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Interference Identification Method of AC Signal
DOI: 10.12677/SG.2021.116040, PDF , HTML, XML, 下载: 255  浏览: 405

Abstract: Full-cycle Fourier algorithm is usually used to analyze discrete AC sampling data in microcomputer protection. When there is interference data in AC signal, it will affect phasor analysis and get wrong system measurement results. In this paper, a method of generating several check sequences using only the original AC sampling sequence is proposed, and based on this, a method of identifying the interference data in AC signal is proposed. The feasibility and validity of the criterion are verified by simulation results. This method can also be applied to the measurement of electrical volume.

1. 引言

2. 大数对全周傅氏算法的影响

3. 单个大数识别算法

3.1. 基本思路

Figure 1. Sampling data set, calibration data set schematic diagram

Table 1. Results of full-cycle Fourier algorithm for primitive, odd and even sequences

3.2. 判据的门槛值

3.2.1. 傅氏算法截断误差分析

$|{E}_{Tra}|=|{\int }_{a}^{b}f\left(x\right)-Tra\left(N\right)|\le \frac{b-a}{12}{\left(\frac{b-a}{N}\right)}^{2}M$ (1)

${a}_{1}=\frac{2}{T}{\int }_{0}^{T}x\left(t\right)\mathrm{cos}\left(\omega t\right)\text{d}t$ (2)

${b}_{1}=\frac{2}{T}{\int }_{0}^{T}x\left(t\right)\mathrm{sin}\left(\omega t\right)\text{d}t$ (3)

${X}_{1}=\sqrt{\frac{{a}_{1}^{2}+{b}_{1}^{2}}{2}}$ (4)

${a}_{1N}=\frac{1}{N}\left[2{\sum }_{k=1}^{N}{x}_{k}\mathrm{sin}\left(k\frac{2\pi }{N}\right)\right]$ (5)

${b}_{1N}=\frac{1}{N}\left[{x}_{0}+2{\sum }_{k=1}^{N-1}{x}_{k}\mathrm{cos}\left(k\frac{2\pi }{N}\right)+{x}_{N}\right]$ (6)

${x}_{k}=A\mathrm{sin}\left(\omega {t}_{k}+\phi \right)=A\mathrm{sin}\left(\frac{2\pi }{T}\frac{k}{N}T+\phi \right)=A\mathrm{sin}\left(\frac{2\pi }{N}k+\phi \right)$ (7)

${x}_{k}$ 代入式(5)，并对其进行化简，可以得到：

$\begin{array}{c}{a}_{1N}=\frac{1}{N}\left[2{\sum }_{k=1}^{N}{x}_{k}\mathrm{sin}\left(k\frac{2\pi }{N}\right)\right]\\ =\frac{A}{N}\left[{\sum }_{k=1}^{N}2\mathrm{sin}\left(k\frac{2\pi }{N}+\phi \right)\mathrm{sin}\left(k\frac{2\pi }{N}\right)\right]\\ =\frac{A}{N}\left[{\sum }_{k=1}^{N}\mathrm{cos}\left(\phi \right)-\mathrm{cos}\left(k\frac{4\pi }{N}+\phi \right)\right]\\ =A\mathrm{cos}\left(\phi \right)-\frac{A}{2\pi }{\sum }_{k=1}^{N}\frac{2\pi }{N}\mathrm{cos}\left(2\frac{2\pi }{N}k+\phi \right)\end{array}$ (8)

${\int }_{0}^{2\pi }\mathrm{cos}\left(2\theta +\phi \right)\text{d}\theta =\underset{N\to \infty }{\mathrm{lim}}{\sum }_{k=1}^{N}\frac{2\pi }{N}\mathrm{cos}\left(2\frac{2\pi }{N}k+\phi \right)=0$ (9)

$\begin{array}{c}|{E}_{g}|=|{\int }_{0}^{2\pi }\mathrm{cos}\left(2\theta +\phi \right)\text{d}\theta -{\sum }_{k=1}^{N}\frac{2\pi }{N}\mathrm{cos}\left(2\frac{2\pi }{N}k+\phi \right)|\\ =|-{\sum }_{k=1}^{N}\frac{2\pi }{N}\mathrm{cos}\left(2\frac{2\pi }{N}k+\phi \right)|\\ \le \frac{2\pi }{12}{\left(\frac{2\pi }{N}\right)}^{2}4=\frac{2\pi }{3}{\left(\frac{2\pi }{N}\right)}^{2}\end{array}$ (10)

$A\mathrm{cos}\left(\phi \right)-\frac{A}{3}{\left(\frac{2\pi }{N}\right)}^{2}\le {a}_{1}\le A\mathrm{cos}\left(\phi \right)+\frac{A}{3}{\left(\frac{2\pi }{N}\right)}^{2}$ (11)

$A\mathrm{sin}\left(\phi \right)-\frac{A}{3}{\left(\frac{2\pi }{N}\right)}^{2}\le {b}_{1}\le A\mathrm{sin}\left(\phi \right)+\frac{A}{3}{\left(\frac{2\pi }{N}\right)}^{2}$ (12)

$\left\{\begin{array}{l}{X}_{1}=\sqrt{\frac{{a}_{1}^{2}+{b}_{1}^{2}}{2}}\le \frac{1}{\sqrt{2}}\sqrt{{\left(A\mathrm{cos}\left(\phi \right)+\frac{A}{3}{\left(\frac{2\pi }{N}\right)}^{2}\right)}^{2}+{\left(A\mathrm{sin}\left(\phi \right)+\frac{A}{3}{\left(\frac{2\pi }{N}\right)}^{2}\right)}^{2}}\\ {X}_{1}=\sqrt{\frac{{a}_{1}^{2}+{b}_{1}^{2}}{2}}\ge \frac{1}{\sqrt{2}}\sqrt{{\left(A\mathrm{cos}\left(\phi \right)-\frac{A}{3}{\left(\frac{2\pi }{N}\right)}^{2}\right)}^{2}+{\left(A\mathrm{sin}\left(\phi \right)-\frac{A}{3}{\left(\frac{2\pi }{N}\right)}^{2}\right)}^{2}}\end{array}$ (13)

$\frac{A}{\sqrt{2}}{\left(1-\frac{2\sqrt{2}}{3}{\left(\frac{2\pi }{N}\right)}^{2}\mathrm{sin}\left(\phi +\frac{\pi }{4}\right)\right)}^{\frac{1}{2}}\le {X}_{1}\le \frac{A}{\sqrt{2}}{\left(1+\frac{2\sqrt{2}}{3}{\left(\frac{2\pi }{N}\right)}^{2}\mathrm{sin}\left(\phi +\frac{\pi }{4}\right)\right)}^{\frac{1}{2}}$ (14)

3.2.2. 校验集相对误差分析

$\frac{A}{\sqrt{2}}{\left(1-\frac{2\sqrt{2}}{3}{\left(\frac{2\pi }{N}\right)}^{2}\mathrm{sin}\left(\phi +\frac{\pi }{4}\right)\right)}^{\frac{1}{2}}\le {X}_{Q}\le \frac{A}{\sqrt{2}}{\left(1+\frac{2\sqrt{2}}{3}{\left(\frac{2\pi }{N}\right)}^{2}\mathrm{sin}\left(\phi +\frac{\pi }{4}\right)\right)}^{\frac{1}{2}}$ (15)

$\frac{A}{\sqrt{2}}{\left(1-\frac{2\sqrt{2}}{3}{\left(\frac{4\pi }{N}\right)}^{2}\mathrm{sin}\left(\phi +\frac{\pi }{4}\right)\right)}^{\frac{1}{2}}\le {X}_{R}\le \frac{A}{\sqrt{2}}{\left(1+\frac{2\sqrt{2}}{3}{\left(\frac{4\pi }{N}\right)}^{2}\mathrm{sin}\left(\phi +\frac{\pi }{4}\right)\right)}^{\frac{1}{2}}$ (16)

$\frac{{X}_{R.\mathrm{min}}-{X}_{Q.\mathrm{max}}}{{X}_{Q.\mathrm{max}}}\le {\delta }_{R}=\frac{{X}_{R}-{X}_{Q}}{{X}_{Q}}\le \frac{{X}_{R.\mathrm{max}}-{X}_{Q.\mathrm{min}}}{{X}_{Q.\mathrm{min}}}$ (17)

$\left\{\begin{array}{l}{\delta }_{R}\le {\delta }_{R.\mathrm{max}}=\frac{\sqrt{1+4h\left(N,b\right)}-\sqrt{1-h\left(N,b\right)}}{\sqrt{1-h\left(N,b\right)}}\\ {\delta }_{R}\ge {\delta }_{R.\mathrm{min}}=\frac{\sqrt{1-4h\left(N,b\right)}-\sqrt{1+h\left(N,b\right)}}{\sqrt{1+h\left(N,b\right)}}\end{array}$ (18)

$\left\{\begin{array}{l}{\delta }_{R.\mathrm{max}}\left(N,\phi \right)=\frac{{c}_{1}\mathrm{sin}\left({c}_{0}\phi +{c}_{2}\right)+{c}_{3}}{N}\\ {\delta }_{R.\mathrm{min}}\left(N,\phi \right)=\frac{{c}_{4}\mathrm{sin}\left({c}_{0}\phi +{c}_{5}\right)+{c}_{6}}{N}\end{array}$ (19)

4. 验证

Figure 2. ${\delta }_{R}\cdot N$ distribution without single large number

${\delta }_{R}$ 分布的上下限根据最小二乘法可以求得 ${c}_{0}~{c}_{6}$ 的具体数值，得到区间边界的表达式如式(20)所示。经校验，该表达式下最小二乘拟合的 ${R}^{2}\ge 0.995$。改变数据进行重复仿真校验，利用最小二乘法求得的上下限表达式始终为相对误差 ${\delta }_{R}$ 分布的包络线。

$\left\{\begin{array}{l}{\delta }_{R.\mathrm{max}}\left(N,\phi \right)=\frac{0.4975\mathrm{sin}\left(2\phi +1.412\right)-0.027}{N}\\ {\delta }_{R.\mathrm{min}}\left(N,\phi \right)=\frac{0.4798\mathrm{sin}\left(2\phi +1.325\right)-0.081}{N}\end{array}$ (20)

$\left\{\begin{array}{l}{\delta }_{S.\mathrm{max}}\left(N,\phi \right)=\frac{0.4975\mathrm{sin}\left(2\phi +\frac{2\pi }{N}+1.412\right)-0.027}{N}\\ {\delta }_{S.\mathrm{min}}\left(N,\phi \right)=\frac{0.4798\mathrm{sin}\left(2\phi +\frac{2\pi }{N}+1.325\right)-0.081}{N}\end{array}$ (21)

Figure 3. ${\delta }_{R}\cdot N$ distribution diagram with single large number ( ${x}^{\prime }\left(3\right)=1.2x\left( 3 \right)\right)$

5. 结语

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