导致极端不稳定的 PID 控制器积分项

Question

我有一个在机器人上运行的 PID 控制器，旨在使机器人转向罗盘方向。 PID 校正以 20Hz 的速率重新计算/应用。

虽然 PID 控制器在 PD 模式下运行良好（即积分项为零），但即使是最轻微的积分也会迫使输出不稳定，从而将转向执行器推向左侧或右极端。

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);
    }

有谁知道为什么会发生这种情况或如何解决它？

Answer 1

看起来您将时基应用于积分三次。 误差已经是自上次采样以来的累积误差，因此您不需要将 deltaTime 乘以它。所以我会将代码更改为以下内容。

SteadyError += error ;

SteadyError 是误差的积分或总和。

所以积分应该只是 SteadyError * Ki

float integral = Ki * SteadyError;

编辑：

我再次检查了您的代码，除了上述修复之外，我还需要修复其他几项。

1) 您不希望增量时间以毫秒为单位。在正常的采样系统中，增量项将是 1，但您为 20Hz 的速率输入了 50 之类的值，这具有将 Ki 增加该因子并将 Kd 减小 50 因子的效果。如果您担心抖动，则需要将增量时间转换为相对采样时间。我会改用公式。

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

500000.0 是每个样本的预期滴答数，对于 20Hz 为 50ms。

2) 将积分项保持在一个范围内。

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

3) 更改以下代码，以便当错误在 -180 左右时，您只需稍作更改就不会出错。

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

4) 验证 Steering.Degree 正在接收正确的分辨率和符号。

5) 最后，您可以将 deltaTime 全部删除，然后按以下方式计算微分项。

float derivative = Kd * (error - PrevError);

所有这些你的代码都变成了。

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

你在初始化SteadyError（奇怪的名字……为什么不是“积分器”）？如果它在启动时包含一些随机值，它可能永远不会返回接近零 (1e100 + 1 == 1e100)。

您可能正在遭受integrator windup 的困扰，它通常应该会消失，但如果减少所需的时间比您的车辆完成完整旋转（并再次启动积分器）所需的时间更长，则不会。简单的解决方案是对积分器施加限制，但如果您的系统需要，也有 more advanced solutions (PDF, 879 kB)。

Ki 的符号是否正确？

我会强烈不鼓励将浮点数用于 PID 参数，因为它们的精度是任意的。使用整数（可能是fixed point）。您将不得不进行限制检查，但这比使用浮点数要明智得多。

Answer 3

积分项已经随时间累积，乘以 deltaTime 将使其以时间平方的速率累积。事实上，由于 SteadyError 已经通过将误差乘以 deltaTime 计算得出错误，因此是时间立方的！

在 SteadyError 中，如果您试图补偿非周期性更新，最好修复非周期性。但是，无论如何，计算都是有缺陷的。您已经以错误/时间为单位进行了计算，而您只需要错误单位。如果确实需要，补偿时序抖动的算术正确方法是：

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

如果 deltaTime 仍以毫秒为单位，并且标称更新速率为 20Hz。但是，如果您尝试检测的是时序抖动，则 deltaTime 最好计算为浮点数或根本不转换为毫秒；您不必要地丢弃了精度。无论哪种方式，您都需要通过标称时间与实际时间的比率来修改误差值。

一个很好的阅读是PID without a PhD

Answer 4

我不确定您的代码为什么不起作用，但我几乎可以肯定您也无法对其进行测试以了解原因。您可以注入一个计时器服务，以便模拟它并查看发生了什么：

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，将两个对DateTime.Now.Ticks 的调用替换为GetCurrentTicks()，然后您可以单步执行代码并查看值的样子。