自动控制原理(中英文对照李道根)习题3.题解

1、 Solutions13 SolutionsSolutionsP3.1 Tlie unit step response of a ceitaiii system is given by c(f)= i+k-严，tn。(a) Detenniiie the impulse response of the system(b) Detenniiie the transfer fimction C(s)/R(s) oftlie systemSolution: Tlie impulse response is the differential of coirespondiiig step response

2、, i.e.g =豊2 =灭）-戶+茲fAs we know that the transfer fimction is the L叩lace transfbmi of conespondiiig impulse response, i.e.# Solutions# SolutionsP3.2 Consider the system desci ibed by the block diagram shown ill Fig. P3.2(a). Detenniiie tlie polarities of two feedbacks for each of the following step r

3、esponses shown in Fig. P3.2(b), where “(T indicates tliat tlie feedback is open.(a) Block diagramW)Parabolic(b) Unit-step resp onsesFigure P32Solution: In general we haveC(s) _¥2R(s) s2 土 k=s 土 k出o 2 o 1 2Note that tlie chai acteristic polynomial isA(s) = s2 ±ks±kxk0.0.wiiere the sign

4、 of k2s is depended on the outer feedback and the sign of kxk2 is depended on the inter feedback.Case (1). Tlie response presents a sinusoidal. It means that tlie system has a pair of pure iinagiiiaiy roots, i.e. the chaiacteristic polynomial is in tlie fonn of 力(s) = s2 + kxk2 Obviously, the outlet

5、 feedback is " - "and tlie iimer feedback is "0：Case (2). Hie response presents a diveied oscillation. Tlie system has a pair of complex coiijugate roots witli positive real parts, i.e. tlie characteristic polynomial is in the fonn of (s) = s? - £s + 上他.Obviously, the outlet feed

6、back isand tlie inner feedback is "一二Case (3). Hie response presents a converged oscillation. It means that tlie system has a pair of complex coiijugate roots witli negative real parts, i.e. the characteristic polynomial is in the fonn of 4(s) = s2 + k2s + Obviously, both the outlet and inner f

7、eedbacks are “ 一二Case (4). In fact tliis is a ramp response of a fnst-order system Hence、the outlet feedback is "CT to prochice a ramp signal and tlie iimer feedback is “ -二Case (5) Consideiiiig that a parabolic fimction is the integral of a ramp fimction, both the outlet and inner feedbacks ar

8、e "(F.P3.3 Consider each of tlie following closed-loop transfer fimction. By consideiiiig the location of tlie poles on tlie complex plane, sketch tlie unit step response, explaining tlie results obtained.20金 2 +12金 + 206s + +lls + 6(d)(s) =12.53 +2s + 5)(s + 5)Solution: (a) (s)= =+12S + 20(s +

9、 2)(s + 10)By inspection, the characteristic roots ai e -2, -10. Tliis is an overdamped second-order system. Tlierefbre, consideruig tliat the closed-loop gain is = , its unit step response can be sketched as shown.(b) (s)= ；=八 +6“ +11S + 6 (s + l)(s + 2)(s + 3)By inspection, the characteristic root

10、s aie 一1 , 一2 , -3 Obviously, all tliree transient components are decayed exponential tenns. Hierefore, its unit step response, witli a closed-loop gain J = 1, is sketched as shown.4(S + l)2 +1Tliis is an underdamped second-order system, because its characteristic roots are -1± j Hence, baiisie

11、nt component is adecayed sinusoid. Noting that tlie closed-loop gain is '% = 2、tlie unit step response can be sketched as shown12.5_12.53 +2s + 5)(s + 5)(s+12 +2?(s + 5)By inspection, the characteristic roots are -1± j2 , 一5 Since(d) 0(s)|- 0.1 «5|, tliere is a pair of dominant poles,

12、-1 ± j2 , for this system. The unit step response, with a closed-loop gain = 0.5、 is sketdied as shown.P3.4 Tlie open-loop transfer fimction of a unity negative feedback system isG(s) =s(s +1)Detenuiiie tlie rise time, peak time, percent overshoot and setting time (using a 5% setting criterion)

13、.Solution: Writing he closed-loop transfer fimctions? +2阿$+此we get = 1, (; = 0.5 . Since this is an underdainped second-order system with $ = 0.5 , the system peiionnaiice can be estimated as followsRising timen 一 arccosq 叫Jl*龙 一 aiccos05« 2.42 sec.15 Solutions# SolutionsPeak time tp = =-= 

14、1; 3.62 sec (Dn Ji* 1 a/1-0.52Percent ovei'shoot ap =严 W xlOO% = e0-51-05" xlOO%«16.3%Setting time t «= 6 sec. (using a 5% setting aiterion)ga)n 0.5x1P35 A second-order system gives a unit step response shown ill Fig. P35 Find the open-loop transfer fimction if the system is a tui

15、it negative-feedback system.Solution: By inspection we have5 = 13"1x100% = 30%p 1Figure P3.5Solving tlie fbnnula for calculating tlie overshoot,=e 陷"r = 0.3 , we have« 0362+ lir 6Since tp = lsec., solving the fonnula for calculating the peak time, tp =-=, we geta)n = 33.7 rad / secHen

16、ce, the open-loop transfer fiinction is塚 1135.7G(s)=s(s + 2阿)s(s + 24.4)P36 A feedback system is shown in Fig. P3.6(a and its unit step response cuive is sliowii in Fig. P36(b). Determine the values of kx, 込，and a.Solution: Tlie transfer fiinction between tlie input and output is given by %)% k2R(s)

17、 s? + +匕The system is stable and we have, fi om the response curve,lim c(t) = lim s )"血= = 2FTCOSTO $+0$+上2 SBy inspection we have=218211x100% = 9%p 2.00Solving the fonnula for calculating the overshoot, ap -enQ '卜' =0.09, we haveU 0.608+ ln2 apSince tp = 0.8 sec ，solving the fonnula fo

18、r calculating the peak time, tp =, we geta)n = 4.95 rad / secThen、compaiiiig the chai acteristic polynomial of tlie system vvitli its standard fbnn, we haves2 + as+ lc2 =s2 +2阿$+ 研k2 =此= 4.952 =24.5a = 2£叫=2 xO.608x4.95 = 6.02P3.7 Aimity negative feedback system has the open-loop transfer fiinc

19、tionG®= J宓(a) Detemiiiie the percent overshoot.(b) For what range of k tlie setting time less than 0.75 s (using a 5% setting criterion).Solution: (a) For tlie closed-loop transfer fiinction we have金 2 + JI云+± S2 + 彳17 Solutionshence, by inspection, we get(t)n=4k rad / sec ,歹=返/2Tlie perce

20、nt overshoot is a)n = 0.818厂ad / sec.xl00%=432%(b) Since = V2/2 < 0.9 , letting33.J Q=< 0.75 sec. (using a 5% setting aiterion)阿 0.5xV2Fresults in、2,i.e. Ar >32(0.75QP38 For the servomechanism system shown ill Fig. P3& detenuiiie tlie values of k and a that satisfy the following closed-

21、loop system design reqiiiiements(a) Maximum of 40% overshoot.(b) Peak time of 4s.Solution: For the closed-loop ti aiisfer fiinction we have皿p+kas + k+2a)ns + a)hence, by inspection, we get0,lcco； = k、2阿=ka, and a = =k (onTakmg considei-ation of ”厂沙x 100 % =40% results in"0.280 hi this case, to

22、satisfy tlie requii ement of peak time, tp/r=4 , we haveHence, the values of k and a are detenuiiied asj2ck = 3； = 0.67 , cl = = 0.68 %P39 Tlie open-loop transfer fiinction of a unity feedback system isG(s) =ks(s + 2)18 SolutionsA step response is specified as: peak time tp = 1.Is, and percent overs

23、lioot ap = 5% (a) Detenuiiie whether botli specifications can be met simultaneously(b) If the specifications caimot be met simultaneously, detenniiie a compromise value for k so tliat the peak time and percent overshoot are relaxed the same percentage.# Solutions# Solutions%s2 +2w+諒# Solutions19 Sol

24、utions(a) Assuming tliat the peak time is satisfied=llsecwe get k = 9.16 Then、we have 歹=0.33 andop =exg W X1OO%=33%>5%Obviously, tliese two specifications caimot be met simultaneously(b) hi order to reduce ap tlie gain must be reduced Choosing tpl = 2tp = 2.2 sec resultsk、= 3.04 , G = 0.57 , dpi

25、= 11.3% >=10%Rechoosing = 2.1tn = 2.31 sec results in DRP处=2.85,= 0.59 , apl = 10.0% < 2.1crp =10.5%Letting tp3 = 2050 = 2.255 sec results ill七 3 =2.941, $3 =0.583, ap3 = 10.5 %« 2.05cr? =10.25%Ill this way、a compromise value is obtained asAr =2.941P3.10 A conti ol system is represented b

26、y tlie transfer fiinctionC(s) 033R(s) (s +2.56)3 + 0.4s+ 0.13)Estimate the peak time, percent overshoot, and setting time (/ = 5%), using tlie dominant pole method, if it is possibleSolution: Rewriting the haiisfer fiinction asC(s) 033R(s) ( + 2.56)( + 0.2)2 +0.3打we get the poles of the system: L2 =

27、 -0.2 ± 丿0.3、 s3 = -2.56 . Then、s】？ can be considered as a pair of dominant poles, because |Re(lt2)| « Re(3)|.Method 1 After reducing to a second-order system, the transfer fimction becomes (Note:R(S) s +04s + 013sto R(s)wliich results ill con = 0.36 rad / sec and 歹=0.55 . Hie specificatio

28、ns can be detenuiiied asxl00%=12.6%20 Solutions# SolutionsMethod 2 Taking consideration of the effect of non-domiiiaiit pole on tlie transient components cause by the dominant poles, we haveIt Z(S Sq)tp = =10.84 sec 一歹2J = i 闻 &広 xl00%=13.6%1-ln=23.6 secP3.11 By means of the algebraic criteria d

29、etenniiie the stability of systems that have the following characteristic equations.(a) s5 + 20s，+ 9s + 20 = 0(b) 3s4+102+5“+s + 2 = 0(c) s+2s4+9s+10s?+s + 2 = 0Solution: (a) s3+ 20“ + 9s+ 20 = 0. All coefficients of the characteristic equation are positive. Using L-C cnterion,20 20= = 160 > 019T

30、liis system is stable.(b) 3s4 +10s3 +5s? +5 + 2 = 0. All coefficients of the characteristic equation are positive. Using LC criterion,1010P3 =352=-153<00101Tliis system is unstable.# Solutions21 Solutions(c) s' +2s4 +9s3 +102 + s + 2 = 0 (Its better to use Routh system.) All coefficients of t

31、lie chai act eristic equation aie positive. Establish tlie Routh array as shown.Tliere are two changes of sign ill die first column, this system is unstable.systems aiecriterion for a higlier-order12410-0.8910020P3.12 Tlie chaiacteristic equations for ceitaiii detenuiiie the number of chai acteristi

32、c roots in the right-half s -plane and the number of pure imaginary roots.(a) s' - 3s十 2 = 0(b) s3+10s2+16s + 160 = 0(c) 2+3s°+12s?+24s?+32s + 48 = 0(d) s' + 2s4 + 3s + 6s2 - 4s - 8 = 0Solution: (a) s'-3s + 2 = 0 Hie Routli array shows that there are two changes of sign in the first

33、 column. So that tliere are two characteristic roots ill the riglit-half plane (b) s3+10s?+16s + 160 = 0 The -row is ail all-zero one and ail aiixiliaiy equation is made based oilgiven below. In10 n w>o3w 2each case,-32s" -rows? +16 = 0Taking derivative witli respect to s yields2s = 0Tlie co

34、efficient of tliis new equation is iiiseited in the s1 row. and tlie Routli array is tlien completed. By inspection, there are no changes of sign in the first column, and tlie system has no characteristic roots in the right-half auxiliaiy are s = ± j4 , tlie system has a pak of pure imagiiiaiy

35、roots.110 => 10 => 21616160 => 160s -plane. Tlie solution of the(c) s' + 3s° +122 + 24s2 +32 + 48 = 0. The Routh array is established as follows The -row is ail all-zero one and ail auxiliaiy equation based on s2 -row iss2 +4 = 0Taking derivative witli respect to s yields 2s = 0The

36、 coefiicient of this new equation is112323 => 124 => 848 => 16s34 => 116 => 44116 => 40 => 204iiiseited in the sl row, and the Routh array is tlien completed. By inspection, there are no changes of sign in the first column, and the system has no characteristic roots in the right

37、-half22 Solutionss -plane. Tlie solution of the aiixiliaiy aie s = ±j2、tlie system has a pair of pure imaginary roots.(d) s5 + 2s4 + 32 + 6s? - 4s - 8 二 0 The Routli array is established as follows. Tlie s3 -row13-42 => 16n3-8 n 40 => 40=>603-850/308is an all-zero one and an auxiliaiy

38、equationbased on -row iss+3s -4 = 0sTaking derivative witli respect to s yieldss'4s+6s = 0sTlie coefficient of tliis new equation is iiiseited ill tlie s3 row, and the Routli array o s is then completed. By inspection, the sign in tlie first column is changed one time, and the system has one roo

39、t in the riglit-half s -plane. Tlie solution of the aiixiliaiy ai e L2 = ±1 3 4 =±j2 , tlie system has one pair of pure iinagiiiaiy roots.P3.13 Tlie chaiacteristic equations for ceitaiii systems are given below. In each case, detenuiiie the value of k so that tlie coirespondiiig system is

40、stable It is assumed that k is positive number.(a) s4+2s3+10s?+2s + t = 0(b) s'+(上 + 05)$2+4后+ 50 = 0Solution: (a) J + Is3 +10s? + 2s + Ar = 0 .Tlie system is stable if and only if>>0V2 2 0D3 =1 10 k0 2 2>0=> Ar <9i.e. tlie system is stable wlien 0 < Ar <9 .(b) s' + (Ar

41、+ 0.5)2 + 4ks + 50 = 0 The system is stable if and only if 上+05 >0,504k>0=>4上？+2上一50 >0 =>4(七 + 38)(上一33)>0«Ar + 0.51i.e. tlie system is stable when >3.3.P3.14 Tlie open-loop transfer fimction of a negative feedback system is given byG(s) =.s(001s+02笄 + 1)Detenuiiie tlie ra

42、nge of K and 歹 in which the closed-loop system is stable.Solution: The chaiacteristic equation is0.01s' +0.2 分2 +s + K = 0Tlie system is stable if and only if23 Solutions> >0, 0.2$ >00.2c KD =>0=>02001疋>0 n k<2Qc-0.01 1Hie requiied range is 20$ > K > 0 .P3.15Tlie open-

43、loop tiTuisfer fimction of negative feedback system is givenG(s)R(s)=K(s +1)s+ l)(2s + l)Tlie parameters K and T may be represented in a plane with K as the horizontal axis and T as tlie vertical axis. Detenuiiie tlie region ill wiiich tlie closed-loop system is stable.Solution: The characteristic e

44、quation is+(T + 2)s2 +(K + l)s+K = 0Since all coefficients are positive, the system is stable if and only ifT+2 K6 =>0=> (:T + 2)(K + l)>0 IT K + l2K-KT+T+2>0K(2- T) + (T-2) + 4 >0 => (T-2)(AT-1)<4 Hie system is stable in the region (T一2)(&-1) <4 , wiiich is plotted as sh

45、own(Letting T = T_2 and £ = results ill TK <4 .)P3.16 Aunity negative feedback system has an open-loop transfer fimctionG(s) =K(Ts + l)(nTs + l)(trTs + l)where 0<n <1, K >0 y T is a positive constant, (a) Detenuiiie the range of K and n so tliat the system is stable, (b) Detenniiie

46、the value of K reqiiiied for stability for n = 1,0.5, 0.1, 0.01, and 0. (c) Discuss tlie stability of the closed-loop system as a fimction of n for a constait Solutions SolutionsSolution: The closed-loop characteristic equation is(Ts + l)(nTs + l)(n2Ts + r)-K = 0i.e. n3T3s3 + (n3 +n2 +n)T2s2 +(n2 +n

47、 + l)7+A?+l = 0(a) Tlie system is stable if and only ifn(n2 +n + l)72n3T3K + l(n2 +n + l)T>0 SolutionsK<一 1 =>K<zr+n + 1 .+1YlK <(n- +n + l + n)=>+n + l1zK<(n + l)n(n2 +n + l)3T3 -(A? + 1)T3 >0=>(n2 +? + !)-n2 -n2A?>0 Solutions23 Solutions0 < AT<0? + l)2+ l)+K = 0

48、1 14k_26hence, the system is stable when(b) Tlie value of K requiied for stability for n = 1, 0.5, 0.1, 0.01, and 0 are calculated as shown.0vK<8 for n = l.0 <<11.5 for n = 0.5 ,0 <r <122.21 for n = 0.1,0< <10202 for « = 0.01,Q < K <s for ? = 0 .(c) For a constant K,

49、the stability of the closed-loop system is related to the value of n , tlie laigei the value of n ,tlie easier the system to be stable. (Stagger piiiiciple.)P3.17 A unity negative feedback system has an open-loop timsfer fimctionG(s) =Detenu iiie tlie range of k requii ed so that tliere are no close

50、d-loop poles to tlie riglit of the line s = 1 Solution: Tlie closed-loop characteristic equation is=> s(s + 3)(s + 6) + 18K = 0 i.e. s3 +9s2 +18S + 18恳=0Lettiiig s = s-l resulting ill(?-l)(?+2)(?+5) + 18AT = 0=?3+6?2 +3+(18110) = 0Using Lienaid-Chipait criterion, all closed-loop poles locate in t

51、he right-half ? -plane, i.e. to the riglit of the line s = -1, if and only if=>£>5/9<6 18K 10,Dr =>0=>28 1.8Q0 => AT <14/9-13/Hie requiied range is 二 < K < 一 , or 0.56 < K <1.5699P3.18 A system has the chai'acteristic equations' +10s? + 29s + Ar = 0Deten

52、uiiie the value of k so tliat tlie real part of complex roots is -2, using tlie algebraic aiterion Solution: Substituting s = F - 2 into the chaiacteristic equation yields (壬一 2)3 +10($ 2乂 +29($ 2)+k = 03s3 + 4s2 + ? + (r - 26) = 0Tlie Routii airay is established as shown.If there is a pair of compl

53、ex roots witli real pait of -2, then'上-26 = 0s°i.e. = 30. In the case of 七=30, we have the solution of the auxiliaiy equation s =±j , i.e. s = 2 土 j .P3.19 Au automatically guided vehicle Detenniiie the value of r requiied for stability. (b) Detenniiie the value of r when one root of t

54、he characteristic equation is s = -5 , and the values of the remaining roots for tlie selected r . (c) Find the response of tlie system to a step command for the r selected in (b).is represented by tlie system ill Fig. P3.19. (a)Fig. P3.19Solution: The closed-loop tiaiisfer fiinction is恥）筒10S? +LOS2

55、 +10ZJ + 10(a) Tlie closed-loop chaiacteristic equation iss3 +102 +10 + 10 = 0Since all coefficients are positive, the system is stable if and only if 10 10Dr =>0=> r >0.11 lOr(b) Substihitiiig s = F - 5 into the characteristic equation yields(壬一5)3+10(彳一5)2 +10r (壬一5) + 10 = 0 s3 -5s2 + (l

56、Or-25)G + (135-50r) = 0hi tlie case of 135-50r = 0 , i.e. r = 2.7 , we have 瓦=0 , i.e.打=一5 Solving tlie characteristic equation witli r = 2.7 , i.e. ?3 -5?3 + 2?+ = 0 results ill 為=456 and ?3 = 0.44 Hence tlie remaining roots are s2 = -0.44 and s3 = -4.56 .(c) Tlie closed-loop transfer iimction tor r = 2.7(s) =10(s + 044)(s + 456)(s + 5)The unit step response of the system isC(s) =10 1 1 1.21 1.21 1.00. 一十(s + 044)(s + 4.56)(s + 5) s s s + 0.44