自控第四章作业答案

P4.1 Sketch the root loci for the following open-loop transfer functions when ∞<<k 0. (a) )3)(2()5()()(+++=s s s s k s H s G ,(b) )4)(3)(1()5()()(++++=s s s s s k s H s G(c) 84)3()()(2+++=s s s k s H s G ,(d) )20020()20()()(2+++=s s s s k s H s G(e) 2)3()()(+=s s ks H s GHere, “sketch ” means that it is not necessary to find the exact positions of the possible breakaway point and the intersection with j-axis, and the exact values of relative angles.Solution: (a) )3)(2()5()()(+++=s s s s k s H s G . By inspection, wehave 01=p , 22-=p , 33-=p and 51-=z . The intersection of asymptotes is013)5()320(=-----=a σand the angles of asymptotes are213)12(ππϕ±=-+=l aThere is a breakaway point on the real-axis segment )0 2(，-. Considering that 5-==∑∑i i p s , due to 2=-m n , the root locus does not intersect with j-axis. The root locus is plotted as shown.(b) )4)(3)(1()5()()(++++=s s s s s k s H s G . By inspection, we have 01=p , 12-=p , 33-=p ,4-=p and 5-=z . The intersection of asymptotes is(c) 2222)2()3(84)3()()(+++=+++=s s k s s s k s H s G . Byinspection, we have 2221j p ±-=， and 51-=z . The root locus in the complex plane 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.(d) ]10)10[()20()20020()20()()(222+++=+++=s s s k s s s s k s H s G . By inspection, we have 01=p ,10103,2j p ±-= and 201-=z . The intersection of asymptotes is013)20()10100(=-----=a σand the angles of asymptotes are213)12(ππϕ±=-+=l aThe characteristic equation is 020********3=+++s s s . If the rootlocus has breakaway points, the closed-loop system will have a thrice root, i.e. a root will satisfy 403-=s and 20003-=s simultaneously. Obviously, it is impossible. The root locus is plotted as shown.The angles of departure of the root locus from the complex poles are 0)()(18012122=-∠--∠+=p p z p p θ, 02=p θ (e) 2)3()()(+=s s ks H s G . By inspection, we have 01=p ,332-==p p . The intersection of asymptotes is13330-=--=a σ and the angles of asymptotes areπππϕ ,33)12(±=+=l aThere is a breakaway point on the real-axis segment )0 3(，-. The root locus is plotted as shown.P4.2 C onsider a unity feedback system with 54)1()(2+++=s s s k s G(a) Find the angles of departure of the root locus from the complex poles. (b) Find the entry point for the root locus as it enters the real axis.Solution: 1)2()1(54)1()(22+++=+++=s s k s s s k s G . By inspection, we have 122 ,1j p ±-= and11-=z .(a) The angles of departure of the root locus from the complex poles are22590135180)()(18021111=-+=-∠--∠+=p p z p p θ, 2252-=p θ (b) The breakaway-point equation is given by012 11544222=-+⇒+=+++s s s s s sSolving this equation yields 212 ,1±-=s . There is only one breakaway point 21--=b s . P4.3 A unity feedback system has a plant transfer function)14.005.0()(2++=s s s Ks G Sketch the root locus as K varies.Solution:]2)4[()208()14.005.0()(2222++=++=++=**s s K s s s K s s s Ks G . Byinspection, we have 01=p and 243 ,2j p ±-=. The intersection of asymptotes is67.2383440-=-=--=a σand their angles areπππϕ ,33)12(±=+=l aSolving the breakaway-point equation020163 0 20882122=++⇒=++++s s s s s swe get two breakaway points 33.31-=b s , 22-=b sThe characteristic equation is 020823=*+++K s s s . Substituting ωj s =into this equationyields ⎪⎩⎪⎨⎧=+-=*+-0200832ωωωK , we get the intersection of root locus with the j-axis47.452±=±=c ω, 160=*cK The root locus is plotted as shown.P4.5 A unity feedback system has )5)(2()(++=s s s ks GFind (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 6=k . (d) Sketch the root locus. Solution: (a) Solving the breakaway-point equation 051211=++++s s swe find the breakaway point 88.0-=b s . (Another solution of this equation is not on the root locus.) The gain for this point is06.45 2=+⋅+⋅=b b b b s s s k(b) The characteristic equation is 010723=+++k s s sSubstituting ωj s =into this equation yields ⎪⎩⎪⎨⎧=+-=+-0100732ωωωk , we get the intersection of rootlocus with the j-axis16.310±=±=c ω, 70=c khence, when two roots lie on the imaginary axis, the gain is 70=k and the roots are 10j s ±=.(c) In the case of 6=k , the characteristic equation is 06107)(23=+++=s s s s ∆Since b k k >=6, there are one real root on segment )5 ,(--∞and a pair of complex roots, it is not difficult to find that 0)34.5(≈-∆, i.e. 34.51-=s is a root of this equation. Denoting jb a s ±=2 ,1, we have83.0 7321-=⇒-=++a s s s 66.0 6321=⇒-=b s s si.e., in this case the closed-loop roots are 34.51-=s and 66.083.03 ,2j s ±-=. (d) The root locus is plotted as shown, where for the asymptotes we have33.2352-=--=a σ, ππϕ ,3±=aP4.7 The transfer functions of a negative feedback system are given by)5)(2()(2++=s s s ks G and 1)(=s H(a) Sketch the root locus for this system. (b) The transfer function of the feedback loop is now changed to 12)(+=s s H . Indicate the crossing points of the locus on the imaginary axis and the corresponding value of k at these points. Determine the stability of the modified system as a function of k . Investigate the effect on the root locus due to this change in )(s H .Solution: (a) There are four open-loop poles: 021==p p , 23-=p and 54-=p , and there is no zero. For the asymptotes we have75.147-=-=a σ, 135 ,45±±=a ϕ Solving the breakaway-point equation051212=++++s s swe get 25.11-=s , 13.42-=s , where 13.4-=b s is a breakaway point. Since the closed-loop system is constructional unstable, the root locus does not intersect with j-axis for 0>k . 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()()(22+++=+++=*s s s s k s s s s k s H s G ,k k 2=*The asymptotes are centered at 17.235.6-=-=a σwith angles 801 ,60±=a ϕ. Investigating the breakaway-point equation5.01512122+=++++s s s s 0105.2016323=+++s s swe know that the root locus does not have breakaway point.Substituting ωj s = into the characteristic equation⎪⎩⎪⎨⎧=+-=+-⇒=++++****705.010 05.0107324234ωωωωk k k s k s s syields 55.25.6±=±=c ω, 5.45=*ck . The root locus is plotted as shown. Now, the system is stable only when 5.450<<*k , i.e. 75.220<<k .As we see, the stability of the closed-loop system is improved due to the change in )(s H . P4.8 The characteristic equation of a feedback control system is given by042)5()(2=++++=k s k s s ∆Sketch the root locus as a function of k (positive k only) for this system. Solution: Rewriting the characteristic polynomial as )2(4542)5()(22++++=++++=s k s s k s k s s ∆ yields a equivalent transfer function)4)(1()2(45)2()(2+++=+++=s s s k s s s k s G e The root locus is plotted as shown.P4.10 The open-loop transfer function of a control system with positive feedback is given by)44()()(2++=s s s ks H s GSketch the root locus for this system when ∞<<k 0. Solution: The root-locus equation is given by1)2(0)()(12-=+⇒=-s s ks H s G using the rules for plotting 0 root locus, we have 33.134-=-=a σ, 120 ,0±=a ϕThe root locus has neither breakaway point nor intersection with j-axis. The root locus is plotted as shown.。