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我需要有关ENA 5071C绝对测量的帮助。
我指的是直接访问绝对R1(n)R2(n)... A(n),B(n)测量值(n是应用刺激的端口)(测量菜单)。 我尝试使用该参数执行一些“特殊”校准,并与仪器的数学功能一起使用,而不是使用正常的S参数。不幸的是,我的算法没有像我预期的那样工作。 经过调查,我发现我认为绝对测量结果有些不一致。 即1)我只使用端口1和端口2(因此R1,A,R2,B是所涉及的测量)2)使用50欧姆电缆(1米)连接端口1和2,并在大约100 MHz(几个)进行测量 指出要轻松验证结果)3)我测量了R1(1),A(1),R2(1)和B(1),并将4测量标准化为第一个(R1(1))的值,这样 ... -R1(1)/ R1(1)为1(显而易见) - 正如预期的那样B(1)/ R1(1)的模数接近1(信号通过电缆传播并进入端口2) - 如 预期A(1)/ R1(1)很小(表示反射波返回端口1),这意味着端口2相当匹配(负载匹配) - 因为我没想到R2(1)/ R1( 1)导致模数约为0.4而不是相当小(如A(1)/ R1(1))..如果R2(1)/ R1(1)与从波特2输出的波成比例......我可以 明白这至少可以不小...(这意味着端口不匹配)但在这种情况下我也期望A(1)/ R1(1)同样不小如w 使用其他端口并在所有可能的方向上发生相同的事情“(刺激位置)由于这种不一致现在很清楚为什么我的算法不能正常工作。 是否有人可以解释R2(1)的行为或建议文档和有关此绝对参数的更多信息? 在ENA 5071C手册中,对这些测量只有一个简短的提及,只有很少的解释。还有另一个谜......为什么上述4个参数在扫描到扫描的相位不同? 我可以验证相对相位是否稳定(归一化后相位从扫描到扫描是稳定的),但不清楚为什么它们不稳定它们每个的绝对相位......“可能是一个架构问题”我 告诉自己.....先谢谢大家......关心Aleberto69 以上来自于谷歌翻译 以下为原文 I need help about Absolute Measurement with ENA 5071C. I refer to the direct access to the absolute R1(n) R2(n) ... A(n), B(n) measurements ( n being the port where the stimulus is applied) ( measurement menu). I tried to perform some " special" calibrations using that parameters and together with the math capability of the instrument, instead of using normal S parameters.. Unfortunately my algorithms didn't work as I would have expected. After investigations I found some inconsistency about what I suppose that the absolute measurement are. Namely 1) I used just port 1 and port 2 ( so that R1,A,R2,B are the involved measurements) 2) Connect port 1 and 2 with a 50 ohm cable ( 1m) and perform measurement at about 100 MHz ( few points to easily verify the results) 3)I measured R1(1), A(1),R2(1) and B(1) and normalize the 4 measurement to the value of the first one ( R1(1)) so that... -R1(1)/R1(1) is 1 (obvious) - as expected B(1)/R1(1) is in modulus close to 1 ( signal propagating through the cable and entering in port 2) - as expected A(1)/R1(1) is small ( representing the reflected wave coming back into port 1) supposing this means that port 2 is rather matched (load match) - as I didn't expect R2(1)/R1(1) result in modulus about 0.4 instead to be rather small ( as A(1)/R1(1)).. If R2(1)/R1(1) is proportional to the wave outgoing from the port2 ... I could understand that this could at least be not small...( It means that the port is not well matched) but in this case I also expect A(1)/R1(1) to be similarly not small as well .. Identical things happens using other ports and in all the possible Direction" ( stimulus position) With this inconsistency is it now clear why my algorithm is not working properly. Is there somebody that could explain the behavior of R2(1) or suggest documentation and more information about this absolute parameter? In the ENA 5071C manual there is just a brief mention to those measurement with very few explanation.. There is also another mystery... Why the above 4 parameters are not constant in phase from sweep to sweep? I could verify that the relative phase is stable ( after normalization the phase is stable from sweep to sweep), but is not clear why it isn't stable the absolute phase of each of them... "May be an architectural question" I told myself ..... Thank you in advance to everybody... Regards Aleberto69 |
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4个回答
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首先,确保R2在端口1处定义源。我不确定ENA的确切语法;
对于PNA,它是R2_1(但R2的默认值是R2_2,意味着在反向扫描期间测量R2)。 其次,请注意,如果端口1通过电缆连接到端口2,则信号进入端口2,从内部源反弹并重新反射出......这由R2_1测量。 所以,我们不希望它为零。 第三:没有网络分析仪测量具有相干扫描到扫描的aboslute相位,并且大多数不用于点对点 - 在SMC +相位模式下接受PNA或PNA-X。 在这种模式下,我们可以使源和接收器的相位同步,从而使单通道相位测量具有可重复性和相干性; 这意味着我们可以将它们校准出来。 比率测量在VNA中始终是一致的.... 以上来自于谷歌翻译 以下为原文 First, make sure R2 is define with source at port 1. I'm not sure the exact syntax for ENA; For PNA it is R2_1 (but the default for R2 is R2_2 meaning measure R2 during a reverse sweep). Second, note if port 1 is connected to port 2 through a cable then signal goes into port 2, bounces off the internal source and re-reflects out...this is measured by R2_1. So, we don't expect it to be zero. Third: No network analyzer measures aboslute phase with coherency sweep-to-sweep and most don't for point-to-point --EXCEPT the PNA or PNA-X in SMC+phase mode. In that mode, we have the abilty to synrchronize the phase of the source and receviever so that single-channel phase measurements are repeatable and coherent; which means we can calibrate them out. Ratio measurements are always coherent in a VNA.... |
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脑洞大赛9 发表于 2019-1-22 10:44 感谢Dr_joel的回答..我联系了安捷伦支持人员并了解问题所在。 我的算法不起作用,因为我下载和处理的数据是由仪器以数学方式操作的。 对于我的NA型号E5071C,绝对测量在数学上被校正,因为它具有更好的“虚拟方向性”,这是“帮助用户”。 不幸的是,当通过另一个端口提供激励时,这种校正不适用于每个端口的参考接收器,例如R2(1)由于这个原因,数据对于执行依赖于8项误差模型的外部校准是无用的(例如:TRL) 校准)。 对于在12项误差模型(例如TOSM_)上进行中继的所有校准都没有问题,因为它们不需要Rn(m)的信息。 要在依赖于8term错误模型的算法中执行数据的外部使用,必须在关闭“虚拟桥”和其他校正之前使Rn(m)数据变得有用。 http://ena.tm.agilent.com/e5071c/manuals/webhelp/eng/遗憾的是,手册中描述的程序不起作用,因为仪器无法识别这些SCPI命令。 我再次询问并等待安捷伦的支持。 如果您对此信息感兴趣,请与我们联系。 其他一些仪器如E5071C 14和20 GHz没有这个问题,因为它们没有数学校正,我想这也是高端仪器的情况,如PNA和PNAX。 看你的答案我反对“第二”点不同意,因为当源在端口1时,端口2的溢出关闭且端口匹配得很好......因此很少有功率应该返回并从R2测量(1 )。 我们在R2(1)中有一个重要的衡量标准,因为“上面提到的”方向性差,使得这个接收器对“进入”已经由B(1)接收机测量的端口的波敏感。 这也发生在端口1中,但仪器校正A(1)和R1(1)的测量值以改善差的方向性但不正确的R2(1)。 感谢干杯亚历山德罗的“第三”信息 以上来自于谷歌翻译 以下为原文 Thank you Dr_joel for your answer.. I contacted the Agilent support and understood what the problem was. My algorithm is not working because the data I downloaded and process are mathematically manipulated by the instrument. For my NA model, E5071C, the absolute measurement are mathematically corrected tho have a better "virtual directivity" and this is to "help the user". Unfortunately this correction doesn't apply to reference receiver of each port when the stimulus is provided through another port eg R2(1) For this reasons the data are not useful to perform external calibration that rely on the 8 term error model ( eg: TRL calibration). No problem for all those calibrations that relay on the 12 term error model ( eg TOSM_) because they no need the information of Rn(m). To perform external use of the data in algorithms that rely in the 8term error model you must before turn OFF the "vitual bridge" and other corrections so that also the Rn(m) data became useful. http://ena.tm.agilent.com/e5071c/manuals/webhelp/eng/ Unfortunately the procedure described in the manual doesn't work because the instrument doesn't recognize those SCPI commands. I'm again asking and waiting from Agilent support on that. Let me know if you are interested on this information. Some other instrument as E5071C 14 and 20 GHz have not this problem because they have no mathematical correction and I suppose this is also the case of high end level instrument as PNA and PNAX. Looking to your answer I anyway disagree with "second" point because when the source is on port 1 the surce in port 2 is off and the port is quite well matched..so that few power should come back and get measured from R2(1). We have a significant measure in R2(1) because of the "above mentioned" poor directivity that make this receiver sensitive also to the wave "entering" the port that is already measured by the B(1) receiver. This happens also in port 1 but the instrument correct the measurement of A(1) and R1(1) to improve the poor directivity but not correct R2(1). Thank you for the "third" information Cheers Alessandro |
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60user18 发表于 2019-1-22 10:54 > {quote:title = aleberto69写道:} {quote} >>看你的答案我反对不同意“第二”点,因为当源端口1在端口1时,端口2的端口关闭,端口匹配得很好。 那么很少有力量应该回来并从R2(1)进行测量。 我们在R2(1)中有一个重要的衡量标准,因为“上面提到的”方向性差,使得这个接收器对“进入”已经由B(1)接收机测量的端口的波敏感。 >这也发生在端口1中,但仪器校正A(1)和R1(1)的测量值以改善差的方向性但不正确的R2(1)。 >感谢您提供“第三”信息>干杯> Alessandro事实上你是对的,你可以有大的R2-因为端口2的负载匹配不佳(在R通道耦合点之后)或者因为差 端口2桥接电路中的方向性。 ENA虚拟网桥稳定,成本非常低,但它没有任何固有的方向性,因此R2信号很大。 我总是忘记对ENA的纵向纠正...... 以上来自于谷歌翻译 以下为原文 > {quote:title=aleberto69 wrote:}{quote} > > Looking to your answer I anyway disagree with "second" point because when the source is on port 1 the surce in port 2 is off and the port is quite well matched..so that few power should come back and get measured from R2(1). We have a significant measure in R2(1) because of the "above mentioned" poor directivity that make this receiver sensitive also to the wave "entering" the port that is already measured by the B(1) receiver. > This happens also in port 1 but the instrument correct the measurement of A(1) and R1(1) to improve the poor directivity but not correct R2(1). > Thank you for the "third" information > Cheers > Alessandro In fact you are correct, you can have large R2 -either- because of poor match in the load of the port 2 (after the R channel coupling point) or because of poor diretivity in the port 2 bridge circuit. the ENA virtual bridge is stable, and very low cost, but it doesn't have any inherent directivity, thus the large R2 signal. I always forget about the diretivity correction on the ENA... |
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脑洞大赛9 发表于 2019-1-22 11:14 “实际上你是正确的,你可以有大的R2 - 或者 - 因为端口2的负载(在R通道耦合点之后)匹配不佳或者因为端口2桥接电路中的导向性差。” 如果在端口2处遇到匹配不佳的情况,您还应该有一个返回到端口1的波形,这样您就应该看到一个重要的A(1)。 差的方向性是问题,我不明白为什么他们纠正他们没有照顾纠正R2(1)。 没关系..当您关闭任何数学校正时,我的算法运行良好,如网络帮助手册中所述。 我遇到的最后一个问题是建议的SCPI命令没有工作的事实......琐碎的问题......我只是有一个太旧的FW版本。 我今天非常满意....算法它是TRL校准的一种变体,我想用它来实现PA的负载拉动优化它让我实时知道在我改变时提供给DUT的负载阻抗(滑动) 短线和延迟线),无需断开DUT的负载。 您有一些信息要分享/您对负载拉动技术和PA设计感兴趣吗? 干杯亚历山德罗 以上来自于谷歌翻译 以下为原文 "In fact you are correct, you can have large R2 -either- because of poor match in the load of the port 2 (after the R channel coupling point) or because of poor diretivity in the port 2 bridge circuit." If it was the case of poor match at the port 2, you should also have a wave that travel back into port1 and so that you should have seen a significant A(1). The poor directivity is the problem and I do not really understand why they correct for them not taking care of correcting R2(1). Never mind .. my algorithm works very well when you turn off any mathematical correction as explained also in the web help manual. The last problem I had was related to the fact that the suggested SCPI command didn work... Trivial issue... I just had a too old FW version.!!! I'm very satisfied today....the algorithm It's a variant of TRL calibration that I want to use for load pull optimization of PA and It let me know in real time the load impedence provided to the DUT while I change it (sliding stub and delay line) without the need to disconnect the load from the DUT. Have you some information to share/are you interested on load pull technic and PA designing? Cheers Alessandro |
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