浙大自控作業(yè)三章六周答案

第二次作業(yè)(如時(shí)間緊張 P659.3.8(b),(d)Solvethefollowingdifferientialequations.Assumezeroinitial(b)D2x+4Dx+3x=9;(d)D3x+3D2x+4Dx+2x第二次作業(yè)(如時(shí)間緊張 P659.3.8(b),(d)Solvethefollowingdifferientialequations.Assumezeroinitial(b)D2x+4Dx+3x=9;(d)D3x+3D2x+4Dx+2x=10sketchtheSolution: TherearetwomethodstosolvetheMethod1:UsingLaplace9 ∵Dx+4Dx+3x=9;\X(s)s(s2= s+ s++4s+ s+ s+Method2:Solvingdifferential ∵x=xss+xtThecharacteristicequation +4l+3=0,l1=-1,l2=-Thenthetransientresponseis:x(t)t=a1e-+a2e-=09Thesteadystateresponseis:x(t)===3 AConsideringzeroinitialconditions,wex(0)=a1+a2+3=x￠(0)=-a1-3a2=0, thengot:a1=-4.5,a2Thereforethesolutionoftheproblemisassameasmethodx(t)=xss+xt=3+1.5e-3t-4.5e-Solution:3.8(d):Alsotherearetwomethodstosolvetheproblem.Usingfirst1∵X(s)=s3+3s2+4s+2s2+ (s+1)(s2+2s+2)(s2+ABs+Ds+=++s+(s+1)2+s+Byresidue’stheorem:A=0.99,B=C=-1,D=0.01andE=-X(s)=0.99-(s+1)+0.01s-s+(s+1)2+s+TheinversetransformofX(s),thesolutionx(t)=0.99e-t-e-tcost+0.01cos10t-=0.99e-t-e-tcost+0.01cos(10t+2.P659.3.9ForProb.2.2,solvefori2(t),where:E=10V,L=1h,C1=C2=0.001F,Rx(t)=0.99e-t-e-tcost+0.01cos10t-=0.99e-t-e-tcost+0.01cos(10t+2.P659.3.9ForProb.2.2,solvefori2(t),where:E=10V,L=1h,C1=C2=0.001F,R1=10W,11Solution:3.9TheloopequationsofProb.2.2:E=(R)i- 112CC11111 )i+(R+LD 2CC C112DuetotheloopequationsofProb.2.2,thetransferfunctionofi2(t)toe(t)canbeI2(D)= C2 R1C1C2LD3+(R1R2C1C2+LC2)D2+(R1C2+R1C1+R2C2)D+D==0.01D3+1.15D2+35D+115D2+3500D+UsingLaplacetransformationmethod:I(s)==2 s3+115s2+3500s s3+115s2+3500s=(s+88.2)(s2+26.8sWecanusethetableofLaplacetransformpairsinP636togetthe i2(t)=L-1[I2(s)]==L-}(s+88.2)[(s+13.4)2+3123（胡壽P.1343-4）已知二階系統(tǒng)的單位階躍響應(yīng)h(t)=10-12.5e-1.2tsin(1.6t+53.1D試求系統(tǒng)的超調(diào)量s%、峰值時(shí)間tp和調(diào)節(jié)時(shí)間ts=12.5；1x20.8；故x1-=1.2；代入x0.6wn2wdwn1x2n以下即可直接求得系統(tǒng)的超調(diào)量s%、峰值時(shí)間tp和調(diào)節(jié)時(shí)間ts-1-x（1）因?yàn)椋撼{(diào)量s%而xcosbcos53.1D-s100%=p(2)因?yàn)閠pdwdp=故：其t=1.9625pwd1x2中可知w(3)因?yàn)?1-x（1）因?yàn)椋撼{(diào)量s%而xcosbcos53.1D-s100%=p(2)因?yàn)閠pdwdp=故：其t=1.9625pwd1x2中可知w(3)因?yàn)閠；從已知w1.6sdnnn 故= 0.6n4（胡壽P.1343-6）已知控制系統(tǒng)的單位階躍響應(yīng)h(t)=1+0.2e-60t-1.2e-10t)試確定系統(tǒng)的阻尼比x和自然頻率wn解：因?yàn)椋合到y(tǒng)的單位脈沖響k(t)的象函數(shù)為系統(tǒng)閉環(huán)傳遞函數(shù)φ(s)，故可以通過求拉氏變換得到系統(tǒng)閉環(huán)傳遞函數(shù)φ(s)，k(t)與單位階躍響應(yīng)成關(guān)系+12e-=12(e--e-60t11故G(sL[k(t]= -s+ s+ss2+70s+ s2+2xwns+n可見：wn=60024.5x7024.5=5（P.1343-7）3-42是簡化的飛行控制系統(tǒng)結(jié)構(gòu)圖，試K1Kt使系統(tǒng)wn6,x1s(sG(s)s(s+0.8)+25K1KtG(s)=s(s+0.8)+25K1Kts s2+s(0.8G(s)s(s+0.8)+25K1KtG(s)=s(s+0.8)+25K1Kts s2+s(0.8+25K1Kt)36K36=w2則11而2xw=120.825KKK120.8n1tK6.P.671.Problem6.2(a)C(s)(e)ForProb.s[s(s+2)(s2+4s+20)+KSolution:6.2(a)Duetothedrivingfunctionisstepfunction,suchthatthecharacteristicequations(s+2)(s2+4s+20)+K=s4+6s3+28s2+40s+K=sss16 K6·28-40=621.3·40-stheresponsec(t)isWhenK=142.2(thes1row),formtheauxiliaryequationfromprecedingrow:21.3s2+142.22=0Therearetworootss1,2=–j2.58locationontheimaginaryaxisyieldsustainedoscillations.Solution:6.2(e),From5.24(b),weP=1P1=1s4+2s3+(K+1)s2+Ks+DThecharacteristicequationis:s4+2s3+(K+1)s2+Ks+1=sss12K+1K12·(K+1)-K=K+2K+2K- K+s2AccordingtoRouth’sstabilitycriterion,combinedrestrictionsonKforstabilitysss12K+1K12·(K+1)-K=K+2K+2K- K+s2AccordingtoRouth’sstabilitycriterion,combinedrestrictionsonKforstabilityofthesystem,whenK>1.236,theresponsec(t)isstable.WhenK=1.236(thes1row),formtheauxiliaryequationfromprecedingrow:=–j0.786,whicharelocatedontheimaginaryaxisyieldTheroots解：(1)圖（a）所示的系統(tǒng)閉11sG(s)s2+顯然wn=1,x且系統(tǒng)單位階躍響應(yīng)為h(t)=1-cos呈等幅振蕩，自然頻率為1(2)圖（b）所示的系統(tǒng)為比例＋微分控制系統(tǒng)，其閉環(huán)傳遞函數(shù)G(s)=s+ s2+s+wn=1,xd0.5,z=1（z為增加的零點(diǎn)1s因?yàn)閤d0.50xd0.5<1。h(t)=1+re-xdwnt 1-x2t+ 2 -2wxz+z=e/1-nnd y=-p+arctg[wn1-xd/(z-xdwn)]+arctg[1-xd/xd]=-h(t)=1+1.155e-0.5tsin(0.866t-60D)=1-1.155e-0.5tsin(0.866t+120D由xd0.5zxdwn2P983-24可得tr0.9bd=1x2x60D=1.047弧 =1.047*2=2.42=-t=pw2wn1-d3+0.5ln(z2-2xdwnz+w2n)+lnz-0.5(1-x2ts= d=bd=1x2x60D=1.047弧 =1.047*2=2.42=-t=pw2wn1-d3+0.5ln(z2-2xdwnz+w2n)+lnz-0.5(1-x2t