嗨BS，你在测量螺旋线的LC吗？
如果是这样，有两种选择：1。将螺旋及其外壳放入EMPro本征模求解器中。
LC = 1 /（2πf0）^ 2，其中f0是第一谐振频率。
不需要激励。
2.如果两端未接地，则在螺旋的两端放置一个50欧姆的端口。
获得2x2 S矩阵结果。
将S矩阵变换为Y矩阵和Z矩阵。
从Y11开始你可以获得循环L.从Z11你可以得到C到地面。
根据您将螺旋解释为电路拓扑的方式，计算可能会有所不同。
由于我们将S转换为Y和Z矩阵，因此将消除激励电阻。
Martin编辑：myuan于2013年5月23日上午9:29
以上来自于谷歌翻译
以下为原文
Hi BS,
Are you measuring the LC of the helix? If so there are two options:
1. Put the helix and its housing in EMPro eigenmode solver. LC = 1 / (2 pi f0)^2, where f0 is the first resonant frequency. No excitation is needed.
2. Put a 50-Ohm port at each end of the helix, if both ends are not grounded. Obtain the 2x2 S matrix result. Transform S matrix to Y matrix and Z matrix. From Y11 you could get loop L. From Z11 you could get C to ground. Depending on how you interpret the helix to a circuit topology, the calculation could be different. Since we transformed S to Y and Z matrices, the excitation resistance would be eliminated.
Martin
Edited by: myuan on May 23, 2013 9:29 AM
嗨BS，你在测量螺旋线的LC吗？
如果是这样，有两种选择：1。将螺旋及其外壳放入EMPro本征模求解器中。
LC = 1 /（2πf0）^ 2，其中f0是第一谐振频率。
不需要激励。
2.如果两端未接地，则在螺旋的两端放置一个50欧姆的端口。
获得2x2 S矩阵结果。
将S矩阵变换为Y矩阵和Z矩阵。
从Y11开始你可以获得循环L.从Z11你可以得到C到地面。
根据您将螺旋解释为电路拓扑的方式，计算可能会有所不同。
由于我们将S转换为Y和Z矩阵，因此将消除激励电阻。
Martin编辑：myuan于2013年5月23日上午9:29
以上来自于谷歌翻译
以下为原文
Hi BS,
Are you measuring the LC of the helix? If so there are two options:
1. Put the helix and its housing in EMPro eigenmode solver. LC = 1 / (2 pi f0)^2, where f0 is the first resonant frequency. No excitation is needed.
2. Put a 50-Ohm port at each end of the helix, if both ends are not grounded. Obtain the 2x2 S matrix result. Transform S matrix to Y matrix and Z matrix. From Y11 you could get loop L. From Z11 you could get C to ground. Depending on how you interpret the helix to a circuit topology, the calculation could be different. Since we transformed S to Y and Z matrices, the excitation resistance would be eliminated.
Martin
Edited by: myuan on May 23, 2013 9:29 AM
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