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仪器:66311b,66319d,GPIB 82357B要求:每秒读取200个样本,总时间为1分钟。
程序脚本:inst.WriteString(“VOLT 3.7”); inst.WriteString(“OUTP ON”):inst.WriteString(“SENS:SWE:tiNT 0.005”):inst.WriteString(“SENS:SWE:POIN 2000”):字符串值; for(int i = 0; i {inst.WriteString(“MEAS:ARR:CURR?”):values + = inst.ReadString();} 0.005 * 2000 = 10秒,问题是“.WriteString(”MEAS: ARR:CURR?“)”和“ReadString()”用了15秒,我们在循环之间丢失了5秒。编辑:truer于2014年8月31日上午10:31 以上来自于谷歌翻译 以下为原文 Instrument: 66311b, 66319d, GPIB 82357B Requirment: Read 200 samples per second, total times is 1 minute. Program scripts: inst.WriteString("VOLT 3.7"); inst.WriteString("OUTP ON"): inst.WriteString("SENS:SWE:TINT 0.005"): inst.WriteString("SENS:SWE:POIN 2000"): string values; for(int i = 0; i< 6; i++) { inst.WriteString("MEAS:ARR:CURR?"): values += inst.ReadString(); } 0.005*2000=10 seconds, the problem is ".WriteString("MEAS:ARR:CURR?")" and "ReadString()" took 15 seconds,we lost 5 seconds between loops. Edited by: truer on Aug 31, 2014 10:31 AM |
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3个回答
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我知道缓冲区只能容纳4096个样本,因此一次读取时不能超过4096个点,但我们希望每秒监测200个样本,总时间为1分30秒,5分钟等。阅读响应必须
需要很多时间,专家请告诉我如何阅读整个1分钟和大于4096分的分数。 我们因为这个问题已经多次受到打扰,非常感谢,期待你的回复。编辑:更真实的2014年8月31日上午10:29 以上来自于谷歌翻译 以下为原文 I know the buffer will only hold 4096 samples, so it can not be greater than 4096 points in once read, but we want to monitor 200 samples per second and total times is 1 minutes,30 seconds, 5 minutes etc. Read response have to take many time, could experts please tell me how to read entire 1 minitues and points greater than 4096 points. we have been disturbed many time because of the problem, thanks a lot, Look forward to your reply. Edited by: truer on Aug 31, 2014 10:29 AM |
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h63542 发表于 2018-10-29 13:24 对不起,电源的采集系统设计为一次扫描一次。 无法将多个扫描组合在一起。 如果你进行多次扫描,你将在扫描之间失去一点时间。 对不起,我帮不了多忙。 以上来自于谷歌翻译 以下为原文 I am sorry, that power supply's acquisition system is designed to do one sweep at a time. There is no way to combine multiple sweeps together. IF you do multiple sweeps, you are going to lose a little bit of time between the sweeps. Sorry that I cannot help more. |
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我用Dlog软件14565B来测量电流波形,发现采样率是5ms对等样本。 那怎么办? 因为cpci命令花了大约20ms来交易到pc.Edited:qkhhyga于2015年8月3日凌晨4:18 以上来自于谷歌翻译 以下为原文 I used Dlog of software 14565B to measure current waveform, found that the sampling rate is 5ms peer sample. so How to do it? Becuse the cpci command took about 20ms to transact to pc. Edited by: qkhhyga on Aug 3, 2015 4:18 AM |
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