加速度和固有频率

PEN

There was a question about how high the natural frequency of the actuator and load should be.   My answer was that the natural frequency at least four times greater than the frequency of acceleration. This is true when using a standard hydraulic motion control.

有一个关于执行器和负载的固有频率应该有多高的问题。 我的答案是，固有频率至少比加速频率高四倍。 使用标准液压运动控制时，这是正确的。

What is the frequency of acceleration?

加速的频率是多少？

The frequency of acceleration is the frequency at which the actuator and load accelerate. A simple example is that if the actuator and load are being moved following a sinusoid at 5 Hz the frequency of acceleration is 5 Hz.  If the actuator is just making a simple move from one position to another, the frequency acceleration is 0.5 seconds / acceleration time. The same can be done for the deceleration time. This assumes the acceleration ramps are sine waves where there is one fundamental frequency with no harmonics. Linear ramps have the fundamental frequency and many harmonics that may cause problems. Most motion controllers express the acceleration as mm/s^2. For instance 2500 mm/s^2 is about one forth the acceleration due to gravity. The acceleration time to 250 mm/s takes 0.1 second at 2500 mm/s^2, so the frequency of acceleration is 5 Hz. If the acceleration is to 100 mm/s the acceleration time is 0.04s so the frequency of acceleration is 12.5 Hz.

加速频率是执行器和负载加速的频率。 一个简单的例子是，如果执行器和负载在5Hz的正弦波下移动，则加速度的频率为5Hz。 如果执行器只是简单地从一个位置移动到另一个位置，则频率加速度为0.5秒/加速时间。 减速时间也可以做同样的事情。 这假设加速斜坡是没有谐波的一个基本频率的正弦波。 线性斜坡具有可能导致问题的基频和许多谐波。 大多数运动控制器将加速度表示为mm / s ^ 2。 例如2500mm / s ^ 2大约是由重力引起的加速度的四分之一。 250mm / s的加速时间在2500mm / s ^ 2时需要0.1秒，因此加速度的频率为5Hz。 如果加速度为100毫米/秒，则加速时间为0.04秒，因此加速度的频率为12.5赫兹。

I have been asked many times why the closed loop gains for short distance moves need to be different from the closed loop gains for long distance moves.  The reason is that short moves have very short acceleration times so the frequency of acceleration is higher.  When making short moves the acceleration time should not be too small.

In my example above, the 12.5 Hz frequency of acceleration may be too high.

The best option might be to slow the acceleration ramp down to 1000 mm/s^2 so the acceleration time is still 0.1 seconds.

我多次被问及为什么短距离移动的闭环增益需要与长距离移动的闭环增益不同。 原因在于短时间移动的加速时间非常短，因此加速的频率较高。 在短时间移动时，加速时间不宜太短。

在我上面的例子中，12.5Hz的加速频率可能太高。

最好的选择可能是将加速斜坡降低到1000 mm / s ^ 2，所以加速时间仍然是0.1秒。