- 易迪拓培训,专注于微波、射频、天线设计工程师的培养
CST中算出的偶合系数怎么这么大?
录入:edatop.com 点击:
Tchebychev Filter
===================
Order = 4
Bandwidth = 8 MHz
Passband ripple = 0.01 dB (1.100747 VSWR)
Return loss = -26.3828 dB
Normed g values:
-------------------------------------------
g1 = 0.7129
g2 = 1.2004
g3 = 1.3213
g4 = 0.6476
g5 = 1.1008
Corresponding coupling coefficients in MHz:
-------------------------------------------
k_E = 11.22
k1_2 = 8.65
k2_3 = 6.35
k3_4 = 8.65
k_out = 11.22
Group Delay Time
----------------
t_d1 = 56.727 ns
t_d2 = 95.519 ns
t_d3 = 161.87 ns
t_d4 = 147.054 ns
t_d5 = 224.313 ns
===================
Order = 4
Bandwidth = 8 MHz
Passband ripple = 0.01 dB (1.100747 VSWR)
Return loss = -26.3828 dB
Normed g values:
-------------------------------------------
g1 = 0.7129
g2 = 1.2004
g3 = 1.3213
g4 = 0.6476
g5 = 1.1008
Corresponding coupling coefficients in MHz:
-------------------------------------------
k_E = 11.22
k1_2 = 8.65
k2_3 = 6.35
k3_4 = 8.65
k_out = 11.22
Group Delay Time
----------------
t_d1 = 56.727 ns
t_d2 = 95.519 ns
t_d3 = 161.87 ns
t_d4 = 147.054 ns
t_d5 = 224.313 ns
你检查哈,结果第一个本征模的谐振频率是不是0哦?
我以前就出现过耦合系数大于1的情况,仔细看原来是谐振频率不对
简易你多加几个模式计算
或者用HFSS
非常感谢指点上面数据是由后处理模板直接算出的,是归一化的吗?
Corresponding coupling coefficients in MHz:
-------------------------------------------
k_E = 11.22
k1_2= 8.65
k2_3= 6.35
k3_4= 8.65
k_out = 11.22
用上面的值处以中心频率就是所需要的耦合系数了
申明:网友回复良莠不齐,仅供参考。如需专业解答,请学习易迪拓培训专家讲授的CST视频培训教程。
上一篇:CST仿真圆极化器设置问题
下一篇:研究光纤的仿真软件有哪些