电磁兼容导论第9章、第10章部分答案VIP免费

9.4.10RepeatProblem9.4.8fortheribboncableofProblem9.4.3whereRS=RL=RNE=RFE=63.25Ω.[VNE,max=0.791,VFE,max=0]解：为1MHz,5V,占空比50%，上升/下降时间为50ns远近端串扰系数：此脉冲串的“摆动速率”为：近端远端9.4.11RepeatProblem9.4.8fortheproblemoftwowiresoveragroundplaneofproblem9.4.4.[VNE,max=10.7mV,VFE,max=2.67mV]解：导线上总电感总电容远端和近端串扰系数此脉冲串的“摆动速率”为：近端远端9.7.2ForthelineinProblem9.7.1,determinethefrequencywheretheshieldwillaffecttheinductivecouplingwhenitisgroundedatbothends.[970Hz]解：设导线长为拐点频率即所求频率为9.7.3IftheshieldinProblem9.7.1isgroundedatbothends,thelinelengthis2mandRS=0,RL=1KΩ,RNE=100Ω,∧RFE=50Ωdeterminethenear-endcrosstalktransferratioat100Hz,1KHz,100KHz,and10MHz.[0.38x10−6,2.67x10−6,3.72x10−6,3.72x10−6]解：因为两端接地，所以容性耦合为0拐点频率所求为所求为：当时，感性耦合和频率的变化无关，所以时，所求值不在变化，均为9.7.4FortheribboncableshowninFig.P9.3.4assumethetotalmutualinductanceandmutualcapacitancetobeLM=0.4μHandCm=400pF.IfVS(t)isa1MHzsinusoidofmagnitude1V,andtheterminationimpedancesareRS=RL=RNE=RFE=50Ω,determinethenear-endcrosstalkifashieldisplacedaroundthereceptorwireandtheshieldisonlyconnectedtothenearendofthereferencewire.[12.57mV]Howmuchdoestheshieldreducethecrosstalk?[10.88dB]解：因为接收导线和参考导线近端相连，故容性融合为0感性融合系数：此脉冲串的“摆动速率”为：则屏蔽层能减少干扰10.2.4Computethereflectionlossandabsorptionlossfora20milsteel(SAE1045)barrierat30MHz,100MHz,and1GHz,assumingafar-fieldsource.[53.23dB,3656.6dB,48dB,6676.0dB,38dB,21,111dB]解：时，反射损耗和吸收损耗分别为：时，反射损耗和吸收损耗分别为：时，反射损耗和吸收损耗分别为：10.3.1Computethereflectionlossandabsorptionlossfora20-milsteel(SAE1045)barrierat10kHz,100kHz,and1MHzforanear-fieldelectricsourcethatisadistanceof5cmfromtheshield.[188.02dB,66.76dB,158.02dB,211.11dB,128.02dB,667.6dB]解：近电场情况下时，反射损耗和吸收损耗分别为：时，反射损耗和吸收损耗分别为：时，反射损耗和吸收损耗分别为：10.3.2Computethereflectionlossandabsorptionlossfora20-milsteel(SAE1045)barrierat10kHz,100kHz,and1MHzforanear-fieldmagneticsourcethatisadistanceof5cmfromtheshield.[211.45dBuseR=0,66.76dB,21.45dBuseR=0,211.11dB,8.55dB,667.6dB]解：近磁场情况下时，反射损耗和吸收损耗分别为：时，反射损耗和吸收损耗分别为：时，反射损耗和吸收损耗分别为：