Riemannian Metric Based Performance Monitoring and Diagnosis for a Class of Feedback Control Systems
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摘要: 针对复杂工业系统对性能衰退的容忍度低等问题, 提出基于系统性能预测的一类反馈控制系统过程监测方法, 通过黎曼度量对控制性能衰退程度进行预测与监测, 并给出发生故障的类型, 以提升过程监测系统的实时性与准确性. 首先, 利用系统的实时数据, 计算系统性能衰退的预测指标; 其次, 利用黎曼度量对系统性能衰退程度进行预测与监测, 并利用随机算法给出对应的阈值来诊断系统性能衰退; 最后, 通过计算各类引发系统性能衰退的故障的性能预测指标集合的中心和阈值, 实现故障的实时定位. 所提出的方法通过三容水箱仿真实验平台进行验证.Abstract: Associated with the issue that the industrial processes have low tolerance in performance degradation, a performance monitoring and diagnosis scheme for a class of feedback control systems is developed based on system performance prediction. The Riemannian metric is then adopted to predict and monitor the degree of the control performance degradation, and classify the fault type, so as to improve the real-time ability and accuracy of the performance monitoring systems. Firstly, a prediction indicator is introduced to predict the system performance degradation using online process data. Then, the Riemannian metric is implemented to predict and monitor the degree of the control performance degradation, and the randomized algorithm is adopted to set the threshold for diagnosing the performance degradation. Finally, by computing the center and threshold for the performance predictor set caused by all the possible faults, the fault localization can be achieved. The effectiveness of the proposed method is verified by a case study on a three-tank system.
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Key words:
- Performance monitoring /
- Riemannian metric /
- performance diagnosis /
- fault location
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表 1 水箱DTS200的参数
Table 1 Parameters of tank DTS200
参数 符号 值 单位 水箱面积 $ {\cal{A}} $ 154 $ \mathrm{cm}^{2} $ 水箱间管道面积 $s_{{n} }$ 0.5 $ \mathrm{cm}^{2} $ 水箱最高水位 $ H_{\mathrm{max}} $ 62 $ \mathrm{cm} $ 泵 1 的最大进水量 $ Q_{1_{\mathrm{max}}} $ 120 $ \mathrm{cm}^{3}/\mathrm{s} $ 泵 2 的最大进水量 $ Q_{2_{\mathrm{max}}} $ 120 $ \mathrm{cm}^{3}/\mathrm{s} $ 管道 1 水流系数 $ a_{1} $ 0.45 管道 2 水流系数 $ a_{2} $ 0.60 管道 3 水流系数 $ a_{3} $ 0.45 表 2 水箱堵塞故障隔离
Table 2 Isolation of pipe plugging
故障 $ d_R^2({\boldsymbol P},{\boldsymbol P}_{z,1}) $ $ d_R^2({\boldsymbol P},{\boldsymbol P}_{z,2}) $ $ d_R^2({\boldsymbol P},{\boldsymbol P}_{z,3}) $ 故障隔离 $a_1 = 0.30$ $ 0.0388 $ $ 0.1088 $ $ 0.1193 $ 故障簇 1 $ a_2 = 0.35 $ $ 0.0894 $ $ 0.0270 $ $ 0.0810 $ 故障簇 2 $ a_3 = 0.28 $ $ 0.1258 $ $ 0.1099 $ $ 0.0478 $ 故障簇 3 $ a_1 = 0.27 $ $ 0.0636 $ $ 0.1325 $ $ 0.1409 $ 故障簇 1 $a_2 = 0.40$ $ 0.0809 $ $ 0.0139 $ $ 0.0732 $ 故障簇 2 -
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