另一个测试。关闭循环

蜻蜓

PEN 发表于 2019-6-20 10:41
你需要解决这个问题
解决Kp，Kd和K2
(s+ λ)^3=(s^3+(5*K2+2)*s^2+(5*Kd+5)*s+5*Kp)

3λ=5*K2+2
3λ^2=5*Kd+5
λ^3=5*Kp
(5*K2+2)* (5*Kp) ^(1/3)= 5*Kd+5
三个方程，四个未知数，
假设λ=1，那么K2=0.2，Kd= -0.4，Kp=0.2
假设λ=2  那么K2=0.8，Kd=7/5= 1.4,   Kp=8/5=1.6
λ可以是负实轴多个数。

PEN

蜻蜓 发表于 2019-6-20 11:30
3λ=5*K2+2
3λ^2=5*Kd+5
λ^3=5*Kp

是的！很好！
理论上，您可以根据需要将闭环极移到左侧。没有像以前的闭环控制方法那样的理论限制。

使用RMC的增益计算器时，有一个向上或向下移动的滑块。向上移动滑块会增加λ。这增加了闭环增益的带宽。理论上可以将λ增加到非常高的数量但是由于反馈的分辨率，气缸和负载的固有频率以及阀的频率响应，存在实际限制。使用模拟器时，更新时间也有限制。

现在你知道大多数人不知道的了！

YES! Very good!
In theory, you can move the closed loop poles as far to the left as you want. There isn’t a theoretical limitation like there is with the previous closed loop control methods.

When using the gain calculator of the RMC, there is a slider that is moved up or down. Moving the slider up increases λ. This increases the bandwidth can the closed loop gains. In theory it is possible to increase λ to very high numbers but There is a practical limit due to the resolution of the feedback, the natural frequency of the cylinder and load and the frequency response of the valve. When using the simulator, there is also a limit from the update time.

Now you know what most people do not know!

蜻蜓

以前看过您讲过解微分增益提供电阻尼，可以控制低固有频率的油缸，（降低活塞杆直径，节省能源），因为不知道原因，所以印象不深刻，看完就忘了。在这个主题开始时，我对此也是一无所知，现在明白了一些，

我想看到这个主题的之后人，会像我一样明白（以前不理解）您之前介绍过的性能。

刚才通电找了一下这个自动增益计算器，自动增益调整功能是这个吧？

希望PEN老师继续讲解。

PEN

是的，这是增益计算器。
看这个：
https://deltamotion.com/peter/Videos/Optimizing%20Gain-%20Chinese.mp4

现在看这个：
https://deltamotion.com/peter/Videos/NF-FOA.mp4
在7:00，您可以看到阻尼系数约为0.15，固有频率约为6.5Hz。
延伸时的开环增益约为14.5（mm / s）/％，收缩时为12.6（mm / s）/％。

现在你知道为什么这个系统很难优化。 该系统的设计使得即使使用PID也无法将闭环极移动到稳定性好的区域。 必须使用PID2。

Yes, that is the gain calculator.
Watch this:
https://deltamotion.com/peter/Videos/Optimizing%20Gain-%20Chinese.mp4

Now watch this:
https://deltamotion.com/peter/Videos/NF-FOA.mp4
At 7:00 you can see the damping factor is about 0.15 and the natural frequency is about 6.5Hz.
The open loop gain is about 14.5(mm/s)/% when extending and 12.6(mm/s)/% when retracting.

Now you know why this system is difficult to optimize. The system was designed so it is impossible to move the closed loop poles to the area of ​​nice stability even when using a PID. The PID2 must be used.

蜻蜓

I see 6.42HZ, 0.105 gain: 14.2,
The natural frequency and damping are very small, and the words are very small.

我看到是6.42HZ，   0.105    gain：14.2，
固有频率和阻尼很小， 字也很小。

PEN

7:00
KP=9.998
文=14.2
我将开环增益转换为（英寸/秒）/伏特到（毫米/秒）/％控制

当您向上移动滑块时，您正在移动使λ更大。 这会导致错误更快地衰减。EXP（-λ* t）的

可以使用Windows放大使屏幕更大。 屏幕以1920x1080记录。

At 7:00
Kp=9.998
Ki=14.2
I converted the open loop gain in (inches/sec)/volt to (mm/s)/%control

When you move the slider up you are moving making λ bigger.  This causes the errors to decay faster.   exp(-λ*t)

It is possible to use the Windows magnify to make the screen bigger.  The screen is recorded at 1920x1080.

蜻蜓

Ga(s)=K*ω^2/(s*(s^2+2*ζ*ω*s+ ω^2 ))
Gc(s)=Kp+Kd*s+K2*s^2
What are the symbolic formulas for Kp, Kd and K2?
(s+ λ)^3=(s^3+( K*ω^2*K2+2*ζ*ω)*s^2+( K*ω^2*Kd+ ω^2)*s+ K*ω^2*Kp)


λ^3=K*ω^2*Kp   
3λ=K*ω^2*K2+2*ζ*ω
3λ^2= K*ω^2*Kd+ ω^2

然后，人工计算多个不同“ω”，输入到DETLA中, 再通过“自动调整增益”改变“λ”的值，优化（ Kp, Kd 和 K2）增益？是这样吗？这里"K "是固定的？


Then, several different "_" are calculated manually and input into DETLA. Then, the value of "lambda" is changed by "auto-adjusting gain" to optimize (Kp, Kd and K2) gains. Is that so? Is "K" fixed here?

PEN

Ga（s）= K *ω^ 2 /（s *（s ^ 2 + 2 *ζ*ω* s +ω^ 2））

GC（S）= K P+的Kd* S+ K2* S^ 2

Ga（s）是开环传递函数。 我很一致。
K是开环增益。（毫米/秒）/％控制。
除非负载或系统压力发生显着变化，否则开环增益通常是恒定的
Kp =λ^ 3 /（K *ω^ 2）
Kd =（3 *λ^2-ω^ 2）/（K *ω^ 2）
K2 =（3 *λ-2 *ζ*ω）/（（K *ω^ 2））

向上移动滑块会增加λ并通过将3个闭环极点移动到s平面的左侧来增加带宽

在-λ处总会有3个闭环极点。 这是非常安全的，因为3个闭环极点位于稳定性好的区域中间。

我做的测试是为了检查稳健性。
我用假设的数字计算Kp，Kd和K2。 通常我用
K =10（毫米/秒）/％控制
ζ= 1/3
ω=2 *π*10

然后使用我为Kp，Kd，K2计算的值来控制开环系统
K，ζ和ω的值变化+/- 20％。

Ga(s)=K*ω^2/(s*(s^2+2*ζ*ω*s+ ω^2 ))
Gc(s)=Kp+Kd*s+K2*s^2

Ga(s) is the open loop transfer function.  I am consistent.
K is the open loop gain. (mm/s)/%control.
The open loop gain is usually constant unless the load or system pressure changes significantly
Kp= λ^3/( K*ω^2)
Kd = (3* λ^2- ω^2)/( K*ω^2)
K2 = (3* λ-2*ζ*ω) /(( K*ω^2))

Moving the slider up increases λ and increases the bandwidth by moving the 3 closed loop poles to the left in the s -plane.

There will always be 3 closed loop poles at – λ.  This is very safe because the 3 closed loop poles are in the middle of the area of nice stability.

A test I do to check for robustness.
I calculate Kp, Kd and K2 using assumed numbers.  Usually I use
K=10(mm/s)/%control
ζ =1/3
ω=2*π*10

Then use the values I calculated for Kp, Kd, K2 to control the open loop system where
The values of K, ζ , and  ω change by +/- 20%.

蜻蜓

PEN 发表于 2019-6-23 11:40
Ga（s）= K *ω^ 2 /（s *（s ^ 2 + 2 *ζ*ω* s +ω^ 2））

GC（S）= K P+的Kd* S+ K2* S^ 2

我没有化简公式，DELTA控制器 应该会计算Kp Kd    K2增益，
I don't have a simplified formula. The DELTA controller should calculate the Kp Kd K2 gain.


A test I do to check for robustness.
I calculate Kp, Kd and K2 using assumed numbers.  Usually I use
K=10(mm/s)/%control
ζ =1/3
ω=2*π*10

Then use the values I calculated for Kp, Kd, K2 to control the open loop system where
The values of K, ζ , and  ω change by +/- 20%.

您说的是Python模拟或者RMC轴模拟器，还是真实开环控制？

翻译过来的意思像是控制真实的系统，

我这样理解：

已经有真实系统，但是不知道ω和ζ数值，通过一个

假设一个ω=2*π*10，ζ =1/3 值，计算前馈增益，然后进行开环控制，通过多次改变ω，ζ的值，找到输出误差最小时候的参数，这时的ω和ζ和K的值，就是这个系统的模型得参数。放到这个等式里(s^3+( K*ω^2*K2+2*ζ*ω)*s^2+( K*ω^2*Kd+ ω^2)*s+ K*ω^2*Kp)得到数学模型。这是不是系统辨识？

Are you talking about Python simulation or RMC axis simulator orreal open-loop control?

The translated meaning is to control the real system.

I understand this:

There's a real system, but we don't know the values of_and_, and wepass through it.

Assuming a value of_=2*pi*10 and_=1/3, the feed-forward gain iscalculated, and then the open-loop control is carried out. By changing thevalue of_and_several times, the parameters of the minimum output error can befound. At this time, the values of_and_and K are the parameters of the systemmodel. In this equation(s^3+( K*ω^2*K2+2*ζ*ω)*s^2+( K*ω^2*Kd+ ω^2)*s+ K*ω^2*Kp)  the mathematical model is obtained. Isthis system identification?

蜻蜓

以前说过检查伺服阀零位，液压缸粗糙度。您要加大负载测试稳定性？
It has been said before to check the zero position of servo valve and the roughness of hydraulic cylinder. Do you want to increase load test stability?