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>
目前,我们进行仅THRU校准,然后执行带有校正的扫描,以在给定量程上测量S21。
我们想知道之后是否有可能以某种方式进行校准校正,如果是,那么如何计算插值校准系数。 1)测量扫描的中心和跨度将不同于校准扫描的中心和跨度,测量扫描以较少的点完成。 如何在这些情况下完成校准插值? 是否只是在一个已知的Ett值之间进行线性外推的情况? 例如,如果我们知道Ett在10.1MHz的Real和Imaginary值分别为0.96和0.11,而Ett在10.2MHz的实数和虚数值分别为0.97和0.10,我们只需计算Ett在10.13MHz的实数和虚数值 是0.963和0.107? 或者是以其他方式计算的插值校准系数? 另外,校准扫描和测量扫描之间的IFBW值的变化是否有任何影响(我不这么认为,但我只是想确定)? 2)还有一个更普遍的问题:为什么不应该以这种方式进行校准校正? 谢谢,我感谢任何人对此有任何意见。 如果有什么我想念或应该知道的,请告诉我。编辑:Cheenu于2014年1月7日下午3:36 以上来自于谷歌翻译 以下为原文 Currently, we do a THRU-only calibration and then execute a sweep with correction on to measure S21 over a given span. We were wondering if it would be possible to somehow do the calibration correction afterwards, and if so, how to calculate the interpolated calibration coefficients. 1) The center and span of the measurement sweep will differ from the center and span of the calibration sweep, and the measurement sweep is done with fewer points. How is the calibration interpolation done in cases like these? Is it simply the case that one linearly extrapolates between known values of Ett? For example, if we know the Real and Imaginary values of Ett at 10.1MHz are 0.96 and 0.11 and the Real and Imaginary values of Ett at 10.2MHz are 0.97 and 0.10, would we simply calculate the Real and Imaginary values of Ett at 10.13MHz to be 0.963 and 0.107? Or are the interpolated calibration coefficients calculated in some other manner? Additionally, does a change in the IFBW value between the calibration sweep and the measurement sweep have any effect (I wouldn’t think so, but I just wanted to be sure)? 2) And one more general question: are there any reasons why the calibration correction should not be done in this manner? Thanks, and I appreciate any input any of you would have about this. If there’s anything I’m missing or should be aware of, please let me know. Edited by: Cheenu on Jan 7, 2014 3:36 PM |
|
相关推荐
5个回答
|
|
在8753上,默认情况下,校准插值已激活。
因此,转到最宽的范围,将点数设置为1601,进行直通响应校准。 无论你喜欢什么点数,都要改变你的窄幅。 注意,如果通过先更改中心然后更改量程进行复位,则可能会设置一个条件,其启动或停止在校准的频率范围之外,因此cal将关闭。 以上来自于谷歌翻译 以下为原文 On an 8753, by default, calibration interpolation is already activated. So go to your widest span, set the point count to 1601, do a thru response calibration. Change to your narrow span with whatever point count you like. Caution, if you reset by first changing center and then changing span, you might set a condition that has the start or stop out of the calibrated frequency range, and so cal will turn off. |
|
|
|
> {quote:title = Dr_joel写道:} {quote}>在8753上,默认情况下,校准插值已经激活。 >>>所以转到最宽的范围,将点数设置为1601,进行直通响应校准。 无论你喜欢什么点数,都要改变你的窄幅。 注意,如果通过先更改中心然后更改量程进行复位,则可能会设置一个条件,其启动或停止在校准的频率范围之外,因此cal将关闭。 嗨Dr_joel,我理解校准的一部分,但问题是我们希望在后处理中进行校准校正,而不是通过8753ES进行校准校正。 我们希望下载校准系数阵列,然后当我们进行测量(初始校准跨度内的起始和终止频率)时,我们可以计算内插校准系数+ +扫描的内容,然后获得 校准校正扫描。 所以我们想知道如何实际计算插值。 它只是对2个最近点的线性外推,还是会更复杂? 谢谢,Cheenu 以上来自于谷歌翻译 以下为原文 > {quote:title=Dr_joel wrote:}{quote} > On an 8753, by default, calibration interpolation is already activated. > > So go to your widest span, set the point count to 1601, do a thru response calibration. Change to your narrow span with whatever point count you like. Caution, if you reset by first changing center and then changing span, you might set a condition that has the start or stop out of the calibrated frequency range, and so cal will turn off. Hi Dr_joel, I understand that part of calibration, but the thing is that we'd like to do the calibration correction in post-processing instead of having the calibration correction done by the 8753ES. Our hope was to download the calibration coefficients array, then when we take a measurement (with the start and stop frequencies inside of the initial calibration span), we could calculate what the interpolated calibration coefficients would be for +that+ sweep, and then obtain a calibration-corrected sweep. So we're wondering how the interpolation is actually calculated. Is it simply a linear extrapolation to the 2 nearest points, or would it be more complicated that that? Thanks, Cheenu |
|
|
|
瓦德瓦155 发表于 2019-3-6 18:44 我不明白为什么你不希望8753只是自动插值校准,但无论如何。 不,这不是简单的线性插值。 8753使用复杂的圆弧插补程序(细节是专有的),可以最大限度地减少基于VNA类型响应的误差。 如果原始校准的每点频率间隔遵循delta-freq 以上来自于谷歌翻译 以下为原文 I'm not understanding why you don't want the 8753 to just automatically interpolate the calibration but whatever. No, it is not simple linear interpolation. The 8753 uses a sophisticated circular interpolation routine (details are proprietary) that minimizes error based for VNA type response. If the frequency spacing per-point of the original calibration follows delta-freq<=12/(cable length in meters) then the interpolation error is typically small then 0.02 dB. Linear interpolation can be far worse (I'd have to do a bit of math to figure it out, but maybe 10x or 20 x worse). It's all based on complex-vector geometry. |
|
|
|
脑洞大赛9 发表于 2019-3-6 18:52 > {quote:title = Dr_joel写道:} {quote}>我不明白为什么你不希望8753只是自动插入校准,但无论如何。 >>不,这不是简单的线性插值。 8753使用复杂的圆弧插补程序(细节是专有的),可以最大限度地减少基于VNA类型响应的误差。 >>如果原始校准的每点频率间隔遵循delta-freq 以上来自于谷歌翻译 以下为原文 > {quote:title=Dr_joel wrote:}{quote} > I'm not understanding why you don't want the 8753 to just automatically interpolate the calibration but whatever. > > No, it is not simple linear interpolation. The 8753 uses a sophisticated circular interpolation routine (details are proprietary) that minimizes error based for VNA type response. > > If the frequency spacing per-point of the original calibration follows delta-freq<=12/(cable length in meters) then the interpolation error is typically small then 0.02 dB. Linear interpolation can be far worse (I'd have to do a bit of math to figure it out, but maybe 10x or 20 x worse). It's all based on complex-vector geometry. Hi Dr_joel, I was afraid of that. The idea was to minimize measurement time on our tester for multiple dies, as we could apply calibration afterwards (and while our tester is measuring other dies). But given that we can't do the interpolation ourselves, we'll just let the network analyzer handle it and look to optimize our code elsewhere, if we can. Thanks for your expertise on this matter -- I do greatly appreciate your help, if only to rule out this option. |
|
|
|
> {quote:title = Cheenu写道:} {quote} >> 2)还有一个更普遍的问题:为什么不应该以这种方式进行校准校正?
仅供参考,使用典型设置进行的S21响应校准可能会产生大约1 dB或甚至更多的误差。 当然,如果您通过重新测量校准,它将不会显示错误,但如果您测量不同长度的直通线,则错误将显示为纹波。 以上来自于谷歌翻译 以下为原文 > {quote:title=Cheenu wrote:}{quote} > > 2) And one more general question: are there any reasons why the calibration correction should not be done in this manner? And just for reference, an S21 response calibration made using a typical setup can have an error on the order of 1 dB or even a little more. Of course, if you re-measure the calibration thru, it won't show the error as but if you measure a thru line of a different length, the error will show up as ripple. |
|
|
|
只有小组成员才能发言,加入小组>>
1193 浏览 0 评论
2335 浏览 1 评论
2133 浏览 1 评论
2009 浏览 5 评论
2881 浏览 3 评论
925浏览 1评论
关于Keysight x1149 Boundary Scan Analyzer
682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
785浏览 0评论
1193浏览 0评论
小黑屋| 手机版| Archiver| 电子发烧友 ( 湘ICP备2023018690号 )
GMT+8, 2024-11-8 05:19 , Processed in 1.360415 second(s), Total 84, Slave 67 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