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最近,一位客户退回了一个电缆组件(15英尺长),声称电缆的阻抗为53欧姆。
我们使用N5225A PNA进行测试,使用85052D校准套件将其设置为10 MHz至8.01 GHz的低通TDR校准,801点。 见附图。 客户使用E5071C ENA,我们也有一个,所以我们以同样的方式设置它,并在情节上有显着差异。 主要区别在于,在PNA上,在小于t = 0的时间,绘图为-10 mU,在ENA上,它为+ 10mU,为电缆提供2欧姆的差异。 如果没有连接电缆或测量短电缆(6英寸),小于t = 0的反射为0(无偏移)。 为什么在t = 0以下存在参考偏移?为什么2个图不能给出相同的结果? 电缆的实际阻抗是多少? 谢谢,David Cable TDR N5225A.png45.8 KB 以上来自于谷歌翻译 以下为原文 Recently a customer returned a cable assembly (15 ft long) claiming the impedance of the cable was 53 Ohms. We tested using the N5225A PNA setting it up for a Low Pass TDR calibration from 10 MHz to 8.01 GHz with 801 points using an 85052D cal kit. See the attached plot. The customer uses a E5071C ENA and we have one also so we set it up the same way and got a significant difference in the plot. The main difference is that on the PNA, at time less than t=0, the plot is at -10 mU and on the ENA, it is +10mU giving a 2 Ohm difference for the cable. Without the cable attached or while measuring a short cable (6 inches), the reflection at less than t=0 is 0 (no offset). Why is there a reference offset below t=0 and why are the 2 plots not giving the same result? What is the actual impedance of the cable? Thanks, David 附件
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4个回答
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它可能值得附加Touchstone文件。
别人可以分析它们。 戴夫 以上来自于谷歌翻译 以下为原文 It's probably worth attaching the Touchstone files. The someone else can analyze them. Dave |
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我不再有电缆组装,我没有保存数据,只是图表。 以上来自于谷歌翻译 以下为原文 I no longer have the cable assembles and I did not save the data, just the plots. |
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T = 0时的偏移是由几个因素造成的,主要因素是用于校准系统的负载,以及使用的点间距。
因为不测量DC值,所以从低频RF数据推断出。 + -1欧姆的偏移表示约-40dB量级的回波损耗误差。 由于电缆远端的终端阻抗,以及时域变换(必须重复)如何导致数据的别名,您也可以出于复杂的原因获得此偏移。 避免这种情况的一种方法是在电缆之前使用短长度良好定义的阻抗线,例如航空公司,以独立于校准或参考负载设置参考。 从零线偏移是数学和设置的一种伪像。 如果电缆很长,你真的需要有足够的点,以允许足够长的跨度到达电缆末端以获得合理的结果。 对于15米长的电缆,可能有20纳秒的延迟,2倍的反射,意味着点间距大约为10 MHz,以获得良好的结果。 以上来自于谷歌翻译 以下为原文 The offset at T=0 is due to several factors, the main one being the load that is used to calibrate the system, and the point spacings used. Because a DC value is not measured, it is inferred from the low frequency RF data. The offset of +-1 ohm represents a return loss error on the order of about -40 dB. You can get this offset also for complicated reasons due to the termination impedance at the far end of the cable, and how the time domain transform (which must be repetitive) causes an alias of the data. One way to avoid this is to use a short length of well defined impedance line, such as an airline, before the cable to set the reference independent of the calibration or the reference load. The offset from zero line is a kind-of artifact of the math and settings. If the cable is very long, you really need to have enough points to allow a long enough span to go to the end of the cable to get a reasonable result. for a 15 meter cable, with maybe 20 nsec of delay, times 2 for reflection, means point spacings on the order of 10 MHz to get good results. |
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脑洞大赛9 发表于 2018-9-13 15:52 乔尔,谢谢你的解释。 我们有航空公司可以作为参考,并将使用它再次尝试测量。 我们还将确保根据您的指南获得足够数量的积分以获得所需的间距。 大卫 以上来自于谷歌翻译 以下为原文 Joel, Thank you for the very thorough explanation. We do have airlines we can use as a reference and will try the measurement again using it. We will also ensure that we have adequate number of points to get the spacing required based on your guidelines. David |
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