自控第三章作业答案 - P3.4 The open-loop transfer function of a unity negative feedback system is G(s)?1s(s?1) Determine the rise time, peak time, percent overshoot and setting time (using a 5% setting criterion). Solution: Writing he closed-loop transfer function ?(s)?1s?s?12-n222s?2-ns-n ?0.5we get ?n?1, -0.5. Since this is an underded second-order system with ?system performance can be estimated as follows. Rising time tr?, the -arccos-n1-2-?arccos0.51?1?0.52?2.42 sec. Peak time tp-?n1-2-1?1?0.52?3.62 sec. Percent overshoot Setting time ts-3p?e-?1-2?100 %?e0.5?1?0.52?100 %?16.3 % 30.5?1-?6 sec. (using a 5% setting criterion) c(t)1.3nP3.5 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 ?p1.0?1.3?112?100 %?30 % 00.1t(s)Figure P3.5 Solving the formula for calculating the overshoot, ?p ?e-1-?0.3, we have --ln?2p2?0.362 -n1-2?ln?pSince tp?1 sec., solving the formula for calculating the peak time, ?n?33.7 rad/sec Hence, the open-loop transfer function is G(s)-n2tp?, we get s(s?2-n)?1135.7s(s?24.4) 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 k1, k2, and a. c(t)R(s)k1?k2s(s?a)C(s)2.182.00t0(a)Figure P3.6 0.8(b) Solution: The transfer function between the input and output is given by C(s)R(s)?sk1k22 ?as?k2k1k2s2The system is stable and we have, from the response curve, limc(t)?lims?t-s?0?1s?as?k2?k1?2 By inspection we have ?p?2.18?2.112.00?100 %?9 % pSolving the formula for calculating the overshoot, ? --ln?2p2?e-1-2?0.09, we have ?0.608 -n1-2?ln?pSince tp?0.8 sec., solving the formula for calculating the peak time, tp?, we get ?n?4.95 rad/sec Then, paring the characteristic polynomial of the system with its standard form, we have 2 s2?as?k2?s2?2-ns-n 22 k2-n?4.95?24.5 a?2-n?2?0.608?4.95?6.02 P3.8 For the servomechanism system shown in Fig. P3.8, determine the values of k and a that 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 ?(s)?ks22?nR(s)?ks2C(s)1?asFigure P3.8 ?k?s?k?s2?2-ns-2-k22n hence, by inspection, we get 2?k, 2-n?k?, and ? ?n?n?2-n results in -n1-2Taking consideration of -0.280. ?p?e-1-?100 %?40 %In this case, to satisfy the requirement of peak time, ?n?0.818 rad/sec. tp-4, we have Hence, the values of k and ?a are determined as ?0.682 k-n?0.67, ?2-n P3.10 A control system is represented by the transfer function C(s)R(s)?(s?2.56)(s0.332 ?5%?0.4s?0.13)Estimate the peak time, percent overshoot, and setting time (?method, if it is possible. Solution: Rewriting the transfer function as C(s)R(s)?0.33(s?2.56)[s(?0.2)2), using the dominant pole ?0.3]2 we get the poles of the system: s1，, s3-2.56. Then, s1， can be considered as 2 2-0.2?j0.3a pair of dominant poles, because Re(s1， 2)-Re(s3). Method 1. After reducing to a second-order system, the transfer function bees C(s)R(s)?s0.132?0.4s?0.13 (Note: k-limC(s)R(s)s?0?1) which results in ?n?0.36 rad/sec and ? tp-0.551-2. The specifications can be determined as ?100 %?12.6 %-n1-210.4 2sec, ?p?e- ts?1-n?1?ln-?1-?2-?20.67 sec -?Method 2. Taking consideration of the effect of non-dominant pole on the transient ponents cause by the dominant poles, we have tp--(s1?s3)?n1-2?10.8 4sec ? p?s3s1?s3e-1-2?100 %?13.6 % ts?1-n?2s3ln--s?s13--23.6 sec -P3.13 The characteristic equations for certain systems are given below. In each case, determine the value of k so that the corresponding system is stable. It is assumed that k is positive number. (a) s4?2s3?10s2?2s?k?0 (b) s3?(k?0.5)s2?4ks?50?0 Solution: (a) s4?2s3?10s2?2s?k?0.The system is stable if and only if ?k?0?2-?D3?1?0?21020k?0 ? k ?920?k?9 i.e. the system is stable when . (b) s3?(k?0.5)s2?4ks?50?0. The system is stable if and only if ?k?0.5?0, k?0?k?0.550?2D-0 ? 4k?2k?50?0 ? 4(k?3.8)(k?3.3)?02?14k?k?3.3. i.e. the system is stable when Ks(0.01s2P3.14 The open-loop transfer function of a negative feedback system is given by G(s)-0.2?s?1) Determine the range of K and ? in which the closed-loop system is stable. Solution: The characteristic equation is 0.01s3?0.2?s2?s?K?0 The system is stable if and only if ?k?0, 0.2-0?0.2?K?D-0 ? 0.2-0.01K?0 ? K?20?2?0.011?20-K?0 The required range is Ks(s3?1)(. P3.17 A unity negative feedback system has an open-loop transfer function G(s)?s6 ?1)kDetermine the range of s-1. s3s6 required so that there are no closed-loop poles to the right of the line Solution: The。