完善资料让更多小伙伴认识你,还能领取20积分哦, 立即完善>
我们使用N5230C测量电缆组件上的反射相位,电缆组件的一端有TNC连接器,另一端没有连接。
关于我们应该向客户指定的容差,我们提出了一个问题。 它们要求最终电缆组件+/- 1度,但由于我们为它们提供“往返”阶段,现在测得的公差是+/- 2度还是+/- 1度? 这里有很多讨论,提出了争论的两个方面。 谢谢。 以上来自于谷歌翻译 以下为原文 We are using an N5230C to measure reflected phase on cable assemblies that have a TNC connector on one end and are unconnectorized on the other. A question has come up as to the tolerance we should be specifying to our customer. They are requesting +/- 1 degree for the final cable assembly, but since we are supplying them with the 'round trip' phase, is the measured tolerance now +/- 2 degrees or does it remain +/- 1 degree? There is quite a bit of discussion here presenting both sides of the argument. Thanks. |
|
相关推荐
1个回答
|
|
如果您指定传输阶段,则会有S21 = sqrt(S11),因此我建议将相位响应除以2.这相当于将限制设置为+ -2。
但是输入反射错误会导致更大的涟漪。 这是自动夹具移除的完美应用,它可以给出电缆的S参数,仅打开测量,并消除不匹配的影响。 以上来自于谷歌翻译 以下为原文 If you are specifying the transmission phase, you would have S21=sqrt(S11), so I would suggest dividing the phase response by 2. That would be equivalent to setting the limit at +-2. But input reflection errors can cause much bigger ripples. This is a perfect application for Automatic Fixture Removal, which would give the S-parameters of the cable, with an open only measurement, and remove the effects of mismatch. |
|
|
|
只有小组成员才能发言,加入小组>>
1226 浏览 0 评论
2348 浏览 1 评论
2159 浏览 1 评论
2024 浏览 5 评论
2906 浏览 3 评论
970浏览 1评论
关于Keysight x1149 Boundary Scan Analyzer
703浏览 0评论
N5230C用“CALC:MARK:BWID?”获取Bwid,Cent,Q,Loss失败,请问大佬们怎么解决呀
804浏览 0评论
1226浏览 0评论
小黑屋| 手机版| Archiver| 电子发烧友 ( 湘ICP备2023018690号 )
GMT+8, 2024-11-25 02:10 , Processed in 1.551437 second(s), Total 77, Slave 60 queries .
Powered by 电子发烧友网
© 2015 bbs.elecfans.com
关注我们的微信
下载发烧友APP
电子发烧友观察
版权所有 © 湖南华秋数字科技有限公司
电子发烧友 (电路图) 湘公网安备 43011202000918 号 电信与信息服务业务经营许可证:合字B2-20210191 工商网监 湘ICP备2023018690号