楼主: PEN
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优化控制器参数
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#36 很好。
计算前馈需要良好的运动轨迹软件和液压系统模型。 #36 is very good. Calculating feed forwards require a good motion trajectory software and a model of the hydraulic system. | |
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a = F/m
a = (Pa*Aa-Pb*Ab)/m |
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IP卡
狗仔卡
PEN 发表于 2021-10-3 23:58 There are two kinds of feedforward, one is the feedforward of the controlled output, which is an open-loop structure. The output is calculated according to the model, and it is directly given to the plant to improve the dynamic response of the system. There is also a kind of feedforward that compensates for external disturbances ITAE is the knowledge of optimization control theory, I am not familiar with frequency response | |
发表于 2021-10-4 09:52:22
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本帖最后由 cn_young 于 2021-10-4 11:26 编辑
PEN 发表于 2017-12-29 05:36 Teacher Pen Are these curves driving actual loads or only simulated in theory?这些曲线是驱动实际负载还是仅在理论上模拟？ | |
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#1、#7、#11、#23、#28 中的曲线由 RMC 模拟器生成，用于训练或指导。 我正在展示如何调整系统。 我的办公桌旁没有液压系统。 我的照片通常是模拟的。
这是我的系统在我们的液压培训区的视频。 如您所见，这是真实的。 https://deltamotion.com/peter/Videos/NF-FOA.mp4 The curves in #1, #7, #11, #23, #28 are generated by the RMC simulator for training or instruction. I am showing how to tune a system. I do not have a hydraulic system by my desk. My pictures are usually simulated. This is a video of my system down in our hydraulic training area. This is real as you can see. https://deltamotion.com/peter/Videos/NF-FOA.mp4 | |
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a = F/m
a = (Pa*Aa-Pb*Ab)/m |
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back2049 发表于 2017-12-23 14:54 速度前馈比加速度前馈更起作用？ | |
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cn_young 发表于 2021-10-27 15:37 应使用速度、加速度和加加速度前馈。 要使用加加速度前馈运动轨迹，如#36，是必要的。 这是非常好的。 Velocity, acceleration and jerk feed forwards should be used. To use jerk feedforward a motion trajectory, like in #36, is necessary. This is very good. | |
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a = F/m
a = (Pa*Aa-Pb*Ab)/m |
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cn_young 发表于 2021-10-28 12:29 很难说。 最好显示控制信号。 看起来没有使用加速和加加速度前馈。 时间 0 处的加速度的导数不像 #36 那样为 0。 #36 中的运动曲线很难计算。 如果用户决定在运动过程中改变轨迹，那就更难了。 该数学运算无法在 PLC 上完。 It is hard to tell. It is best to display the control signal. It doesn’t look like acceleration and jerk feedforward are used. The derivative of the acceleration at time 0 is not 0 like in #36. The motion profile in #36 is very hard to compute. It is even harder if the user decides to change the trajectory during motion. This math cannot be done on a PLC. | |
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a = F/m
a = (Pa*Aa-Pb*Ab)/m |
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cn_young 发表于 2021-10-28 12:29 避免使用正弦信号作起步, 因为那是速度最快, 也就是阀要开得大, 但因阀在100%命令时频率是最低, 总增益很大, 产生超调。 解决办法是用余弦信号, 因为起步时速度最慢, 误差不大, 不会产生振荡。 | |
发表于 2021-10-30 16:48:15
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本帖最后由 cn_young 于 2021-10-30 22:20 编辑
游勇 发表于 2021-10-30 16:48 游兄，我把加速度信号从正弦改为余弦改的对吗？怎么超调更大了？ | |
