ITAE。 综合时间绝对误差。

蜻蜓

我刚才发个帖子，可能是字少，又被审核，显示不出来，为了尽快得到答案，我现在不得不重新发个帖子，为了不被审核，我不得不说点废话，凑点字数，管理员液压哥：我从不灌水，最近总被审核，能否设置一下，免除安检，给个绿色通道，让我不被审核？


我想请教的问题是如下：
What is the gain in this Byrd chart and why does it decrease over 10HZ?
What is the gain in this bode diagram and why does it go down over 10HZ?
7#这个伯德图中的增益是什么增益，为什么超过10HZ 下降了？

auca

PEN 发表于 2018-6-22 11:32
由于ITAE应该用于寻找最佳闭环极点位置，因此我想从具有最少闭环极点的系统开始。
由于液压缸的开环传递函 ...

如果用积分器,为什么会增加一个极点，又该如何计算ITAE？

PEN

见上面的＃7
基本等式
H（s）= 1 /（A3 * s ^ 3 + A2 * s ^ 2 + A1 * s + A0）
计算闭环极点位置，但带宽仅为每秒1弧度。
为了增加响应，“s”必须除以λ。
为了使带宽大约为10Hz，我设置λ= 2 *π* 10，我使用了这个修正的方程
H（s）= 1 /（A3 *（s /λ）^ 3 + A2 *（s /λ）^ 2 + A1 *（s /λ）+ A0）
要么
H（s）=λ^ 3 /（s ^ 3 + A2 *λ* s ^ 2 + A1 *λ^ 2 * s +λ^ 3）
要增加带宽，请增加λ。
这使闭环极点偏离原点（0，j0）

我想重复一遍。放置闭环极的ITAE方法并不完美。当有3个闭环极点时，两个复极点不在稳定性好的区域。
有时，闭环极点的最佳位置存在差异。

看看＃9。
你应该能够理解这些公式从何而来
λ^ 3 = K *ω^ 2 * Kp                  ; s ^ 0求解比例增益
A1 *λ^ 2 =ω^ 2 + K *ω^ 2 * Kd  ; s ^ 1求解微分增益（速度）
A2 *λ= 2 *ζ*ω+ K *ω^ 2 * K2      ; s ^ 2求解二阶导数增益（加速度）

我们正在迅速前进。
你应该能够理解图片ppt4pg61，png
我已经说过这是一个非常重要的理解图片。
到目前为止，您应该能够将ppt4pg61，png中的等式分为开环传递函数和控制器传递函数。

ppt4pg61.png中的等式非常重要，因为它基本上表明需要a来加速。我称之为二阶导数增益或加速度增益。
在图片中，加速度增益为Kaf。



See #7 above
The basic equation
H(s) = 1/(A3*s^3+A2*s^2+A1*s+A0)
Computes closed loop pole locations, but the bandwidth is only about 1 radian per second.
To increase the response the “s” must be divided by λ.
To make the bandwidth about 10Hz I set λ=2*π*10 and I used this modified equation
H(s) = 1/(A3*(s/ λ)^3+A2*(s/ λ)^2+A1*(s/ λ)+A0)
or
H(s)= λ^3/(s^3+A2*λ*s^2+A1*λ^2*s+λ^3)
To increase the bandwidth, increase λ.
This move the closed loop poles away from the origin (0,j0)

I want to repeat. The ITAE method of placing the closed loop poles is not perfect. When there are 3 closed loop poles, the two complex poles are not in the area of nice stability.
Sometimes there is a difference in opinion where the optimal location of the closed loop poles.

Look at #9.
You should be able to understand where these formulas come from now
λ^ 3=K *ω^ 2*Kp                  ; s^0  solve for the proportional gain
A1 *λ^ 2 = ω^2+K*ω^2 * Kd        ; s^1   solve for the derivative gain ( velocity )
A2 *λ=2 *ζ*ω+ K *ω^ 2 * K2       ; s^2 solve for the second derivative gain ( acceleration )

We are moving ahead quickly.
You should be able to understand the picture ppt4pg61,png
I have said this is a very important picture to understand.
By now you should be able to separate the equation in ppt4pg61,png into the open loop transfer function and the controller transfer function.

The equation in ppt4pg61.png is significant because it basically shows that a for the acceleration is required.  I call that a second derivative gain or acceleration gain.
In the picture the acceleration gain is Kaf.

PEN

auca 发表于 2019-6-12 09:28
如果用积分器,为什么会增加一个极点，又该如何计算ITAE？

好问题。 我想稍后在闭环主题中讨论这个问题，但我将从这里开始
PI控制器的拉普拉斯变换是：
Gc(s)=Ki/s+Kp
执行器的开环传递功能是：
Ga(s)=K*ω^2/(s*(s^2+2*ζ*ω*s+ω^2)）
闭环传递函数是：
CLTF(s)=Gc(s)*Ga(s)/(1+Gc(s)*Ga(s))）
闭环传递函数是
CLTF(s)=(Ki* K*ω^2)/(s^4+2*ζ*ω*s^3+ ω^2*s^2+Kp* K*ω^2*s+Ki* K*ω^2)
这是一个四阶方程，因此必须使用四阶ITAE。 但是，ITAE4不能与PI控制器一起使用。 控制器每个闭合极需要一个增益。



Good question.  I wanted to cover this later in the closed loop topic, but I will start here
The Laplace transform for the PI controller is:
Gc(s)=Ki/s+Kp
The open loop transfer function for the actuator is:
Ga(s)=K*ω^2/(s*(s^2+2*ζ*ω*s+ω^2)
The closed loop transfer function is:
CLTF(s)=Gc(s)*Ga(s)/(1+Gc(s)*Ga(s))
The closed loop transfer function is
CLTF(s)=(Ki* K*ω^2)/(s^4+2*ζ*ω*s^3+ ω^2*s^2+Kp* K*ω^2*s+Ki* K*ω^2)
This is a 4th order equation so a 4th order ITAE must be used.  However, ITAE4 cannot be used with a PI controller.  The controller needs one gain for every closed pole.

蜻蜓

PEN 发表于 2019-6-15 11:32
见上面的＃7
基本等式
H（s）= 1 /（A3 * s ^ 3 + A2 * s ^ 2 + A1 * s + A0）

We are moving ahead quickly.

我知道(Kfa)。您可以在闭环的主题里，像用“速度和加速度前馈”加大输出那样，加快主题速度，我看几遍会理解。

我的理解，做个比喻：
DELTA 控制器，集成Kv，Kd，K2(Kfa)， pid，
因为在运行时，DELTA 控制器的 Kv，Kd，K2，
提供一个大的控制信号，并且运行时参数是可以变化的，
所以Kv，Kd，K2类似于（PID控制器）中的（比例P）作用，
而DELTA 控制器中“pid”消除小的误差，类似（PID控制器）中的（积分I）作用 ，,
Kd补偿阻尼，K2双向(增加和吸收)补偿能量，相当D。
所以，DELTA 控制器, 相当于具有一大和一小,两个PID.。（PID和pid）

所以DELTA =P I D+pid        （+ Other functions）
我想证明上面这个等式。
我有一个问题，假如系统过阻尼，ξ大于0.707时，Kd能减小阻尼吗？

I know (Kfa). You can speed up the theme in the closed-loop theme, just like using "speed and acceleration feed-forward" to increase the output. I will understand it several times.


My understanding, make a metaphor:
DELTA controller, integrated Kv, Kd, K2 (Kfa), pid,
Because at runtime, the delta controller Kv, Kd, K2,
Provides a large control signal, and runtime parameters can be changed.
So Kv, Kd, K2 are similar to (proportional P) in (PID controller).
The "pid" in the DELTA controller eliminates small errors, similar to the (integral I) effect in the (PID) controller.
Kd compensates for damping, and K2 compensates energy in two directions (increasing and absorbing), which is equivalent to D.
Therefore, the DELTA controller, equivalent to a large and a small, two PIDs. (PID and pid)

So DELTA = P I D + pid     (+ Other functions)
I want to prove the equation above.
I have a question, if the system damping, factor is greater than 0.707, Kd can decrease damping?


The ITAE and PID algorithms here, as well as the function information of the controller scattered among other topics, should be put together to make a PDF specification of the functions inherent

PEN

你要我写一本书。我被问过很多次。
写书的时间还不对。我被问过很多次。我拒绝了。

主持人应该将这些主题永久化。

我有一台Apple电脑。它非常擅长保存主题。我保存了许多主题，包括许多已被删除的主题。太多的美国液压论坛已经关闭。除非我保存最好的主题，否则所有信息都会丢失。因此，我保存了一切。我有一个非常大的文件，我用英文写的。我还有一个保存为iyeya.cn主题的目录，保存为webarchives。这样我的信息就不会丢失。有时我的帖子一开始不被接受。我需要保存所有内容，以便稍后发布相同的信息。我真的应该把这些信息放在我自己的论坛上。
我在这里透露了很多信息。但是，我向你提供的信息远远超过那些不知道他们不知道的人。

RMC是灵活的。通常只有一个闭环控制对任何一个轴都有效。
有时级联循环是必要的。这是外环产生速度或位置的地方。

You want me to write a book. I have been asked many times.
The time to write a book is not right yet. I have been asked many times. I have refused.

The moderator should make these topics permanent.

I have an Apple computer. It is very good at saving topics. I have saved many topics including many that have been deleted. Too many US hydraulic forums have shut down. All information is lost unless I save my best topics. Therefore, I save everything. I have a very big file with what I have written in English. I also have a directory of saved iyeya.cn topics that are saved as webarchives. This way my information is not lost.  Sometimes my posts are not accepted at first.  I need to save everything so I can post the same information later.  I really should put this information on my own forum.
I have revealed much information here. However, I reveal just enough information to you know much more than those that do not know what they do not know.

The RMCs are flexible. Usually only one closed loop control is active for any one axis.
Sometimes cascaded loops are necessary. This is where an outer loop generates the velocity or position.


The time to write a book is not right yet.  I have been asked many times. I have refused
I have revealed much information here.  However, I reveal just enough information to you know much more than those that do not know what they do not know.

The RMC are flexible.  Usually only one closed loop control is active for any one axis.
Sometimes cascaded loops are necessary.  This is where an outer loop generates the velocity or position for the inner loop.
When the damping factor for the open loop system is high,  a PID with velocity and acceleration feedforwards is good to provide excellent control.  The problem is that if the damping factor is high,  too much energy is wasted.  If the damping factor is low then the PID2 with velocity,  acceleration and jerk feed forwards must be used.  As far as I know,  only the RMCs can do this.  Only RMCs have auto tuning.
This is because finding the coefficients for the open loop transfer function is difficult.  It is also difficult to computer the actual velocity and acceleration.

We have not covered how to use the integrator gain yet.

蜻蜓

不是写书。软件翻译不正确。我的意思是，你在这里的内容很有价值，应该被收集和组织在一起。任何人都可以整理成pid，，然后放在一个主题里，供人学习和浏览，避免一遍一遍答复的重复劳动。

Not writing books. The software translationis incorrect. I mean, your content here is valuable and should be collected andorganized. Anyone can organize the documents of PDF valves, and I can also putthem in a topic for people to learn and browse, so as to avoid repetitive workof replying over and over again.

无锡李工

赞一个。
ITAE虽然可以得到误差最少的PID参数，AMESIM也有此功能，但是实用性可以商榷，个人谈点拙见：
1.应注意被控对象并不一定是定常系统，比如液压伺服系统中油源压力和负载的变化可能会使得ITAE整定出来的参数引起系统稳定性的问题；
2.大部分液压伺服系统相对简单，抓住油缸或作动器度的开环速度增益，确定反馈增益，根据闭环系统带宽要求或者斜坡输入跟踪误差要求，在稳定裕度满足的条件，很容易确定Kp。由于作动器本身就是一个积分环节，故一般液压伺服系统中Ki更多是为了弥补控制流量的泄漏而存在。
3.确定Kp和Ki的单位和软硬件接口，控制软件中根据PID计算结果的值域可以确定编程时Kp和Ki的数码值。

PEN

ITAE是一种放置闭环极点的方法。 它不是直接计算控制器增益。
液压系统确实会发生变化。 闭环极点位置也将改变，但不应改变太多。 如果正确使用ITAE，则系统永远不会不稳定。 我看到的许多论文都没有正确使用ITAE。
应使用ITAE查找闭环极点位置。 然后，计算控制器增益以实现计算出的闭环极点位置。 由于将有四个闭环极点，因此必须有四个控制器增益才能正确放置每个闭环极点。 直接使用ITAE计算控制器增益是错误的。

ITAE is a method of placing closed loop poles.  It is not for calculating controller gains directly.
Hydraulic systems do change.  The closed loop pole locations will also change but they should not change much.  The system should never be unstable if ITAE is used properly.  Many papers I see do not use ITAE properly.
ITAE should be used to find the closed loop pole positions.  Then the controller gains are calculated to achieve the calculated closed loop pole positions.  Since there will be four closed loop poles, there must be four controller gains to place each closed loop pole correctly. Using ITAE to calculate the controller gains directly is wrong.