怎么在DSPIC30F4011中实现PID

Himaybe不是“集成”而是“AcMUTE”。给定函数的“积分”是由函数图、x轴和积分极限所限制的区域。如果函数是一个变化的电压，则只需要在很小的时间间隔对它进行采样（ADC），并添加（AcMutt）ADC结果。PS//E.维基百科. org／.



      以下为原文

    Hi
 
Maybe you don't "integrate" but "acumulate" instead.
 
The "integral" of a given function is the area limited by the function graph, the "X" axis and the integration limits.
If the function is a varying voltage, you just need to sample it (ADC) at small intervals and add (acumulate) the ADC results.
 
https://en.wikipedia.org/...ral_approximations.svg
 
 
HIH
 
Best regards
Jorge
 

怎样才能找到一个信号的微分。



      以下为原文

    How can I find D differentiation of a signal.
 

再次回到基础！在给定的点上函数的导数表示在该点上函数的切线的斜率。因此，在2个非常接近的点上得到函数的值，并计算这2点通过的行的斜率。http://e.维基百科。org/wiki／文件：TangeTeNoto-Ay曲线。可变电压（V）信号只得到2个ADC读数，用非常小的时间（t）分隔，计算（V2-V1）/（T2-T1）。



      以下为原文

    Hi
 

Back to the basics again!
The derivative of a function at a given point represents the slope of the tangent to the graph of the function at that point.
So, got the value of the function in 2 very close points and calculate the slope of the line passing by those 2 points.
 
https://en.wikipedia.org/wiki/File:Tangent_to_a_curve.svg
 
For a varying voltage (V) signal just get 2 ADC readings separated by a very small amount of time (t) and calculate (V2-V1)/(t2-t1).
 
 
Gosh...looks like I'm back in high school.
 
 
Best regards
Jorge