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嗨,我使用安捷伦MSO7104A示波器测量了一些数据(两个输出信号之间的延迟)。
我将结果保存在USB(到csv文件),现在我想分析它们。 当我打开保存的文件时,我看到第二个,是否定的! 怎么会发生..任何人都可以帮助我如何显示数据:这里有一些数据:x.axis X1 X3 1秒Volt Volt 2 -500.00E-06 -18.75000E-03 + 0.0E + 00 3 -499.00E-06 + 12.50000E-03 -31.25000E-03 4 -498.00E-06 + 12.50000E-03 -31.25000E-03 5 -497.00E-06 -18.75000E-03 + 31.25000E-03 6 -496.00 E-06 + 43.75000E-03 + 0.0E + 00 7 -495.00E-06 -18.75000E-03 + 0.0E + 00 8 -494.00E-06 + 12.50000E-03 + 0.0E + 00 9 -493.00E- 06 + 12.50000E-03 + 0.0E + 00 10 -492.00E-06 + 12.50000E-03 -0.0E + 00 11 -491.00E-06 + 12.50000E-03 -31.25000E-03 12 -490.00E-06 + 12.50000E-03 + 31.25000E-03 13 -489.00E-06 + 12.50000E-03 -31.25000E-03 14 -488.00E-06 -18.75000E-03 + 31.25000E-03 15 -487.00E-06 -18.75000E -03 -0.0E + 00 16 -486.00E-06 + 12.50000E-03 -31.25000E-03 17 -485.00E-06 -18.75000E-03 + 31.25000E-03 18 -484.00E-06 + 12.50000E-03 -31.25000E-03 19 -483.00E-06 + 12.50000E-03 + 31.25000E-03 20 -482.00E-06 + 12.50000E-03 -31.25000E-03 21 -481.00E-06 + 43.75000E-03 +0.0 E + 00 BR,Mozhdeh 以上来自于谷歌翻译 以下为原文 Hi, I measured some data (delay between two output signal) with agilent MSO7104A Oscilloscope . I save the result on USB( to csv file), and now I want to analyse them. when I open the saved file, I saw the second, are negative!! how it can happen .. can anyone help me how I can show the data : Here is some of the data : x.axis X1 X3 1 second Volt Volt 2 -500.00E-06 -18.75000E-03 +0.0E+00 3 -499.00E-06 +12.50000E-03 -31.25000E-03 4 -498.00E-06 +12.50000E-03 -31.25000E-03 5 -497.00E-06 -18.75000E-03 +31.25000E-03 6 -496.00E-06 +43.75000E-03 +0.0E+00 7 -495.00E-06 -18.75000E-03 +0.0E+00 8 -494.00E-06 +12.50000E-03 +0.0E+00 9 -493.00E-06 +12.50000E-03 +0.0E+00 10 -492.00E-06 +12.50000E-03 -0.0E+00 11 -491.00E-06 +12.50000E-03 -31.25000E-03 12 -490.00E-06 +12.50000E-03 +31.25000E-03 13 -489.00E-06 +12.50000E-03 -31.25000E-03 14 -488.00E-06 -18.75000E-03 +31.25000E-03 15 -487.00E-06 -18.75000E-03 -0.0E+00 16 -486.00E-06 +12.50000E-03 -31.25000E-03 17 -485.00E-06 -18.75000E-03 +31.25000E-03 18 -484.00E-06 +12.50000E-03 -31.25000E-03 19 -483.00E-06 +12.50000E-03 +31.25000E-03 20 -482.00E-06 +12.50000E-03 -31.25000E-03 21 -481.00E-06 +43.75000E-03 +0.0E+00 BR, Mozhdeh |
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2个回答
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现代数字示波器可以捕获预触发数据。
触发时间定义为T = 0,因此在此之前的任何时间都是负的,即在触发事件之前。 以上来自于谷歌翻译 以下为原文 Modern digital scopes can capture pre-trigegr data. The trigger time is defined as T = 0, so any timne befoire that is negative, i.e. before the trigger event. |
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啊,是的。 凯文让我走在正确的轨道上......第一列是每个样本的时间。 第二列是通道1的电压(参见第一行中的X1?)第三列是通道3上的电压。触发器通常位于中心,时间= 0,每个时间戳位于左侧 这将是消极的。 如果您正在寻找从通道1到通道3的延迟,那么负时间戳不应该导致任何问题。 扫描通道1,寻找边缘。 如果找到它,请记下时间(T1)。 现在扫描通道3寻找边缘,并记下时间(T3)。 差异(T3-T1)是延迟。 边缘在哪里无关紧要,数学将始终有效。 如果有多个边缘,或T3有时可能在T1之前,您可能必须更改扫描算法,但数学将保持不变。 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技术呼叫中心。 安捷伦论坛在“可用”的基础上进行监控,并不一定是解决技术问题的最快方式。 以上来自于谷歌翻译 以下为原文 Aah, yes. Kevin put me on the right track... The first column is the time of each sample. The second column is the voltage of channel 1 (see the X1 in the first row?) The third column is the voltage on channel 3. The trigger is generally in the center, at Time = 0, and every time stamp to the left of that will be negative. If you are looking for the delay from Channel 1 to Channel 3, then negative time stamps shouldn't cause any problems. Scan through Channel 1, looking for an edge. If you find it, note the time (T1). Now scan through Channel 3 looking for an edge, and note the time (T3). The difference (T3-T1) is the delay. It doesn't matter where the edge is, the math will always work. If there are multiple edges, or T3 might sometimes be before T1, you may have to change your scanning algorithm, but the math will stay the same. 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. |
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