内部衰减器踩踏输出无法使用

声明“幅度分辨率：10位”表示在任意波形发生器中使用的DAC是10位DAC。
“由于内部衰减器步进而在输出端无法获得全分辨率”这一说法是警告用户，如果函数发生器发出的信号不是使用整个DAC范围产生的，则分辨率不会是10位。
考虑以下。
假设DAC输出，输出放大器增益和函数发生器的衰减器是这样的，为了提供10Vp-p输出，整个DAC范围使用0dB衰减。
在这种情况下，10Vp-p信号将具有10位分辨率。
现在让我们将输出幅度更改为9Vp-p。
为了产生这个9Vp-p信号，电路必须将DAC输出衰减0.9倍。
如果没有这样的衰减器，则唯一的选择是对DAC进行编程，以输出小于0.9的信号。
该信号不会使用DAC的满量程范围，因此不具有10位分辨率。
事实上，函数发生器中可用的衰减器步骤有限，这意味着DAC输出必须经常缩放，并且函数发生器信号不能总是具有10位分辨率。



     以下为原文

  The statement “Amplitude Resolution: 10 bits “, indicates the DAC used in the arbitrary waveform generator is a 10bit DAC.
The statement “Full resolution is not available at output due to internal attenuator stepping “, is to warn users that if the signal coming out of the function generator was not generated using the full range of the DAC, the resolution will not be 10bits.

Consider the following. Let us say the DAC output, output amplifier gain and attenuators of the function generator are such that in order to provide a 10Vp-p output, the full DAC range is used with 0dB of attenuation. In this case, the 10Vp-p signal would have 10bits of resolution. Now let us change the output amplitude to 9Vp-p. To generate this 9Vp-p signal, the circuit must attenuate the DAC output by a factor of 0.9. If no such attenuator is available, then the only option is to program the DAC to output a signal smaller by a factor of 0.9. This signal will not be using the full scale range of the DAC and, thus, not have 10bits of resolution.  The fact that there are limited attenuator steps available in the function generator means the DAC output must often be scaled and the function generator signal cannot always have 10bits of resolution.

嗨，Fungible：你说的是有意义的，也同意我目前正在进行的实验。
谢谢！
现在，只是为了好奇，我想知道是否有一种方法可以预测，给定所需的幅度和下载的任意波形的值，将产生什么样的分辨率。
而且，更重要的是，预期的最坏情况解决方案是什么。
你觉得这有可能吗？
Franco Web：http：//www.flanguasco.org



     以下为原文

  Hi, Fungible:
what you says makes sense of it and also agrees with the experiments that I am currently running.
Thank you!

Now, just for the sake of curiosity, I wonder if there is a way to predict, given the requested amplitude and the values of the downloaded arbitrary waveform, what is going to be the resulting resolution.
And, even more important, what is the expected worst case resolution.

Do you think this is possible?

Franco

Web: http://www.flanguasco.org

嗨，Fungible：你说的是有意义的，也同意我目前正在进行的实验。
谢谢！
现在，只是为了好奇，我想知道是否有一种方法可以预测，给定所需的幅度和下载的任意波形的值，将产生什么样的分辨率。
而且，更重要的是，最糟糕的情况解决方案是什么。
你觉得这有可能吗？
Franco Web：http：//www.flanguasco.org



     以下为原文

  Hi, Fungible:
what you says makes sense of it and also agrees with the experiments that I am currently running.
Thank you!

Now, just for the sake of curiosity, I wonder if there is a way to predict, given the requested amplitude and the values of the downloaded arbitrary waveform, what is going to be the resulting resolution.
And, even more important, what is the worst case resolution.

Do you think this is possible?

Franco

Web: http://www.flanguasco.org

你究竟想做什么，你是否担心决议？
这是一种低成本的信号发生器。
还有更多的错误来源会导致您比步骤解决更麻烦。
幅度精度为+/- 2％，仅在1KHz时指定。
DC模式下的DC精度为+/- 1.5％+ / - 3mV。
也就是说，大多数示波器衰减器使用1-2-5步进。
不能保证这里使用的是什么，但如果是，那么最坏的情况可能是最佳情况的2.5倍。
请注意，即使您知道步长，也无法明确知道DAC切换衰减的位置。
我通过考虑具体案例获得了2.5倍的数字。
例如，如果你假设DAC使用全范围来获得200mV，你想要201 mV，那么它将切换到下一个范围500mV，并使用40.2％的DAC范围来获得201 mV（非常接近）
。
免责声明：为了获得更可靠的响应，您应该考虑致电当地的安捷伦威廉希尔官方网站 呼叫中心。
安捷伦论坛在“可用”的基础上进行监控，并不一定是解决威廉希尔官方网站 问题的最快方式。



     以下为原文

  Exactly what are your trying to do, that you are that concerned about the resolution?

This is a low-cost signal generator. There are many more sources of error that will cause you more trouble than the step resolution. The amplitude accuracy is +/- 2%, and only specified at 1KHz. The DC accuracy in DC mode is +/- 1.5% +/- 3mV.

That said, most scope attenuators use 1-2-5 stepping. There is no guarantee that's what's being used here, but if it is, then the worst case could be as much as 2.5x the best case. Note that there is no way to know, definitively, where the DAC switches attenuation, even if you know the step size.

I got the 2.5X number by considering a specific case. If you assume, for instance, that the DAC uses full range to get 200mV, and you want 201 mV, then it will switch to the next range, 500mV, and use 40.2% of the DAC range to get 201 mV (very approximately).

Al

Disclaimer: For more reliable response, you should consider calling your local Agilent Technical Call Center. The Agilent Forums are monitored on an "as available" basis, and aren't necessarily the fastest way to get technical questions answered.

嗨，Al：>你究竟想做什么，你是否关注决议？
我使用3000-X WaveGen任意波函数生成PRN激励信号，用于测量双端口网络（*）的矢量阻抗。
在测量电感时，我注意到，叠加到响应信号，很多小脉冲，然后，挖掘激励信号，我发现这些脉冲是由信号的阶梯形状（分辨率）引起的电感导数
影响。
> ...例如，如果你假设DAC使用全范围得到200mV，你想要201 mV，那么它将切换到下一个范围，500mV，并使用40.2％的DAC范围得到201
mV ......绝对正确：我还发现，对于要求波幅的非常小的变化，脉冲幅度突然发生了很大的变化。
例如，如果我将WaveGen幅度从1.25 Vpp更改为1.26 Vpp，则信号的阶梯步长从1.05 mV变为3.1 mV;
对我来说，这意味着，在第一种情况下，我有1190个可用电平（大约10位分辨率，考虑到测量误差），在第二种情况下，我有406级（9位？）。
有趣的是，如果我使用正弦（或斜坡）波而不是任意波，则会发生同样的情况：DAC和衰减器/放大器电路是相同的，不是吗？
正如我之前所说的那样，我的问题只是由我的病态好奇心所决定的（我喜欢在所有细节中知道这个野兽），因为这些虚假的脉冲不会让我感到烦恼，因为我可以很容易地将它们过滤掉并且仍能获得良好的阻抗
测量。
希望已经足够清楚，因为我的个人英语*分辨率*仅为4位:-)）Ciao并感谢您的关注。
Franco（*）我所做的是在PC程序中准备8192个所需波形的样本，值从-1.0到1.0，用（“WGEN：ARBitrary：DATA”，fArbWave）下载到示波器
命令然后用（“：WAVeform：DATA？”）命令读回激励和响应信号。
网址：http：//www.flanguasco.org



     以下为原文

  Hi, Al:

> Exactly what are your trying to do, that you are that concerned about the resolution?

I am using the 3000-X WaveGen arbitrary wave function to generate a PRN excitation signal for measuring the vector impedance of a two ports network (*).
While measuring inductances I noticed, superimposed to the response signal, a lot of small pulses and then, digging into the excitation signal, I discovered that these pulses were caused by the staircase shape of the signal (resolution) as a result of the inductance derivative effect.

> ... If you assume, for instance, that the DAC uses full range to get 200mV, and you want 201 mV,
> then it will switch to the next range, 500mV, and use 40.2% of the DAC range to get 201 mV ...

Absolutely right: I also found that the pulse amplitude was suddenly changing quite a lot for very small changes of the requested wave amplitude.
As an example, if I change the WaveGen amplitude from 1.25 Vpp to 1.26 Vpp the staircase steps of the signal change from 1.05 mV to 3.1 mV; this means, to me, that in the first case I have 1190 available levels (roughly 10 bits of resolution, considering the measurements errors) and in the second case 406 levels (9 bits?).

Interesting enough the same happens if I use the Sine (or Ramp) wave  instead of the Arbitrary: the DAC and the attenuator/amplifier circuits are the same, isn't it?. 

As I previously said, my question was only dictated by my pathological curiosity (I like to know this beast in all its intimate details) because these spurious pulses don't bother me to much since I can easily filter them out and still get good impedance measurements.

Hope to have been clear enough, because my personal English *resolution* is 4 bits only :-))

Ciao and thanks for the attention.

Franco

(*) What I do is to prepare in a PC program a 8192 samples of the desired waveform, with values going from -1.0 to 1.0, download them to the oscilloscope with the (":WGEN:ARBitrary:DATA ", fArbWave)  command and then read back the excitation and response signals with the (":WAVeform:DATA?")  command.

Web: http://www.flanguasco.org