自动控制原理(中英文对照-李道根)习题4.题解.pdf

1、Solutions31SolutionsP4.1 Sketch the root loci for the following open-loop transfer functions when k0.(a)3)(2()5()()(ssssksHsG,(b)4)(3)(1()5()()(sssssksHsG(c)84)3()()(2sssksHsG,(d)20020()20()()(2ssssksHsG(e)2)3()()(ssksHsGHere,“sketch”means that it is not necessary to find the exact positions of the

2、possible breakaway point and the intersection with j-axis,and the exact values of relative angles.Solution:(a)3)(2()5()()(ssssksHsG.By inspection,we have 01p,22p,33pand 51z.The intersection of asymptotes is013)5()320(aand the angles of asymptotes are213)12(laThere is a breakaway point on the real-ax

3、is segment)02(，.Considering that 5iips,due to 2mn,the root locus does not intersect with j-axis.The root locus is plotted as shown.Solving the breakaway-point equation015)25)10(0152510513121123ssssssssssyields the breakaway point88.0bs.(b)4)(3)(1()5()()(sssssksHsG.By inspection,we have 01p,12p,33p,4

4、4pand 51z.The intersection of asymptotes is114)5()4310(aand the angles of asymptotes are,314)12(laThere are three breakaway points on the real-axis segments:)01(，,)34(，and)5(，.The root locus is plotted as shown.Solving the breakaway-point equationSolutions32060190139353514131111234ssssssssswe have44

5、.01bs,7.32bs,8.53bs.The characteristic equation is0)2(198234ksksssSubstituting js into this equation yields0)2(8019324kkwe get the intersection of root locus with the j-axis 53.3c,where 4.87ck.(c)2222)2()3(84)3()()(ssksssksHsG.By inspection,we have 2221jp，and 51z.The root locus in the complex plane

6、is a part of a circle with the center at the zero and a radius of the length from the zero to one pole.There is a breakaway point on the real-axis segment,)3(，.The root locus is plotted as shown.Solving the breakaway-point equation04631844222sssssswe have 24.5bs.Or,511 pz,24.553bs.The angles of depa

7、rture of the root locus from the complex poles are4.15390232arctan180)()(18021111ppzpp,4.1532p(d)10)10()20()20020()20()()(222ssskssssksHsG.By inspection,we have 01p,10103,2jpand 201z.The intersection of asymptotes is013)20()10100(aand the angles of asymptotes are213)12(laThe characteristic equation

8、is 020004004023sss.If the root locus has breakaway points,the closed-loop system will have a thrice root,i.e.a root will satisfy 403sand 20003ssimultaneously.Obviously,it is impossible.The root locus is plotted as shown.The angles of departure of the root locus from the complex poles are0)()(1801212

9、2ppzpp,02pj010-20-10Solutions33(e)2)3()()(ssksHsG.By inspection,we have 01p,332 pp.The intersection of asymptotes is13330aand the angles of asymptotes are,33)12(laThere is a breakaway point on the real-axis segment)03(，.The root locus is plotted as shown.Solving the breakaway-point equation0321ss,we

10、 have 1bs.The characteristic equation is 09623ksss.Substituting js into this equation yields090632k,we get the intersection of root locus with the j-axis3c,54ck.P4.2 Consider a unity feedback system with54)1()(2sssksG(a)Find the angles of departure of the root locus from the complex poles.(b)Find th

11、e entry point for the root locus as it enters the real axis.Solution:1)2()1(54)1()(22ssksssksG.By inspection,we have 122,1jpand 11z.(a)The angles of departure of the root locus from the complex poles are22590135180)()(18021111ppzpp,2252p(b)The breakaway-point equation is given by01211544222ssssssSol

12、ving this equation yields 212,1s.There is only one breakaway point 21bs.P4.3 A unity feedback system has a plant transfer function)14.005.0()(2sssKsGSketch the root locus as Kvaries.Solution:2)4()208()14.005.0()(2222ssKsssKsssKsG.By inspection,we have 01pand 243,2jp.The intersection of asymptotes is

13、Solutions3467.2383440aand their angles are,33)12(laSolving the breakaway-point equation020163020882122sssssswe get two breakaway points33.31bs,22bsThe characteristic equation is 020823Ksss.Substituting js into this equation yields 0200832K,we get the intersection of root locus with the j-axis47.452c

14、,160cKThe root locus is plotted as shown.P4.4 A control system has the open-loop transfer function)4()84()()(22ssssksHsGIt is desired that the damping ratio of the dominant poles is equal to 0.5.Using the root locus,show that 35.7kis required and the dominant poles are 2.23.1js.Solution:At first we

15、plot the root locus.Since)4(2)2()4()84()()(22222ssskssssksHsGwe get 021 pp,43pand 222,1jz.The characteristic equation is084)423kkssks（It can be proven that there are neither breakaway point nor intersection with j-axis for the root locus.The angles of arrival of the root locus to the complex zeros a

16、re45)()(180213111zzpzjiz,452zThe root locus is plotted as shown.Assuming the line 5.0intersects with the root locus at xjxjyxs31,we have)12()()(311211lpszsjjii)12(43arctan322223arctan223arctanlxxxxxx5.0Solutions35xxxxxx43arctan3223arctan223arctan04424331433223223122322323xxxxxxxxxxxxxxx3.1x,2.23.1js

17、Hence,2.23.1jsare a pair of complex poles with 5.0.Using magnitude condition we get that the gain for 5.0is34.708.3)42.27.0(2.27.2)2.23.1(2)2(4222222222.23.1222jsssskFrom the characteristic equation we get 34.11)4(321ksss,74.83sTherefore,the dominant poles are 2.23.1js,because)Re(13ss.P4.5 A unity f

18、eedback system has)5)(2()(sssksGFind(a)the breakaway point on the real axis and the gain for this point,(b)the gain and the roots when two roots lie on the imaginary axis,and(c)the roots when 6k.(d)Sketch the root locus.Solution:(a)Solving the breakaway-point equation051211ssswe find the breakaway p

19、oint88.0bs.(Another solution of this equation is not on the root locus.)The gain for this point is06.452bbbbsssk(b)The characteristic equation is010723ksssSubstituting js into this equation yields 0100732k,we get the intersection of root locus with the j-axis16.310c,70ckhence,when two roots lie on t

20、he imaginary axis,the gain is 70kand the roots are 10js.(c)In the case of 6k,the characteristic equation is06107)(23ssssSince bkk 6,there are one real root on segment)5,(and a pair of complex roots,it is Solutions36not difficult to find that 0)34.5(,i.e.34.51sis a root of this equation.Denoting jbas

21、2,1,we have83.07321asss66.06321bsssi.e.,in this case the closed-loop roots are 34.51sand 66.083.03,2js.(d)The root locus is plotted as shown,where for the asymptotes we have33.2352a,3aP4.6 The open-loop transfer function of a unity feedback system is given by)()1()(2asssksGDetermine the values of as

22、o that the root locus will have zero,one,and two breakaway points,respectively,not counting the one at 0s.Sketch the root loci for k0for all three cases.Solution:Solving the breakaway-point equation02)3(211122asassasswehave4)9)(1()3(413)3()3(2aaaaaas(a)In the case of 9a,the equation has two real roo

23、ts and both are breakaway points.For example,letting 10aresulting 5.21bsand 42bs.The root locus when 10ais plotted as shown,where5.4a,90a.(b)In the case of 9a,the equation has a twice root 3sand it is a twice breakaway point.The root locus when 9ais plotted as shown,where4a,90a.(c)In the case of 91

24、a,the equation has no real roots For example,letting 8aj0-1-10-4.5(a)Solutions37results in the root locus as shown,where5.3a,90a.(d)In the case of 1a,the solution of the equation is 1s.In fact,in this case a branch of the root locus becomes a point at 1s.The root locus when 10ais plotted as shown,wh

25、ere5.4a,90a.(e)In the case of 1a,the equation has two real roots,but neither is breakaway point.For example,letting 5.0ayields two roots 6.31sand 39.12s,which are not on the root locus.The root locus when 5.0ais plotted as shown,where25.0a,90a.P4.7 The transfer functions of a negative feedback syste

26、m are given by)5)(2()(2sssksGand 1)(sH(a)Sketch the root locus for this system.(b)The transfer function of the feedback loop is now changed to 12)(ssH.Indicate the crossing points of the locus on the imaginary axis and the corresponding value of kat these points.Determine the stability of the modifi

27、ed system as a function of k.Investigate the effect on the root locus due to this change in)(sH.Solution:(a)There are four open-loop poles:021 pp,23pand 54p,and there is no zero.For the asymptotes we have75.147a,135,45aSolving the breakaway-point equation051212ssswe get 25.11s,13.42s,where 13.4bsis

28、a breakaway point.Since the closed-loop system is constructional unstable,the root locus does not intersect with j-axis for 0k.The root locus is plotted as shown.(b)In this case,the open-loop transfer function is given by)5)(2()5.0()5)(2()12()()(22sssskssssksHsG,kk2j0-1-8-3.5(c)j0-12.5-0.5(e)j0-1(d)

29、Solutions38The asymptotes are centered at 17.235.6awith angles 801,60a.Investigating the breakaway-point equation5.01512122ssss0105.2016323ssswe know that the root locus does not have breakaway point.Substituting js into the characteristic equation 0705.01005.0107324234kkksksssyields 55.25.6c,5.45ck

30、.The root locus is plotted as shown.Now,the system is stable only when 5.450k,i.e.75.220 k.As we see,the stability of the closed-loop system is improved due to the change in)(sH.P4.8 The characteristic equation of a feedback control system is given by042)5()(2kskssSketch the root locus as a function

31、 of k(positive konly)for this system.Solution:Rewritingthe characteristic polynomial as)2(4542)5()(22sksskskssyields a equivalent transfer function)4)(1()2(45)2()(2sssksssksGeThe root locus is plotted as shown.P4.9A unity feedback system has a plant)1()(25.0)(2ssassGSketch the root locus as s functi

32、on of a(positive aonly)for this system.Solution:Rewritingthe characteristic polynomial asasssassss25.025.0)(25.0)1()(232yields a equivalent transfer function22)5.0()25.0(25.0)(ssasssasGeUsing the rules for plotting root locus,we have33.031a,180,60a17.061bs;5.0cThe root locus is plotted as shown.-5j0

33、-2j0-0.5Solutions39P4.10The open-loop transfer function of a control system with positive feedback is given by)44()()(2sssksHsGSketch the root locus for this system when k0.Solution:The root-locus equation is given by1)2(0)()(12ssksHsGusing the rules for plotting 0root locus,we have33.134a,120,0aThe root locus has neither breakaway point nor intersection with j-axis.The root locus is plotted as sh