CLTF(s)=(5Kp+5KdS)/(S^3+3S^2+5*(1+Kd)S+5Kp),不知道分子中的S怎么消掉。
蜻蜓 发表于 2019-6-4 15:23
Mr. Pen teaches a lot of knowledge. Thank you.
Yes,the open loop model requires 3 gains to pla ...
First you must know where the closed loop poles need to be.
That is in the area of nice stability as shown above.
It is even better if the poles are located closer to the negative real axis.
The characteristic equation is:
s^3+4*s^2+5*s+5*Kp
There are 3 poles. One pole can be placed at -a and two poles can be place at -b.
The poles are real poles and will cause errors to decay at exp(-a*t) and exp(-b*t)
The response will have no overshoot because there are not complex poles.
So the desired character is equation is (s+a)*(s+b)^2
So
s^3+4*s^2+5*s+5*Kp=(s+a)*(s+b)^2
What is the value of Kp,a,and b that solves this equation?
This part can be done by hand in 5 minutes.
Anybody is allowed to answer.
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我不知道为什么我的签名在文本中。
该网站无法正常运作。
我稍后会再试一次。
The copy and paste does not work with Microsoft Edge.I am not able to edit.
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我昨天回复了10＃PD控制器，失败了
没显示。我计算结果a=2，b=1，Kp=2，不知对不对，另外5# 的ce= # the coefficients of the characteristic equation，特征方程的系数，为什么不是ce= ？
蜻蜓 发表于 2019-6-8 05:38
我昨天回复了10＃PD控制器，失败了
没显示。我计算结果a=2，b=1，Kp=2，不知对不对，另外5# 的ce=
CE应该是
S^ 3+ 4* S^ 2+ 5* S+ 5 *的Kp
Kp需要低于2。
The CE should be
s^3+4*s^2+5*s+5*Kp
Kp needs to be lower than 2.
import numpy as np # import the numerical library for roots
Kp=2 # initialize the proportional gain
ce= # the coefficients of the characteristic equation
poles = np.roots(ce) # compute the poles or roots
for i in range(poles.size):# print out the poles one at a time.
print("Pole{} = {:.2f}".format(i,complex(poles)))
如果您不喜欢python，请尝试Scilab或Octave。 他们都非常接近Matlab。
他们是免费的。
If you do not like python then try Scilab or Octave.They are both very close to Matlab.
They are free.
本帖最后由 蜻蜓 于 2019-6-8 06:14 编辑
我不是不喜欢python，可能是电脑配置原因，前段时间安装python不成功，我先学习您帖子，以后再尝试安装python。a=2，b=1，
Kp=0.4对吧？
本帖最后由 PEN 于 2019-6-8 08:56 编辑
蜻蜓 发表于 2019-6-8 06:06
我不是不喜欢python，可能是电脑配置原因，前段时间安装python不成功，我先学习您帖子，以后再尝试安装pyth ...
你是对的。
两个在-1和一个在-2非常慢。
拉普拉斯逆变换可用于显示对10mm的突然变化的响应。
蜻蜓 发表于 2019-6-7 18:42
CLTF(s)=(5Kp+5KdS)/(S^3+3S^2+5*(1+Kd)S+5Kp),不知道分子中的S怎么消掉。
这几乎是正确的
CLTF(s)=(5*Kp+5*Kd*s)/(s^3+4*s^2+5*(1+Kd)*s+5*Kp)
您可能犯了一个复制错误。 应该没有3。
好的，到目前为止还不错。
现在计算控制器增益。
现在我们必须选择闭环极点位置。
首先，我们将尝试将所有3个极点放在-Q。 这意味着所需的特征方程是
(s+Q)^3
这必须等于
s^ 3+ 4*s^2+ 5 *(1+Kd)*s+ 5*Kp
所以
(s+Q)^3= s^3 + 4* s^2+ 5 *(1 + Kd)* s+ 5 *Kp
Kp，Kd和Q的值是多少？
三个闭环极点位于-Q。
比较3个闭环极点的位置与使用导数增益和3个闭环极点仅使用比例增益。
This is almost right
CLTF(s)=(5*Kp+5*Kd*s)/(s^3+4*s^2+5*(1+Kd)*s+5*Kp)
You probably made a copying error.There should be no 3.
OK, good so far.
Now compute the controller gains.
Now we must chose the closed loop pole positions.
First we will try placing all 3 poles at -Q. This means the desired characteristic equation is
(s+Q)^3
This must be equal to
s^3+4*s^2+5*(1+Kd)*5*s+5*Kp
so
(s+Q)^3= s^3 + 4* s^2+ 5 *(1 + Kd)* s+ 5 *Kp
What are the values of Kp, Kd and Q?
The three closed loop poles are at -Q.
Compare the location of the 3 closed loop poles with the using the derivative gain with the 3 closed loop poles using only the proportional gain.
本帖最后由 蜻蜓 于 2019-6-9 07:28 编辑
我用手算的， Kd=1/15=0.066 Kp=64/135=0.474 ，Q=4/3=1.333, 假如正确的话，有两个极点与仅用应Kp=0.4相比，远离虚轴。用Kd和Kp放置的最佳极点位置，应该与仅用Kp放置的最佳极点位置相比。
蜻蜓 发表于 2019-6-9 07:10
我用手算的， Kd=1/15=0.066 Kp=64/135=0.474 ，Q=4/3=1.333, 假如正确的话，有两个极点与仅用应Kp=0.4 ...
是！！ 你是对的！
在继续之前，您应该尝试找到开环传递函数的控制器增益和极点位置
5/(s*(s^2+2*s+5)
控制器的解决方案仅限Kp
如果你正确地解决问题，将会有一对复杂的极点。
复杂的极点不会处于良好的稳定区域。
解决控制器的问题是Kp + Kd * s
仍然会有3个真正的极点。
你的结论是什么？
Yes!! You are correct!
Before going on you should try to find the controller gains and pole locations for when the open loop transfer function is
5/(s*(s^2+2*s+5)
Solve for Kp only
If you solve the problem correctly, there will be a pair of complex poles.
The complex poles will not be in the nice stability zone.
Solve for Kp+Kd*s
There will still be 3 real poles.
What is your conclusion?
Compare.