CST中算出的偶合系數(shù)怎么這么大，盼資深工程師指點(diǎn)

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

你檢查哈，結(jié)果第一個(gè)本征模的諧振頻率是不是0哦?
我以前就出現(xiàn)過耦合系數(shù)大于1的情況，仔細(xì)看原來是諧振頻率不對
簡易你多加幾個(gè)模式計(jì)算
或者用HFSS

非常感謝指點(diǎn) 上面數(shù)據(jù)是由后處理模板直接算出的，是歸一化的嗎?

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
用上面的值處以中心頻率就是所需要的耦合系數(shù)了

非常感謝！