自动控制原理(中英文对照李道根)习题4.题解

1、 SolutionsSolutionsP4.1 Sketch the root loci for the following open-loop transfer fiinctions when 0 < Ar < oo k(s + 5)二心2心3)'(c) GH(s)= $2 ,+ 4s + 8(e) G(s)H(s)=*s(s + 3)-k(s + 5)(b) G(s)H(s) =")(» 3)(” 令(d)呦阳蔦;31 SolutionsHere, "sketch" means that it is not necessaiy t

2、o find the exact positions of the possible breakaway point and the intersection with jaxis, and the exact values of relative angles.Solution: (a)G(s)H(s)=上(s + 5)s(s + 2)(s + 3)By inspection、wehave Pi = 0 , p2 = -2 , p3 = -3 and 二=一5 Hie intersection of asymptotes is(0-2-3)-(-5) A6 = 03-1and the ang

3、les of asymptotes are(Pa =(2/+lpr _ £3±3# SolutionsTliere is a breakaway point on the real-axis segment (-2,0) Consideiiiig tliat另 Si =乞 Pi = 5 , due to n - ni = 2 , tlie root locus does not intersect vvitli j-axis Tlie root locus is plotted as shownSolving tlie breakaway-poiiit equation-+

4、 11=> s' +lQs2 + 25s+ 15 = 0=>(s + 10)s + 25)s + 15 = 0s s+2 s + 3 s + 5yields the breakaway point sb = -0.88 (b) G(s)H(s)二k(s + 5)s(s + l)(s + 3)(s + 4)By inspection, we have = 0 , p2 p3 = -3,p4 = -4 and 二】=-5 The intersection of asymptotes is(0_l_3_4)_(_5)4-1and the angles of asymptotes

5、are(Pa =+l);r4-1n=±TTliere aie three breakaway points on the real-axis segments: (一1,0), (-4,-3) and (-s,- 5) Tlie root locus is plotted asshown.Solving the breakaway-point equation# Solutions-+= =>3s° +35s +139s? + 190s+ 60 = 0s s + 1 s + 3 s + 4 s + 5we have sbl = -0.44, sb2 = -3.7 ,

6、sb3 = -5.8 Tlie characteristic equation iss" +8s3 +19s? +(r+2)s + ± = 0Substituting s = ja into this equation yields(xf -19” + = 0一 8血3 +(任+ 2)69=0we get tlie intersection of root locus with the j-axis coc = ±3.53 , where kc = 87.4 .k(s + 3)k(s + 3)(c) Gs)H(s) = = By inspection, we ha

7、ve pL2 = -2± j2 ands + 4s + 8 (s + 2) * + 2 *:=一5 Tlie root locus in the complex plane is a pait of a cii cle vitli the center at the zero and a radius of the lengtli fioin the zero to one pole. Tliere is a breakaway point oil the real-axis segment, (-s 3) The root locus is plotted as shown.Sol

8、ving the breakaway-poiiit equation=> s2 +6s + 4 = 02s + 4_1s' +4s + 8 s + 3we have sb = -5.24 . Oi; -P = 4 sb = 一3-VJ = -5.24 . Tlie angles of deparhire of the root locus from tlie complex poles are2Or =180°+ZQ?!-p J = 180° +arctan90° =153.4°,=-153.4°3-2厂By inspection

9、, we have p】=0 ,(d) G(s)H(s)=严 TO)二饥s+严,十 20$ 十 200) s(s 十 10) +10-,_ (0-10-10)-(-20) “ a = Up? 3 = -10 ± J10 and 二=-20 The iiitei section of asymptotes is3-1and the angles of asymptotes are “(2/+l)%=TT- = ±TThe characteristic equation is s3 + 40s2 + 400s + 2000 = 0 . If the root locus has

10、 breakaway points, the closed-loop system will have a thrice root, i.e. a root will satisfy 3s = -40 and s3 = 2000 simultaneously. Obviously、it is impossible Tlie root locus is plotted as sliowii.Tlie angles of departure of the root locus fi om the complex poles are&P2 =180° +Z(p2 -S!)-Z(/?

11、2 - Pl) = 0°, 0p2 = 0° Solutions(e) G(s)H(s)=7 . By inspection, we have p】=0 ,s(s + 3)-p?=卩3 = 3. Tlie intersection of asymptotes is 0-3-3.咕一-=-iand the angles of asymptotes are(21 +1)7： n从一 = ±亍，Tliere is a breakaway point on the real-axis segment (一3.0) Tlie root locus is plotted as

12、 sliowii.12Solving the breakaway-point equation += 0 , we have sb =-l. The characteristic-6ar +k = Q一 / +9a)=Qs s + 3equation is s3 +6s2 +9s + k = 0 . Substituting s = js into this equation yields we get tlie intersection of root locus witli the j-axis q = ±3, kc =54 P4.2 Consider a unity feedb

13、ack system withG(沪冲-s +4s + 5(a) Find the angles of departure of tlie root locus fi om tlie complex poles, (b) Find the entry point for the root locus as it enters the real axis.Ar(s +1) lc(s + 1).Solution: G(s)= =-.By inspection, we have p, 9 = -2 ± jl and/ +4s + 5 (s + 2) +1-1 = -1 (a) Tlie a

14、ngles of depaitiire of tlie root locus fi om tlie complex poles ai e% =180°+Z(p1 一二J一上(pi -p2) = 180o+135°-90° = 225°,=-225°(b) The breakaway-point equation is given by2s + 41o r c=> 匸 + 2s-1 = 0匸 +4s + 5 s + 1Solving this equation yields s2 = 一1 土血.Tliere is only one bre

15、akaway point sb = -1 -血.P4.3 A unity feedback system has a plant transfer fiinctionG(s) =ss(005s +0.4S + 1)Solution: G(s)=Sketch the root locus as K varies-_- t- - By inspection, wes(005s，+0.45 + 1)s(s* +8s + 20) s(s + 4) +2"have Pi = 0 mid p? 3 = -4 土 丿2 Tlic intersection of asymptotes is0 4 4

16、8oa = -2.6733and their angles are(2Z+1)nSolving the breakaway-point equation1 2s + 8= 0 =>3s + 16s+ 20 = 0 s 厂 + 8s+ 20we get two breakaway points=-333, sb2 = -2Tlie characteristic equation is s' + 8s? + 20s += 0yields <- Zgt + K* = 0.,we get the intersection of root locus with the j-axis-

17、 ft? +2069= 0G)c = ±2y5 = ±4.47 , K； = 160 Hie root locus is plotted as shown.P4.4 A contiol system has tlie open-loop transfer fimetion、k© +4s + 8)G(s)HG)=s (s + 4)It is desiied that the damping ratio g of the dominant poles is equal to 0.5. Using the root locus, show that k = 7.35 i

18、s reqiiiied and tlie dominant poles are s = 13 土 j2.2 .Solution: At fust we plot the root locus Sincek(s2 +4s + 8)灿G + 2)? +22 s?(s + 4)s"s + 4)we get Pi = j92 = 0 , p3 = -4 and 2 = -2± jl . The characteristic equation iss' + (上+ 4)s? +4好+ 8上=0It can be proven that there aie neither br

19、eakaway point nor intersection witli j-axis for the root locus Tlie angles of arrival of the root loais to tlie complex zeros areg =180° + £z(z1-jpl)-Z(21-z2) = 45°, % 一45° 丿=1Hie root locus is plotted as shown.Assuming the line $ = 0.5 intersects witli the root locus at = -x+ jy

20、 = -x+ j4x , we have刀 Z($i -二i)-另 - Pj)= (21 +1)兀1=1 ;=1arctan + arctan + - 2 x- arctan= (2/ +1)2x2x34xa/Jx + 2 n 込x =+ arctan2-x34-x77任arctan :+ arctail5 =42 - => x3 -2x2 +4x-4 = 0 V3x-2 V3x + 2 爲3x2-x2-x4-xx = 1.3, s = -1.3 + J2.2Hence, s = 1.3 土 j2.2 are a pair of complex poles with = 0.5 .Usi

21、ng magnitude condition we get that tlie gain for 歹=0.5 is35 Solutions# Solutions($+2)2 +2?J(0.72 -2.22 +4)2 +3.0821513+/2.2From tlie characteristic equation we getS + S3 = (Jc + 4) = 11.34 9 S3 = 8.74Tlierefore. the dominant poles are s = 13 土 j2.2 , because |31 »|Re(S )| P4.5 A unity feedback

22、system hass(s + 2)(s + 5)Find (a) tlie breakaway point 011 the real axis and the gain for tliis point, (b) tlie gain and the roots when two roots lie on the imagiiiaiy axis, and (c) the roots when 世=6 (d) Sketch the root locus.Solution: (a) SoKing the breakaway-point equation# Solutionss s+2 s+5we f

23、iud the breakaway point sb = -0.88 (Another solution of this equation is not on tlie root locus.) The gain for tliis point is心*b| 卡 b+2|Sb+5| = 4.06(b) The characteristic equation is s' +7s? +10s + t = 0Substituting s = ja into this equation yields <-+1069= 0,we get the intersection of rootlo

24、cus with tlie j-axisa)c = ±710 = ±3.16 , kc =70lienee, Wien two roots lie 011 tlie imagiiiaiy axis, the gain is k = 70 and the roots aie s = ± 7-0 .(c) In the case of t = 6 , the characteristic equation is($) = $' +7s2 +10 + 6 = 0Since k = 6、tliere are one real root on segment (-c

25、o,-5) and a pair of complex roots, it isnot diffiailt to find that 力(一534)« 0, i.e. s = -5.34 is a root of this equation. Denoting 2 jb , we have© + $2 + S3 = -7=> a = -0.83= -6= b = 0.66i.e., in this case the closed-loop roots are s】=-5.34 and s? 3 = -0.83 士 j0.66 (d) Tlie root locus i

26、s plotted as shown、vvliere for the asymptotes we have2 571aa= = 一233,(pa =± , 7133P4.6 Tlie open-loop transfer fimction of a unity feedback system is given bylc(s 十 1)G(s)=斗-(s + d)Detenniiie the values of a so that the root locus will have zero, one, and two breakaway points, respectively, not

27、 counting the one at s = 0 Sketch the root loci for 0 <r < co fbr all three casesSolution: Solving the breakaway-point equation2 1 1 ,一 += => 2厂 + (d + 3)s + 2d = 0s s + a s + 1we have_ _(a + 3)± J(a + 3)2 _3d _ -(d + 3)±J(d_l)(a_9)s =44(a) I11 the case of a > 9, the equation h

28、as two real roots and botli ai e breakaway points. For example, letting a = 10 resulting sbl = -2.5 and sb2 = -4. The root locus when a =10 is plotted as shown, where aa = -4.5 , (pa = ±90°.j(a)(b) hi the case of a = 9 .the equation has a twice root s = -3 and it is a twice breakaway point

29、. Hie root louus wiien a = 9 is plotted as shown、 vvliei e = 一4 , % = ±90° I11 tlie case of 1 < d < 9, tlie equation has no real roots For example, letting d = 836 Solutionsresults in the root locus as shown, where aa = -3.5 , (pa = ±90° 37 Solutions# Solutions(c) hi tlie c

30、ase of a = 1, the solution of die equation is s = 1 In fact, in tliis case a braicli of the root locus becomes a point at s = 1 Tlie root locus when cr = 10 is plotted as shown, where oa = -4.5 , (pa = ±90°.(d) In the case of a v 1, the equation has two real roots, but neither is breakaway

31、 point. For example, letting a = 0.5 yielck two roots sx = -3.6 and s2 = -139 , which are not on tlie root locus. The root locus when a = 0.5 is plotted as shown, where aa = 0.25 , (pa = ±90°.P4.7 Tlie transfer fiinctions of a negative feedback system are given byifG(s)=and H(s) = ls-(s+2)

32、(s + 5)(a) Sketch tlie root locus for this system(b) Tlie transfer fimction of the feedback loop is now changed to H(s) = 2s +1 Indicate the crossing points of the locus oil tlie iinagiiiaiy axis and tlie con espondiiig value of k at tliese points. Detemiiiie tlie stability of the modified system as

33、 a fimction of k. Investigate the effect on the root locus chie to this change in H(s).Solution: (a) Tliere are four open-loop poles: Pi = P? = 0、P3 = -2 and p4 9 andthere is no zero. For the asymptotes we haveoa =_7/4 = -1.75, % =±45°,±135°Solving the breakaway-point equation2 1

34、 1+= 0s s+2s+5we get S = -1.25 , s2 = -4.13, where sb = -4.13 is a breakaway point. Since tlie closed-loop system is constnictional unstable, tlie root locus does not intersect vvitli j-axis for k >0. The root locus is plotted as sliowii.(b) I11 tliis case, tlie open-loop tiaiisfer fimction is gi

35、ven byG(s)H(s)=严“)=f(")s(s + 2)(s + 5) (s + 2)(s + 5)Tlie asymptotes are centered at aa = 一 6.5/3 = -2.17 with angles (pa = ±60o,180° . Investigating the breakaway-poiiit equation2 1 1 1H1=s- s + 2 s + 5s + 0.53八 +16s? +20.5s + 10 = 0we know that the root locus does not have breakaway

36、 point.Substituting s= (t into tlie characteristic equations4 +7s? +10S? + Ar * + 0.5 Ar* = 0q410q2 +0.5Ar* =0-7 a)3 +0byields a)c = ±77 = ±2.55、 k： = 45.5 Tlie root locus is plotted as shown.Now, the system is stable only when 0 < f v 45.5 , i.e. Q <k < 22.75 .As we see, the stability of tlie closed-loop system is improved due to the change ill H(s)P4.8 Tlie characteristic equation of a feedback conh ol system is given byA(s) = s2 + (£ + 5)s + 2世 + 4 = 0 Sket