[问答]

89441A矢量信号分析仪+抖动测量问题

问答对人有帮助，内容完整，我也想知道答案 0 嗨我在另一个类别中发布了相同的消息，但我相信这是正确的。我对旧的89441A矢量信号分析仪有一些疑问。我知道这个设备已经老了，但希望有人可以帮我解决困扰我的事情。我使用带有RF部分的VSA来测量650MHz振荡器的相位噪声。有一个非常有用的安捷伦分步教程来执行相位噪声测量：http：//www.home.agilent.com/upload/cmc_upload/All/PhaseNoiseMeasurementsonthe89441A.pdf我想使用带有噪声减法的相位调制方法和要插入仪器的数学方程是F3 = PSD1-（D1 / K2）（第13页），其中K2 = 0.05 - 更多细节在教程中。但是，应用说明“Agilent PN 894400-2”第4页描述了相同的方法，相同的公式，其中K2 = 20。 http://cp.literature.agilent.com/litweb/pdf/5091-7193E.pdf还有其他人发现了这个问题吗？哪个是正确的常量？我试图在逻辑上找到解决方案，但不能。我的第二个问题是关于抖动。在使用相位解调的情况下，我拍摄了功率谱密度图。测量带宽为5MHz。然后，我将解调信号的时域用于抖动测量。考虑附加图像。使用带功率标记我得到rm ^ 2的均方根相位偏差。然后，rms时间抖动的计算就像对rms相位偏差进行平方并除以2 * pi * f一样简单，对吗？另外，为了得到p-p抖动，我应该将两个标记设置为最小和最大相位偏差，得到p-p相位偏差并除以2 * pi * f？困扰我的一种情况是计算抖动的另一种方法是在示波器中采用直方图。在过零时设置触发，示波器（Infiniium）需要许多周期到样本并将命中存储为直方图。在示波器中很容易观察到峰值到峰值作为直方图的终点。我遇到的问题是直方图所取的值因计算（p-p相）/ 2 * pi * f而大不相同。我弄错了吗？谢谢，我可以提供更多信息，如果你想。编辑：igouzouasis于2013年4月6日上午10:32 以上来自于谷歌翻译以下为原文 Hi I posted the same message in another category, but I believe this is the right one. I have some questions on the old 89441A Vector Signal Analyzer. I know that the device is old, but hopefully someone may give me a little help on the matters that trouble me. I used the VSA with the RF section to measure phase noise of an oscillator at 650MHz. There is a very useful Agilent step by step tutorial to perform phase noise measurement: http://www.home.agilent.com/uplo ... entsonthe89441A.pdf I want to use the Phase Modulation method with noise subtraction and the math equation to be inserted in the instrument is F3 = PSD1-(D1/K2) (page 13), where K2=0.05 - more details are in the tutorial. However, application note "Agilent PN 894400-2" page 4, describes the same method, same equation, where K2=20. http://cp.literature.agilent.com/litweb/pdf/5091-7193E.pdf Has anyone else identified this problem? Which is the right constant to use? I tried to find a solution logically, but couldn't. The second question I have is about jitter. Ather using Phase Demodulation, I took a plot of Power Spectral Density. Measurement bandwidth is 5MHz. I then took the time domain of the demodulated signal for a jitter measurement. Consider attached image. Using band power markers I get the rms phase deviation in rad^2. Then, a computation to rms time jitter is as simple as squaring the rms phase deviation and dividing by 2pif, right? Also, in order to get the p-p jitter, should i set two markers to min and max phase deviation, get the p-p phase deviation and divide by 2pif? A situation that troubles me is that another way of computing jitter is by taking a histogram in an oscilloscope. Setting trigger at zero crossing, the oscilloscope (an Infiniium) takes many period-to-period samples and store the hits as histogram. Peak to peak is easy to be observed in the oscilloscope as the end points of the histogram. The problem I have is that the value taken by the histogram is considerably different by the computation (p-p phase)/2pif. Am i getting something wrong? Thanks, I can provide more information if want to. Edited by: igouzouasis on Apr 6, 2013 10:32 AM 附件 650MJITT.tif179.8 KB 0
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答案对人有帮助，有参考价值 0 其中K2 = 0.05的PSD1-（D1 / K2）是正确的。 PSD1-（D1 * K2），其中K2 = 20也可以使用应用笔记中的公式不正确，我已通知可以纠正此问题的人。然后，我采用解调信号的时域进行抖动测量。考虑附加图像。使用波段功率标记，我得到rm ^ 2的均方根相位偏差。+你可以通过选择标记功能，波段功率标记，rms sqrt（pwr）以rms为单位显示波段功率。这将为您提供频段功率标记之间的平均值。 +然后，rms时间抖动的计算就像平方rms相位偏差并除以2 * pi * f一样简单，对吧？这似乎是正确的方法。如果选择rms sqrt（pwr），则不需要取平方根。 +另外，为了得到p-p抖动，我应该将两个标记设置为最小和最大相位偏差+ +，得到p-p相位偏差并除以2 * pi * f？+这似乎也是正确的方法。 +令我困扰的一个情况是另一种计算抖动的方法是在示波器中采用直方图。在过零时设置触发，示波器（Infiniium）需要许多周期到样本并将命中存储为直方图。在示波器中很容易观察到峰值到峰值作为直方图的终点。我的问题是直方图所取的值因计算（pp阶段）/2pif而大不相同。我怀疑你在这里看到的差异是范围会显示数据超过相对于您在89441A上显示的1或2 mS测量的更长时间。鉴于较长的时间，您可能会看到更高的p-p值。以上来自于谷歌翻译以下为原文 PSD1-(D1/K2) where K2=0.05 is correct. PSD1-(D1K2) where K2=20 could also be used The formula in application note is incorrect and I have notified the folks who can correct this. +I then took the time domain of the demodulated signal for a jitter measurement. Consider attached image. Using band power markers I get the rms phase deviation in rad^2.+ You can display the band power in units of rms by selecting Marker Function, band power markers, rms sqrt(pwr). This will provide you with the average of the values between the band power markers. +Then, a computation to rms time jitter is as simple as squaring the rms phase deviation and dividing by 2pif, right?+ This seems like the correct approach. Taking the square root will not be required if you select rms sqrt (pwr). +Also, in order to get the p-p jitter, should i set two markers to min and max phase deviation+ +, get the p-p phase deviation and divide by 2pif?+ This also seems like the correct approach. +A situation that troubles me is that another way of computing jitter is by taking a histogram in an oscilloscope. Setting trigger at zero crossing, the oscilloscope (an Infiniium) takes many period-to-period samples and store the hits as histogram. Peak to peak is easy to be observed in the oscilloscope as the end points of the histogram. The problem I have is that the value taken by the histogram is considerably different by the computation (p-p phase)/2pi*f.+ I suspect that the difference you are seeing here is that the scope will show you data over a much longer period of time relative to the 1 or 2 mS measurement you are showing on the 89441A. Given a longer period of time you are likely to see higher p-p values.

2019-7-22 07:43:21 评论举报宁晓蕾