群延迟测量不确定度

解释这一点的最简单方法是增加孔径有点类似于增加平均值。
因此，增加平均值的方式相同，可以减少残留误差（提高精度），增加孔径尺寸也是一样的。



     以下为原文

  The simplest way to explain this is that increasing the aperture is somewhat similar to increasing averaging.  So the same way increasing averaging reduces the residual error (increased accuracy), increasing the aperture size does the same thing.

> {quote：title = jeff_philips001写道：} {quote}>我刚刚下载了不确定度计算器并绘制了三种不同Sii的群延迟精度。
我想看看不匹配的终端如何影响群延迟测量的准确性：它们似乎几乎没有差别。
>>>不确定度计算器绘制组延迟的精度（ns）与孔径（MHz）的关系曲线。
也许垂直轴应该被标记为“错误”而不是“精确度”。
我惊讶于误差随着频率孔径的增加而下降。
群延迟基本上是频率相位的导数。
只有当delta频率接近零时，导数才变得精确。
据我所知，当delta频率接近于零时，delta相位也变小，与相位测量中的误差相比可能不再大，但由于相邻相位测量值的差异，系统相位误差应该下降。
>我在本书的第292-295页详细介绍了这一点，但其实质是：在大多数工具中，群延迟不是通过取一个导数来计算的，而是通过采用delta相/ delta-的有限差分来计算
频率。
存在与相位测量相关的噪声，并且噪声随频率几乎恒定，因此Δ相中的噪声贡献几乎恒定并且是误差的主要来源。
但随着分母变小，噪声除以较小的数字会使延迟误差增大。
因此，apeture越小，由于噪声，误差越大。
另一方面，由于源和负载匹配导致的误差本质上是测量上的纹波，它不是随机的（如噪声），因此当apeture变小时，f1和f2处的误差（f2-f1为delta）
频率）变得几乎相同，并从差异中消失。
随着apeture变大，源和负载匹配误差增加。
因为这个erorr的影响是纹波，并且随着电缆越来越长，纹波变得越来越快，源和负载匹配变得不受影响的点很难确定，所以我们不指定群延迟但只给出不确定性的指示
分析。
我不记得确切的细节，但我认为我们假设源和负载匹配在大于100 MHz的频率下不相关，并且在小于10 MHz的频率下几乎相关。
一旦源和负载匹配误差完全不相关，它们就会变得恒定，delta频率超过，因此，更大的delta意味着更低的误差。
>我的直觉显然是不正确的。
我的基本问题是为什么群延迟误差会随着频率孔径的增加而下降？
我正在寻找一个简短，直观的答案，如果存在的话。
>>请不要在此花太多时间。
>只有3页。
>谢谢，杰夫



     以下为原文

  > {quote:title=jeff_philips001 wrote:}{quote}
> I just downloaded the Uncertainty Calculator and plotted Group Delay Accuracy for three different Sii.  I wanted to see how mismatched terminations affect the accuracy of a group delay measurement: they seemed to make little to no difference.
> 
> The Uncertainty Calculator plots Accuracy (ns) versus Aperture (MHz) for Group Delay.  Perhaps the vertical axis should've been labelled "Error" instead of "Accuracy."  I was surprised that the error declines with increasing frequency aperture.  Group delay is basically the derivative of phase with frequency.  The derivative only becomes exact when delta frequency approaches zero.  I understand that as delta frequency approaches zero, delta phase also becomes small and that it may no longer be large compared to the errors in the phase measurement, but the systemic phase error should fall out because the difference of adjacent phase measurements is being taken.
> 
This is covered in great detail on pages 292-295 of my book, but the essence is this:
In most instruments, the group delay is not computed by taking a derivative, but by taking a finite difference of delta-phase/delta-freq.
There is noise assoicated with the phase measurement, and the noise is nearly constant with frequency, so the noise contribution in the delta-phase is nearly constant and a major source of error. But as the denominator gets smaller, the noise divided by a smaller number grows makes the delay error grow. Thus the smaller the apeture the greater the error, due to noise.

On the other hand, the error due to source and load match essentially causes are ripple on the measurement, it is not random (like noise is) so as the apeture become small, the error at f1 and f2 (f2-f1 being the delta frequency) becomes almost identical, and disappears from the difference.  As the apeture becomes large, the source and load match error increases.  Because the effect of this erorr is ripple, and with longer cables the ripple becomes faster, the apeture at which the source and load match becomes uncorrleated is hard to determine, and so we do NOT specify group delay but give only an indication in the uncertainty analysis.  I don't recall the exact detail but I think we presume that the source and load match are not correlated at apetures greater than 100 MHz, and are nearly correlated at apetures less than 10 MHz.  Once the source and load match errors are fully uncorrelated, they become constant with delta frequency beyond than, so again larger delta means lower error.

> My intuition is apparently incorrect.  My basic question is why does Group Delay Error decline with increasing frequency aperture?  I'm looking for a brief, intuitive answer if one exists.
> 
> Please don't spend too much time on this.
> 
Just 3 pages is all.
> Thanks, Jeff