PID控制器积分术语导致极端不稳定

Question

我有一个PID控制器在机器人上运行，该机器人旨在使机器人转向指南针。 PID校正以20Hz的速度重新计算/应用。

尽管PID控制器在PD模式下运行良好（即，具有整体术语为零），即使是最小的积分也会迫使输出不稳定，以使转向执行器将转向执行器推向左或右极端。

代码：

        private static void DoPID(object o)
    {
        // Bring the LED up to signify frame start
        BoardLED.Write(true);

        // Get IMU heading
        float currentHeading = (float)RazorIMU.Yaw;

        // We just got the IMU heading, so we need to calculate the time from the last correction to the heading read
        // *immediately*. The units don't so much matter, but we are converting Ticks to milliseconds
        int deltaTime = (int)((LastCorrectionTime - DateTime.Now.Ticks) / 10000);

        // Calculate error
        // (let's just assume CurrentHeading really is the current GPS heading, OK?)
        float error = (TargetHeading - currentHeading);

        LCD.Lines[0].Text = "Heading: "+ currentHeading.ToString("F2");

        // We calculated the error, but we need to make sure the error is set so that we will be correcting in the 
        // direction of least work. For example, if we are flying a heading of 2 degrees and the error is a few degrees
        // to the left of that ( IE, somewhere around 360) there will be a large error and the rover will try to turn all
        // the way around to correct, when it could just turn to the right a few degrees.
        // In short, we are adjusting for the fact that a compass heading wraps around in a circle instead of continuing
        // infinity on a line
        if (error < -180)
            error = error + 360;
        else if (error > 180)
            error = error - 360;

        // Add the error calculated in this frame to the running total
        SteadyError = SteadyError + (error * deltaTime);

        // We need to allow for a certain amount of tolerance.
        // If the abs(error) is less than the set amount, we will
        // set error to 0, effectively telling the equation that the
        // rover is perfectly on course.
        if (MyAbs(error) < AllowError)
            error = 0;

        LCD.Lines[2].Text = "Error:   " + error.ToString("F2");

        // Calculate proportional term
        float proportional = Kp * error;

        // Calculate integral term
        float integral = Ki * (SteadyError * deltaTime);

        // Calculate derivative term
        float derivative = Kd * ((error - PrevError) / deltaTime);

        // Add them all together to get the correction delta
        // Set the steering servo to the correction
        Steering.Degree = 90 + proportional + integral + derivative;

        // We have applied the correction, so we need to *immediately* record the 
        // absolute time for generation of deltaTime in the next frame
        LastCorrectionTime = DateTime.Now.Ticks;

        // At this point, the current PID frame is finished
        // ------------------------------------------------------------
        // Now, we need to setup for the next PID frame and close out

        // The "current" error is now the previous error
        // (Remember, we are done with the current frame, so in
        // relative terms, the previous frame IS the "current" frame)
        PrevError = error;

        // Done
        BoardLED.Write(false);
    }

有人知道为什么会发生这种情况或如何解决吗？

Answer 1

看来您将时间表应用于整数三次。错误已经是累积错误，因为最后一个示例，因此您不需要乘以Deltatime Time。因此，我会将代码更改为以下内容。

SteadyError += error ;

SteadyError是误差总和。

因此，积分应该是稳定的 * ki

float integral = Ki * SteadyError;

编辑：

我再次浏览了您的代码，除上述修复程序外，还有其他一些项目还可以修复。

1）您不想以毫秒为单位的三角洲时间。在正常的采样系统中，增量项将是一个，但对于20Hz速率，您将其视为50，这具有增加Ki的效果，而Ki则增加了KI，并且KD也将KD降低50倍。如果您担心抖动，则需要将增量时间转换为相对样本时间。我将使用公式。

float deltaTime = (LastCorrectionTime - DateTime.Now.Ticks) / 500000.0

500000.0是每个样品的预期刻度数，为20Hz为50ms。

2）将整数术语保持在范围内。

if ( SteadyError > MaxSteadyError ) SteadyError = MaxSteadyError;
if ( SteadyError < MinSteadyError ) SteadyError = MinSteadyError;

3）更改以下代码，以便当错误约为-180左右时，您不会在较小的更改中获得错误。

if (error < -270) error += 360;
if (error >  270) error -= 360;

4）验证转向。Degree正在接收正确的分辨率和符号。

5）最后，您可能只能将deltatime全部放在一起，并以以下方式计算差异术语。

float derivative = Kd * (error - PrevError);

所有这些都变成了代码。

private static void DoPID(object o)
{
    // Bring the LED up to signify frame start
    BoardLED.Write(true);

    // Get IMU heading
    float currentHeading = (float)RazorIMU.Yaw;


    // Calculate error
    // (let's just assume CurrentHeading really is the current GPS heading, OK?)
    float error = (TargetHeading - currentHeading);

    LCD.Lines[0].Text = "Heading: "+ currentHeading.ToString("F2");

    // We calculated the error, but we need to make sure the error is set 
    // so that we will be correcting in the 
    // direction of least work. For example, if we are flying a heading 
    // of 2 degrees and the error is a few degrees
    // to the left of that ( IE, somewhere around 360) there will be a 
    // large error and the rover will try to turn all
    // the way around to correct, when it could just turn to the right 
    // a few degrees.
    // In short, we are adjusting for the fact that a compass heading wraps 
    // around in a circle instead of continuing infinity on a line
    if (error < -270) error += 360;
    if (error >  270) error -= 360;

    // Add the error calculated in this frame to the running total
    SteadyError += error;

    if ( SteadyError > MaxSteadyError ) SteadyError = MaxSteadyError;
    if ( SteadyError < MinSteadyError ) SteadyError = MinSteadyError;

    LCD.Lines[2].Text = "Error:   " + error.ToString("F2");

    // Calculate proportional term
    float proportional = Kp * error;

    // Calculate integral term
    float integral = Ki * SteadyError ;

    // Calculate derivative term
    float derivative = Kd * (error - PrevError) ;

    // Add them all together to get the correction delta
    // Set the steering servo to the correction
    Steering.Degree = 90 + proportional + integral + derivative;

    // At this point, the current PID frame is finished
    // ------------------------------------------------------------
    // Now, we need to setup for the next PID frame and close out

    // The "current" error is now the previous error
    // (Remember, we are done with the current frame, so in
    // relative terms, the previous frame IS the "current" frame)
    PrevError = error;

    // Done
    BoardLED.Write(false);
}

Answer 2

您是初始化吗？ SteadyError （怪异名称...为什么不“集成器”）？如果它在启动中包含一些随机值，则可能永远不会返回到零接近（1e100 + 1 == 1e100).

你可能遭受 集成商定盘, ，通常应该消失，但如果要减小的时间比您的车辆完成完整旋转（并再次打开集成商），则不得减少。微不足道的解决方案是对积分器施加限制，尽管有 更高级的解决方案 （PDF，879 KB）如果您的系统需要。

做 Ki 有正确的标志吗？

我会 强烈 由于其任意精度，因此阻止将浮子用于PID参数。使用整数（也许 固定点）。您将不得不进行限制检查，但是比使用浮子要更理智。

Answer 3

整数术语已经累积了，随着时间的推移，乘以三角洲将使它以时间平方的速度积累。实际上，由于已经通过将误差乘以三角洲的误差已经错误地计算出来，因此已被时间内！

在SteadyError中，如果您试图弥补上的周围更新，最好修复Aperiodicition。但是，无论如何，计算都是有缺陷的。您已经以错误/时间单位进行了计算，而您只需要错误单位。如果真正需要的话，算术的正确方法可以补偿定时抖动：

SteadyError += (error * 50.0f/deltaTime);

如果三角洲保留为毫秒，名义更新率为20Hz。但是，如果您要检测到的正时抖动，则会更好地计算出浮子或根本不转换为毫秒。您不必丢弃精度。无论哪种方式，您需要的是通过名义时间与实际时间的比率修改错误值。

一个很好的阅读是 没有博士学位的PID

Answer 4

我不确定您的代码为什么不起作用，但是我几乎是肯定的，您也无法测试它以了解原因。您可以注入计时器服务，以便您嘲笑它，看看发生了什么：

public interace ITimer 
{
     long GetCurrentTicks()
}

public class Timer : ITimer
{
    public long GetCurrentTicks() 
    {
        return DateTime.Now.Ticks;
    }
}

public class TestTimer : ITimer
{
    private bool firstCall = true;
    private long last;
    private int counter = 1000000000;

    public long GetCurrentTicks()
    {
        if (firstCall)
            last = counter * 10000;
        else
            last += 3500;  //ticks; not sure what a good value is here

        //set up for next call;
        firstCall = !firstCall;
        counter++;

        return last;
    }
}

然后，将两个呼叫替换为 DateTime.Now.Ticks 和 GetCurrentTicks(), ，您可以浏览代码，看看值的外观。