电路理论基础(陈希有)课后题答案第15章剖析

u200103Z===500Ωci+400N=2c=2R+Z132cu=u++Nu+=(1+)200103=246.15kV222213端反射系数N=Lc=2Z+Z3Lc1Z+Zdvu15Vi=i+=1==30mAZ0c2l4l2l4lvv1212vv212121233i()=1=15mAR(30mA551212d1i+Ni+2td3dNNi+2dxl46tdd图题15.2ScSc刻到达始端。由于R=Z，所以到达始端后不再产生第二次反射NRZciNiNi.1A2R+Z1800+600222212c12(c(cc1)ZZZ0.10123456t/10μs有Z=Z=R+RZ2c2c111R+Zc2111ZZRR+Zc2221c1RZR+Zc21c111Z=R+=R+cRZZ=R+=R+cc21c1cc1c2将R代入式(3)得1即R=1RZc2Z2ZZ+Zcc1c2c1Z2Z2ZZZ化简得Z=c1c1c2=c1c1c2即R=ZRZZ2c2ccZZ2uu+Nu+=(1+N)u+222222u7kVN=21=1=0.42ZZN=c1c2Z+Zc1c2LZ=0=175，C1C0sm0v002L=8.75106H/m,C=109F/m07CuR=OC===Zii2i+i+cuCCC案15.7但由于Z=R，所以在始端不再产生第二次反射。根据彼德生法则，得到t>1mscS+LuLucRuZcuete(t0.001)5e(t0.002)]VR+ZSScR=RZc=300200=120ΩT=l=1siiic所以解：终端入射波电压象函数为：U+(s)=ZI(s)=Z，cSc终端反射系数的复频域形式为：ccR+sL+Zcc2解波阻抗c00c00始端输入电阻RLcR47.223业RLcR47.223业iRsh(yl)+Rch(yl)cLcUic1L0000000j(RjL)(GjC)0.0533j0.066(1/km)000022puUexcos(tx)iImxcos(txi)umij0.044j0.046(1/km)0.046Y331l11ZZ/th(l)ic2iZsh(yl)+Zch(yl)cL1c111i211c11Y2YY解：0000a0.11878cc00000，，1m1mc线路终端电压1m线路终端电流21mn2222P10021b1m1mu(xII(x)xoi(x)15.14iZcosblc2cl2/2,cosl0,sinl14Zc702452ZZjZ2cZ35j35j707070526.6iZjZc7035j35c2cl/4时l/2,cosl0,sinl1Z2ZZc(1)iR2ici2若R0则Z2i解21082mv31083Lic1RZS11iSi1Ll24.53，cosl1，sinl03ZZcosljZsinlLcZZ500iZcosljZsinlcLcL11ic1LiL2100j50011c1l=2fl=1.5，cosl=0，sinl=1vRcosl+jZsinlZ2iZcosl+jRsinlcRZ=iZcosl+jRsinlcRC221i1c1c11ZZccLccC20ic1R+ZiSiLC002Z,=一jZctg(bl))wicZ=jX2Cc2cI1=Z1=jX1iCic11ZcXCX11XC1ZcZcX1XCCCX11C，ZX