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嘿那里,我正在使用DSO-X 3034A。
我在正常模式下测量信号,垂直分辨率为200mV / div。 使用此分辨率,测量信号(输入信号)不会剪辑(不超过显示)。 噪声非常高(峰值到峰值约1000mV),我想测量的信号(特定频率的正弦)约为10mV(我使用数学函数FFT并检测我所需频率的电平)。 在平均模式(2048平均值)中,只有我所需频率的信号(正弦)可见(无噪声),幅度约为10mV。 为了更好地查看/分辨此信号,我将垂直分辨率调整为20mV / div。 我的问题是:平均模式下垂直分辨率的变化是否真的不会导致输入信号出现任何削波问题? 我是否会在此配置中导致任何错误? 以上来自于谷歌翻译 以下为原文 Hey there, i am using a DSO-X 3034A. I measure a signal in normal mode with a vertical resolution of 200mV/div. With this resolution the measured signal (inputsignal) doesn't clip (doesn't exceed the display). The noise is very high (about 1000mV peak to peak) and the signal (a sine at a specific frequency) i want to measure is about 10mV (I use the math function FFT and detect the level at my desired frequency). In averaging mode (2048 averages), only the signal (a sine) at my desired frequency is visible (no noise) and the amplitude is about 10mV. To get a better view/resolution of this signal, i adjust the vertical resolution to 20mV/div. My question is: Does the change of the vertical resolution in averaging mode really not cause any clipping problems of my inputsignal? Do i cause any errors in this configuration? |
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4个回答
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哦,是的,你得到了各种各样的剪辑。
您必须考虑“采集”和“显示”采集处于“正常”模式,然后进行一些数学运算,显示屏和仅显示屏处于“平均”模式。 当您从200mV / div切换到20mV / div时,您将更改示波器前端的衰减器。 如果“正常”数据被剪切,那么剪切的数据将被传递到平均函数中。 如果您知道您的“真实”信号很小,并且您知道噪声是真正随机的,没有DC分量,那么您应该能够获得良好的结果。 这假设在正方向和负方向上正在剪切相同数量的信号。 使用良好的滤波器来消除大部分宽带噪声可能会更好。 作为旁注,随着时间的推移,可能会出现示波器前端的问题。 示波器的最大值假定您处于高衰减模式,如1V / div。 当您进入低衰减模式时,如此,您将前端过度驱动10倍或更多,这可能会导致问题。 这不是好习惯。 人 以上来自于谷歌翻译 以下为原文 Oh, yeah, you're getting all kinds of clipping. You have to think about the 'acquisition' and the 'display' The acquisition is in 'normal' mode, then some math is done, and the display, AND ONLY THE DISPLAY, is in 'averaged' mode. When you switch from 200mV/div to 20mV/div, you are changing the attenuators in the front end of the scope. If the 'normal' data is clipped, then that clipped data is what's getting passed into the averaging function. If you KNOW that your 'real' signal is small, and you KNOW that you noise is truly random, with no DC component, then you should be able to get good results. This assumes that the same amount of signal is being clipped in both the positive and negative direction. You would probably be better off with a good filter to remove most of the wideband noise. As a side note, there is a possibility of having problems with the front end of the scope over time. The maximum values for the scope assume that you are in a high attenuation mode, like 1V/div. When you go to a low attenuation mode, like this, you are overdriving the front end by a factor of 10 or more, and that may cause problems. It's not good practice. Al |
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nvywyerwer 发表于 2019-4-29 16:24 谢谢你的回复这是一个坏消息。 如果我以200mV / div运行示波器并且所需信号大约为10mV(事实上,当我想测量背景信号时它可以更低,以及它可以升高到1000mV),那么量化误差是巨大的。 1位表示6.25mV ...(200mV * 8/2 ^ 8Bit)平均> = 256次,12位垂直分辨率仍然有一个令人不快的错误。 1位代表0.39mV。 因此,如果我想以4位或更多位开始测量,则信号必须大于1.56mV。 在前面已经有一个放大器(G = 1000),带通带行为。 但是,如果有人通过测量麦克风走路或将笔放在桌子上,则会产生很高的噪音:P。 然后我得到短峰值电平,我的所需信号仍然覆盖。 如果它剪辑,它不是一件好事。 但实际上,在使用平均模式时,我的显示器仍然有一个良好的信号。 以上来自于谷歌翻译 以下为原文 Thanks for your reply That's some bad news. If i run the oscilloscope in 200mV/div and the desired signal is about 10mV (well in fact it can be lower when i want to measure the backgroundsignal, aswell as it can rise up to 1000mV), then the quantization error is huge. 1 Bit represents 6.25mV... (200mV*8/2^8Bit) Averaging >= 256 times and 12 Bit vertical resolution still has an unpleasant error. 1 Bit represents 0.39mV. So if i want to start measuring with 4 or more bits, the signal has to be greater than 1.56mV. There is already an amplifier (G=1000), with bandpass behavior, preceding. But the high noise levels are occuring if someone is walking by the measuring-microphone or dropping his pen on the table :P. Then i get short peak-levels where my desired signal is still overlaid. If it clips, its not a good thing. But in practice i still have a good signal on my display when using averaging mode. |
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uiuisky 发表于 2019-4-29 16:32 >如果我以200mV / div运行示波器并且所需信号大约为10mV(事实上,当我想测量背景信号时它可以更低,以及它可以升高到1000mV),那么量化误差是巨大的。 1位表示6.25mV ...(200mV * 8/2 ^ 8Bit)>平均> = 256次,12位垂直分辨率仍然有一个令人不快的错误。 1位代表0.39mV。 因此,如果我想以4位或更多位开始测量,则信号必须大于1.56mV。 几乎是真的,但不可避免...... 4个量化等级是2比特,我相信...>已经有一个放大器(G = 1000),前面有带通行为。 但是,如果有人通过测量麦克风走路或将笔放在桌子上,则会产生很高的噪音:P。 然后我得到短峰值电平,我的所需信号仍然覆盖。 如果它剪辑,它不是一件好事。 但实际上,在使用平均模式时,我的显示器仍然有一个良好的信号。 存在大噪声时捕获小信号的标准问题。 通常,您应该尽一切努力通过在安静的环境中工作或通过过滤来降低噪音。 如果这些都不适合您,那么平均可能会有所帮助,如果您有重复的波形。 只要噪声与信号具有相同的平均DC偏移,平均值就会产生良好的结果,但正如我所说,你会冒着对示波器前端施加压力的风险,这将导致过早失效。 我总是尽一切努力小心对待我的乐器。 人 以上来自于谷歌翻译 以下为原文 > If i run the oscilloscope in 200mV/div and the desired signal is about 10mV (well in fact it can be lower when i want to measure the backgroundsignal, aswell as it can rise up to 1000mV), then the quantization error is huge. 1 Bit represents 6.25mV... (200mV*8/2^8Bit) > Averaging >= 256 times and 12 Bit vertical resolution still has an unpleasant error. 1 Bit represents 0.39mV. So if i want to start measuring with 4 or more bits, the signal has to be greater than 1.56mV. Almost true, but unavoidable... 4 quantization levels is 2 bits, I believe... > There is already an amplifier (G=1000), with bandpass behavior, preceding. But the high noise levels are occuring if someone is walking by the measuring-microphone or dropping his pen on the table :P. Then i get short peak-levels where my desired signal is still overlaid. If it clips, its not a good thing. But in practice i still have a good signal on my display when using averaging mode. A standard problem of capturing small signals in the presence of big noise. Generally you should make every attempt to reduce the noise, either by working in a quiet environment, or by filtering. If neither of those works for you, then averaging might be helpful, IF you have a repetitive waveform. As long as the noise has the same Average DC offset as the signal, averaging should produce good results, but, as I said, you run the risk of stressing the front end of the scope, which will lead to premature failure. I always make every attempt to treat my instruments carefully. Al |
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nvywyerwer 发表于 2019-4-29 16:39 是4个量化级别,而不是位。 我的错。 以上来自于谷歌翻译 以下为原文 Yes 4 quantization levels, not bits. My fault. |
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