线性控制系统分析与设计期末考试(大学).docx
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1、一.(a)求位置y(t)与力f(t)有关的微分方程;(b)画出机械网络图;(C)确定传递函数G(D)=y/f。ri(a)nodexa(K2MyD2)xa-K2xh=fnodeXb(/C2+K3+BD+M2D2)xh-K2xa=O(c)G(D)=-4v7D4+BD3+KD2+BKa+KaKh-KwhereKa=KI+K?,Kb=K2+KyiK=Ka+Kb二、So1vethefo11owingdifferentia1equations.Assumezeroinitia1conditions.Sketchtheso1utions.D2x+6x=(1),尸1,k=0,HUO.q=k-w=0Theste
2、adystateoutputistherefore:xss=b0GXSSz0.Insertingtheseva1uesintopreviousequation(1):16XSS=16f1XSS=b匠(2)16Thehomogeneousequationisformedby1ettingtherightsideofthedifferentia1equationequa1zero:D2xz+16xz=0(3)thetransientresponseistheso1utionofthehomogeneousequation,isobtainedbyassumingaso1utionoftheform
3、(4)xt=Atwheremisaconstantyettobedeterminedthecharacteristicequationofsystem:/+i6m=/+16=0加尸4上忻一4jva1uesofmarecomp1ex,byusingtheEu1eridentityeij,t,=cosdtjsndtandthencombiningterms,transientso1utionsare(6)x1=A1e4j,+A2e4j,=B1cos4r+B2sin4z=Asin(4r+)x=xt+XSS=ASin(4,+。)+一16Assumezeroinitia1conditions,i.e.,
4、t=0,x(O)=Q,DX(O)=Q、insertingtheseva1uesintopreviousequation(7):X(O)=ASin+=0,Dx(O)=4ACOSo=O16gjA=-1216x=xt+xss=-sin(4r+)+16216三、Writethe1ap1acetransformsofthefo11owingequationsandso1vefoex);theinitia1conditionsaregiventotheright.万户2.8ZZv+4产10X(O)=2,以(0)=3The1ap1acetransformsoftheequations(s)-sx(0)-(0
5、)+2.8(SX(S)-X(O)+4X(S)=IOSTX(s)(s2+2.8s+4)-(2s+8.6)=IOs1X(s)(s2+2.8s4)=2a+8,65+10XG)=s2j-8.65102s+8.6s+10S(S2+2.85+4)S($+1.4)2+(04)2)=A+As+B522.8+4Theinverse1ap1acetransformsoftheequation(p637,appendxA36)x(t)=2.51.69e,4zsin(2X)4r-17.25)rvr1252+8.65+10A=sX(S)1=SZ8s+4=2.55=0XX=21+8.6s+102.5-0.55+1.65(
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