..9.4.10RepeatProblem9.4.8fortheribboncableofProblem9.4.3whereRS=R解：為1MHz,5V,占空比50%,上升/下降時(shí)間為50ns遠(yuǎn)近端串?dāng)_系數(shù)：此脈沖串的"擺動(dòng)速率"為：近端遠(yuǎn)端9.4.11RepeatProblem9.4.8fortheproblemoftwowiresoveragroundplaneofproblem9.4.4.[VNE,max解：導(dǎo)線上總電感總電容遠(yuǎn)端和近端串?dāng)_系數(shù)此脈沖串的"擺動(dòng)速率"為：近端遠(yuǎn)端9.7.2ForthelineinProblem9.7.1,determinethefrequencywheretheshieldwillaffecttheinductivecouplingwhenitisgroundedatbothends.[970Hz]解：設(shè)導(dǎo)線長(zhǎng)為拐點(diǎn)頻率即所求頻率為9.7.3IftheshieldinProblem9.7.1isgroundedatbothends,thelinelengthis2mandRS=0,RL=1KΩ,RNE解：因?yàn)閮啥私拥?所以容性耦合為0拐點(diǎn)頻率所求為所求為：當(dāng)時(shí),感性耦合和頻率的變化無(wú)關(guān),所以時(shí),所求值不在變化,均為9.7.4FortheribboncableshowninFig.P9.3.4assumethetotalmutualinductanceandmutualcapacitancetobeLM=0.4μHandCm解：因?yàn)榻邮諏?dǎo)線和參考導(dǎo)線近端相連,故容性融合為0感性融合系數(shù)：此脈沖串的"擺動(dòng)速率"為：則屏蔽層能減少干擾10.2.4Computethereflectionlossandabsorptionlossfora20milsteel<SAE1045>barrierat30MHz,100MHz,and1GHz,assumingafar-fieldsource.[53.23dB,3656.6dB,48dB,6676.0dB,38dB,21,111dB]解：時(shí),反射損耗和吸收損耗分別為：時(shí),反射損耗和吸收損耗分別為：時(shí),反射損耗和吸收損耗分別為：10.3.1Computethereflectionlossandabsorptionlossfora20-milsteel<SAE1045>barrierat10kHz,100kHz,and1MHzforanear-fieldelectricsourcethatisadistanceof5cmfromtheshield.[188.02dB,66.76dB,158.02dB,211.11dB,128.02dB,667.6dB]解：近電場(chǎng)情況下時(shí),反射損耗和吸收損耗分別為：時(shí),反射損耗和吸收損耗分別為：時(shí),反射損耗和吸收損耗分別為：10.3.2Computethereflectionlossandabsorptionlossfora20-milsteel<SAE1045>barrierat10kHz,100kHz,and1MHzforanear-fieldmagneticsourcethatisadistanceof5cmfromtheshield.[211.45dBuseR=0,66.76dB,21.45dBuseR=0