电磁兼容导论第9章、第10章部分答案.doc

9.4.10 Repeat Problem 9.4.8 for the ribbon cable of Problem 9.4.3 where RS=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

9.4.11 Repeat Problem 9.4.8 for the problem of two wires over a ground plane of problem 9.4.4.[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

9.7.2 For the line in Problem 9.7.1, determine the frequency where the shield will affect the inductive coupling when it is grounded at both ends.[970 Hz]
解： 设导线长为
拐点频率

即所求频率为

9.7.3 If the shield in Problem 9.7.1 is grounded at both ends, the line length is 2m and RS=0, RL=1KΩ, RNE=100Ω, and RFE=50Ω determine the near-end crosstalk transfer ratio at 100 Hz, 1 KHz, 100 KHz, and 10 MHz.[0.38 x 10-6,2.67 x 10-6, 3.72 x 10-6, 3.72 x 10-6]
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

9.7.4 For the ribbon cable shown in Fig. P9.3.4 assume the total mutual inductance and mutual capacitance to be LM=0.4μH and Cm=400pF. If VSt is a 1 MHz sinusoid of magnitude 1 V, and the termination impedances are RS=RL=RNE=RFE=50Ω, determine the near-end crosstalk if a shield is placed around the receptor wire and the shield is only connected to the near end of the reference wire. [12.57mV] How much does the shield reduce the crosstalk? [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

10.2.4 Compute the reflection loss and absorption loss for a 20 mil steel (SAE1045) barrier at 30 MHz, 100 MHz, and 1 GHz, assuming a far-field source. [53.23 dB, 3656.6 dB, 48 dB, 6676.0 dB, 38 dB, 21,111 dB]
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：

10.3.1 Compute the reflection loss and absorption loss for a 20-mil steel (SAE1045) barrier at 10 kHz, 100 kHz, and 1 MHz for a near-field electric source that is a distance of 5 cm from the shield. [188.02 dB, 66.76 dB, 158.02 dB, 211.11 dB, 128.02 dB, 667.6 dB]
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：

10.3.2 Compute the reflection loss and absorption loss for a 20-mil steel (SAE 1045) barrier at 10 kHz, 100 kHz, and 1 MHz for a near-field magnetic source that is a distance of 5 cm from the shield. [211.45 dB use R = 0, 66.76 dB, 21.45 dB use R =0, 211.11 dB, 8.55 dB, 667.6 dB]
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：