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我的业余无线电俱乐部有人问我是否可以测量某些同轴电缆的阻抗。
他是从业余无线电设备零售商那里购买的。 它以50欧姆的价格出售,但对我们来说看起来都像便宜的75欧姆电视哄。 就个人而言,我会把它扔进垃圾桶,但我说我会为他测量它。 我有几米的东西。 测量阻抗的最佳方法是什么? 我知道可以用VNA上的时域选项做事,但我不确定使用什么频率范围。 两个问题让我觉得从直流到光是不理想的1)我知道同轴电缆的阻抗高于高频值,所以收集低于10 MHz的数据会导致我给他错误的标称值。 2)我也知道这个人不会使用432 MHz以上的这个。 那么在低于10 MHz或高于432 MHz的频率范围内是否有任何一点席卷呢? 这些东西看起来相当便宜和可怕,这让我觉得以高频率测试它是不明智的。 如果它在10 GHz时为1欧姆或1000欧姆,那对他来说没有任何区别。 但我只有一个几米长的样本。 我随心所欲。 * 8720D带时域选项。 但8720D(50 MHz)的频率下限高于他感兴趣的频率。 * 8753ES。 这个频率范围(300 kHz到3 GHz)更接近于他使用这个东西的频率范围。 但是8753ES缺乏时域选项。 但当然我可以在VNA之外进行逆傅立叶变换 - 低成本VWNA的软件将从Touchstone文件中进行IFT。 戴夫 以上来自于谷歌翻译 以下为原文 Someone at my amateur radio club asked me if I could measure the impedance of some coax. which he purchased from a retailer of amateur radio equipment. It was sold as 50 Ohm, but looks like cheap 75 Ohm TV coax to us both. Personally I would chuck it in the bin, but I said I will measure it for him. I have a few metres of the stuff. What is the best way to measure the impedance of this? I know its possible to do things with the time-domain option on a VNA, but I'm not sure what frequency range to use. Two issues make me think that going from DC to light is not ideal 1) I know the impedance of coax rises above the high frequency value, so would collecting data at less than 10 MHz result in me giving him the wrong nominal value. 2) I also know that this guy will not use this above 432 MHz. So is there any point sweeping it over a frequency range below 10 MHz, or above 432 MHz? The fact the stuff looks pretty cheap and horrible, makes me think it is unwise to test it at a high frequency. If it is 1 Ohm or 1000 Ohms at 10 GHz, it is going to make no difference to him. But I only have a sample that is a few metres long. I have at my disposal. * 8720D with time domain option. But the lower frequency limit of the 8720D (50 MHz) is higher than the frequency of interest to him. * 8753ES. The frequency range of this (300 kHz to 3 GHz) more closely matches the frequency range he would use this stuff at. But the 8753ES lacks the time-domain option. But of course I could do the inverse Fourier Transform outside the VNA - the software for the low-cost VWNA will do the IFT from a Touchstone file. Dave |
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问题形成不良,或者更确切地说,问题有很多答案,即:一个同轴电缆的阻抗沿同轴电缆变化,同轴电缆的每个位置随频率变化;
因此,分配单个值意味着对结果进行一些平均。 时域变换将显示沿着电缆长度的变化平均阻抗(在用于生成时域的频率上的平均值)。 通过众所周知的公式转换为阻抗的S11回波损耗将显示平均阻抗(沿着电缆长度的一部分平均,该部分取决于电缆的S21损耗),前提是电缆端接在 阻抗负载与电缆匹配。 或者,您可以调整测量系统的源和负载阻抗(现代分析仪具有端口阻抗变换)以最小化S11响应,并将源和负载阻抗重新归类为电缆的阻抗(它是最匹配的阻抗) 电缆最小回损)。 以上来自于谷歌翻译 以下为原文 The question is poorly formed, or rather, the question has many answers, namely: The impedance of a piece of coax changes along the coax, and at each position in the coax, changes with frequency; thus, assigning a single value implies some averaging of results. The time domain transform will show you the variation average impedance (averaged across the frequency used to generate the time domain) along the lenght of the cable. The S11 return loss, converted to impedance through well known formulas, will show you the average impedance (averaged along a portion of the length of the cable, that portion being dependent on the S21 loss of the cable) provided that the cable is terminated in a impedance load matched to the cable. Alternatively, you can adjust the source and load impedance of your measurement system (modern analyzers have port impedance transformation) to minimize the S11 response, and reprort that source and load impedance as the impedance of the cable (It is the impedance which best matches the cable for minimum return loss). |
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