电磁兼容导论第9章、第10章部分答案精编版.doc

9.4.10 Repeat Problem9.4.8 forthe ribboncable ofProblem 9.4.3 where[]解： Lm 1 HCm250 pFRsRLRNERFE 63.25VS (t) 为 1MHz,5V,占空比50%，上升 / 下降时间为 50ns远近端串扰系数：M NEINDRNELm11*10 63.96*10 9RNERFERSRL2 63.25*2M NECAPM FECAPRNE RFERL Cm= 63.25 250*10 12=3.95*10 9RNERFERSRL22INDRFELm11*10 63.95*109M FERNERFE RSRL263.25*2此脉冲串的“摆动速率”为：dVS (t)5V0.1*10 9dt50ns近端V(t )(M INDM CAP ) dVS (t)NENENEdtVNE ,max0.79V远端VFE (t )( M FEINDM FECAP ) dVS (t)dtVFE ,max0 wires over a ground plane of problem ]解 ： lm2nH / m cm0.6 pF / m2mR 0R 50R100SLNERFE 200导线上总电感Lm l m 4*10 9 H总电容Cm cm 1.2*10 12 F1远端和近端串扰系数M NEINDRNELm1004*10 92.67*10 11RNERFE RSRL100200 050CAPCAPRNE RFERL Cm=20012=8*1011M NEM FERNERFE RSRL*1.2*103M FEINDRFELmRL8*1095.33*1011RNERFERS150此脉冲串的“摆动速率”为：dVS (t)5V0.1*10 9dt50ns近端V (t )( MINDM CAP ) dVS (t)NENENEdtVNE ,max0.01067V10.67mV远端VFE (t )( M FEINDM FECAP ) dVS (t)dtVFE ,max0.00267V 2.67 mV解： rs 4.43m / m l s 0.727 H / m 设导线长为拐点频率 f SHf SHRSHrsrs4.43*103970Hz2 LSH2 l s2 l s2*3.14*0.727.10 6即所求频率为 970 Hzdetermine the near-end crosstalk transfer ratio at100 Hz, 1 KHz, 100 KHz, and 10 MHz.[ , , , ]解： r4.43m/ m ls0.727 H / m lGS0.4575mH / m2mR0 R 1ksSLRNE100RFE502因为两端接地，所以容性耦合为 0拐点频率fSHRSHrsrs4.43*103970Hz2L2 l2l2*3.14*0.727.10 6ssSHf100HzfSHSF1VNERNELGR*100 2 2*0.4575*106j 2fSFj 260.383j *10VSRNERFERS RL310006所求为 0.383*10f1kHzf SHSFRSH所求为：j LSHVNEj 2fRNELGRRSH6VSRFE RSRLj LSH3.72*10RNE当 ff SH 时，感性耦合和频率的变化无关，所以f100kHzf10MHz 时，所求值不在变化，均为63.72*10capacitance to beand. Ifis a 1 MHz sinusoid of magnitude 1V, and the termination impedances are, determine the near-endcrosstalk if a shield is placed around the receptor wire and the shield is only connected to thenear end of the reference wire. [12.57mV]How much does theshield reduce the crosstalk?[10.88dB]解： Lm 0.4 H Cm400 pF RSRLRNERFE50因为接收导线和参考导线近端相连，故容性融合为0M NECAP0感性融合系数：M NEINDRNELm10.4*10 60.002*10 6RNERFERSRL25050此脉冲串的“摆动速率”为：dVS (t)2*10 6 COS(2*10 6 t)dt则VNE (t )INDCAPdVS (t)(M NEM NE)dt3V (t)(MINDMCAP ) dVS (t )0.002*10 6 *2 *10 60.01256V 12.56mVNENENEdtVNE (t)(M NEINDM NECAP ) dVS (t)3.5dtRdB20log 10 3.5 10.88dB屏蔽层能减少干扰10.88dB解： t 20mil 20*10 3 inchr1000r0.1f30MHz30*10 6 Hz 时，反射损耗和吸收损耗分别为：RdB16810log 10 (r) 168 10log 10 (0.16)53.23dBr f1000*30*10AdB3.338tfrr3.338*20*10330*10 6 *1000*0.13656.6dBf 100MHz100*10 6 Hz 时，反射损耗和吸收损耗分别为：RdB16810log 10 (r) 168 10log 10 (0.16 )48.0dBr f1000*100*10A3.338tfrr3.338*20*103100*10 6 *1000*0.16676.0dBdBf1GHz 1*10 9 Hz 时，反射损耗和吸收损耗分别为：RdB 16810log 10 (r ) 16810log 10 (0.19 )38.0dBrf1000*1*10AdB3.338tfrr3.338*20*1031*10 9 *1000*0.121111.4dB解： t 20mil 20*10 3 inch r 5*10 2 m r 1000 r 0.1 近电场情况下f 10KHz 10*10 3 Hz 时，反射损耗和吸收损耗分别为：4Re, dB 322 10log 10 (r2 ) 322 10log10 (0.12 ) 188.02 dB3r1000*(10*103)3(5*102)r fA3.338tfrr3.338*20*10310*10 3 *1000*0.。