -K o A A= 0 0 State diagram: K I nN A 0 0 1 = KK - i b JRa 0 0 -1 0.016 0 0 1 0 0 -11.767 0 B = 0 = K K i a JRa 0 0 8333.33 66 (b) Characteristic equation of A: s+ sI - A = Ko A - K I nN A s 0 0, 0 -1 s+ Ki K b JRa = s s + 0 0 Ko s + Ki K b = s ( s +1)( s + 11.767) A JRa Eigenvalues of A: -1, -11.767. (c) Controllability: S = B (d) Observability: (1) C = 1 0 0 AB 0 133.33 0 0 A B = 8333.33 - 98058 8333.33 -98058 1153848 2 S 0. The system is controllable. : ' ' V = C ' AC -1 1 1 0 0.016 -0.016 (A ) C = 0 0 0.016 ' 2 ' V is nonsingular. The system is observable. (2) C = 0 1 0 : ' ' V = C (3) C = 0 0 ' AC 0 0 0 1 0 (A ) C = 0 0 1 -11.767 ' 2 ' V is singular. The system is unobservable. 1 : ' ' V = C ' AC 0 0 0 0 (A ) C = 0 0 1 -11.767 138.46 ' 2 ' V is singular. The system is unobservable. 5-41 (a) Characteristic equation: ( s ) = s I - A = s 4 - 25 . 92 s 2 =0 Roots of characteristic equation: -5.0912, 5.0912, 0, 0 (b) Controllability: 67 S = B AB A B 2 - 0.0732 0 - 1.8973 0 -0.0732 0 -1.8973 0 3 A B = 0 0.0976 0 0.1728 0.0976 0 0.1728 0 B S is nonsingular. Thus, A , is controllable. (c) Observability: (1) C = 1 0 0 0 V = C ' A C ' ' (A ) C ' 2 ' 1 0 ' 3 ' (A ) C = 0 0 0 1 0 0 25.92 0 0 0 25.92 0 0 0 S is singular. The system is unobservable. (2) C = 0 1 0 0 V = C ' A C ' ' (A ) C ' 2 ' 0 671.85 0 25.92 1 0 25.92 0 ' 3 ' (A ) C = 0 0 0 0 0 0 0 0 S is singular. The system is unobservable. (3) C = 0 0 1 0 V = C ' A C ' ' (A ) C ' 2 ' 0 0 ' 3 ' (A ) C = 1 0 0 0 0 1 - 2.36 0 0 0 - 2.36 0 0 0 S is nonsingular. The system is observable. (4) C = 0 0 0 1 68 V = C ' A C ' ' (A ) C ' 2 ' 0 -61.17 0 -2.36 0 0 -2.36 0 ' 3 ' (A ) C = 0 0 0 0 1 0 0 0 S is singular. The system is unobservable. 5-42 The controllability matrix is 0 - 384 0 -1 0 -16 -1 0 -16 0 -384 0 0 0 0 16 0 512 S= 0 512 0 0 0 16 0 1 0 0 0 0 1 0 0 0 0 0 5-43 (a) Transfer function: v ( s ) R (s ) = J vs 2 Rank of S is 6. The system is controllable. (J KI H G s + K Ps + K I + K N 2 ) State diagram by direct decomposition: State equations: & x ( t ) = Ax ( t ) + B r ( t ) 1 0 1 0 - ( KI + KN JG 0 0 A = 0 0 0 0 0 ) 2 (b) Characteristic equation: Jvs 2 (J 0 1 - KP JG 0 0 0 B= 0 1 G s + K P s + K I + KN = 0 ) 69 & 5-44 (a) State equations: x ( t ) = Ax ( t ) + B u 1 ( t ) A= -3 1 0 - 2 B= 0 1 S = [B AB ] = 0 1 1 -2 S is nonsingular. A , B is controllable. C = -1 1 Output equation: y 2 = C x V = C (b) With feedback, u 2 ' AC ' ' = - kc 2 , & the state equation is: x ( t ) -1 3 = 1 -3 = Ax ( t ) V is singular. The system is unobservable. + B u1 ( t ) . 1 0 1+k S= 1 -2 -3 - 2 k A = 1+ g 0 State diagram: 1+ k -2 1 0 B= 1 S is nonsingular for all finite values of k. The system is controllable. Output equation: y2 = Cx V = D' -1 1 + k -1 1 + K ' ' AD= 1 1+ k C= 1+ k 1 (1 + k ) 3 + 2k - 2 (1 + k ) 3 + 2k 2 V is singular for any k. The system with feedback is unobservable. 5-45 (a) S = [B V = C (b) u = - k 1 k2 x ' AB ] = 1 2 2 -7 1 - 1 ' ' AC = 1 -2 S is nonsingular. System is controllable. V is nonsingular. System is observable. A c = A - BK = 0 1 k1 - -1 -3 2 k1 1 - k2 -k1 = 2 k2 - 1 - 2 k1 - 3 - 2 k 2 k2 70 S = [B For controllabillity, k 2 A cB ] = - 11 2 ' ' 1 - k1 - 2 k2 + 2 2 - 7 - 2k - 4k 1 2 S = - 11 - 2 k2 0 V = C For observability, V ' A cC - 1 - 1 - 3 k1 = 1 -2 - 3 k 2 = -1 + 3k 1 - 3 k 2 0 71 Chapter 6 6-1 (a) Poles are at s STABILITY OF LINEAR CONTROL SYSTEMS = 0 , - 1. 5 + j 1. 6583 , - 1. 5 - j 1. 6583 One poles at s = 0. Marginally stable . Two poles on j axis. Marginally stabl e . Two poles in RHP. Unstable . All poles in the LHP. Stable . Two poles in RHP. Unstable . Two poles in RHP. Unstable . (b) Poles are at s = - 5, - j 2 , j 2 (c) Poles are at s = - 0 .8688 , 0 .4344 + j 2 . 3593 , 0 .4344 - j 2 . 3593 (d) Poles are at s = - 5, - 1 + j , - 1 - j (e) Poles are at s = - 1.3387 , 1. 6634 + j 2 . 164, 1. 6634 - j 2 .164 (f) Poles are at s = - 22 . 8487 j 22 . 6376 , 21 . 3487 j 22 . 6023 6-2 (a) s 3 + 25 3 s 2 + 10 s + 450 = 0 Roots: - 25 . 31 , 0 .1537 + j 4.214, 0 .1537 - 4.214 Routh Tabulation: s s s 1 25 250 10 450 2 1 - 450 25 = -8 Two sign changes in the first column. Two roots in RHP. 0 s 0 450 s 2 (b) s 3 + 25 3 + 10 s + 50 = 0 Roots: - 24. 6769 , - 0 .1616 + j 1.4142 , - 0 .1616 - j 1.4142 Routh Tabulation: s s s 1 25 250 10 50 2 1 - 50 25 =8 No sign changes in the first column. No roots in RHP. 0 s 0 50 s 2 (c) s 3 + 25 3 + 250 s + 10 = 0 Roots: - 0 . 0402 , - 12 .48 + j 9 . 6566 , - j 9 . 6566 Routh Tabulation: s s s 1 25 6250 250 10 2 1 - 10 = 249 No sign changes in the first column. No roots in RHP. .6 0 25 s 0 10 (d) 2s 4 + 10 s 3 + 5 . 5 s + 5 . 5 s + 10 = 0 2 Roots: - 4.466 , - 1.116 , 0 .2888 + j 0 . 9611 , 0 .2888 - j 0 . 9611 Routh Tabulation: 71 s s s s 4 2 10 55 5 .5 5 .5 10 3 2 - 11 = 4.4 = - 75 . 8 10 1 10 24.2 - 100 4.4 s 0 10 Two sign changes in the first column. Two roots in RHP. (e) s 6 + 2 s + 8 s + 15 5 4 6 s 3 + 20 s 2 + 16 s + 16 = 0 Roots: - 1.222 j 0 .8169 , 0 . 0447 j 1.153 , 0 .1776 j 2 .352 Routh Tabulation: s s s 1 2 16 8 15 20 16 16 5 4 - 15 2 = 0 .5 40 - 16 2 = 12 s 3 - 33 -396 + 24 -33 - 541 .1 + 528 11 .27 - 48 = 11 .27 = -1.16 16 s 2 s 1 0 s 0 0 Four sign changes in the first column. Four roots in RHP. (f) s 4 + 2 s + 10 3 4 s 2 + 20 s + 5 = 0 Roots: -0 .29 , - 1. 788 , 0 . 039 + j 3 .105 , 0 . 039 - j 3 .105 Routh Tabulation: s s s 1 2 20 10 20 5 3 2 - 20 2 =0 5 s 2 20 5 Replac e 0 in last row by s 1 - 10 - 10 Two sign changes in first column. Two roots in RHP. s 0 5 3 6-3 (a) s 4 + 25 s + 15 s 2 + 20 s + K = 0 Routh Tabulation: 72 s s s s 4 1 25 375 15 20 K 3 2 - 20 25 = 14.2 K K K 1 284 - 25 = 20 - 1. 76 20 - 1. 76 >0 K >0 or K < 11 . 36 14.2 s 0 K K Thus, the system is stable for 0 < K < 11.36. When K = 11.36, the system is marginally stable. The auxiliary equation equation is A ( s ) = 14.2 s frequency of oscillation is 0.894 rad/sec. 2 + 11 . 36 = 0 . The solution of A(s) = 0 is s 2 = - 0 .8 . The (b) s 4 + Ks + 2 s + ( K + 1) s + 10 = 0 3 2 Routh Tabulation: s s s 4 1 K 2K K 2 10 K K 3 +1 10 >0 >1 2 2 - K -1 K 2 = K -1 K s 1 -9 K - 1 K -1 -9 K -1 > 0 s 0 10 2 2 The conditions for stability are: K > 0, K > 1, and - 9 K - 1 > 0 . Since K is always positive, the last condition cannot be met by any real value of K. Thus, the system is unstable for all values of K. (c) s 3 + ( K + 2 ) s + 2 Ks + 10 = 0 2 3 Routh Tabulation: s s 1 K 2K 2K 10 K 2 +2 2 > -2 2 s 1 + 4 K - 10 K +2 K + 2 K -5 > 0 s 0 10 The conditions for stability are: K > -2 and K + 2 K - 5 > 0 or (K +3.4495)(K - 1.4495) > 0, or K > 1.4495. Thus, the condition for stability is K > 1.4495. When K = 1.4495 the system is 2 marginally stable. The auxiliary equation is A ( s ) The frequency of oscillation is 1.7026 rad/sec. = 3.4495 s 2 + 10 = 0 . The solution is s 2 = -2 .899 . (d) s 3 + 20 s 2 + 5 s + 10 K =0 Routh Tabulation: 73 s s s 3 1 20 100 5 10 K K 2 1 - 10 20 = 5 - 0 .5 K 5 - 0 .5 K > 0 >0 or K < 10 s 0 10 K K The conditions for stability are: K > 0 and K < 10. Thus, 0 < K < 10. When K = 10, the system is marginally stable. The auxiliary equation is A ( s ) equation is s 2 = 20 s 2 + 100 = 0 . The solution of the auxiliary = -5 . 2 The frequency of oscillation is 2.236 rad/sec. K (e) s 4 + Ks + 5 s + 10 s + 10 3 =0 10 K K Routh Tabulation: s s s 4 1 K 5K 5 10 10 K 3 >0 - 10 > 0 or K 2 - 10 K 5K >2 50 K s 1 - 100 K 5K - 10 K 2 - 10 K = 50 K - 100 - 10 5K K 3 - 10 5K - 10 - K > 0 3 s 0 10 K K >0 The last condition is written as The conditions for stability are: K > 0, K > 2, and 5 K K + 2 . 9055 2 K 2 - 2 . 9055 K + 3 .4419 - 10 - K > 0 . 3 <0. The second-order term is positive for all values of K. Since these are contradictory, the system Thus, the conditions for stability are: K > 2 and K < is unstable for all values of K. -2.9055. (f) s 4 + 12 . 5 s + s + 5 s + K = 0 3 Routh Tabulation: s s s 4 1 12 . 5 12 . 5 1 5 K 3 2 -5 = 0 .6 = 5 - 20 .83 K 12 . 5 s 1 3 - 12 . 5 K 0 .6 K 5 - 20 . 83 K > 0 or K < 0 .24 s K K >0 The condition for stability is 0 < K < 0.24. When K = 0.24 the system is marginally stable. The auxiliary 0 equation is A ( s ) = 0 . 6 s oscillation is 0.632 rad/sec. 2 + 0 .24 = 0 . 2 The solution of the auxiliary equation is s 2 = - 0 .4. The frequency of 6-4 The characteristic equation is Ts 3 + ( 2T +1) s + ( 2 + K ) s + 5 K = 0 Routh Tabulation: 74 s s 3 T 2T ( 2T K +2 5K T T >0 > -1 / 2 2 +1 s 1 + 1 )( K + 2 ) - 5 KT 2T +1 K (1 - 3 T ) K + 4T + 2 > 0 s 0 5K >0 The conditions for stability are: T > 0, K > 0, and K < 4T 3T +2 -1 . The regions of stability in the T-versus-K parameter plane is shown below. 6-5 (a) Characteristic equation: s 5 + 600 s 4 + 50000 s 3 + Ks 2 + 24 Ks + 80 K =0 Routh Tabulation: s s 5 1 600 3 50000 K 7 24 K 80 K 4 s 3 10 -K 00 K 7 14320 K 600 K < 3 10 7 600 s 2 214080 3 -K 2 10 - K 16 80 K 11 K < 214080 2 00 s 1 - 7 .2 10 + 3 .113256 10 600( 214080 00 K ) - 14400 K 2 -K K - 2 .162 10 7 K + 5 10 12 <0 s 0 80 K K >0 Conditions for stability: 75 From the s From the s 3 2 1 row: row: K 2 From the s row: Thus, From the s 0 < 3 10 7 K < 2 .1408 10 7 12 - 2 .162 10 K + 5 10 < 0 K 7 or (K - 2 . 34 10 7 5 )( K - 2 .1386 10 7 ) <0 2 . 34 K>0 10 5 < K < 2 .1386 10 10 5 row: Thus, the final condition for stability is: When K When K 2 . 34 < K < 2 .1386 10 7 = 2 . 34 10 7 = 2 .1386 10 5 = 10 . 6 rad/sec. = 188 . 59 rad/sec. 2 (b) Characteristic equation: s 3 + ( K + 2 ) s + 30 Ks + 200 K =0 Routh tabulation: s s s 3 1 K 30 K 30 K 200 K K K K 2 +2 2 > -2 > 4. 6667 >0 =0. The solution is s 2 1 - 140 +2 K s 0 200 K K > 4.6667 K Stability Condition: When K = 4.6667, the auxiliary equation is A ( s ) The frequency of oscillation is 11.832 rad/sec. = 6 . 6667 s 2 + 933 . 333 = - 140 . (c) Characteristic equation: s 3 + 30 s 2 + 200 s +K =0 Routh tabulation: s s s 3 1 30 6000 30 200 K 2 1 -K K < 6000 >0 2 s 0 K 0 K Stabililty Condition: < K < 6000 = 30 s 2 When K = 6000, the auxiliary equation is A ( s ) The frequency of oscillation is 14.142 rad/sec. + 6000 = 0 . The solution is s = - 200 . (d) Characteristic equation: s 3 + 2 s + ( K + 3) s + K + 1 = 0 2 Routh tabulation: s s s 3 1 2 K K +3 2 K +1 K 1 +5 30 > -5 > -1 s 0 K +1 K> K Stability condition: -1. When K = -1 the zero element occurs in the first element of the 76 s row. Thus, there is no auxiliary equation. When K = -1, the system is marginally stable, and one of the three characteristic equation roots is at s = 0. There is no oscillation. The system response would increase monotonically. 0 6-6 State equation: Open-loop system: A= 1 -2 10 0 1 10 - k 1 -k 2 2 s + k2 or k 2 & x ( t ) = Ax ( t ) + B u ( t ) B= Closed-loop system: & x ( t ) = ( A - BK ) x ( t ) -2 0 1 A - BK = Characteristic equation of the closed-loop system: sI - A + BK = Stability requirements: s -1 -10 + k1 k 2 = s + ( k 2 - 1 ) s + 20 - 2 k1 - k 2 = 0 2 -1 > 0 >1 or k 2 20 - 2 k1 - k 2 > 0 < 20 - 2 k 1 Parameter plane: 6-7 Characteristic equation of closed-loop system: s -1 0 sI - A + BK = 0 k1 Routh Tabulation: s k2 + 4 -1 s + k3 + 3 = s + ( k 3 + 3 ) s + ( k 2 + 4 ) s + k1 = 0 3 2 s s s s 3 1 k3 + 3 k2 + 4 k1 k3 +3>0 or k3 > - 3 2 1 (k 3 + 3) ( k 2 + 4 ) - k 1 k3 + 3 (k k >0 1 3 + 3 )( k + 4) - k > 0 2 1 0 k 1 Stability Requirements: k 3 > - 3, 6-8 (a) k 1 > 0, (k 2 3 + 3) ( k 2 + 4 ) - k 1 > 0 Since A is a diagonal matrix with distinct eigenvalues, the states are decoupled from each other. The second row of B is zero; thus, the second state variable, x is uncontrollable. Since the uncontrollable 77 state has the eigenvalue at -3 which is stable, and the unstable state x with the eigenvalue at -2 is 3 controllable, the system is stabilizable. (b) 6-9 Since the uncontrollable state x has an unstable eigenvalue at 1, the system is no stabilizable. 1 The closed-loop transfer function of the sysetm is Y (s) R (s ) The characteristic equation is: = 1000 s + 15.6 s + ( 56 + 100 K t ) s + 1000 3 2 s 3 + 15 . 6 s + ( 56 + 100 2 K )s t + 1000 = 0 Routh Tabulation: s s s 3 1 15 . 6 873 . 6 56 + 100 1000 K t 2 1 + 1560 Kt - 1000 15 .6 s 0 1560 K t - 126 .4 >0 1000 K t Stability Requirements: > 0 . 081 6-10 The closed-loop transfer function is Y (s) R( s) The characteristic equation: = K(s s 3 + 2 )( s + ) + Ks 3 2 + ( 2 K + K - 1 ) s + 2 K 2 s + Ks + ( 2 K + K - 1) s + 2 K = 0 Routh Tabulation: s s 3 1 K (2 2K + K - 1 2 K K 2 >0 - 1 - 2 > 0 s 1 + ) K - K - 2 K 2 (2 + )K K s 0 2 K K >0 > 0, K Stability Requirements: > 0 , K -versus- Parameter Plane: > 1 + 2 + . 2 78 6-11 (a) Only the attitude sensor loop is in operation: K t = 0. The system transfer function is: K s 2 ( s) r ( s ) If KK If KK s s = G (s) p 1+ K G ( s) s p = - + KK s >, , the characteristic equation roots are on the imaginary axis, and the missible will oscillate. the characteristic equaton roots are at the origin or in the right-half plane, and the system is unstable. The missile will tumble end over end. (b) Both loops are in operation: The system transfer function is ( s) r ( s ) For stability: When K t = G (s) p 1 + K sG t p (s) + K sG p ( s ) = K s 2 + KK t s + KK s - KK and KK =0 s t s > 0, >, t KK s - > 0 . the characteristic equation roots are on the imaginary axis, and the missile will oscillate back and forth. For any KK - if KK < 0, the characteristic equation roots are in the right-half plane, and the system If KK is unstable. The missile will tumble end over end. > 0 , and KK < , the characteristic equation roots are in the right-half plane, and the system is t t unstable. The missile will tumble end over end. 6-12 Let s 1 = s + , 1 then when s = -, s 1 = 0. This transforms the s = - axis in the s-plane onto the imaginary axis of the s -plane. (a) F (s) Or = s 1 s 2 2 + 5s +3 = 0 Le t s = s1 - 1 We get (s 1 -1) + 5 ( s1 -1) + 3 = 0 2 + 3 s1 - 1 = 0 s1 2 1 3 -1 Routh Tabulation: s s 1 1 0 1 -1 Since there is one sign change in the first column of the Routh tabulation, there is one root in the region to the right of s = -1 in the s-plane. The roots are at -3.3028 and 0.3028. (b) F (s) = s 3 + 3s + 3s +1 = 0 2 Let s = s 1 -1 We get ( s1 -1)3 + 3( s1 -1) + 3 ( s1 -1) + 1 = 0 2 79 Or s 3 1 = 0. 3 2 The three roots in the s -plane are all at s 1 1 = 0. Thus, F(s) has three roots at s = -1. 2 (c) F (s) Or = s s + 4 s + 3 s + 10 = 0 + s 1 - 2 s1 + 10 = 0 2 Let s = s 1 -1 We get (s 1 -1) + 4 ( s1 -1) + 3 ( s1 -1) +10 = 0 3 3 1 s s 3 1 2 1 1 1 1 -2 10 Routh Tabulation: s1 s 1 - 12 10 0 Since there are two sign changes in the first column of the Routh tabulation, F(s) has two roots in the region to the right of s = -1 in the s-plane. The roots are at -3.8897, -0.0552 + j1.605, and -0.0552 - j1.6025. 2 (d) F (s) Or = s 3 +4s +4s +4 = 0 3 Let s = s 1 -1 We get (s 1 -1) + 4 ( s1 -1) + 4 ( s1 -1) + 4 = 0 3 2 s 1 + s1 - s1 + 3 = 0 2 s s 3 1 2 1 1 1 1 -1 3 Routh Tabulation: s1 s 1 -4 3 0 Since there are two sign changes in the first column of the Routh tabulation, F(s) has two roots in the region to the right of s = -1 in the s-plane. The roots are at -3.1304, -0.4348 + j1.0434, and -0.4348 -j1.04348. 6-13 (a) Block diagram: (b) Open-loop transfer function: G ( s) = H ( s) E (s ) Closed-loop transfer function: H (s) R( s) = s (R a Js + K i K b ) ( As + K o ) K a K i nK I N = 16.667 N s( s + 1)( s + 11.767) = 1 G (s) +G ( s) = s 16 .667 N 3 + 12 . 767 s 2 + 11 . 767 s + 16 . 667 N (c) Characteristic equation: 80 s 3 + 12 . 767 11 . 767 s 2 + 11 . 767 s + 16 . 667 N =0 Routh Tabulation: s s s 3 1 12 .767 150 .22 2 16 . 667 N N 1 50.22 1 - 16 . 667 - 16.667 N >0 or N <9 12 . 767 s 0 16 .667N 0 N >0 Stability condition: < N <9. 6-14 (a) The closed-loop transfer function: H (s) R( s) = s ( 0 .06 s 250 N + 0 . 706 )( As + 50 ) + 250 N The characteristic equation: 0 . 06 As 3 + ( 0 . 706 A + 3) s + 35 . 3 s + 250 2 N =0 Routh Tabulation: s s s 3 0 . 06 A 0 . 706 A 24. 92 A 35 .3 250 N A >0 +3 > 0 + 105 .9 2 +3 - 15 NA +3 0 . 706 A 24.92 A 1 + 105 .9 07 . 06 A s 0 - 15 NA >0 250N 1 N > 0 When A From the s row, N < 1.66 + 7.06/A N max 1. 66 Thus, N max = 1. (b) For A = 50, the characteristic equation is 3 s + ( 35.3 + 0.06 K o ) s + 0.706 Ko s + 250 N = 0 3 2 Routh tabulation 81 s s 3 3 35 . 3 0706 K K 250 N o 2 + 0 . 06 2 o K o > - 588 . 33 s 1 0 . 0424 K o + 24. 92 K o K - 750 N 35 . 3 s 0 + 0 . 06 N < 0.0000 5653 K o N 2 + 0 . 03323 K o 250 N >0 (c) N = 10, A = 50. The characteristic equation is s + ( 35.3 + 0.06 K o ) s + 0.706 Ko s + 50 KI = 0 3 2 Routh Tabulution: s s 3 1 35 . 3 0 . 706 K K 50 K Ko Ko o 2 + 0 . 06 2 o I K o > -588 . 33 s 1 0 . 04236 K o + 24. 92 + 0 . 06 - 50 KI 35 .3 s 0 KI < 0 . 000847 >0 2Ko 2 + 0 .498 Ko 50 K I K I 6-15 (a) Block diagram: (b) Characteristic equation: 2 Ms + K s + K + K = 0 D s P 500 s 2 + K D s + 500 + K P = 0 82 (c) For stability, K D > 0, 0 .5 + K P > 0. Thu s, K P > - 0 .5 Stability Region: 6-16 State diagram: = 1 + s + Ks Characteristic equation: s 2 2 +s+K =0 Stability requirement: K>0 83 Chapter 7 TIME-DOMAIN ANALYSIS OF CONTROL SYSTEMS n 2 (b) 0 . 707 n 2 7-1 (a) 0 . 707 rad / sec 0 rad / sec (c) 0 . 5 1 n 5 rad / sec (d) 0 .5 0 . 707 n 0 . 5 rad / sec 7-2 (a) 7-3 (a) (b) (c) (d) (e) (f) Type 0 (b) Type 0 (c) Type 1 (d) Type 2 (e) Type 3 (f) = lim = lim = lim = lim = lim = lim Type 3 K p = lim = lim = lim = lim = lim = lim s0 G ( s) = 1000 = = = = Kv = lim = lim = lim = lim = lim = lim s 0 sG ( s ) =0 =1 =K = =1 = Ka s0 s G (s) 2 =0 =0 =0 =1 =0 = K K p s0 G ( s) Kv s 0 sG ( s ) Ka s0 s G (s) 2 K p s0 G ( s) Kv s0 sG ( s ) Ka s0 s G (s) 2 K p s0 G ( s) Kv s 0 sG ( s ) Ka s0 s G ( s) 2 K K p p s0 G ( s) Kv Kv s 0 s 0 sG ( s ) sG ( s ) Ka Ka s0 s0 s G (s) s G (s) 2 2 s0 G ( s) = 84 7-4 (a) Input Error Constants Steady -state Error ________________________________________________________________________________ u (t ) s K K K p v a = 1000 =0 =0 1 1001 tu ( t ) s t u (t ) / 2 s 2 0 (b) Input Error Constants Steady -state Error ________________________________________________________________________________ u (t ) s K K K p v a = =1 =0 tu ( t ) s 1 t u (t ) / 2 s 2 0 1/ K (c) Input Error Constants Steady -state Error ________________________________________________________________________________ u (t ) s K K K p v a = =K =0 tu ( t ) s t u (t ) / 2 s 2 The above results are valid if the value of K corresponds to a stable closed-loop system. (d) (e) The closed-loop system is unstable. It is meaningless to conduct a steady-state error analysis. Input Error Constants Steady -state Error ________________________________________________________________________________ u (t ) s K K K p v a = 0 1 tu ( t ) s =1 =0 t u (t ) / 2 s 2 0 0 1/ K (f) Input Error Constants Steady -state Error ________________________________________________________________________________ u (t ) s K K K p v a = =K tu ( t ) s = t u (t ) / 2 s 2 The closed-loop system is stable for all positive values of K. Thus the above results are valid. 7-5 (a) K H = H (0) =1 M (s) a = G (s) 1+ G ( s)H ( s) a 1 = 2, s s 3 +1 2 0 = 3, 1 = 3, a 2 = b + 2 s + 3s +3 = 1, b 1 = 1. 0 Unit-step Input: ess = Unit-ramp input: a 0 b0 KH 1 - a KH 0 2 =3 Thus e ss - b0 K H = 3 - 1 = 2 0. - b0 K H = 2 0 and a = . e Unit-parabolic Input: a 0 1 - b1 K H = 1 0. Thus ss = . (b) K H = H (0) =5 M (s) = G (s) 1+ G ( s)H ( s) = 1 s 2 +5s + 5 a 0 = 5, a 1 = 5, b 0 = 1, b 1 = 0. 85 Unit-step Input: ess = Unit-ramp Input: i e b0K H 1 - a KH 0 1 a a 1 0 = 1 5 1 - = 0 5 5 i = 0: ss - b0 K H = 0 = 5 25 = 1: a 1 - b1 K H = 5 0 = - b1 K H a K 0 = 1 5 H Unit-parabolic Input: e ss = (c) K H = H (0) =1/5 M (s) = G (s) 1+ G ( s)H ( s) a 0 = s s 4 3 +5 2 = 1, a 1 = 1, + 15 s + 50 s + s + 1 a = 50 , a = 15 , b = 5 , 2 3 0 The system is stable. b 1 =1 Unit-step Input: ess = Unit-ramp Input: i e b0K H 1 - a KH 0 1 a a 1 0 5/5 = 5 1 - 1 = 0 i = 0: ss - b0 K H = 0 = 1-1 / 5 1/5 = 1: a 1 - b1 K H = 4 /5 0 = - b1 K H a K 0 =4 H Unit-parabolic Input: e ss = (d) K H = H (0) = 10 M (s) = a G (s) 1+ G ( s)H ( s) 0 = 1 s = 10 , a 1 = 5, + 12 s + 5 s + 10 a = 12 , b = 1, 2 0 3 2 The system is stable. b 1 = 0, b 2 =0 Unit-step Input: ess = Unit-ramp Input: i e b0K H 1 - a KH 0 1 a a 1 0 1 10 = 10 1 - 10 = 0 i = 0: ss - b0 K H = 0 = 5 100 = 1: a 1 - b1 K H = 5 0 = - b1 K H a K 0 = 0 . 05 H Unit-parabolic Input: e 7-6 (a) M (s) = s s 4 +4 2 ss = K a H + 16 s 3 + 48 s + 4 s + 4 a = 4, a = 4, 0 1 ess = 1 =1 = 48 , a 3 The system is stable. 2 = 16 , b 0 = 4, b 1 = 1, b 2 = 0, b 3 =0 Unit-step Input: b0K H 1 - a KH 0 a 0 4 = 1 - 4 = 0 i Unit-ramp input: i = 0: -b0 KH = 0 = 1: a 1 - b1 K H = 4 - 1 = 3 0 86 e ss = a 1 - b1 K H a K 0 = 4 -1 4 = 3 4 H Unit-parabolic Input: e K(s s 3 2 ss = K a (b) M (s) = + 3) + 3s + ( K + 2)s + 3K a = 3K , a = K + 2, 0 1 ess = H =1 = 3, b 0 The system is stable for K > 0. 2 =3K, b 1 =K Unit-step Input: b0K H 1 - a KH 0 1 a a 1 0 3K = 1 - 3 K = 0 i Unit-ramp Input: i e = 0: ss - b0 K H = 0 = K 3K = 1: = 2 a 1 - b1 K H = K + 2 - K = 2 0 = - b1 K H a K 0 +2 -K H 3K Unit-parabolic Input: e ss = The above results are valid for K > 0. (c) M (s) = s s 4 +5 + 50 s + 10 s a = 0 , a = 10 , 0 1 2 + 15 s 3 H ( s) a = 10 s 2 = s+5 50 , a K H = lim b 0 H (s) s b 1 3 = 15 , s 0 =2 =1 = 5, Unit-step Input: ess = Unit-ramp Input: a 2 - b1K H a KH 1 1 e ss 1 50 - 1 2 = 2 10 = 2.4 = = K Unit-parabolic Input: e K(s s 4 ss (d) M (s) = + 5) s 2 + 17 s 3 + 60 a 0 + 5 Ks + 5 K a 1 H =1 a The system is stable for 0 < K < 204. = 5K , = 5K , a 2 = 60 , 3 = 17 , b 0 = 5K, b 1 = K Unit-step Input: ess = Unit-ramp Input: i e b0K H 1 - a KH 0 1 a a 1 0 5K = 1 - 5 K = 0 i = 0: ss -b0 KH = 0 = e 5K = 1: 4 5 a 1 - b1 K H = 5 K - K = 4 K 0 = - b1 K H a K 0 -K H 5K = Unit-parabolic Input: ss = The results are valid for 0 < K < 204. 87 7-7 G (s) = Y (s) E ( s) = KG 1 p ( s ) 20 s + K tG p( s) K p = K 100 K 20 s ( 1 + 0 .2 s + 100 , K Type-1 system. K ) t Error constants: = , v = 5K 1 + 100 K t a =0 (a) (b) (c) r(t ) = u s ( t ): e ss = = 1 1+ K 1 K v p =0 1 + 100 K 5K t r(t ) = tu s ( t ): =t 2 e 1 K a ss = r(t ) u ( t ) / 2: s e ss = = 7-8 G p (s) = (1 100 + 0 .1 s )( 1 + 0 . 5 s ) 100 K G (s) = Y (s) E (s) = KG 20 s 1 p (s) + K tG p ( s ) G (s) = 20 s ( 1 + 0 .1 s )( 1 + 0 . 5 s ) + 100 K Kt 5K Error constants: p = , K v = 1 + 100 K , t K a =0 (a) (b) (c) r(t ) = u s ( t ): e ss = = 1 1+ K 1 K v p =0 1 + 100 K 5K t r(t ) = tu s ( t ): =t 2 e 1 K a ss = r(t ) u ( t ) / 2: s e ss = = t Since the system is of the third order, the values of K and K stable. The characteristic equation is 3 2 must be constrained so that the system is s + 12 s + ( 20 + 2000 K t ) s + 100 K = 0 20 Routh Tabulation: s s s 3 1 12 240 + 2000 100 K K t 2 1 + 24000 12 Kt - 100 K s 0 100 K Stability Conditions: K>0 12 ( 1+ 100 K t ) - 5 K > 0 or 1 + 100 K t 5K > 1 12 Thus, the minimum steady-state error that can be obtained with a unit-ramp input is 1/12. 88 7-9 (a) From Figure 3P-19, o ( s) r ( s ) 1+ = 1+ K1 K 2 Ra + La s + K1 K 2 Ra + La s + (R K i K b + KK1 K i K t a + La s ) ( Bt + J t s ) + s ( Ra + La s )( Bt + J t s ) KK s K 1K i N (R K i K b + KK1 K i K t a + La s ) ( Bt + J t s ) o ( s) r ( s ) = L a J t s + ( L a Bt + Ra J t + K 1 K 2 J t ) s + ( Ra Bt + K i K b + K Ki K1K t + K 1 K 2 Bt ) s + KK s K 1K i N 3 2 s [( Ra + La s ) ( Bt + Jt s ) + K1 K2 ( Bt + Jt s ) + Ki Kb + KK1 K i K t ] r ( t ) = u s ( t ), r ( s ) = 1 s e lim s ( s ) s 0 e =0 Provided that all the poles of s ( s ) are all in the left -half s-plane. (b) For a unit-ramp input, r ( s ) = 1 / e ss s . lim s ( s ) s 0 e 2 = t lim e (t ) = = R B a t + K 1 K 2 B t + K i K b + KK KK s 1 K K i t K K N 1 i if the limit is valid. 7-10 (a) Forward-path transfer function: [n(t) = 0]: K (1 + 0.02s ) G ( s) = Y ( s) E (s ) = 2 K (1 + 0.02s ) s ( s + 25) = 2 K Kt s s s + 25 s + KKt 1+ 2 s ( s + 25) ( ) Type-1 system. Error Constants: K p = , R( s) K v = 1 , 1 , K t K a =0 = lim sE ( s ) For a unit-ramp input, r ( t ) = tu s ( t ), = s 2 e ss = t lim e ( t ) s 0 = 1 K v = K t Routh Tabulation: s s s s 3 1 25 25 K ( Kt + 0.02) - K 25 K K >0 =1/ K s. KK t + 0.02 K K 2 1 0 Stability Conditions: 25 ( Kt + 0.02 ) - K > 0 or K t > 0.02 (b) With r(t) = 0, n ( t ) = u s ( t ), N (s) System Transfer Function with N( s) as Input: Y (s ) N (s ) = K s ( s + 25) = 3 2 K (1 + 0.02 s) K Kt s s + 25 s + K ( K t + 0.02 ) s + K 1+ 2 + 2 s ( s + 25) s ( s + 25) 2 89 Steady -State O utput due to n ( t): y ss = t lim y ( t ) = lim s0 sY ( s ) =1 if the limit is valid. 7-11 (a) n(t ) = 0, r (t ) = tu s ( t ). G ( s) = Y ( s) E (s ) n =0 Forward-path Transfer function: = K ( s + )( s + 3) s s -1 2 Ramp-error constant: Steady -state error: Characteristic equation: s + Ks Routh Tabulation: 3 2 Kv e = lim = 1 K v ( ) Type-1 system. s 0 sG ( s ) = -3 K ss =- 1 3K v + [ K ( 3 + ) - 1] s + 3 K = 0 3K s s s 3 1 K K (3K + K - 1 3 K 2 1 + K - 1 ) - 3 K K 3 K 3K s 0 Stability Conditions: + K - 1 - 3 > 0 or K > 1+ 3K 3 + K > 0 (b) When r(t) = 0, n ( t ) = u s ( t ), N (s) =1/ s. K ( s + 3) Transfer Function between n ( t) and y( t): Y (s ) N (s ) r =0 = 1+ 2 Ks ( s + 3) s -1 = K ( s + )( s + 3) s 3 + Ks 2 + [ K ( s + ) - 1]s + 3 K s s -1 2 ( ) Steady -State Output due to n ( t): y ss = t lim y ( t ) = lim s0 sY ( s ) =0 if the limit is valid. 7-12 Percen t maxi mum ov Thus ershoo t = 0 .25 - =e 2 1 - 2 Solving for 1- 2 = - ln0.25 = 1.386 = 0.404. sec. Thus, = 1.922 1 - 2 ( 2 ) = 343 2 from the last equation, we have t Peak T ime max = n 1- = 0 .01 2 n = 0 . 01 1 - ( 0 .404 ) .4 rad / sec Transfer Function of the Second-order Prototype System: Y (s) R( s) = n 2 s 2 + 2 n s + n 2 = 117916 s 2 + 277 .3 s + 117916 7-13 Closed-Loop Transfer Function: Characteristic equation: 90 Y (s ) R (s ) = 25 K s + ( 5 + 500 Kt ) s + 25 K 2 s + ( 5 + 500 Kt ) s + 25 K = 0 2 For a second-order prototype system, when the maximum overshoot is 4.3%, = 0 . 707 . n = Rise Time: [Eq. (7-104)] t r 25 K , 2 2 n = 5 + 500 = 0 .2 K t = 1.414 25 K = = = 1 - 0 .4167 + 2 . 917 n 2 = 2 .164 n 5 sec Thu s n = 10 . 82 rad / sec Thus, With K K n 2 = ( 10 . 82 ) 25 and K = 4. 68 t + 500 K 25 4. 68 t = 1.414 n = 15 . 3 r func tion i s Thus K t = 10 . 3 500 = 0 . 0206 = 0 . 0206 , the sy stem t ransfe Y (s) R( s) = s 117 2 + 15 . 3 s + 117 Unit-step Response: y = 0.1 at t = 0.047 sec. y = 0.9 at t = 0.244 sec. t = 0 .244 - 0 . 047 = 0 .197 r sec. y max = 0 . 0432 ( 4. 32% max. overs hoot) 7-14 Closed-loop Transfer Function: Y (s ) R (s ) = 25 K s + ( 5 + 500 Kt ) s + 25 K 2 Characteristic Equation: s + ( 5 + 500 Kt ) s + 25 K = 0 2 When Maximum overshoot = 10%, 1- 2 = - ln0.1 = 2.3 = 5.3 1 - 2 2 ( 2 ) Solving for , we get = 0.59. The Natural undamped frequency is n = 2 25 K Thus, 5 + 500 K t = 2 n = 1.18 n n = 17 . 7 rad / sec Rise Time: [Eq. (7-114)] t r = = 1 - 0 .4167 + 2 . 917 n = 0 .1 = = 1. 7696 n sec. Th us K n 2 = 12 . 58 t 25 With K = 12.58 and K = = 0 . 0318 500 0 . 0318 , the system transfer function is Thus K t 15 . 88 Y (s) R( s) = s 313 2 + 20 .88 s + 314. 5 Unit-step Response: y = 0.1 when t = 0.028 sec. y = 0.9 when t = 0.131 sec. 91 t r = 0 . 131 - 0 . 028 = 0 .103 = 1.1 sec. y max ( 10% max. overs hoot ) 7-15 Closed-Loop Transfer Function: Y (s) R( s) Characteristic Equation: = 25 K s 2 + ( 5 + 500 K )s t + 25 K 1- 2 s + ( 5 + 500 Kt ) s + 25 K = 0 2 When Maximum overshoot = 20%, = - ln0.2 = 1.61 = 2.59 1 - 2 2 ( 2 ) Solving for , we get = 0 .456 . 25 K 2 The Natural undamped frequency n = 5 + 500 K t = 2 n = 0 . 912 n sec. Thus, Rise Time: [Eq. (7-114)] t r = = 1 - 0 .4167 + 2 . 917 n = 0 . 05 = 1.4165 n n = K 1.4165 0 . 05 = 28 .33 = 32 .1 5 + 500 K = 0 . 912 = 25 . 84 Thus, t n 25 With K = 32.1 and K = 0 . 0417 , the system transfer function is K t n 2 t = 0 . 0417 Y (s) R( s) = 802 . 59 s 2 + 25 . 84 s + 802 . 59 Unit-step Response: y = 0.1 when t = 0.0178 sec. y = 0.9 when t = 0.072 sec. t = 0 . 072 - 0 . 0178 = 0 . 0542 r sec. y max = 1.2 ( 20% max. overs hoot ) 7-16 Closed-Loop Transfer Function: Y (s ) 25 K = 2 R ( s ) s + ( 5 + 500 Kt ) s + 25 K From Eq. (7-102), Delay time t d Characteristic Equation: s + ( 5 + 500 Kt ) s + 25 K = 0 2 1.1 + 0 .125 + 0 .469 n . t d 2 = 0 .1 sec. When Maximum overshoot = 4.3%, 2 2 = 0 . 707 = 1.423 n = 0 .1 sec. Thus n = 14.23 K rad/sec. n = 14.23 = 8.1 5 + 500 K = 2 = 1.414 = 20.12 K = t n n 5 5 With K = 20.12 and K t Thus t = 15 .12 500 = 0 . 0302 = 0 . 0302 , the system transfer function is Y (s) R( s) = s 202 . 5 2 + 20 .1 s + 202 .5 Unit-Step Response: 92 When y = 0.5, t = 0.1005 sec. Thus, t = 0 .1005 sec. d y max = 1. 043 ( 4. 3% max. overs hoot ) 7-17 Closed-Loop Transfer Function: Y (s ) 25 K = 2 R ( s ) s + ( 5 + 500 Kt ) s + 25 K From Eq. (7-102), Delay time t 2 2 Characteristic Equation: s + ( 5 + 500 Kt ) s + 25 K = 0 2 d 1.1 + 0 .125 + 0 .469 n 5 2 = 0 . 05 = 1. 337 n Thus, n = 1. 337 0 . 05 = 26 . 74 K = n = 26.74 = 28.6 5 5 t + 500 K t = 2 n = 2 0 . 59 26 . 74 = 31 . 55 r func tion i s Thus K t = 0 . 0531 With K = 28.6 and K = 0 . 0531 , the sy stem t ransfe Y (s) R( s) = s 715 2 + 31 . 55 s + 715 Unit-Step Response: y = 0.5 when t = 0.0505 sec. Thus, t = 0 . 0505 sec. d y max = 1.1007 ( 10 . 07% max. overs hoot ) 7-18 Closed-Loop Transfer Fu nction: Y (s ) 25 K = 2 R ( s ) s + ( 5 + 500 Kt ) s + 25 K For Maximum overshoot = 0.2, = 0.456 From Eq. (7-102), Delay time t d Characteristic Equation: s + ( 5 + 500 Kt ) s + 25 K = 0 2 . 2 = 1.1 + 0 .125 + 0 .469 n = 1.2545 n = 0 . 01 K = t sec. 2 Natural Undamped Frequency 5 n = 1.2545 0 . 01 = 125 .45 .45 rad/sec. Thus, + 500 K t = 2 n = 2 0 .456 125 t = 114.41 Thus, K = 0.2188 n = 15737.7 = 629.5 25 5 With K = 629.5 and K = 0.2188 , the system transfer function is Y (s) R( s) = 15737 . 7 s 2 + 114.41 s + 15737 .7 Unit-step Response: y = 0.5 when t = 0.0101 sec. 93 Thus, t y d = 0 . 0101 sec. x. overs hoot ) max = 1.2 ( 20% ma 7-19 Closed-Loop Transfer Function: Y (s ) 25 K = 2 R ( s ) s + ( 5 + 500 Kt ) s + 25 K = 0 .6 2 = 5 + 500 n From Eq. (7-109), settling time t s Characteristic Equation: s + ( 5 + 5000 Kt ) s + 25 K = 0 2 K t = 1.2 n 3.2 n 3 .2 n -5 = 0 .6 = 0 .1 n sec. Thus, n = 3 .2 0 . 06 = 53 . 33 rad / sec K t = 1.2 = 0 .118 Y (s) R( s) K = n 2 = 113 . 76 500 25 System Transfer Function: = 2844 s 2 + 64 s + 2844 y(t) reaches 1.00 and never exceeds this value at t = 0.098 sec. Thus, t = 0 . 098 sec. s Unit-step Response: 7-20 (a) Closed-Loop Transfer Function: Y (s ) 25 K = 2 R ( s ) s + ( 5 + 500 Kt ) s + 25 K For maximum overshoot = 0.1, = 0 . 59 . Settling time: t s Characteristic Equation: s + ( 5 + 500 Kt ) s + 25 K = 0 2 5 + 500 sec. K t = 2 n = 2 0 . 59 n = 1.18 n n = 3 .2 0 . 05 = 3 .2 n t = 3 .2 0 .59 n = 0 . 05 0 . 59 = 108 .47 K = 1.18 n -5 500 = 0 .246 Y (s) R( s) K = n 2 = 470 . 63 25 System Transfer Function: = 11765 . 74 s 2 + 128 s + 11765 . 74 Unit-Step Response: y(t) reaches 1.05 and never exceeds this value at t = 0.048 sec. Thus, t = 0 . 048 sec. s 94 (b) For maximum overshoot = 0.2, Settling time t s = 0 .456 3 .2 0 .456 K . 5 + 500 K t = 2 n = 0 . 912 n n = 3 .2 0 .456 = 3 .2 n = n = 0 . 01 n -5 sec. 0 . 01 = 701 . 75 rad / sec t = 0 . 912 = 1.27 500 System Transfer Function: Y (s) R( s) = 492453 s 2 + 640 s + 492453 y(t) reaches 1.05 and never exceeds this value at t = 0.0074 sec. Thus, t = 0 . 0074 sec. This is less s Unit-Step Response: than the calculated value of 0.01 sec. 7-21 Closed-Loop Transfer Function: Y (s ) 25 K = 2 R ( s ) s + ( 5 + 500 Kt ) s + 25 K Damping ratio Characteristic Equation: s + ( 5 + 500 Kt ) s + 25 K = 0 2 = 0 . 707 . Settling time t s = 4. 5 n = 3 .1815 n = 0 .1 sec. Thus, n = 31 .815 n 2 rad/sec. 5 + 500 K t = 2 n = 44. 986 Y (s) R( s) Thus , K t = 0 . 08 K = 2 = 40 .488 System Transfer Function: = 1012 .2 s 2 + 44. 986 s + 1012 .2 s Unit-Step Response: The unit-step response reaches 0.95 at t = 0.092 sec. which is the measured t . 95 7-22 (a) When = 0 . 5 , the rise time is t r 1 - 0 .4167 + 2 . 917 n 2 2 = 1. 521 n =1 sec. Thus n = 1. 521 rad/sec. The second-order term of the characteristic equation is written s 2 + 2 n s + n = s 2 s 2 + 1. 521 s + 2 . 313 = 0 3 The characteristic equation of the system is Dividing the characteristic equation by s + ( a + 30 )s 2 + 30 as +K =0 + 1. 521 s + 2 . 313 , we have For zero remainders, 28 .48 a = 45 . 63 Thus, a = 1. 6 69 . 58 K = 65 .874 + 2 . 313 a = 69 . 58 Forward-Path Transfer Function: G (s) = s( s + 1.6 )( s + 30 ) Unit-Step Response: y = 0.1 when t = 0.355 sec. y = 0.9 when t = 1.43 sec. Rise Time: t = 1.43 - 0 . 355 r = 1. 075 sec. (b) The system is type 1. seconds ss (i) (ii) For a unit-step input, e For a unit-ramp input, = 0. K v = lim s0 sG ( s ) = K 30 a = 60 . 58 30 1. 6 = 1.45 e ss = 1 K v = 0 .69 7-23 (a) Characteristic Equation: s 3 + 3s + (2 + K ) s - K = 0 2 Apply the Routh-Hurwitz criterion to find the range of K for stability. Routh Tabulation: 96 s s s 3 1 3 6 2 +K 2 -K 1 +4K 3 s 0 -K -1.5 < K < 0 This simplifies the search for K for two equal roots. When K = -0.27806, the characteristic equation roots are: -0.347, Stability Condition: -0.347, and -2.3054. (b) Unit-Step Response: ( K = - 0.27806) (c) Unit-Step Response ( K = - 1) The step responses in (a) and (b) all have a negative undershoot for small values of t. This is due to the zero of G(s) that lies in the right-half s-plane. 7-24 (a) The state equations of the closed-loop system are: = -6 x 1 dt dt The characteristic equation of the closed-loop system is 1 2 dx = - x1 + 5 x 2 dx - k1 x 1 - k 2 x 2 + r = s +1 6 + k1 -5 s + k2 = s + ( 1 + k2 ) s + ( 30 + 5 k1 + k 2 ) = 0 2 97 For n = 10 rad / sec, 30 + 5 k 1 + k 2 = n = 100 2 . k Thu s 5 k 1 + k 2 = 70 (b) For = 0 . 707 , 2 n = 1+ k2. Th us 2 n = 1 + 2 . 1.414 n = 2 (1 + k ) 2 2 and = 30 + 5k1 + k 2 , 1 Thus k 2 = 59 + 10 k1 2 (c) For n = 10 rad / sec 5k 1 1 = 0 . 707 an d + k 2 = 100 k 1 Solving for k , we hav e = 11 . 37 . + k 2 = 2 n = 14. 14 Th us k 2 = 13 .14 (d) The closed-loop transfer function is Y (s) R (s ) For a unit-step input, = 5 s + ( k 2 + 1) s + ( 30 + 5 k1 + k 2 ) 2 = 5 s + 14.14 s + 100 2 lim y ( t ) t = lim sY ( s ) s0 = 5 100 = 0 . 05 (e) For zero steady-state error due to a unit-step input, 30 + 5 k1 + k 2 = 5 1 2 Thus 5k 1 + k 2 = -25 Parameter Plane k versus k : 7-25 (a) Closed-Loop Transfer Function Y (s) 100 ( K P + K D s ) = 2 R ( s ) s + 100 K D s + 100 K P The system is stable for K > 0 and P (b) For (b) Characteristic Equation: s K 2 + 100 K D s + 100 K P =0 D > 0. = 1, 2 n = 100 K P K D . 2 n n = 10 (c) (d) T hus = 100 KD = 20 KP K D = 0 .2 K P See parameter plane in part (g) . See parameter plane in part (g) . 98 (e) Parabolic error constant s 0 K 2 a = 1000 s 0 sec -2 K a = lim s G ( s ) = lim100 ( K P + K D s ) = 100 K P = 1000 (f) Natural undamped frequency Thus K P = 10 n = 50 K P rad/sec. n = 10 (g) When K P = 50 Th us K P = 25 = 0, G (s) = 100 K s 2 D s = 100 K s D (pole-zero cancellation) 7-26 (a) Forward-path Transfer Function: G ( s) = When r ( t ) Y ( s) E (s ) K v = s [ Js(1 + Ts ) + K i K t ] sG ( s ) KKi K K t = s 0.001 s + 0.01 s + 10 K t 2 ( 10 K K K ) = tu s ( t ), = lim s0 = e ss = 1 K v = t (b) When r(t) = 0 Y (s ) Td ( s) For T ( s ) d = 1 + Ts s [ Js(1 + Ts) + Ki Kt ] + KKi lim y ( t ) t = s 0.001 s + 0.01 s + 10 Kt + 10 K 2 ( 1 + 0.1s ) = 1 s = lim sY ( s ) s 0 = 1 if the system is stable. 10 K (c) The characteristic equation of the closed-loop system is 0 . 001 s 3 + 0 . 01 s + 0 .1 s + 10 2 K =0 The system is unstable for K > 0.1. So we can set K to just less than 0.1. Then, the minimum value of the steady-state value of y(t) is 1 + =1 - 10 K K =0 .1 However, with this value of K, the system response will be very oscillatory. The maximum overshoot will be nearly 100%. (d) For K = 0.1, the characteristic equation is 0 . 001 s 3 + 0 . 01 s + 10 2 K s t +1 = 0 or s 3 + 10 s 2 + 10 4 K s t + 1000 = 0 For the two complex roots to have real parts of -2/5. we let the characteristic equation be written as 99 ( s + a) ( s Then, a The three roots are: s 2 + 5s + b = 0 a ) or s + ( s + 5)s + (5a+ b) s+ ab = 0 3 2 + 5 = 10 = - a = -5 =5 ab = 1000 b = 200 s = - a = -5 s = -2 . 5 j13 . 92 5a + b = 10 4 K t K t = 0 . 0225 7-27 (a) K t = 10000 oz - in / rad The Forward-Path Transfer Function: G ( s) = s s + 5000 s + 1.067 10 s + 50.5 10 s + 5.724 10 4 3 7 2 9 ( 9 10 K 12 12 ) = Routh Tabulation: s s s s 5 9 10 K 12 s ( s + 116)( s + 4883)( s + 41.68 + j3178.3)( s + 41.68 - j 3178.3) 10 10 7 1 500 0 5.7 2. 895 16 . 6 1.067 50.5 5 5.7 24 9 9 10 12 12 4 9 10 0 K 3 10 8 5 . 72 7 10 9 12 - 1. 8 10 12 K 2 10 + 1. 579 10 13 K 10 K s 1 10 + 8 .473 10 29 12 K - 2 . 8422 10 K 9 K 2 + 1. 579 10 12 s 0 9 1 K From the s row, the condition of stability is 165710 or K 2 + 8473 (K K - 2 . 8422 K - 2981 .14 K - 58303 .427 <0 or + 19 .43)( >0 - 3000 . 57 ) < 0 K 2 Stability Condition: 0 < K < 3000.56 The critical value of K for stability is 3000.56. With this value of K, the roots of the characteristic equation are: -4916.9, -41.57 + j3113.3, -41.57 + j3113.3, -j752.68, and j752.68 (b) K L = 1000 oz-in/rad. The forward-path transfer function is G ( s) = s s + 5000 s + 1.582 10 s + 5.05 10 s + 5.724 10 4 3 6 2 9 ( 9 10 K 11 11 ) = 9 10 K 11 s (1 + 116.06)( s + 4882.8)( s + 56.248 + j1005)( s + 56.248 - j1005) (c) Characteristic Equation of the Closed-Loop System: s 5 + 5000 s 4 + 1. 582 10 6 s 3 + 5 . 05 10 9 s 2 + 5 . 724 10 11 s + 9 10 10 10 6 11 K =0 5.7 24 9 Routh Tabulation: s s s s 5 1 500 0 5.72 4.6503 1 .582 5. 05 5 10 11 11 4 9 10 0 K 3 10 7 5 .724 6 10 11 - 1.8 10 11 8 K 2 10 + 1. 5734 10 K 9 10 K 100 s 1 26 . 618 10 18 + 377 .43 7 10 11 15 K - 2 . 832 10 6 14 K 2 4. 6503 s 0 10 + 1. 5734 10 9 K 10 K 26 . 618 or (K From the s row, the condition of stability is Or, K 2 1 10 4 - 1332 . 73 K - 93990 < 0 - 1400 )( K + 3774. 3 K - 2 .832 + 67 .14 ) < 0 K 2 >0 Stability Condition: 0 < K < 1400 The critical value of K for stability is 1400. With this value of K, the characteristic equation root are: -4885.1, -57.465 + j676, -57.465 - j676, j748.44, and -j748.44 (c) K L = . G ( s) = = nK s K i K s L a J T s + ( Ra J T + R mL a ) s + R a B m + K i K b 2 Forward-Path Transfer Function: JT = J m + n J L 2 s s + 5000 s + 566700 2 ( 891100 K ) = 891100 K s ( s + 116)( s + 4884) Characteristic Equ ation of the Closed-Loop System: s 3 + 5000 s 2 + 566700 s + 891100 K =0 Routh Tabulation: s s s s 3 1 5000 5 66700 5667 00 89 1100 K .22 K 2 1 - 178 0 89 91100 K 1 From the s row, the condition of K for stability is 566700 - 178.22K > 0. Stability Condition : 0 < K < 3179.78 The critical value of K for stability is 3179.78. With K = 3179.78, the characteristic equation roots are -5000, When the motor shaft is flexible, K becomes stiffer, K K L L j752.79, and -j752.79. is finite, two of the open-loop poles are complex. As the shaft =, L increases, and the imaginary parts of the open-loop poles also increase. When the shaft is rigid, the poles of the forward-path transfer function are all real. Similar effects L are observed for the roots of the characteristic equation with respect to the value of K . 7-28 (a) G (s) c =1 G (s) = 100( s s 2 + 2) -1 K p = lim G (s) s0 = - 200 When d(t) = 0, the steady-state error due to a unit-step input is e ss = 1 1+ K p = 1 1 - 200 =- 1 199 = - 0 . 005025 101 (b) Gc (s ) = (c) s + s G (s ) = 100( s + 2)( s + ) s s -1 2 ( ) Kp = e ss = 0 =5 = 50 = 500 m aximum ma ximum overs hoot oversh oot = 5.6% = 22% = 54.6% max imum o versho ot As the value of increases, the maximum overshoot increases because the damping effect of the zero at s = - becomes less effective. Unit-Step Responses: (d) r(t ) =0 and G (s) c = 1. d (t ) = u s(t ) D (s) = 1 s System Transfer Function: ( r = 0) Y ( s) D (s) r =0 = 100( s s 3 + 2) + 100 s 2 + ( 199 + 100 ) s + 200 100( s Output Due to Unit-Step Input: Y ( s) = s s + 2) 3 + 100 s 2 + ( 199 + 100 ) s + 200 sY ( s ) y ss = lim y ( t ) t = lim s0 = 200 200 = 1 (e) r(t ) = 0, d (t ) = us(t ) G (s) c = s + s System Transfer Function [ r( t) = 0] Y (s) D ( s) r =0 = 100 s ( s s 3 + 20 + 100 s 2 + ( 199 + 100 ) s + 200 D (s) = 1 s 102 y ss = t lim y ( t ) = lim s0 sY ( s ) =0 (f) =5 Y (s) D ( s ) r =0 Y (s) D ( s) Y ( s) D (s) r =0 r =0 = 100 s ( s s 3 + 2) s + 100 s 2 + 699 + 1000 = 50 = 100 s ( s s 3 + 2) s + 100 s 2 + 5199 + 10000 = 5000 = 100 s ( s s 3 + 2) s + 100 s 2 + 50199 + 100000 Unit-Step Responses: (g) As the value of increases, the output response y(t) due to r(t) becomes more oscillatory, and the overshoot is larger. As the value of increases, the amplitude of the output response y(t) due to d(t) becomes smaller and more oscillatory. 7-29 (a) Forward-Path Transfer function: G (s) Characteristic Equation: = H (s) E (s) = 10 N s( s + 1 )( s + 10 ) s N s(s 2 + 1) s 2 +s+N =0 n = 1 n 1 N=1: Characteristic Equation: - 1 - 2 + s +1 = 0 (16.3%) Peak time t = 0 .5 = rad/sec. Maximu m over shoot = e = 0 .163 s 2 max - = 3 . 628 2 sec. N=10: Characteristic Equation: - + s = 10 = 0 Peak time t = 0 . 158 = n 1 n = 10 = 1.006 2 rad/sec. Maximum overshoot =e 1 - 2 = 0 . 605 (60.5%) max sec. - (b) Unit-Step Response: N=1 103 Second-order System Maximum overshoot Peak time 0.163 3.628 sec. Third-order System 0.206 3.628 sec. Unit-Step Response: N = 10 Second-order System Maximum overshoot Peak time 0.605 1.006 sec. Third-order System 0.926 1.13 sec. 7-30 Unit-Step Responses: When T z is small, the effect is lower overshoot due to improved damping. When T z z is very large, the overshoot becomes very large due to the derivative effect. T derivative control or a high-pass filter. improves the rise time, since 1 +Tz s is a 104 7-31 Unit-Step Responses The effect of adding the pole at s less stable. When T p =- 1 T p to G(s) is to increase the rise time and the overshoot. The system is > 0 . 707 , the closed-loop system is stable. 7-32 (a) N=1 Closed-loop Transfer Function: M H (s) = Y ( s) R( s) = 10 s 3 + 11 s + 10 s + 10 2 = 1 1 + s + 1. 1 s + 0 .1 s 2 3 Second-Order Approximating system: M M M d e f f (s) (s) l L (s) = 1 1+ d s 1 2 + d2s = 2 H L = 1+ d s 1 + d2s 2 1+ m s 1 +m2s 2 2 1+ s + 1.1 s + 0 .1 s l 2 3 1+ l s 1 + l2 s + l3s 3 1 = m1 = f 2 d 2 = m2 2 1 =1 2 = 1.1 l 3 = 0 .1 e f 4 2 = 2m 2 2 - m1 = 2 d 2 - d1 2 = = f 4 = 2 m 4 - 2 m 1m 3 + m 2 = m 2 = d 2 2 2 2 4 2 2 = 2 l 2 - l1 = 2 1.1 - 1 = 1.2 = 2 l6 - 2 l 1 l 5 + 2 l 2 l 4 - l 3 = - l3 = - 0 . 01 2 4 2l - 2 l1 l 3 + l 2 = -2 1 0 .1 + ( 1.1 ) = 1. 01 2 6 Thus, f 2 = 1.2 = -0 . 01 1 e 2 = 1.2 = d 2 1 2d 2 - d1 2 f 4 = 1. 01 1 e 4 = 1. 01 = d 2 2 Thus d 2 = 1. 005 f M (s) 6 = 2 d 2 - 1.2 = 0 .81 = 0 . 995 s 2 Thus d = 0 .9 G L ( s) L = 1 + 0 . 9 s + 1. 005 s 2 + 0 . 8955 s + 0 . 995 = 0 . 995 s( s + 0 .895) Roots of Characteristic Equation: Third-order System Second-order System -10.11 -0.4457 + j0.8892 -0.4457 - j0.8892 -0.4478 + j0.8914 -0.4478 - j0.8914 105 The real root at -10.11 is dropped by the second-order approximating system, and the two complex roots are slightly preturbed. Unit-Step Responses: (b) N = 2: Closed-loop Transfer Function: M (s) H = 20 s l 2 3 + 11 s + 10 s + 20 2 = = 1 1 + 0 .5 s 2 + 0 . 55 f 2 s 2 + 0 . 05 2l 2 s 2 3 l 1 = 0 .5 = d2 = 2 l f 4 2 = 0 . 55 3 = 0 . 05 e 2 2d 2 - d1 = = - l1 = 2 0 . 55 - ( 0 . 5) d 2 2 2 = 0 .85 d 2 e d 4 2 1 = -2 l1 l 3 + l 2 = -0 . 05 + 0 . 3025 = 0 .2525 d 1. 99 s 2 1 Thus = f 4 = 0 .2525 = 0 . 5025 = 2d 2 - e 2 = 2 0 . 5025 - 0 .85 = 0 .155 1 1 = 0 . 3937 G L(s) M L (s) = + 0 . 3937 s + 0 . 5025 s 2 = + 0 . 7834 s + 1. 99 = s(s 1. 99 + 0 . 7835) Roots of Characteristic Equations: Third-order System Second-order System - 10.213 -0.3937 + j1.343 -0.3937 - j1.343 -0.3917 + j1.3552 -0.3917 - j1.3552 The real root at -10.213 is dropped by the second-order approximating system, and the two complex roots are slightly preturbed. Unit-Step Responses: 106 (c) N=3 Closed-Loop Transfer Function: M (s) H = 30 s 3 + 11 s + 10 s + 30 2 = 1 1 + 0 . 333 s + 0 . 3667 s 2 + 0 . 0333 s 3 l e 2 1 = 0 . 3333 2 l f 2 2 = 0 . 3667 2 2 1 l 3 = 0 . 0333 e 4 = 2 d 2 - d1 = 2 2 = 2 l 2 - l1 = 0 . 7333 - 0 .1111 = 0 . 6222 d d 1 = d2 = 2 f 4 = -2 l1 l 3 + l 2 = 0 .1122 2 Thus, d = f 4 = 0 . 1122 = 0 .2186 = 2 d2 - f 2 = 2 0 . 335 - 0 . 6222 = 0 . 0477 d 2 = 0 . 335 2 . 985 G L(s) M L (s) = 1 1 + 0 .2186 s + 0 . 335 s 2 = s 2 + 0 . 6524 s + 2 .985 = s( s 2 . 985 + 0 . 6525) Roots of Characteristic Equation: Third-order System Second-order System -10.312 -0.3438 + j1.6707 -0.3438 - j1.6707 -0.3262 + j1.6966 -0.3262 - j1.6966 The real root at -10.312 is dropped by the second-order approximating system, and the complex roots are slightly preturbed. Unit-Step Responses: (d) N= 4 Closed-Loop Transfer Function: 107 M H (s) = 40 s 3 + 11 s + 10 s + 40 2 = 1 1 + 0 .25 s + 0 .275 l 4 3 s 2 + 0 . 025 s 3 l e 2 1 = 0 .25 2 l 2 = 0 .275 e 4 2 1 = 0 . 025 f 2 4 = 2d d 2 2 - d1 = 2 f 2 = 2 l 2 - l 1 = 0 .4875 d 1 1 2 =d2 = d = -2 l1 l 3 + l 2 = 0 . 06313 2 Thus, 2 = M f 4 = 0 . 06313 (s) = 0 .2513 = = 2d - f 2 = 0 . 5025 - 0 .4875 = 0 . 015 G (s) d 3 . 98 1 = 0 .1225 L = 3 . 98 s 2 + 0 .1225 s + 0 .2513 s 2 + 0 .4874 s + 3 .98 L = s( s + 0 .4874 ) Roots of Characteristic Equation: Third-order System Second-order System - 10.408 -0.2958 + j1.9379 -/2958 - j1.9379 -0.2437 + j1.98 -0.2437 - j1.98 The real root at -10.408 is dropped by the second-order approximating system, but the complex roots are preturbed. As the value of N increases, the gain of the system is increased, and the roots are more preturbed. Unit-Step Responses: (e) N= 5 Closed-loop Transfer Function: M (s) H = 50 s 3 + 11 s + 10 s + 50 2 = 1 l 1 + 0 .2 s + 0 .22 s 2 + 0 . 02 s 3 l e e 2 1 = 0 .2 2 l 2 2 = 0 .22 2 3 = 0 . 02 = 2 d1 - d 2 = = d2 = 2 f = 2 l 2 - l1 = 0 .44 - 0 .04 = 0 .4 2 4 f 4 = -2 l1 l 3 + l 2 = -0 . 008 + 0 . 0484 = 0 . 0404 108 Thus, d d M L 2 2 2 1 = = f 4 = 0 .0404 2 d 2 = 0 .201 d 1 2d - e 2 = 0 .402 - 0 .4 = 0 . 002 s 2 = 0 . 04472 G L(s) (s) = 1 1 + 0 . 0447 s + 0 .201 = 1 s 2 + 0 .2225 s + 4. 975 = 4. 975 s(s + 0 .2225) Roots of Characteristic Equation: Third-order System Second-order System - 10.501 -0.2494 + j2.678 -0.2494 -j2.678 -0.1113 + j2.2277 -0.1113 - j2.2277 The real root at -10.501 is dropped by the second-order approximating system, and the complex roots are changed, especially the real parts. Unit-Step Responses: 7-33 (a) K=1 Forward-path Transfer Function: G ( s) = s s + 5000 s + 566700 2 ( 891100 ) = 891100 s ( s + 116)( s + 4884) Closed-Loop Transfer Function : M (s) H = l 891100 s 3 + 5000 s 2 + 566700 l 2 2 s + 891100 -3 = 1 1 + 0 . 636 l 3 s + 5. 611 10 -6 -3 s 2 + 1.1222 10 -6 s 3 1 = 0 . 636 = d2 = 2 = 5. 611 10 2 2 2 = 1.122 10 -3 -6 e e Thus, 2 = 2 d 2 - d1 = f 4 f = 2 l 2 - l1 = 1.1222 10 - 0 .4045 = -0 . 3933 + ( 5 . 611 10 -3 4 = - 2 l1 l3 + l 2 = -2 0 . 636 1.1222 10 ) 2 = 0 . 000030 06 109 d d 2 2 2 1 = 0 . 000030 = 2 d2 - f 2 06 d 2 = 0 .005482 d 1 = 0 . 01096 + 0 . 3933 = 0 .4042 = 182 .4 s 2 = 0 . 6358 G L ( s) M L (s) = 1 1 + 0 . 6358 s + 0 . 005482 s 2 + 115 . 97 s + 182 = 182 .4 s( s .4 + 115 . 97 ) Roots of Characteristic Equations: -1.595 -114.4 -4884 Third-order System -1.5948 -114.38 Second-order System The real root at -4884 is dropped by the second-order approximating system. the other two roots are hardly preturbed. Unit-Step Responses (b) K = 100 Closed-loop Transfer Function: M H (s) = 891100 s 3 00 00 + 5000 l 1 s 2 + 566700 + 891100 l 2 = 1 1 + 0 . 00636 s -5 -5 + 5. 611 10 -5 -8 s 2 + 1.1222 10 -8 s 3 = 0 .00636 2 = 5. 611 10 2 2 l 3 = 1. 1222 10 -5 -8 e e Thus, d d 2 = 2 d 2 - d1 = = d2 = 2 f 2 = 2 l 2 - l1 = 11 .222 10 - 4. 045 10 = 0 . 000071 8 -5 4 2 2 2 1 f 4 = 2 l 4 - 2 l1 l 3 + l 2 = - 2 0 . 00636 1.1222 10 -9 + ( 5 . 611 10 ) 2 = 3 . 0056 10 -9 = f 4 = 3 . 0056 10 f 2 d 6 2 = 0 . 000054 8 8 8 d = 2 d2 - = 0 . 000109 1 - 0 . 000071 = s = 0 . 000054 1 = 0 . 007403 (s) M L (s) = 1 18248 2 + 0 . 007403 s + 0 . 000054 8s 2 + 135 .1 s + 18248 G L = 18248 s( s + 135 . 1) Roots of Characteristic Equations: Third-order System Second-order System 110 -4887.8 -56.106 + j122.81 -56.106 - j122.81 -67.55 + j114.98 -67.55 - j114.98 Unit-Step Responses (c) K = 1000 Closed-loop Transfer Function: M (s) H = 891100 s 3 000 s + 5000 l 1 s 2 + 566700 + 891100 l 2 2 = 000 1 1 + 0 . 000636 -6 s + 5 . 611 10 -6 -9 s 2 + 1.1122 10 -9 s 3 = 0 . 000636 2 = 5 . 611 10 l 3 = 1.1222 10 404 -9 e e Thus, d d 2 = 2 d 2 - d1 = = d2 = 2 f 2 = 2 l 2 - l1 = 11 .222 10 2 -6 - 0 .000000 = 0 . 000010 82 -6 4 2 2 2 1 f 4 = - 2 l1 l3 + l 2 = -2 0 . 000636 1.1222 10 -11 + ( 5 . 611 10 482 ) 2 = 3 . 0056 10 -11 = = f 4 = 3 . 0055 10 2 d 965 2 = 0 . 000005 = 0 . 000000 182415 .177 2d - f 2 = 0 . 000010 1 - 0 . 000010 = 818 147 d 1 = 0 . 000382 G L (s) 9 182415 .177 s( s M L (s) = 1 + 0 . 000382 9s + 0 . 000005 482 s 2 s 2 + 69 . 8555 s + 182415 = .177 + 69 . 8555) Roots of Characteristic Equations: Third-order System Second-order System - 4921.6 -39.178 + j423.7 -39.178 - j423.7 -34.928 + j425.67 -34.928 - j425.67 111 Unit-step Re sponses 7-34 Forward-path Transfer Function G (s) Closed-loop Transfer Function M (s) = K(s s(s - 1) + 1 )( s + 2 ) 1 = -s +1 s 3 +3s + s +1 2 Second-order System: M L (s) 2 = + c1 s 1 M H (s ) M L (s ) = l ( - s + 1) (1 + d s + d s ) 1 + ( d - 1) s + ( d - d ) s - d s = ( s + 3s + s + 1) ( 1+ c s ) 1 + ( c + 1) s + ( c + 3) s + ( 3c + 1) s + c s 2 3 1 2 1 2 1 2 3 2 2 3 1 1 1 1 1 1+ d s + d2s 2 4 1 = 1 + c1 1 l m 2 = 3 + c1 = d 2 - d1 l m 3 = 1 + 3 c1 = -d 2 2 l 4 = c1 m = d1 -1 2 2 3 e2 = f 2 = 2m 2 - m1 = 2 ( d 2 - d 1 ) - ( d 1 - 1) = 2 d 2- d 1 - 1 2 = 2l2 - l1 = 2 ( 3 + c1 )- ( 1+ c1 ) = 5 - c1 2 2 2 2 e4 = f 4 = 2m 4 - 2m1 m3 + m 2 = -2 ( d 1 - 1) ( -d 2 ) + ( d 2 - d1 ) = d 2 - 2d 2 + d 1 2 2 2 = 2l4 - 2l1l3 + l2 = - 2 ( 1 + c1 ) (1 + 3c1 ) + ( 3 + c1 ) = 7 - 2c1 - 5c1 2 2 2 e6 = f 6 = 2m 6 - 2 m1m 5 + 2 m 2 m4 - m 3 = -m 3 = - ( -d 2 ) = -d 2 2 2 2 2 2 2 = 2l6 - 2l1l5 + 2l 2l4 - l3 = 2l2 l4 - l3 = 2 ( 3 + c1 ) c1 - (1 + 3c1 ) = - 1 - 7c1 2 2 Simultaneous equations to be solved: Solutions: 112 2d 2 - d1 = 6 -c1 2 2 c1 d d 1 = -1. 0408 = 0 . 971 = 2 . 93 G L(s) d d 2 2 2 2 2 = 1 + 7 c1 2 2 - 2 d 2 + d 1 = 5 - c1 = 1 1 2 M L (s) - 1. 0408 s s 2 + 0 . 971 s + 2 . 93 = -0 .3552( s 2 s - 0 . 9608 s ) + 0 . 3314 + 0 .3413 = -0 . 3552( s( s s - 0 . 9608 ) + 0 . 69655) Roots of Characteristic Equations: -2.7693 -0.1154 + j0.5897 -0.1154 - j0.5897 Unit-step Responses Third-orde r System -0.1657 + j0.5602 -0.1657 -j0.5602 Second-order System 7-35 (a) K = 10 Closed-loop Transfer Function: M (s) H = 10 s 4 + 23 s 3 + 62 s 2 + 40 s + 10 = 1 1+ 4 s + 6 .2 s + 2 . 3 s + 0 .1 s 2 3 4 Second-order System Approximation: M (s) L = l 1 1+ d s 1 + d2s 2 l e e e 2 1 =4 2d 2 2 2 2 2 = 6 .2 2 l 3 = 2 .3 2 l 4 = 0 .1 = = = f f f 2 = - d 1 = 2 l 2 - l1 = 12 .4 - 16 = -3 . 6 2 2 2 4 4 = d 2 = 2 l 4 - 2 l1 l 3 + l 2 = 0 .2 - 2 4 2 . 3 + ( 6 .2 ) = 20 .24 = 2 d 6 - 2 d 1 d 5 + 2 d 2 d 4 - d 3 = - d 3 = 2 l 2 l 4 - l3 = - 4. 05 d d 2 2 2 1 6 6 Thus, = = 20 .24 2d d f 2 = 4. 5 = 3 . 55 2 - 2 = 9 + 3. 6 = 12 . 6 d 1 113 M L (s) = 1 1 + 3 .55 s + 4. 5 s 2 = 0 .2222 s 2 + 0 . 7888 s + 0 .2222 G L(s) = 0 .2222 s( s + 0 . 7888 ) Roots of Characteristic Equations: Fourth -order system Second-order system -2.21 -20 -0.3957 + j0.264 -0.3957 -j0.264 (b) K = 10 Third-order System Approximation: 1 1+ d s 1 2 -0.3944 + j0.258 -0.3944 -j0.258 M L (s) = M 2 3 H L (s) (s) + d 2s + d3s M d 2 3 = 1 1 + d1 s + d 2 s + d 3 s 2 2 3 3 + 4 s + 6 .2 s + 2 . 3 s + 0 .1 s Thus d 4 e e 2 = = f f 2 = 2d 2 - d 1 = -3 . 6 2 = -f6 = 2 4. 05 d 2 3 = 2 . 0125 4 4 = 2 d 4 - 2 d 1 d 3 + d 2 = - 2 d 1 d 3 + d 2 = - 4. 025 d 1 +d2 = f 4 = 20 .24 Thus, 2 1 = 0 . 5 d 1 - 1.8 2 d , 2 2 Solving for d , d 1 = 3 .9528 - 3 . 0422 = 0 .25 d 1 - 1. 8 d 1 + 3 .24 = 20 .24 + 4. 025 4 2 d , - 0 .4553 + d 1 j 2334, . d 2 - 0 .4553 - 2 1 j 2334. Selecting the positive and real solution, we have 1 1 = 3 . 9528 = = 0 . 5 d 1 - 1.8 = 6 . 0123 0 .4969 M L (s) = + 3. 9528 s + 6 . 0123 s 2 + 2 . 0125 s 3 s 3 + 2 . 9875 s 2 + 1. 964 s + 0 .4969 GL ( s) = Roots of Characteristic Equations: Fourth -order System s s + 2.9875 s + 1.964 2 ( 0.4969 ) Third-order System -2.21 -20 -0.39565 + j0.264 -0.39565 - j0.264 -2.1963 -0.3956 + j0.264 -2.1963 - j0.264 Unit-step Response 114 (c) K = 40 Closed-loop Transfer Function: M (s) H = l 40 s 4 + 23 s 3 + 62 l 2 s 2 + 40 s + 40 l 3 = 1 1+ s + 1. 55 s l 2 + 0 . 575 = 0 . 025 s 3 + 0 . 025 s 4 1 =1 = 1. 55 = 0 . 575 4 Second-order System Approximation: M (s) L = 1 1+ d s 1 + d2s 2 e e Thus, 2 = = f f 2 = 2 d2 - d1 = 2 2 2l 2 - l 1 = 3 .1 - 1 = 2 2 2 .1 4 4 = d 2 = 2 l 4 - 2 l1 l3 + l 2 = 0 . 05 - 1.15 + 2 .4025 = 1. 3025 d d 2 2 2 1 = 1. 3025 = 2 d2 - f 2 d 2 = 1.1413 1 = 2 1.1413 - 2 . 1 = 0 .1825 = 0 .8762 s 2 d = 0 .4273 G L ( s) M L (s) = 1 1 + 0 .4273 s + 1.1413 s 2 + 0 . 3744 s + 0 .8762 = 0 .8762 s( s + 0 . 3744 ) Roots of Characteristic Equations: Fourth -order System Second-order System -2.5692 -19.994 -0.2183 + j0.855 -0.2183 - j0.855 Third-order system Approximation: M e2 e4 e 6 -0.1872 + j0.9172 -0.1872 - j0.9172 L (s) = 1 1 + d 1s + d 2 s + d 3 s 2 3 = = = f2 f4 f 6 = 2 d2 - d1 = 2 2l2 2 - l 1 = 3 .1 - 1 = 2 2 .1 2 = -2 d 1 d 3 + d 2 = -1. 0062 2 2 d1 + d 2 = 2 l 4 - 2 l1 l 3 + l 2 = 1. 3025 2 = - d 3 = 2 l 2 l 4 - l3 = - 0 .2531 Equations to be Solved Simultaneously: 2d d 2 2 2 - d 1 = 2 .1 2 d 2 2 2 = -1. 00623 d 1 = 1. 3025 d 4 1 Thus d 2 1 d 2 = 0 .5 d 1 + 1. 05 2 1 = 0 .25 d 4 1 + 1. 05 d 1 + 1.1025 Th us + 4.2 - 4. 0249 d - 0 .8 = 0 115 The roots of the last equation are: d 1 = - 0 .1688 d 1 , 0 . 9525 , . - 0 .392 + d d j 2 .196 , - 0 . 392 - j 2 .196 Selecting the positive real solution, we have d d 2 2 2 3 = 0 .9525 = 1. 00623 = 0 .2531 (s) d 1 + 1.3025 = 2 .261 Th us T hus 2 = 1. 5037 = 0 . 5031 1. 9876 3 M L = 1 1 + 0 . 9525 s + 1. 5037 s 2 + 0 . 5031 s 3 = s 3 + 2 . 9886 s 2 + 1. 8932 s + 1.9876 GL ( s) = s s + 2.9886 s + 1.8932 2 ( 1.9876 ) = 1.9876 s ( s + 2.0772)( s + 0.9114) -2.5692 -19.994 -0.2183 + j0.855 -0.2183 - j0.855 Unit-step Responses Roots of Characteristic Equati ons: Fourth -order System -2.552 -0.2183 j 0 .8551 Third-order System 116 Chapter 8 8-1 (a) P (s) 4 3 ROOT LOCUS TECHNIQUE 2 = s + 4 s + 4 s +8 s 0, Q (s) = s +1 Finite zeros of P( s): Finite zeros of Q( s): -3.5098, -0.24512 j1.4897 60 o -1 , 180 o Asymptotes: K > 0: Intersect of Asymptotes: , 300 o K < 0: 0 , o 120 o , 240 o 1 = (b) P( s) -3 . 5 - 0 .24512 - 0 .24512 - ( -1 ) 4 -1 = -1 = s + 5s + s 3 2 Q (s) 0, = s +1 Finite zeros of P( s) : Finite zeros of Q( s): -1 -4.7912, -0.20871 90 o Asymptotes: K > 0: Intersect of Asymptotes: , 270 o K < 0: 0 , o 180 o 1 = (c) P( s) -4. 7913 - 0 .2087 - ( - 1) 3 -1 = -2 =s 2 Q (s) = s + 3s + 2s +8 3 2 Finite zeros of P( s): Finite zeros of Q( s): Asymptotes: 0, 0 - 3.156 , K > 0: 0 . 083156 180 o j 1. 5874 K < 0: 0 o (d) P (s) = s + 2s + 3s 3 2 Q(s ) = s - 1 2 ( ) ( s + 3) Finite zeros of P( s): Finite zeros of Q( s): Asymptotes: -1 1, -1, -3 0, j 1.414 There are no asymptotes, since the number of zeros of P( s) and Q( s) are equal. Q (s) 0, 0, (e) P (s) = s + 2 s + 3s 5 4 3 = s +3s + 5 2 Finite zeros of P( s): Finite zeros of Q( s): 0, o - 1.5 -1 , j 1.414 o o j 1. 6583 60 180 , 300 K < 0: 0 , o Asymptotes: K > 0: Intersect of Asymptotes: 120 o , 240 o 1 = (f) P (s) -1 - 1 - ( -1. 5) - ( -1. 5) 5 -2 = 1 3 = s + 2 s + 10 4 2 Q (s) = s +5 j 1.4426 , o o Finite zeros of P( s): Finite zeros of Q( s): - 1.0398 -5 1. 0398 o j1.4426 K < 0: o o o Asymptotes: K > 0: Intersect of Asymptotes: 60 , 180 , 300 0 , 120 , 240 120 1 = 8-2 (a) Angles of departure and arrival. K > 0: -1. 0398 - 1. 0398 + 1. 0398 + 1. 0398 - ( - 5) 4 -1 = -5 3 - 1 - 2 - 3 + 4 = -180 - 1 - 90 - 45 + 90 1 = 135 o o o o o o o o o = -180 K < 0: - 1 - 90 - 45 + 90 1 = -45 o =0 o (b) Angles of departure and arrival. K > 0: K < 0: - 1 - 2 - 3 + 4 = -180 - 1 - 135 o o - 90 + 90 o o o =0 o 1 = -135 (c) Angle of departure: K > 0: - 1 - 2 - 3 + 4 = -180 - 1 - 135 1 = -90 o o o - 90 o o - 45 o = -180 (d) Angle of departure K > 0: - 1 - 2 - 3 - 4 = -180 - 1 - 135 o o o - 135 o o - 90 o = -180 1 = -180 121 (e) Angle of arrival K < 0: 1 + 6 - 2 - 3 - 4 - 5 = -360 1 + 90 - 135 1 = -108 .435 o o o o - 135 o - 45 - 26 . 565 o = -360 o o 8-3 (a) (b) (c) (d) 122 8-4 (a) Breakaway-point Equation: Breakaway Points: 2s 5 + 20 s 4 + 74 s 3 + 110 s 2 + 48 s = 0 - 0 . 7275 6 , 4 - 2 . 3887 + 100 s 3 (b) Breakaway-point Equation: 3 s + 22 Breakaway Points: s 5 + 65 s + 86 s 2 + 44 s + 12 = 0 - 1, - 2 . 5 6 (c) Breakaway-point Equation: 3 s + 54 Breakaway Points: 6 s 5 + 347 .5 s 4 + 925 s 3 + 867 .2 s 2 - 781 .25 s - 1953 = 0 - 2 .5 , 5 1. 09 s 4 (d) Breakaway-point Equation: - s - 8 s - 19 Breakaway Points: + 8 s + 94 3 s 2 + 120 s + 48 = 0 - 0 . 6428 +8) , 2 .1208 8-5 (a) G ( s)H ( s) = o K(s s(s + 5)( s + 6 ) 270 o Asymptotes: K > 0: Intersect of Asymptotes: 90 and K < 0: 0 o and 180 o 1 = Breakaway-point Equation: 2s 3 0 - 5 - 6 - ( -8 ) 3 s , 2 -1 s = - 1.5 + 35 + 176 + 240 = 0 - 9 . 7098 Breakaway Points: Root Locus Diagram: - 2 .2178 - 5 . 5724, 8-5 (b) 123 G ( s)H ( s) = o K s(s , + 1 )( s + 3)( s + 4 ) o Asymptotes: K > 0: Intersect of Asymptotes: 45 135 , 225 0 o , 315 o K < 0: 0 , o 90 o , 180 o , 270 o 1 = Breakaway-point Equation: Breakaway Points: Root Locus Diagram: 3 -1 -3 - 4 4 2 = -2 4 s + 24 s -0 .4189 , + 38 s + 12 = 0 - 2 , - 3 . 5811 8-5 (c) G ( s)H ( s) = K(s s (s 2 +4) + 2) o 2 Asymptotes: K > 0: 60 , Intersect of Asymptotes: 180 o , 300 0 o K < 0: 0 , o 120 o , 240 o 1 = Breakaway-point Equation: Breakaway Points: Root Locus Diagram: 3s 0, 4 + 0 - 2 - 2 - ( -4 ) 4 3 -1 2 =0 + 24 s + 52 s + 32 s = 0 - 1. 085 , - 2 , - 4. 915 124 8-5 (d) G ( s)H ( s) = K(s s( s 2 + 2) + 2s + 2) o Asymptotes: K > 0: 90 , Intersect of Asymptotes: 270 o K < 0: 0 0 , o 180 o 1 = Breakaway-point Equation: Breakaway Points: 2s 3 - 1 - j - 1 - j - ( -2 ) 3 2 -1 =0 +8 s +8 s + 4 = 0 The other two solutions are not breakaway points. - 2 .8393 Root Locus Diagram 125 8-5 (e) G ( s) H( s) = s s + 2s + 2 2 ( K ( s + 5) o ) K < 0: 0 , o Asymptotes: K > 0: 90 , Intersect of Asymptotes: 270 o 180 o 1 = Breakaway-point Equation: Breakaway Points: 3 0 - 1 - j - 1 - j - ( -5) 3 2 -1 = 1. 5 2 s + 17 s - 7.2091 + 20 s + 10 = 0 The other two solutions are not breakaway points. 8-5 (f) G ( s) H( s) = s ( s + 4 ) s + 2s + 2 2 ( K ) 126 Asymptotes: K > 0: 45 , Intersect of Asymptotes: o 135 o , 225 0 o , 315 o K < 0: 0 , o 90 o , 180 o , 270 o 1 = Breakaway-point Equation: Breakaway Point: 3 - 1 - j -1 + j - 4 4 2 = -1. 5 4 s + 18 s -3 . 0922 + 20 s + 8 = 0 The other solutons are not breakaway points. 8-5 (g) G ( s)H ( s) = K(s s (s 2 + 4) + 8) 90 o 2 2 Asymptotes: K > 0: , 270 o K < 0: 0 , o 180 o Intesect of Asymptotes: 1 = Breakaway-point Equation: Breakaway Points: 0, s 5 0 + 0 - 8 - 8 - ( -4 ) - ( -4 ) 4 s 4 -2 3 + 20 + 160 s + 640 s 2 + 1040 s =0 -4, -8, -4 - j4, -4 + j4 127 8-5 (h) G ( s)H ( s) = K s (s o 2 + 8) 2 Asymptotes: K > 0: 45 , Intersect of Asymptotes: 135 o , 225 o , 315 o K < 0: 0 , o 90 o , 180 o , 270 o 1 = Breakaway-point Equation: Breakaway Point: s 0, 3 -8 - 8 4 2 = -4 s + 12 s + 32 -4, -8 =0 8-5 (i) 128 G ( s) H( s) = K s + 8 s + 20 2 ( ) 0 , o s ( s + 8) 2 2 Asymptotes: K > 0: 90 , Intersect of Asymptotes: o 270 o K < 0: 180 o 1 = Breakaway-point Equation: Breakaway Points: s 5 -8 - 8 - ( - 4 ) - ( -4 ) 4 s 4 -2 s 3 = -4 + 1280 s + 20 + 128 + 736 s 2 =0 - 4, -8, - 4 + j 4. 9 , -4 - j 4. 9 (j) G ( s) H( s) = (s Ks 2 2 -4 ) Since the number of finite poles and zeros of G ( s ) H ( s ) are the same, there are no asymptotes. Breakaway-point Equation: 8 s = 0 Breakaway Points: s=0 129 8-5 (k) G ( s) H( s) = Asymptotes: (s K s -4 2 2 ( +1 90 )( s o ) 2 +4 270 ) K < 0: 0 , o K > 0: , o 180 o Intersect of Asymptotes: 1 = -2 + 2 4 -2 s 0, 6 =0 - 8 s - 24 4 Breakaway-point Equation: Breakaway Points: s 3 .2132 , =0 - 3 .2132 2 , j 1. 5246 , - j1. 5246 8-5 (l) 130 G ( s) H( s) = Asymptotes: K > 0: (s K ( s - 1) 2 2 +1 90 )( s o 2 +4 270 ) K < 0: 0 , o , o 180 o Intersect of Asymptotes: 1 = Breakaway-point Equation: Breakaway Points: s 5 -1 + 1 4 -2 3 =0 - 2s - 9 s = 0 2 . 07 , - 2 . 07 , - j 1.47 , j 1.47 (m) G ( s)H ( s) = K(s + 1 )( s + 2 )( s + 3) s (s 3 - 1) 131 Asymptotes: K > 0: 180 6 o K < 0: 5 4 0 s 2 o Breakaway-point Equation: s + 12 s + 27 s Breakaway Points: + 2 s - 18 3 =0 0.683, 0, 0 - 1.21, -2.4, -9.07, (n) G ( s)H ( s) = K (s s (s 3 + 5)( s + 40 )( s ) ) o + 250 60 o + 1000 o Asymptotes: K > 0: , 180 , 300 K < 0: 0 , o 120 o , 240 o 132 Intersect of asymptotes: 1 = Breakaway-point Equation: 3750 s Breakaway Points: 6 0 + 0 + 0 - 250 - 1000 - ( -5) - ( -40 5 s 0 5 ) -2 8 = - 401 . 67 10 + 335000 0, + 5 .247 10 s 4 + 2 . 9375 10 s 3 + 1. 875 10 11 s 2 =0 - 7.288, -712.2, 8-5 (o) G ( s)H ( s) = K(s s( s - 1) + 1 )( s + 2 ) 90 o Asymptotes: K > 0: , 270 o K < 0: 0 , o 180 o Intersect of Asymptotes: 133 1 = Breakaway-point Equation: Breakaway Points; s 3 -1 - 2 - 1 3 -1 = -2 -3 s -1 = 0 1.879 -0.3473, -1.532, 8-6 (a) Q(s ) = s + 5 P (s ) = s s + 3s + 2 = s (s +1)( s + 2) 2 ( ) Asymptotes: K > 0: 90 , Intersect of Asymptotes: o 270 o K < 0: 0 , o 180 o 1 = Breakaway-point Equation: s 3 -1 - 2 - ( -5) 3 2 -1 =1 + 9 s + 15 s + 5 = 0 134 Breakaway Points: -0.4475, -1.609, -6.9434 8-6 (b) Q (s) = s +3 P (s) = s s 2 + s+2 o o Asymptotes: K > 0: 90 , 270 K < 0: 0 , o 180 o Intersect of Asymptotes: 1 = Breakaway-point Equation: Breakaway Points: s 3 -1 - ( -3) 3 2 -1 =1 +5s + 3s + 3 = 0 The other solutions are not breakaway points. - 4.4798 135 8-6 (c) Q (s) = 5s P (s) = s 2 + 10 180 o Asymptotes: K > 0: K < 0: 5s 2 0 o Breakaway-point Equation: Breakaway Points: - 50 = 0 3.162 - 3.162, 136 8-6 (d) Q(s ) = s s + s + 2 2 ( ) P( s) = s + 3s + s + 5s +10 4 3 2 Asymptotes: K > 0: 180 o K < 0: s 6 5 4 0 o Breakaway-point Equation: Breakaway Points: + 2 s + 8 s + 2 s - 33 3 s 2 - 20 s - 20 = 0 -2, 1.784. The other solutions are not breakaway points. 137 8-6 (e) Q(s ) = s -1 2 ( ) ( s + 2) P( s) = s s + 2 s + 2 2 ( ) Since Q ( s ) and P ( s ) are of the same order, there are no asymptotes. Breakaway-point Equation: Breakaway Points: 6s 3 + 12 s 2 + 8s + 4 = 0 -1.3848 138 8-6 (f) Q(s ) = (s + 1)(s + 4) Asymptotes: K > 0: P (s ) = s s - 2 2 ( ) K < 0: 0 s 2 180 o o Breakaway-point equations: Breakaway Points: s 4 + 10 s 3 + 14 0.623 -8 = 0 -8.334, 139 8-6 (g) Q(s ) = s + 4s + 5 2 P( s) = s 90 o 2 (s 270 2 + 8 s + 16 o ) K < 0: 0 , o Asymptotes: K > 0: , 180 o Intersect of Asymptotes: 1 = Breakaway-point Equation: Breakaway Points: s 5 -8 - ( -4 ) 4 s 4 -2 + 42 - 4, 140 s = -2 3 + 10 + 92 -2 + s 2 + 80 s =0 -2 - j 2 .45 0, - 2, j 2 .45 , 8-6 (h) Q(s ) = s - 2 2 ( )( s + 4) P (s ) = s s + 2s + 2 2 ( ) Since Q ( s ) and P ( s ) are of the same order, there are no asymptotes. Breakaway Points: - 2, 6.95 141 8-6 (i) Q (s) = ( s + 2 )( s + 0 . 5) K > 0: 180 o P( s) =s - s 2 1 Asymptotes: K < 0: s 4 3 2 0 o Breakaway-point Equation: Breakaway Points: +5s + 4s -1 = 0 -4.0205, 0.40245 The other solutions are not breakaway points. 142 8-6 (j) Q(s ) = 2 s + 5 Asymptote s: K > 0: P (s ) = s 60 o 2 (s , 2 +2s +1 = s o ) 2 ( s + 1) 2 , 180 o 300 K < 0: 0 , o 120 o , 240 o Intersect of Asymptotes; 1 = Breakaway-point Equation: 6s 4 0 + 0 - 1 - 1 - ( - 2 . 5) 4 s 3 -1 s 2 = =0 0 .5 3 = 0 .167 + 28 + 32 143 + 10 s Breakaway Points: 0, -0.5316, -1, -3.135 8-7 (a) Asymptotes: K > 0: 45 o , 135 o , 225 o , 315 o Intersect of Asymptotes: 1 = Breakaway-point Equation: Breakaway Points: When 4s 5 -2 - 2 - 5 - 6 - ( -4 ) 5 s 4 -1 s 3 = - 2 . 75 s 2 + 65 + 396 -5.511 + 1100 + 1312 s + 480 = 0 = 0 . 707 -0.6325, K = 13.07 (on the RL) , 144 8-7 (b) Asymptotes: K > 0: 45 o , 135 o , 225 0 o , 315 o Intersect of Asymptotes: 1 = Breakaway-point Equation: When = 0 . 707 , K = 61.5 4s 3 - 2 - 5 - 10 4 2 = -4.25 + 100 = 0 + 51 s + 160 s 145 8-7 (c) Asymptotes: K > 0: 180 o Breakaway-point Equation: Breakaway Points: When s 4 + 4 s + 10 3 s 2 + 300 s + 500 = 0 = 0 . 707 -1.727 K = 9.65 (on the RL) , 146 8-7 (d) K > 0: 90 o , 270 o Intersect of Asymptotes: 1 = When -2 - 2 - 5 - 6 4 -2 = -7 . 5 = 0 . 707 , K = 8.4 147 8-8 (a) Asymptotes: K > 0: 60 o , 180 o , 300 0 o Intersect of Asymptotes: 1 = 2 - 10 - 20 3 = -10 Breakaway Point: (RL) Breakaway-point Equation: 3 s + 60 s + 200 = 0 -4.2265, K = 384.9 148 (b) Asymptotes: K > 0: 45 o , 135 o , 225 0 o , 315 o Intersect of Asymptotes: 1 = Breakaway-point Equation: 4 s + 27 s Breakaway Points: (RL) -0.4258 3 2 -1 - 3 - 5 4 = -2 .25 -4.2537 K = 12.95 + 46 s + 15 = 0 K = 2.879 , 149 150 8-10 P (s) = s + 25 3 s 2 + 2 s + 100 K t Q (s) 90 o = 100 o s Asymptotes: > 0: , 270 Intersect of Asymptotes: 1 = Breakaway-point Equation: Breakaway Points: (RL) s 3 -25 - 0 3 -1 2 = -12 . 5 + 12 . 5 s - 50 = 0 -12.162 -2.2037, 8-11 Characteristic e quation: s 3 +5s + K ts + K = 0 2 151 (a) Kt = 0 : P (s ) = s 2 ( s + 5) o Q (s) = 1 180 o Asymptotes: K > 0: 60 , Intersect of Asymptotes: , 300 o 1 = Breakaway-point Equation: 3s 2 -5 - 0 3 s = -1. 667 Breakaway Points: + 10 =0 0, -3.333 8-11 (b) P (s) = s + 5 s + 10 = 0 3 2 Q (s) o = s Asymptotes: K > 0: 90 , Intersect of Asymptotes: o 270 152 1 = 3 -5 - 0 2 -1 =0 Breakaway-point Equation: 2 s + 5 s - 10 = 0 There are no breakaway points on RL. 8-12 P( s) = s + 116 2 . 84 s + 1843 J L Q (s) 180 o 4 = 2 . 05 s ( s 2 + 5) s 2 Asymptotes: = 0: Breakaway-point Equation: - 2 . 05 s Breakaway Points: (RL) - 479 0, s 3 - 12532 - 37782 s =0 -204.18 153 8-13 (a) P (s ) = s s - 1 2 ( ) Q (s ) = ( s+ 5)( s + 3) K > 0: 180 s 4 Asymptotes: o Breakaway-point Equation: Breakaway Points: (RL) + 16 s 3 + 46 s 2 - 15 = 0 0.5239, -12.254 154 8-13 (b) P (s ) = s s + 10s + 29 2 ( ) Q (s ) = 10(s + 3) o Asymptote s: K > 0: 90 , Intersect of Asymptotes: 270 o 1 = 3 0 - 10 - ( -3) 3 -1 2 = -3 . 5 s Breakaway-point Equation: 20 s + 190 s There are no breakaway points on the RL. + 600 + 870 = 0 155 8-14 (a) P( s) = s ( s + 12 . 5)( s + 1 ) Q (s) = 83 .333 60 o Asymptotes: N > 0: Intersect of Asymptotes: , 180 o , 300 0 o 1 = Breakaway-point Equation: 3 s Breakaway Point: (RL) - 12 . 5 - 1 3 = - 4. 5 + 27 -0.4896 2 s _12 .5 =0 156 8-14 (b) P( s) = s + 12 . 5 s + 833 2 .333 Q (s) = 0 .02 s (s 2 + 12 . 5) A > 0: 180 o 4 Breakaway-point Equation: 0 . 02 s Breakaway Points: (RL) 0 + 0 . 5 s + 53 .125 3 s 2 + 416 . 67 s =0 157 8-14 (c) P (s) = s + 12 . 5 s + 1666 . 67 = ( s + 17 . 78 Q ( s ) = 0 . 02 s ( s + 12 . 5) 3 2 )( s - 2 . 64 + j 9 . 3)( s - 2 . 64 - j 9 . 3) Asymptotes: K o > 0: 180 o Breakaway-point Equation: 0 . 02 s + 0 . 5 s + 3 .125 s Breakaway Point: (RL) -5.797 4 3 2 - 66 .67 s - 416 . 67 =0 158 8-15 (a) A = K o = 100 : P (s) = s ( s + 12 . 5)( s + 1 ) 60 o Q (s) o = 41 . 67 Asymptotes: N > 0: Intersect of Asymptotes: 180 0 o 300 1 = Breakaway-point Equation: Breakaway Points: (RL) 3s 2 - 1 - 12 . 5 3 = - 4. 5 + 27 s + 12 . 5 = 0 -0.4896 159 8-15 (b) P( s) = s + 12 . 5 + 1666 2 . 67 = ( s + 6 .25 + j 40 . 34 )( s + 6 .25 - j 40 . 34 ) Q (s) = 0 . 02 s (s 2 + 12 . 5) 180 o Asymptotes: A > 0: Breakaway-point Equation: Breakaway Points: (RL) 0 . 02 s 0 4 + 0 . 5 s + 103 3 .13 s 2 + 833 . 33 s =0 160 8-15 (c) P (s) = s + 12 . 5 s + 833 3 2 . 33 = ( s + 15 . 83)( s - 1. 663 + j 7 . 063)( s - 1. 663 - j 7 .063) Q (s) = 0 . 01 s ( s + 12 . 5) K o Asymptotes: > 0: 180 o Breakaway-point Equation: Breakaway Point: (RL) 0 . 01 s 4 + 0 .15 s 3 + 1. 5625 s 2 - 16 . 67 s - 104. 17 =0 -5.37 161 8-16 (a) P (s) =s 2 (s + 1 )( s + 5) Q (s) =1 45 o Asymptotes: K > 0: Intersect of Asymptotes: , 135 o , 225 0 o , 315 o 1 = Breakaway-point Equation: 4s 3 +0 -1- 5 4 s = -1. 5 Breakaway point: (RL) 0, -3.851 + 18 s 2 + 10 =0 162 (b) P (s) =s 2 (s + 1 )( s + 5) Q (s) = 5s +1 60 o Asymptotes: K > 0: Intersect of Asymptotes: , 0 180 o , 300 o 1 = 4 + 0 - 1 - 5 - ( -0 .2 ) 4 3 -1 2 =- 5. 8 3 = -1. 93 Breakaway-point Equation: 15 s + 64 s + 43 s + 10 s = 0 Breakaway Points: (RL) -3.5026 8-17 P( s) Q (s) =s = 10 2 (s s + 1 )( s + 5) + 10 = ( s + 4. 893)( T d s + 1.896 o )( s - 0 . 394 + j 0 .96 )( s - 0 . 394 + j 0 . 96 ) Asymptotes: > 0: 60 o , 180 o , 300 Intersection of Asymptotes: There are no breakaway points on the RL. 1 = -4. 893 - 1. 896 + 0 . 3944 + 0 . 3944 4 -1 = -2 163 8-18 (a) K = 1: P (s) =s 3 (s + 117 .23)( s K L + 4882 .8 ) 90 o Q (s) , 270 .23 o = 1010( s + 1. 5948 )( s + 114.41 )( s + 4884 ) Asymptotes: > 0: Intersect of Asymptotes: 1 = Breakaway Point: (RL) 0 -117 - 4882 .8 + 1. 5948 + 114.41 + 4884 5 -3 = -0 .126 8-18 (b) K = 1000: P( s) Q (s) =s 3 (s + 117 s s o 3 .23)( s s + 4882 2 .8 ) 5 = 1010( = 1010( + 5000 o + 5 . 6673 10 + 39 .18 + s + 891089 110 ) .7 ) + 4921 270 . 6 )( s j 423 . 7 )( s + 39 .18 - 423 Asymptotes: K L > 0: 90 , Intersect of Asymptotes: 1 = -117 .23 - 4882 .8 + 4921 5 .6 + 39 .18 + 39 .18 -3 = -0 . 033 164 Breakaway-point Equation: + 5.279 10 Breakaway points: (RL) 0, -87.576 2020 s 7 7 + 2 .02 10 s 6 10 s 5 + 1. 5977 10 13 s 4 + 1. 8655 10 16 s 3 + 1. 54455 10 18 s 2 =0 8-19 Characteristic Equation: P( s) 3 2 s 3 + 5000 s 2 + 572 , 400 s + 900 s ,000 +JL . 6 )( s 10 s 3 + 50 ,000 .8 ) s 2 = 0 = s + 5000 s + 572 , 400 s + 900 ,000 = ( s + 1. 5945)( Since the pole at -5000 is very close to the zero at -4882.8, P ( s ) and P( s) + 115 + 4882 Q (s) = 10 s (s 2 + 5000 ) Q ( s ) can be approximated as: ( s + 1. 5945)( s + 115 .6 ) 2 Q ( s) 10 .24 s s 2 Breakaway-point Equation: 1200 s + 3775 =0 Breakaway Points: (RL): 0, -3.146 165 8-20 (a) = 12 : P ( s) = s (s 2 + 12 ) o Q (s) 270 = s +1 o Asymptotes: K > 0: 90 , Intersect of Asymptotes: K < 0: 0 0 , o 180 o 1 = Breakaway-point Equation: 2s 3 + 0 - 12 - ( -1 ) 3 2 -1 s = -5 . 5 Breakaway Points: 0, + 15 s + 24 =0 -2.314, -5.186 8-20 (b) = 4 : P( s) =s 2 (s +4) o Q (s) 270 o = s +1 o Asymptotes: K > 0: 90 , Intersect of Asymptotes: K < 0: 0 , 180 o 1 = Breakaway-point Equation: 2 s + 7 s 3 2 0 + 0 - 4 - ( - 10 3 -1 = - 1. 5 +8 s = 0 Breakaway Points: K > 0 0. None for K < 0. 166 (c) Breakaway-point Equation: 2 2s 2 + ( + 3) s + 2 s = 0 Solutions: s =- The +3 4 ( + 3) - 16 2 , s =0 4 For one nonzero breakaway point, the quantity under the square-root sign must equal zero. Thus, - 10 + 9 = 0 , = 1 or = 9 . The a nswer is cancellation in the equivalent G ( s ). When = 9, = 9. the nonzero breakaway point is at s = 1 solution represents pole-zero = - 3. 1 = -4. 8-21 (a) P (s) =s 2 (s + 3) Q ( s) = s + 3 Breakaway-point Equation: 2 s + 3 (1 + ) s + 6 = 0 The roots of the breakaway-point equation are: s = -3 ( 1 + ) 4 9( 1 + ) - 48 2 4 167 For no breakaway point other than at s = 0 , set 9( 1 + ) 2 - 48 < 0 or - 0 . 333 < < 3 Root Locus Diagram with No Breakaway Point other than at s = 0. 8-21 (b) One breakaway point other than at s = 0: = 0 . 333 , Breaka way po int at s = -1. 168 8-21 (d) Two breakaway points : > 3: 169 8-22 Let the angle of the vector drawn from the zero at s be = j12 to a point s on the root locuss near the zero 1 . 1 = 2 = 3 = 4 = Let angle angle angle angle of th e vect or dra wn fro m the of th e vect or dra wn fro m the pole a t j 10 to s . 1 pole a t 0 to s . 1 of th e vect or dra wn fro m the of th e vect or dra wn fro m the pole a t zero a t - - j 10 to s1 . j 12 to s . 1 Then the angle conditions on the root loci are: = 1 - 2 - 3 + 4 = 1 = 2 = 3 = 4 = 90 o odd m ultipl es of Thus, 180 = 0 o o The root loci shown in (b) are the correct ones. 170 Chapter 9 FREQUENCY DOMAIN ANALYSIS 9-1 (a) K=5 n = n = 5 = 2 .24 rad / sec = 4.48 6.54 = 1.46 = 0 . 707 M r =1 = 2 r = 0 1 1 rad / sec (b) K = 21.39 21 . 39 = 4. 62 1- rad / sec = 9.24 6.54 M r =1 - 2 r = n (c) 9-2 (a) (b) (c) (d) (e) (f) (g) (h) 9-3 Maximu M m over shoot 1 2 1- 2 2 = 3 .27 = 20 6.54 rad / sec K = 100 n = 10 ( 9 . 38 dB) rad / sec = 0 . 327 M r = 1. 618 r = 9 .45 ra d / sec M M M M M M r = 2 . 944 = 15 . 34 = 4.17 =1 r = 3 rad r =4 / sec BW = 4.495 rad / sec r ( 23 . 71 dB) rad / sec BW = 6.223 rad / sec r ( 12 .4 dB) r = 6 .25 ra rad / sec d / sec BW = 9.18 r ad / sec r ( 0 dB) r =0 BW = 0.46 r ad / sec r = 1. 57 ( 3 . 918 dB) r = 0 .82 r = 1. 5 rad r = 1.25 ra rad / sec BW = 1.12 r ad / sec r = ( unstab = 3 . 09 = 4.12 le) / sec BW = 2.44 r ad / sec M M r ( 9 . 8 dB) d / sec BW = 2.07 r ad / sec r ( 12 . 3 dB) r = 3 . 5 rad / sec BW = 5.16 r ad / sec = 0.1 t T hus, = 0.59 r = = 1. 05 n = 17 . 7 n r = 1 - 0 .416 + 2 . 917 n M 2 = 0 .1 sec Thus, minimum rad / sec Maxi mum r = 1. 05 = 20.56 rad/sec Minimum BW = ( ( 1 - 2 2 )+ 4 - 4 + 2 4 2 ) 1/2 9-4 Maximu m over shoot = 0.2 Thus, 0 .2 - =e 1 - 2 = 0.456 171 M r = 2 1 1- 2 = 1.232 t r = 1 - 0 .416 + 2 . 917 n 2 = 0 .2 Thus, minimum n = 14. 168 rad/sec Maximum M r = 1.232 Minimum BW = ( (1 - 2 =e 2 )+ 2 4 - 4 + 2 4 2 ) 1/2 = 18.7 rad/sec 9-5 Maximum overshoot = 0.3 1 2 1- 2 - Thus, 0 . 3 1 - 0 .416 1 - = 0 . 358 2 M r = = 1.496 t r = + 2 . 917 n = 0 .2 Thus, minimum n = 6 .1246 rad/sec Maximum M r = 1.496 Minimum BW = ( (1 - 2 2 )+ 4 - 4 + 2 4 2 ) 2 1/2 = 1.4106 rad/sec 9-6 M - r = 1.4 = 2 1 Thus, 1- 2 = 0.387 Maximum overshoot = e 1 - = 0.2675 (26.75%) r = 3 rad t / sec = n = 2 1 - 2 2 = 0 .8367 n rad/sec n = 3 0 . 8367 = 3 . 586 r ad/sec max = n 1 3 . 586 1 - - ( 0 . 387 = 0 .95 ) 2 sec At = 0, M = 0 .9 . This indicates that the steady-state value of the unit-step response is 0.9. Unit-step Response: 9-7 T BW (rad/sec) Mr ________________________________________________________________ 0 0.5 1.0 2.0 1.14 1.17 1.26 1.63 1.54 1.09 1.00 1.09 172 3.0 1.96 4.0 2.26 5.0 2.52 _________________________________________________________________ 1.29 1.46 1.63 9-8 T BW (rad/sec) Mr _________________________________________________________________ 0 1.14 0.5 1.00 1.0 0.90 2.0 0.74 3.0 0.63 4.0 0.55 5.0 0.50 _________________________________________________________________ 1.54 2.32 2.65 2.91 3.18 3.37 3.62 9-9 (a) L( s ) = 20 s ( 1 + 0 . 1 s )( 1 + 0 .5 s ) P = 1, P =0 When = 0: L ( j ) = -90 o 20 L( j ) = 2 When = : L ( j ) = -270 o 2 L( j ) =0 =0 L ( j ) = -0.6 + j 1 - 0.05 2 2 ( 2 ) = 20 -0.6 - j 1 - 0.05 0.36 + 4 2 2 (1 - 0.05 ) ( ) 2 Setting Im L ( j ) 1 - 0 .05 = 0 o Thus , = 4.47 rad / sec L ( j 4.47 ) = -1. 667 360 180 o o 11 = 270 = ( Z - 0.5 P - P ) 180 = ( Z - 0.5 )180 o o Thus, Z = =2 The closed-loop system is unstable. The characteristic equation has two roots in the right-half s-plane. Nyquist Plot of L ( j ): 173 (b) L( s ) = 10 s ( 1 + 0 . 1 s )( 1 + 0 .5 s ) Based on the analysis conducted in part (a), the intersect of the negative real axis by the L ( j ) plot is at -0.8333, and the corresponding -P is 4.47 rad/sec. o 11 = -90 o = - Z 0 . 5 P 180 o = 180 Z - 90 Thus, Z = 0 . The closed-loop system is stable. Nyquist Plot of L ( j ): (c) L( s ) = 100( 1 + s ) s ( 1 + 0 . 1 s )( 1 + 0 .2 s )( 1 + 0 . 5 s ) L( j0 ) o P = 1, P = 0. L ( j ) = -270 o When When = 0 : L ( j 0 ) = -90 o = : L ( j ) = -270 = When = : When L ( j ) o =0 L( j ) L( j ) =0 = : L ( j ) = -270 =0 174 L ( j ) = ( 0.01 100(1 + j ) 4 - 0.8 2 ) + j (1 - 0.17 ) 2 4 = 100(1 + j ) 0.01 - 0.8 4 ( 0.01 ( 4 - 0.8 2 ) - j (1 - 0.17 ) ) + ( 1 - 0.17 ) 2 2 2 2 2 2 Setting Im L ( j ) Thus, =0 0 .01 - 0 .8 2 - 1 + 0 .17 2 = 0 4 - 63 2 - 100 = 0 = 64. 55 2 = 8 . 03 rad/sec 100 ( 0.01 4 - 0.8 2 ) + 2 (1 - 0.17 2 ) L ( j 8.03) = = - 10 ( 0.01 2 - 0.8 2 )2 + 2 (1 - 0.17 2 ) 2 =8.03 11 = 270 = ( Z - 0.5 P - P ) 180 = ( Z - 0.5 )180 o o o Thus, Z = 2 The closed-loop system is unstable. The characteristic equation has two roots in the right-half s-plane. Nyquist Plot of L ( j ): (d) L( s ) = 10 s ( 1 + 0 .2 s )( 1 + 0 . 5 s ) 2 P =2 P =0 When = 0: L ( j ) = -180 o 10 4 L ( j ) = 4 2 When 3 = : L ( j ) = -360 o L ( j ) =0 L ( j ) = ( 0.1 - 2 ) - j 0.7 3 = 10 0.1 - + j0.7 ( 0.1 ( 4 - 2 ) ) 2 + 0.49 6 Setting Im L ( j ) origin where = . = 0 , = . The Nyquist plot of L ( j ) does not intersect the real axis except at the 175 11 = ( Z - 0.5 P - P )180 = ( Z - 1) 180 o o Thus, Z = 2 . The closed-loop system is unstable. The characteristic equation has two roots in the right-half s-plane. Nyquist Plot of L ( j ): 9-9 (e) L(s ) = s s + 3s + 1 3 ( 3( s + 2) ) P =1 P=2 When = 0 : L ( j 0 ) = -90 o 3( j + 2) 4 L( j0 ) = When 4 2 = : L ( j ) = -270 o L ( j) =0 = 0, L ( j ) = ( - 3 2 ) + j or = 3( j + 2) - 3 (4 ( 4 - 3 2 ) ) - j 2 2 + Setting Im L ( j ) - 3 - 2 = 0 4 2 = 3 . 56 2 o = 1.89 o rad/sec. L ( j 1.89 ) o =3 11 = ( Z - 0.5 P - P )180 = ( Z - 2.5 ) 180 = - 90 Thus, Z = 2 The closed-loop system is unstable. The characteristic equation has two roots in the right-half s-plane. Nyquist Plot of L ( j ): 176 9-9 (f) L(s ) = 0.1 s ( s + 1) s + s + 1 2 ( ) P =1 L( j0 ) P=0 When = 0 : L ( j 0 ) = -90 o 0.1 = When = : L ( j ) = -360 o 2 L ( j) =0 =0 L ( j ) = = or ( 4 - 2 2 2 ) + j (1 - 2 ) 2 = 0.1 - 2 4 ( o ( 4 - 2 2 ) - j (1 - 2 ) ) + ( 1 - 2 ) 2 2 2 2 2 Settiing Im L ( j ) = 0 .5 = 0 . 707 o rad / sec L ( j 0 . 707 ) o = -0 .1333 The closed-loop system is stable. 11 = ( Z - 0.5 P - P )180 = ( Z - 0.5 ) 180 = - 90 Nyquist Plot of L ( j ): Thus, Z = 0 9-9 (g) L(s ) = 100 s ( s + 1) s + 2 2 ( ) P =3 P=0 177 When = 0 : L ( j 0 ) = -90 o o L( j0 ) = rad/sec. o When = : L ( j ) = -360 o o o L ( j) =0 o o The phase of L ( j ) is discontinuous at = 1.414 11 = 35.27 + 270 - 215.27 o ( o ) = 90 11 = ( Z - 1.5) 180 = 90 Thus, P 11 = 360 180 = 2 The closed-loop system is unstable. The characteristic equation has two roots in the right-half s-plane. Nyquist Plot of L ( j ): 9-9 (h) L( s ) = 10( s s(s + 10 ) ) + 1 )( s + 100 P =1 P =0 When = 0 : L ( j 0 ) = -90 o L ( j ) = L( j0 ) = = When = : L ( j ) = -180 o 2 L ( j) =0 10( j + 10) -101 + j (100 - ) 2 2 10( j + 10) - 101 - j 100 - 10201 + 4 2 (100 - ) 2 ( 2 ) 2 Setting Im L ( j ) = 0, = 0 is the only solution. Thus, the Nyquist plot of L ( j ) does not intersect the real axis, except at the origin. 11 = ( Z - 0.5 P - P )180 = ( Z - 0.5 ) 180 = - 90 o o o Thus, Z = 0 . The closed-loop system is stable. Nyquist Plot of L ( j ): 178 9-10 (a) L( s ) = s(s K + 2 )( s + 10 ) P =1 o P =0 o For stability, Z = 0. 11 = -0 . 5 P 180 = -90 This means that the (-1, j0) point must not be enclosed by the Nyquist plot, or 0 < 0 .004167 K Nyquist Plot of L ( j ): < 1. Thus, 0 < K < 240 9-10 (b) L( s ) = s(s K(s + 1) + 2 )( s + 5)( s + 15) 11 = -0 . 5 180 0 o P =1 o P =0 For stability, Z = 0. = -90 9K This means that the (-1, j0) must not be enclosed Thus, by the Nyquist plot, or < 0 . 000517 <1 0 < K < 1930.9 Nyquist Plot of L ( j ): 179 9-10 (c) L ( j ) = K s (s 2 + 2 )( s + 10 ) P =2 P o =0 = -180 o For stability, Z = 0. 11 = -0 . 5 P 180 For all K > 0 11 = 180 o , not o - 180 for o . Thus, the system is unstable for all K > 0. For K < 0, the critical point is (1, j0), all K < 0. Thus, the system is unstable for all values of K. 11 = 0 Nyquist Plot of L ( j ): 9-11 (a) G (s) = K (s + 5) 2 P =0 P =0 180 G ( j 0 ) = 0 G ( j ) o (K o > 0) G ( 0 ) = 180 G ( j ) = 0 o (K <0) G ( j0) G ( j ) = K 25 = -180 (K > 0) o (K < 0) =0 11 = - ( 0.5 P + P ) 180 = 0 o For stability, Z = 0. o 0 K < K < 11 = 0 o Stable o < -25 K 11 = 180 < 0 11 = 0 o Unstable - 25 < Stable The system is stable for - 25 < K < . 9-11 (b) G (s) = (s K + 5) 3 P =0 P =0 K 125 G ( j 0 ) = 0 G ( j ) o (K o > 0) G ( 0 ) = 180 o (K o < 0) G ( j0 ) G ( j ) = =0 = -270 (K > 0) G ( j ) = 270 (K < 0) For stability, Z = 0. 11 = - ( 0.5 P + P ) 180 = 0 o o 0 < K < 1000 K K 11 = 0 o o o Stable Unstable Unstable Stable > 1000 < - 125 11 = 360 11 = 180 11 = 0 o -125 < K < 0 The system is stable for - 125 < K < 0. 9-11 (c) G (s) = K (s + 5) 4 P =P =0 K 625 G ( j 0 ) = 0 G ( j ) o (K > 0) G ( 0 ) = 180 o (K o < 0) G ( j0 ) G ( j ) = =0 =0 o (K > 0) G ( j ) = 180 (K < 0) 181 For stability, Z = 0. 11 = - ( 0.5 P + P ) 180 = 0 o o 0 K < K < 2500 11 = 0 o o o Stable Unstable Unstable Stable > 2500 < - 625 11 = 360 11 = 180 11 = 0 o K -625 < K < 0 The system is stable for - 625 < K < 2500. 9-12 s s + 2s + s +1 + K s + s + 1 = 0 3 2 2 ( Leq ( s) = K s + s +1 2 s s + 2s + s + 1 3 2 ( ( ) ( ) 2 ) Leq ( j ) = K ( eq (1 - 2 4 - ) + j = K - ( ) + j (1 - 2 ) ( 2 ) P =1 P =0 6 L eq ( j 0 ) = - 90 4 o L eq ( j ) 2 = 0 180 o + - 4 4 2 ) - j ( - 2 + 1) ) + (1 - 2 ) 2 2 2 2 Setting Im L ( j ) =0 4 - 2 2 + 1 = 0 Thus, = 1 = -K rad/sec are the real solutions. L eq ( j 1) For stability, 11 = - ( 0 . 5 P + P )180 o = - 90 o When K = 1 the system is marginally stable. K K >0 <0 11 = -90 11 = +90 o o Stable Unstable Routh Tabulation 182 s s s 4 1 2 K K K +1 +1 K K 3 2 +1 2 K > -1 s 1 K 2 - 2K +1 K +1 = (K K - 1) +1 2 s 0 K 1 K>0 2 When K = 1 the coefficients of the s row are all zero. The auxiliary equation is s + 1 = 0 The solutions are = 1 rad/sec. Thus the Nyquist plot of L eq ( j ) intersects the -1 point when K = 1, when = 1 rad/sec. The system is stable for 0 < K < , except at K = 1. s 0 9-13 Parabolic error constant K a = lim s G ( s ) = lim10 ( K P + K D s ) = 10 K P = 100 2 s 0 Thus K P = 10 Characteristic Equation: 10 K s 2 s 2 + 10 P K D s + 100 = 0 G eq ( s) = D s + 100 P =2 =0 For stability, 11 = - ( 0 . 5 P + P )180 o = -180 D o The system is stable for 0 < K < . 9-14 (a) 1 The characteristic equation is G (s) 2 +G ( s) -G ( s) - 2 = 1- 2 2 G (s) 2 =0 2 G eq ( s ) = -2 G (s) = -2 K (s + 4) + G 2 (s + 5) 2 2 P =0 = P =0 2 G eq ( j ) = = ( 400 - 120 K 2 -2 K 2 4 -2 K ) + j ( 360 - 18 ) 2 ( 400 - 120 eq ( 400 - 120 2 + 2 ) - j ( 360 - 18 2 ) 2 + =0 2 ) + ( 360 - 18 ) 2 2 2 G eq ( j0) 180 o 200 eq ( j ) = 0 180 o Setting Im G ( j ) 183 =0 and = 4.47 r ad / sec G eq ( j 4.47 ) = K 2 800 For stability, 11 = - ( 0.5 P + P ) 180 = 0 o o The system is stable for K or K 2 < 200 < 200 Characteristic Equation: s 4 + 18 s 3 + 121 s s 4 s 2 + 360 s + 400 - 2 K 2 =0 - 2K 2 Routh Tabulation 1 18 101 29160 - 36 K 101 s 0 2 121 360 400 400 3 s s 2 -2K 2 1 29160 + 36 K 2 >0 400 -2K 2 K 2 < 200 Thus for stability, K < 200 9-15 (a) G (s) = 83 . 33 N s(s Nyquist Plot ) + 2 )( s + 11 . 767 For stability, N < 3.89 Thus N < 3 since N must be an integer. (b) G (s) = s ( 0 . 06 s 2500 + 0 . 706 )( As + 100 3 ) Characteristic Equation: 0 . 06 As + ( 6 + 0 . 706 A )s 2 + 70 . 6 s + 2500 = 0 184 G eq (s) = As 6s 2 2 ( 0 . 06 s + 0 . 706 ) + 70 . 6 s + 2500 Since G eq ( s ) has more zeros than poles, we should sketch the Nyquist plot of 1 / G eq ( s ) for stability study. G eq ( 2500 - 6 ) + j70.6 = ( 2500 - 6 ) + j70.6 ( -0.706 + j 0.06 ) = ( j ) A ( -0.706 - j0.06 ) A ( 0.498 + 0.0036 ) 1 2 2 2 3 2 3 4 6 1 / G eq ( j 0 ) = - 180 2 o 1 / G eq ( j ) = 0 - 90 rad/sec o Setting 1 G eq Im -4.23 A =0 G eq ( j ) 1 100 .156 For stability, - 0 . 36 = 0 = 16 . 68 ( j16 . 68 ) = 11 = - ( 0.5 P + P ) 180 = -180 o o For A > 4.23 < A 11 = 180 o Unstable o For 0 < 4.23 11 = -180 Stable The system is stable for 0 < A < 4.23. (c) G (s) = 2500 s ( 0 . 06 s + 0 . 706 )( 50 s +Ko) s ( 0 . 06 s ) Characteristic Equation: K s ( 0 . 06 s o + 0 . 706 P )( 50 s + K o ) + 2500 = 0 G eq ( j 0 ) G eq ( s) = + 0 . 706 2 3s 3 + 35 . 3 s + 2500 3 P =0 =0 2 = 0 90 o G eq ( j ) = 0 - 90 3 o G eq ( j ) = ( 2500 - 35.3 ) - j3 2 K o ( -0.06 + 0.706 j ) 3 = K o - 0.06 + 0.706 j ( ) ( 2500 - 35.5 ) + j 3 ( 2500 - 35.3 ) + 9 2 2 2 6 Setting Im G G eq ( j 7 .18 ) eq ( j ) =0 Ko + 138 4 .45 2 - 9805 . 55 =0 = 51 . 6 2 = 7 .18 rad/sec = -0 . 004 185 For stability, 11 = - ( 0.5 P + P ) 180 = 0 o o For stability, 0 < K o < 217 .4 9-16 (a) K t = 0: G ( s) = Y ( s) E (s ) = s s + 10 s + 10000 Kt 2 ( 10000 K ) = 10000 K s ( s + 10) 2 The (-1, j0) point is enclosed for all values of K. The system is unstable for all values of K. (b) K t = 0 . 01 : G ( s) = s s + 10 s + 100 2 ( 10000 K ) G( j ) = 10000 K - 10 - j 100 - 2 100 + 4 2 (100 - ) 2 ( 2 ) 2 186 Setting Im G ( j ) = 0 2 = 100 G ( j 10 ) = 10 rad / sec = -10 K The system is stable for 0 < K < 0.1 (c) K t = 0 .1: G ( s) = s s + 10 s + 1000 2 ( 10000K ) G ( j ) = 10000 K -10 - j 1000 - 2 100 + 4 2 (1000 - ) 2 ( 2 ) 2 Setting Im G ( j ) = 0 2 = 100 = 31 . 6 rad/sec G ( j 31 . 6 ) = -K For stability, 0 < K < 1 9-17 The characteristic equation for K = 10 is: s G eq ( s ) 3 + 10 s 2 + 10 ,000 =0 K s t + 100 ,000 =0 = 10 ,000 K t s s 3 + 10 s 2 + 100 ,000 P P =2 4 G eq ( j ) = 10,000 K t j 100,000 - 10 - j 2 3 = 10,000 K t - + j 10,000 - 10 (10,000 - 10 ) 2 ( 2 ) 2 + Setting Im G 6 eq ( j ) =0 187 = 0 , = 10 ,000 2 = 100 For stability, rad / sec G eq ( j 100 ) = -Kt 11 = - ( 0.5 P + P ) 180 = -360 o o The system is stable for K > 0. t 9-18 (a) Let G ( s ) = G 1 ( s )e 100 2 - Td s Then G1 ( s ) = s s + 10 s + 100 100 ( ) 2 Let - 10 + j 100 - 2 ( ) 2 =1 or 100 Thus 100 + 4 -1 2 The real solution for are = 1 rad/sec. (100 - ) 2 100 4 + 2 (100 - 2 )2 6 4 1/2 =1 = 10,000 - 100 + 10,000 - 10,000 = 0 2 G1 ( j 1) = - tan Td Equating =1 ( 264.23 = = 84.23 180 100 - 2 o -10 = 264.23 =1 o - 180 o ) 180 = 1.47 rad Thus the maximum time delay for stability is T d = 1.47 sec. (b) T d =1 sec. G ( s) = s s + 10 s + 100 2 ( 100 Ke -s ) G( j ) = rad/sec. 100 Ke 2 - j 2 - 10 + j 100 - At the intersect on the negative real axis, = 1.42 ( ) 188 G ( j 1.42 ) = -0 . 7107 K. The system is stable for 0 < K < 1.407 9-19 (a) K = 0.1 G ( s) = s s + 10 s + 100 2 ( 10 e - Tds ) = G1 (s )e =1 -Td s 10 Let 2 - 10 + j 100 - 6 4 ( 2 ) or 2 10 100 4 + 2 (100 - 2 ) 2 The real solutions for 1/2 =1 Thus - 100 + 10 ,000 - 100 = 0 -1 is = 0 .1 rad/sec. G1 ( j 0.1) = - tan Td = Equate ( 269.43 100 - 2 o = 269.43 -10 =0.1 o - 180 o o ) =0.1 = 1.56 rad We have T 180 d = 15 . 6 sec. We have the maximum time delay for stability is 15.6 sec. 9-19 (b) T d = 0 .1 sec. G ( s) = s s + 10 s + 100 2 ( 100 Ke -0.1 s ) G( j ) = 100 Ke 2 -0.1 j 2 - 10 + j 100 - ( ) At the intersect on the negative real axis, = 6 . 76 rad/sec. G ( j 6 . 76 ) = -0 .1706 K 189 The system is stable for 0 < K < 5.86 9-20 (a) The transfer function (gain) for the sensor-amplifier combination is 10 V/0.1 in = 100 V/in. The velocity of flow of the solution is v = 10 in 3 / sec = 100 d in/sec / v sec. The loop transfer function is 0 .1 in The time delay between the valve and the sensor is T G (s) =D = 100 Ke s 2 - Td s + 10 s + 100 190 191 9-21 (a) The transfer function (gain) for the sensor-amplifier combination is 1 V/0.1 in = 10 V/in. The velocity of flow of the solutions is v = 10 in 3 The time delay between the valve and sensor is T G (s) d = 100 in / sec 0 .1 in = D / v sec. The loop transfer function is -T d s / sec = 10 Ke s 2 + 10 s + 100 - j Td (b) K = 10: G ( s) = G1 (s ) e 100 2 - Td s G ( j ) = (100 - ) + j10 2 2 2 100 e Setting ( 100 - ) + j10 =1 (100 - ) + 100 = 10,000 2 Thus, - 100 = 0 4 2 -1 Real s olutio ns: = 0, = 10 rad / sec G1 ( j 10) = - tan Thus, 10 T Thus, T d d 10 = - 90 o 100 - 2 =10 o = 20 90 o = 2 rad 180 = = = 0 .157 sec Maximum D vT d = 100 0 .157 = 15 . 7 in (c) D = 10 in. T d v 100 s + 10 s + 100 The Nyquist plot of G ( j ) intersects the negative real axis at = 12 . 09 rad/sec. G ( j ) For stability, the maximum value of K is 12.94 . 2 = D = 10 = 0 .1 sec G (s) = 10 Ke -0 . 1 s = -0 . 0773 K 9-22 (a) 192 G H ( s) = s s + 6 s + 12 2 ( 8 ) rad / sec, G L ( s) = 2 . 31 s( s + 2 . 936 M ) M r = 1, r = 0 BW = 1.02 r ad / sec r = 1, r = 0 rad / sec, BW = 1.03 r ad / sec (b) G H ( s) = s 1 + 0.5455 s + 0.0455s rad / sec, ( 0.909 2 ) G L ( s) = 0 . 995 s ( 1 + 0 .4975 s ) M r = 1, r = 0 BW = 1.4 ra d / sec M r = 1, r = 0 rad / sec, BW = 1.41 r ad / sec (c) G H ( s) = s 1 + 0.75 s + 0.25 s rad / sec, ( 0.5 2 ) G L ( s) = 0 . 707 s ( 1 + 0 . 3536 s ) M r = 1, r = 0 BW = 0.87 r ad / sec M r = 1, r = 0 rad / sec, BW = 0.91 r ad / sec (d) G H ( s) = s 1 + 0.00283 s + 8.3056 10 s rad / sec, BW = 119.74 ( 90.3 -7 2 ) G L (s) = s (1 92 . 94 + 0 . 002594 s) M r = 1, r = 0 rad / sec M r = 1, r = 0 rad / sec B W = 118.76 rad / sec (e) G H ( s) = s 1 + 0.00283 s + 8.3056 10 s rad / sec, BW = 270.55 ( 180.6 -7 2 ) G L (s) = 189 . 54 s (1 + 0 . 002644 s) M r = 1, r = 0 rad / sec M r = 1, r = 0 rad / sec, BW = 268.11 rad / sec (f) G H ( s) = s 1 + 0.00283 s + 8.3056 10 s . 67 rad / sec ( 1245.52 -7 2 ) G L (s) = 2617 . 56 s (1 + 0 . 0053 s) M r = 2 . 96 , r = 666 M r = 3 . 74, r = 700 rad / sec BW = 1054.4 1 rad / sec BW = 1128.7 4 rad / sec 9-23 (a) (b) M r = 2 . 06 , r = 9 . 33 rad / sec, BW = 15.2 r ad / sec 193 M r = 2 1 - 2 1 1- 2 2 = 2 . 06 - + 0 . 0589 = 0 4 2 The solution for < 0 . 707 is = 0.25. r = 9 . 33 rad n 2 / sec Thus n = 9 . 33 0 .9354 = 9 . 974 rad / sec GL ( s) = 9-24 (a) (b) M s ( s + 2 n ) = 99.48 s( s + 4.987) = 19.94 s(1 + 0.2005 s) BW = 15.21 rad/sec r = 2 . 96 , r = 666 = 2 1 1- 2 2 . 67 rad / sec, BW = 1054.4 1 rad / sec M r = 2 . 96 - + 0 . 0285 = 0 4 2 The solution for < 0 . 707 is = 0.1715 r 1 - 2 = 666 n 2 . 67 rad / sec Thus n = 666 . 67 0 . 97 = 687 .19 rad / sec GL ( s) = 9-25 (a) G (s) s ( s + 2 n ) = 472227.43 s( s + 235.7) = 2003.5 s(1 + 0.00424 s) BW = 1079.28 rad/sec = 5 s ( 1 + 0 . 5 s )( 1 + 0 .1 s ) 9-25 (b) G (s) = 10 s ( 1 + 0 . 5 s )( 1 + 0 .1 s ) 194 (c) G (s) = 500 (s + 1.2 )( s + 4 )( s + 10 ) (d) G (s) = s(s 10( s + 1) ) + 2 )( s + 10 9-25 (e) 195 G ( s) = s s + s +1 2 ( 0.5 ) (f) G ( s) = 100 e 2 -s s s + 10 s + 50 ( ) (g) G ( s) = 100 e 2 -s s s + 10 s + 100 ( ) 9-25 (h) 196 G ( s) = s s + 5s + 5 2 ( 10( s + 5) ) 9-26 (a) G (s) = K s ( 1 + 0 .1 s )( 1 + 0 . 5 s ) The Bode plot is done with K = 1. GM = 21.58 dB For GM = 20 dB, K must be reduced by -1.58 dB. Thus K = 0.8337 PM = 60 .42 . For PM = 45 K should be increased by 5.6 dB. Or, K = 1.91 o o (b) G (s) s ( 1 + 0 .1 s )( 1 + 0 .2 s )( 1 The Bode plot is done with K = 1. GM = 19.98 dB. For GM = 20 dB, K 1. PM = 86 . 9 . For PM = 45 K should be increased by 8.9 dB. Or, K = 2.79. o o = K(s + 1) + 0 .5 s ) 9-26 (c) G ( s) = (s K + 3) 3 The Bode plot is done with K = 1. GM = 46.69 dB PM = infinity. 197 For GM = 20 dB K can be increased by 26.69 dB or K = 21.6. For PM = 45 deg. K can be increased by 28.71 dB, or K = 27.26. (d) G (s) = K (s + 3) 4 The Bode plot is done with K = 1. GM = 50.21 dB PM = infinity. For GM = 20 dB K can be increased by 30.21 dB or K = 32.4 For PM = 45 deg. K can be increased by 38.24 dB, or K = 81.66 (e) G ( s) = Ke -s 2 s 1 + 0.1s + 0.01s ( ) The Bode plot is done with K = 1. GM=2.97 dB PM = 26.58 deg For GM = 20 dB K must be decreased by -17.03 dB or K = 0.141. For PM = 45 deg. K must be decreased by -2.92 dB or K = 0.71. 9-26 (f) G ( s) = K (1 + 0.5s ) s s + s +1 2 ( ) The Bode plot is done with K = 1. GM = 6.26 dB PM = 22.24 deg For GM = 20 dB K must be decreased by - 13.74 dB or K = 0.2055. For PM = 45 deg K must be decreased by - 3.55 dB or K = 0.665. 9-27 (a) 198 G (s) = 10 K s ( 1 + 0 .1 s )( 1 + 0 . 5 s ) The gain-phase plot is done with K = 1. GM = 1.58 dB PM = 3.95 deg. For GM = 10 dB, K must be decreased by - 8.42 dB or K = 0.38. For PM = 45 deg, K must be decreased by -14 dB, or K = 0.2. For M = 1.2 , K must be decreased to 0.16. r (b) G (s) = 5K ( s + 1) + 0 .5 s ) s ( 1 + 0 .1 s )( 1 + 0 .2 s )( 1 The Gain-phase plot is done with K = 1. GM = 6 dB PM = 22.31 deg. For GM = 10 dB, K must be decreased by -4 dB or K = 0.631. For PM = 45 deg, K must be decrease by -5 dB. For M = 1.2 , K must be decreased to 0.48. r 9-27 (c) G ( s) = 10 K 2 s 1 + 0.1s + 0.01s ( ) The gain-phase plot is done for K = 1. GM = 0 dB M = r PM = 0 deg For GM = 10 dB, K must be decreased by -10 dB or K = 0.316. For PM = 45 deg, K must be decreased by - 5.3 dB, or K = 0.543. For M = 1.2 , K must be decreased to r 0.2213. (d) 199 G ( s) = s 1 + 0.1s + 0.01s ( Ke -s 2 ) The gain-phase plot is done for K = 1. GM = 2.97 dB M = 3 . 09 r PM = 26.58 deg For GM = 10 dB, K must be decreased by - 7.03 dB, K = 0.445. For PM = 45 deg, K must be decreased by - 2.92 dB, or K = 0.71. For M = 1.2 , K = 0 . 61 . r 9-28 (a) rad/sec Gain crossover frequency = 2.09 PM = 115.85 deg Phase crossover frequency = 20.31 rad/sec Gain crossover frequency = 6.63 rad/sec Phase crossover frequency = 20.31 rad/sec Gain crossover frequency = 19.1 rad/sec Phase crossover frequency = 20.31 rad/sec GM = 21.13 dB PM = 72.08 deg GM = 15.11 dB PM = 4.07 deg GM = 1.13 dB (b) (c) (d) (e) For GM = 40 dB, reduce gain by (40 - 21.13) dB = 18.7 dB, or gain = 0.116 nominal value. For PM = 45 deg, the magntude curve reads -10 dB. This means that the loop gain can be increased by 10 dB from the nominal value. Or gain = 3.16 nominal value. The system is type 1, since the slope of G ( j ) is GM = 12.7 dB. PM = 109.85 deg. The gain crossover frequency is 2.09 rad/sec. The phase margin is 115.85 deg. Set (f) (g) (h) -20 dB/decade as 0. T d = 2 . 09 T d = Thus, the maximum time delay is T d 115 .85 180 o o = 2 .022 rad = 0 .9674 sec. 9-29 (a) The gain is increased to four times its nominal value. The magnitude curve is raised by 12.04 dB. Gain crossover frequency = 10 rad/sec PM = 46 deg Phase crossover frequency = 20.31 rad/sec GM = 9.09 dB (b) (c) (d) The GM that corresponds to the nominal gain is 21.13 dB. To change the GM to 20 dB we need to increase the gain by 1.13 dB, or 1.139 times the nominal gain. The GM is 21.13 dB. The forward-path gain for stability is 21.13 dB, or 11.39. The PM for the nominal gain is 115.85 deg. For PM = 60 deg, the gain crossover frequency must be moved to approximately 8.5 rad/sec, at which point the gain is -10 dB. Thus, the gain must be increased by 10 dB, or by a factor of 3.162. With the gain at twice its nominal value, the system is stable. Since the system is type 1, the steady-state error due to a step input is 0. (e) 200 (f) (g) With the gain at 20 times its nominal value, the system is unstable. Thus the steady-state error would be infinite. With a pure time delay of 0.1 sec, the magnitude curve is not changed, but the the phase curve is subject to a negative phase of - 0 .1 rad. The PM is .85 PM = 115 - 0 .1 gain c rossov er fre quency = 115.85 - 0 .209 = 115 . 64 deg The new phase crossover frequency is approximately 9 rad/sec, where the original phase curve is reduced by -0.9 rad or -51.5 deg. The magnitude of the gain curve at this frequency is -10 dB. Thus, the gain margin is 10 dB. (h) When the gain is set at 10 times its nominal value, the magnitude curve is raised by 20 dB. The new gain crossover frequency is approximately 17 rad/sec. The phase at this frequency is Thus, setting -30 deg. sec. T d = 17 T d = 30 o o = 0 . 5236 Thus T 180 d = 0 .0308 9-30 (a) Bode Plot: 201 For stability: 166 (44.4 dB) < K < 7079 (77 dB) Phase crossover frequencies: 7 rad/sec and 85 rad/sec Nyquist Plot: 9-30 (b) Root Loci. 202 9-31 (a) Nquist Plot 203 9-31 Bode Plot 9-31 (b) Root Loci 9-32 Bode Diagram 204 When K = 1, GM = 68.75 dB, PM = 90 deg. The critical value of K for stability is 2738. 9-33 (a) Forward-path transfer function: 205 G (s) where = L ( s ) E (s) = K G (s) a p = K K ( Bs a i +K ) o o = 0.12 s ( s + 0.0325 ) s + 2.5675 s + 6667 2 ( 2 ) = s 0.12 s + 0.312 s + 80.05 s + 26 3 ( ) ) G ( s) = (b) Bode Diagram: s s + 2.6 s + 667.12 s + 216.67 3 2 ( 43.33( s + 500) Gain crossover frequency = 5.85 rad/sec Phase crossover frequency = 11.81 rad/sec PM = 2.65 deg. GM = 10.51 dB 9-33 (c) Closed-loop Frequency Response: 206 M r = 17 . 72 , r = 5 . 75 rad / sec, BW = 9.53 r ad / sec 9-34 (a) When K = 1, the gain crossover frequency is 8 rad/sec. (b) When K = 1, the phase crossover frequency is 20 rad/sec. (c) (d) (e) (f) (g) (h) (i) When K = 1, GM = 10 dB. When K = 1, PM = 57 deg. When K = 1, M When K = 1, r = 1.2. rad/sec. r = 3 When K = 1, BW = 15 rad/sec. When K = -10 dB (0.316), GM = 20 dB When K = 10 dB (3.16), the system is marginally stable. The frequency of oscillation is 20 rad/sec. 207 (j) 9-35 The system is type 1, since the gain-phase plot of G ( j ) approaches infinity at -90 deg. Thus, the steady-state error due to a unit-step input is zero. When K = 5 dB, the gain-phase plot of G ( j ) is raised by 5 dB. (a) (b) (c) (d) (e) The gain crossover frequency is 10 rad/sec. The phase crossover frequency is 20 rad/sec. GM = 5 dB. PM = 34.5 deg. M r =2 rad/sec (f) r = 15 (g) (h) 9-36 (a) BW = 30 rad/sec When K = -30 dB, the GM is 40 dB. d The phase margin with K = 1 and T produces a phase lag of Thus, =0 sec is approximately 57 deg. For a PM of 40 deg, the time delay -17 deg. o The gain crossover frequency is 8 rad/sec. T d = 17 T = 17 o o = 0 .2967 rad / sec T hus = 8 rad / sec 180 d = 0 .2967 8 = 0 . 0371 sec (b) With K = 1, for marginal stability, the time delay must produce a phase lag of Thus, at = 8 rad/sec, -57 deg. T T d = 57 9-37 (a) =0 o = 57 o o = 0 . 9948 rad 180 d = 0 . 9948 8 = 0 .1244 sec The phase margin with K = 5 dB and T delay must produce a phase lag of d is approximately 34.5 deg. For a PM of 30 deg, the time The gain crossover frequency is 10 rad/sec. Thus, -4.5 deg. T d = 4. 5 = (b) o 4. 5 o o = 0 . 0785 rad Thus T 180 d = 0 . 0785 10 = 0 .00785 sec With K = 5 dB, for marginal stability, the time delay must produce a phase lag of Thus at -34.5 deg. = 10 rad/sec, T d = 34. 9-38 5 o = 34. 5 180 o o = 0 . 602 rad Thus T d = 0 . 602 10 = 0 .0602 sec For a GM of 5 dB, the time delay must produce a phase lag of -34.5 deg at = 10 rad rad/sec. Thus, T d = 34. 5 o = 34. 5 180 o o = 0 . 602 Thus T d = 0 . 602 10 = 0 .0602 sec 208 9-39 (a) Forward-path Transfer Function: G (s) = Y (s) E ( s) = e -2 s ( 1 + 10 s )( 1 + 25 s) From the Bode diagram, phase crossover frequency = 0.21 rad/sec GM = 21.55 dB gain crossover frequency = 0 rad/sec PM = infinite (b) G ( s) = (1 + 10s )( 1 + 25s ) ( 1 + 2 s + 2 s gain crossover frequency = 0 rad/sec 1 2 ) PM = infinite From the Bode diagram, phase crossover frequency = 0.26 rad/sec GM = 25 dB (c) G ( s) = 1- s (1 + s ) (1 + 10 s )( 1 + 2s ) From the Bode diagram, phase crossover frequency = 0.26 rad/sec GM = 25.44 dB gain crossover frequency = 0 rad/sec PM = infinite 9-39 (continued) Bode diagrams for all three parts. 209 9-40 (a) Forward-path Transfer Function: 210 ( 1 + 10 s )( 1 + 25 s ) From the Bode diagram, phase crossover frequency = 0.37 rad/sec GM = 31.08 dB gain crossover frequency = 0 rad/sec PM = infinite G (s) = e -s (b) G ( s) = (1 + 10s ) ( 1 + 25s ) (1 + s + 0.5s ( 1 - 0 .5 s ) 1 2 ) From the Bode diagram, phase crossover frequency = 0.367 rad/sec gain crossover frequency = 0 rad/sec GM = 30.72 dB PM = infinite (c) G (s) ( 1 + 10 s )( 1 + 25 s )( 1 + 0 .5 s ) From the Bode diagram, phase crossover frequency = 0.3731 rad/sec gain crossover frequency = 0 rad/sec = GM = 31.18 dB PM = infinite 9-41 Sensitivity Plot: 211 S M G max = 17 .15 max = 5 .75 rad/sec 212 Chapter 10 DESIGN OF CONTROL SYSTEMS 10-1 Forward-path Transfer Function: G (s) = 1 M ( s) -M (s) = K s 3 + ( 20 + a ) s + ( 200 + 20 2 a) s + 200 a -K For type 1 system, 200 a Ramp-error constant: -K = 0 K v Thus sG ( s ) K = 200a = lim s 0 = 200 K + 20 a = 200 a 200 + 20 a =5 Thus a = 10 K = 2000 The forward-path transfer function is The controller transfer function is G ( s) = s s + 30 s + 400 2 ( 2000 ) Gc ( s) = G ( s) Gp (s ) = 20 s + 10 s + 100 2 (s ( 2 + 30 s + 400 ) ) The maximum overshoot of the unit-step response is 0 percent. 10-2 Forward-path Transfer Function: G (s) = 1 M ( s) -M (s) = K s 3 + ( 20 + a ) s + ( 200 + 20 2 a) s + 200 a -K For type 1 system, 200 a Ramp-error constant: -K = 0 K v Thus sG ( s ) K = 200a = lim s 0 = 200 K + 20 a = 200 a 200 + 20 a =9 Thus a = 90 K = 18000 The forward-path transfer function is The controller transfer function is G ( s) = s s + 110 s + 2000 2 ( 18000 ) Gc ( s) = G ( s) Gp (s ) = 180 s + 10 s + 100 2 2 ) ( s + 110 s + 2000 ) ( The maximum overshoot of the unit-step response is 4.3 percent. From the expression for the ramp-error constant, we see that as a or K goes to infinity, K Thus the maximum value of K v v approaches 10. that can be realized is 10. The difficulties with very large values of K and a are that a high-gain amplifier is needed and unrealistic circuit parameters are needed for the controller. 10-3 (a) Ramp-error Constant: K v = lim s s 0 1000 ( K P + K D s ) s ( s + 10) 2 = 1000K P 10 = 100 K P = 1000 Thus K P = 10 Characteristic Equation: s + 10 + 1000 K D s + 1000 K P = 0 ( ) n = Thus 1000 K P = K 10000 90 1000 = 100 = 0 . 09 rad/sec 2 n = 10 + 1000 K D = 2 0 . 5 100 = 100 D = 213 10-3 (b) For K v = 1000 and = 0 . 707 K , and from part (a), n = 100 rad/sec, 131 .4 1000 2 n = 10 + 1000 = 1000 and D = 2 0 . 707 100 = 141 .4 Thus K D = = 0 . 1314 (c) For K v K = 1.0 , and from part (a), n = 100 rad/sec, 190 1000 2 n = 10 + 1000 D = 2 1 100 = 200 Thus K D = = 0 . 19 10-4 The ramp-error constant: K v = lim s s 0 1000 ( K P + K D s ) s ( s + 10) = 100 K P = 10,000 1000 (100 + K D s ) s( s + 10) Mr 13.5 1.817 1.291 1.226 1.092 1.06 Thus K P = 100 The forward-path transfer function is: G ( s) = KD 0 0.2 0.4 0.6 0.8 1.0 PM (deg) 1.814 36.58 62.52 75.9 81.92 84.88 GM BW (rad/sec) 493 525 615 753 916 1090 D Max overshoot (%) 46.6 41.1 22 13.3 8.8 6.2 The phase margin increases and the maximum overshoot decreases monotonically as K increases. 10-5 (a) Forward-path Transfer Function: G ( s) = G c ( s ) G p ( s) = 4500 K ( K D + K P s ) s ( s + 361.2) = lim sG ( s ) Ramp Error Constant: K v s0 = 4500 KK 361 .2 P = 12 .458 KK P e ss = 1 K v = 0 . 0802 KK P 0 . 001 Thus KK P 80 .2 Let K P =1 and K = 80 .2 Attributes of Unit-step Response: KD 0 0.0005 0.0010 0.0015 0.0016 0.0017 0.0018 0.0020 Select K 0 . 0017 D tr (sec) 0.00221 0.00242 0.00245 0.0024 0.00239 0.00238 0.00236 0.00233 ts (sec) 0.0166 0.00812 0.00775 0.0065 0.00597 0.00287 0.0029 0.00283 Max Overshoot (%) 37.1 21.5 12.2 6.4 5.6 4.8 4.0 2.8 (b) BW must be less than 850 rad/sec. 214 KD 0.0005 0.0010 0.0015 0.0016 0.0017 0.00175 0.0018 Select K D GM PM (deg) 48.45 62.04 73.5 75.46 77.33 78.22 79.07 D Mr 1.276 1.105 1.033 1.025 1.018 1.015 1.012 BW (rad/sec) 827 812 827 834 842 847 852 . A larger K 0 . 00175 would yield a BW larger than 850 rad/sec. 10-6 The forward-path Transfer Function: N = 20 G ( s) = 200 ( K P + K D s ) s ( s + 1)( s + 10) To stabilize the system, we can reduce the forward-path gain. Since the system is type 1, reducing the gain does not affect the steady-state liquid level to a step input. Let K = 0 . 05 G ( s) = Unit-step Response Attributes: KD 200 ( 0.05 + K D s ) s ( s + 1)( s + 10) ts P (sec) Max Overshoot (%) When K D 0.01 5.159 12.7 0.02 4.57 7.1 0.03 2.35 3.2 0.04 2.526 0.8 0.05 2.721 0 0.06 3.039 0 0.10 4.317 0 = 0 . 05 the rise time is 2.721 sec, and the step response has no overshoot. 10-7 (a) For e ss = 1, K v = lim sG ( s ) = lim s s 0 s 0 200 ( K P + K D s ) s ( s + 1)( s + 10) = 20 K P = 1 Thus K P = 0 . 05 Forward-path Transfer Function: G ( s) = Attributes of Frequency Response: KD 0 0.01 0.02 0.05 0.09 0.10 0.11 0.20 0.30 0.50 200 ( 0.05 + K D s ) s ( s + 1)( s + 10) PM (deg) 47.4 56.11 64.25 84.32 93.80 93.49 92.71 81.49 71.42 58.55 D GM (deg) 20.83 Mr 1.24 1.09 1.02 1.00 1.00 1.00 1.00 1.00 1.00 1.03 BW (rad/sec) 1.32 1.24 1.18 1.12 1.42 1.59 1.80 4.66 7.79 12.36 For maximum phase margin, the value of K is 0.09. PM = 93.80 deg. GM = , M r = 1, 215 and BW = 1.42 rad/sec. (b) Sensitivity Plots: The PD control reduces the peak value of the sensitivity function S M G ( j ) 10-8 (a) Forward-path Transfer Function: 100 K P + 2 G ( s) = K KI s s + 10 s + 100 I For K v = 10, K v = lim sG ( s ) = lim s s 0 s 0 100 ( K P s + K I ) s s + 10 s + 100 2 ( ) = K I = 10 Thus = 10 . s (b) Let the complex roots of the characteristic equation be written as 2 = - + j 15 an d s 2 = - - j 15 . The quadratic portion of the characteristic equation is 3 2 s + 2 s + + 225 = 0 ( ) The characteristic equation of the system is s + 10 s + (100 + 100 KP ) s + 1000 = 0 The quadratic equation must satisfy the characteristic equation. Using long division and solve for zero remainder condition. s + (10 - 2 ) s + 2 s + + 225 s + 10 s + (100 + 100 K P ) s + 1000 2 2 3 2 s + 2 s + + 225 s 3 2 2 ( ) (10 - 2 ) s + 100 KP - - 125 s + 1000 2 2 (10 - 2 ) s 2 ( ) + ( 20 - 4 ) s + (10 - 2 ) ( s 2 2 + 225 ) 216 (100 K For zero remainder, 2 and The real solution of Eq. (1) is 3 P + 3 - 20 - 125 s + 2 - 10 2 3 ) 2 + 450 -1250 (1) (2) - 10 2 + 450 - 1250 = 0 100 K P + 3 - 20 - 125 = 0 2 = 2 .8555 K . From Eq. (2), P = s 125 + 20 - 3 100 2 = 1. 5765 - 2 . 8555 - j 15 , and s The characteristic equation roots are: = - 2 .8555 + j 15 , = -10 + 2 = -4.289 (c) Root Contours: G eq ( s ) = 100 K P s s + 10 s + 100 s + 1000 3 2 = ( s + 10) ( s 100 K P s 2 + 100 ) Root Contours: 217 10-9 (a) Forward-path Transfer Function: 100 K P + 2 G ( s) = KI s s + 10 s + 100 For K v = 10, K v = lim sG ( s ) = lim s s 0 s 0 100 ( K P s + K I ) s s + 10 s + 100 2 ( ) = K I = 10 Thus the forward-path transfer function becomes G ( s) = s 1 + 0.1s + 0.01s ( 10 (1 + 0.1K P s ) 2 ) Attributes of the Frequency Response: KP 0.1 PM (deg) 5.51 GM (dB) 1.21 Mr 10.05 BW (rad/sec) 14.19 218 0.5 0.6 0.7 0.8 0.9 1.0 1.5 1.1 1.2 When K 22.59 25.44 27.70 29.40 30.56 31.25 31.19 31.51 31.40 and K 6.38 8.25 10.77 14.15 20.10 2.24 1.94 1.74 1.88 1.97 2.00 1.81 2.00 1.97 15.81 16.11 16.38 16.62 17.33 18.01 20.43 18.59 19.08 P = 1.1 I = 10 , K v = 10 , the phase margin is 31.51 deg., and is maximum. The corresponding roots of the characteristic equation roots are: -5 .4, - 2 .3 + j 13 .41 , and - 2 .3 - j 13 .41 Referring these roots to the root contours in Problem 10-8(c), the complex roots corresponds to a relative damping ratio that is near optimal. (b) Sensitivity Function: In the present case, the system with the PI controller has a higher maximum value for the sensitivity function. 10-10 (a) Forward-path Transfer Function: G ( s) = For K v 100 ( K P s + K I ) s s + 10 s + 100 2 ( ) 100 ( K P s + K I ) s s + 10 s + 100 2 = 100 , K v = lim sG ( s ) = lim s s 0 3 s 0 ( ) = K I = 100 Thus K I = 100 . (b) The characteristic equation is s + 10 s + (100 + 100 KP ) s + 100 KI = 0 2 Routh Tabulation: 219 s s s s 3 1 10 100 K P 100 + 100 K 10,000 P 2 For stability, 100 K 1 - 900 P - 900 > 0 Thus K P >9 0 0 10,0 00 Root Contours: G eq ( s ) = 100 K s 3 P s s + 10 s 2 + 100 + 10 ,000 = 100 K P s (s + 23 . 65)( s - 6 .825 + j 19 .4 )( s - 6 . 825 - j19 .4 ) (c) K I = 100 G ( s) = 100 ( K P s + 100 ) s s + 10 s + 100 2 ( ) P The following maximum overshoots of the system are computed for various values of K KP ymax 15 1.794 20 1.779 22 1.7788 24 1.7785 25 1.7756 26 1.779 30 1.782 . 40 1.795 100 1.844 1000 1.859 When K P = 25, minimum ymax = 1.7756 10-11 (a) Forward-path Transfer Function: 220 G ( s) = 100 ( K P s + K I ) s s + 10 s + 100 2 ( ) For K v = 100 K 100 I = 10 , K I = 10 (b) Characteristic Equation: Routh Tabulation: s s s s 3 s3 + 10 s2 + 100 K + 1 s + 1000 = 0 P 100 + 100 K 1000 ( ) 1 10 100 K 1000 P P 2 For stability, 1 KP > 0 0 0 Root Contours: G eq (s) = 100 K s 3 P s s + 10 s 2 + 100 + 1000 (c) KP ymax The maximum overshoots of the system for different values of K computed and tabulated below. P ranging from 0.5 to 20 are 0.5 1.393 1.0 1.275 1.6 1.2317 1.7 1.2416 1.8 1.2424 1.9 1.2441 2.0 1.246 3.0 1.28 5.0 1.372 10 1.514 20 1.642 When K P = 1.7, maximum ymax = 1.2416 10-12 Gc (s ) = K P + K D s + where K P KI s K = K Ds + K P s + K I 2 s + K D1K I 2 = (1 + K D1 s ) K P2 + K D K I2 s I = P2 = K D1K P 2 K = K I2 Forward-path Transfer Function: 221 G ( s) = G c ( s ) G p ( s) = Thus K I 100 ( 1 + K D1 s ) ( K P2 s + K I 2 ) s s + 10 s + 100 2 ( ) Kv = lim s0 sG ( s ) = K I 2 = 100 = K I 2 = 100 D1 Consider only the PI controller, (with K =0) Characteristic Equation: Forward-path Transfer Function: G ( s) = For stability, K 100 ( K P 2 s + 100 ) s s + 10 s + 100 2 ( ) = 10 for fast rise time. s + 10 s + (100 + 100 KP 2 ) s + 10,000 = 0 3 2 P2 > 9. Select K P2 G ( s) = When K K D1 1000 ( 1 + K D 1s ) ( s + 10 ) s s + 10 s + 100 2 ( ) = 0 .2 , the rise time and overshoot requirements are satisfied. K D = K D1 K P2 = 0 .2 10 = 2 P = K P2 + K D 1 K I 2 = 10 + 0 .2 100 = 30 G (s) c = 30 + 2 s + 100 s Unit-step Response 10-13 Process Transfer Function: Gp ( s) = (a) PI Controller: KP + G ( s) = G c ( s ) G p ( s) KI 200 ( K P s + K I ) Y ( s) U (s ) = e -0.2s 1 + 0.25 s (1 + 0.25s ) (1 + 0.2s + 0.02s 1 2 ) (1 + 0.25s ) ( = lim s0 s 2 1 + 0.2 s + 0.02 s 200 K 4 I ) = s ( s + 4 ) s + 10s + 50 2 ( ) K I For K v = 2, K v sG ( s ) = 50 = K I =2 Thus =2 Thus G ( s) = s ( s + 4 ) s + 10s + 50 2 200 ( 2 + K P s ) ( ) ts ts The following values of the attributes of the unit-step response are computed for the system with various values for K . P KP Max overshoot (%) (sec) (sec) 222 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 1.0 19.5 13.8 8.8 4.6 1.0 0 0 0 0 0.61 0.617 0.615 0.606 0.5905 0.568 0.541 0.5078 0.44 2.08 1.966 1.733 0.898 0.878 0.851 1.464 1.603 1.618 223 224 225 Unit-step Response. 226 The unit-step response shows a maximum overshoot of 26%. Although the relative damping ratio of the complex roots is 0.707, the real pole of the third-order system transfer function is at -0.667 which adds to the overshoot. (c) G ( s) = G c ( s ) G p ( s) = = 100 = 0.00667 (1 + K D1 s ) ( K P2 s + K I 2 ) s s + 0.00667 2 ( ) For K v , K I2 K I = 100 . Let us select K P2 = 50 . Then G ( s) = For a small overshoot, K 0.00667 (1 + K D1 s ) ( 50 s + 100 ) s s + 0.00667 2 ( ) D1 must be relatively large. When K D1 = 100 , the maximum overshoot is approximately 4.5%. Thus, KP K K D I = = K K P2 D1 + K D 1 K I 2 = 50 + 100 100 = 10050 K P2 = 100 50 = 5000 = 100 3 2 System Characteristic Equation: s + 33 . 35 s + 67 . 04 s + 0 .667 = 0 Roots: - 0.01, - 2.138, - 31.2 Unit-step Response. 227 10-16 (a) G (s) p = Z (s) F (s) = 1 Ms 2 +Ks = 1 150 s 2 +1 = 0 . 00667 s 2 + 0 . 00667 The transfer function G p ( s ) has poles on the j axis. The natural undamped frequency is n = 0 . 0816 (b) PID Controller: rad/sec. G ( s) = G c ( s ) G p ( s) = Kv 0.00667 K D s + K P s + K I 2 s s + 0.00667 2 ( ( ) ) = lim s 0 sG ( s ) = KI = 100 T hus KI = 100 Characteristic Equation: For s + 0.00667 K D s + 0.00667 (1 + K P ) s + 0.00667 K I = 0 3 2 = 1 and n = 1 rad/sec, the second-order term of the characteristic equation is s 2 + 2 s + 1. Dividing the characteristic equation by the seond-order term. s + ( 0.00667 K D - 2 ) s + 2s + 1 s + 0.00667 K D s + ( 0.00667 + 0.00667 K P ) s + 0.00667 K I 2 3 2 s + 2s + s 3 2 ( 0.00667 K ( 0.00667 K ( 0.00667 K For zero remainder, 0 . 00667 K D - 2 ) s + ( 0.00667 KP - 0.99333 ) s + 0.00667 K I 2 D - 2 ) s + ( 0.01334 K D - 4 ) s + 0.00667 K D - 2 2 P - 0.01334 K D + 3.00667 ) s + 0.00667 K I - 0.00667 K D + 2 - 0 . 01334 + 0 . 00667 K K -0 . 00667 From Eq. (2), P D D I + 3. 00667 = 0 +2 =0 (1) (2) K 228 0 . 00667 K From Eq. (1), 0 . 00667 K D = 0 . 00667 = 0 . 01334 K K I + 2 = 2 . 667 Thus K D = 399 K . 85 P D - 3 . 00667 = 2 .3273 Th us P = 348 . 93 Forward-path Transfer Function: G ( s) = Characteristic Equation: 0.00667 399.85 s + 348.93 s + 100 2 ( s s + 0.00667 2 ( ) ) ( s + 0.667 ) = 0 s 3 + 2.667 s 2 + 2.334 s + 0.667 = ( s + 1) Roots: Unit-step Response. 2 - 1, - 1, - 0.667 The maximum overshoot is 20%. 10-17 (a) Process Transfer Function: G ( s) Forward-path Transfer Function p = 4 s 2 G ( s) = G c ( s ) G p ( s) = s 2 4 ( KP + KD s) s +1 = 0 2 Characteristic Equation: K P + 4K Ds + 4 K P = s 2 + 1.414 s for = 0 .707 , n =1 rad/sec = 0 .25 a nd K D = 0 . 3535 Unit-step Response. 229 Maximum overshoot = 20.8% (b) Select a relatively large value for K The closed-loop zero at s = -K P D and a small value for K / K D P so that the closed-loop poles are real. is very close to one of the closed-loop poles, and the system D dynamics are governed by the other closed-loop poles. Let K The following results show that the value of K Kp 0.1 0.05 0.2 P = 10 and use small values of K P . is not critical as long as it is small. tr ts Max overshoot (%) 0 0 0 (sec) 0.0549 0.0549 0.0549 (sec) 0.0707 0.0707 0.0707 (c) For BW 40 rad/sec and M r = 1, we can again select K D = 10 and a small value for K P . The following frequency-domain results substantiate the design. Kp 0.1 0.05 0.2 Mr 1 1 1 PM (deg) 89.99 89.99 89.99 BW (rad/sec) 40 40 40 10-18 (a) Forward-path Transfer Function: G ( s) = G c ( s ) G p ( s) = Routh Tabulation: Characteristic Equation: s 3 10,000 ( K P + K D s ) s 2 ( s + 10) + 10 s 2 + 10 ,000 K D s + 10 ,000 K P =0 230 s s s s 3 1 10 10,000 K 10 ,000 K 10, 000 K D 2 P The system is stable for K 1 D - 1000 K P K P >0 and K D > 0 .1 K P P 0 0 10,000 (b) Root Locus Diagram: G ( s) = 10,000 K P s 2 ( s + 10 ) Root Contours: 0 K D < , = K P = 0 . 001 , D 0 . 002 , 0 .005 , 0 . 01 . s K P G eq (s) 10 ,000 K s 3 + 10 s 2 + 10 ,000 231 (c) The root contours show that for small values of K means that if we choose K P P the design is insensitive to the variation ofK P . This to be between 0.001 and 0.005, the value of K P for a relative damping ratio of 0.707. Let K The forward-path transfer function becomes = 0 . 001 and K D = 0 . 005 D can be chosen to be 0.005 . G (s) c = 0 . 001 + 0 . 005 s. G ( s) = Since the zero of G ( s ) is at s = 10 (1 + 5 s ) s 2 ( s + 10 ) -0.2, which is very close to s = 0, G(s) can be approximated as: G (s) 50 s ( s + 10 ) . Using Eq. (7-104), the rise time is obtained as For the second-order system, = 0 . 707 t r = 1 - 0 .4167 + 2 . 917 n 2 = 0 . 306 sec Unit-step Response: 232 (d) Frequency-domain Characteristics: G ( s) = 10 (1 + 5 s ) s PM (deg) 63 2 ( s + 10 ) Mr 1.041 GM (dB) BW (rad/sec) 7.156 10-19 0 25.92 A = 0 -2.36 0 0 0 0 0 1 0 0 0 1 0 0 -1 0 0 s -25.92 s 0 0 sI - A = 0 0 s - 1 2.36 0 0 s s 0 0 -25.92 0 2.36 s 2 0 s 0 0 0 -1 = s s 2 = sI - A = s 0 s - 1 + 0 0 s (s 0 2 - 25.92 ) s3 2 1 25.92 s -1 ( sI - A ) = - 2.36 s 2 -2.36 s s 0 0 3 2 -2.36 s - 25.92 s s - 25.92 3 -2.36 s 0 s - 25.92s 3 233 ( sI - A ) -1 -0.0732 s 2 0 -0.0732 1 -0.0732 s 3 -1 = B = ( sI - A ) 0 0.0976 s 2 - 2.357 0.0976 3 0.0976 s - 2.357 s Y ( s) = D sI - A Characteristic Equation: ( ) 4 -1 B = [0 0 3 1 0 ] sI - A ( ) -1 B= 2 0.0976 s - 24.15 2 s 2 (s D ( 2 - 25.92 ) ) s + 0.0976 s + ( 0.0976 KP - 25.92 ) s - 2.357 K D s - 2.357 KP = 0 3 1 The system cannot be stabilized by the PD controller, since the s and the s terms involve K require opposite signs for K D which . 10-20 Let us first attempt to compensate the system with a PI controller. G (s) c = K P + K s I Then G ( s) = Gc ( s ) G ( s) = p 100 ( K P s + K I ) s s + 10 s + 100 2 ( ) Since the system with the PI controller is now a type 1 system, the steady-state error of the system due to a step input will be zero as long as the values of K and K are chosen so that the system is stable. Let us choose the ramp-error constant K v = 100 I P I . Then, K performance characteristics are obtained with K KP 10 20 30 40 50 100 = 100 I = 100 . The following frequency-domain P and various value of K ranging from 10 to 100. Mr 29.70 7.62 7.41 8.28 8.45 11.04 P PM (deg) 1.60 6.76 7.15 6.90 6.56 5.18 GM (dB) BW (rad/sec) 50.13 69.90 85.40 98.50 106.56 160.00 The maximum phase margin that can be achieved with the PI controller is only 7.15 deg when K = 30 . Thus, the overshoot requirement cannot be satisfied with the PI controller alone. Next, we try a PID controller. Gc (s ) = K P + K Ds + KI s = (1 + K s ) (K D1 P2 s + KI2 ) = (1 + K s ) (K D1 P2 s + 100 ) Based on the PI-controller design, let us select K becomes P2 = 30 . s s Then the forward-path transfer function G ( s) = 100 ( 30 s + 100 )( 1 + K D1s ) s s + 10 s + 100 2 ( ) The following attributes of the frequency-domain performance of the system with the PID controller are obtained for various values of K ranging from 0.05 to 0.4. D1 K D1 0.05 0.10 0.20 0.30 PM (deg) 85.0 89.4 90.2 90.2 GM (dB) Mr 1.04 1.00 1.00 1.00 BW (rad/sec) 164.3 303.8 598.6 897.0 234 0.40 We see that for values of K D1 90.2 D1 1.00 1201.0 greater than 0.2, the phase margin no longer increases, but the . Thus we choose bandwidth increases with the increase in K K K D1 = 0 .2, K P2 K I = K I2 = 100 , K D = K D1 K P2 = 0 .2 30 = 6 , P = + K D 1 K I 2 = 30 + 0 .2 100 = 50 G (s) c The transfer function of the PID controller is s The unit-step response is show below. The maximum overshoot is zero, and the rise time is 0.0172 sec. = 50 + 6 s + 100 10-21 (a) Gp (s ) = 4 s 2 G ( s) = Gc ( s ) G p ( s ) = 4 (1 + aTs ) s (1 + Ts ) 2 G eq (s) = 4 aTs Ts 3 + s +4 2 Root Contours: ( T is fixed and a varies) 235 Select a value for a. Let T = 0.02 small value for T and a large and a = 100. G (s) c = 1+ 2 s 1 + 0 . 02 s s 3 G ( s) = + 50 2 400 ( s + 0.5) s 2 ( s + 50) The characteristic equation is The roots are: The system transfer function is s + 400 s + 200 = 0 -0.5355, -9.3, -40.17 Y (s) R (s ) = ( s + 0.5355) ( s + 9.3) ( s + 40.17 ) 400 ( s + 0.5 ) Since the zero at -0.5 is very close to the pole at -0.5355, the system can be approximated by a secondorder system, Y (s) R (s ) = 373.48 ( s + 9.3)( s + 40.17 ) = 0 . 6225 sec t The unit-step response is shown below. The attributes of the response are: Maximum overshoot = 5% t s r = 0 .2173 sec Unit-step Response. 236 The following attributes of the frequency-domain performance are obtained for the system with the phase-lead controller. PM = 77.4 deg GM = infinite Mr = 1.05 BW = 9.976 rad/sec 10-21 (b) The Bode plot of the uncompensated forward-path transfer function is shown below. The diagram shows that the uncompensated system is marginally stable. The phase of G ( j ) is frequencies. For the phase-lead controller we need to place -180 deg at all m at the new gain crossover frequency to realize the desired phase margin which has a theoretical maximum of 90 deg. For a desired phase margin of 80 deg, a = 1 + sin 80 1 - sin 80 o o = 130 The gain of the controller is 20 log 10 a = 42 dB. The new gain crossover frequency is at G ( j ) G ( j ) =- 42 2 = -21 2 dB Or = 4 m 2 = 0 . 0877 Thus = 45 . 61 = 6 . 75 = 0 . 013 rad/sec 1 T 1 aT = a = 130 6 . 75 = 77 = 1. 69 Thus T = 0 . 592 1 + aTs 1 + Ts Thus aT Gc (s ) = Bode Plot. = 1 + 1.702 s 1 + 0.0131s G( s) = s 4 (1 + 1.702s ) 2 (1 + 0.0131s ) 237 10-22 (a) Forward-path Transfer Functi on: aT G ( s) = G c ( s ) GP ( s) = = s ( s + 10 ) (1 + Ts ) 1 s ( s + 10 ) s + T 1000 (1 + aTs ) 1000a s + 1 238 Set 1/aT = 10 so that the pole of G ( s ) at s = becomes s 2 -10 is cancelled. 1 T s The characteristic equation of the system + + 1000 a =0 Thus a = 40 and T = 0.0025 n = 1000 a 2 n = 1 T =2 1000 a Controller Transfer Function: G (s) c Forward-path Transfer Function: = 1 + 0 . 01 s 1 + 0 . 0025 s G ( s) = 40,000 s ( s + 400 ) The attributes of the unit-step response of the compensated system are: Maximum overshoot = 0 t r = 0 . 0168 sec t s = 0 . 02367 sec (b) Frequency-domain Design The Bode plot of the uncompensated forward-path transfer function is made below. G (s) = 1000 s ( s + 10 ) The attributes of the system are PM = 17.96 deg, GM = infinite. Mr = 3.117, and BW = 48.53 rad/sec. To realize a phase margin of 75 deg, we need more than 57 deg of additional phase. Let us add an additional 10 deg for safety. Thus, the value of for the phase-lead controller is chosen to be m 67 deg. The value of a is calculated from a = 1 + sin 67 1 - sin 67 log 10 o o = 24. 16 = 27 . 66 dB. The new gain crossover frequency The gain of the controller is 20 log is at 10 a = 20 24. 16 G ( j ' m ) =- 27 . 66 2 = - 13 . 83 dB From the Bode plot 1 T m ' is found to be 70 rad/sec. Thus, = aT = 24. 16 70 = 344 = 1 + aTs or T = 0 . 0029 s aT = 0 . 0702 1 + Ts 1 + 0 . 0029 s The compensated system has the following frequency-domain attributes: PM = 75.19 deg GM = infinite Mr = 1.024 BW = 91.85 rad/sec Thus G (s) c = 1 + 0 . 0702 The attributes of the unit-step response are: Rise time tr = 0.02278 sec Settling time ts = 0.02828 sec Maximum overshoot = 3.3% 239 10-23 (a) Forward-path Transfer Function: ( N = 10) G ( s) = G c ( s ) G p ( s) = 200 ( 1 + aTs ) s ( s + 1) ( s + 10 ) ( 1 + Ts ) Starting with a = 1000, we vary T first to stabilize the system. The following time-domain attributes are obtained by varying the value of T. 240 T 0.0001 0.0002 0.0003 0.0004 0.0005 0.0006 0.0007 0.0008 0.0009 0.0010 Max Overshoot (%) 59.4 41.5 29.9 22.7 18.5 16.3 15.4 15.4 15.5 16.7 tr ts 5.205 2.911 1.83 1.178 1.013 0.844 0.699 0.620 0.533 0.525 0.370 0.293 0.315 0.282 0.254 0.230 0.210 0.192 0.182 0.163 The maximum overshoot is at a minimum when T = 0.0007 or T = 0.0008. The maximum overshoot is 15.4%. Unit-step Response. ( T = 0.0008 sec a = 1000) 10-23 (b) Frequency-domain Design. Similar to the design in part (a), we set a = 1000, and vary the value of T between 0.0001 and 0.001. The attributes of the frequency-domain characteristics are given below. T 0.0001 0.0002 0.0003 0.0004 0.0005 0.0006 0.0007 PM (deg) 17.95 31.99 42.77 49.78 53.39 54.69 54.62 GM Mr 3.194 1.854 1.448 1.272 1.183 1.138 1.121 (dB) 60.00 63.53 58.62 54.53 51.16 48.32 45.87 BW (rad/sec) 4.849 5.285 5.941 6.821 7.817 8.869 9.913 241 0.0008 0.0009 0.0010 53.83 52.68 51.38 43.72 41.81 40.09 1.125 1.140 1.162 10.92 11.88 12.79 The phase margin is at a maximum of 54.69 deg when T = 0.0006. The performance worsens if the value of a is less than 1000. 10-24 (a) Bode Plot. The attributes of the frequency response are: PM = 4.07 deg GM = 1.34 dB Mr = 23.24 BW = 4.4 rad/sec 10-24 (b) Single-stage Phase-lead Controller. G ( s) = 6 (1 + aTs ) s (1 + 0.2s ) (1 + 0.5s ) (1 + Ts ) We first set a = 1000, and vary T. The following attributes of the frequency-domain characteristics are obtained. T 0.0050 0.0010 0.0007 0.0006 PM (deg) 17.77 43.70 47.53 48.27 Mr 3.21 1.34 1.24 1.22 242 0.0005 0.0004 0.0002 0.0001 48.06 46.01 32.08 19.57 1.23 1.29 1.81 2.97 The phase margin is maximum at 48.27 deg when T = 0.0006. Next, we set T = 0.0006 and reduce a from 1000. We can show that the phase margin is not very sensitive to the variation of a when a is near 1000. The optimal value of a is around 980, and the corresponding phase margin is 48.34 deg. With a = 980 and T = 0.0006, the attributes of the unit-step response are: Maximum overshoot = 18.8% tr = 0.262 sec T = 0.0006) ts = 0.851 sec (c) Two-stage Phase-lead Controller. ( a = 980, G( s) = 6 ( 1 + 0.588s ) 1 + bT s 2 s (1 + 0.2s )(1 + 0.5 s ) (1 + 0.0006 s ) 1 + T s 2 ( ) ( ) Again, let b = 1000, and vary T . The following results are obtained in the frequency domain. 2 T2 PM (deg) Mr 0.0010 93.81 1.00 0.0009 94.89 1.00 0.0008 96.02 1.00 0.0007 97.21 1.00 0.0006 98.43 1.00 0.0005 99.61 1.00 0.0004 100.40 1.00 0.0003 99.34 1.00 0.0002 91.98 1.00 0.0001 73.86 1.00 Reducing the value of b from 1000 reduces the phase margin. Thus, the maximum phase margin of 100.4 deg is obtained with b = 1000 and T 2 = 0.0004. The transfer function of the two-stage phaselead controller is Gc ( s) = (c) Unit-step Responses. ( 1 + 0.588s ) (1 + 0.4s ) (1 + 0.0006s ) (1 + 0.0004s ) 243 10-25 (a) The loop transfer function of the system is G ( s) H( s) = 10 K pK aK e Ns (1 + 0.05 s ) s 3 1 + R2C s RCs 1 2 = 68.76 s (1 + 0.05s ) -4 1 + R2 10 s 2s .6 -6 The characteristic equation is For root locus plot with R 2 + 20 s + 6 .876 10 R s 2 + 687 =0 as the variable parameter, we have -4 G eq ( s ) = 6.876 10 R2 s s + 20 s + 687.6 3 2 = ( s + 21.5) ( s - 0.745 + 6.876 10 R2 s -4 j 5.61) ( s - 0.745 - j 5.61 ) Root Locus Plot. 244 When R 2 = 2 . 65 10 5 , the roots are at -6 . 02 j 7 . 08 , and the relative damping ratio is 0.65 which is maximum. The unit-step response is plotted at the end together with those of parts (b) and (c). (b) Phase-lead Controller. G ( s) H( s) = 68.76 (1 + aTs ) s (1 + 0.05 s ) (1 + Ts ) Characteristic Equation: Ts + 1 + 20 T s + 20 + 1375.2 aT s + 1375.2 = 0 3 2 ( ) ( ) With T = 0.01, the characteristic equation becomes 3 2 s + 120 s + ( 2000 + 1375.2 a ) s + 137520 = 0 The last equation is conditioned for a root contour plot with a as the variable parameter. Thus 1375 .2 as G eq ( s ) = 3 2 s + 120 s + 2000 s + 137 ,520 From the root contour plot on the next page we see that when a = 3.4 the characteristic equation roots are at - 39 .2 , - 40 .4 + j 43 . 3, and - 40 .4 - j 43 . 3 , and the relative damping ratio is maximum and is 0.682. Root Contour Plot ( a varies). 245 Unit-step Responses. 10-25 (c) Frequency-domain Design of Phase-lead Controller. For a phase margin of 60 deg, a = 4.373 and T = 0.00923. The transfer function of the controller is G (s) c = 1 + aTs 1 + Ts = 1 + 0 . 04036 s 1 + 0 . 00923 s 246 10-26 (a) Time-domain Design of Phase-lag Controller. Process Transfer Function: Gp ( s) = 200 s ( s + 1 )( s + 10 ) = - 0 .475 + j 0 .471 For the uncompensated system, the two complex characteristic equation roots are at s and -0 .475 - j 0 .471 which correspond to a relative damping ratio of 0.707, when the forward path gain is 4.5 (as against 200). Thus, the value of a of the phase-lag controller is chosen to be a Then G (s) c = 4. 5 200 = 0 . 0225 1 + aTs 1 + Ts 1 Select T = 1000 which is a large number. = = + 22 . 5 s 1 + 1000 s G ( s) = Gc ( s ) G ( s) = p s ( s + 1 )( s + 10 )( s + 0.001) 4.5 ( s + 0.0889) ) Unit-step Response. Maximum overshoot = 13.6 Bode Plot (with phase-lag controller, a = 0.0225, T = 1000) tr = 3.238 sec ts = 18.86 sec 247 PM = 59 deg. GM = 27.34 dB Mr = 1.1 BW = 0.6414 rad/sec 10-26 (b) Frequency-domain Design of Phase-lag Controller. For PM = 60 deg, we choose a = 0.02178 and T = 1130.55. The transfer function of the phase-lag controller is 1 + 24. 62 s G (s) = GM = 27.66 dB Mr = 1.093 BW = 0.619 rad/sec c 1 + 1130 . 55 s Unit-step Response. Max overshoot = 12.6%, tr = 3.297 sec ts = 18.18 sec 10-27 (a) Time-domain Design of Phase-lead Controller Forward-path Transfer Function. G ( s) = Gc ( s ) G ( s) = p K (1 + aTs ) s ( s + 5) 2 (1 + Ts ) K v = lim sG ( s ) s 0 = K 25 = 10 Thus K = 250 248 With K = 250, the system without compensation is marginally stable. For a > 1, select a small value for T and a large value for a. Let a = 1000. The following results are obtained for various values of T ranging from 0.0001 to 0.001. When T = 0.0004, the maximum overshoot is near minimum at 23%. T 0.0010 0.0005 0.0004 0.0003 0.0002 0.0001 Max Overshoot (%) 33.5 23.8 23.0 24.4 30.6 47.8 (sec) 0.0905 0.1295 0.1471 0.1689 0.1981 0.2326 tr (sec) 0.808 0.6869 0.7711 0.8765 1.096 2.399 ts As it turns out a = 1000 is near optimal. A higher or lower value for a will give larger overshoot. Unit-step Response. (b) Frequency-domain Design of Phase-lead Controller G ( s) = s 250 ( 1 + aTs ) 2 ( s + 5 ) (1 + Ts ) 2 Setting a = 1000, and varying T, the following attributes are obtained. T 0.00050 0.00040 0.00035 0.00030 0.00020 PM (deg) 41.15 42.85 43.30 43.10 38.60 Mr 1.418 1.369 1.355 1.361 1.513 BW (rad/sec) 16.05 14.15 13.16 12.12 10.04 When a = 1000, the best value of T for a maximum phase margin is 0.00035, and PM = 43.3 deg. As it turns out varying the value of a from 1000 does not improve the phase margin. Thus the transfer function of the controller is 249 G (s) c = 1 + aTs 1 + Ts = 1 1 + 0 . 35 s and s + 0 . 00035 G ( s) = 250 ( 1+ 0.35s ) s ( s + 5) 2 (1 + 0.00035s ) 10-27 (c) Time-domain Design of Phase-lag Controller Without compensation, the relative damping is critical when K = 18.5. Then, the value of a is chosen to be 18 . 5 a = = 0 . 074 250 We can use this value of a as a reference, and conduct the design around this point. The value of T is preferrably to be large. However, if T is too large, rise and settling times will suffer. The following performance attributes of the unit-step response are obtained for various values of a and T. a 0.105 0.100 0.095 0.090 0.090 0.090 0.090 0.090 0.090 0.090 T 500 500 500 500 600 700 800 1000 2000 3000 Max Overshoot (%) 2.6 2.9 2.6 2.5 2.1 1.9 1.7 1.4 0.9 0.7 tr 1.272 1.348 1.422 1.522 1.532 1.538 1.543 1.550 1.560 1.566 ts 1.82 1.82 1.82 2.02 2.02 2.02 2.02 2.22 2.22 2.22 As seen from the results, when a = 0.09 and for T 2000, the maximum overshoot is less than 1% and the settling time is less than 2.5 sec. We choose T = 2000 and a = 0.09. The corresponding frequency-domain characteristics are: PM = 69.84 deg GM = 20.9 dB Mr = 1.004 BW = 1.363 rad/sec (d) Frequency-domain Design of Phase-lag Controller G ( s) = 250 ( 1 + aTs ) s ( s + 5) 2 (1 + Ts ) a <1 The Bode plot of the uncompensated system is shown below. Let us add a safety factor by requiring that the desired phase margin is 75 degrees. We see that a phase margin of 75 degrees can be realized if the gain crossover is moved to 0.64 rad/sec. The magnitude of G ( j ) at this frequency is 23.7 dB. Thus the phase-lag controller must provide an attenuation of the new gain crossover frequency. Setting 20 log a -23.7 dB at 10 = -23 .7 dB we have a = 0.065 We can set the value of 1/aT to be at least one decade below 0.64 rad/sec, or 0.064 rad/sec. Thus, we get T = 236. Let us choose T = 300. The transfer function of the phase-lag controller becomes G (s) c = 1 + aTs 1 + Ts = 1 + 19 . 5 s 1 + 300 s 250 The attributes of the frequency response of the compensated system are: 251 PM = 71 deg GM = 23.6 dB Mr = 1.065 BW = 0.937 rad/sec The attributes of the unit-step response are: Maximum overshoot = 6% tr = 2.437 sec ts = 11.11 sec Comparing with the phase-lag controller designed in part (a) which has a = 0.09 and T = 2000, the time response attributes are: Maximum overshoot = 0.9% tr = 1.56 sec ts = 2.22 sec The main difference is in the large value of T used in part (c) which resulted in less overshoot, rise and settling times. 10-28 Forward-path Transfer Function (No compensation) G ( s) = G p ( s) = 6.087 10 3 2 7 6 8 s s + 423.42 s + 2.6667 10 s + 4.2342 10 ( ) The uncompensated system has a maximum overshoot of 14.6%. The unit-step response is shown below. (a) Phase-lead Controller G (s) c = 1 + aTs 1 + Ts (a > 1) By selecting a small value for T, the value of a becomes the critical design parameter in this case. If a is too small, the overshoot will be excessive. If the value of a is too large, the oscillation in the step response will be objectionable. By trial and error, the best value of a is selected to be 6, and T = 0.001. The following performance attributes are obtained for the unit-step response. Maximum overshoot = 0% tr = 0.01262 sec ts = 0.1818 sec However, the step response still has oscillations due to the compliance in the motor shaft. The unitstep response of the phase-lead compensated system is shown below, together with that of the uncompensated system. (b) Phase-lead and Second-order Controller The poles of the process G p ( s ) are at -161.3, order term is s 2 -131+ j1614.7 and -131 - j1614.7. The second- + 262 s + 2 ,624, G ( s) 417 .1 . Let the second-order controller transfer function be s 2 c1 = + 262 s 2 s + 2 ,624, 417 .1 2 + 2 p n s + n rad/sec, so that the steady-state error is not affected. The value of n is set to 2 ,624, 417 .1 = 1620 and 2 Let the two poles of G c1 ( s ) be at s = - 1620 = s s - 1620 s s . Then, p = 405 . G c1 ( s ) + 262 + 3240 + 2 ,624, + 2 ,624, 417 .1 417 .1 10 2 G ( s) = G c ( s ) G c ( s ) G p ( s) = 1 s ( s + 161.3 ) s + 3240 s + 2,624,417.1 ( 1 + 0.001s ) 2 ( 6.087 10 (1 + 0.006 s ) ) The unit-step response is shown below, and the attributes are: 252 Maximum overshoot = 0.2 The step response does not have any ripples. tr = 0.01012 sec ts = 0.01414 sec Unit-step Responses 10-29 (a) System Equations. e a = R a i a + K b m T m = K i ia d L dt Tm = Jm d m dt + Bmm + KL ( m - L ) + BL ( m - L ) K L ( m - L ) + BL ( m - L ) = J L State Equations in Vector-matrix Form: d L dt 0 K d L - L dt J L d = 0 m dt K L d m J m dt 1 - BL JL 0 BL Jm - 0 KL JL 0 KL Jm 0 BL L 0 JL L + 0 ea 1 m Ka B + Bl Ki K b m - m - Ra J m Jm J m Ra 0 253 State Diagram: Transfer Functions: m ( s) Ea ( s ) L ( s) E a (s ) m ( s) Ea ( s ) L ( s) E a (s ) = Ki s + BL s + KL / Ra 2 J m J L s + ( K e J L + BL J L + BL J m ) s + ( J m K L + J L K L + K e BL ) s + K L K e 2 2 ( ) = J m J L s + ( K e J L + BL J L + B L J m ) s + ( J m K L + J L K L + K e BL ) s + K L K e 3 2 K i ( B Ls + K L ) / R a = 133.33 s + 10 s + 3000 2 3 2 ( ) s + 318.15 s + 60694.13 s + 58240 = 133.33 s + 10s + 3000 2 ( s + 0.9644) ( s + 158.59 + ( j187.71) ( s + 158.59 - j187.71) ) = ( s + 0.9644 ) ( s + 158.59 + 1333.33 ( s + 300 ) j187.71 ) ( s + 158.59 - j187.71) (b) Design of PI Controller. 1333.33 K P s + = G ( s) = L (s ) E (s ) s ( s + 0.9644 ) s + 317.186s + 60388.23 2 ( ( s + 300 ) KP KI ) Thus Kv = lim s0 sG ( s ) = 1333 .33 0 . 9644 300 K I 60388 . 23 = 6 . 87 KI = 100 KI = 14 .56 With K I = 14 .56 , KP 20 18 we study the effects of varying K P . The following results are obtained. tr ts (sec) 0.00932 0.01041 (sec) 0.02778 0.01263 Max Overshoot (%) 4.2 0.7 254 17 16 15 10 With K I 0.01113 0.01184 0.01303 0.02756 0.01515 0.01515 0.01768 0.04040 0 0 0 0.6 = 14 .56 and K P ranging from 15 to 17, the design specifications are satisfied. Unit-step Response: (c) Frequency-domain Design of PI Controller ( KI = 14.56) G ( s) = s s + 318.15 s + 60694.13s + 58240 3 2 ( 1333.33 ( K P s + 14.56 )( s + 300 ) ) tr ts The following results are obtained by setting K I KP 20 18 17 16 15 10 8 7 6 5 = 14 .56 and varying the value of K P . PM (deg) 65.93 69.76 71.54 73.26 74.89 81.11 82.66 83.14 83.29 82.88 GM (dB) Mr 1.000 1.000 1.000 1.000 1.000 1.005 1.012 1.017 1.025 1.038 BW (rad/sec) 266.1 243 229 211.6 190.3 84.92 63.33 54.19 45.81 38.12 Max Overshoot (%) 4.2 0.7 0 0 0 0.6 1.3 1.9 2.7 4.1 (sec) 0.00932 0.01041 0.01113 0.01184 0.01313 0.0294 0.04848 0.03952 0.04697 0.05457 (sec) 0.02778 0.01263 0.01515 0.01515 0.01768 0.0404 0.03492 0.05253 0.0606 0.0606 255 From these results we see that the phase margin is at a maximum of 83.29 degrees when K P = 6 . However, the maximum overshoot of the unit-step response is 2.7%, and M r is slightly greater than one. In part (b), the optimal value of K P from the standpoint of minimum value of the maximum overshoot is between 15 and 17. Thus, the phase margin criterion is not a good indicator in the present case. 10-30 (a) Forward-path Transfer Function Gp ( s) = K m ( s) Tm ( s ) = 100 K s + 10 s + 100 2 3 2 s s + 20 s ( ( ( = + 2100 s + 10,000 ) s ( s + 4.937 ) ( s ts = 0.4949 sec ) 10,000 s + 10 s + 100 2 2 ) + 15.06 s + 2025.6 ) The unit-step response is plotted as shown below. The attributes of the response are: Maximum overshoot = 57% tr = 0.01345 sec (b) Design of the Second-order Notch Controller The complex zeros of the notch controller are to cancel the complex poles of the process transfer function. Thus s s 2 2 G (s) c = + 15 . 06 s + 2025 + 90 p s + 2025 .6 and .6 ( G ( s) = G ( s ) G ( s) = s ( s + 4.937 ) ( s c p 10,000 s + 10 s + 100 2 2 ) + 90 p s + 2025.6 ) The following results are obtained for the unit-step response when various values of The maximum overshoot is at a minimum of 4.1% when p p are used. = 1.222 . The unit-step response is plotted below, along with that of the uncompensated system. p 2.444 2.333 2.222 1.667 1.333 1.222 1.111 1.000 2 200 210 200 150 120 110 100 90 n Max Overshoot (%) 7.3 6.9 6.5 4.9 4.3 4.1 5.8 9.8 Unit-step Response 256 10-30 (c) Frequency-domain Design of the Notch Controller The forward-path transfer function of the uncompensated system is ( G ( s) = s ( s + 4.937 ) ( s 10000 s + 10 s + 100 2 2 ) + 15.06 s + 2025.6 ) -22 dB or 0.0794 The Bode plot of G ( j ) is constructed in the following. We see that the peak value of G ( j ) is approximately 22 dB. Thus, the notch controller should provide an attentuation of at the resonant frequency of 45 rad/sec. Using Eq. (10-155), we have G ( j 45) c = z p = 0 .167 p = 0 . 0794 Thus p = 2 .1024 Notch Controller Transfer Function s s 2 2 Forward-path Transfer Function G (s) c = + 15 .06 s + 2025 .216 s .6 .6 + 189 + 2025 G (s ) = ( s ( s + 4.937 ) (s 10,000 s + 10 s + 100 2 2 ) + 189.22s + 2025.6 ) Bode Plots 257 Attributes of the frequency response: PM = 80.37 deg GM = infinite M r = 1. 097 BW = 66.4 rad/sec Attributes of the frequency response of the system designed in part (b): PM = 59.64 deg GM = infinite M r = 1. 048 BW = 126.5 rad/sec 10-31 (a) Process Transfer Function Gp ( s) = s s + 10 s + 1000 2 ( 500 ( s + 10 ) ) M The Bode plot is constructed below. The frequency-domain attributes of the uncompensated system are: PM = 30 deg GM = infinite r = 1.86 and BW = 3.95 rad/sec The unit-step response is oscillatory. (b) Design of the Notch Controller 258 For the uncompensated process, the complex poles have the following constants: n = 1000 = 31 .6 rad / sec 2 n = 10 Thus = 0.158 The transfer function of the notch controller is s 2 G (s) c = + 2 z n s + n 2 2 s + 2 p s + n 2 For the zeros of G ( s ) to cancel the complex poles of G p ( s ) , c z = = 0 .158 . From the Bode plot, we see that to bring down the peak resonance of G ( j n ) in order to smooth out the magnitude curve, the notch controller should provide approximately -26 dB of attenuation. Thus, using Eq. (10-155), z p -26 = 10 20 = 0 . 05 Thu s p = 0 .158 0 . 05 = 3 .1525 The transfer function of the notch controller is G (s) c = s s 2 2 + 10 s + 1000 . 08 s + 199 + 1000 G ( s) = G c ( s ) G p ( s) = s s + 199.08 s + 1000 2 ( 500 ( s + 10 ) ) The attributes of the compensated system are: PM = 72.38 deg GM = infinite t M r =1 t s BW = 5.44 rad/sec Maximum overshoot = 3.4% r = 0 . 3868 sec = 0 .4848 sec Bode Plots 259 Step Responses 10-31 (c) Time-domain design of the Notch Controller 260 With z = 0 .158 and n = 31 . 6 , the forward-path transfer function of the compensated system is G ( s) = G c ( s ) G p ( s) = s s + 63.2 p s + 1000 2 ( 500 ( s + 10 ) ) p. (sec) 0.5859 0.5657 0.5455 0.5253 0.5152 0.4840 0.4848 ts tr The following attributes of the unit-step response are obtained by varying the value of p 1.582 1.741 1.899 2.057 2.215 2.500 3.318 When 2 n Max Overshoot (%) 0 0 0 0 0.2 0.9 4.1 (sec) 0.4292 0.4172 0.4074 0.3998 0.3941 0.3879 0.3884 100 110 120 130 140 158.25 209.7 p = 2 .5 the maximum overshoot is 0.9%, the rise time is 0.3879 sec and the setting time is 0.4840 sec. These performance attributes are within the required specifications. 10-32 Let the transfer function of the controller be Gc ( s) = 20,000 s + 10 s + 50 2 ( ) ( s + 1000 ) 2 Then, the forward-path transfer function becomes G ( s) = Gc ( s ) G ( s) = p For G (s) 20,000 K s + 10 s + 50 2 s s + 10 s + 100 2 6 ( ( ) ( s + 1000) ) 2 cf = 1, K v = lim K sG ( s ) s0 = 10 K 8 = 50 = 6000 Thus t he nom inal K = 5000 10 and For 20% variation in K, min = 4000 K max . To cancel the complex closed-loop poles, we let G cf ( s ) = 50 ( s + 1 ) s + 10 s + 50 2 where the (s + 1) term is added to reduce the rise time. Closed-loop Transfer Function: Y (s) R (s ) = s s + 10 s + 100 2 ( )( s + 1000 ) + 20,000K ( s 2 7 2 8 10 K ( s + 1) 6 2 + 10 s + 50 ) Characteristic Equation: K = 4000: s 5 Roots: - 97 . 7 , 5 + 2010 s + 1,020 , 100 s + 9 . 02 10 s + 9 10 - 648 . 9 , - 1252 . 7 , - 5 . 35 + j 4. 6635 , 4 3 Max overshoot K = 5000: s 6.7% + 4 10 = 0 - 5 . 35 - j 4. 6635 9 9 Rise time < 0.04 sec 3 8 2 9 Roots: -132 + 2010 .46 , s 4 + 1,02010 587 .44, Max overshoot 4% 0 s + 1.1 10 s + 1.1 10 s + 5 10 - 1279 . 6 , - 5 .272 + j 4. 7353 , - 5 .272 =0 - j 4. 7353 Rise time < 0.04 sec 261 K = 6000 Roots: -176 s 5 + 2010 . 77 , s , 100 s + 1. 3 10 s + 1.3 10 s + 6 10 = 0 - 519 . 37 , - 1303 .4, - 5 .223 + j 4. 7818 , - 5 .223 - j 4. 7818 4 3 8 2 9 9 + 1,020 Max overshoot 2.5% Rise time < 0.04 sec Thus all the required specifications stay within the required tolerances when the value of K varies by plus and minus 20%. Unit-step Responses 10-33 Let the transfer function of the controller be Gc ( s) = The forward-path transfer function becomes 200 s + 10 s + 50 2 ( ) ( s + 100 ) 2 G ( s) = G c ( s ) G p ( s) = 200,000 K s + 10 s + 50 2 ( ) s ( s + a ) ( s + 100 ) 2 For a = 10, K v = lim sG ( s ) s 0 = 10 7 K 5 = 100 K = 100 Thus K =1 10 Characteristic Equations: ( K = 1) a = 10: Roots: a = 8: + 210 s + 2 .12 10 s + 2 .1 10 s + 10 = 0 -4. 978 + j 4. 78 , - 4. 978 - j 4. 78 , - 100 + j 447 s 4 3 5 2 6 7 4 3 5 2 6 7 .16 - 100 - j 447 .16 s + 208 s + 2 .116 10 s + 2 . 08 10 s + 10 = 0 Roots: - 4. 939 + j 4. 828 , - 4. 939 - j 4. 828 , - 99 . 06 + j 446 .97 , - 99 . 06 s 4 - j 446 . 97 a = 12: + 212 s 3 + 2 .124 10 + 2 .12 10 5 6 s + 10 7 =0 262 Roots: - 5. 017 + j 4. 73 , - 5 . 017 - j 4. 73 , - 100 . 98 + j 447 . 36 , - 100 . 98 - j 447 . 36 Unit-step Responses: All three responses for a = 8, a = 10, and 12 are similar. 10-34 Forward-path Transfer Function: G ( s) = Y ( s) E (s ) Characteristic Equation: = K s ( s + 1 )( s + 10 ) + K Kt s 3 2 Kv = lim sG( s ) = s 0 3 2 K =1 10 + KKt s + 11s + (10 + KKt ) s + K = s + 11 s + Ks+ K = 0 K ( s + 1) s 2 For root loci, G eq ( s ) = ( s + 11) 263 Root Locus Plot ( K varies) The root loci show that a relative damping ratio of 0.707 can be realized by two values of K. K = 22 and 59.3. As stipulated by the problem, we select K = 59.3. 10-35 Forward-path Transfer Function: G ( s) = 10 K s ( s + 1) ( s + 10 ) + 10 K t s K v = lim s 0 sG ( s ) = 10 10 K + 10 K = t K 1+ K t =1 Thus K t = K -1 264 Characteristic Equation: When K = 5.93 and K s 3 2 s ( s + 1 )( s + 10 ) + 10 Kt + 10 K = s + 11 s +10 Ks + 10 K = 0 3 2 t = K - 1 = 4. 93 s , the characteristic equation becomes + 11 s + 10 . 046 - 10 . 046 , + 4. 6 = 0 j 0 .47976 , The roots are: - 0 .47723 + - 0 .47723 - j 0 .47976 10-36 Forward-path Transfer Function: G ( s) = K (1 + aTs ) s ( (1 + Ts ) s + 10 s + KK t 2 ( ) 2 K v = lim s 0 sG ( s ) = 1 K t = 100 Thu s K t = 0 . 01 Let T = 0.01 and a = 100. The characteristic equation of the system is written: s + 110 s + 1000s + K 0.001s + 101s +100 = 0 4 3 2 ( ) To construct the root contours as K varies, we form the following equivalent forward-path transfer function: G eq ( s ) = 0.001 K s + 101,000 s + 100,000 2 ( s 2 ( s + 10) ( s + 100 ) ) = 0.001K ( s + 1)( s + 50499 ) s 2 ( s + 10) ( s + 100) From the root contour diagram we see that two sets of solutions exist for a damping ratio of 0.707. These are: K = 20: Complex roots: - 1.158 + j 1.155 , - 1.158 - j1 .155 K = 44.6: Complex roots: - 4. 0957 + j 4. 0957 , - 4. 0957 - j 4. 0957 The unit-step responses of the system for K = 20 and 44.6 are shown below. Unit-step Responses: 265 10-37 Forward-path Transfer Function: G ( s) = K s K K Ki N 1 s J t L a s + ( Ra J t + L a Bt + K 1 K 2 J t ) s + R a Bt + K 1 K 2 Bt + K b K i + K K Ki Kt 1 2 G ( s) = s s + 3408.33s + 1,204,000 + 1.5 10 KK t 2 8 ( 1.5 10 K 7 ) Ramp Error Constant: K v = lim s 0 sG ( s ) = 1.204 15 K + 150 KK = 100 t Thus 1.204 + 150 KK t = 0 .15 K The forward-path transfer function becomes G ( s) = Characteristic Equation: s s + 3408.33s + 150,000 K 2 ( 1.5 10 K 7 3 ) + 1. 5 10 7 s + 3408 . 33 s + 150 ,000 Ks K =0 When K = 38.667 the roots of the characteristic equation are at - 0 .1065 , - 1. 651 + j 1. 65 , - 1.651 - j1. 65 ( 0 . 707 for the complex roots) The forward-path transfer function becomes G ( s) = s s + 3408.33 s + 5.8 10 2 ( 5.8 10 8 6 ) 266 Unit-step Response Unit-step response attributes: Maximum overshoot = 0 Rise time = 0.0208 sec Settling time = 0.0283 sec 10-38 (a) Disturburnce -to-Output Transfer Function Y (s) TL ( s) For T ( s ) L r =0 == = 1/ s s (1 + 0.01 s)(1 + 0.1 s ) + 20 K 2 (1 + 0.1s ) G (s) c =1 10-38 (b) 0 . 01 Thus 10 K Performance of Uncompensated System. K = 10, G ( s ) = 1 lim y ( t ) sY ( s ) t s 0 c = lim = 1 K 10 G ( s) = 200 s (1 + 0.01s ) (1 + 0.1s ) -5.19 dB. The Bode diagram of G ( j ) is shown below. The system is unstable. The attributes of the frequency response are: PM = -9.65 deg GM = (c) Single-stage Phase-lead Controller Design To realize a phase margin of 30 degrees, a = 14 and T = 0.00348. G (s) c = 1 + aTs 1 + Ts = 1 + 0 . 0487 s 1 + 0 . 00348 s 267 The Bode diagram of the phase-lead compensated system is shown below. The performance attributes are: PM = 30 deg GM = 10.66 dB M = 1. 95 BW = 131.6 rad/sec. r (d) Two-stage Phase-lead Controller Design Starting with the forward-path transfer function G ( s) = s (1 + 0.1s ) (1 + 0.01s ) (1 + 0.00348 s ) 200 ( 1+ 0.0487 s ) The problem becomes that of designing a single-stage phase-lead controller. For a phase margin or 55 degrees, a = 7.385 and T = 0.00263. The transfer function of the second-stage controller is G ( s) c1 = 1 + aTs + Ts = 1 1 + 0 . 01845 + 0 . 00263 s s 1 Thus G ( s) = s (1 + 0.1s ) (1 + 0.01s ) (1 + 0.00348s )( 1 + 0.00263s ) 200 ( 1+ 0.0487 s ) (1 + 0.01845 s ) The Bode diagram is shown on the following page. The following frequency-response attributes are obtained: PM = 55 deg GM = 12.94 dB M r = 1.11 BW = 256.57 rad/sec Bode Plot [parts (b), (c), and (d)] 268 10-39 (a) Two-stage Phase-lead Controller Design. The uncompensated system is unstable. PM = -43.25 deg and GM = -18.66 dB. With a single-stage phase-lead controller, the maximum phase margin that can be realized affectively is 12 degrees. Setting the desired PM at 11 deg, we have the parameters of the single-stage phaselead controller as a = 128.2 and T = 0 . 00472 . The transfer function of the single-stage controller 1 is G c1 ( s) = 1 + aT 1 s + T1 s = 1 + 0 . 6057 s s 1 1 + 0 . 00472 Starting with the single-stage-controller compensated system, the second stage of the phase-lead controller is designed to realize a phase margin of 60 degrees. The parameters of the second-stage controller are: b = 16.1 and T = 0 . 0066 . Thus, 2 G c2 (s) = 1 + bT s 2 1+T s 2 = 1 + 0 .106 s 1 + 0 . 0066 s 1 G (s) c = G c 1 ( s )G c 2 ( s ) = 1 + 0 . 6057 s + 0 .106 s 1 + 0 . 00472 s 1 + 0 . 0066 s Forward-path Transfer Function: 269 G ( s) = G c1 ( s ) G c2 ( s ) G p ( s) = s ( s + 2)( s + 5 ) ( s + 211.86 ) ( s + 151.5 ) = 1. 08 BW = 65.11 rad/sec 1,236,598.6 ( s + 1.651 )( s + 9.39 ) Attributes of the frequency response of the compensated system are: GM = 19.1 dB PM = 60 deg M r The unit-step response is plotted below. The time-response attributes are: Maximum overshoot = 10.2% t s = 0 .1212 sec t r = 0 . 037 sec (b) Single-stage Phase-lag Controller Design. With a single-stage phase-lag controller, for a phase margin of 60 degrees, a = 0.0108 and T = 1483.8. The controller transfer function is 1 + 16 . 08 s G (s) = c 1 + 1483 . 8 s The forward-path transfer function is G (s) = G c ( s )G p ( s ) = s s + + + + 6 .5 s s 0 . 0662 s 2 5 0 . 000674 BW = 1.07 rad/sec From the Bode plot, the following frequency-response attributes are obtained: PM = 60 deg GM = 20.27 dB M r = 1. 09 The unit-step response has a long rise time and settling time. The attributes are: Maximum overshoot = 12.5% t s = 12 . 6 sec t r = 2 .126 sec (c) Lead-lag Controller Design. For the lead-lag controller, we first design the phase-lag portion for a 40-degree phase margin. The result is a = 0.0238 and T = 350 . The transfer function of the controller is 1 1 + 350 s The phase-lead portion is designed to yield a total phase margin of 60 degrees. The result is b = 4.8 and T = 0 .2245 . The transfer function of the phase-lead controller is 2 G c1 ( s) = 1 + 8 . 333 s 1 + 0 .2245 s The forward-path transfer function of the lead-lag compensated system is G c2 (s) = 1 + 1.076 s G ( s) = Frequency-response attributes: Unit-step response attributes: s ( s + 2 ) ( s + 5 ) ( s + 0.00286 ) ( s + 4.454 ) PM = 60 deg GM = 13.07 dB M = 1. 05 BW = 3.83 rad/sec r Maximum overshoot = 5.9% t s 68.63 ( s + 0.12 )( s + 0.929 ) = 1. 512 sec t r = 0 . 7882 sec 270 Unit-step Responses. 10-40 (a) The uncompensated system has the following frequency-domain attributes: PM = 3.87 deg The Bode plot of G GM = 1 dB M r = 7 .73 BW = 4.35 rad/sec p ( j ) shows that the phase curve drops off sharply, so that the phase-lead controller would not be very effective. Consider a single-stage phase-lag controller. The phase margin of 60 degrees is realized if the gain crossover is moved from 2.8 rad/sec to 0.8 rad/sec. The attenuation of the phase-lag controller at high frequencies is approximately -15 dB. Choosing an attenuation of -17.5 dB, 20 log 10 we calculate the value of a from a = -17 . 5 dB Thus a = 0.1334 The upper corner frequency of the phase-lag controller is chosen to be at 1/aT = 0.064 rad/sec. Thus, 1/T = 0.00854 or T = 117.13. The transfer function of the phase-lag controller is 1 + 15 .63 s G (s) = c 1 + 117 .13 The forward-path transfer function is G ( s) = G c ( s ) G p ( s) = From the Bode plot of G ( j ) , the following frequency-domain attributes are obtained: PM = 60 deg GM = 18.2 dB M r s (1 + 0.1 s ) (1 + 0.5 s ) (1 + 117.13 s ) (1 + 0.05 s ) = 1. 08 5 ( 1 + 15.63 s ) (1 - 0.05 s ) BW = 1.13 rad/sec The unit-step response attributes are: maximum overshoot = 10.7% t s = 10 .1 sec t r = 2 .186 sec Bode Plots 271 10-40 (b) Using the exact expression of the time delay, the same design holds. The time and frequency domain attributes are not much affected. 272 10-41 (a) Uncompensated System. Forward-path Transfer Function: G ( s) = The Bode plot of G ( j ) is shown below. 10 (1 + s ) (1 + 10 s )( 1 + 2s ) (1 + 5s ) GM = The performance attributes are: PM = -10.64 deg The uncompensated system is unstable. -2.26 dB (b) PI Controller Design. Forward-path Transfer Function: G ( s) = Ramp-error Constant: K v = lim sG s 0 10 K p s + K I s (1 + s ) (1 + 10 s )( 1 + 2s ) (1 + 5s ) ( s ) = 10 K I = 0 .1 Thu s K I = 0 . 01 0.1 (1 + 100K P s ) ( ) G ( s) = s (1 + s ) (1 + 10 s )( 1 + 2s ) (1 + 5s ) P The following frequency-domain attributes are obtained for various values of K KP 0.01 0.02 0.05 0.10 0.12 0.15 0.16 0.17 0.20 . Mr 2.54 2.15 1.52 1.17 1.13 1.14 1.16 1.18 1.29 PM (deg) 24.5 28.24 38.84 50.63 52.87 53.28 52.83 51.75 49.08 GM (dB) 5.92 7.43 11.76 12.80 12.23 11.22 10.88 10.38 9.58 BW (rad/sec) 0.13 0.13 0.14 0.17 0.18 0.21 0.22 0.23 0.25 The phase margin is maximum at 53.28 degrees when K P = 0 .15 . The forward-path transfer function of the compensated system is G ( s) = s (1 + s ) (1 + 10 s )( 1 + 5 s ) ( 1 + 2 s ) 0.1 (1 + 15 s ) The attributes of the frequency response are: PM = 53.28 deg GM = 11.22 dB M r = 1.14 BW = 0.21 rad/sec The attributes of the unit-step response are: Maximum overshoot = 14.1% t r = 10 . 68 sec t s = 48 sec Bode Plots 273 10-41 (c) Time-domain Design of PI Controller. 274 By setting K I = 0 . 01 and varying K P we found that the value of K P that minimizes the maximum overshoot of the unit-step response is also 0.15. Thus, the unit-step response obtained in part (b) is still applicable for this case. 10-42 Closed-loop System Transfer Function. Y (s) R (s ) = 1 s + ( 4 + k 3 ) s + ( 3 + k 2 + k3 ) s + k1 3 2 1 2 For zero steady-state error to a step input, k =1. For the complex roots to be located at -1 +j and -1 and solve for zero remainder. - j, we divide the characteristic polynomial by s s + ( 2 + k2 ) 3 +2s+ 2 s + 2 s + 2 s + ( 4 + k3 ) s + ( 3 + k2 + k3 ) s + 1 2 2 s + 3 2s 3 2 + 2s 2 2 3 ( 2 + k ) s + (1 + k + k ) s + 1 ( 2 + k ) s + ( 4 + 2k ) s + 4 + 2k 2 3 3 3 ( -3+k For zero remainder, 2 - k 3 ) s - 3 - 2 k3 Thus Thus k 3 - 3 - 2k3 = 0 -0.5. = -1. 5 k 2 -3 + k 2 - k 3 = 0 = 1. 5 The third root is at steady-state error. Not all the roots can be arbitrarily assigned, due to the requirement on the 10-43 (a) Open-loop Transfer Function. G ( s) = X 1 (s ) E (s ) = k3 s s + (4 + k 2 ) s + 3 + k 1 + k 2 2 1 2 3 Since the system is type 1, the steady-state error due to a step input is zero for all values of k , k , and k that correspond to a stable system. The characteristic equation of the closed-loop system is 3 2 s + ( 4 + k2 ) s + ( 3 + k1 +k 2 )s + k 3 = 0 and For the roots to be at -1 + j, -1 - j, -10, the equation should be: 3 2 s + 12 s + 22 s + 20 = 0 2 Equating like coefficients of the last two equations, we have 4 + k = 12 3 Thus Thus Thus k + k 1 + k 2 = 22 k 3 2 1 =8 = 11 k = 20 k3 = 20 (b) Open-loop Transfer Function. Y (s ) Gc ( s) 20 = = 2 E ( s ) ( s + 1) ( s + 3 ) s s + 12 s + 22 ( ) Thus Gc ( s) = 20 ( s + 1 ) ( s + 3 ) s s + 12 s + 22 2 ( ) 10-44 (a) 275 0 25.92 A = 0 -2.36 1 0 0 0 0 0 0 0 0 0 1 0 0 -0.0732 B = 0 0.0976 The feedback gains, from k 1 to k 4 : -2.4071E+03 The A -4.3631E+02 - B K -8.4852E+01 -1.0182E+02 matrix of the closed-loop system -1.5028E+02 0.0000E+00 0.0000E+00 2/3258E+02 -3.1938E+01 1.0000E+00 -6.2112E+00 0.0000E+00 4.2584E+01 0.0000E+00 0.0000E+00 -7.4534E+00 0.0000E+00 1.0000E+00 8.2816E+00 9.9379E+00 The B vector -7.3200E-02 0.0000E+00 0.0000E+00 9.7600E-02 Time Responses: 10-44 (b) The feedback gains, from k 1 to k 2 : -9.9238E+03 The A -1.6872E+03 - B K -1.3576E+03 -8.1458E+02 matrix of the closed-loop system 276 -7.0051E+02 0.0000E+00 0.0000E+00 9.6621E+02 -1.2350E+02 1.0000E+00 0.0000E+00 1.6467E+02 -9.9379E+01 0.0000E+00 0.0000E+00 -5.9627E+01 0.0000E+00 1.0000E+00 1.3251E+02 7.9503E+01 The B vector -7.3200E-02 0.0000E+00 0.0000E+00 9.7600E-02 Time Responses: 10-45 The solutions are obtained by using the pole-placement design option of the linsys program in the ACSP software package. (a) The feedback gains, from k 1 to k 2 : -6.4840E+01 -5.6067E+00 2.0341E+01 2.2708E+00 The A - B K matrix of the closed-loop system -3.0950E+02 -4.6190E+02 0.0000E+00 0.0000E+00 1.0000E+00 0.0000E+00 0.0000E+00 1.1463E+02 1.4874E+01 0.0000E+00 0.0000E+00 1.0000E+00 -3.6724E+01 1.7043E+02 1.477eE+01 -3.6774E+01 The B vector -6.5500E+00 0.0000E+00 0.0000E+00 277 -6.5500E+00 (b) Time Responses: x ( 0 ) = 0 .1 0 0 0 ' With the initial states x ( 0 ) = 0 .1 0 0 0 , the initial position of ' from its stable equilibrium position. The steel ball is initially pulled toward the magnet, so x 1 or y1 is preturbed downward x 3 = y 2 is x 3 or y 2 is preturbed negative at first. Finally, the feedback control pulls both bodies back to the equilibrium position. With the initial states x ( 0 ) = 0 0 0 .1 0 , the initial position of ' downward from its stable equilibrium. For t > 0, the ball is going to be attracted up by the magnet toward the equilibrium position. The magnet will initially be attracted toward the fixed iron plate, and then settles to the stable equilibrium position. Since the steel ball has a small mass, it will move more actively. 10-46 (a) Block Diagram of System. 278 u = -k1 x1 + k 2 ( - x1 + w1 ) dt State Equations of Closed-loop System: dx1 dt -2 - k1 1 x1 0 + = dx2 - k2 0 x2 k2 dt Characteristic Equation: 1 w1 0 w2 sI - A = For s = s + 2 + k1 k2 s2 -1 s = s + ( 2 + k1 ) s + k2 = 0 2 -10, -10, X (s) sI -A = = + 20 s + 200 = 0 -2 Thus k1 = 18 and k2 = 200 = X 1 (s) 200 W 1 ( s ) s 1+ 2 s -1 + s -1W 2 ( s ) -1 + 18 = s + 200 s -2 = 200 W 1 ( s ) s 2 + sW 2 ( s ) + 20 s + 200 W1 ( s ) = 1 s W2 (s) W2 s W2 = const ant X ( s) = s s + 20 s + 200 2 ( 200 + W2s ) lim x( t) = lim sX ( s) = 1 t s 0 10-46 (b) With PI C ontroller: Block Diagram of System: 279 Set K P = 2 and KI = 200 . X (s ) = (K P s + K I ) W1 ( s ) + sW2 ( s ) s + 20 s + 200 2 = ( 2s + 200 ) W ( s ) + sW ( s ) 1 2 s + 20 s + 200 2 Time Responses: 280 Chapter 11 THE VIRTUAL LAB Part 1) Solution to Lab questions within Chapter 11 11-5-1 Open Loop Speed 1. Open loop speed response using SIMLab: a. +5 V input: The form of response is like the one that we expected; a second order system response with overshoot and oscilla tion. Considering an amplifier gain of 2 and K b = 0.1 , the desired set point should be set to 2.5 and as seen in the figure, the final value is approximately 50 rad/sec which is armature voltage divided by K b . To find the above response the systems parameters are extracted from 11-3-1 of the text and B is calculated from 11-3 by having m as: m = Ra J m , Ra B + k b k m B= Ra J m - k b k m m = 0.000792kg m 2 / sec Ra m b. +15 V input: 282 c. 10 V input: 2. Study of the effect of viscous friction: 283 The above figure is plotted for three different friction coefficients (0, 0.001, 0.005) for 5 V armature input. As seen in figure, two important effects are observed as the viscous coefficient is increased. First, the final steady state velocity is decreased and second the response has less oscillation. Both of these effects could be predicted from Eq. (11.1) by increasing damping ratio . 284 3. Additional load inertia effect: As the overall inertia of the system is increased by 0.005 / 5.2 2 and becomes 1.8493 10 -3 kg.m2 , the mechanical time constant is substantially increased and we can assume the first order model for the motor (ignoring the electrical sub-system) and as a result of this the response is more like an exponential form. The above results are plotted for 5 V armature input. 285 4. Reduce the speed by 50% by increasing viscous friction: As seen in above figure, if we set B=0.0075 N.s/m the output speed drop by half comparing with the case that B=0 N.s/m. The above results are plotted for 5 V armature input. 286 5. Study of the effect of disturbance: Repeating experiment 3 for B=0.001 N.s/m and T L =0.05 N.m result in above figure. As seen, the effect of disturbance on the speed of open loop system is like the effect of higher viscous friction and caused to decrease the steady state value of speed. 287 6. Using speed response to estimate motor and load inertia: Using first order model we are able to identify system parameters based on unit step response of the system. In above plot we repeated the experiments 3 with B=0.001 and set point voltage equal to 1 V. The final value of the speed can be read from the curve and it is 8.8, using the definition of system time constant and the cursor we read 63.2% of speed final value 5.57 occurs at 0.22 sec, which is the system time constant. Considering Eq. (11-3), and using the given value for the rest of parameters, the inertia of the motor and load can be calculated as: J= m ( Ra B + K m K b ) 0.22(1.35 0.001 + 0.01) = = 1.8496 10 - 3 kg.m2 Ra 1.35 We also can use the open loop speed response to estimate B by letting the speed to coast down when it gets to the steady state situation and then measuring the required time to get to zero speed. Based on this time and energy conservation principle and knowing the rest of parameters we are able to calculate B. However, this method of identification gives us limited information about the system parameters and we need to measure some parameters directly from motor such as Ra , K m , K b and so on. So far, no current or voltage saturation limit is considered for all simulations using SIMLab software. 288 7. Open loop speed response using Virtual Lab: a. +5 V: 289 b. +15 V: c. 10 V: 290 Comparing these results with the part 1, the final values are approximately the same but the shape of responses is closed to the first order system behavior. Then the system time constant is obviously different and it can be identified from open loop response. The effect of nonlinearities such as saturation can be seen in +15 V input with appearing a straight line at the beginning of the response and also the effects of noise and friction on the response can be observed in above curves by reducing input voltage for example, the following response is plotted for a 0.1 V step input: 291 8. Identifying the system based on open loop response: Open loop response of the motor to a unit step input voltage is plotted in above figure. Using the definition of time constant and final value of the system, a first order model can be found as: G( s ) = 9 , 0.23s + 1 where the time constant (0.23) is found at 5.68 rad/sec (63.2% of the final value). 292 11-5-2 Open Loop Sine Input 9. Sine input to SIMLab and Virtual Lab (1 V. amplitude, and 0.5, 5, and 50 rad/sec frequencies) a. 0.5 rad/sec (SIMLab): 293 b. 5 rad/sec (SIMLab): c. 50 rad/sec (SIMLab): 294 d. 0.5 rad/sec (Virtual Lab): e. 5 rad/sec (Virtual Lab): 295 f. 50 rad/sec (Virtual Lab): 10. Sine input to SIMLab and Virtual Lab (20 V. amplitude) a. 0.5 rad/sec (SIMLab): 296 b. 5 rad/sec (SIMLab): c. 50 rad/sec (SIMLab): 297 d. 0.5 rad/sec (Virtual Lab): e. 5 rad/sec (Virtual Lab): 298 f. 5 rad/sec (Virtual Lab): In both experiments 9 and 10, no saturation considered for voltage and current in SIMLab software. If we use the calculation of phase and magnitude in both SIMLab and Virtual Lab we will find that as input frequency increases the magnitude of the output decreases and phase lag increases. Because of existing saturations this phenomenon is more sever in the Virtual Lab experiment (10.f). In this experiments we observe that M = 0.288 and = -93.82 o for = 50. 299 11-5-3 Speed Control 11. Apply step inputs (SIMLab) In this section no saturation is considered either for current or for voltage. a. +5 V: 300 b. +15 V: c. -10 V: 301 12. Additional load inertia effect: a. +5 V: b. +15 V: 302 c. -10 V: 13. Study of the effect of viscous friction: 303 As seen in above figure, two different values for B are selected, zero and 0.0075. We could change the final speed by 50% in open loop system. The same values selected for closed loop speed control but as seen in the figure the final value of speeds stayed the same for both cases. It means that closed loop system is robust against changing in system's parameters. For this case, the gain of proportional controller and speed set point are 10 and 5 rad/sec, respectively. 14. Study of the effect of disturbance: Repeating part 5 in section 11-5-1 for B=0.001 and T L =0.05 N.m result in above figure. As seen, the effect of disturbance on the speed of closed loop system is not substantial like the one on the open loop system in part 5, and again i is shown the robustness of closed loop system against disturbance. Also, to study the effects of t conversion factor see below figure, which is plotted for two different C.F. and the set point is 5 V. 304 By decreasing the C.F. from 1 to 0.2, the final va lue of the speed increases by a factor of 5. 15. Apply step inputs (Virtual Lab) a. +5 V: 305 b. +15 V: c. 10 V: 306 As seen the responses of Virtual Lab software, they are clearly different from the same results of SIMLab software. The nonlinearities such as friction and saturation cause these differences. For example, the chattering phenomenon and flatness of the response at the beginning can be considered as some results of nonlinear elements in Virtual Lab software. 11-5-4 Position Control 16. 160 o step input (SIMLab) 307 17. 0.1 N.m step disturbance 18. Examine the effect of integral control 308 In above figure, an integral gain of 1 is considered for all curves. Comparing this plot with the previous one without integral gain, results in less steady state error for the case of controller with integral part. 19. Additional load inertia effect (J=0.0019, B=0.004): 309 20. Set B=0: 21. Study the effect of saturation 310 The above figure is obtained in the same conditions of part 20 but in this case we considered 10 V. and 4 A. as the saturation values for voltage and current, respectively. As seen in the figure, for higher proportional gains the effect of saturations appears by reducing the frequency and damping property of the system. 22. Comments on Eq. 11-13 After neglecting of electrical time constant, the second order closed loop transfer function of position control obtained in Eq. 11-13. In experiments 19 through 21 we observe an under damp response of a second order system. According to the equation, as the proportional gain increases, the damped frequency must be increased and this fact is verified in experiments 19 through 21. Experiments16 through 18 exhibits an over damped second order system responses. 23. In following, we repeat parts 16 and 18 using Virtual Lab: Study the effect of integral gain of 5: 311 312 Ch. 11 Problem Solutions Part 2) Solution to Problems in Chapter 11 11-1. In order to find the current of the motor, the motor constant has to be separated from the electrical component of the motor. The response of the motor when 5V of step input is applied is: a) The steady state speed: 41.67rad/sec b) It takes 0.0678 second to reach 63% of the steady state speed (26.25rad/sec). This is the time constant of the motor. c) The maximum current: 2.228A 313 11.2 The steady state speed at 5V step input is 50rad/sec. a) It takes 0.0797 seconds to reach 63% of the steady state speed (31.5rad/sec). b) The maximum current: 2.226A c) 100rad/sec 314 11-3 a) b) c) d) 50rad/sec 0.0795 seconds 2.5A. The current When Jm is increased by a factor of 2, it takes 0.159 seconds to reach 63% of its steady state speed, which is exactly twice the original time period . This means that the time constant has been doubled. 315 11-4 Part 1: Repeat problem 11-1 with TL = -0.1Nm a) It changes from 41.67 rad/sec to 25 rad/sec. b) First, the speed of 63% of the steady state has to be calculated. 41.67 - (41.67 - 25) 0.63 = 31.17 rad/sec. The motor achieves this speed 0.0629 seconds after the load torque is applied c) 2.228A. It does not change Part 2: Repeat problem 11-2 with TL = -0.1Nm a) It changes from 50 rad/sec to 30 rad/sec. b) The speed of 63% of the steady state becomes 50 - (50 - 30) 0.63 = 37.4 rad/sec. The motor achieves this speed 0.0756 seconds after the load torque is applied c) 2.226A. It does not change. 316 Part 3: Repeat problem 11-3 with TL = -0.1Nm a) It changes from 50 rad/sec to 30 rad/sec. b) 50 - (50 - 30) 0.63 = 37.4 rad/s The motor achieves this speed 0.0795 seconds after the load torque is applied. This is the same as problem 11-3. c) 2.5A. It does not change d) As TL increases in magnitude, the steady state velocity decreases and steady state current increases; however, the time constant does not change in all three cases. 317 11-5 The steady state speed is 4.716 rad/sec when the amplifier input voltage is 5V: 11-6 a) 6.25 rad/sec. b) 63% of the steady state speed: 6.25 0.63 = 3.938 rad/sec It takes 0.0249 seconds to reach 63% of its steady state speed. c) The maximum current drawn by the motor is 1 Ampere. 318 11-7 a) 9.434 rad/sec. b) 63% of the steady state speed: 9.434 0.63 = 5.943 rad/sec It takes 0.00375 seconds to reach 63% of its steady state speed. c) The maximum current drawn by the motor is 10 Amperes. d) When there is no saturation, higher Kp value reduces the steady state error and decreases the rise time. If there is saturation, the rise time does not decrease as much as it without saturation. Also, if there is saturation and Kp value is too high, chattering phenomenon may appear. 11-8 a) The steady state becomes zero. The torque generated by the motor is 0.1 Nm. b) 6.25 - (6.25 - 0) 0.63 = 2.31 rad/sec. It takes 0.0249 seconds to reach 63% of its new steady state speed. It is the same time period to reach 63% of its steady state speed without the load torque (compare with the answer for the Problem 11-6 b). 11-9 The SIMLab model becomes The sensor gain and the speed input are reduced by a factor of 5. In order to get the same result as Proble m 11-6, the Kp value has to increase by a factor of 5. Therefore, Kp = 0.5. The following graphs illustrate the speed and current when the input is 2 rad/sec and Kp = 0.5. 319 11-10 320 a) 1 radian. b) 1.203 radians. c) 0.2215 seconds. 11-11 a) The steady state position is very close to 1 radian. b) 1.377 radians. c) 0.148 seconds. It has less steady state error and a faster rise time than Problem 11-10, but has larger overshoot. 321 11-12 Different proportional gains and the ir corresponding responses are shown on the following graph. As the proportional gain gets higher, the motor has a faster response time and lower steady state error, but if it the gain is too high, the motor overshoot increases. If the system requires that there be no overshoot, Kp = 0.2 is the best value. If the system allows for overshoot, the best proportional gain is dependant on how much overshoot the system can have. For instance, if the system allows for a 30% overshoot, Kp = 1 is the best value. 322 11-13 Let Kp = 1 is the best value. As the derivative gain increases, overshoot decreases, but rise time increases. 323 11-14 324 11-15 There could be many possible answers for this problem. One possible answer would be Kp = 100 Ki= 10 Kd= 1.4 The Percent Overshoot in this case is 3.8%. 325 11-16 0.1 Hz 0.2 Hz 326 0.5 Hz 1 Hz 327 2Hz 5Hz 328 10Hz 50Hz 329 As frequency increases, the phase shift of the input and output also increase. Also, the amplitude of the output starts to decrease when the frequency increases above 0.5Hz. 11-17 As proportional gain increases, the steady state error decreases. 330 11-18 Considering fast response time and low overshoot, Kp =1 is considered to be the best value. 331 11-19 It was found that the best Kp = 1 As Kd value increases, the overshoot decreases and the rise time increases. 332 Appendix H GENERAL NYQUIST CRITERION H-1 (a) L( s ) = 5( s s(s - 2) + 1 )( s - 1 ) P =1 P =1 When = 0 : L ( j 0 ) = -90 o o L( j0 ) = =0 When = : L ( j ) = -180 L ( j) L ( j ) = 5( j - 2) - j 1 + ( 2 ) = - 5( + 2 j ) 1+ ( 2 ) When L ( j ) = 0 , = . The Nyquist plot of L ( j ) does not intersect the real axis except at the origin. o o o 11 = ( Z - 0.5 P - P )180 = ( Z - 1.5 ) 180 = - 90 Thus, Z=1. The closed-loop system is unstable. The characteristic equation has 1 root in the right-half s-plane. Nyquist Plot of L ( j ): (b) L ( j ) = s( s 50 + 5)( s - 1 ) P =1 o P =1 When = 0 : L ( j 0 ) = 90 L( j0) o = =0 When = : L ( j ) = -270 L ( j) 333 L ( j ) = 50 2 -4 - j 5 + ( 2 ) = 50 -4 + j 5 + 2 16 + 4 2 (5+ ) 2 ( 2 ) 2 For Im L ( j ) = 0 , = . Thus, the Nyquist plot of L(s) intersects the real axis only at the origin where = . 11 = ( Z - 0.5 P - P )180 = ( Z - 1.5 ) 180 = 90 o o o Thus, Z = 2 The closed-loop system is unstable. The characteristic equation has two roots in the right-half s-plane. The Nyquist plot of L ( j ): (c) L ( s ) = s s + 3s + 1 3 ( 3( s + 2) ) P =1 P=2 When = 0 : L ( j 0 ) = -90 o o L( j0 ) = =0 4 2 When = : L ( j ) = -270 3( j + 2) 4 L ( j) L ( j ) = ( - 3 2 ) + j or = 3( j + 2) - 3 (4 ( 4 - 3 2 ) ) - j 2 2 + Setting Im L ( j ) = 0, - 3 - 2 = 0 4 2 = 3 . 56 2 o = 1.89 rad/sec. o o L ( j 1.89 ) =3 11 = ( Z - 0.5 P - P )180 = ( Z - 2.5 ) 180 = - 90 Thus, Z = 2 The closed-loop system is unstable. The characteristic equation has two roots in the right-half s-plane. 334 Nyquist Plot of L ( j ): (d) L ( s ) = 100 s ( s + 1) s + 2 2 ( ) P =3 o o P=0 L( j0 ) When = 0 : L ( j 0 ) = -90 = When = : L ( j ) = -360 L ( j) = 0 The phase of L ( j ) is discontinuous at = 1.414 rad/sec. 11 = 35.27 + 270 - 215.27 o o ( o ) = 90 o 11 = ( Z - 1.5) 180 = 90 o o Thus, P 11 = 360 180 o o = 2 The closed-loop system is unstable. The characteristic equation has two roots in the right-half s-plane. Nyquist Plot of L ( j ): 335 (e) L(s ) = s s + 2 s + 2 s + 10 3 2 ( s - 5s + 2 2 ) L( j0 ) P =1 P=2 When = 0 : L ( j 0 ) = -90 o o = =0 2 4 2 2 When = : L ( j ) = -180 2 L ( j) ( 2 - ) - j5 ( - 2 ) - j (10 - 2 ) ( 2 - ) - j5 L ( j ) = = - 2 ) + j (10 - 2 ) ( ( - 2 ) + (10 - 2 ) 4 2 2 4 2 2 2 2 2 Setting L( j) = 0, 4 ( 2 - )(10 - 2 ) + 5( 2 2 4 - 2 2 )= 0 / sec or - 3 .43 + 2 .86 = 0 2 Thus, = 1.43 o r 2.0 1. = 1.2 rad / sec or L ( j1.2 ) = - 0 . 7 L ( j1.42 ) = -0 . 34 2 o o 1.42 r ad 11 = ( Z - 0.5 P - P )180 = ( Z - 2.5 )180 = -90 o Thus, Z = 2 . The closed-loop system is unstable. The characteristic equation has two roots in the right-half s-plane. Nyquist Plot of L ( j ): (f) L(s ) = - s -1 2 2 ( ) ( s + 2 ) = -0.1s - 0.2 s + 0.1s + 0.2 s ( s + s + 1) s ( s + s + 1) 3 2 2 P =1 P=0 When = 0 : L ( j 0 ) = -90 o o L( j0 ) = = 0 .1 When = : L ( j ) = 180 L( j) 336 ( 0.2 + 0.2 ) + 0.1 j ( 1 + ) = (1+ ) ( 0.2 + j 0.1 ) - - j ( 1- ) L ( j ) = - + j (1 - ) + ( 1- ) 2 2 2 2 2 2 2 4 2 2 2 Setting Im L ( j ) = 0 , = 2 . 2 o Thus, o = 1.414 rad/sec o L ( j 1.414 ) = - 0 .3 11 = ( Z - 0.5 P - P )180 = ( Z - 0.5 ) 180 = - 90 The closed-loop system is stable. Nyquist Plot of L ( j ): Thus, Z = 0 H-2 (a) L(s ) = K ( s - 2) s s -1 2 ( ) P =1 P =1 o o For stability, Z = 0. 11 = - ( 0.5 P + P ) 180 = -270 For K > 0, 11 = -90 . For K < 0, 11 = +90 o o The system is unstable. o -270 The system is unstable. Thus the system is unstable for all K. 337 Nyquist plot (b) L( s ) = s(s K + 10 )( s - 2) P =1 P =1 For stability, Z = 0. 11 = - ( 0.5 P + P ) 180 = -1.5 180 = - 270 o o o For K > 0, 11 = 90 o . o The system is unstable. -270 o For K < 0, 11 = -90 Nyquist Plot of L ( j ): . The system is unstable for all values of K . 338 (c) L ( s ) = s s + 3s + 1 3 ( K ( s + 1) ) o o P =1 P = 2 For stability, Z = 0. 11 = - ( 0.5 P + P- 1 ) 180 = - 450 o o For K > 0, 11 = -90 For K < 0, 11 = +90 . The system is unstable. when K > -1. 11 = -270 o when K < -1. The Nyquist plot of L ( j ) crosses the real axis at K, and the phase crossover frequency is 1.817 rad/sec. The system is unstable for all values of K. Nyquist Plot of L ( j ): (d) L ( s ) = K s - 5s + 2 2 s s + 2 s + 2 s + 10 3 2 ( ( ) ) P =1 P = 2 For stability, Z = 0. 11 = - ( 0.5 P + P ) 180 = -2.5 180 = - 450 o o o The Nyquist plot of L ( j ) intersects the real axis at the following points: = : L ( j 0 ) = 0 = 1.42 rad / sec: L ( j 1.42 ) = - 0 . 34 = 1.2 rad / sec: L ( j 1.4 ) = -0 . 7 K K 339 0 1.43 < < K K < 1.43 < 2 . 94 11 = -90 11 = 270 11 = -90 11 = 90 o o o o 2 . 94 K <K <0 Since 11 does not equal to - 450 Nyquist Plot of L ( j ): o for any K, the system is unstable for all values of K . (e) L(s ) = K s -1 2 2 ( )( s + 2) s ( s + s + 1) L ( j ) P =1 P=0 For stability, Z = 0. The Nyquist plot of 11 = - ( 0.5 P + P ) 180 = - 90 o o intersects the real axis at: 340 = : = 2 rad / sec: L ( j ) L( j = K 2 ) = 3K K>0 K 11 = 90 11 = 90 1 3 o o o Unstable Unstable Unstable Stable < -1 -1 < K < - - 1 3 11 = 270 11 = -90 <K <0 o The system is stable for - 1 3 <K <0. (f) L(s ) = ) s ( s + 1) ( s + 4 ) K s - 5s+1 2 2 ( P =3 P=0 For stability, Z = 0. 11 = - ( 0.5 P + P ) 180 = -270 o o The Nyquist plot of L ( j ) plot intersects the real axis at = : < = 0 . 92 0 K K K L ( j ) rad / sec: =0 L ( j 0 . 92 ) = -1. 3 K o o < 0 . 77 11 = -90 - 180 11 = 90 o o = - 270 o Stable Unstable Unstable > 0 . 77 <0 11 = 270 The system is stable for 0 < K < 0.77. 341 Nyquist Plot: H-3 (a) = ( Z - 0.5 P - P )180 = ( Z - 1.5 ) 180 = 90 o o 11 o Thus, Z = 2 The closed-loop system is unstable. The characteristic equation has two roots in the right-half s-plane. (b) 11 = ( Z - 0.5 P - P )180 = ( Z - 1 ) 180 = - 180 o o o Thus, Z = 0 . The closed-loop system is stable. (c) = ( Z - 0.5 P - P )180 = ( Z - 2 )180 = -360 o o 11 o Thus, Z = 0 . The closed-loop system is stable. (d) = ( Z - 0.5 P - P )180 = Z 180 = 360 o o 11 o Thus, Z = 2 The closed-loop system is unstable. The characteristic equation has two roots in the right-half s-plane. (e) = ( Z - 0.5 P - P )180 = ( Z - 1 ) 180 = - 162 - 18 o o o 11 o Thus, Z = 0 . The closed-loop system is stable. (f) 11 = ( Z - 0.5 P - P )180 = ( Z - 0.5 ) 180 = - 90 o o o Thus, Z = 0 The closed-loop system is stable. H-4 (a) The stability criterion for 1 / L ( s ) is the same as that for o L ( s ). o Thus, 11 = ( Z - 0.5 P - P )180 = ( Z - 1.5 ) 180 From Fig. HP-4 (a), 11 = -270 o . Thus , Z = 0. The closed-loop system is stable. 342 (b) The stability criterion for 1 / L ( s ) is the same as that for 11 = ( Z - 0.5 P - P )180 = ( Z - 1 ) 180 o L ( s ). o Thus, From Fig. HP-4 (b), 11 = -180 H-5 o . Thus, Z = 0. The closed-loop system is stable. Thus, 11 = -270 o (a) = ( Z - 0.5 P - P )180 = ( Z - 1.5 ) 180 11 o For stability, Z = 0. 0 < K K K < 0 .5 11 = 90 o o Unstable Stable Unstable o > 0 .5 <0 11 = -270 11 = -90 The system is stable for K > 0.5. (b) = ( Z - 0.5 P - P )180 = ( Z - 1 ) 180 o 11 o For stability, Z = 0. Thus, 11 = -180 o 0 K K < K < 0 .5 11 = 90 o o Unstable Stable Unstable o > 0 .5 <0 11 = -270 11 = -90 The system is stable for K > 0.5. 343 (c) = ( Z - 0.5 P - P )180 = ( Z - 0.5 ) 180 o 11 o For stability, Z = 0. Thus, 11 = -90 o o o o 0 < K < 0 .5 K 11 = -90 11 = 270 11 = -90 11 = 90 o Stable Unstable Stable Unstable 0 .5 K K < >1 <1 <0 The system is stable for K > 1. (d) = ( Z - 0.5 P - P )180 = ( Z - 2 ) 180 o 11 o For stability, Z = 0. Thus, 11 = -360 o o o 0 K K < K < 0 .5 11 = 0 11 = 0 Unstable Stable Unstable > 0 .5 <0 11 = -360 o The system is stable for K > 0.5. H-6 (a) + 4 Ks + ( K + 5) s + 10 = 0 Ks ( 4 s + 1 ) L (s) = P =0 eq 3 s + 5 s + 10 s 3 2 P =2 =0 =0 When = 0 : L eq ( j 0 ) = 90 When = : Leq ( j ) = 4 2 o o L eq ( j0 ) eq L eq ( j ) = - 90 2 L ( j ) 2 K ( j - 4 ) 10 + j (5 - ) 2 2 = K ( j - 4 ) 10 - j 5 - 100 + Thus, o 2 ( (5- ) 2 2 ) 2 Setting Im L eq ( j ) =0 - 5 - 2 .5 = 0 = 5 .46 = 2 .34 rad/sec. o L eq ( j 2 . 34 ) = -2 . 18 K For stability, 11 = - ( 0.5 P + P ) 180 = -360 Thus for the system to be stable, the - 1 point should be encircled by the Nyquist plot once. The system is stable for K > 1/2.18 = 0.458. 344 Routh Tabulation: s s s 3 1 4K 4K 2 K +5 10 K K 2 2 >0 1 + 20 4K K - 10 + 5 K - 2 .5 > 0 + 5 .45)( K s o 10 (K - 0 .458 ) >0 For system stability, K > 0.458. (b) s 3 + K ( s 3 + 2s 2 +1 ) = 0 Leq ( s) = K s + 2s + 1 3 ( ) s 3 P = 3 o P =0 L eq When = 0: When = : Leq ( j ) = 1 L eq ( j 0 ) = - 270 L eq ( j ) = 0 o ( j0 ) = L eq ( j ) =K Setting =K shows that its real part is Im L ( j ) K 1 - 2 ( 2 ) - j 3 3 - j K 3 + j ( 1 - 2 2 ) = 3 L eq ( j 0 . 707 ) eq =0 - 2 3 2 =0 = , = 0 . 707 rad/sec The Nyquist plot is a straight line, since the equation of always equal to K for all values of . For stability, = - ( 0.5 P + P ) 180 = -270 o 11 o L eq ( j ) K>0 K < 0 11 = 90 11 = -90 o o Unstable Unstable The system is unstable for all values of K. Routh Tabulation: s s s 3 1+ K 2K 0 K K K 2 >0 < -1 >0 1 -1 - K 2K s o K K The system is unstable for all values of K . 345 (c) s ( s + 1) ( s2 + 4) + K ( s 2 + 1) = 0 Leq ( s) = K s +1 2 s ( s + 1) s + 4 2 ( ( ) ) P = 3 P =0 For stability, 11 = - ( 0.5 P + P ) 180 = -270 o o 0 K < K < 11 = - 270 11 = 270 o o Stable Unstable <0 (d) Leq ( s) = Leq ( j ) = s s + 2 s + 20 2 ( 10 K ) P =1 2 P =0 10 K - 2 + j 20 - 2 ( 2 ) = 10 K -2 - j 20 - 4 + 4 2 ( 20 - ) 2 ( 2 ) 2 Setting L eq Im L eq ( j ) K 4 =0 = 4.47 ( j 4.47 ) =- For stability, 11 = - ( 0.5 P + P ) 180 = - 90 o o K 0 >4 < K 11 = 270 <4 11 = -90 o o Unstable Stable The system is stable for 0 < K < 4. Routh Tabulation s s s 3 1 2 40 20 10 K K K 2 1 - 10 2 <4 >0 For stability, 0 < K < 4 s o 10 K K 346 (e) s ( s + 2s + s +1) + K ( s + s + 1) = 0 3 2 2 Leq ( s) = s s + 3s + 3 3 ( K ( s + 2) ) P =1 = K P=2 L eq ( j 0 ) = - 90 o L eq ( j ) = 0 - 270 o Leq ( j ) = Setting ( K ( j + 2) 4 - 3 ( j ) 2 ) + j 3 ( 2 4 - 3 2 ) + j ( 4 - 3 2 - 6 ) ( 4 - 3 2 ) 2 + 4 2 Im L eq =0 4 - 3 2 - 6 = 0 The real positive solution is = 2 . 09 rad/sec L eq ( j 2 . 09 ) = 0 . 333 K For stability, 11 = - ( 0.5 P + P ) 180 = -450 o o K K 0 >0 < -3 > K 11 = -90 11 = +90 > -3 o o o Unstable Unstable Unstable 11 = -270 The system is unstable for all values of K . Routh Tabulation s s s 4 1 3 3 + K 2 K 3 3 2 -3- K -3 - K 2 2K K < -3 s 1 -( 3 + K ) - 2 K -( 3 + K ) s o 1 2K K >0 The conditions contradict. The system is unstable for all values of K. 347 Appendix I DISCRETE-DATA CONTROL SYSTEMS I-1 (a) F (z) = (b) (z-e ) -3 ze-3 2 F (z ) = -2 j ze 2 j ze - 2 2 j ( z - e 2 j ) ( z - e- 2 j ) 2 1 (c) z z F (z ) = z - e - (2- j ) - z - e - (2+ j ) 2j 1 (d) F (z ) = ze (z + e ) (z - e ) -2 -2 -2 3 I-2 (a) F (z ) = - j 2T ze j 2 T ze - 2 2 2 j ( z - e j 2T ) ( z - e - j 2T ) T (b) F (z) =1 - z -1 +z -2 -z -3 +z -4 -L Add Eqs. (1) and (2). We have (1 + z ) F ( z ) = 1 -1 (1) z -1 F (z ) = z -1 -z -2 +z -3 -z -4 +L (2) Or, F (z) =z (z + 1) I-3 (a) Z 2 -5 T 2 -10T T e z T e z 1 ( -1) 2 2 z = = + 2 -aT 2 ( s + 5) 3 2 ! a z - e a = 5 2 ( z - e -5T ) ( z - e -5T )3 (b) F (s) = = 1 s (s z z 3 + 1) - (z = Tz 1 s - 1 s 2 + 2 1 s 3 - 1 s +1 - z z F (z) -1 - 1) = - 2 + T z(z + 1) 3 2( z - 1) 10 -e 2 -T (c) F (s) = s(s 10 10 25 s + 5) 2 - 10 z 25 ( s + 5) - - (s + 5) -5T -5 T 2 F (z ) = 10 z 25( z - 1) 25 z - e ( -5 T ) (z -e ) 2 Tze 2 (d) F (s ) = s s +2 2 ( 5 ) = 2.5 s - 2.5 s s +2 2 F (z) = 2 .5 z z -1 - 2 .5 z ( z z 2 - cos T ) - 2 z cos T + 1 348 I-4 (a) F (z) = 10 z (z - 1 )( z - 0 .2 ) = 12 . 5 z z -1 - 12 . 5 z z - 0 .2 f (k ) = 12 . 5 1 - ( 0 .2 ) k k = 0 , 1, 2, L (b) F (z ) = ( z - 1) ( z z 2 + z +1 ) = ( 0.1667 + j 0.2887 ) z ( 0.1667 - j 0.2887) z - 3( z - 1 ) ( z + 0.5 + j0.866) ( z + 0.5 - j0.866 ) z - - (0.1667 - j 0.2887)e j 2 k 3 f ( k ) = 0.333 - (0.1667 + j 0.2887) e = 0.333 - 0.333cos - j 2 k 3 2 k 3 - 0.576sin 2 k 3 = 0.333 + 0.666cos 2 k + 240 0 k = 0,1, 2 , L 3 (c) F (z) = z (z - 1 )( z + 0 . 85) = 0 . 541 z z -1 - 0 . 541 z z + 0 .85 f (k ) = 0 .541 1 - ( -0 . 85) k k = 0 , 1, 2 , L (d) F (z) = 10 (z - 1 )( z - 0 . 5) = 20 z -1 - 20 z - 0 .5 f (0) =0 f (k ) = 20 1 - ( 0 .5) k -1 k = 0 , 1, 2 , L I-5 (a) lim f ( k ) = lim 1 - z k z 1 ( -1 ) F ( z ) = lim z z 1 0.368 2 - 1.364 z + 0.732 z -1 =1 Expand F(z) into a power series of k 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 f(k) 0.000000 0.000000 0.368000 0.869952 1.285238 1.484260 1.451736 1.261688 1.026271 0.844277 0.768362 0.798033 0.894075 1.003357 1.082114 . k 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 f(k) 1.109545 1.089310 1.041630 0.991406 0.957803 0.948732 0.960956 0.984269 1.007121 1.021225 1.023736 1.016836 1.005586 0.995292 0.989487 349 (b) F (z) = 10 z (z - 1 )( z + 1 ) The function 1 - z ( -1 ) F ( z) has a pole at z = -1 , so the final-value theorem cannot be applied. The response f(k) oscillates between 0 and 10 as shown below. k 1 2 3 4 5 6 7 8 9 10 f(k) 0.000000 10.000000 0.000000 10.000000 0.000000 10.000000 0.000000 10.000000 0.000000 10.000000 k 11 12 13 14 15 16 17 18 19 20 z (z z f(k) 0.000000 10.000000 0.000000 10.000000 0.000000 10.000000 0.000000 10.000000 0.000000 10.000000 (c) k lim f (k ) = lim z 1 - 1 z -1 F (z) = lim z 1 - 0 . 5) = 2 F(z) is expanded into a power series of k 1 2 3 4 5 6 7 8 9 10 f(k) 1.000000 1.500000 1.750000 1.875000 1.937500 1.968750 1.984375 1.992188 1.996094 1.998047 z (z -1 by long division. k 11 12 13 14 15 16 17 18 19 20 f(k) 1.999023 1.999512 1.999756 1.999878 1.999939 1.999969 1.999985 1.999992 1.999996 1.999998 (d) F (z) = - 1 )( z - 1. 5) F (z) has a pole outside the unit circle. F ( z ) . F ( z ) is f (k ) is unbounded. The final-value theorem cannot be applied to z -1 expanded into a power series of . f(k) 0.000000 1.000000 2.500000 4.750000 8.125000 13.187500 k 11 12 13 14 15 16 f(k) 113.33078 170.995117 257.492676 387.239014 581.858521 873.787781 k 1 2 3 4 5 6 350 7 8 9 10 20.781250 32.171875 49.257813 74.886719 x (0) 17 18 19 20 = x (1 ) 1311.681641 1968.522461 2953.783691 4431.675781 Taking the z-transform on both z I-6 (a) x( k + 2 ) - x ( k + 1) + 0 .1 x ( k ) = u s ( k ) -z - zx ( 1) - zX z 2 =0. sides, z X (z) 2 2 x(0) (z) + zx ( 0 ) + 0 .1 X ( z ) = 10 z z -1 + 1.455 z z - 0.1127 - z X ( z) = transform, x( k ) ( z - 1) ( z - z + 0.1 ) = -1 11.455z z - 0.8873 Taking the inverse z- = 10 + 1.455 ( 0 .1127 ) k - 11 .455 = 1, ( 0 . 8873) k k = 0 , 1, 2 , L (b) x( k + 2) - x (k ) = 0 x (0) x (1 ) =0 2 Taking the z-transform on both sides, we have z X ( z ) - z x(0) - zx(1) - X ( z ) = 0 2 2 X ( z) = z 2 z -1 x ( k ) = cos k 2 k = 0,1,2, L I-7 (a) P (1 ) P( 2) = (1+ r) P (0) - u = ( 1 + r ) P ( 1) - u M where P (1 ) P(0 ) = = amoun t owed P0 after the f irst p eriod. = amoun t borr owed i nitial ly P (k + 1 ) = (1 + r ) P ( k ) - u u = amoun t paid each period inclu ding p rincip al and inter est. (b) By direct substitution, P (2) P ( 3) = ( 1 + r ) P (1 ) - u = (1 + r ) = (1 + r ) P ( 2 ) - u = ( 1 + r ) = (1 + r ) N 2 P (0) P (0) -(1+ r)u - u - (1 + r ) 2 3 u - (1 + r ) u - u P (N ) P (0) -u (1+ r) M N -1 + (1 + r ) N -2 + L + (1 + r ) + 1 = 0 Solving for u from the last equation, we have u = (1+ r) N -1 N P 0 (1 + r ) k =0 = (1 (1 +r ) + r) N N P r 0 k -1 + 1 ) = (1 + r ) P ( k ) - u , (c) Taking the z-transform on both sides of the difference equation, we have P( k k =0 P( k + 1) z -k = (1 + r ) P( k ) z k =0 -k - uz z -1 351 Or, zP ( z ) - zP 0 = (1 + r ) P ( z ) - uz z -1 Thus, P (z) = zP z 0 -(1+ r) - uz (z - 1) z - (1 + r ) Taking the inverse z-transform and setting k = N, we have P (N ) = P 0 (1 + r ) N + u r - u (1 r + r) N =0 Solving for u from the last equation, we have u = (1 (1 + r) + r) N N P0 r -1 (d) For u P 0 = $15 (1 (1 ,000 , r N N = 0 . 01 , = ( 1. 01 ) and N 48 = 48 -1 month s, = +r ) + r) P0 r ( 15000 )( 0 . 01 ) 48 -1 = ( 1. 612 )( 15000 )( 0 . 01 ) 1. 612 ( 1. 01 ) -1 = $395 .15 I-8 (a) G( z) = 5.556 z -1 - 5.556 z - 0.1 (b) G( z ) = (c) G( z ) = (d) G( z ) = 2z z - 1 (z - 0.5) 2 -1 20 z z - 0.5 4z z -1 - - 20z z - 0.8 4z - + 80 z z -1 2z z - 0.5 2z 2 ( z - 0.5 ) 2 - I-9 k lim f (k ) = lim z 1 - 1 z F (z) = lim z (z z 1 - 0 . 5) = z 2 -1 F(z) is expanded into a power series of k 1 2 3 4 5 6 7 8 9 10 f(k) 1.000000 1.500000 1.750000 1.875000 1.937500 1.968750 1.984375 1.992188 1.996094 1.998047 by long division. k 11 12 13 14 15 16 17 18 19 20 f(k) 1.999023 1.999512 1.999756 1.999878 1.999939 1.999969 1.999985 1.999992 1.999996 1.999998 (d) 352 F (z) = z (z - 1 )( z - 1. 5) F (z) has a pole outside the unit circle. f (k ) is unbounded. The finalvalue theorem cannot be applied to k 1 2 3 4 5 6 7 8 9 10 F ( z ) . F ( z ) is expanded into a power series of z -1 . f(k) 0.000000 1.000000 2.500000 4.750000 8.125000 13.187500 20.781250 32.171875 49.257813 74.886719 x (0) k 11 12 13 14 15 16 17 18 19 20 = x (1 ) f(k) 113.33078 170.995117 257.492676 387.239014 581.858521 873.787781 1311.681641 1968.522461 2953.783691 4431.675781 2-18 (a) sides, x( k + 2 ) - x ( k + 1) + 0 .1 x ( k ) = u s ( k ) -z - zx ( 1) - zX z 2 = 0 . Taking the z-transform on both z z X (z) X (z ) 2 2 x(0) (z) + zx ( 0 ) + 0 .1 X ( z ) = + 1.455 z z = - z 1 z - z + 0 .1 k = 10 z z -1 - 0 .1127 - z -1 11 .455 z z - 0 . 8873 Taking the inverse z- transform, x( k ) = 10 + 1.455 ( 0 .1127 ) - 11 .455 = 1, ( 0 . 8873) k k = 0 , 1, 2 , L (b) have x( k + 2) - x (k ) = 0 x (0) x (1 ) =0 Taking the z-transform on both sides, we z z 2 2 z X (z) 2 -z 2 x ( 0 ) - zx ( 1 ) - X ( z ) =0 X (z) = -1 x (k ) = cos k 2 k = 0 , 1, 2 , L 2-19 (a) P (1 ) P( 2) = (1+ r) P (0) - u = ( 1 + r ) P ( 1) - u M where P (1 ) P(0 ) = = amoun t owed P0 after the f irst p eriod. = amoun t borr owed i nitial ly P (k + 1 ) = (1 + r ) P ( k ) - u P (2) P ( 3) u = amoun t paid each period inclu ding p rincip al and inter est. (b) By direct substitution, = ( 1 + r ) P (1 ) - u = (1 + r ) = (1 + r ) P ( 2 ) - u = ( 1 + r ) = (1 + r ) N 2 P (0) P (0) -(1+ r)u - u - (1 + r ) 2 3 u - (1 + r ) u - u P (N ) P (0) -u (1+ r) M N -1 + (1 + r ) N -2 + L + (1 + r ) + 1 = 0 353 Solving for u from the last equation, we have u = (1+ r) N -1 N P 0 (1 + r ) k =0 = (1 (1 +r ) + r) N N P r 0 k -1 + 1 ) = (1 + r ) P ( k ) - u , (c) Taking the z-transform on both sides of the difference equation, we have P( k k =0 P( k + 1) z -k = (1 + r ) P( k ) z k =0 -k - uz z -1 uz (z Or, zP ( z ) - zP 0 = (1 + r ) P ( z ) - uz z -1 Thus, P (z) = zP z 0 -(1+ r) - - 1) z - (1 + r ) Taking the inverse z-transform and setting k = N, we have P (N ) = P 0 (1 + r ) N + u r - u (1 r + r) N =0 Solving for u from the last equation, we have u = (1 (1 + r) + r) N N P0 r -1 (d) For u P 0 = $15 (1 (1 ,000 , r N N = 0 . 01 , = ( 1. 01 ) and N 48 = 48 -1 month s, = +r ) + r) P0 r ( 15000 )( 0 . 01 ) 48 -1 = ( 1. 612 )( 15000 )( 0 . 01 ) 1. 612 ( 1. 01 ) -1 = $395 .15 I-9 I-10 354 I-11 355 I-12 (a) Discrete state equations: 0 1 -2 -3 -1 A= B= 0 1 ( sI - A ) = e -e -t s -1 2 s + 3 - 2t ( sI - A ) -1 = 1 s + 3 1 ( s +1)( s + 2) -2 s -e + 2 e e -T ( t) = L 2 e - t - e -2 t ( sI - A ) -1 = - t -2 e + 2 e-2 t T = 1 sec. 2 e - T - e -2 T (T ) = -t -2 t -T -2 T - e + 2e -2 e + 2 e 0.2325 0.6 (T ) = -0.465 -0.0972 (T ) = -e -2 T -T -2 T ( t) = L -1 -t -2 t 1 1 e - e ( sI - A)-1 B = L-1 = -t ( s + 1)( s + 2) s - e + 2 e -2 t 0.2325 T = 1 sec. (T ) = - 0.0972 e - T - e -2 T - e -T + 2 e -2 T (b) 2 e -NT - e -2 NT ( NT ) = - NT -2 NT - 2e + e e -e -NT - NT -2 NT + 2e -e -2 NT N -1 k=0 2 e -N - e -2 N = -N -2 N - 2e + e e -e -N -N -e -2 N -2 N + 2e x ( NT ) = ( NT )x (0) + [ ( N - k - 1) T ] (T )u (kT ) I-13 (a) Discrete state equations: 1 1 s s2 B= = ( sI - A ) = ( s I - A ) -1 = 1 s 0 s 0 s 0 1 s 1 t 1 T 1 0.001 ( t) = (T ) = 0 1 0 1 = 0 1 1 s 2 t T 0.001 -1 -1 -1 ( t) = L ( sI - A) B = L = (T ) = 1 1 = 1 1 s 0 s -1 1 s 1 ( NT ) = A= 0 0 0 1 (b) 1 0 N -1 k=0 NT 1 0.001N = 0 1 1 x ( NT ) = ( NT )x (0) + [ ( N - k - 1) T ] (T )u (kT ) 356 [( N - k - 1) T ] = 1 ( N - k - 1) T 1 0.001( N - k - 1) 0 = 0 1 1 0.001( N - k ) [( N - k - 1) T ] (T ) = 1 I-14 (a) Transfer function: X( z ) U (z ) = [ zI - (T ) ] (T ) = -1 z - 0.6 0.465 1 z + 0.0972 0.2325 0.2325 0.2325 = z + 0.0972 - 0.0972 ( z ) -0.465 z - 0.6 - 0.0972 -0.2325 -1 X( z ) U (z ) = 0.2325 z -0.0972( z + 0.6673) ( z ) 1 ( z ) = z - 0 . 5028 2 ( z ) = z - 0.5028 z + 0.04978 2 (b) Characteristic equation: z + 0 . 04978 = 0 I-15 (a) Transfer function: X( z ) U (z ) = [ zI - (T ) ] (T ) = -1 1 0.001z z - 1 0.001 0.001 0 1 = (z ) z - 1 (z ) z -1 1 2 ( z ) = ( z - 1) (b) Characteristic equation: ( z ) = ( z -1) = 0 2 I-16 State diagram: ( z ) = z -1 -z -1 -6z -2 -z -2 + 5z -2 - 10 z -3 - 5z -3 = -2 z -2 - 15 z -3 =0 Characteristic equation: 2z + 15 = 0 357 I-17 State diagram: State equations: x ( k + 1) = 2 x ( k ) + x ( k ) 1 2 3 x (k 2 Output equation: y (k ) + 1) = x 3 ( k ) + 1) = -0 . 1 x 1 ( k ) - 0 .2 x 2 ( k ) - 0 .1 x 3 ( k ) + r ( k ) Y (z) R( z ) z 1 + 0 .1 z -1 -2 = x1( k ) x (k 3 Transfer function: = +z -3 -2 + 0 .2 z + 0 .2 z -3 = z z 3 2 +2 + 0 .1 z + 0 .2 z + 0 .2 I-18 Open-loop transfer function: G( z ) = Closed-loop transfer function: Y (z) R( z ) Y ( z) E( z) = 1- z ( -1 ) Z s ( s + 1) = z - 0.368 1 0.632 = G (z) 1+ G ( z) = 0 . 632 z - 0 .264 + 0 . 632 Discrete-data state equation: x [( k + 1 ) T ] = 0 .264 x ( kT ) r ( kT ) I-19 (a) = z + 1. 5 z - 1 = 0 This is a second-order system, n = 2. F(1) = 1.5 > 0 F(-1) = -1.5 < 0 Thus, for n = 2 = even, F(-1) < 0. The system is unstable. The characteristic equation roots are at z = 0.5 and z = -2. F(z) has one root outside the unit circle. F (z) 2 Let Or z = w w +1 -1 . The characteristic equation becomes ( w + 1) F (z) 3 2 2 + 1.5 w - 1 - ( w - 1) = 0 or 1.5 w + 4 w - 1.5 = 0 2 2 2 ( ) 2 w + 1 + 1.5 w + 1 - 1 = 0 w-1 w-1 The last coefficient of the last equation is negative. Thus, the system is unstable. (b) = z + z + 3 z + 0 .2 = 0 This is a third-order system, n = 3. 358 Let z = w w 3 +1 -1 . The characteristic equation becomes 2 w + 1 + w + 1 + 3 w + 1 + 0.2 = 0 w-1 w-1 w-1 Or + 1 ) + ( w + 1 ) ( w - 1 ) + 3 ( w + 1 )( w - 1 ) + 0 .2( w - 1 ) = 0 or 5 .2 w + 0 .4 w - 0 .4 w + 2 . 8 = 0 Since there is a negative sign in the characteristic equation in w, the equation has at least one root in the right-half w-plane. (w 3 2 2 3 3 2 Routh Tabulation: w w w w 3 5 .2 0 .4 - 0 .4 2 .8 2 1 - 36 .8 2 .8 0 Since there are two sign changes in the first column of the Routh tabulation, F(z) has two roots outside the unit circle. The three roots are at: -0.0681, -0.466 + j1.649 and -0.466 - j1.649. (c) = z - 1.2 z - 2 z + 3 = 0 w +1 Let z = . The characteristic equation becomes w -1 F (z) 3 2 3 2 w + 1 - 1.2 w + 1 - 2 w + 1 + 3 = 0 w-1 w -1 w-1 Or Routh Tabulation: w w w w 3 0 .8 w 3 - 5 .2 w + 15 .2 w - 2 .8 = 0 2 0 .8 15 .2 2 - 5.2 14. 77 - 2 .8 1 0 - 2 .8 There are three sign changes in the first column of the Routh tabulation. Thus, F(z) has three roots outside the unit circle. The three roots are at -1.491, 1.3455 + j0.4492, and 1.3455 - j0.4492. (d) = z - z - 2 z + 0 .5 = 0 w +1 Let z = . The characteristic equation becomes w -1 F (z) 3 2 w + 1 - w + 1 - 2 w + 1 + 0.5 = 0 w-1 w-1 w-1 3 2 Or - 1.5 w + 2 . 5 w + 7 . 5 w - 0 . 5 = 0 3 2 359 Routh Tabulation: 3 w - 1. 5 w w w 2 7 .5 2 .5 7 .2 - 0 .5 1 0 - 0 .5 Since there are two sign changes in the first column of the Routh tabulation, F(z) has three zeros outside the unit circle. The three roots are at z = 1.91, -1.1397, and 0.2297. I-20 Taking the z-transform of the state equation, we have zX ( z ) = ( 0 . 368 - 0 . 632 X (z) R (z ) K ) X ( z ) + KR ( z ) K Or, = z z The characteristic equation is Stability Condition: 0 . 368 3 2 - 0 .368 + 0 . 632 - 0 .368 + 0 . 632 K K K =0 -1 < K The root is z = 0.368 - 0.632K < 2 .165 - 0 . 632 <1 or I-21 z + z + 1. 5 Kz - ( K + 0 . 5) = 0 w +1 Let z = . The characteristic equation becomes w -1 w + 1 + w + 1 + 1.5 K w + 1 - ( K + 0.5) = 0 w -1 w -1 w-1 3 2 Or ( 1. 5 + 0 . 5 K Routh Tabulation: w w w 3 )w 3 + ( 5. 5 + 1. 5 K ) w + ( 0 . 5 - 4. 5 K ) w + ( 2 . 5 K + 0 . 5) = 0 2 1.5 5 .5 1 + 0 .5 K + 1. 5 K K 0 .5 - 4. 5 K + 0 .5 (K 2 2 .5 K 2 1 - 14 5 .5 -4K + 0 .5 + 1. 5 K - 0.07)( K + 3.57) < 0 w 0 2 .5 K 2.5 K + 0 .5 > 0 Stability Condition: -0 .2 < K < 0 . 07 I-22 (a) Forward-path transfer function: G ho Gp ( z ) = 1 - z ( -1 )Z s = 1 - z -1 K Z 0.667 - 0.444 + 0.444 = K (0.00476 z + 0.004527) ) s2 ( 2 s s + 1.5 z - 1.8607 z + 0.8607 ( s + 1.5) K 2 Characteristic equation: z 2 2 For T = 0.1 sec. z - 1.8607 z + 0 . 8607 + 0 .00476 Kz + 0 . 004527 + ( 0 . 00476 K - 1.8607 ) z + 0 .8607 + 0 . 004527 K K =0 =0 360 Let z = w w +1 -1 . The characteristic equation becomes 0 . 009287 w 2 (b) + ( 0 .2786 - 0 .009054 K ) w + 3 .7214 - 0 . 000233 K = 0 For stability, all the coefficients of the characteristic equation must be of the same sign. Thus, the conditons for stability are: 0.2786 - 0.009054K > 0 or K < 30.77 3.7214 - 0.000233K > 0 or K < 15971.67 Thus, for stability, 0 < K < 30.77 T = 0.5 sec. Forward-path transfer function: K ( 0 . 09883 z + 0 . 07705) G G (z ) = ho p 2 z - 1.4724 z + 0 .47237 Characteristic equation: z 2 + ( 0 . 09883 K - 1.4724 )z + ( 0 . 07705 K + 0 .47237 ) =0 Let z = w w +1 -1 . The characteristic equation becomes 0 .17588 Kw 2 + ( 1. 05526 - 0 .1541 K )w + 2 .94477 - 0 . 02178 K =0 For stability, all the coefficients of the characteristic equation must be of the same sign. Thus, the conditions for stability are: 1. 05526 2 . 9447 - 0 . 1541 - 0 . 02178 < 6 . 8479 K K > 0 or > 0 or K K < 6 . 8479 < 135 .2 Stability condition: 0 < K (c) T = 1 sec. Forward-path transfer function: G G (z ) ho p = K ( 0 .3214 z z 2 + 0 .19652 z ) - 1.2231 + 0 .22313 + ( 0 .22313 + 0 .19652 =0 Characteristic equation: z 2 + ( 0 . 3214 K - 1.2231 )z K ) Let z = w w +1 -1 . The characteristic equation becomes 2 0 . 5179 Kw + ( 1. 55374 - 0 . 393 K )w + 2 .4462 - 0 .12488 K =0 For stability, all the coefficients of the characteristic equation must be of the same sign. Thus, the conditions for stability are : 1. 55374 2 .4462 - 0 . 393 K > 0 - 0 .12488 K > 0 < 3 . 9535 or or K K < 3 . 9535 < 19 . 588 Stability condition: 0 < K I-23 (a) (b) Roots: -1.397 -0.3136 + j0.5167 Unstable System. Roots: 0.3425 -0.6712 + j1.0046 Unstable System. -0.3136 j0.5167 -0.06712 j1.0046 361 (c) (d) Roots: -0.4302 Unstable System. Roots: 0.5 -0.8115 Stable System. -0.7849 + j1.307 -0.7849 j1.307 -0.0992 j0.7708 -0.0992 + j0.7708 I-24 (a) Forward-path Transfer Function: T = 0.1 sec. 5 = 1 - z -1 Z 2.5 - 1.25 + 1.25 = 0.02341z + 0.021904 -1 G ( z ) = (1 - z ) Z 2 ) s 2 s s + 2.5 z 2 - 1.8187z + 0.8187 ( s ( s + 2) Closed-loop Transfer Function: Y (z) R( z ) = G (z) 1+ G ( z) = 0 . 0234 z z 2 + 0 . 021904 z - 1. 7953 + 0 .8406 (b) Unit-Step Response y(kT) (c) Forward-path Transfer Function: T = 0.05 sec. 5 = 0.0060468 z + 0.0058484 -1 G ( z ) = (1 - z ) Z 2 s ( s + 2) z 2 - 1.9048z + 0.9048 Closed-loop Transfer Function: Y (z) R( z ) = 0 . 006046 z 2 z + 0 . 005848 z 4 - 1. 8988 + 0 . 91069 362 Unit-step Response y(kT) I-25 (a) Y ( z) = 1 - z ( -1 )Z s 1 3 U ( z) - K T t U ( z ) = 10 E( z) - Kt 1 - z U (z) Th us U (z) ( -1 )Z s - 1) 1 2 U ( z) U (z) = 10 E (z) 2 z -1 = 10( z z -1+ K tT E (z) Y ( z) = 1 - z G( z) = Y ( z) ( -1 )T z ( z + 1) 10( z -1) 3 2( z - 1) 2 z - 1 + K tT E ( z) = 5T ( z + 1) E( z) Error Constants: K p ( z - 1) ( z - 1 + K T ) t = lim K z 1 G (z) 1 T 2 = lim ( z z 1 K v = 1 T lim ( z z 1 -1 )G ( z ) = 10 T K T t 2 2 = 10 K t a = - 1) 2 G (z) =0 Closed-loop Transfer Function: Y ( z) R (z ) = z + K tT + 5T - 2 z + 5T - K tT + 1 2 2 2 (b) Forward-path Transfer Function: G( z) = 5T ( z + 1) 2 ( z - 1) ( z - 1 + K T ) t ( 5T ( z + 1) 2 ) ( ) (c) Characteristic Equation: T = 0.1 sec. F (z ) = z + Kt T + 5T - 2 z + 5T - K tT + 1 = z + ( 0.1Kt - 1.95) z + ( 1.05 - 0.1Kt ) = 0 2 2 2 2 ( ) ( ) 363 For stability, from Eq. (6-62), where Thus, a 0 F (1 ) a 1 > 0, F ( -1 ) > 0, a0 < a2 or K = 1. 05 - 0 .1 K t F (1 ) a0 = 0 .1 K t - 1. 95 F ( - 1) = 0 .1 > 0 1. 05 = 4 - 0 .2 K t > 0 or 0.5 < K t t < 20 = - 0 .1 K t < a 1 = 1 < 20 . 5 Stability Condition: 0 .5 < K t < 20 = 0 . 05 ( z z 2 (d) Unit-step Response: K t =5 Y (z) R( z) + 1) - 1.45 z + 0 . 55 (e) Unit-step Response K t =1 = 0 . 05 ( z z 2 Y (z) R( z ) +1) - 1. 85 z + 0 . 95 364 I-26 (a) T = 0.1 sec. Gp (s ) = s + 37.06 s + 141.2 E (z ) ( s + 37.06 z + 141.2 ) m ( z) -1 - 30 + 14.2 + 29.7 + 0.418 = (1 - z ) Z 2 E( z) s s + 4.31 s + 32.7 s 1000( z + 20) m ( z) 2 = 1- z ( -1 )Z s 100( s + 20) 2 = 1- z ( -1 ) -30 z z -1 2 2 + 14.2 z ( z -1) z 2 + 29.7 z z - 0.65 = + z - 0.038 0.418 z 2 = 3 . 368 z z 3 + 1. 7117 - 0 . 3064 z 3 . 368 z (z + 1. 7117 z z - 0 . 3064 - 0 . 038 ) - 1. 6876 z + 0 . 71215 - 0 . 024576 - 1 )( z - 0 . 65)( (b) Closed-loop Transfer Function: T = 0.1 sec. m ( z ) R (z ) = 3. 368 z z 3 3 2 + 1. 7117 z z 2 2 z - 0 . 3074 z z + 1. 6805 + 2 .424 - 0 . 331 Characteristic Equation: z + 1. 6805 + 2 .424 - 0 . 331 = 0 - 0 . 903 - j 1. 3545 Roots of Characteristic Equation: z = 0 .125 , - 0 . 903 + j1. 3545 , z The complex roots are outside the unit circle = 1, so the system is unstable. 365 (c) T = 0.01 sec. m ( z ) E (z) 0 . 04735 z z 3 2 Forward-path Transfer Function Closed-loop Transfer Function = = + 0 . 005974 z 2 2 3z z - 0 . 03663 - 0 . 69032 - 0 . 03663 - 0 . 72695 z - 2 . 6785 z 3 + 2 . 3689 2 m ( z ) R (z ) 0 . 047354 z + 0 . 005875 z 3z - 2 . 6312 + 2 . 3748 Characteristic Equation: - 2 . 6312 Characteristic Equation Roots: z 3 z 2 + 2 . 3748 + j 0 .27 , z - 0 . 72695 = 0 0 . 921 z = 0 . 789 , 0 . 921 - j 0 .27 T = 0.001 sec. Forward-path Transfer Function m ( z ) -1.43 10 = E (z) -6 z 3 + 4. 98 10 3 -4 2 z 2 - 2 . 384 10 z -7 z - 4. 89 10 -4 z -6 - 2 . 9635 z + 2 .9271 z 2 - 0 .9636 -7 Closed-loop Transfer Function m ( z ) -1.43 10 = R (z ) z 3 + 4. 98 10 z 3 -4 2 - 2 . 384 10 z z - 4. 89 10 -4 - 2 . 963 z + 2 . 9271 - 0 . 9641 Characteristic Equation: z 3 - 2 .963 z 2 + 2 . 9271 z - 0 . 9641 = 0 Characteristic Equation Roots: z = 0 . 99213 , 0 . 98543 + j 0 . 02625 , 0 . 98543 - j 0 . 02625 (d) Error Constants: K p = lim z 1 G ho G p ( z ) = lim 3 . 368 z (z 2 + 1.7117 z z - 0 . 3064 - 0 . 038 2 z 1 - 1)( z - 0 .65)( = - lim = ) z Kv = 1 T lim ( z z 1 - 1) 2 G ho G p ( z ) 3. 368 z (z + 1. 7117 z - 0 . 3064 ) z 1 - 0 . 65)( 2 - 0 . 038 = 14. 177 T Ka = 1 T 2 lim ( z - 1) Gho Gp ( z) = 2 z 1 1 T 2 ( 3.368 z lim z 1 + 1.7117 z - 0.3064 ( z - 1)( z - 0.038) ) ( z - 1) = 0 Steady-state Errors: Step Input: Ramp Input: e ss = = 1 1+ K 1 K v p =0 T 14. 177 e ss = 366 Parabolic Input: e ss = 1 K a = I-27 (a) Forward-path Transfer Function: (no zero-or der hold) T = 0.5 sec. G (z) = z 0 .1836 zK 2 - 1. 0821 z + 0 . 0821 = (z 0 .1836 zK - 1 )( z - 0 . 0821 ) T = 0.1 sec. G (z) = 0 . 0787 zK z 2 - 1. 6065 z + 0 . 6065 = 0 . 0787 zK (z - 1)( z - 0 .6065) 367 (b) Open-loop Transfer Function: (with zero-order hold) T = 0.5 sec. G (z) = K ( 0 . 06328 z z 2 + 0 . 02851 + 0 . 08021 ) - 1. 0821 = 0 . 06328 K ( z (z + 0 .4505) ) z - 1 )( z - 0 . 0821 T = 0.1 sec. G (z) = K ( 0 . 00426 z z 2 + 0 .003608 z ) - 1. 6065 + 0 . 6065 = 0 . 00426 K ( z (z + 0 .8468 ) - 1 )( z - 0 .6065) I-28 Forward-path Transfer Function: G( z) = 0.0001546 K z + 3.7154 z + 0.8622 2 z z - 2.7236 z 3 ( ( 2 ) + 2.4644 z - 0.7408 ) 368 I-29 (a) P (z ) = z 3 -1 Q (z) = z 2 + 1. 5 z - 1 = ( z - 0 . 5)( z + 2) The system is unstable for all values of K . 369 I-30 (a) Bode Plot: The system is stable. (b) Apply w-transformation, z = 2 2 + wT - wT = G ho G ( z ) G (w ) 2 + wT 2 - wT Then G ho G(w) z= = 10( 1 - 0 .0025 w (1 + w ) w ) 2 The Bode diagram of G ho is plotted as shown below. The gain and phase margins are determined as follows: GM = 32 dB PM = 17.7 deg. 370 Bode plot of G ho G (w ) : I-31 2 16.67 N 1 - e -Ts = 0.000295 ( z + 3.39 z + 0.714) G hoG ( z ) = Z s s ( s + 1)( s + 12.5) ( z - 1) ( z - 0.9486 )( z - 0.5354 ) The Bode plot of G ho G(z) is plotted as follows. The gain margin is 17.62 dB, or 7.6. Thus selecting an integral value for N, the maximum number for N for a stable system is 7. Bode Plot of G ho G ( z ) 371 I-32 (a) G (s) c =2+ 200 s Backward-rectangular Integration Rule: Gc ( z ) = 2 + 200T z -1 = 2 z - 2 + 200T z -1 = 2 + ( 200T - 2 ) z 1- z -1 -1 Forward-rectangular Integration Rule: Gc ( z ) = 2 + 200Tz z -1 = ( 2 + 200T ) z - 2 ( 2 + 200T ) - 2 z z -1 = 1-z -1 -1 Trapazoidal Integration Rule: Gc ( z ) = 2 + 200T ( z + 1) 2 ( z - 1) = ( 4 + 200T ) z + 200T 2 ( z - 1) -2 = ( 4 + 200T ) + ( 200T - 2 ) z 2 1-z -1 ( -1 ) (b) G (s) c = 10 + 0 .1 s The controller transfer function does not have any integration term. The differentiator is realized by backward difference rule. G c ( z ) = 10 + 5 s 0.1 ( z - 1) Tz = (10T + 0.1) z - 0.1 z = (10T + 0.1) - 0.1z -1 (c) G (s) c = 1 + 0 .2 s + Backward-rectangular Integration Rule: Gc ( z ) = 1 + 0.2 ( z - 1) Tz + 5T = ( z - 1) ( T + 0.2 ) - 0.2 z T -1 + 5Tz -1 -1 1- z Forward-rectangular Integration Rule: Gc ( z ) = 1 + 0.2 ( z - 1) Tz + 5Tz z -1 = ( T + 0.2 ) - 0.2 z T -1 + 5T 1- z -1 Trapezoidal Integration Rule: 0.2 ( z - 1) Tz 5T ( z + 1) 2 ( z - 1) Gc ( z ) = 1 + + = ( T + 0.2 ) - 0.2z T -1 + 5T 1 + z ( ) 2 (1 - z ) -1 -1 372 I-33 (a) G (s) c = 10 s + 12 T = 0.1 sec Gc (z ) = 1 - z ( -1 ) Z s ( s + 12) = 0.8333( 1 - z ) Z s - s + 12 10 -1 1 1 = 0.8333 1 - z 10 ( s + 1.5 ) s + 10 ( -1 ) z -1 - z - e z z -1.2 = 0.5825 z - 0.301 (b) G c ( s ) = T = 1 sec Gc (z ) = 1 - z = 1- z s s ( -1 ) Z s ( s + 10) = ( 1 - z ) Z -1 10 ( s + 1.5 ) 8.5 1.5 s + s + 10 8.5 ( -1 ) z -1 + z - e 1.5 z T = 0.1 sec -1 = 10 ( z - 0.9052) z - 0.368 (c) G (s) c = + 1. 55 Gc (z ) = 1 - z ( -1 ) Z s + 1.55 = ( 1 - z ) z - e 1 -1 z -0.155 = z -1 z - 0.8564 (d) G (z) c = 1 + 0 .4 s 1 + 0 . 01 s Gc (z ) = 1 - z ( -1 )Z -1 0.025 z 0.975z z - 0.975 = 40 ( 1 - z ) Z z - 1 + z - e-10 = 40 z - 0.0000454 s (1 + 0.01 s ) 1 + 0.4 s b I-34 (a) Not physically realizable, since according to the form of Eq. (11-18), (b) Physically realizable. (c) Physically realizable. (d) Physically realizable. (e) Not physically realizable, since the leading term is 0.1z. (f) Physically realizable. (a) G (s) c 0 0 but a 0 = 0. I-35 = 1 + 10 s K P =1 K D = 10 Thus Gc ( z ) = ( T + 10) z - 10 Tz 373 G ho Gp ( z ) = 1 - z ( -1 ) Z s4 = 2T ( z + 1) 2 G ( z ) = G c ( z ) Gho G p ( z ) = ( z - 1) 2T ( z + 1) [(T + 10 ) z - 10 ] z ( z - 1) 3 2 2 By trial and error, when T = 0.01 sec, the maximum overshoot of When T = 0.01 sec, G( z) = 0.02 ( z + 1) (10.01 z - 10 ) z ( z - 1) 2 y ( kT ) is less than 1 percent. Y ( z) R (z ) = 0.02 ( z + 1)( 10.01 z - 10 ) z - 1.7998 z + 1.0002 z - 0.2 3 2 When the input is a unit-step function, the output response y ( kT ) is computed and tabulated in the following for 40 sampling periods. The maximum overshoot is 0.68%, and the final value is 1. Sampling Periods k y ( kT ) ------------------------------ I-36 (a) G ho Gp ( z ) = ( 1 - z -1 ) Z 1 s3 2 = 2T ( z + 1) ( z - 1 )2 G ( z ) = G c ( z ) Gho G p ( z ) = K PTz + KD ( z - 1 ) 2T Tz 2T ( z + 1) ( z - 1) 2 2 2 = ( K PT + KD ) z2 + K P Tz - K D z ( z - 1) Characteristic Equation: z + 2 K PT + K DT - 1 z + 2K PT + 1 z - 2 K DT = 0 3 2 2 2 2 ( ) ( ) For two roots to be at z = 0.5 and 0.5, the characteristic equation should have z - z + 0 .25 as a 2 factor. Dividing the characteristic equation by z - z + 0 .25 and solving for zero remainder, we get 374 4K T P 2 + 2 K D T - 0 .25 = 0 = 0 . 0139 T 2 2 a nd - 0 . 5 K P T - 2 .5 K D T + 0 .25 = 0 2 Solving for K P and K D from these two equations, we have P K K D = 0 . 0972 T f or T The third root is at z = 1 - 2 K P T - 2 K D T = 0 . 7778 = 0 . 01 sec The forward-path transfer function is G( z) = 0.2222 ( z + 1) ( z - 0.8749) z ( z - 1) 0 .2222 z z 3 2 2 Y (z) R( z ) = + 0 . 0278 z 2 z - 0 .1944 z - 1. 7778 + 1. 0278 - 0 .1944 Unit-step Response: (b) (b) KP = 1, T = 0.01 sec G (z) = G c ( z ) G ho G p ( z ) = 2T = 0 . 02 + + - - + + - T K D z 2 Tz K D z z 1 2 0 . 01 KD z z z 2 0 . 01 z - KD 1 2 The unit-step response of the system is computed for various values of tabulated below to show the values of the maximum overshoot. KD Max overshoot (%) 1.0 14 5.0 0.9 6.0 0.67 7.0 0.5 8.0 0.38 9.0 0.31 K D . The results are 9.1 0.31 9.3 0.32 9.5 0.37 10.0 0.68 375 I-37 (a) Phase-lead Controller Design: G ( z ) = G ho Gp ( z ) = 1 - z ( -1 )Z s 4 3 = 0.02 ( z + 1) z ( z - 1) 2 T = 0.1 sec Closed-loop Transfer Function: Y (z ) R (z ) = Gho Gp ( z) 1 + G ho Gp ( z ) = 0.02 ( z + 1) z - 1.98 z + 1.02 2 The system is unstable. 4 (1 - 0.05 w ) w 2 2 With the w-transformation, z = T T +w 2 = 20 20 +w -w G ( w) = -w From the Bode plot of G ( j w ) the phase margin is found to be -5.73 degrees. For a phase margin of 60 degrees, the phase-lead controller is G (w ) c = 1 + aTw 1 + Tw = 1 1 + 1.4286 + 0 . 0197 w w The Bode plot is show below. The frequency-domain characteristics are: PM = 60 deg GM = 10.76 dB M r = 1.114 The transfer function of the controller in the z-domain is Gc ( z ) = 21.21 ( z - 0.9222 ) ( z + 0.4344) (b) Phase-lag Controller Design: Since the phase curve of the Bode plot of G ( j w ) is always below -180 degrees, we cannot design a phase-lag controller for this system in the usual manner. 376 Bode Plots for Part (a): G ho G p ( z ) = - + 1 z -1 Z 4500 K s 361 .2 s 2 + = - - K 0 . 002008 z 1 z z 0 . 001775 0 . 697 377 I-38 (a) Forward-path Transfer Function: Kv = 1 T lim z 1 [ ( z - 1) G ho G p ( z ) = lim z 1 K ( 2.008 z + 1.775) z - 0.697 = 1000 Thus K v = 80 .1 (b) Unit-step Response: Maximum overshoot = 60 percent. (c) Deadbeat-response Controller Design: (K = 80.1) G ho Gp ( z ) = -1 -1 0.16034 z + 0.14217 ( z - 1) ( z - 0.697 ) -1 G ho G p ( z -1 ) = Q(z P(z ) ) = 0 . 16034 z + 0 .14217 + 0 . 697 z z -2 1 - 1. 697 z -1 -2 Q (1 ) = 0 . 3025 Digital Controller: Gc ( z ) = P (z ) Q (1) - Q (z ) -1 -1 = = 1 - 1.697 z + 0.697 z -1 -1 -2 -2 0.3025 - 0.16034 z - 0.14217z 3.3057 ( z - 1 ) ( z - 0.697 ) z - 0.53 z - 0.47 2 Forward-path transfer function: G ( z ) = G c ( z ) Gho G p ( z ) = 3.3057 ( z - 1 ) ( z - 0.697 ) z - 0.53 z - 0.47 2 Closed-loop system transfer function: Unit-step response: Y (z) M (z) = 0 . 53 z z + 0 .47 2 = 0 . 53 z -1 +z -2 +z -3 +L 378 Deadbeat Response: I-39 G p (s) = 2500 s(s + 25) -1 T = 0.05 sec G hoG ( z ) = 1 - z ( )Z -1 -1 2.146 z + 1.4215 = ( z - 1 ) ( z - 0.2865 ) s ( s + 25 ) 2500 2 -1 -2 G ho ( ) = 2.146 z + 1.4215 z G( z ) = P ( z ) 1 - 1.2865 z + 0.2865z -1 Q z -1 -2 Q (1 ) = 3 . 5675 Deadbeat Response Controller Transfer Function: Gc z ( )=Q -1 ( ) = 1 - 2.865z + 0.2865z (1) - Q ( z ) 3.5675 - 2.146 z - 1.4215 z P z -1 -1 -2 -1 -1 -2 G (z) c = (z - 1 )( z - 0 .285) 2 3 . 5675 z - 2 .146 z - 1.4215 Forward-path Transfer Function: G (z) = G c ( z ) G ho G p ( z ) = 2 .146 z 3 . 5675 z 2 + 1.4215 z - 2 . 146 - 1.4215 Closed-loop System Transfer Function: M (z) = 0 . 6015 z -1 + 0 . 3985 2 z Unit-step response: Y (z) = 1 + 0 . 6015 z +z -2 +z -3 +L 379 I-40 The characteristic equation is z + ( - 1.7788 + 0.1152 k1 + 22.12 k 2 ) z + 0.7788 + 4.8032 k1 - 22.12 k 2 = 0 2 For the characteristic equation roots to be at 0.5 and 0.5, the equation should be z 2 - z + 0 .25 = 0 Equating like coefficients in the last two equations, we have - 1.7788 + 0 .1152 0 . 7788 k k 1 + 22 .12 k 2 = - 1 + 4. 8032 1 - 22 . 12 k 2 = 0 .25 k 1 Solving for the value of k and k 1 2 from the last two equations, we have = 0 . 058 and k 2 = 0 . 035 . 380 ...
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Automatic Control Systems 8ed by Kuo and Golnaragh -Solutions Manual

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