我有一个带有“16位”DAC集成arbs的MXG演示。
当我关闭波形缩放，ALC等时，我的幅度仅由I和Q值控制，该单元表现得好像DAC是14位。
通过电话和
电子邮件与安捷伦进行的讨论没有得到任何令人满意的解释。DAC是否为14位？
即如果我要看DAC
电路的原理图，它会是14位吗？
不是，“哦，插值后它的性能就像一个16位DAC”如果波形是直流怎么办？
如果我保持Q = 0并将I从-32767改为+32767，我会看到2 ^ 16个不同的输出吗？
因为这就是我所做的，看起来DAC是14位。
如果我使I = sin（x）和Q = cos（x）使我的振幅恒定，我可以改变x并获得16位的相位分辨率吗？
以上来自于谷歌翻译
以下为原文
I had a demo of an MXG with integrated arbs with "16 bit" DACs. When I turned off waveform scaling, ALC, etc, so that my amplitude is only controlled by the I and Q values, the unit behaved as if the DACs are 14 bits. Discussion with Agilent via phone and email did not result in any sa
tisfactory explanation.
Are the DACs 14 bits? i.e. if I were to see a schematic of the DAC circuit, would it be 14 bits? Not, "oh, after interpolation it performs like a 16 bit DAC" What if the waveforms are DC? Would I see 2^16 different outputs if I kept Q=0 and changed I from -32767 to +32767? Because that's what I did, and it looks like the DACs are 14 bits. If I made I=sin(x) and Q=cos(x) so that my amplitude is constant, can I vary x and get 16 bits of phase resolution?
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