[足式机器人]Part2 Dr. CAN学习笔记-自动控制原理Ch1-8Lag Compensator滞后补偿器

2024-01-03 06:34:32

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Dr. CAN学习笔记-自动控制原理Ch1-8Lag Compensator滞后补偿器


稳态误差入手(steady state Error)
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误差 Error E ( s ) = R ( s ) ? X ( s ) = R ( s ) ? E ( s ) ? K G ( s ) ? E ( s ) ( 1 + K G ( s ) ) = R ( s ) ? E ( s ) = 1 1 + K G ( s ) R ( s ) = R ( s ) 1 1 + K N ( s ) D ( s ) = 1 s 1 1 + K N ( s ) D ( s ) E\left( s \right) =R\left( s \right) -X\left( s \right) =R\left( s \right) -E\left( s \right) \cdot KG\left( s \right) \Rightarrow E\left( s \right) \left( 1+KG\left( s \right) \right) =R\left( s \right) \Rightarrow E\left( s \right) =\frac{1}{1+KG\left( s \right)}R\left( s \right) =R\left( s \right) \frac{1}{1+K\frac{N\left( s \right)}{D\left( s \right)}}=\frac{1}{s}\frac{1}{1+K\frac{N\left( s \right)}{D\left( s \right)}} E(s)=R(s)?X(s)=R(s)?E(s)?KG(s)?E(s)(1+KG(s))=R(s)?E(s)=1+KG(s)1?R(s)=R(s)1+KD(s)N(s)?1?=s1?1+KD(s)N(s)?1?

单位阶跃unit step R ( s ) = 1 s R\left( s \right) =\frac{1}{s} R(s)=s1?
稳态误差Steady State Error——FVT终值定理
e s s = lim ? t → ∞ e ( t ) = lim ? s → o s E ( s ) = lim ? s → o s ? 1 s 1 1 + K N ( s ) D ( s ) = 1 1 + K N ( 0 ) D ( 0 ) = D ( 0 ) D ( 0 ) + K N ( 0 ) ess=\underset{t\rightarrow \infty}{\lim}e\left( t \right) =\underset{s\rightarrow o}{\lim}sE\left( s \right) =\underset{s\rightarrow o}{\lim}s\cdot \frac{1}{s}\frac{1}{1+K\frac{N\left( s \right)}{D\left( s \right)}}=\frac{1}{1+K\frac{N\left( 0 \right)}{D\left( 0 \right)}}=\frac{D\left( 0 \right)}{D\left( 0 \right) +KN\left( 0 \right)} ess=tlim?e(t)=solim?sE(s)=solim?s?s1?1+KD(s)N(s)?1?=1+KD(0)N(0)?1?=D(0)+KN(0)D(0)?

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文章来源:https://blog.csdn.net/LiongLoure/article/details/135351025
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