启用平均功能并设置平均计数但N5230A仅扫描一次

触发器立即命令对平均值一无所知，因此只需要一次扫描。
当您打开扫描平均并将平均因子设置为n时，您还需要扫描通道n次。
你可以通过发送INIT：IMM命令n次（需要在每次扫描之间进行串行轮询或* OPC？）或者你可以使用：SENSe：SWEep命令来做到这一点。
在你的例子中，平均计数设置为2，你需要先发送以下命令：SENSe：SWEep：GROups：COUNt 2然后发送：SENSe：SWEep：MODE GROups这样你只需串行轮询或做
* OPC？
一次完成平均所需的所有扫描。
预先警告，如果您正在进行慢速扫描和/或大量平均值的组合，并且您正在使用* OPC ?,则可能会遇到超时问题。
在这种情况下，最好使用串行轮询。



     以下为原文

  the trigger immediate command doesn't know anything about averaging, so it only takes one sweep.  when you turn on sweep averaging and set the averaging factor to n, then you need to also sweep the channel n times.  You can do that by sending the INIT:IMM command n times (need to do serial poll or *OPC? in between each sweep) or you can use the the :SENSe:SWEep commands.

in your example where the average count is set to 2, you'll need to first send the following command:

SENSe:SWEep:GROups:COUNt 2

then send:

SENSe:SWEep:MODE GROups

this way you only need to serial poll or do *OPC? once for all the sweeps required to complete the averaging.  be forewarned that if you are doing a combination of a slow sweep and/or high number of averages, and you are using *OPC?, you are likely to run into timeout issues.  in situations like this it is always best to use serial polling.

嗨，非常感谢你的回答。
我试图扫描n次进行平均。
我将平均计数设置为4，并在测量前将INIT：IMM +暂停4次。
仪器似乎扫描了4次，最终的图是平均波形。
然后它测量截止频率一次。
我将它与另一种方法进行比较，即我扫描一次并进行测量，重复3次并平均3次测量值。
我发现第二种方法比第一种方法更准确。
你知道为什么会这样吗？
谢谢！



     以下为原文

  Hi,
Thank you so much for your answers. I tried to sweep n times for the averaging.

I set the average count for 4 and make INIT:IMM + pause for 4 times before measurement. It seems that the instrument sweep 4 times and the final plot is the averaged waveform. Then it measures the cutoff frequency once. I compare it with another method which is that I sweep once and do a measurement, repeat it 3 times and average the 3 measured values. I found that the second method is more accurate then the first method. Do you know why it would be like this?

Thank you!

我个人不倾向于使用平均值，而是降低IF带宽以降低噪声。
但我看到平均值有一些优势 - 例如，室温的波动会被平均。
我很少使用平均值的主要原因是不知道在获取数据之前需要等待多少次扫描。
由于平均是一个无限脉冲响应滤波器，理论上一个是等待无限时间使数据正确。
也许你应该给数据更多的时间来解决，直到抓住它。
也许你应该等待徘徊戴夫



     以下为原文

  I personally don't tend to use averaging,  but instead reduce the IF bandwidth to lower noise. But I cab see some advantages in averaging - for example fluctuations in room temperature would be averaged. 

My main reason for rarely using averaging is not knowing how many sweeps to wait before 
taking the data. Since the averaging is an infinite impulse response filter,  in theory one was to wait an infinite time for the data to be right. Maybe you should give the data more time to settle until grabbing it. 


Maybe you should wait linger

Dave

> {quote：title = zhaoqy写道：} {quote}>嗨，>非常感谢您的回答。
我试图扫描n次进行平均。
>>我将平均计数设置为4并在测量前使INIT：IMM +暂停4次。
仪器似乎扫描了4次，最终的图是平均波形。
然后它测量截止频率一次。
我将它与另一种方法进行比较，即我扫描一次并进行测量，重复3次并平均3次测量值。
我发现第二种方法比第一种方法更准确。
你知道为什么会这样吗？
>>谢谢！
你怎么说“更准确”？
如果你只是平均4次，那么数据开始时一定不会那么嘈杂，所以我不确定平均数据和非平均数据之间是如何存在显着差异的，更不用说两种不同版本的平均值之间。
可能发生的事情是你没有停留足够长的时间来完成每次扫描，因此一个INIT：IMM正在中止前一个，因此你没有得到正确的平均结果。
您正在描述的两种技术之间唯一的另一个区别是，当您测量S参数（我假设您正在测量滤波器的S21），并且您打开平均值时，我们执行矢量平均值（我们平均值
复数），但是当您进行离线平均时，我假设您从测量中提取格式化的幅度数据以及程序中数据的平均值。
这将被视为标量平均值。
矢量平均可以产生一个明显不同的标量平均答案的唯一时间是矢量的一个分量（幅度或相位）比另一个更嘈杂。
对于滤波器S21测量，我想不出可能发生的物理解释。
我们只在测量是非比率接收器测量（如测量功率）时采用标量平均，在这种情况下，相位是完全随机的，但幅度不是。



     以下为原文

  > {quote:title=zhaoqy wrote:}{quote}
> Hi,
> Thank you so much for your answers. I tried to sweep n times for the averaging.
> 
> I set the average count for 4 and make INIT:IMM + pause for 4 times before measurement. It seems that the instrument sweep 4 times and the final plot is the averaged waveform. Then it measures the cutoff frequency once. I compare it with another method which is that I sweep once and do a measurement, repeat it 3 times and average the 3 measured values. I found that the second method is more accurate then the first method. Do you know why it would be like this?
> 
> Thank you!

how do you mean "much more accurate"?  if you are only averaging 4 times, the data must not be that noisy to begin with, so I am not sure how there could be a significant difference between averaged and unaveraged data, much less between two different versions of averaging.  what could be happening is that you are not pausing long enough for the each sweep to complete so one INIT:IMM is aborting the previous one and therefore you are not getting the correctly averaged result.  the only other difference between the two techniques that you are describing is that when you are measuring an S-Parameter (I am assuming you are measuring the S21 of the filter), and you turn on averaging, we perform vector averages (we average the complex numbers), but when you do the averaging offline, I assume that you pull the formatted magnitude data from the measurement and the average that data in your program.  this would be considered scalar averaging.  the only time where vector averaging could yield a significantly different answer to scalar averaging, is if one component of the vector (magnitude or phase) is much more noisy than the other.  for a filter S21 measurement, I can't think of physical explanation of how that could happen.

we employ scalar averaging only when the measurement is an un-ratio receiver measurement (like measuring power), in which case, the phase is completely random, but the magnitude is not.

daras，我似乎现在不能找到它，但我敢肯定，乔尔写道一旦*所有的平均功能*惠普和安捷伦的VNA使用无限脉冲响应滤波器，用New_Data_Displayed =（NEW_DATA / N）+ Old_displayed_data
*（N-1）/ N（我认为这将应用于原始复杂数据，而不是经过处理以找到角度和dB的东西）这在旧的VNA上非常有效且实用。
如果相反，平均算法更明显：New_Data_Displayed =（data + data_-1 + data_-2 + data_-3 ....）/ N它将要求所有先前的数据都已存储在内存中，这将具有
在旧机器上不切实际。
我的8720D平均可以做到1000次，但我确信它不能存储1000套S参数。
即使要删除繁琐的适配器，也需要将一组S参数存储到软盘中，因为我认为机器确实有足够的内存来同时存储两个集合，因此1000个完全不存在。
我不是数学家，所以我不知道有多少样本这个IIR滤波器需要稳定到95％或99％，但在我看来，使用的New_Data_Displayed =（NEW_DATA / N）+ Old_displayed_data *（
N-1）/ N意味着VNA需要大大超过4次扫描才能得到平均值的合理估计值，平均值设置为4.这就是我有点不愿意使用平均值的原因，并且更喜欢缩小
IF带宽，因为我不确定等待收集数据需要多长时间。
也许原始海报可以将平均值设置为4，但在从VNA收集平均数据之前执行20次扫描。
我怀疑通过采取一些数据的平均值，可能会更好地与我认为他/她正在做的事情一致。
使用平均而不是窄的IF带宽会降低室温小波动的影响，因此我可以看到它的一些优势。
DaveEdited：drkirkby于2014年9月20日上午8:19



     以下为原文

  daras, 
I can't seem to find it now, but I'm sure that Joel wrote once that the averaging function of *all* HP and Agilent VNAs used an infinite impulse response filter, with 

New_Data_Displayed=(New_Data/N)+Old_displayed_data*(N-1)/N

(I would assume this would be applied to the raw complex data, rather than something that has been processed to find angles and dBs)

This is very memory efficient and practical on old VNAs. 

If instead the averaging algorithm was the more obvious:

New_Data_Displayed=(data+data_-1+data_-2+data_-3....)/N

it would have required that all previous data had been stored in memory, which would have been impractical on older machines. My 8720D can do at least 1000 averages, but I'm sure it could not store 1000 sets of S-parameters. Even to do the tedious adapter removal, a set of S-parameters needs to be stored to floppy disk, as I assume the machine does have enough memory to store two sets at the same time, so 1000 would be totally out. 

I'm not a mathematician, so I don't know what how many samples this IIR filter requires to settle to 95% or 99%, but it seems to me that the use of 

New_Data_Displayed=(New_Data/N)+Old_displayed_data*(N-1)/N

means the VNA needs considerably more than 4 sweeps to get a reasonable estimate of an average, with averaging set to 4. 

This is the reason I'm a bit reluctant to use averaging, and prefer to just narrow the IF bandwidth, as I'm not really sure how long to wait to collect the data. 

Perhaps the original poster could set the averaging to 4, but perform 20 sweeps before collecting the averaged data from the VNA. I suspect that might agree somewhat better with what I assume he/she is doing, by taking a mean of some data. 

Using averaging, rather than narrow IF bandwidth, would tend to reduce the effect of small fluctuations in room temperature, so I can see some advantage in it. 

Dave

Edited by: drkirkby on Sep 20, 2014 8:19 AM

是的 - 实际平均值即使在新模型中描述也是如此。
我们没有维护所有先前扫描的数据集。
给定所有可能的通道和跟踪数量，即使对于具有基于Windows的新CPU的VNA，也会造成巨大的内存负担。
通常，对于s参数（或其他比率）测量，平均和带宽减少产生相同的效果。
因此，在这种特殊情况下，海报可以简单地减少IFBW并进行单次扫描并完成。
但是他/她问了一个关于平均和触发的问题，因此问我的答案。
有一些特殊情况涉及非比率测量，其中带宽减少实际上降低了测量。
在这些情况下，平均是降低噪音的唯一可行方法。



     以下为原文

  Yes - the actual averaging is done as you describe it even in the new models.  we do not maintain a dataset for all the previous sweeps.  given all the possible number of channels and traces, it would be a tremendous memory burden, even for VNAs with new Windows based CPUs.

in general, for s-parameter (or other ratio) measurements, averaging and bandwidth reduction yield the same effect.  so in this particular case, the poster can simply reduce the IFBW and take a single sweep and be done.  but he/she asked a question about averaging and triggering and hence my answer.  there are some special cases dealing with non-ratio measurements where bandwidth reduction actually degrades the measurement.  in those situations, averaging is the only viable way to reduce noise.

我试图降低此测量的IF带宽。
完成此测量需要更多时间，但结果没有太大改善。
之后，我尝试运行此测试四到五次并平均测量值，结果改善了很多。
所以这就是我想在NWA中使用平均功能的原因。
但我错了的是，NWA中的内置平均值是一个垂直平均值，如果我只是平均截止频率值，那么它实际上是不同的。
我有一个Spec来比较理想值和测量值。
对于相同的设置，有时我可以衡量真正不合理的价值，但有时它是好的。
这是随机的，我无法找到关于这些失败的任何结论。
（例如，系统在某些特定条件下始终失败）。
可能这个系统或探头精度存在一些问题。
但我仍然需要了解它。
非常感谢您的评论。
它帮助我更清楚地了解这个功能。



     以下为原文

  I tried to reduce the IF bandwidth for this measurement. It takes more time to finish this measurement but the results did not improved much. After that, I tried to run this test for four or five times and average the measured value, the results improved a lot. So that's why I want to use the average function in the NWA. But what I got wrong is that the build-in average in NWA is a vertor average which is really different if I just average the cut-off frequency values. I have a Spec to compare the ideal value and a measured value. For a same settings, sometimes I could measure really unreasonable value but sometimes it is good. It is random and I can not find any conclusion about these failures. (For example, the system is always failed at some specific conditions). Maybe there is some problems with this system or the probe accuracy. But I still need to find out about it. Really thanks for your comments. It helped me get a more clearly mind of this function.

实际上，您发现使用扫描平均值而不是IF BW减少或点平均值的关键原因之一。
IF BW减少消除了IF BW采集的时间孔径内的固有噪声，例如1 kHz IFBW的1毫秒。
如果您遇到随机噪声具有长时间常数变化的情况（例如您的探测情况），则扫描到扫描平均将消除这些噪声影响。
它类似于激光的相干长度。
您必须对变化进行平均（或采用IFBW数据）足够长的时间，以使变化无法平均以减少其影响。



     以下为原文

  In fact, you found one of the key reasons to use sweep average rather than IF BW reduction or point average.

IF BW reduction removes inherent noise over the time aperture of the IF BW acquision, for example, 1 msec for a 1 kHz IFBW.  IF you have a situation where the random noise has a variation with a long time constant (such as your probing case) then sweep to sweep averaging will remove these noise effects.  It is similar to the coherence length of lasers.  You have to average (or take IFBW data) long enough for the variation to become incorherent for averaging to reduce its effect.

> {quote：title = zhaoqy写道：} {quote}>嗨，>目前我正在研究一个使用N5230A测试低通滤波器截止频率的项目。
由于探测器有时测量不稳定，我想使用平均功能扫描2-3次并得到平均频率值。
在仔细阅读您的帖子后，我相信您错误地使用了VNA。
均衡和降低IF带宽均可降低随机噪声，这在测量高衰减值时尤为重要。
这两种技术都不能纠正不稳定的探针。
我认为你正在以错误的方式解决这个问题。
如果计算滤波器的3 dB点，应该有足够的信号，不需要降噪技术，如平均或降低IF带宽。
我不想劫持你的步伐，所以创造了另一个：HP / Agilent VNA中的平均功能的响应时间，但我很确定你需要找到一种方法，使你的测量稳定，或忽略不良数据点，而不是
而不是试图平均好坏点。
也许如果您知道截止频率大约为1 GHz，则忽略任何显示不同值的数据集，而不是尝试对它们进行平均。
在不知道你的探测器是什么的情况下，很难提出任何建议，但我认为我说你错误地使用VNA中的平均功能是正确的，这可能解释了为什么它不像你想要的那样表现。
只是我的2p。
戴夫



     以下为原文

  > {quote:title=zhaoqy wrote:}{quote}
> Hi,
> Currently I am working on a project which is using N5230A to test a cutoff frequency of a low pass filter. Since the measurements sometimes are not stable because of the probes, I want to use the average function to sweep 2-3 times and get the averaged frequency value. 
After reading your post more carefully, I believe you are mis-using the VNA. 
Both averaging and reducing the IF bandwidth reduce the random noise, which is particularly important if measuring high values of attenuation. Neither technique will correct for unstable probes. I think you are tacking the problem the wrong way.
If computing the 3 dB point of a filter, there should be plenty of signal and no need for noise reduction techniques like averaging or reducing IF bandwidth. 
I did not wish to hijack your tread, so created another:
Response time of the averaging function in HP/Agilent VNAs

but I'm pretty sure you need to find a way of either making your measurement stable, or ignoring bad data points, rather than trying to average good and bad points. Maybe if you know the cutoff frequency is around 1 GHz, you ignore any data sets which show a significantly different value, rather than trying to average them. 

Without knowing what your probes are, it is hard to suggest anything, but I think I am right in saying you are really mis-using the averaging function in the VNA, which probably explains why it is not behaving as you want it to. 

Just my 2p. 

Dave

是。
我想我终于对这个功能有了清晰的认识。
我还在研究另一篇论文，讨论如何改善动态范围。
使用平均和降低IF带宽是改善动态范围的两种方法。
如果我们使用这两个功能，噪声基底将会减少。
以前当我测量不好时，我试图降低IF带宽，结果没有改善。
所以我认为这意味着如果我使用平均的内置函数，我可能会得到相同的结果，因为这些不良测量与噪声无关。
我应该尝试一种新的方法来分析Excel数据模板中的数据，而不仅仅是尝试启用平均功能。



     以下为原文

  Yes. I think I finally got a clear idea about this function. I also look into another paper which is talking about how to improve the dynamic range. Using averaging and decreasing IF bandwidth are two methods to improve dynamic range. The noise floor will be decreased if we using these two function. Previously when I got bad measurements, I tried to decrease the IF bandwidth and the result did not improve. So I think that means if I use the build-in function of averaging, it is likely that I will get the same result because these bad measurements are not related to the noise. I should try to get a new way to analysis the data in Excel data template, not just try to enable the average function.

是平均和IF带宽减少都降低了测量的本底噪声和跟踪噪声，但是如果你有某种瞬态（时变）信号，那么两者都没有帮助。
因此，如果您正在使用探头进行测量并且正在移动探头并持续扫描，那么平均可能最终会在非常多次扫描之后并且仅在您停止移动探头之后接近正确的答案。



     以下为原文

  yes averaging and IF Bandwidth reduction both reduce the noise floor and the trace noise of the measurement, but if you have some kind of transient (time varying) signal, then neither is going to help.  so if you are making a measurement with a probe and you are moving the probe around and continuously sweeping, then averaging might eventually give you close to the right answer after very many sweeps and only after you stop moving the probe.

是平均和IF带宽减少都降低了测量的本底噪声和跟踪噪声，但是如果你有某种瞬态（时变）信号，那么两者都没有帮助。
因此，如果您正在使用探头进行测量并且正在移动探头并持续扫描，那么平均可能最终会在非常多次扫描之后并且仅在您停止移动探头之后接近正确的答案。



     以下为原文

  yes averaging and IF Bandwidth reduction both reduce the noise floor and the trace noise of the measurement, but if you have some kind of transient (time varying) signal, then neither is going to help.  so if you are making a measurement with a probe and you are moving the probe around and continuously sweeping, then averaging might eventually give you close to the right answer after very many sweeps and only after you stop moving the probe.

是平均和IF带宽减少都降低了测量的本底噪声和跟踪噪声，但是如果你有某种瞬态（时变）信号，那么两者都没有帮助。
因此，如果您正在使用探头进行测量并且正在移动探头并持续扫描，那么平均可能最终会在非常多次扫描之后并且仅在您停止移动探头之后接近正确的答案。



     以下为原文

  yes averaging and IF Bandwidth reduction both reduce the noise floor and the trace noise of the measurement, but if you have some kind of transient (time varying) signal, then neither is going to help.  so if you are making a measurement with a probe and you are moving the probe around and continuously sweeping, then averaging might eventually give you close to the right answer after very many sweeps and only after you stop moving the probe.