自动控制原理中英文对照李道根习题3题解.pdf

Solutions 13 Solutions P3.1 The unit step response of a certain system is given by tt eetc 2 1)( , 0t (a) Determine the impulse response of the system. (b) Determine the transfer function )()(sRsCof the system. Solution: The impulse response is the differential of corresponding step response, i.e. tt eet t tc tk 2 2)( d )(d )( As we know that the transfer function is the Laplace transform of corresponding impulse response, i.e. 23 24 2 2 1 1 12)( )( )( 2 2 2 ss ss ss eetL sR sC tt P3.2 Consider the system described by the block diagram shown in Fig. P3.2(a). Determine the polarities of two feedbacks for each of the following step responses shown in Fig. P3.2(b), where “0” indicates that the feedback is open. Solution: In general we have )(sR)(sC s k2 (a) Block diagram s k1 00 0. 1 0 t )(tc 0. 1 0 t )(tc 0. 1 0 t )(tc 0. 1 0 t )(tc Asymptotic line 0 .1 0 t )(tc Parabolic (b) Unit-step responses (1)(2)(3) (4)(5) Figure P3.2 Solutions 14 21 0 2 0 2 21 )( )( kksks kk sR sC Note that the characteristic polynomial is 21 0 2 0 2 )(kkskss where the sign of sk2is depended on the outer feedback and the sign of 21k kis depended on the inter feedback. Case (1). The response presents a sinusoidal. It means that the system has a pair of pure imaginary roots, i.e. the characteristic polynomial is in the form of 21 2 )(kkss. Obviously, the outlet feedback is “”and the inner feedback is “0”. Case (2). The response presents a diverged oscillation. The system has a pair of complex conjugate roots with positive real parts, i.e. the characteristic polynomial is in the form of 212 2 )(kkskss. Obviously, the outlet feedback is “+” and the inner feedback is “”. Case (3). The response presents a converged oscillation. It means that the system has a pair of complex conjugate roots with negative real parts, i.e. the characteristic polynomial is in the form of 212 2 )(kkskss. Obviously, both the outlet and inner feedbacks are “”. Case (4). In fact this is a ramp response of a first-order system. Hence, the outlet feedback is “0” to produce a ramp signal and the inner feedback is “”. Case (5). Considering that a parabolic function is the integral of a ramp function, both the outlet and inner feedbacks are “0”. P3.3 Consider each of the following closed-loop transfer function. By considering the location of the poles on the complex plane, sketch the unit step response, explaining the results obtained. (a) 2012 20 )( 2 ss s,(b) 6116 6 )( 23 sss s (c) 22 4 )( 2 ss s, (d) )5)(52( 5 .12 )( 2 sss s Solution: (a) )10)(2( 20 2012 20 )( 2 ss ss s By inspection, the characteristic roots are 2, 10. This is an overdamped second-order system. Therefore, considering that the closed-loop gain is 1 k, its unit step response can be sketched as shown. (b) )3)(2)(1( 6 6116 6 )( 23 sss sss s By inspection, the characteristic roots are 1, 2, 3. Obviously, all three transient components are decayed exponential terms. Therefore, its unit step response, with a closed-loop gain 1 k, is sketched as shown. 0 . 1 0 t )(tc 0 . 1 0 t )(tc Solutions 15 (c) 1) 1( 4 22 4 )( 22 sss s This is an underdamped second-order system, because its characteristic roots are j1. Hence, transient component is a decayed sinusoid. Noting that the closed-loop gain is 2 k, the unit step response can besketched as shown. (d) )5(21( 5 .12 )5)(52( 5 .12 )( 222 sssss s ） By inspection, the characteristic roots are 21j, 5. Since 51 . 0, there is a pair of dominant poles,21j, for this system. The unit step response, with a closed-loop gain 5 . 0 k, is sketched as shown. P3.4 The open-loop transfer function of a unity negative feedback system is ) 1( 1 )( ss sG Determine the rise time, peak time, percent overshoot and setting time (using a 5% setting criterion). Solution: Writing he closed-loop transfer function 22 2 2 21 1 )( nn n ssss s we get 1 n , 5 . 0. Since this is an underdamped second-order system with 5 . 0, the system performance can be estimated as follows. Rising time .sec42. 2 5 . 011 5 . 0arccos 1 arccos 22 n r t Peak time .sec62 . 3 5 . 0111 22 n p t Percent overshoot %3 .16%100%100 22 5 . 015 . 01 ee p Setting time .sec6 15 . 0 33 n s t (using a 5% setting criterion) P3.5 A second-order system gives a unit step response shown in Fig. P3.5. Find the open-loop transfer function if the system is a unit negative-feedback system. Solution: By inspection we have %30%100 1 13 . 1 p Solving the formula for calculating the overshoot, (s) 0 .1 3 .1 01.0 t )(tc Figure P3.5 5 .0 0 t )(tc 0 . 2 0 t )(tc Solutions 16 3 . 0 2 1 e p , we have 362. 0 ln ln 22 p p Since .sec1 p t, solving the formula for calculating the peak time, 2 1 n p t, we get sec/7 .33rad n Hence, the open-loop transfer function is )4 .24( 7 .1135 )2( )( 2 ssss sG n n P3.6 A feedback system is shown in Fig. P3.6(a), and its unit step response curve is shown in Fig. P3.6(b). Determine the values of 1 k, 2 k, and a. Solution: The transfer function between the input and output is given by 2 2 21 )( )( kass kk sR sC The system is stable and we have, from the response curve, 2 1 lim)(lim 1 2 2 21 0 k s kass kk stc st By inspection we have %9%100 00. 2 11 . 2 18. 2 p Solving the formula for calculating the overshoot, 09 . 0 2 1 e p , we have 608. 0 ln ln 22 p p Since .sec8 . 0 p t, solving the formula for calculating the peak time, 2 1 n p t, we get sec/95 . 4 rad n Then, comparing the characteristic polynomial of the system with its standard form, we have 00.2 18.2 08 .0 t )(tc )(sR)(sC )( 2 ass k 1 k (a)(b) Figure P3.6 Solutions 17 22 2 2 2 nns skass 5 .2495 . 4 22 2 n k 02 . 6 95 . 4 608. 022 n a P3.7A unity negative feedback system has the open-loop transfer function )2( )( kss k sG (a) Determine the percent overshoot. (b) For what range of kthe setting time less than 0.75 s (using a 5% setting criterion). Solution: (a) For the closed-loop transfer function we have 22 2 2 22 )( nn n ssksks k s hence, by inspection, we get sec/radk n , 22 The percent overshoot is %32 . 4 %100 2 1 e p (b) Since 9 . 022, letting .sec75. 0 25 . 0 33 k t n s (using a 5% setting criterion) results in 2 275 . 0 6 k, i.e. 32k P3.8 For the servomechanism system shown in Fig. P3.8, determine the values of kand athat satisfy the following closed-loop system design requirements. (a) Maximum of 40% overshoot. (b) Peak time of 4s. Solution: For the closed-loop transfer function we have 22 2 2 2 )( nn n ssksks k s hence, by inspection, we get k n 2 , k n 2, and n n k 22 Taking consideration of %40%100 2 1 e p results in 280 . 0 . In this case, to satisfy the requirement of peak time, 4 1 2 n p t, we have )(sR)(sC 2 s k as1 Figure P3.8 Solutions 18 .sec/818 . 0 rad n Hence, the values of kand aare determined as 67. 0 2 n k, 68 . 0 2 n P3.9 The open-loop transfer function of a unity feedback system is )2( )( ss k sG A step response is specified as: peak time s1 . 1 p t, and percent overshoot %5 p . (a) Determine whether both specifications can be met simultaneously. (b) If the specifications cannot be met simultaneously, determine a compromise value for kso that the peak time and percent overshoot are relaxed the same percentage. Solution: Writing the closed-loop transfer function 22 2 2 22 )( nn n sskss k s we get k n and k1. (a) Assuming that the peak time is satisfied sec1 . 1 1 1 2 k t n p we get 16. 9k. Then, we have 33 . 0 and %5%33%100 2 1 e p Obviously, these two specifications cannot be met simultaneously. (b) In order to reduce p the gain must be reduced. Choosing sec2 . 22 1 pp ttresults in 04 . 3 1 k, 57. 0 1 , %102%3 .11 1 pp Rechoosing sec31 . 2 1 . 2 2 pp ttresults in 85 . 2 1 k, 59. 0 1 , %10.51 . 2%0 .10 1 pp Letting sec255 . 2 05 . 2 3 pp ttresults in 941 . 2 3 k, 583. 0 3 , %10.2505 . 2 %5 .10 3 pp In this way, a compromise value is obtained as 941 . 2 k P3.10A control system is represented by the transfer function )13 . 0 4 . 0)(56. 2( 33. 0 )( )( 2 sss sR sC Estimate the peak time, percent overshoot, and setting time (%5), using the dominant pole method, if it is possible. Solution: Rewriting the transfer function as Solutions 19 3 . 0)2 . 0)(56. 2( 33. 0 )( )( 22 ss sR sC we get the poles of the system: 3 . 02 . 0 21 js ， , 56 . 2 3 s. Then, 21， scan be considered as a pair of dominant poles, because )Re()Re( 321 ss ， . Method 1. After reducing to a second-order system, the transfer function becomes 13. 04 . 0 13 . 0 )( )( 2 ss sR sC (Note: 1 )( )( lim 0 sR sC k s ) which results insec/36 . 0 rad n and55 . 0 . The specifications can be determined as sec0.421 1 2 n p t, %6 .12%100 2 1 e p sec67.20 1 1 ln 1 2 n s t Method 2. Taking consideration of the effect of non-dominant pole on the transient components cause by the dominant poles, we have sec0.841 1 )( 2 31 n p ss t %6 .13%100 2 1 31 3 e ss s p sec6 .23 2 ln 1 31 3 ss s t n s P3.11 By means of the algebraic criteria, determine the stability of systems that have the following characteristic equations. (a) 020920 23 sss (b) 025103 234 ssss (c) 021092 2345 sssss Solution: (a) 020920 23 sss. All coefficients of the characteristic equation are positive. Using L-C criterion, 0160 91 2020 2 D This system is stable. (b) 025103 234 ssss. All coefficients of the characteristic equation are positive. Using L-C criterion, 0153 1100 253 0110 3 D This system is unstable. Solutions 20 (c) 021092 2345 sssss. (Its better to use Routh criterion for a higher-order system.) All coefficients of the characteristic equation are positive. Establish the Routh array as shown. There are two changes of sign in the first column, this system is unstable. P3.12 The characteristic equations for certain systems are given below. In each case, determine the number of characteristic roots in the right-half s-plane and the number of pure imaginary roots. (a) 023 3 ss (b) 01601610 23 sss (c) 0483224123 2345 sssss (d) 084632 2345 sssss Solution: (a) 023 3 ss. The Routh array shows that there are two changes of sign in the first column. So that there are two characteristic roots in the right-half s-plane. (b) 01601610 23 sssThe 1 s-row is an all-zero one and an auxiliary equation is made based on 2 s-row 016 2 s Taking derivative with respect to syields 02 s The coefficient of this new equation is inserted in the 1 srow, and the Routh array is then completed. By inspection, there are no changes of sign in the first column, and the system has no characteristic roots in the right-half s-plane. The solution of the auxiliary are 4js, the system has a pair of pure imaginary roots. (c) 0483224123 2345 sssss. The Routh array is established as follows. The 1 s-row is an all-zero one and an auxiliary equation based on 2 s-row is 04 2 s Taking derivative with respect to syields 02 s The coefficient of this new equation is inserted in the 1 srow, and the Routh array is then completed. By inspection, there are no changes of sign in the first column, and the system has no characteristic roots in the right-half 5 s191 4 s2102 3 s40 2 s102 1 s-0.80 0 s2 3 s1-3 2 s0 02 1 s 23 0 0 s2 3 s116 2 s10116016 1 s020 0 s16 5 s11232 4 s312484861 3 s41164 2 s41164 1 s0 20 0 s4 Solutions 21 s-plane. The solution of the auxiliary are 2js, the system has a pair of pure imaginary roots. (d) 084632 2345 sssss. The Routh array is established as follows. The 3 s-row is an all-zero one and an auxiliary equation based on 4 s-row is 043 24 ss Taking derivative with respect to syields 064 3 ss The coefficient of this new equation is inserted in the 3 srow, and the Routh array is then completed. By inspection, the sign in the first column is changed one time, and the system has one root in the right-half s-plane. The solution of the auxiliary are1 21 ， s2 43 js ， , the system has one pair of pure imaginary roots. P3.13 The characteristic equations for certain systems are given below. In each case, determine the value of kso that the corresponding system is stable. It is assumed that kis positive number. (a) 02102 234 kssss (b) 0504)5 . 0( 23 kssks Solution: (a) 02102 234 kssss.The system is stable if and only if 90 220 101 022 0 3 kkD k i.e. the system is stable when 90 k. (b) 0504)5 . 0( 23 kssks. The system is stable if and only if 0) 3 . 3)(8 . 34(050240 41 505 . 0 0, 05 . 0 2 2 kkkk k k D kk i.e. the system is stable when 3 . 3k. P3.14 The open-loop transfer function of a negative feedback system is given by ) 12 . 001. 0( )( 2 sss K sG Determine the range of Kand in which the closed-loop system is stable. Solution: The characteristic equation is 02 . 001 . 0 23 Ksss The system is stable if and only if 5 s13-4 4 s2163-84 3 s04060 2 s3 -8 1 s 350 0 0 s-8 Solutions 22 200010200 101 . 0 2 . 0 02 . 0, 0 2 KK. K D k The required range is 020 K. P3.15The open-loop transfer function of negative feedback system is given ) 12)(1( ) 1( )()( sTss sK sHsG The parameters Kand Tmay be represented in a plane with Kas the horizontal axis and Tas the vertical axis. Determine the region in which the closed-loop system is stable. Solution: The characteristic equation is 0) 1()2(2 23 KsKsTTs Since all coefficients are positive, the system is stable if and only if 0) 1)(2(0 12 2 2 KT KT KT D 022TKTK 04)2()2(TTK4) 1)(2(KT The system is stable in the region 4) 1)(2(KT, which is plotted as shown. (Letting 2TTand 1KK results in 4 K T.) P3.16A unity negative feedback system has an open-loop transfer function ) 1)(1)(1( )( 2 TsnnTsTs K sG where 10 n, 0K, Tis a positive constant. (a) Determine the range of Kand nso that the system is stable. (b) Determine the value of Krequired for stability for 1n, 0.5, 0.1, 0.01, and 0. (c) Discuss the stability of the closed-loop system as a function of nfor a constant K. Solution: The closed-loop characteristic equation is 0) 1)(1)(1( 2 KTsnnTsTs i.e. 01) 1()( 22223333 KTsnnsTnnnsTn (a) The system is stable if and only if 0 ) 1( 1) 1( 233 22 2 TnnTn KTnnn D i.e. 0) 1(0) 1() 1( 2223322 KnnnnTKTnnn 1 1 1 1 1 1 22 2 2 n nn n nn K n nn K 2 2 2 2 2 1 1) 1( 11 )1( n nK n n nn nnnK T T K K 0 0 2 2 2 2 4 4 4 4 6 6 1 1 1 1 Solutions 23 hence, the system is stable when 2 2 1 1) 1(0 n nK. (b) The value of Krequired for stability for 1n, 0.5, 0.1, 0.01, and 0 are calculated as shown. 80 Kfor 1n, 5 .110 Kfor 5 . 0n, 21.1220 Kfor 1 . 0n, 102020 Kfor 01 . 0 n, K0for 0n. (c) For a constant K, the stability of the closed-loop system is related to the value of n, the larger the value of n,the easier the system to be stable. (Stagger principle.) P3.17 A unity negative feedback system has an open-loop transfer function ) 1 6 )(1 3 ( )( ss s K sG Determine the range of krequired so that there are no closed-loop poles to the right of the line 1s. Solution: The closed-loop characteristic equation is 018)6)(3(0) 1 6 )(1 3 (KsssK ss s i.e. 018189 23 Ksss Letting 1 ssresulting in 0)1018( 3 6 018)5 )(2 )(1 ( 23 KsssKsss Using Lienard-Chipart criterion, all closed-loop poles locate in the right-half s -plane, i.e. to the right of the line 1s, if and only if 91408 . 1820 31 10186 95, 01018 2 KK K D KK The required range is 9 14 9 5 K, or 56 . 1 0.56 K P3.18 A system has the characteristic equation 02910 23 ksss Determine the value of kso that the real part of complex roots is 2, using the algebraic criterion. Solution: Substituting 2 ssinto the characteristic equation yields 02 292 102 23 ksss）（）（）（ 0)26( 4 23 ksss The Routh array is established as shown. If there is a pair of complex roots with real part of 2, then 026 k 3 s11 2 s426k 1 s 0 s Solutions 24 i.e. 30k. In the case of 30k, we have the solution of the auxiliary equation js , i.e. js 2. P3.19 An automatically guided vehicle is represented by the system in Fig. P3.19. (a) Determine the value of required for stability. (b) Determine the value of when one root of the characteristic equation is 5s, and the values of the remaining roots for the selected . (c) Find the response of the system to a step command for the selected in (b). Solution: The closed-loop transfer function is 101010 10 )( )( )( 23 sss sR sC s (a) The closed-loop characteristic equation is 0101010 23 sss Since all coefficients are positive, the system is stable if and only if 1 . 00 101 1010 2 D (b) Substituting 5 ssinto the characteristic equation yields 0105 105 105 23 ）（）（）（sss 0)50135( )2510( 5 23 sss In the case of 050135, i.e. 7 . 2, we have 0 1 s, i.e. 5 1 s. Solving the characteristic equation with 7 . 2, i.e. 0 2 5 23 sssresults in 56. 4 2 sand 44 . 0 3 s. Hence the remaining roots are 44. 0 2 sand 56 . 4 3 s. (c) The closed-loop transfer function for 7 . 2is )5)(56. 4)(44 . 0 ( 10 )( sss s The unit