自动控制原理英文版课后全部_答案

自动控制原理英文版课后全部_答案 Module3 Problem 3.1 (a) When the input variable is the force F. The input variable F and the output variable y are related by the equation obtained by equating the moment on the stick: ydy2lFlkcl,，.2 33dt Taking Laplace transforms, assuming initial conditions to be zero, k4FYcsY,， 33 leading to the transfer function Yk3, Fcks1(4)， ,where the time constant is given by 4c,, k (b) When F = 0 The input variable is x, the displacement of the top point of the upper spring. The input variable x and the output variable y are related by the equation obtained by the moment on the tick: s yydy2kxlklcl().2,,， 333dt Taking Laplace transforms, assuming initial conditions to be zero, 3(24)kXkcsY,， leading to the transfer function Y32, Xcks1(2)， ,where the time constant is given by 2c,, k Problem 3.2 P54 Determine the output of the open-loop system aG(s) = 1，sT to the input r(t) = t Sketch both input and output as functions of time, and determine the steady-state error between the input and output. Compare the result with that given by Fig3.7 . Solution: While the input r(t) = t , use Laplace transforms, 1Input r(s)= 2 s ,, ,,1TTaa ,，Output c(s) = r(s) G(s) = = ,,221s,，(1)Ts,,s，ss,,T the time-domain response becomes ,tTataT,,1 c(t) = ，，e Problem 3.3 3.3 The massless bar shown in Fig.P3.3 has been displaced a distance and is subjected to x0 ,a unit impulse in the direction shown. Find the response of the system for t>0 and sketch the result as a function of time. Confirm the steady-state response using the final-value theorem. Solution: The equation obtained by equating the force: ，kxcxt，,,() 00 Taking Laplace transforms, assuming initial condition to be zero, K+Cs=1 XX00 leading to the transfer function 11X1 == KCKCs，Fs()s，C The time-domain response becomes C1,tK x(t)= eC The steady-state response using the final-value theorem: 111lim()xtlim=s= t,,s,0sKKCs， ，KxxCxtKxXKCs()()()1;，，,,，，,,00 111,,KxKx00？,,,X CCsKK，s，1K Kt,1,Kx0Cxte(),,C According to the final-value theorem: 1,Kxs0 lim()limlim0xtsX,,,,,tss,,,,00CKs，1K Problem 3.4 Solution: 1.If the input is a unit step, then 1Rs(), s 1RsCs()() ,,,,,,,,,s，1 leading to, 1, Cs(),，(1)ss taking the inverse Laplace transform gives, ,t, cte()1,, as the steady-state output is said to have been achieved once it is within 1% of the final value, we can solute ―t‖ like this, ,t,cte()199%1,,,， (the final value is 1) hence, ,t,e,0.01(the time constant =10s) , ,ts,，,4.60546.05 2.the numerical value of the numerator of the transfer function doesn’t affect the answer. See this equation, If CsA() ,,Gs(),，Rss()1 then A Cs(),,(1)，ss giving the time-domain response ,t, ctAe()(1),, as the final value is A, the steady-state output is achieved when, ,t, ctAeA()(1)99%,,,， solute the equation, t=4.605=46.05s , the result make no different from that above, so we said that the numerical value of the numerator of the transfer function doesn’t affect the answer. R(t) C(t) Fig 3.7 t If a<1, as the time increase, the two lines won`t cross. In the steady state the output lags the input by a time by more than the time constant T. The steady error will be negative infinite. R(t) C(t) t If a=1, as the time increase, the two lines will be parallel. It is as same as Fig 3.7. R(t) C(t) t If a>1, as the time increase, the two lines will cross. In the steady state the output lags the input by a time by less than the time constant T. The steady error will be positive infinite. R(t) C(t) t Problem 3.5 Solution: 61296145s，6，R(s)=, Y(s)==,，， 222sssss51，ss,，(51) ,t/5 ？,,，ytte()62929 so the steady-state error is 29(-30). To conform the result: t,5lim()lim(62929);ytt,,，,, ,,,,tt s，6. lim()lim()lim()limytysYs,,,,,tsss,,,,,000ss(51)， eetSESSYSRS,,,,,,lim()lim()lim[()()]ss,,,,00tss 161 ,,,,,,,，lim[()1]()lim(1)()SGSRSS2,,00ss51SSS， ,,30 Therefore, the solution is basically correct. Problem 3.6 ，23yyx，, since input is of constant amplitude and variable frequency , it can be represented as: ,jtX, eA0 as we know ,the output should be a sinusoidal signal with the same frequency of the input ,it can also be represented as: jt,yye, 0 hence jtjtjt,,, 23j,，,yyeeeA000 y10 ,,，32jA0 2wA0 ,,, tan,y20394，, A0Its DC(w?0) value is ,y0,,30 11Requirement ,yy0w,022 311AA00 ?w, ,，2223294，, ,,1,,,,tan1while phase lag of the input: ,4 Problem 3.7 One definition of the bandwidth of a system is the frequency range over which the amplitude of the output signal is greater than 70% of the input signal amplitude when a system is subjected to a harmonic input. Find a relationship between the bandwidth and the time constant of a first-order system. What is the phase angle at the bandwidth frequency, Solution: From the equation 3.41 r0Ar,,0.7 (1) 0,,,,1， ,,,and (2) 1.02so ,,,0, 1.02,Bso the bandwidth ,, from the equation 3.43 ,,,11，,,,,,ctantan1.02,,the phase angle 04 Problem 3.8 3.8 Solution According to generalized transfer function of First-Order Feedback Systems CKGK ,, RKGHKs11,，，， ？the steady state of the output of this system is 2.5V . C2.511？if s0, . From this ,we can get the value of K, that is . ,,,K,R1043 10Since we know that the step input is 10V, taking Laplace transforms,the input is . SThen the output is followed 1103,， Cs()1S,，，1s3 Taking reverse Laplace transforms, ,t 4,,4/3t3,C,,,2.52.5e2.5(1),e From the figure, we can see that when the time reached 3s,the value of output is 86% of the steady state. So we can know ,348,,,,,,,,,23(2)*, 439,3, ,4/3t,10.8642,,,,e, t,3 31The transfer function is 46，s 128，s s,,1.5,2/3Let 12+8s=0, we can get the pole, that is Problem 3.9 Page 55 Solution: The transfer function can be represented, vsvsvs()()()oom Gs(),,,vsvsvs()()()imi While, vs()1o,vssRC()1，m 11,, //R，,,vs()sCsC,,m,vs()，，11,,iRR，，//,,,,sCsC,,，， Leading to the final transfer function, 1 Gs(),213()，，sRCsRC And the reason: the second simple lag compensation network can be regarded as the load of the first one, and according to Load Effect, the load affects the primary relationship; so the transfer function of the combination doesn’t equal the product of the two individual lag transfer function Module4 Problem4.1 4.1The closed-loop transfer function is 10SS，(6)C10,,102R，1 ，，Ss610SS，(6) Comparing with the generalized second-order system,we get W,10n 26EW,n 310 E,10 2WWE11,, ,dn Problem4.3 4.3Considering the spring rise x and the mass rise y. Using Newton’s second law of motion ..dxy(),myKxyc,,，() dt Taking Laplace transforms, assuming zero initial conditions 2mYsKXKYcsXcsY,,，, resulting in the transfer funcition where YcsK，, 2XmscsK，， And c,,mk 2 5c1.26*10, Problem4.4 Solution: The closed-loop transfer function is K1,CKSS，2 ,,2K1RSSK，，21，,SS，2 Comparing the closed-loop transfer function with the generalized form, 2,Cn ,222，，Rss,,,nn it is seen that 2K,,n And that 1 ; 22,,,,,nK The percentage overshoot is therefore ,,,,,1k1,,21,,k,100ePOe,100 Where PO,10% 1.2,KWhen solved, gives (2.86) When K takes the value 1.2, the poles of the system are given by 2ss，，,21.20 Which gives s=-11.36j, sj,,,10.45 Im 0.45 Re -1 -0.45 Problem4.5 4.5 A unity-feedback control system has the forward-path transfer function KG= (s) S(s，10) Find the closed-loop transfer function, and develop expressions for the damping ratio And damped natural frequency in term of K Plot the closed-loop poles on the complex Plane for K = 0,10,25,50,100.For each value of K calculate the corresponding damping ratio and damped natural frequency. What conclusions can you draw from the plot? Solution: GS()K Substitute G(s)= into the feedback formula : Φ(s)=.And in unit ss(10)，1()，HGS feedback system H=1. K Result in: Φ(s)= 2ssK，，10 , So the damped natural frequency =, Kn 105damping ratio == . , 2kk 2sThe characteristic equation is +10S+K=0. ,,,525KWhen K25,s=; , ,,,525iKWhile K>25,s=; ,The value of and corresponding to K are listed as follows. ,n K 0 10 25 50 100 ,，515,，553iSPole 1 0 -5 -5+5i 1 ,,515SPole 2 -10 -5 -5-5i 2 ,,553i ,102 0 5 5 10 n 2.50.5, 1 0.5 , Plot the complex plane for each value of K: We can conclude from the plot. When k25,poles distribute on the real axis. The smaller value of K is, the farther poles is , away from point –5. The larger value of K is, the nearer poles is away from point –5. When k>25,poles distribute away from the real axis. The smaller value of K is, the further (nearer) poles is away from point –5. The larger value of K is, the nearer (farther) poles is away from point –5.And all the poles distribute on a line parallels imaginary axis, intersect real axis on the pole –5. Problem4.6 tvdv1bb iiiivdtC,，，,，，,RLCbRLdtoTaking Laplace transforms, assuming zero initial conditions, reduces this equation to 11,,ICsV,，， 0b,,RLs,, VRLsb, 2ILsRRLCs，，0 Since the input is a constant current i, so 0 1I, 0s then, RL ，，CsV,,b2LsRRLCs，， Applying the final-value theorem yields limlim0ctsCs,, ，，，，ts,,,0 indicating that the steady-state voltage across the capacitor C eventually reaches the zero ,resulting in full error. Problem4.7 4.7 Prove that for an underdamped second-order system subject to a step input, the percentage overshoot above the steady-state output is a function only of the damping ratio . Fig .4.7 Solution The output can be given by 2,nCs(),22sss(2)，，,,,nn (1) s，21,,n,,222ss()(1)，，,,,,,nn ,the damped natural frequency can be defined as d 2,,1,,= (2) nd substituting above results in s，,,,,1nnCs(),,, (3) 2222sss()()，，，，,,,,,,ndnd taking the inverse transform yields ,,,tnectt()1sin(),,，,,d21,, where(4) 21,,tan,, , the maximum output is ,,,tne ctt()1sin(),,，,,pdp21,, (5) ,,,,tp2,1,d,,n so the maximum is 2,,,,,/1cte()1,， p the percentage overshoot is therefore 2,,,,,/1POe,100 Problem4.8 Solution to 4.8: Considering the mass m displaced a distance x from its equilibrium position, the free-body diagram of the mass will be as shown as follows. P a m cdxkx kx Using Newton’s second law of motion, ，，，pkxcxmx,,,2 ，，，mxcxkxp，，,2 Taking Laplace transforms, assuming zero initial conditions, 2 XmscskP(2)，，, results in the transfer function 2 XPmscmskm/(1/)/((/)2/),，， 2 ,，，(2/)(2/)((/)2/)kkmscmskmAs we see 2 XmscskP(2)，，, As P is constant 1 So X . ,2mscsk，，2 c,5When s,,,,，6.25102m 25mscsk，，,210 ，，min 410 X,,0.1max510 This is a second-order transfer function where 2 ,,2/kmn and ,,,cwmckm/2/22n The damped natural frequency is given by 22,,,,,,,12/1/8kmckm dn 2,,2/(/2)kmcm Using the given data, 46,,，，，,,2510/2100.050.2236 n 250,4 ,,,，2.7950106，，，22100.05 2,4 ,,，,，,0.223612.7950100.2236，，d With these data we can draw a picture ,T,,14.050p,d 1 ,,16000,e,, T,,4.673600se, Xkm112/1,d,,,,,222ck2Pmscskkms，，，，22()2,,,,dnd，，ssmm 222kckcc其中,,,,,,,,,,,,,dnn2mmmm428km p,,,tn ？,xetsind,md, p,t,,np，？xett,,，,(sincos)0,ttndpddpp,,,,,md, d？,,,,,tan7.030.02ttxm,dppp,n,, Problem4.10 4.10 solution: The system is similar to the one in the book on PAGE 58 to PAGE 63. The difference is the connection of the spring. So the transfer function is 2,wln, 222，，swsw,,dnn kkkN,paml,2()RJsRCsRkkNk，，，,damp 22,，,，,;JNJJCNcc mlml N,,lmmN,,,Nmll,, KKKpamw, n,NJR cNR,,damping ratio ,KKKJ2pam But the value of J is different, because there is a spring connected. JJs1,JJ,，， m22NN Because of final-value theorem, ,wln, 2,,d Module5 Problem5.4 5(4 The closed-loop transfer function of the system may be written as 10K 2KK10(1)，SS，，610C ,, R210K161010，，，，KSSK1，2SS，，610 The closed-loop poles are the solutions of the characteristic equation ,,,，6364(1010)KSKJ,,,,，3110 2 WK,，10(1)n 210(1)6EK，, 3E, 10(1)，K In order to study the stability of the system, the behavior of the closed-loop poles when the gain K increases from zero to infinte will be observed. So when 30SJ,,,321K,2 E,110 3110SJ,,,3101K,10 E,2110 210SJ,,,3201K,20 E, 370 双击下面可以看到原图 21J Re -3 Problem5.5 ,21J Solution The closed-loop transfer function is K 2CKKs ,,,22KRsKassaKsK，，，，(1)1(1)，,，as2s Comparing the closed-loop transfer function with the generalized form, 2,Cn ,222，，Rss,,,nn Leading to ,,Kn aK,,2 The percentage overshoot is therefore ,,,21,, POe,,10040% Producing the result (0.28) ,,0.869 And the peak time , Ts,,4P21,,,n Leading to ,,1.586(0.82) n Problem5.7 5.7 Prove that the rise time T of a second-order system with a unit step input is given by r ,,,1,2d11-1 -1 T= tan= tanr ,,n,d,d,d Plot the rise against the damping ratio. Solution: ,,,,tnAccording to (4.33):c(t)=1-. 4.33 ett(cossin)，,,dd21,, When t=,c(t)=1.substitue c(t)= 1 into (4.33) Tr Producing the result ,,,12,d11-1 -1 = tan= tanTr,,n,d,d, Plot the rise time against the damping ratio: Problem5.9 Solution to 5.9: As we know that the system is the open-loop transfer function of a unity-feedback control system. GHSGS,So ，，，， Given as 4KGHs, ，，ss,，25，，，， The close-loop transfer function of the system may be written as Gs，，CK4 s,,，，RGHsssK1254，,，，，，，，，， The characteristic equation is 2 ssKssK,，，,,，，,,254034100，，，， According to the Routh’s method, the Routh’s array must be formed as follow 2sK1410, s30 0sK410, For there is no closed-loop poles to the right of the imaginary axis 41002.5KK,,,, Given that ,,0.5 ,,,410Kn 3,,,K4.75,2410K, When K=0, the root are s=+2,-5 According to the characteristic equation, the solutions are 349sK,,,,4 24 K,3.0625while , we have one or two solutions, all are integral number. Or we will have solutions with imaginary number. So we can draw K=10 K=2.5 2 Open-loop poles K=3.0625 -5 K=0 Closed-loop poles K=10 K=0 Problem5.10 5.10 solution: ,,0.6 wrads,2/n according to ,,wtne1,,，,,cwt1sin() ，，d22,1, 4,1.2t,,tanet,，,sin(1.6)0.4, 3 finally, t is delay time: ts,1.23(0.67) Module6 Problem 6.3 First we assume the disturbance D to be zero: ？ eRC,, 101 CKe,,,,ss，1 ess(1)，Hence: ,RKss10(1)，， Then we set the input R to be zero: 10e10 CKeDe,,，,,,(),,,ss(1)，DKss10(1)，， Adding these two results together: ss(1)10， eRD,,,,10(1)10(1)KssKss，，，， ？ 11Ds(),Rs(), ; 2ss ？ ss，,1109 e,,,22210(1)10(1)100(1)KsssKssssss，，，，，， the steady-state error: 2sss,,99ese,,,,,,limlimlim0.09 ss322,,,sss000sssss，，，，100100 Problem 6.4 Determine the disturbance rejection ratio(DRR) for the system shown in Fig P.6.4 Td + —2+ —0.2 0.10.05s，— __ — + fig.P.6.4 solution: from the diagram we can know : Km,0.2 R K,1 v c,0.05 so we can get that KK()0.21,，,mmOLv,,，,，,115DRR ()0.05,cR,nCL 21,, so c=0.025, DRR=9 0.10.050.050.025ss，， Problem 6.5 6.5 Solution For the purposes of determining the steady-state error of the system, we should get to know the effect of the input and the disturbance along when the other will be assumed to be zero. First to simplify the block diagram to the following patter: Td 1,, 20 2doJs s1， 10 Allowing the transfer function from the input to the output position to be written as ,2001 ,2Jss，，220,d 2020240 ,,,,,*01d222JssJsssJsss，，，，，，220220(220) According to the equation E=R-C: 2202(2)Js，,, esss()lim[()()]lim[(1)]lim0.2,,,,,,ssrd022,,,,,,sss000sJssJss，，，，220220 问题; 1. 系统型为2，对于阶跃输入，稳态误差为0. 2. 终值定理写的不对。 and the transfer function from the input to the output position to be written as ,102 ,2TJss，，220d 11 ,,,T02d22JsssJss，，，，220(220) According to the equation e= -c 11,ess, ()lim[()]lim[]lim0.05,,,,,,,ssfd22,,,,,,000sssJssJss，，，，220220 eee,，,,,0.20.050.15so the state error should be: ssssrssf Problem 6.9 Solution: The transfer function of the inside loop is 10 10s(s，1)Q,,2 10kss，(10k，1)s1， s(s，1) And the transfer function of the whole system is C10, 2Rs，(10k，1)s，10 wWe get the value of and , from the above equation : n W,10 n 101k， ,,210 And the value of PO is 2,,,1,, PO,100e According to the former equation: 10k，1PO,10%,,0.6,, 210 The final result is 1.210,1k,,0.28 10 Module7 Problem 7.5 Determine the closed-loop transfer function and the percentage overshoot for a system described by the pole-zero map shown in Fig.P7.5, assuming the steady state gain of the closed-loop transfer function is unity. 1 -3 -2 -1 SOLUTION: From the figure, we can get the Bode form function Cs1，, ,22Rss，，21,,,nn 122,,,,,,,2,,,，,215 ，，nn3 13, ,,？,,,0.8940.9,2,,n From Fig.7.6 on page 117, we can see that, under the effect of closed-loop zero, while 3 and , the Percent overshoot is less than 1%. ,,,,0.92 C1,22Rsss(1)(21)，，，,,,,nn ,,,,,1/31/3,, ,,,,,,2,52.242/50.894,,,,,nn C15,,2Rsss(3)(45)，，， Problem 7.7 7.7 A system has a transfer function that may be written in the form C1 ,2Rssasb(1)()，，， It is known that for the second-order term, =0.2. Investigation of the unit step response , reveals an overshoot of 5% of the input. Calculate the constants a and b, plot the closed-loop system poles on the complex plane, and comment on the reducability of the system. Solution: The third-order system with =0.2 would have an overshoot of 5% if the additional pole , were located such that the parameter . ,,2.25 1,1And from the known parameters,, , then 11,,,,2.25, (看图似不超过2.1) 0.2,,,nn The time-domain response parameters may be calculate as ,,4 rad/s and =0.2 ,n So 22,, a==1.6 and b=,=16 nn then we will plot closed-loop system poles on the complex plane: Im 3.92 -1 Re -0.8 The other real pole can not be neglected, the effect of the real-axis pole contribution to the response will be to make it more sluggish. Problem 7.9 Solution: Examination of the plant reveals the following time-domain parameters: 1,,,,,radsb; 2/1n, Examining Fig7.6, the system with would have an overshoot of 5% if the zero is ,,1 located such that the parameter . This related to the parameter b by ,,0.6 12,,,,0.6, b,,n b,1.2 If the input is a unit step, the steady-state output may be obtained by using the final-value theorem s，1.2lim()lim()lim0.3ctCs,,,s 2tss,,,,00s，，44s As there is a 5% overshoot, the maximum value of the system is (15%)lim()0.3105%0.315，，,，,ct t,, Problem 7.11 ks(1)， Gks(1)，sA ; s=-1; ,()s,,,ks(1)，1(1)，，，GHksk1，s GkB ; no zeros; ,()s,,11(1)，，，GHks (1)(2)ss，， Gsss(1)(2)(4)，，，ss(3)，C; s=-1,-2,-4 ,()s,,,(1)(2)(5)sss，，，1(3)(4)(1)(2)(5)，，，，，，，GHssssss1，sss(3)(4)，， s 2Gsss(210)，，(1)s，D ; s=0,-1+3i; ,()s,,,2s1(1)(210)，，，，，GHssss1，2(1)(210)sss，，， Module8 Problem 8.3 'CGG,,8.3(1) RGHG11'，， Resulting in GG'' G, 书上有结果(8.62)G,1(1)，，H1(1)，,GH 101G,H,Take , into 2ss，，25s，1 We get that 1010s，'G, 32sss，,，335 (2)According to Table8.1 ''KGs,,lim()2 s,0 It’s a type 0 system , the steady-state error due to a unit step input 111et,,,() ss'，，KK113p (3)For a unit step input 1Rs(), s According to 8.59 1(1)12*(11)1，,，,GH et()lim,,,ss,0s112*13，，GH Problem 8.3 Solution 110G(s) = , H(s ) = 21，sss，，25 Hence,the closed-loop transfer function is 10 2Gs()10(1)s，ss，，25 = ,2101(1)(25)10sss，，，，1()()，GsHs1，2sss，，，251 ,GSuppose the unity-feedback system has the open-loop transfer function (s),and its closed-loop transfer function is ,Gs()10(1)s， = 2,1()，Gs(1)(25)10sss，，，， Hence 10(1)s，,G(s) = 32sss，,，335 For a unit step input 1 R(s) = s The true steady-state error is sRs()11 分母上应该是1，GH，不论是C/R ,,,,esEslim()limlimss3,,,000sss，，1()1()GsGs E/R，分母都是闭环特征式,1，开环传递函数 Problem8.4 Solution: Take Laplace transform, the ramp input of slope 4 is, 4Rs(), 2s For type 1 system, mKz1151K,i,1vi,,，,，,，()lim5Gs q,0s，1sssssp,k,1k The velocity error coefficient is, K,5 v When H(s)=1, The corresponding steady-state error is, 24/s ,,ets()lim0.8ss,s0，1/Ksv (a) H(s)=4; GG5 G',,,21(1)1315，,，，，GHGss The system is of type 0, as show in Table 8.1 The corresponding steady-state error is, e,,ss (b) H(s)=s G5 G',,21(1)65，,，,GHss The system is of type 0, as show in Table 8.1 The corresponding steady-state error is, e,,ss 1(c) H(s)= 1，s Gs5(1)， G',,21(1)(24)，,，,GHsss The system is of type 1, 5K,, v4 The corresponding steady-state error is, 24/16s,,,,, ets()lim3.2ss,s0，1/5Ksv s，1(d) H(s)= s，2 Gs5(2)， G',,1(1)(1)(2)5，,，，,GHsss The system is of type 0, as show in Table 8.1 The corresponding steady-state error is, e,, ss Problem 8.5 ,,sRsp()ss(2)，2p,n, limlimsThe steady state error: ess(t)==; 222s,,s00,,,，，，1()2Gssss,nnn 2,nThe transfer function: ; ()s,,22ss，，2,,,nn 2pp,nIf R(s)=, then; (),Cs2222s(2)，，sss,,,nn The inverse transform gives the time response: ,5,t2npe21,,2p,(2cossin)tt，C(t)=pt-+; ,,,nn2,,1,n,n 2p,When the system reach the steady state, C(t)=pt-; ,n cThen the time for the input to reach the value C can be shown by , t,1p c2,And the time for the output to reach the value C can be shown by so the ,，t,2p,n 2,time is that:. ,,,tt,21,n Problem8.6 1 2KKJsaa, ,,G=321JsKs，sv，,Ks1v2Js KKKKpapa, GKG,,,,p322()JsKssJsK，，vv It is a type 2 system R,,,lim eKsssp,0s1，G 1Rs(),i.) Unit step input s Kp ,,lim0ess,0sKa1，32，JsKsv 1ii.) Ramp step input Rs(), 2s Kp,,lim0 ess,0sKa，s2，JsKsv iii.) Acceleration input KKpv,,lime sss,0KKK2apas，，JsKv Problem 8.8 8.8 Solution: The first question (1) When a unit step change in input R(assuming T constant): since disturbance is constant, the effect of it tends to be steady. So at this time, we only need to calculate the effect of the change in R. The system has unity feedback, and the open-loop transfer function is given by K G(s)=, and the system is of type 1. Jss(1)， Since the input is a unit step and the system type is 1,so =0 essr (2) When a unit step change in input T(assuming R constant): since the input is constant, the effect of it tends to be steady. So at this time, we only need to calculate the effect of the change in T. escscs()()(),,fdff Because the expected output of the disturbance is zero, so escs()(),,ff Cs，1The transfer function ,()s,,,,efTKJss，，(1) The effect of the disturbance ，，s，1K,1 ,etssTs()lim()()lim1,,,,,,,,,ssfef,,,,ss,,00KJss，，(1)，， So the steady-state error eee,，,,1ssssfssr The second question: ，，s，11As the conclusion we get above, eee,，,,,,limssssfssr,,,,0sKJssK，，(1)，， The resulting change in speed is less than 10% of the disturbance value, so 1,,,0.110K K Problem 8.10 Solution: The transfer function between error and input : e1, R1，GGH12 And the transfer function between error and the disturbance : GHe2,, T1，GGH12 When the R and T are unit step functions 1,GH2e,lim sss,01，GGH12 When the order of is lower than that of , the error is zero. GHGGH212 KszKszKsz()()()，，，,,,123ikmikmGGH,,,,,12qqq312sspsspssp()()()，，，,,,jlnjln KKszszKKzz()()，，,,2323kmkmkmkm,,11,,qq，qq，2323sspsps()()，，pp,,lnlnln,ln,E,,KKKszszsz()()()，，，KKKzzz,,123ikm123ikmikm,,ikm,,1，1，qq，qq，q，q，231231sspspsp()()()，，，sppp,,jlnjlnjln,,jln,, qqqq，，qqq，2323111spppsKKzzpspsppzzKK,,(),,,,,jlnkmjjlnkm2323jlnkmjjlnkm,,,,,,,,qq，q，231spppKKKzzzKKKzzz，,,,jlnikmikm123123jlnikmikm,,,,,, q1sp,j1j,,,,KzGslim(),11is,0 i q1 is the type(not order) of G1, when q1>0, the steady state error is zero. Please note the difference between type and order Discussion: eGHGGHWhen the order of is equal to the one of , will depend on the specific ss212 eGGHcondition. And when the order of is higher than the order of , will have no GHss122 limit. Module9 Problem 9.1 Solution Im Re Problem 9.2 Solution: In this subject, there’re two open-loop poles at s=0, -4 and no zeros. Inspecting the real axis, it smay be seen that the argument equation is satisfied only for -4 < < 0. Figure SP9.21 t shows the locus so far. Im Fig SP9.21Re Since there are two poles and no zeros, there must be two branches of the locus going to ,40 sinfinity. A trial point is indicated in Fig SP9.22 and the argument of each component of t the open-loop transfer function is shown. In this subject, the argument equation becomes K GH(s)= s(s+4) K 0GH(s)==-180，， s(s+4)st Im 0=Ks(s+4)=-180， , ， , ， 0 =-s(s+4)2=-180， , ， ,, , ，(s+4)，s Fig SP9.22 Re 0,40-s(s+4)2=-180， , ， ,, , 0Figure SP9.23 shows in which directions the two branches of the locus go to infinity. ？,90, Im Re Fig SP9.23 ,40Promblem 9.3 Solution In Fig.P9.3, we can see that the open-loop poles are 0, ,3,,4 So the open- loop transfer function is K G(s)H(s) = 错了 sss(3)(4),, To calculate the gain, the magnitude equation can be used K = sss,,34 When s = -22j, K 96.4 17.89 ,, When s = - 2.5, K 89.4 1.875 , Promblem 9.4 Solution: The open-loop transfer function is K GHs(),sss(5)(6)，， Draw the root locus. ,,We should have ,,,，，,,，,n360180123 Imagining s to be on different portions of the real axis, only when s,,6 or ,,,50s, ttt the argument equation will be satisfying. In order to determine the direction that the branches go to infinity, a circle of infinity radius centered at the origin is considered. Only when the roots go to infinity in a ,60direction at to the real axis, the phase angle equation will be satisfied, that is Consider K as a positive number: ,,,2(a) When the real part Substituting gives sbj,,，2 22 (2)(3)(4)245(2),，，，,,,，,,bjbjbjbbbj Thus 22 ,,,,,,2450;(2)0bbb b=0 Then the character equation will be 22 (2)(3)(4)245(2)24,，，，,,,，,,,,bjbjbjbbbj So the point is s=-2 The gain at this location is K,24 ,,,1(b) When the imaginary part ,,1When Substituting gives saj,， 322 ()(5)(6)112711(32229)ajajajaaaaaj，，，，，,，，,，，， Thus 322 aaaaa，，,,，，,1127110;322290 a=-1.72 ,,,1When , see the picture, the answer will be the same as that above. So the point is sj,,,1.72 The gain at this location is KK,,,,1 32aaa，，,11271129.99 K,30 Problems 9.5 9.5 A unity-feedback position control system shown in Fig.P9.5 has an actuator and load represented by the transfer function K Gs(),ss(2)， Write down the characteristic equation of the system, solve it, and plot the locus of its roots as the gain K varies from zero to infinity. Compare the result with that obtained from the root locus approach utilizing the open-loop transfer function of the system. Solution: 2The characteristic equation is:. ssK，，,20 When , the roots are on the real axis. K,1 2. Kss,,,2 When K>1,the roots have value 1 for the real part, and they are conjugate. . sKjsKj,,，,,,,,11,1112 We can plot the root locus: Use the locus approach. The open-loop poles are 0 and –2.The asymptotes intersect the pz,,,ii,2,,,,,1real axis at . The angle between the 2 asymptotes is 180 degree. anm,2 Value of K gets larger from poles to middle of them at point –1, and then goes up and down. (Shown in the picture above.)Both method get the same result and the same picture. Problem 9.6 (c)根轨迹是圆形 ksa()，2,,,,,，,，,1,0,0()0abskbsakssb(), 22,xykbxak,，,，,()0 sxyi,，,,2()0xykby，,,, 222()xayaba，，,， (d)中复极点处根轨迹的出射角是负值; 出射角,(各零点指向本极点的方向角),(其他极点指向本极点的方向角)，反向,, 吴麒《自动控制原理》p249 Problem 9.8 solution We begin by drawing the complex plane and marking in the open-loop (by a cross)and the open loop zeros(by a circle).In this problem there are three coincident open-loop poles at s=0 and double zeros at s=-0.5. Inspecting the real axis ,it may be seen that the argument equation ,,,,S0is satisfied only for . t And the imaginary-axis crossing points can be get through following steps 2ks(0.5)，10，,The close-loop characteristic equation 3s dk3s,0,,,s1.5Then k,,2(0.5)s，ds And we can get the value of the imaginary-axis crossing point by following steps 2ks(0.5)，3210(0.5)0，,,，，,sks 3s 2w,,0.5kw,,0.5Let s=jw, then and So the final locus is following Figure A (-0.5,) 0.53 0.5 0(threee) -0.5(two) -1.5 Problem 9.9 a. The characteristic equation is: s(s+3)(s+4)+K=0 32sssK，，，,7120 Substituting yields the following equation: sj,, 2K,,70, 3120,,,,And 2K,84,,12Which will be solved to yield and . Yields the crossing points at . sj,,23 b. To check the previous result graphically, we will plot the root locus that have been shown in the Fig.1. Fig.1 c. The characteristic equation is 32sssK，，，,7120 Yields the Routh array 3s112 2sK7 K1s12,70sK Since we have to determine the maximum value of K consistent with stability, according to the Routh’s criterion, we will yield: KK,0120,, and . 7 084,,KSolving them yields: . K,84When , the system will be critically stable. Thereby, we can confirm the previous results. Module10 Problem 10.2 Plot the root locus for the feedback control system specified by the open-loop transfer function 2Kss(413)，， GHs(),2(0.5)(22)sss，，， Solution : 2Root #1: Here we have two complex zeros located at the root of , which are (413)ss，， s,,0.5. There are three open-loop poles, and . sj,,,23sj,,,1 s,,0.5Rule #2: For there is only one pole on the real axis at , the real-axis segment ,0.5forming part of the locus is the part on the left of . ,,360360,Rule #3: From the formula: , we can get the angle between ,,,,360,,nm32 adjacent asymptotes. We can see that there is only one asymptote. ,,,pz,,，0.524iiRule #4: The asymptote intersect the real axis at , ,,,1.5anm,,32 Rule #5: Applying this rule to the zero at and measuring the angles from the sj,,，23 ,formula:. This is the angle of entry into the ,,,，，，,,180(116.6116.6104)9067.2 zero from complex poles. And the angle of emergence from the pole sj,,，23 , is . sj,,，1,,,，，，,,180(1166490)7614 Rule #6: The characteristic equation for this problem becomes 32 sKsKsK，，，，，，,(2.5)(34)(113)0 Substituting sj,, yields the following equations: 2,，，,,340K 2 and ,，，，,(2.5)1130KK, 2 the solution is and Kj,,1.274,,，35.10j so we can conclude that the locus has no intersection with imaginary axis. Rule #7: Normally, there should be a real axis breakaway point on the left of the open pole at s,,2. From the formula: 22Kss(413)，，(2)(22)sss，，，, we can get : 10，,K,,22(2)(22)sss，，，(413)ss，， But at the negative real axis, we cannot calculate the maximum number of K, so there is no breakaway point on the real axis. This is the root locus drawing by matlab, and the parameters we got from matlab accord well with the numbers we counted. Problem 10.5,NO.4 Finished Solution: The open-loop transfer function of the system is : KGH,2sss，，，146.25，，，， There are three poles at s=-1 ,s= -2+1.5j ,s=-2-1.5j. Using Rules #1 and #2 ,it may be seen that the negative real-axis segment for s<-1from part of the locus. Since there are three more poles than zeros, Rule #3 indicates there are three branches of the locus going to infinity and the negative real axis is one of them. The point where the asymptotes intersect the real axis is given by Rule #4 and may be calculated from ,,,2215,,,,a303, 。360。,603And the angle between adjacent asymptotes is The locus up to this point is shown in the Fig. The angle of emergence from the complex pole located at s= -2+1.5j may be evaluated by using Rule #5 as follows: ,,,，,,180(90123.7)33.7k The final plot may be sketched as follows: Using Rule #10, it may be seen that there are three more open-loop poles the open-loop zeros, so if all the closed-loop poles have the same real part, the real part a can be calculate as follows: -2+(-2)+(-1)=3a; 5a,,3Leading to ; The required value K can be calculated as: 255552522Ksss,，，，,,，,，，,(1)(4)(1)[()4*)1.5743334 These are the homework of Linear Control Systems Engineering，module . K的计算结果是正确的，但是应该根据规则9计算。按照题中的解法，K的值可能会为 负。此外，式中应该是4*(-5/3)。 Problem 10.5,NO.5 Finished 10.5 Shown in Fig. P10.5 is an open-loop pole-zero map of a feedback control system in which it is necessary to ensure that all closed-loop poles have the same real part. Determine the required value of gain and the real part of all the closed-loop poles. Solution: X is the same real part of closed-loop ploes. According to Rule #10: XXX，，,,,,122 Result in X=-5/3. Write down the characteristic equation: 32sssk，，，，,510.256.250 ssjsj,,,,，,,,5/3,5/3,5/3.,,Roots are: 123 Because ssssss，，,10.25122331 23Solve the equation.. ,,12 Because sssK,,,6.25.123 155K==. sss,6.2512327 155So the required value of gain is ,real part of all the closed-loop poles is –5/3. 27 K的求解过程是正确的，但是结果不正确。最后一步的解答有问题，应该是，,,sss6.25123 正确结果是1.5741。还可以参考规则9。 Problem 10.10 Solution: (1)The block may now be simplified to 10K GHs,，，2ssss(8)(24)，，， s,,8(2)For this system there are four poles at and one at the origin and two complex at . sj,,,13 (3)The point where the asymptotes intersect the real axis is given by Rule #4 and may be calculated from 081313,,，,,jj, ,,,2.5a40, The angle between adjacent asymptotes and real axis is ,,(21)(21)kk，，,, ,anm,4 ,,,,35kkkk,,,,,,,,,,,0,;1,;1,;1,,,,,. aaaa4444 (4)The characteristic equation of the system closed-loop is 2fsssssk()(8)(24)100,，，，，, 432fsssssk()102032100,，，，，, 1432 kssss,,，，，(102032)10 dk132,,，，，,(4304032)0sss ds10 3221520160sss，，，, Since the system characteristic equation is fourth order ,any stationary values of K will be determined graphically. sk,,,5,28.5 sk,,,6,33.6 sk,,,7,27.3When sk,,6.5,32.42 sk,,6.1,33.6226 sk,,6.05,33.6257 s,,6.05Determine ,k=33.6257. The angle between the directions of emergence of q coincident poles on the real axis is given by (21)(21)kk，，, ,,,dl2 ,, ,d2 (5)The point where the locus crosses the imaginary axis may be obtained by substituting into the characteristic equation and solving for . sj,,, 432()10()20()32()100jjjjk,,,,，，，，, 433(2010)(3210)0,，，,,kj,,,, 42,,,，,20100k ,332100,,,, 4,,0 or ,,, 5 ,,0,k=0; 4,,,,k=5.376. 5 -8 -6.05 -2.5 应该计算两点的出射角。 sj,,,13 ,,既然已经计算出，为什么在图中不是这样表示。 ,d2 Problem 10.10 (1) First of all, the inner loop may be represented by the closed-loop transfer function: C2()s, 2Bss，，24 Then, the open-loop transfer function of the complete system may be written as 7K,10K GHs(),2sss(8，，，)(24)s 2ss，，24Here we have two complex poles located at the roots of , which are ,and two poles on the real axis. sj,,12,3,,13,j The real-axis segment forming part of the locus in only between the poles at the origin and s,,8. ,90There are four asymptotes inclined at to each other. They obey the symmetry rule. ,The asymptotes intersect the real axis at , a pz,,,ii ,,,,2.5where anm, The angle of emergence from complex poles is given by , Qpppz,,，,,，,180()(),,plljlj,jlj ,,,,,,,,1809012014 ,,,46 2The characteristic equation is ssssK(8)(24)70，，，，, Substituting yields the following equations: sj,, 42 ,,,，,2070K 3and 32100,,,, K,7.68,,1.79Solving the equations yields: and . 1432 Kssss,,，，，(102032)7 dK132 ,,，，，(4304032)sssds7 The point at which the locus leaves the real axis is given by Rule 7,which yields the result s,,6.1-6.07 for the break point. Finally, we can plot the root locus shown in Fig.1 Fig.1 (2) K,20There are two complex closed-loop poles and two real closed-loop poles for . s,0s,,6.1The root to the left of the pole at and the right of the point at is best determined. We will use a trial point . st s,,3.4This root is found to be at approximately . s,,6.1s,,8Next, the root to the left of the point at and the right of the pole at is best determined. We will also use a trial point . st s,,7.6This root is found to be approximately . Rule 10 may now be used to find the remaining root. ,,,,,,3.47.610,, ,,0.5, sj,,0.52.26 s,,3.4s,,7.6and the points at and . Module11 Problem 11.1 Using the methed outlined in the previous section for multiloop systems, the inner loop will be closed first using the root locus methed, and these closeloop poles will become the open-loop poles of the outer loop. k2()Gs, when =10 kin21s， the inner closed-loop pole is s=-11 2kk12 wherek=10 Gs(),1out(2)(11)ss，， the outer closed-loop poles are sj,,,6.5719 Problem 11.2 1.GROUP 4 finished Solution: The open-loop transfer function of the system is : 0.01 125，s10.5Gs,，，, ()2500K0.237sssK，，(10.50.237)Vv，1，s10.5 The closed-loop transfer function of the system is: 2,,20202.5，n0,,222sKsss，，，，，(20.474)2.52,,,,ivnn It is seen that ,,,2.51.58n And the damping ratio is 20.474，Kv,,,，0.630.15Kv,2n 01,,KvFor The range of the damping ratio of the closed-loop poles available is 0.630.78,,, These are the homework of Linear Control Systems Engineering，module . 2的求解有问题，应该是3.5。 ,n 2. GROUP 6 finished ,i,0.0101 2500 1 s1，s 2 K 23.7 v 0.07 Solution 1: The inner-loop will be closed and gives the transfer function 0.01 R/C= 1sK，，10.237v2 Thus the outer-loop’s open-loop transfer function is represented as 2500*0.01*0.07 GHs,()1ssK，，(10.237)v2 It’s easy to obtain: ,500 ,2sKs，，，(20.474)3.5,iv ,,3.5n 20.4742，,K,,vn 10.237，Kv ,, 3.5 ？01,,K ,？,,0.5350.661 Solution 2: If we have obtained the open-loop transfer function of the outer loop 2500*0.01*0.073.5 GHs,,()1ssK，，(20.474)vssK，，(10.237)v2 We will use the root locus method: Im Kv=0 3.5 Kv=1 - (1+0.237Kv) 0 - (2+0.474Kv) Re ，, We can calculate that Kv=3.674 at the midpoint between two open-loop poles, so the close-loop poles must be on the vertical line when 0 < Kv < 1. 10.237，Kv,,Then we can calculate geometrically ,,cos 3.5 Which results in 0.535 < , < 0.661 confirming the previous answer. Problem 11.8 As we can see from the diagram block 2121sKs，，，，，，K Gs,，,，，0ssssss，，，，1.521.52，，，，，， Rule #1. Draw the complex plane and mark the m open-loop poles and n zeros. The locus starts at a pole for K=0 and finishes at a zero or infinity when K=. The number of segments going to , infinity is therefore nm, s,,1s,,1.5s,0s,,2Here we have one zero at ,three poles at and Rule #2. Segments of the real axis to the left of an odd number of poles or zeros are segments of the root locus, remembering that complex poles or zeros have no effect. Rule #3. The loci are symmetrical about the real axis since complex roots are always in conjugate pairs. o360The angle between adjacent asymptotes is and to obey the symmetry rule,the negative nm, real axis is one asymptote when is odd. nm, nm,,2As the number of the asymptotes is according to , there will be two asymptotes o,,901o180 to each other. The angle of the asymptotes inclined at o,270,2 Rule #4. pThe asymptotes intersect the real axis at, where is the sum of the real part of the ,,ia zopen-loop poles (including complex roots) and is the sum of the real part of the ,i open-loop zeros(also including complex roots). Substituting into the above equation yields pz,21.501,,,，,,ii, 1.25,,,,a2nm, For we do not have complex poles ,so we can ignore the Rule #5 Rule #6 The point where the locus crosses the imagingary axis maybe obtained by substituting sj,,, into the characteristic equation and solving for . The characteristic equation for this problem becomes 32ssKsK，，，，,3.53220 ，， sj,,Substituting yeilds 23,，，,，，,3.62320,,,,KKj ，，，， we can not gain a real root of ,so we can draw a conclusion that the root locus will not , attach the imaginary axis. Rule #7.The point at which the locus leaves a real-axis segment is found by determining a local maximum value of K, while the point at which the locus enters a real-axis segment is found by determining a local minimum value of K. K,,1.72527We try to figure out suppose we have a impulse input, the output becomes 21Ks，，，ABC Cs,,，，，，ssssss，，，，，，，，1.521.52 24AKBKCK,,,,2then we figure out that 33 which yields the inverse transform 2401.52,,,，,ctKeKeKe2 ，，33 because we know in the question that a. Not more than 15% overshoot ,,,21,,We have the overshoot percentage that e，100% oso we can get that that ,,,acos0.51758.87,,0.517 b. Dominant time constant less than 2s So we know that ,,,,,2n c. Damped natural frequency of oscillation less than 2 rad/s We can get that ,,2d sands,,,,1.31.27So we can estimate the range of K is that while 21Ks，，，K,,,0.303,0.395We know that,so ,,1,,sss，，1.52，，，， 对绘制根轨迹掌握的比较好。 应用规则7求解根轨迹与实轴的交点，计算过程有些模糊。 事实上，求解该题目应该将根轨迹图和闭环传递函数结合起来计算。解题思路是不正确 的。K的取值是在大于0范围内，怎么可能解出负值。而且―Damped natural frequency of oscillation less than 2 rad/s‖的意思是。 ,,2d Module12 Problem 12.6 2 1/221/2M=5/[w(1+w/4)(1+w/64)] o-1-1Φ=-90-tg(0.5w)- tg(0.125w) o W?0: M??, Φ?-90 oW??: M?0, Φ?-270 o o-1-1(1) -180 =-90 –tg(w/2)-tg(w/8) o-1-1 90 =tg(w/2)+tg(w/8) Taking tangents of both sides of this equation yields,then ?=tg[th(w/2)+tg(w/8)] 2 ?=tg(A+B)=(tgA+tgB)/(1-tgAtgB)=(w/2+w/8)/(1-w/16) And therefore w=4 Substituting 21/221/2M=5/[4 (1+4/4)(1+4/64)]=0.5 w=4 w=? -0.5 w=0 2 1/221/2(2) M=5/[w(1+w/4)(1+w/64)]=1 22 1/221/2w (1+w/4)(1+w/64)=25 2446w +w/64+w/4+w/256=6400 642w+68w+256w=6400 222w =7.66 (omitted w=-62.23 and w=-13.43) w=2.77 23.3184=10w/(16- w) tgΦ=3.3184 o o oΦ=-90-73.23= -163.23 Problem 12.9 a, d, e, c, f, b Module13 Problem 13.2 D. Solution: Im [S] G H 1 A Re b F(+) -1 a 1 B -2 F(-) D C E -1 The figure above shows the two poles of the equation. a,5,0,02M,a1b,1010A: ,arctg,1211c,1,,,arctgarctg,a123,arctg2,13 0.1-0.1j a,2,,010M,b,3bB: 6,0,1c,1,0,b,0,2 1/6 ,GHs() ss，，1)(2)( ,，(1)(，,2)0ss ,,,,12sor a,5,,002M,c1b,1010C: 0.1，0.1j ,,arctg,1211c,1,,，arctgarctgc123,,arctg2,3a,22,,00M,db,5o10D: 0.1+0.3j ,,45,11c,1o,,，45arctg1darctg,,2,22 a,2,,002M,ob,1e,,135E: -0.5-0.5j ,21c,1oo,,225,,90,e2 a,1,,00M,,b,0f,F(-): ,0,1oc,1,90,fo,,,90,2 a,1,,00M,,b,0f，F(+): ,0,1,c,1,90,,fo，,90,2 a,2,,002M,ob,1g,135G: -0.5+0.5j ,21oc,1,,,225,,90g,2 a,210,,00M,hb,510oH: 0.1-0.3j ,45,11c,1o,,,45arctg,1harctg,2,22 F(+) Im [F] G D C Re B A H E F(-) Problem 13.3 13.3 (b) N=Z-P=2 P=1 Z=P+N=1+2=3 Module14 Problem 14.1(c) KK,1GHsMtgw(),3,,,,,,332(1)，s(1)，w 0wMK,,,,0:,0 0 wM,,,,,,:0,270 0110,,,,,,,,,1803603tgwtgww ,KMKK,,,,,,188max32(13)， K -1 Problem 14.1(d) Solution: K GHs(),2Im (1)(2)ss，， KMThe magnitude ,22,,,,,1(4)，，-1 K/4 ,,,0,,,arcarctan2tanPhase ,,Re 2Critical point K,,0M, : ,,04 :M,0,,, : ,,,270According to Nyquist stability criterion, magnitude should be less than unity when the phase is -180 ,:,,,,180arctan2arctan , 2 2,,8Which will lead to KThus => 1,K,36max36 Problem 14.2 KAccording to the problem, we can get that GHs,，，2sss，，9，，The magnitude and phase may be written directly as 1K9 M,22,,,,,，1,,,299,, , o,19 ,,,tg90,2,,19 oand we know that ,,,180 ,,3 so we can know that so that we can get that ,so we say that the maximum walue of K consistent with K,9c stability is 9 And if we use the routh metod’s 32DssssK,，，，,90 ，， 3s19 2 sK1 sK9, so we get that K<9 and if we draw the root locus ,then we can get the picture. And if we want to get the position that the root locus attach on the imaginary axis, we should 390,,,,use , then we can get sj,,2K,, ,,3So we get K,9 K,9And it also check the result Problem 14.4 Solution: KGHs(), (1)(12)(1/2)，，，sss Compared to the standard from, we can get the magnitude and phase KM ,22211(2)1(/2),,,，，， ,,,111 ,,,,,,,,tantan(2)tan(/2) ,Then let ,,,180 ,,,,111 ,,,,,180tantan(2)tan(/2),,, ,,1.87This makes When the Nyquist diagram just cross the point (-1, 0j), which means M=1, the system is marginal stability K 1,222,,,11(2)1(/2)，，， ,,1.87In the equation, so we can get K=7.36 11.2 That is the maximum value of K for the system is sable. Problem 14.6 Solution: We can write expressions for the magnitude and phase angle as: 2Kw(1)，M, 2w225(1)w，25 wo,,,，,1802tan2tan w5 oIn order to determine the Gain Margin , we can set the ,180 to and it yields: , woo1801802tan2tan,,,，,w5 0,,w ,MeanwhileM,, To determine the stability of the system, we take Root Locus into account: Poles: s=-5 or 0 Zeros: s=-1 0,,,,,,,222180,12 ,,,,,,,,90,12 And the interception of the asymptotes with the real axis is obtained from the equation: nn pz,,,ii501,，，,,11ii,,,,,4 aIm 21nm,,[S] Now, we can draw the Root Locus as right: We can easily see there is no positive pole of the system. Re So the system is stable forever. -5 -10 And to determine the Phase Margin, we can set E E 2M,1, and it yields: 2Kw(1)， 1,2w225(1)w，25 for K=100 ,,wrads12.5/ Substituting into the phase angle equation gives 00000 ,,,，,,,，,,,1802tan(12.5)2tan(2.5)180170.9136.4145.5 The Phase Margin is obtained from: oo PM= 18034.5，,, Problem 14(7 Solution: 1 GH,112(1)(1)sss，，，55 1For the second-order term: ,,,,5n25 ,1,,,arctanarctan ,,,M225,,,,222，,，1(1)(),55 ,,0M,1: ,,0 :M,0: ,,,,,,270 For gain margin, we turn Φ to -180: ,:,,,,180arctanarctan　 (,,) ,,25,, 这个式子有些问题，反正切函数的值域是,90,90，所以这两个反正切相加不可能是180度，这里是一阶惯性环节(相角0,-90)和二阶振荡环节(相角0,-180)的叠加，二阶振荡环节的相角不能用一个arctan式子来表达的，因为它有,90,,180的部分，对应于,,,的频率部分。上面式子前面的arctanw是锐角，那后面应该是个钝角，所以要用 , ,:,,,,，180arctanarctan)5(　　(,) ,,,25,, 2,,6结果一样但是概念不同了，你们可以把代入你们的式子，等式不能成立的 52,,6Leading to M, then 259 1259So GM,,,3.22M5 1For phase margin, we turn M to unity: =1 ,M2,,222，,，1(1)(),55 2It may seem somehow cumbersome to solve this function, we substitute x,, ,,2Soon we gain x=4 then :So PM,,,,2arctan2126.9 Problem 14.8 Ks(1)，K(s，2)102()GHs,,222(420)ss，s，ss2(1)s，，520 1'()GHs, 2,2s1()，s，2wwnn 5w,,,25,n5 2Kww，1,w,,1011045Mtgtg,,,,，,()180()222w2ww221,w，,(1)202520 0wM,,,,,,0:,180 0wM,,,,,,:0,270 www,,,www,,,,010111555,,,，,，,,,,,,tgtgtgtgw180()180()()()023222www222111,,,202020K12，1K148104,,,,,,,GMK224？,Mn48MK12122n,)，12(12025 w,, 23 w,0 Problem 14.10 4Transfer the function into Bode form GHs(),3ss(2)， 1GHs(), s32(1)s，2 The magnitude and phase may be written as ,1o,1,,,903tan() ,M,23/22,,，2(1/4) (1) Gain margin: ,oo,1,,,,,903tan()180,,2/3 —> ,2 Substituting into the magnitude equation gives 19 M,, 223223/22[1()/4] ， 33 The gain margin is given by 1132GM,,, OAM9 GM>1, the system is stable. (2)Phase margin: 1,,0.462,, M1 —> 23/2,,，2(1/4) Substituting into the phase equation gives 0.462oo,1 ,,,,,,903tan()1292 The phase margin is given by o PM,，,18051, PM>0, the system is stable. Module15 Problem 15.1(a) 10From the open-loop transfer function: GHs(),11(1)(1)，，ss32 (1)Calculate the break frequencies: 11 : : ,,2/rads,,3/radsbb111，s1，s32 (2)Plot the straight-line magnitudes as follows: 10 1 11，s2 1 11，s3 (3)Graphically add all element magnitudes (The figure will be shown later). (4)Plot the straight-line phase approximations as follows: 1 11，s3 1 11，s2 (5)Graphically add all element phases (The figure will be shown later). (6)The complete Bode diagram is shown below: Problem 15.1(b) 16s GHs(),(1)(2)ss，， The bode diagrams is the following: Problem 15.1(d) Solution: s2(1)，10 GHs(),ssss(1)(1)(1)，，，10010002000 Since the transfer function has already been in the form of Bode, then we can calculate the break frequencies. The break points are as follows: s1:10，,wb10 1:1w,bs 1 :100w,bs1，100 1:1000w,bs1，1000 1:2000w,bs1，2000 Now we can draw the picture of magnitude against w and phase against w M(db) w 40db/decade 60db/decade 80db/decade 点击这里放大 1 10 100 1000 Now we have finished the exercise! 0 -90 -180 -270 Problem 15.1(e) The Bode diagrams from Marlab. (补)Problem 15.1 20lg|m| 20lg5 w=1 w=5 w=0.5 0.5 φ lgw w=0 w=0.5 w=50 w=0.5 0.5 w=0.5 (补)Problem 15.5 for 20lg|M|=-10 lg|M|=-1/2 |M|=(10)^-0.5 and wo have |M|=K |M|=K(0.5^2+5^2)^-0.5/{(100-25)^2+(12*5)^2}^-0.5 =>k=6.04 Problem 15.5 Ks(0.5)，Solution: , and assuming H(s)=1, then Gs(),2ss，，12100 'KsKsKs(0.5)0.5(21)(21)，，， GHs(),,,2221212ssss，，12100100(1)(1)，，，，ss100100100100 1''KKKK,,,200 200 212,,,,,10,0.6,, n100,n '2K14，, M,224*0.36,,2(1),，100100 ,,5/radssubstituting into the above equation, then ''KK14*2510，' MK,,,10.420.96254*0.36*252(1),，100100 '20lg(10.42)10K,, 1,'2 10.4210K, ''200KK,KK,,,,,,,0.036 ,,5/radsSo, when a magnitude of -10db at a frequency , the value of K=6. Problem 15.8 Solution: Ks(12)，GH,312200(1)，，ss25100 Mw,，,,20lg20lg220lg102012，sdb: 1 w1003122nM,,20lg()20lg1，，ssdbw2525100: So: 20lgK-20lg200+20lg10+20lg4=-10 Leading to: lg(10*4*K/200)=-0.5 K=1.6 k=6.07 Module16 Problem 16.7(a) Kss(1)(2)，，GHs(), 4s We can’t get the maximum K here. And we will see its root Routh’s array: 这张图奈氏曲线的变化趋势不对:ω?0，φ?-360，曲线沿正实轴延伸;ω??，φ?-180，曲线平行负实轴逆向延伸。 And its root locus: So it’s obviously a unstable system. Problem 16.7(c) Solution kk GHs(),,sssss(0.01)(0.08)，，0.0008(1)(1)s，，0.010.08 The break points will occur at values of 0.01,0.08 rad/s, while setting the Bode gain to , unity, k,1 0.0008 k,0.0008corresponds to putting .The Bode diagram for this value of gain is plotted in 16.7(c).From the Bode diagram the gain and phase margins may be measured as ,PM,90 GMsdB()40, 20lg40N, N,100kThis produces , giving a minimum value of consistent with stability of kN,,0.0008*0.08 To check this with Routh’s array, the characteristic equation may be written as ksss，，，,(0.01)(0.08)0 lesding to 32sssk，，，,0.090.00080 Routh’s array takes the form 30.0008 1s: 20.09k s: 0.0008*0.09,k10 s:0.09 0k0 s: 1The left-hand column element of the row yields s 0.0008*0.090,,k ,5 k,7.2*10Bode diagram: k,0.0008 20lg20lg0.000862kdB,,, 1 T,100,,,0.0111T1 1 T,12.5,,,0.0822T2 ,,1, ,,,,,arg90，，1,,j,,, ,,1 ,,,,,,argarctan100，，，，2,,,1001j，,, ,,1,,, argarctan12.5,,,，，，，3,,12.51j，,,, To see the complete diagram, please stick here. ,2,101,4,31010101010100.0080.0820 0 ,20 ,40 Problem 16.10 Solution : K GH(s)= 2ss(1/3)， 1=(logK-log3) log,1010102 ,, - 45log3- 90log10 - 135log>0 1801010103Yielding K<9.65 1 <0.3 K Yielding K>3.33 Hence 3.33 4420log,,,,166 the higher break frequency is one decade below the new crossover point 10.06,, ,16.67,116.67，s G,c,276712767，s,, 1,0.00036 ,, then a compensated Bode Diagram will be obtained, we will check the phase margin at the new crossover frequency : PMwwww,,,,,，,arctan16.67arctan2767arctan0.2arctan0.49018064 that’s the very phase margin we demand! thus the transfer functions: