Sistemas de control automatico benjamin c kuo 7ed (solucionario)

http://www.elsolucionario.blogspot.com LIBROS UNIVERISTARIOS Y SOLUCIONARIOS DE MUCHOS DE ESTOS LIBROS LOS SOLUCIONARIOS CONTIENEN TODOS LOS EJERCICIOS DEL LIBRO RESUELTOS Y EXPLICADOS DE FORMA CLARA VISITANOS PARA DESARGALOS GRATIS Chapter MATHEMATICAL FOUNDATION 2-1 (a) Poles: s = 0, 0, Zeros: s = −1, −10; (b) Poles: s = −2, ∞, ∞, ∞ −2, −2; Zeros: s = The pole and zero at s = (c) Poles: s = 0, −1 + j, −1 − j; Zeros: s = (d) Poles: s = 0, −1, −2, ∞ −2 2-2 (a) (b) G (s) = ( s + 5) (c) G ( s) = (d) (s 4s (e) G ( s) = −1 cancel each other s ∞ G (s) +4 +4 = ) ∑e + G (s) s+ kT ( s + ) = k =0 −e 2-3 (a) g ( t ) = u s ( t ) − 2u s (t − 1) + u s( t − 2) − u s ( t − 3) + L G (s ) = s (1 − 2e − s + 2e−2 s − 2e −3s + L ) = gT (t ) = u s (t ) − 2us (t − 1) + us (t − 2) GT (s ) = s (1 − 2e − s + e −2s ) = ( − e − s ) 1−e ( −s s 1+ e −s 0≤ t ≤ 2 s ) −T ( s+5 ) = s + 4s +8 ∞ g(t ) ∞ ∑ = g T − 2k )us (t − 2k ) (t G (s) ∑s = k =0 −e (1 −s ) e −2 ks = k =0 −s 1− e s (1 + e −s ) (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 (1 − 2e 2 s −0.5 s + 2e −s − 2e −1.5 s − 0.5 s ( ) + L) = −0.5 s s (1 + e ) 1−e = tu s ( t ) − ( t − 5) u s ( t − 5) + 2( t − ) u s ( t − ) (t ) (1 − 2e−0.5 s + e− s ) = s (1 − e−0.5 s ) GT ( s ) = s 2 ∞ g (t ) = ∑ k=0 G(s ) = ≤ t ≤1 ∞ g T ( t − k )us ( t − k ) ∑ s2 ( −e −0.5 s k=0 ) e − ks = ( −0.5 s ) −0.5s s (1 + e ) 1−e 2-4 g(t ) = ( t + ) u s ( t ) − ( t − ) u s ( t − ) − u s ( t − ) − ( t − ) u s ( t − ) + ( t − 3) u s ( t − 3) + u s ( t − 3) G ( s) = s 2-5 (a) (1 − e − s − e −2 s + e −3 s ) + s (1 −2e − s + e −3 s ) 1 Taking the Laplace transform of the differential equation, we get 1 ( s + 5s + 4) F ( s) = F (s) = s +2 ( s + 1)( s + )( s f (t ) = e −4 t + (b) sX ( s ) 1 e −t − − x1( ) = e −2 t t +4) = 6( s + 4) = X (s) = 2 s X (s) =1 x (0) s( s + s +1 + )( s + ) −1 (s + )( s + ) = + e −t − e 3( s + 1) − 2( s + 2) ≥0 sX (s) − x ( ) = −2 X ( s ) − X ( s ) + = = 2s + −1 s +1 s + +1 − 2( s + 2) s +2 Taking the inverse Laplace transform on both sides of the last equation, we get x (t ) Solving for X1 (s) and X2 (s), we have X ( s) + −2 t t ≥0 x (t ) = −e −t +e −2 t t ≥0 s x (0) =0 2-6 (a) G (s) (b) G (s) = (c) G (s ) = = 3s −2 ( 50 − s (s 20 = s − s s g(t ) = t e −0.5 t + s = s g(t ) +3 s +4 −1 g (t ) + 3) 3( s 30s + 20 +s+2 g (t ) = + 1.069e (e) − s +1 (d) G (s) + 1) + + 2) 2( s + +1 s − + ) e s − e = −2 e −t [ −s −2 t +s +2 − + e −3 t t ≥0 + te g (t ) = 50 − 20e = −t − (t −1) + e −3 t t ≥0 ] us (t − 1) − 30cos2(t − 1) − 5sin2(t − 1) s s Taking the inverse Laplace transform, +s+2 [ sin1.323t + sin (1.323t − 69.3o ) ] = + e−0.5 t (1.447sin1.323t − cos1.323t ) −t t ≥0 2-7  −1  A =  −2     −1 −3 −1  2-8 0  B = 1    0  u (t ) = (a)  u1( t)   u ( t)    (b) Y (s ) = R (s ) 3s + Y (s) s + s +5s + = R (s ) (c) s + 10 s + s + (d) Y (s ) R (s ) = s ( s + 2) Y (s ) s + 10 s + s + s + R (s ) = 1+ 2e −s 2s + s + t≥0 (c) T = 0.01 sec Forward-path Transfer Function Ωm ( z ) E (z) Ωm ( z ) Closed-loop Transfer Function R (z ) = = − 6785 z z z 047354 − 6312 z + 005974 04735 z z − 69032 z + 005875 − 03663 3z + 3748 − 03663 3z + 3689 z − 72695 Characteristic Equation: − 6312 Characteristic Equation Roots: z z = 789 , + 3748 z + 921 T = 0.001 sec Forward-path Transfer Function Ω m ( z ) −1.43 × 10 = −6 z −6 z j 27 , + 98 × 10 −4 z − 9635 z −4 + 98 × 10 R (z ) − 963 z z + 9271 z j 27 − 384 × 10 z −7 z − 89 × 10 z − 89 × 10 −4 − 9636 z − 384 × 10 + 9271 z − 921 E (z) Closed-loop Transfer Function Ω m ( z ) −1.43 × 10 = − 72695 = z −7 −4 − 9641 Characteristic Equation: z − 963 + 9271 z z − 9641 = Characteristic Equation Roots: z = 99213 , + 98543 j 02625 , − 98543 j 02625 (d) Error Constants: K ∗ p = lim ∗ Kv ∗ z →1 = Ka = lim ( z z →1 T T G ho G p ( z ) − 1) = lim z →1 + 1.7117 z − 3064 − 1)( z − 65)( z − 038 368 z (z G ho G p ( z ) 368 z = − lim z →1 lim  ( z − 1) Gho Gp ( z)  = z →1 T − 65)( (z z →1 Steady-state Errors: Step Input: Ramp Input: e e ∗ ss ∗ ss = = 1+ K K ∗ + 7117 ( 3.368 z lim 2 ∗ =0 p = v 366 T 14 177 =∞ ) z z − 3064 − 038 = 14 177 ) T + 1.7117 z − 0.3064 ( z − 1)( z − 0.038) ) ( z − 1) = e Parabolic Input: I-27 ∗ ss = K =∞ ∗ a (a) Forward-path Transfer Function: (no zero-or der hold) T = 0.5 sec G (z) 1836 zK = z z − 0821 z + 0821 1836 zK = (z − )( z − 0821 T = 0.1 sec G (z) = 0787 zK − 6065 z + 6065 367 = 0787 zK (z − 1)( z − 6065) ) (b) Open-loop Transfer Function: (with zero-order hold) T = 0.5 sec G (z) = K ( 06328 z − 0821 + 02851 z + 08021 K ( 00426 z + 003608 z ) = 06328 K ( z (z + 4505) − )( z − 0821 ) T = 0.1 sec G (z) = z I-28 − 6065 Forward-path Transfer Function: G( z) = z + 6065 ( ) = 00426 K ( z (z − )( z − 6065) ) + 2.4644 z − 0.7408 ) 0.0001546 K z + 3.7154 z + 0.8622 ( z z − 2.7236 z 2 368 + 8468 ) I-29 (a) P (z ) = z −1 Q (z) = z + z − = ( z − 5)( The system is unstable for all values of K 369 z + 2) I-30 (a) Bode Plot: The system is stable (b) Apply w-transformation, z Then G = ho + wT − wT G(w) The Bode diagram of G ho = G ho G ( z ) G (w ) z= + wT = 10( − wT − 0025 w (1 + w ) is plotted as shown below The gain and phase margins are determined as follows: GM = 32 dB PM = 17.7 deg 370 w ) Bode plot of I-31 G ho G (w ) : 16.67 N  − 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 Bode Plot of G ho G ( z ) 371 I-32 (a) G (s) c =2+ 200 s Backward-rectangular Integration Rule: Gc ( z ) = + 200T z −1 = z − + 200T z −1 = + ( 200T − ) z 1− z −1 −1 Forward-rectangular Integration Rule: Gc ( z ) = + 200Tz = z −1 ( + 200T ) z − ( + 200T ) − z = z −1 1−z −1 −1 Trapazoidal Integration Rule: 200T ( z + 1) Gc ( z ) = + (b) G (s) c ( z − 1) = ( + 200T ) z + 200T ( z − 1) −2 = ( + 200T ) + ( 200T − ) z ( 1−z −1 ) = 10 + s The controller transfer function does not have any integration term The differentiator is realized by backward difference rule G c ( z ) = 10 + (c) G (s) c = + s + 0.1 ( z − 1) = (10T + 0.1) z − 0.1 Tz z = (10T + 0.1) − 0.1z s Backward-rectangular Integration Rule: Gc ( z ) = + 0.2 ( z − 1) + Tz 5T ( z − 1) = ( T + 0.2 ) − 0.2 z −1 + T 5Tz −1 1− z −1 Forward-rectangular Integration Rule: Gc ( z ) = + 0.2 ( z − 1) + Tz 5Tz z −1 = ( T + 0.2 ) − 0.2 z −1 T + 5T 1− z −1 Trapezoidal Integration Rule: Gc ( z ) = + 0.2 ( z − 1) Tz + 5T ( z + 1) ( z − 1) = 372 ( T + 0.2 ) − 0.2z T −1 + ( ) (1 − z ) 5T + z −1 −1 −1 −1 I-33 (a) G (s) c = 10 s T = 0.1 sec + 12 ( Gc (z ) = − z −1  ( 10 ( s + 1.5 ) s + 10 ( Gc (z ) = − z ( (c) s = G (s) c ( G (z) c = I-35  −1 )  z − − z − e z z −1.2  = 0.5825  z − 0.301 T = sec  10 ( s + 1.5 )  ) Z  s ( s + 10)  = ( − z ) Z    )  z − + z − e 1.5 z 8.5 −1 −1 1.5 s +   s + 10  8.5  = 10 ( z − 0.9052)   z − 0.368 −1 ) Z  s + 1.55  = ( − z )  z − e  z −1   −0.155  = z −1   z − 0.8564 + 01 s Gc (z ) = − z I-34 1 + s ( −1 T = 0.1 sec + 55 s Gc (z ) = − z (d) −1 −1 = 1− z 10  = 0.8333 − z (b) G c ( s ) =  ) Z  s ( s + 12)  = 0.8333( − z ) Z  s − s + 12  −1 ) Z   −1  0.025 z + 0.975z  = 40  z − 0.975  = 40 ( − z ) Z     z − z − e−10   z − 0.0000454   s (1 + 0.01 s )  + 0.4 s (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 = + 10 s K P =1 K D = 10 373 Thus Gc ( z ) = b ≠0 but a = ( T + 10) z − 10 Tz ( ) Z  s4  = 2T ( z + 1) G ho Gp ( z ) = − z −1  G ( z ) = G c ( z ) Gho G p ( z ) = ( z − 1) 2T ( z + 1) [(T + 10 ) z − 10 ] z ( z − 1)  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) Y ( z) R (z ) = y ( kT ) is less than percent 0.02 ( z + 1)( 10.01 z − 10 ) z − 1.7998 z + 1.0002 z − 0.2 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 Sampling Periods k y ( kT ) I-36 1  s3 (a) G ho Gp ( z ) = ( − z −1 ) Z   = 2T ( z + 1)   ( z − )2 G ( z ) = G c ( z ) Gho G p ( z ) = = Characteristic Equation: ( K PTz + KD ( z − ) 2T ( z + 1) ( z − 1) 2 Tz ( K PT + KD ) z2 + K P Tz − K D  2T z ( z − 1) ) ( ) z + K PT + K DT − z + 2K PT + z − K DT = 2 For two roots to be at z = 0.5 and 0.5, the characteristic equation should have z − z + 25 as a factor Dividing the characteristic equation by z − z + 25 and solving for zero remainder, we get 374 4K T P Solving for K P and K + K D T − 25 = D K The third root is at z − K P T − K D T + 25 = a nd from these two equations, we have P = 0139 T K 0972 = D T = − K P T − K D T = 7778 T f or = 01 sec The forward-path transfer function is G( z) = Y (z) R( z ) = 0.2222 ( z + 1) ( z − 0.8749) z ( z − 1) z + 0278 2222 z − 7778 z z + 0278 − 1944 z − 1944 Unit-step Response: (b) (b) KP = 1, T = 0.01 sec G (z) = G c ( z ) G ho G p ( z ) = = 2T δ+ ι + − β− γ δ + ι + β− γ T K z D z z 02 01 Tz K D KD z z z 01 z − KD The unit-step response of the system is computed for various values of K D The results are 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 0.5 375 8.0 0.38 9.0 0.31 9.1 0.31 9.3 0.32 9.5 0.37 10.0 0.68 I-37 (a) Phase-lead Controller Design: ( G ( z ) = G ho Gp ( z ) = − z −1 ) Z  s  = 0.02 ( z + 1)   z ( z − 1)  T = 0.1 sec Closed-loop Transfer Function: Y (z ) R (z ) = Gho Gp ( z) 0.02 ( z + 1) = + G ho Gp ( z ) = z With the w-transformation, T G ( jω w ) +w = T From the Bode plot of The system is unstable z − 1.98 z + 1.02 20 +w 20 −w G ( w) = (1 − 0.05 w ) w −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 = + aTw + Tw = + 1.4286 w + 0197 w The Bode plot is show below The frequency-domain characteristics are: PM = 60 deg M GM = 10.76 dB 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 ( jω 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 ) = Λ ε− ϕΜ Μ Ν β+ z −1 4500 K Z s s Ο β + Π = γΠ Θ β− γβ− K 002008 361 377 z z z 001775 697 γ γ I-38 (a) Forward-path Transfer Function: ∗ Kv = T lim z →1 [ ( z − 1) G ] G p ( z ) = lim ho K ( 2.008 z + 1.775) = 1000 z − 0.697 z →1 Thus K ∗ v = 80 (b) Unit-step Response: Maximum overshoot = 60 percent (c) Deadbeat-response Controller Design: (K = 80.1) G ho Gp ( z ) = G ho G p ( z −1 ) = Q(z P(z −1 −1 ) = 0.16034 z + 0.14217 ( z − 1) ( z − 0.697 ) 16034 z −1 − 697 z ) + 14217 −1 + 697 z z −2 Q (1 ) −2 = 3025 Digital Controller: −1 Gc ( z ) = −1 P (z ) −1 Q (1) − Q (z ) = = − 1.697 z + 0.697 z −1 0.3025 − 0.16034 z − 0.14217z z − 0.53 z − 0.47 M (z) Closed-loop system transfer function: Unit-step response: Y (z) −2 3.3057 ( z − ) ( z − 0.697 ) G ( z ) = G c ( z ) Gho G p ( z ) = Forward-path transfer function: −2 = 53 z 378 −1 +z −2 +z = −3 3.3057 ( z − ) ( z − 0.697 ) z − 0.53 z − 0.47 53 z +L z + 47 Deadbeat Response: I-39 G p (s) 2500 = ( G hoG ( z ) = − z G ho T = 0.05 sec + 25) s(s −1 ) Z   2.146 z + 1.4215 =   s ( s + 25 )  ( z − ) ( z − 0.2865 ) 2500 ( ) = 2.146 z + 1.4215 z G( z ) = P ( z ) − 1.2865 z + 0.2865z Q z −1 −1 −1 −1 −2 −1 Q (1 ) −2 = 5675 Deadbeat Response Controller Transfer Function: ( )=Q Gc z −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 Forward-path Transfer Function: G (z) = G c ( z ) G ho G p ( z ) = 146 z 5675 z + 1.4215 − 146 z 6015 z + 3985 − 1.4215 Closed-loop System Transfer Function: M (z) Unit-step response: Y (z) = = + 6015 z 379 −1 z +z −2 +z −3 +L = (z − )( z − 285) 5675 z − 146 z − 1.4215 I-40 The characteristic equation is z + ( − 1.7788 + 0.1152 k1 + 22.12 k ) z + 0.7788 + 4.8032 k1 − 22.12 k = For the characteristic equation roots to be at 0.5 and 0.5, the equation should be z − z + 25 = Equating like coefficients in the last two equations, we have − 1.7788 + 1152 7788 Solving for the value of k and k + 8032 k k 1 + 22 12 k = − − 22 12 k = 25 from the last two equations, we have 380 k = 058 and k = 035 ... function: Ec ( s) G ( s) = = E (s ) + R 2C s + ( R1 + R ) Cs (b) Block diagram: (c) Forward-path transfer function: Ωm ( s) = E (s ) [1 + ( R K (1 + R 2C s ) + R ) Cs ] ( K b Ki + Ra JL s ) (d) Closed-loop... 5-8 (c) ,  C  V =  CA  =  2 CA   0 − 0    −4  Since V is singular, the OCF transformation cannot be conducted (d) From Problem 5-8(d),  3 1 M =  0    0 Then,  C ... transfer function: Ωm ( s) Fr ( s ) (e) = Gc ( s) = [1 + ( R E c ( s) = Kφ K (1 + R 2C s ) + R2 ) Cs] ( K b K i + Ra J L s ) + Kφ KK e N (1 + R 2C s ) (1 + R C s ) E (s ) Forward-path transfer function: