我有一个PID控制器在机器人上运行，该机器人旨在让机器人转向罗盘标题。以20Hz的速率重新计算/应用PID校正。
尽管PID控制器在PD模式下运行良好（IE，积分项为零），即使是最轻微的积分也会迫使输出不稳定，使得转向执行器被推到左侧或正确的极端。
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这个问题？
答案 0 :(得分：8)
看起来你正在将你的时基用于积分三次。 错误已经是自上一个样本以来的累积误差，所以你不需要将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：
我再次浏览了您的代码，除了上述修复程序之外，我还会修复其他一些项目。
1）您不希望以毫秒为单位的增量时间。在正常的采样系统中，delta项将是1，但是对于20Hz的速率，你输入的值为50，这具有通过该因子增加Ki并且还将Kd减小50倍的效果。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是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）最后你可能只是将deltaTime放在一起并按照以下方式计算差分项。
float derivative = Kd * (error - PrevError);
完成所有这些代码后。
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);
}
答案 1 :(得分：5)
您是在初始化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到接近零（1e100 + 1 == 1e100）。
你可能会遇到integrator windup，它通常应该消失，但是如果它需要更长的时间来减少，而不是你的车辆完成一个完整的旋转（并再次结束积分器）。简单的解决方案是对集成商施加限制，但如果您的系统需要，则more advanced solutions（PDF，879 kB）。
Ki是否有正确的符号？
我会强烈阻止使用浮点数作为PID参数，因为它们具有任意精度。使用整数（可能是fixed point）。你将不得不进行限制检查，但它比使用浮动更加明智。
答案 2 :(得分：4)
积分项已经累积了一段时间，乘以deltaTime将使其以时间平方的速率累积。事实上，由于已经通过将错误乘以deltaTime错误地计算了SteadyError，这是时间立方的！
在SteadyError中，如果您尝试补偿非周期性更新，则最好修复非周期性。但是，无论如何，计算都是有缺陷的。您已经以误差/时间为单位计算，而您只想要误差单位。如果真的有必要，那么可以正确地补偿定时抖动的方法是：
SteadyError += (error * 50.0f/deltaTime);
如果deltaTime保持为毫秒且标称更新速率为20Hz。但是deltaTime可以更好地计算为float，如果它是你试图检测的定时抖动，则根本不会转换为毫秒;你不必要地抛弃精确度。无论哪种方式，您需要的是通过标称时间与实际时间的比率来修改误差值。
好的阅读是PID without a PhD
答案 3 :(得分：1)
我不确定为什么你的代码不能正常工作，但我几乎肯定你无法测试它以查明原因。你可以注入一个计时器服务，这样你就可以嘲笑它，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DateTime.Now.Ticks的两次调用替换为GetCurrentTicks()，您可以单步执行代码并查看值的样子。