https://bbs.21ic.com/icview-570139-1-1.html
所谓“经验”值....
Z0 C = 2 d / v
v / f = λ = 6 d
代入下式
Zc = 1 / (j 2πf C)
得
Zc = -j Z0 / 2
即
ZL = Z0(1 - j/2)
反射系数
Γ = (ZL - Z0)/(ZL + Z0)
= -j/(4 - j)
驻波比
SWR = (1 + |Γ|) / (1 - |Γ|)
= (1 + 1/√17) / (1 - 1/√17)
= 1.64
若取
Z0 C = d / v
那么
SWR = 2.61
注意,这仅仅是考虑SWR。如果再考虑信号带宽(譬如进入微波)和电容本身器件特性,那就不是那么回事了。
此外,警告一句,别轻信仿真!
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